Guest Post by Willis Eschenbach
Short answer? Of course not, that would violate the Second Law of Thermodynamics … BUT it can leave the hot object warmer than it would be if the cold object weren’t there. Let me explain why this is so.
Let me start by introducing the ideas of individual flows and net flows. Suppose I owe you twenty-five dollars. I run into you, but all I have is a hundred dollar bill. You say no problem, you have seventy-five in cash. I give you the hundred, you give me the seventy-five, and the debt is paid.
Now, there are two equally valid ways to describe that transaction. One way looks at both of the individual flows, and the other way just looks at the net flow. Here they are:
Figure 1. Net flows and individual flows. The individual flows are from me to you, $100, and from you to me, $75. The net flow is from me to you, $25.
What does this have to do with cold and warm objects? It points out a very important distinction, that of the difference between individual flows of energy and the net flow of energy, and it relates to the definition of heat.
Looking at Figure 1, instead of exchanging dollars, think of it as two bodies exchanging energy by means of radiation. This is what happens in the world around us all the time. Every solid object gives off its own individual flow of thermal radiation, just as in the upper half of Figure 1. We constantly radiate energy that is then being absorbed by everything around us, and in turn, we constantly absorb energy that is being radiated by the individual objects around us.
“Heat”, on the other hand, is not those individual flows of energy. Heat is the net flow of energy, as represented in the bottom half of Figure 1. Specifically, a heat flux is the net flow of energy that occurs spontaneously as a result of temperature differences.
Now, the Second Law of Thermodynamics is only about net flows. It states that the net flow of thermal energy which we call “heat” goes from hot to cold each and every time without exception. However, the Second Law says nothing about the individual flows of energy, only the net flow. Heat can’t flow from cold to hot, but radiated energy absolutely can.
When an object emits radiation, that radiation goes on until it hits something that absorbs it, whereupon it is converted to thermal energy. The individual temperatures of the emitting and absorbing objects are not significant because these are individual energy flows, and not the net energy flow called “heat”. So there is no violation of the Second Law.
Here’s the thing that keeps it all in balance. If I can see you, you can see me, so there are no one-way energy flows.
Which means that if I am absorbing radiation from you, then you are absorbing radiation from me. If you are warmer than me, then the net flow of energy will always be from you to me. But that says nothing about the individual flows of energy. Those individual flows only have to do with the temperature of the object that is radiating.
So how do we calculate this net energy flow that we call “heat”? Simple. Gains minus losses. Energy is conserved, which means we can add and subtract flows of energy in exactly the same way that we can add and subtract flows of dollars. So to figure out the net flow of energy, it’s the same as in Figure 1. It’s the larger flow minus the smaller flow.
With all of that as prologue, let me return to the question that involves thermal radiation. Can a cold object leave a warm object warmer than it would be without the cold object?
While the answer is generally no, it can do so in the special case when the cold object is hiding an even colder object from view.
For example, if a person walks between you and a small campfire, they hide the fire from you. As soon as the fire is hidden, you can feel the immediate loss of the radiated energy. At that moment, you are no longer absorbing the radiated energy of the fire. Instead, you are absorbing the radiated energy of the person between you and the fire.
And the same thing can happen with a cold object. If there is a block of wood between you and a block of ice, if you remove the wood, you’ll get colder because you will be absorbing less radiation from the ice than you were from the wood. You no longer have the wood to shield you from the ice.
Why is all of this important? Let me offer up another graphic, which shows a simple global energy budget.
Figure 2. Greatly simplified global energy budget, patterned after the Kiehl/Trenberth budget. Unlike the Kiehl/Trenberth budget, this one is balanced, with the same amount of energy entering and leaving the surface and each of the atmospheric layers. Note that the arrows show ENERGY flows and not HEAT flows.
These ideas of individual flows, net flows, and being shielded from radiation are important because people keep repeating over and over that a cold atmosphere cannot warm the earth … and they are right. The temperature and the radiation are related to each other by the Stefan-Boltzmann equation. When we apply the S-B equation to the 321 W/m2 of downwelling “back radiation” shown in the graphic above, it tells us that the effective radiating level is somewhere around freezing, much colder than the surface.
BUT a cold atmosphere can leave the earth warmer than it would be without the atmosphere because it is hiding something even colder from view, the cosmic microwave background radiation that is only a paltry 3 W/m2 …
And as a result, with the cold atmosphere shielding us from the nearly infinite heat sink of outer space, the earth ends up much warmer than it would be without the cold atmosphere.
To summarize …
• Heat cannot flow from cold to hot, but radiated energy sure can.
• A cold atmosphere radiates about 300-plus W/m2 of downwelling radiation measured at the surface. This 300-plus W/m2 of radiated energy leaves the surface warmer than it would be if we were exposed to the 3 W/m2 of outer space.
My best regards to all,
w.
My Usual Request: When you comment, please QUOTE THE EXACT WORDS THAT YOU ARE DISCUSSING, so that we can all understand the nature of your objections.
My Second Request: Please keep it civil. Speculation about the other person’s motives and cranial horsepower are greatly discouraged.
Further Reading: My post entitled “The Steel Greenhouse” looks at how the poorly-named “greenhouse effect” work, based on the principles discussed above.
Math Notes: There’s an excellent online calculator for net energy flow between two radiating bodies here. It also has the general equation used by the calculator, viz:
with the following variables:
and Q-dot (left-hand side of the equation) being the net flow.
Now, when the first object is totally enclosed by the second object, then area A2 is set to a very large number (I used a million) and the view factor F12 is set to 1. This is the condition of the earth completely surrounded by the atmosphere. For the general case, I’ve set area A1 to 1 square metre. Finally, I’ve made the usual simplifying assumption that thermal IR emissivity is 1.0 for the surface and the atmosphere. The emissivity values are greater than 0.9 in both cases, so the error is small. With those usual assumptions, the equation above simplifies as follows, courtesy of Mathematica:
But sigma T ^ 4 is simply the Stefan-Boltzmann radiation for the given temperature. That is why, in the energy budget above, we can simply add and subtract the energy flows to produce the budget and check to see if it is balanced.
SkepticGoneWild November 29, 2017 at 11:15 pm
Either it is a thought experiment, or it is pure fantasy. You can’t have it both ways.
Oh, sure, I’ll just build a planet in space and surround it with a steel shell … in any case, as I pointed out above, ANTHONY DID THE EXPERIMENT.
In addition, we have the example of the new lightbulb which uses some of the reflected IR to heat up the incandescent element … clearly there is ENERGY flowing from cold to hot in that case.
Finally, if as folks keep saying energy cannot flow from cold to hot, that a hot object won’t absorb a “cold” photon, then how can we see cold things? For me to see something means my eyes have to absorb the energy … but lots of folks say that because my eyes are warm, they can’t accept photons from something colder. So how come I can see cold things?
Skeptic, you and others don’t seem to get it. This stuff is very basic thermodynamics. I provided an online calculator so you can check my results. I provided the actual thermodynamic equation, and I’ve shown what it reduces to. You and others have totally ignored the math … poor choice. The answer is in the math.
At this point, it is up to you to FALSIFY my claims … and to date nobody’s done that. Not only that, but the people here who are experts in the field, including folks who teach this stuff for a living, say that my steel greenhouse analysis is 100% correct.
Your move …
w.
“Either it is a thought experiment, or it is pure fantasy” you say, Willis.
Well, talking about thought experiments, vis a vis, pure fantasy….what about this thought experiment which reckons that if the Earth had no atmosphere, (thought experiment), we would have an average “atmospheric?” temperature of about negative 18 deg C. Obviously, at that temperature, all the oceans would be frozen solid…..but the last time I looked at the sea….it wasn’t frozen solid, Willis.
So the only conclusion I can reach is that the thought experiment is pure fantasy.
Oh! did I just say that the atmospheric “greenhouse effect” was pure fantasy. ..wow, sorry, heretical talk.
Should read….average surface “atmospheric?” temperature…
Mack November 30, 2017 at 1:00 am Edit
Mack, one of Einstein’s famous thought experiments started out “Suppose you were chasing a ray of light …”
Now, you and I know that we can’t just chase a ray of light through space … but Einstein wasn’t just pointlessly indulging in pure fantasy. Instead, he was conducting a “thought experiment”. One of the things about thought experiments is that we do them because we cannot do the real experiment. We can’t race a ray of light or build a steel greenhouse the size of a planet.
But that doesn’t mean that we cannot consider what would happen if we did race a ray of light or build a steel greenhouse. The key that makes these “thought experiments” and not pure fantasy is simple. In a thought experiment, once the situation is set up, we don’t suspend the laws of physics.
Instead, we use those natural laws to illuminate, investigate, and hopefully understand something we didn’t understand before … as I did with the steel greenhouse.
Surely you can see the difference between Einstein’s story about chasing a ray of light, and say the story of Sleeping Beauty, which is pure fantasy set in a world where there are witches and spells and the normal laws of physics are in abeyance. As I mentioned above, Maxwell’s Demon and Schrodinger’s Cat are thought experiments which have a solid place in science … while Sleeping Beauty is a fairy tale which has no place in science at all.
I hope that clarifies the difference.
Regards,
w.
“Finally, if as folks keep saying energy cannot flow from cold to hot, that a hot object won’t absorb a “cold” photon, then how can we see cold things? For me to see something means my eyes have to absorb the energy … but lots of folks say that because my eyes are warm, they can’t accept photons from something colder. So how come I can see cold things?“
Well, I can help with this one. Objects at room temperature emit IR, Willis, not visible light, and no, your eyes don’t see IR. You see objects because of the visible light REFLECTED off the objects. Reflected off, not emitted from. The clue is in the title, really, “visible light”. As in, it’s at a wavelength of the EM spectrum that is visible to the human eye. This visible light comes from an object at higher temperature, like the sun. Be that during the day, or reflected off the moon at night. Or it could come from a light bulb; that’s also operating at sufficient temperature to produce visible light.
You see, it doesn’t all have to be a mystery…
http://www.bbc.co.uk/bitesize/ks2/science/physical_processes/how_we_see_things/read/1/
Tony November 30, 2017 at 1:46 am Edit
http://www.bbc.co.uk/bitesize/ks2/science/physical_processes/how_we_see_things/read/1/
“KS2” in your citation means that it is designed for 7- to 11-year-old children … it says nothing of interest to what we are talking about. Stop wasting our time with your grade school explanations, this is an adult discussion.
w.
Yes, I must admit that I found it a little bit funny that you needed something explained to you that most 7 to 11 year olds understand. I’m sorry. I shouldn’t laugh.
Tony, Mr Eshenbach should place an Ice cube in a completely dark room to see if his eyes can see those special photons streaming of it. ROFL.
I put up this post earlier, challenging all the people who deny established radiation. And got crickets.
Now I see that A.C. Osborne is such a smart guy he will be just
A.C. Osborne:
Answer this simple question..
Without changing the subject, and without any need for models or esoteric physics nobody is talking about, answer this simple question…….
Here’s the shot:
Your claim is that IR from a 90 k source incident on a 91 k target, does nothing to the target. If true it has to then either pass through it (albedo of 0) or be reflected from it (albedo of 1). You further claim if that very same source hits the very same target but now the target has a temperature of 89 k it is absorbed, at least partially.
Here’s the chaser:
If you would grant me the liberty of putting that in an equation, it would be albedo= f(T). Do you agree that my equation correctly supports your claim?
Come on. It’s simple question involving just three numbers and an equation with one term. Surely such a smart one as you can answer this in a jiffy….
Wating….
Plus, nobody is saying ENERGY doesn’t flow from cold to hot. HEAT doesn’t flow from cold to hot.
Tony November 30, 2017 at 2:49 am
Say what?? Of course they are saying that, that’s the whole Slayer theory. The claim is that there is no way that the atmosphere could make the ground warmer because the atmosphere is colder than the surface. Here’s Crispin on the subject:
Nickreality says:
Richard Verney thinks the atmosphere can’t leave the ground warmer …
wickedwenchfan says:
and finally, from wickedwenchfan again:
Sorry, Tony … but you’re wrong again. People are indeed claiming that radiant energy cannot flow from cold to hot.
w.
Hey Willis don’t put me in that list! What you copied is from a thread with Stephen Wilde where I was restating HIS position. Read the whole thread.
I am trying to get a 1st law energy balance from him with his slayer atmosphere model. The handwaving is epic
Do not try to get me into trouble with Willis.
I happen to agree with him that a cooler atmosphere can warm a hotter surface.
I am not advancing a Slayer model.
I am just pointing out that the job can be done by conduction and convection without invoking a surface warming effect from DWIR.
The final decision as to which option prevails is open for debate and a lot of contributors here and elsewhere are beginning see how important atmospheric mass can be.
So…you DON’T know the difference between heat and energy?
Paul Bahlin November 30, 2017 at 4:12 am
My bad, Paul, I’ve removed you from the list.
w.
No Willis,
YOU came up with this steel greenhouse proposition. The scientific method puts the responsibility on the one proposing his idea. Your next move is an experiment. You are not even half-way through the scientific process.
Secondly, you don’t even understand what falsfification is!
Third. Who is saying energy cannot flow from cold to hot? Not I. You are the one saying HEAT flows from cold to hot, because that is what is happening in your thought experiment. (The cooler shell warms the sphere)
Our move is to sit back and watch you fumble in the dark and fail.
Tony November 30, 2017 at 1:18 am
While that is true, the claim was that photons could not go from a cold object to a hot object. Are you now saying that some photons can and some photons can’t? Because both visible light and IR are just photons …
You ever see fireflies? How about a glow stick? In addition to IR, objects at room temperature can emit visible light. We are not seeing light REFLECTED from fireflies. Despite your clear claim that it is all reflection from some hotter source, in fact it is light EMITTED FROM the cold fireflies.
As to the “clue is in the title, really, visible light”, that says nothing about whether that light is reflected or emitted, or what the temperature of the emitter is.
Again I have to ask … you ever see a firefly or a glow stick or bioluminescence in the ocean? Are you claiming that that light “comes from an object at higher temperature, like the sun”???
It’s not a mystery. Once a photon is emitted it carries its tiny bit of energy until it hits and is absorbed by something. That energy is then converted into another form, be it thermal, mechanical, or chemical.
My eyes have no way of knowing whether a light of a given color comes from a glow stick (cold) or an incandescent light (hot). Doesn’t matter. When the photon hits it is absorbed regardless of the temperature of the emitter.
So … to return to the question, how do you explain that my warm eyes can see light from a cold firefly? According to you, the “cold” photons are supposed to bounce off my warm eyes or something, I’m not sure just how you explain your claims. You seem to think there’s some essential difference between photons of visible light and photons of IR, and some other difference between “hot” and “cold” photons, I don’t know how you think this works.
So here’s your chance to explain it …
w.
https://en.m.wikipedia.org/wiki/Bioluminescence
https://www.pantone.com/how-do-we-see-color
Tony’s link: “Thus, red is not “in” an apple. The surface of the apple is reflecting the wavelengths we see as red and absorbing all the rest.” I see this so often that it is almost an immutable truth. Instruments measure otherwise.
The writer of that web page obviously has never gone into a lab and actually measured the spectrum of an apple illuminated by daylight. According to his writing the resulting spectrum would show no emission from any wavelength band (“absorbing all the rest”) and a huge peak reflected around 660nm band. That is not what is measured; the actual visible spectrum from the apple shows abundant radiance measured from 300-700nm which one’s brain interprets as the color red.
The natural opaque apple absorbs & emits about 95% of the incident daylight and scatters only about 5% as measured by spectrophotometers.
OK.
https://en.m.wikipedia.org/wiki/Photon_energy
Happy reading!
This one was for your statement:
“Because both visible light and IR are just photons“
Which to me implied that you were not aware that photons of visible light and photons of IR ARE actually different.
Oh, sorry, forgot the glow stick:
https://en.m.wikipedia.org/wiki/Glow_stick
Long story short, objects at room temperature do NOT produce visible light. A glow stick produces visible light due to an exothermic reaction between the chemicals that are mixed together when you bend the stick to start with. Bioluminescence is a related phenomenon (again the visible light is produced by a chemical reaction, this time within the organism).
Since not one of those links discusses the question we’re looking at, as to whether photons can carry energy from a cold object to a warm object, I fear that perhaps you are playing some kind of joke … or perhaps not, but if not, then what are you claiming that these links show? Because they have nothing to do with what we’re talking about here.
I do note that after curiously claiming that we only see reflected light;
and after bizarrely stating that the reflected light always comes from something hotter than our eyes;
after I pointed out that neither one is true and I gave examples showing why they are not true, you’ve gone strangely radio silent … I mean, I can understand your unwillingness to admit that neither claim is even remotely valid, but posting random links won’t cut it …
w.
Oh, sorry, Willis. I assumed you would read the five links I have posted to you before responding. This is from the Pantone “how do we see color” link:
“The human eye and brain together translate light into color. Light receptors within the eye transmit messages to the brain, which produces the familiar sensations of color.
Newton observed that color is not inherent in objects. Rather, the surface of an object reflects some colors and absorbs all the others. We perceive only the reflected colors.
Thus, red is not “in” an apple. The surface of the apple is reflecting the wavelengths we see as red and absorbing all the rest. An object appears white when it reflects all wavelengths and black when it absorbs them all.
Color is made up of red, green and blue lightRed, green and blue are the additive primary colors of the color spectrum. Combining balanced amounts of red, green and blue lights also produces pure white. By varying the amount of red, green and blue light, all of the colors in the visible spectrum can be produced.”
Let me know when you have read the links. I have given you an answer on the “examples” you provided, bioluminescence and the glow stick. Gave you those answers very quickly, in fact. No “radio silence” here.
Tony November 30, 2017 at 2:30 am Edit
I read it. It waxed lyrical about things like “color is made up of red, green and blue light”. Gosh. Who knew? … well, everybody knew, that’s a beginners explanation. However, it said nothing about the subjects under discussion, which are a) can objects at or below room temperature emit light (yes), and b) can photons from a cold object flow to a warm object (yes).
I know how we see. The links you sent are all grade-school explanations of how we see that have nothing to do with this discussion.
w.
Yes, Willis. It also says:
“Newton observed that color is not inherent in objects. Rather, the surface of an object reflects some colors and absorbs all the others. We perceive only the reflected colors.“
Tony November 30, 2017 at 3:06 am
As I said, it’s a grade school explanation so it doesn’t cover everything. For example: when you look at a glowstick, what is the light “reflecting” off of? When you look at a bioluminescent fish, what is the light “reflecting” off of?
In short: if an object does NOT emit light, then your statement about Newton is 100% correct.
But when we’re talking about something that emits light, say a blue LED, that blue is NOT a reflected color. It is an emitted color. Same is true of lasers, and light bulbs, and anything else that emits light.
Sorry, but once again, you’re taking the simplified and thus ultimately incorrect path that doesn’t include all the possibilities.
w.
“In short: if an object does NOT emit light, then your statement about Newton is 100% correct.”
Yes, precisely. I have absolutely no idea why you’re acting like you’re disagreeing with me, whilst agreeing with me. Though of course it wasn’t MY statement. It was a quote from the link. And in fact nowhere have I made the statement that energy from a cold object can’t travel to a warm object, nor that a photon cannot travel from a cold object to a warm object.
A FLIR detector sees infrared.
And it is warmer than the objects it sees, at least some of the time.
Our eyes do not see infrared, but infrared eyes are not impossible.
We have the eyes we have due to evolution giving us the ability to detect the wavelengths that are the most abundant and that therefore contain the most information about our surroundings.
Tony November 30, 2017 at 2:21 am
Bioluminescent creatures produce light at well below room temperature. I see bioluminescence in the waves on the north coast where I live, and the water there is about 50°F. Not only that, but fish living in the deep, black ocean at about 4°C are bioluminescent.
So long story short, once again you are 100% wrong—objects at room temperature and well below room temperature DO produce visible light.
In addition, this question of “room temperature” is a bit of a straw man. The issue was whether light is produced from something colder than my eyes, and the answer is absolutely yes …
Having settled that, you still haven’t explained why these “cold” photons don’t just bounce off my eyes or whatever explanation you might have for your claim that radiated energy cannot flow from a cold object to a hot object.
Because when I look at bioluminescent cold north-coast waves, that’s exactly what’s happening. Photons are flowing from a cold object to a warm object, where they are absorbed. You say this is impossible. Please explain why.
w.
PS—Regarding glow sticks, your own citation says that they “generate negligible heat”, which certainly fits with my experience. They are nowhere near as warm as my eyes … but I can still see them, which you and others have claimed is impossible because you say that photons won’t flow from cold to hot …
HEAT will not and cannot flow from cold to hot … but radiant energy absolutely can. This does NOT violate the second law, because as I explained above, if I can see you, you can see me. By this I mean that if one object is able to radiate to a second object, that means the second object is able to radiate to the first. As a result, there’s no “one-way” radiation from cold to hot, that would indeed violate the second law.
Instead, since both are radiating, whichever object is warmer will radiate more, and as a result the NET flow, not the individual flows but the NET flow which we call a heat flux, always and ever goes from hot to cold.
Again, in the head post I gave the relevant equation, its simplification for the situation of the earth and the atmosphere, and a link to an online calculator so that folks like you could actually play with the numbers and see what happens … let me recommend again that you do that.
The math is well known and tested in thousands of situations, from outer space to the factory floor. It’s not cutting edge. Instead it is cut and dried, bog-standard centuries-old stuff that you will find in any introductory thermo textbook.
You’ve yet to show me where I’ve been wrong. It goes without saying that we can see visible light. As I said, the clue is in the title. We see the visible light from a light source, reflected off objects. That’s why we see the objects. Not because of what the objects themselves emit. If an object emits visible light directly, we will see that too. We can see the sun, we can see a light bulb. What makes you think I would be arguing otherwise? The glow stick itself may be at room temperature, but it’s producing light due to an exothermic reaction. So, it has a source of energy for the light. Same as we can see the light bulb, we can see the glow stick (even with the lights off). It’s a light source. This doesn’t change the fact that we will see other objects, with the light source as a glow stick, light bulb, or sun, due to light REFLECTED off them.
It’s all getting a bit desperate, Willis.
“You’ve yet to show me where I’ve been wrong.”
You are wrong about origins of color & so is the website you linked & many more which can be demonstrated by going into the optics lab or looking up an actual measured spectrum of an apple or the paragon of yellowness, a banana. Not all visible light can be picked up by the eye. Turn off the lights in your room at night and test for that. The objects are now bathed in BB radiation at room temperature and emit at visible wavelengths too dim to be perceived by the eye but not too dim for a spectrophotometer.
OK.
Golly.
The first time I read through a comment thread on this subject, I wound up with a terrible headache.
That was years ago.
It is dismaying to see the exact same conversation again, and again, and again…
Ad nauseum.
All these years later, and I still get the exact same headache.
It is very exasperating.
menicholas, it’s like Stendahl said in “The Red And The Black” about the Church. Like you regarding this issue, someone said the Catholic Church never changed. The priest replied “Oh, it changes … it just does it so slowly that no one lives long enough to see it happening”.
What keeps me going is that I’m not writing to convince Tony. He’s beyond my poor powers to add or detract. Instead, I write for the lurkers, those that read but never comment. Among them are many who are unconvinced, many who are just beginning to be interested in climate, many whose minds are not set in stone.
I know that although I can’t see it, some among those minds are changing when they read the nonsense that comes from folks who don’t think that a GHG atmosphere can leave the surface warmer than if that atmosphere didn’t exist … so I persevere.
Onwards, thanks for the comment,
w.
Yes, the lurkers.
There are many, of that I am sure.
I too make many of the comment I make for the benefit of any who might be reading, and not because I think I will change anyone’s mind. Although I do have in mind planting seeds of doubt.
I come from a large family of stubborn people, and I have seen many instances where an idea from person A is ignored by person B when first raised, but then re-emerges in the mind of the person B some time later, as person B’s own idea.
In my family, it was hard to ever win an argument, but sometimes a suggestion made and then left alone would sprout and grow.
I sometimes ask people if they ever change their mind about anything, and some people admit they never do.
A warmista woman I have known from childhood was recently engaged in a conversation about a climate issue, and she seemed to be gaining some knowledge, but at one point she blurted out ” I will never be a den!er”. I engaged her no further, nor will I ever. Just bowed out gracefully. It was not easy, but I like her.
I just want to say, you and several others here should be in the running for sainthood with the patience you have displayed in this thread.
Keep fighting the good fight!
Onward, yes…and you are very welcome. Thank you for the reply.
Stephen, I am really trying to understand your concept but we have so many places where we diverge I think it would be useful to highlight and resolve them.
First you state:
“S-B gives the temperature achievable by a blackbody at a given level of irradiation.
That assumes all incoming and outgoing energy runs at the speed of light.”
I dispute this claim. S-B says nothing at all about irradiation. It relates the energy RADIATING from a BODY at T when it has no other means to shed the energy it needs to maintain 1st law energy balance. Stated as the inverse you could say that if you observe a radiation level, say X, then you can use X to compute a surface temperature. From a first law view of a body, if it is receiving X+Y input energy then it must be shedding X+Y at steady state. If Y is being ‘stolen’ by conduction then X will come out as radiation.
Note here the important distinction that this has nothing to do with irradiation. The body can be receiving energy by little mice bringing in pails of Joules in buckets. Most importantly, it has no dt at all. Nada. It is relating two steady state properties of an object with a constant emissivity. When emissivity is 1 then it is relating properties of a black body.
Then you say:
“In order to make the surface temperature fall below S-B as proposed by you a method of energy removal faster than light would have to be applied.
Conduction and convection operate at a slower rate and so cannot reduce the temperature of the irradiated surface below the S-B temperature.
Instead, conduction and convection take their energy from the outgoing radiation AFTER the S-B temperature has been achieved and cannot reduce that pre existing surface temperature.”
All of this is totally irrelevant to a discussion of steady state energy flux. No matter the rate differences of the various transfer mechanisms, each part of a 1st law accounting will achieve steady state over reasonable integration periods. If it is not steady state then you can’t even talk about energy balance of the parts because you don’t need to obey 1st law until you reach steady state. The water entering your bathtub won’t equal the water leaving your bathtub until it’s full and that’s OK. But, and this is the important thing, you can’t put more water into the overflowing tub and claim you are still filling it and you can’t ever get more out of it than you put in.
Next you say:
“The same parcel of kinetic energy at the surface cannot both radiate and conduct simultaneously so whatever goes into conduction reduces the number of photons emitted to space. Observed from space the planet would appear to drop below S-B but at the surface it would not have done so.”
First sentence is sort of true but it’s more correct to say that surface energy removed by conduction reduces the surface energy and the remainder is then radiated to maintain balance. It doesn’t reduce photons in the strictest sense. It just reduces the amount of energy that needs to be gotten rid of by radiation. Importantly here, and this a huge area of disagreement, surface observations made before and after the introduction of a conduction loss would most definitely show a temperature drop. You have lowered the energy content of the radiator. The temperature will go down. Not debatable.
Penultimately and this is the nut:
“That being the case the surface temperature must then rise above S-B when KE is released from PE beneath descending columns of air and observed from space the planet would return to S-B despite the surface actually being warmer.”
What you are proposing here is that energy can leave the surface by conduction and that does not result in an energy loss observable by measuring the radiative output of the body. Simultaneously (remember we are talking about steady state) you allow that an equal amount of energy entering the body by conduction as a consequence of the pe-ke pump does raise the energy (and temperature) by an amount exactly equal to what would have been lost by the conduction loss except somehow magic is happening in the atmosphere that prevents the lost energy from being lost.
Lastly:
“That is exactly what we observe in the real world.”
Where has this ever been observed?
My position here is that a surface energy loss due to conduction lowers the energy and subsequently (when you reach steady state) the temperature to a new level. Surface energy gain due to conduction results in more energy and subsequently more temperature. If the conduction loss equals the conduction gain (of energies) then the surface temperature is unchanged from what it would be if there were no atmosphere at all.
If the atmosphere radiates negligibly as you propose then the energy entering your box (atmosphere) can do anything it wants it will all reach steady state returning exactly what was put there in the first place. You can have Jupiter size storms, oceans evaporating, torrential rains, all of that, and you still will get all the energy back when it is integrated over reasonable time periods.
A lot of work there but the nub is whether or not conduction and convection reduces surface temperature when it begins at the initial formation of the atmosphere.
I say it does not because it simply absorbs energy that would otherwise have gone out to space as radiation. You appear to accept that.
The temperature appears to fall when observed from space because from that observation point one can only see the radiating portion and not the conducting portion.
Why does that require a surface temperature fall?
I don’t see how adding the slower process of conduction and convection can accelerate the rate of energy loss from the surface so as to induce a temperature fall to a level lower than could be achieved by the faster process of radiation out.
If it does not produce a temperature fall then surface temperature must rise when the descending column gets back to the surface.
“If it does not produce a temperature fall then surface temperature must rise when the descending column gets back to the surface.”
This is your imagination at work Stephen & is not in accord with test. The low pressure system created by the rise in the convecting column shows high pressure system moving in laterally with surface temperature (blue) fluid just like the windy atm. and NOT your imaginary “when the descending column gets back to the surface.”
First off, it would clarify things if you stopped talking about convection. We have defined a simple system: a surface and a box. The surface is the earth outside layer. The box is the atmosphere.
Convection is inside the box and has no relevancee to this fundamental flux accounting as it never penetrates a boundary. Likewise there is no discussion of molten core or plate tectonics.
Each component must balance and the system must balance. If you Can’t get that part done properly then discussions about quantum physics and photons and typhoons and latent heat and lapse rates are so premature as to be totally irrelevant, right?
Now as to your energy questions….
If energy is removed from the surface by conduction then X joules/s leave and after y sec you have reduced the surface by x*y joules. The temperature will go down. Three days later, mice bring back all those joules and plunk em down by conduction. The temperature will go right back where it was as long as that is the only system change.
The high level model that this represents is fundamental to getting to the internal details you are anxious to delve into. But, you can not let that internal dabbling live inside a falsity of energy imbalance.
You can speed up and slow Down offers to your heart’s content only after you vet past 1st law.
The temperature need not go down because the energy taken up in conduction is immediately replaced by continuing insolation.
That energy comes from a reduction of outgoing radiation and is ADDED to system energy content as PE until it is returned to the surface as KE to reinstate the original outward radiative flow.
You really can’t handle the idea of multiple moving parts can you?
You are using flux and energy in the same argument.
Too many units for ya’?
I used flow not flux, you must be rattled.
And since you replied to Willis with a ‘no’ that makes you the Slayer.
A cool atmosphere obviously lets an irradiated surface become warmer than S-B
The only issue is whether that is caused by DWIR or by convective overturning.
I contend that DWIR is a consequence of the lapse rate structure and not a cause of surface heating but the debate on that is ongoing.
Goodnight.
Yikes! Better check that Stephen. If the ke comes back to restore the original flow then your scenario just blew up because the temperature is right back where you started from.
Oh wait…. That’s what I’ve been saying all along
Only if it dropped in the first place and you haven’t
showed that.
The temperature need not go down because the energy taken up in conduction is immediately replaced by continuing insolation.
That energy comes from a reduction of outgoing radiation and is ADDED to system energy content as PE until it is returned to the surface as KE to reinstate the original outward radiative flow.
You really can’t handle the idea of multiple moving parts can you?
Sorry. Last paragraph is an auto correct nightmare.
Should be something like… speed up and slow down transfers only after 1st law accounting.
No matter, I know what you are trying to say and you are wrong.
The atmosphere subtracts from outgoing and adds to incoming so as to make the surface warmer than the S-B prediction whilst radiation to space matches S-B.
In effect the work done by the atmosphere mimics increased external insolation.
The evidence is the observation that for objects with atmospheres of more tha 0.1 bar the temperatures at the same pressure are similar after accounting for distance from the sun.
Radiative theory has no explanation for that.
I have the only detailed description as to why and how it works.
“Radiative theory has no explanation for that.”
Sure it does. Every atmosphere has GHG. Therefore every atmosphere should experience GHG warming.
What there *is* no explanation for is how work done by a falling atmosphere counts, but the energy input to raise the air to begin with is ignored.
Tim, I also disagree with Stephen as radiative-convective atm. theory is good enough to calculate T(z) for earth, venus to within instrumental accuracy.
I disagree with you though, in that the data to date indicate atm.s with not enough pressure to raise the atm. opacity very much looking up. So that the global surface T is about equal to planet brightness T such as for Mars by example. The global thermometer surface T for Mars is only sparsely measured but from what has been recorded this is close enough to planet brightness T for gov. work on the surface.
So to see if I can clarify things, let me ask a question of Tony and all the rest.
Do you think that a non-GHG atmosphere (say argon) can maintain the surface temperature of a planet at a level higher than the calculated Stefan-Boltzmann temperature of that planet given its incoming solar radiation?
Let’s set aside the question of the particular mechanism for the moment. Do you believe that such a continuing temperature elevation from the action of a non-GHG atmosphere alone is physically possible?
Serious yes or no question … I’m not looking for your explanation, just whether you think it’s possible.
w.
Willis, should I et. al. presume you are looking for serious answers for Earth orbit, similar rotating planet typical atm. parameters, like 1 atm. at surface, same L&O surface emissivity so forth?
Trick November 30, 2017 at 3:59 pm
Those don’t matter, although I’m leaving out planets made of radioactive materials and such. Just a ball floating in space, warmed by a sun, with an argon atmosphere. Some folks think that the pressure of the atmosphere, or the overturning of the atmosphere, or conduction, or some other atmospheric phenomenon, will permanently raise the surface temperature of the planet above the expected S-B temperature.
I’d like to know who those folks are so I can discuss it with them.
w.
“Some folks think that the pressure of the atmosphere…will permanently raise the surface temperature of the planet..
I’m one of those “folks” as the physics (measured in lab & theory) shows an increase in pressure adds to the first order opacity of an atmosphere. A rotating planet like Earth means you can reasonably spread insolation over the entire surface not just the illuminated half for a basic analogue to understand some basic atm. physics.
So I will go with those parameters as observed for Earth for basic calculation.
“Do you think that a non-GHG atmosphere (say argon) can maintain the surface temperature of a planet at a level higher than the calculated Stefan-Boltzmann temperature of that planet given its incoming solar radiation?”
If the water condenses out of current earth atm. as it changes to all argon in such a way that total albedo remains ~0.3, surface emissivity remains near enough 1.0 rounded, and the current atm. spectral emissivity looking up drops from about global median 0.8 to that of argon spectral emissivity (pure guess near 0.01 for calculations as I couldn’t quickly find a measured value), then a basic calculation shows 255.3K global median and with 0.0 argon emissivity 255.0K so the answer is: with pure argon atm. emissivity 0.01 find basic physics shows surface temperature a very little higher than a perfectly transparent atm. If find a measured emissivity applicable for argon Earth atm. then I’ll use it for better est.
“Do you believe that such a continuing temperature elevation from the action of a non-GHG atmosphere alone is physically possible?”
Yes, the 255.3K with argon would remain steady state global median, not be transient above the 255K completely transparent atm., at 0.3 albedo, current insolation (I used 1370), current rotation, current L&O surface at 1 atm. pressure.
I wonder if the various parties to this conversation could come to some sort of agreement of an online text of relevant information that can be referred to by everyone involved as well as those of us following along as best we can?
Maybe something like this, which I just found at random.
Not saying this one, but one.
Is that possible, or are some here of the opinion that the information they have is not written down in any texts on the subject?
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html
menicholas, the problem with the site you link is that it has no cites to source material and the testing that backs the source material. You are better off looking to gain an understanding of climate physics in a beginning college level meteorology text, several of which are on line if you have the chops (pre-req.s) to read them. The best ones willl cite past experimental work or publish their own weather & climate observations and instrumented lab tests.
Trick.
I studied meteorology on college.
And climatology.
And physics.
Interdisciplinary natural sciences.
My degree was physical chemistry.
The idea that there is a separate physics called climate physics seems as if it may be a problem, not a solution.
And I was the guy sitting in the front row getting an A in every one of those classes.
BTW, Trick, you are incorrect.
The site has an index and links to extensive references and such.
Menicholas, why not have this one instead.
http://ozonedepletiontheory.info/primary-problem-with-GG.html
Which contains many references.
I would prefer to have someone discuss not just the Physics, but also the Mechanics and Characteristics of how the CO2 part of the GHG Theory is so important.
“The site has an index and links to extensive references and such.”
Not on the page you linked. And no tags pointing to any of the references.
No
Whoops, Willis is asking about an argon atmosphere so you would not be a slayer just an agw proponent or a lukewarmer.
My answer would be yes on the basis of the conduction issue.
No — a purely non-GHG like argon could not raise the temperature above the S-B predicted temperature.
Paul says no, the atmosphere alone would not heat the surface.
wildeco2014 says yes, it would heat the surface.
Anyone else? Step up, here’s your chance, inter alia science is about taking a stand and defending it.
Regards to everyone, whether you have an opinion on the question you’re willing to share or not.
w.
Note that wildeco2014 and Stephen Wilde are the same. It just depends on which device I am using.
Sorry for any confusion but I thought people would realise.
I am still learning but my thoughts have been evolving from not being sure who to believe to believing that some people are just being hard headed and have confused the issue for people who are not physicists.
Gravitational collapse causes heating of a gas cloud, but there is no reason to think that heat is then stuck.
It seems implausible that an atmosphere with no (GHGs…are these the same as “radiative gasses”?) could just keep getting hotter and have no way to cool itself.
I say no.
I’m in the NO camp — the surface could radiate directly to space just as if there were no atmosphere. Since the atmosphere cannot absorb any energy from space or emit any energy to space, it can only exchange energy with the planet’s surface, so no net transfer to or from the surface.
Ed, how did you find an experiment that an argon gas atm. at STP cannot emit any energy to space? And thus emit no radiation to the surface either.
Ed, no net radiative transfer to or from the surface once the atmosphere is in place but there is a net conducted energy transfer from surface to atmosphere throughout the first convective overturning cycle.
That omission is the elephant in the room for the radiative theory.
Ed,
From conduction no?
So the initial heat from the collapse of the atmosphere would conduct to the surface which would then radiate to space?
The atmosphere would be warmed during the day by the ground, and convect and expand, and then be cooled at night and contract, no?
Seems the cooling part would be slower.
Trick:
A century’s worth of spectroscopic measurements in the laboratory has failed to find any IR emissions for argon gas. All of our theoretical knowledge as well says that there is no mechanism for argon gas to absorb or emit in the IR range.
Stephen and menicholas:
Since a transparent atmosphere has no mechanism to exchange energy with space, it can only exchange energy with the surface. This means that it must receive as much energy from the surface through conduction as it transmits to it — otherwise it would continually diverge in temperature from the surface.
The earth’s surface emits ~250 W/m2 more than the earth system absorbs from the sun (averaged over area and time). The zero-sum exchange between surface and atmosphere cannot remotely make up this difference.
What might have happened once billions of years ago cannot make any difference now. The residual geothermal flux density is less than 0.1 W/m2. There is no way that ancient event could still be providing 250 Joules every second over every square meter of the earth.
“A century’s worth of spectroscopic measurements in the laboratory..”
Cite just one for Ar Ed. Ought have a stacked high pile of ’em then.
NIST seems to indicate otherwise but I need to track down their 1973 ref. Also a chance these tables are in obscure handbooks, I’ll probably end up spending some time in the stacks. All I need is the spectral emissivity for Ar applicable to STP atm., can’t find it on the ‘net. Though I have developed a strange and sudden urge to get all my double glazed windows filled with Ar gas.
Tim Folkerts gives us another “No” vote … Tony, wickedwenchfan, Brett Keane, Steven Wilde, where do you guys stand on this issue?
w.
Dirty Argon with dust, water vapor, and the “usual” 0.65 to 0.75 atmosphere clarity, or laboratory pure clean Argon with nothing at all in it globally above a perfectly clean dust-free smooth surface with no oceans or seas?
(Heck, I’m still trying to find somebody who can re-create the 28 day lunar soil temperature cycle that we DO have measurements for in a “climate” model of a simple sphere in a vacuum. Not a flat-moon average temperature at equilibrium, but the pole-to-pole actual lunar soil surface temperature. )
RACookPE1978 November 30, 2017 at 6:47 pm Edit
Pure, undiluted, clean as clean argon. And water vapor is a GHG, no water vapor. No dust, it acts like a GHG although it’s an aerosol. Just argon.
Not sure why you’d want that, but the calculations are quite simple. The regolith (lunar dust) is a good insulator, so it heats and cools rapidly, and doesn’t transfer heat horizontally. Seems easy enough to do … I’m curious, though. What is it you hope to show with such a model?
And I’m curious about your answer to the argon question as well.
Thanks for the comment, always good to hear from you,
w.
Let me note that I bring this up because lots of folks say that either there is no downwelling radiation from GHGs, or that there is radiation but it doesn’t heat anything.
This leaves them trying to explain why the temperature of the earth stays above the S-B temperature. The usual explanation is that it’s the atmosphere itself that is keeping the earth warmer than S-B says it should be, and that it has nothing to do with GHGs.
So I ask again, can an atmosphere with no GHGs maintain a planetary surface temperature higher than Stefan-Boltzmann predicts from the solar energy radiating the planet.
w.
it’s the hydrosphere that captures the heat that a rock can’t.
the radiation goes deep and then it’s trapped cuz ir & water.
it’s not the atmosphere that heats the ocean. it’s the other way round.
Funny how Oleg Sorokhtin also agree with them about the The adiabatic theory of greenhouse effect.
But he is just some Russion Scientist without your grasp of things.
Oleg G. Sorokhtin – a graduate of the Leningrad Mining Institute in 1951, Dr.,., Academician (RANS), Honored Scientist of Russia, Honorary Polar Explorer. After graduating with honors from the LGI worked Hydroproject – hydrogeological exploration carried out by the “great construction projects of communism” (Kuibyshev HES, Turkmen channel). In 1953 he returned to Moscow and began work at the Institute of Physics of the Earth. Participated in three Antarctic expeditions, carried out a deep drilling and seismic studies, attended the Pole of Cold, the geomagnetic pole, opened the Pole of Inaccessibility. Since 1966 he has been working at the Institute of Oceanology. Shirshov USSR Academy of Sciences (RAS). He has participated in many oceanographic expeditions, fell to the bottom of the ocean, explore the underwater volcanoes, hot springs (black smokers). He has more than 300 scientific publications, including publications in “Reports of the USSR Academy of Sciences / RAS” and “Nature.” He was awarded the Order of Labor Red Banner, medals. He has two sons, two grandsons (the youngest – a year) and a granddaughter. The youngest son – Doctor of sciences, Professor of Murmansk State University (Apatity), corresponding member of Academy of Natural Sciences.
Or how about Hans Jelbring, no he won’t do as you have already made your mind up about him.
A C Osborn December 1, 2017 at 10:19 am Edit
Thanks, A C. Perhaps you decide whether a scientific claim is valid by looking at the resumes of people who agree with it. If you were to follow that to its reasonable conclusion, you’d have to believe Michael Mann, Peter Gleick, and a host of well-credentialed alarmists.
Me, I prefer running the numbers. I did that in my post The Mystery of Equation 8. No one has found flaws with it yet. I invite you to find one.
As to Jelbring, Dr. Robert Brown at Duke who teaches this stuff for a living wrote a very clear proof that Jelbring is wrong. Again, nobody has found any fault in his proof, but you’re welcome to try. That’s what science is about—not resumes, but logic and math.
Let me add that when Jelbring came out with his theory, I believed it and defended it.
However, Dr. Brown showed me I was wrong. So in contradiction to those who endlessly claim that my mind is made up, it’s not. When the facts change, I change my mind. Read Dr. Brown’s proof, it might change yours
w.
But it wasn’t just Oleg, it was quite a few others, with far more expertise than you who do not just believe it, they propose it.
Like I said why don’t you try argueing with all of them and see how far you get on.
And let me add that Nikolov and Zeller say that the answer is definitely “Yes” … and so do Volokin and ReLlez, so that’s four more in the “Yes” column.
w.
Trick, you basically said “Yes, if argon is a GHG”, because you specified that the emissivity of argon is 0.01.
However, the emissivity of argon for thermal IR is zero. It’s monatomic, so it is physically incapable of absorbing any thermal IR.
Given that case, a perfectly IR-transparent atmosphere, what would be your answer?
w.
“However, the emissivity of argon for thermal IR is zero.”
Zero or near zero? Tests show increasing pressure broadens the lines of any atm. gas in several unrelated ways. Can you cite an experiment showing your conclusion for argon emissivity 0.0 looking up at Earth atm. pressures? I get the feeling you are guessing 0.0 just like I guessed 0.01.
This is why I asked about pressure. Atm. so close to transparent as to be unmeasurably different from 0.0 atm. emissivity, then 255K results for no GHE, but there would still be an ATE over the moon as the surface emissivity and albedo sans atm. drive that physics.
Trick, the issue is that when a molecule absorbs IR, the IR is transformed into movement of the bonds between the atoms making up the molecule—they flex, they scissor, they twist, they stretch, depending on the number of atoms in the molecule they can have several flexing modes, and thus several different absorption bands.
However, monatomic gases can’t do that. There’s only one atom, no bands to flex. As a result, their absorption is not just low—monatomic gases can’t absorb at all because they have no bonds for the IR to grab on to and flex.
For example:
Finally, I’m sorry, but what is an “ATE”?
w.
Willis:
An “ATE” is an “atmospheric thermal effect” — it’s what Nikolov and Zeller (or is it Volokin and ReLllez?) assert is the alternative warming mechanism on planets.
Sometimes you’ll see it called the “atmospheric pressure effect” (APE). I like to think of it as the APE-sh*t effect.
Willis 8:55pm, I am unfamiliar with plasma jet spectroscopy, that paper is paywalled. If you think it holds the answer to finding an experiment for the emissivity of argon at Earth surface STP (looking up), I’ll go obtain a copy at the local college library.
”Finally, I’m sorry, but what is an “ATE”?”
The way you write about N&Z (V&R if you prefer), I was under the impression you had read their 2017 paper. Ed Bo is correct. See end of intro. paragraph and page 3, first paragraph under reference temperatures and reference pressure.
”monatomic gases can’t absorb at all”
Then you are the first in science to have discovered a pure white body object in argon gas. If argon can not absorb, it won’t emit and therefore is white (and black). However, I am willing to bet if you construct a very transparent container for some argon gas at say 1000C and place it in deep space, the gas will emit radiation and cool. If argon emits, argon can absorb.
Trick December 1, 2017 at 2:19 am
Trick, I just provided you with a citation saying, word for word,
Since they said it, how could I be the “first in science” to say that? And why would you think that all things MUST absorb and emit thermal radiation? Is there some law of nature I don’t know about, that says “All things must radiate thermal energy”? People mostly say that because that’s what they put in the high school textbooks, but what those textbooks mean is that all SOLID objects radiate thermal energy.
Finally, we are not talking about argon at 1,000°C. We are talking about thermal radiation at Earth-like temperatures. If that was not clear before, it is now. Nobody but you has said anything about gases at a thousand degrees, certainly not me. I suspect that if you torture argon enough it would confess, and that there could certainly be some combination of heat and pressure at which argon will do strange tricks … but argon in our normal environment neither absorbs nor emits thermal radiation.
Best regards,
w.
SpectraCalc.com provides some empirical support for Willis. They list the IR spectra for over 40 different atmospheric gases — but NOT argon (the 4th most common gas after N2, O2, & H2O). The only reason to leave it off is that it’s spectrum is completely negligible — less even than N2 and O2 that are listed.
Willis, it is a minor point debating the difference between steady state at 255.3K (argon .01) and 255.0 (argon 0.0). However, it’s sort of interesting as I don’t yet know the answer. Thanks to Tim F. doing some investigating but not yet finding Ar emissivity from experiment.
You have given a reference for argon plasmajet emissivity. I am not familiar and can’t read the paper; please explain emissivity in Ar plasmajet: is spin constrained? Ar being monatomic in earth atm. can spin and that spin is found quantized not continuous; also velocity differences will cause a doppler shift in the spin lines broadening them among other reasons.
All objects emit at all temperatures at all frequencies all the time, no exceptions. No means no so includes Ar gas. OK, make the very transparent container of 1atm. Ar in deep space start at 255K. I will bet a test will show it will emit and cool. I’ll look later for a ref. showing identified Ar spectrum in deep space and an emissivity test.
No steady state equilibrium earth surface & an Ar atm. is possible until the Ar emission spectrum shifts to regions for which the emissivity is not zero.
—-
Tim 4:12am, “completely negligible” is not zero. Ar emission cannot be identically zero as Willis is writing.
I was about to agree with Trick that it seems doubtful that there are substances that cannot emit any radiation, and so if they were hot and floating in space they would stay hot.
I used to read a lot about gas clouds in space, and how they are detected via various emission and absorption spectra.
So I did a quick search to see if any argon cloud have been discovered in space.
If they have, then they must be either emitting or absorbing.
I found this article, that a few years ago, they found argon in a gas cloud, the one around the crab nebula.
But…oooh…it was only detectable because it had formed a hydride, and unexpected result for a noble gas.
Reading between the lines, we know there is argon in these clouds, there has to be because supernovae produce argon…argon 36. The argon 40 we have a lot of here on earth comes from the decay of potassium.
And they can only see it when it forms an unlikely molecular pairing with a hydrogen ion.
I recall reading of some other instances a long time back that noble gasses can form molecules…it is unlikely but not impossible.
Here:
https://www.universetoday.com/107154/argon-the-first-noble-gas-discovered-in-space/
And then there are the argon windows…very efficient.
And we know why…no emissivity. Or really really low? *shrugs shoulders* Seems not…seems it may be goose eggs.
Golly, you can learn a lot by looking stuff up.
https://www.thebalance.com/cost-benefits-and-drawbacks-of-argon-gas-windows-844558
Trick, in thermodynamics and statistical mechanics, you quickly learn:
1) very few things are completely impossible.
2) it is easy to find things that are so improbably they would likely not occur in several lifetimes of the universe.
So, I am not going to say that argon @ur momisugly 300 K categorically can’t emit thermal radiation. On the other hand, if 1 IR photon came out of a cloud of argon each year, I am not going to worry about that process when calculating the energy balance of earth’s atmosphere.
Trick December 1, 2017 at 6:30 am
All objects emit at all temperatures at all frequencies all the time, no exceptions. No means no so includes Ar gas. OK, make the very transparent container of 1atm. Ar in deep space start at 255K. I will bet a test will show it will emit and cool. I’ll look later for a ref. showing identified Ar spectrum in deep space and an emissivity test.
No steady state equilibrium earth surface & an Ar atm. is possible until the Ar emission spectrum shifts to regions for which the emissivity is not zero.
—-
Tim 4:12am, “completely negligible” is not zero. Ar emission cannot be identically zero as Willis is writing.
No, the energy levels for Ar are well known and will require excitation in the far UV at near ambient temperatures there is no way they will be populated so no emission also no way that Ar can absorb IR, there’re no available energy levels to accept those wavelengths.
Phil., if that is well known about Ar it ought to be easy to find, post up an experiment showing such.
—–
Tim F., “On the other hand, if 1 IR photon came out of a cloud of argon each year..”
Then the answer to Willis’ question:
“Do you think that a non-GHG atmosphere (say argon) can maintain the surface temperature of a planet at a level higher than the calculated Stefan-Boltzmann temperature of that planet given its incoming solar radiation?”
…is calculated YES by 1 IR photon each year. If 0 IR photons ever come out, then the answer is NO.
Trick December 1, 2017 at 12:30 pm
Phil., if that is well known about Ar it ought to be easy to find, post up an experiment showing such.
https://physics.nist.gov/PhysRefData/Handbook/Tables/argontable5.htm
Phil., thanks. I pounded around on many of the live links could not find any emissivity information for Ar or anything else. Maybe you have a link for measured emissivities?
The list of Ar “strong lines” in air shows many wavelengths in IR range though. As I wrote below, once the AR atm. is in place, no steady state equilibrium earth surface & an Ar atm. is possible until the Ar emission spectrum shifts to regions for which the emissivity is not zero.
Mr Eschenbach says “Is there some law of nature I don’t know about, that says “All things must radiate thermal energy”?”
Well you are contradicting Wiki
“Thermal radiation also known as heat is the emission of electromagnetic waves from all matter that has a temperature greater than absolute zero”
Note All Matter, no mention of solids, liquids or gases.
Credit to Willis for calmly narrowing the issue down to my single point.
The whole debate depends on one question and one alone.
Where does an atmosphere in the process of forming acquire its internal energy from?
If it comes from a temporary reduction in outgoing radiation then the pressure hypothesis is correct.
If it comes from a temporary reduction in surface temperature then the radiative theory is correct.
We have here the opportunity to destroy one hypothesis or the other.
Observations of multiple objects with atmospheres support the pressure based theory.
What else do we have either way?
I’ll explain why the only two available options are so brutally stark.
If the developing atmosphere takes the energy it needs whilst rising against gravity from outgoing radiation then there is no drop in surface temperature so that the surface must warm when convective descent begins.
If it takes its energy from the surface then the surface cools during the formation process but warms back up again when convective descent begins
In the first case you have a surface warming effect without DWIR
In the second case you need to add a warming effect from DWIR to get the surface temperature rise.
Stephen:
Here you say…..
“If the developing atmosphere takes the energy it needs whilst rising against gravity from outgoing radiation then there is no drop in surface temperature so that the surface must warm when convective descent begins.
If it takes its energy from the surface then the surface cools during the formation process but warms back up again when convective descent begins”
You are right about this being a crucial piece of logic to your premise that atmosphere alone can raise surface temperature but my position on this, based only on 1st law, is quite different. Here’s why……
If you have planet with no atmosphere, we all agree that EIN=EOUT and all transfer is by radiation. In previous comments, though you never outright said this, you implied that surface temperature is set by insolation. I have to deduce this when you say that even though your ‘developing atmosphere’ is siphoning off energy from the surface, the surface temperature is maintained constant by the ample supply of insolation.
I have two problems with this argument of yours.
One. You are making this argument while the system is in a state of disequilibrium. There is a net positive flow into the system during your `development` where outflow is inflow minus what is going into the new atmosphere. So the system is not in thermal equilibrium and should have no place in a discussion of 1st law energy accounting.
Two. If you insist that we go there, while developing atmosphere, the surface is shedding energy by two methods: radiation and conduction. It will radiate as a function of how much energy it needs to get rid of to maintain balance. By your own statements you acknowledge that outbound energy is being reduced by what is transferring to atmosphere. Radiation of a body is a property of the body as is its temperature. You simply can not claim that a constant temperature is being maintained by insolation. To do that implies that you are using insolation and T in S-B calculation. Surely you would agree that this is an invalid use of that equation.
Until we resolve this fundamental question i respectfully request that you don’t wander off to other planets or convection or lapse rates. There really is no point to going on about this topic (by any participant in the various atmosphere vs LWIR threads) and extending it to deeper considerations if we can’t all agree on something so basic.
Please address points one and two seperately.
Are we, after 4 billion years, in thermal equilibrium on a planet wide basis over a reasonable integration period?
Do you believe surface temp is f(EIN)?
i) It is perfectly reasonable to propose a temporary period of disequilibrium resulting in a stable surface temperature when the inflow remains constant and the outflow is split into two components adding up to the same energy value as the inflow. The point you seem to miss is that during that period the siphoned off energy goes to PE which does not register as thermal energy and so cannot manifest itself as a temperature rise until it gets back to the surface. That PE is ‘hidden’ from temperature sensors until it is returned to the surface as KE on closure of the first cycle. Thus the surface temperature stays stable during cycle one but rises as soon as the adiabatic loop closes.
ii) This is also dealt with by the point about the non thermal nature of PE. PE does not radiate so the temperature as viewed from space can drop whilst PE is building up within the atmosphere even though the surface remains at S-B. The thing is that as long as PE is in the process of being created it is not sufficient to rely on the S-B equation. The non thermal nature of PE breaks the radiative rules. The S-B equation is radiative only and does not deal with non radiative processes that create non thermal PE.
Here is the very important thing you’ve got wrong….
You keep insisting that the developing atmosphere is gaining energy by taking it from the outgoing radiation. If that were the case then you can make a nice looking diagram with a diversion of outgoing energy and a constant temperature surface so say, 240 w/m^2 In, 240 out in two vectors, one to space and one to atmosphere. Surface is in balance, T stays the same.
That is not what is happening. By your own definition you have an atmosphere that does not interact with IR. IT can only interact with surface in your definition.
That being the case, you have to account for the fact that the surface has Joules in it. Your developing atmosphere is not taking energy from radiant flow. It is taking it from the pool of joules in the surface. It is reducing surface energy content. That reduces surface radiation. The reduction of outgoing LW is not a diversion in a flow. It is an outcome of a reduction in surface energy content.
Surface energy content goes down when you fill your DA and it is exactly replaced once your new atmosphere equilibrates with the surface. In balance, after enough time, no change in surface temp and for T analysis atmosphere is irrelevant.
And BTW, I am perfectly ok with a system imbalance while your atmosphere comes to speed. Indeed, I would expect it. And while all that is going on, gravity and convection, and evaporation will all move around lots of energy but eventually the surface atmosphere
Boundary equilibrates.and when it does, it goes away from the math. Zeroes right out.
So you have a pool of joules at the surface which provided a pre atmosphere surface temperature for Earth of 255k as per S-B.
Those joules can either be radiated away by release of photons or conducted away into adjoining molecules. Both cannot occur simultaneously otherwise there is a breach of the first law.
During the formation of the atmosphere joules to the value of 222k release photons to space and joules to the value of 33k get conducted to the atmosphere where they do work in lifting mass up against the downward force of gravity.
How does that reduce surface temperature below 255k?
On completion of the formation process you still have the surface at 255k worth of joules as per S-B but you are also then getting an additional flow of joules arriving at the surface in adiabatic descent worth another 33k to make it 288k.
Thereafter the 288k surface radiates to space at 255k as per S-B and recycles 33k up and down to keep the atmosphere suspended off the surface against gravity indefinitely.
PLEASE take 10 minutes and draw a cartoon with your own ramblimgs. You can not reduce the energy content of something without a temperature drop.
During the formation process there is radiation worth 255 coming in constantly to replace the 222k going out to space and the 33k going into conduction.
How do you reduce the surface below 255k without being in breach of S-B?
The energy going into your atmosphere is not coming from radiation. It is coming from the energy that is contained in the surface.
Joules go down. T goes down. LW out goes down.
During your ‘devolopment’. EIN>EOUT for the system.
During the development of the atmosphere energy in is more than energy out but the excess of energy in is passed to PE which is not heat and which does not radiate. The resultant accumulation of energy in the system continues until the first convective cycle completes and then the additional energy becomes apparent at the surface as a 33k surface temperature enhancement.
Sigh
Suit yourself. I can see when I’m flogging a dead horse.
Imagine a really big bucket full of molten cheese in a mouse house. The mice that live there are very neat but not so bright.
For years, the mice kept putting cheese balls into the bucket until one day it overflowed. The neat mouse dullards scooped up the molten cheese and turned it into more cheese balls. Then they sent these extra cheese balls back to the cheese ball factory.
Since they weren’t too bright, it never occurred to them that stopping the cheese ball orders would have saved them from all the hard work making cheese balls to send back so it went on and on. Cheese balls in cheese balls out.
Then one day a mouse called Albert, said to all the tired cheese ball mouse workers, “I am going to town to buy a bucket. I will put my small bucket under your very large bucket and we will make a hole in yours to catch the extra melted cheese and you won’t have to make cheese balls anymore.”
The cheese ball worker mice were very happy. As soon as they opened the valve to the new bucket their lives improved. It wasn’t perfect because the hole wasn’t big enough to catch all the extra cheese. Each day 10 new cheese balls were delivered but only 9 cheese balls had to be cast from the overflowing messy big bucket. Better than before for sure.
For many months it was 10 cheese balls put into the big bucket and nine returned to the factory. Then one day, to their great dismay, the little bucket started to overflow. The cheese ball worker mice were moaning and cleaning up the mess, ready to turn it into cheese balls.
Happily, Albert came to the rescue with a cleanup plan. Look, if we just take the little bucket overflow and put it back in the big bucket every day, we won’t have to make more cheese balls we can keep on making 9.
So they did it. The next day, they went to the cheese room and found 10 cheese balls worth of melted cheese on the floor anyway. No matter how fast they moved the little bucket overflow into the big bucket, it was the same thing. 10 balls in 10 balls out. They were right back where they were months ago.
The moral of the story?
Don’t be a cheese ball! Just because Albert says it’s so, doesn’t mean you don’t have to do your own figgers’.
Ten cheese balls on a table. ( or 255k ready to go to space)
Send 5 to factory and put 5 in bucket. ( or 222k to space and 33k to atmosphere)
Return the 5 from the bucket to the table at the same time as 5 from the NEXT 10 made go into the bucket so you then have 10 on the table plus 5 in the bucket for a total of 15 ( or 222k + 33k = 255k back on table and 33k still in atmosphere being added to the surface in descent = 288k)
Send 10 to the factory.(255k to space) but since you waited for the next batch of 10 before despatch to the factory you still have 10 on the table PLUS 5 in the bucket
You have 5 in the bucket constantly and the most recent batch of 10 still on the table.
So, on the surface of the Earth you have 255k just arrived having just sent 255 to space and you still have 33k KE recovered from PE in descent at the surface which then warms above S-B.
You see, it is all in the TIMING.
The time taken for the first convective cycle causes the first 5 in the atmospheric bucket to be retained until the next 5 are ready so you always have 15 despite 10 in and 10 out.
That is why you MUST INTEGRATE BOTH radiative and non radiative processes to get the actual surface temperature for a planet with an atmosphere. S-B does not apply if non radiative processes interfere with the radiation budget.
It still applies from a point outside the atmosphere, however.
Can’t see any typos but you never know. It was tricky to formulate but readers should get the point anyway.
Nothing there but cheese, mice, and buckets. Don’t know what those 255,288,33 things are
Too painful to admit that a ten year old would know that energy loss delayed is temperature gained.
Glad to see you have come around.
Well at least you still have a sense of humour 🙂
The energy content of the system as a whole does not drop. It increases due to the creation of PE whilst the atmosphere is forming.
The energy content of the surface does not drop because it is still being fed new insolation at the same rate as pre atmosphere.
You can do a chart if you wish but it looks so simple to me that a ten year old could follow it and at present I don’t know how to get a chart into this thread and don’t intend to spend time learning how.
Ongoing convection is indeed net zero but due to the energy required to establish the atmosphere in the first place the system is always one cycle out of phase.
The result is a shedload of PE in the atmosphere that when it returns to the surface in descending air (half the atmosphere at any given moment) produces KE at the surface which raises the Earth’s surface temperature by 33k
did you read your last paragraph????
You have an atmosphere in thermal equilibrium (no net energy change) maintaining a surface T 33 degrees above what it HAS TO BE based on the inbound energy.
Please, please, please, draw some elementary diagrams and put your own numbers on them. What you are saying simply can not happen.
There is a net energy change during the formation of the atmosphere.
The S-B equation only works for a body without an atmosphere or from a viewpoint outside the atmosphere of a body with one.
“hereafter the 288k surface radiates to space at 255k as per S-B..”
No Stephen, the 288K surface radiates to space at 288K as per S-B.
The Earth has never radiated to space at 288k
If you can find a source that sets the Earth’s radiative loss to space at 288k I’d be interested to see it. I seem to recall that it is 255k.
“The Earth has never radiated to space at 288k”
Geez, Stephen, come on, the earth global surface on avg. radiates direct to space at 288K thermometer temperature in the so called window. Top of earth atm. (say defined by CERES orbit) radiates at brightness temperature 255K.
Well if top of atmosphere radiation to space is 255k then the surface at 288k is not radiating to space at 288k. Something gets in the way and that is conduction and convection. The additional internal energy acquired when the atmosphere formed is heating the surface by 33k.
“Well if top of atmosphere radiation to space is 255k then the surface at 288k is not radiating to space at 288k.”
Apparently, my ref. to “window” needs explanation, as the spectrum(s) posted above show, the atm. can be transparent to LWIR from the surface at certain wavelengths (frequencies) so some of surface radiation is not absorbed by the atm. (never gets to conduct and convect) and escapes to deep space. This radiation is shown in the energy budget arrows as an arrow traced straight up and out, that process is referred to as the window for radiation to escape directly to space.
This is possible at Earth surface pressures and atm. opacity; this window does not exist for Venus (is closed
to IR as surface atm. is totally IR opaque) and is much wider open on Mars due to its much lower total surface pressure.
I’ll explain why the only two available options are so brutally stark.
If the developing atmosphere takes the energy it needs whilst rising against gravity from outgoing radiation then there is no drop in surface temperature so that the surface must warm when convective descent begins.
If it takes its energy from the surface then the surface cools during the formation process but warms back up again when convective descent begins
In the first case you have a surface warming effect without DWIR
In the second case you need to add a warming effect from DWIR to get the surface temperature rise.
Your premise is built upon a ‘developing atmosphere’, right? Well that defines an atmosphere that is a net consumer of energy which means it is not in thermal equilibrium (defined as no net energy flow).
Do you have an energy flow accounting for thermal equilibrium? What do you think is a reasonable integration interval to use that would provide an equilibrated flow solution.
“If the developing atmosphere takes the energy it needs whilst rising against gravity from outgoing radiation …”
… and the only way to “take energy from outgoing radiation” is to absorb some of that radiation, which requires gases (or aerosols or clouds) that can absorb thermal IR. In other words — the greenhouse effect!
It takes it from the ground. Conduction reduces the ability of the surface to radiate to space thus it is taken from outgoing radiation and the surface temperature does not drop because fresh insolation maintains it.
And while this
And while this is going on what is the value of the outbound radiation?
“In the first case you have a surface warming effect without DWIR.”
This warming exists in Stephen’s imagination only until such time as Stephen can produce an observation of nature or lab test supporting his assertion.
The convection experiment above shows the low pressure system created by the rising red dye tracing the fluid is filled in with higher pressure surface” fluid at ambient temperature, no warming. Any red dye that makes it back to surface will be at surface ambient, no warming. The convective process tested has no surface warming effect.
Should be easy for Stephen to set up the test as shown, add some thermometers and prove it one way or the other. Simple logic means there is no surface warming demonstrated in the test and it shows no descending columns.
Yes. Sufficient insolation is needed to make that first rise.
The sun-warmed ground would conduct energy to the gas/ice.
A lapse rate could form, and phase changes at height could assist this. Depends on TSI.
Persistence of this atmosphere would be made harder if too much evaporative cooling was needed eg gas bleeding off into space, taking IR energy with it.
The existence of any atmosphere leads to an average emission height, and increasing energy density below that ie lapse rate. Isothermality seems very hard to achieve in an orbital situation.
Various planets and moons suggest the above, but it would be foolish to be dogmatic on my/our state of knowledge.
I take it that both Brett and Steven Wilde are in the “Yes” crowd. Anyone else have the courage to take a stand on this one? Tony? Anyone?
w.
yeah- but i’ll stand on water. this is a water planet.
it does not radiate like a rock. it can not. it traps radiant energy and then can not reradiate the energy as ir cuz it’s not transparent to ir.
that’s what heats the atmosphere.
Thanks, gnomish. As I said, I’m not interested in the proposed mechanism at the moment, just whether it’s possible. You say yes.
w.
https://www.britannica.com/technology/solar-pond
the question was ” can an atmosphere with no GHGs maintain a planetary surface temperature higher than Stefan-Boltzmann predicts from the solar energy radiating the planet.”
my answer is on a water planet, GHL does do it all day and night.
The Russian Scientist Oleg Sorokhtin also believes in The adiabatic theory of greenhouse effect.
But of course he obvioulsy doesn’t have Mr Eschenbach’s graps of the Science.
Oleg G. Sorokhtin – a graduate of the Leningrad Mining Institute in 1951, Dr.,., Academician (RANS), Honored Scientist of Russia, Honorary Polar Explorer. After graduating with honors from the LGI worked Hydroproject – hydrogeological exploration carried out by the “great construction projects of communism” (Kuibyshev HES, Turkmen channel). In 1953 he returned to Moscow and began work at the Institute of Physics of the Earth. Participated in three Antarctic expeditions, carried out a deep drilling and seismic studies, attended the Pole of Cold, the geomagnetic pole, opened the Pole of Inaccessibility. Since 1966 he has been working at the Institute of Oceanology. Shirshov USSR Academy of Sciences (RAS). He has participated in many oceanographic expeditions, fell to the bottom of the ocean, explore the underwater volcanoes, hot springs (black smokers). He has more than 300 scientific publications, including publications in “Reports of the USSR Academy of Sciences / RAS” and “Nature.” He was awarded the Order of Labor Red Banner, medals. He has two sons, two grandsons (the youngest – a year) and a granddaughter. The youngest son – Doctor of sciences, Professor of Murmansk State University (Apatity), corresponding member of Academy of Natural Sciences.
Plus of course there are massive holes in the current CO2 warming theory that involves the small matter of the number CO2 molecules and the Distance of them Surface that the Radiating molecules are and that the other 70% by Density of Air molecules between them and the Surface and add to that 70% of the Surface is not warmed by LWIR, but by SWIR.
I forgot so do G.V.ChilingarN.O.Sorokhtin, who work with him
Ho, and L. KhilyukM. V. Gorfunkel as well.
Oops I forgot this one S.A. Ushakov
Whoops there goes another one, how about John O Robertson.
My My, perhaps Mr Eschenbach should go and argue with with all those emminent Scientists instead of us LOL
Then od course there is Alberto Miatello, but he doesn’t count because he belongs to PSI.
This is getting embarrassing, look another one Charles R. Anderson, Ph.D., Physics.
Perhaps all you guys supporting Mr Eschenbach should have a rethink, don’t you?
To be fair to Willis, he seems more receptive to the idea now than he used to be unless I’ve misinterpreted his expressed wish to discuss it with us.
Do Scientists debunking AGW count?
https://arxiv.org/pdf/0707.1161v4.pdf
How about Theo Wolters, will he do?
Add M. Liu to the list.
Trick December 1, 2017 at 2:19 am
Why would I do that? I gave up on those jokers long ago, and when they tried the scam with the fake names, that was the last straw. However, reading over my posts from five years ago, I find that at one point I did know what ATE was, their so-called “Atmospheric Thermal Effect” … but obviously, I don’t try to retain such nonsense.
And Trick, if you are impressed by their equation and its ability to emulate the planetary temperatures, then you definitely should read my post entitled “The Miracle of Equation Eight“. It lays out all of the problems with their theory … I’ve written well over 500 scientific posts, and I’d totally forgotten about that one. It’s an interesting analysis, and it has stood the test of time. Not only did I show the problems with their bogus equation, I wrote and tested my own bogus equation which was both simpler and gave more accurate results than theirs. It’s just as meaningless as theirs, but it’s simpler and works better …
Finally, do you agree with N&Z that a non-GHG atmosphere, say argon, can permanently raise the temperature of a planet?
w.
I tend to agree with Willis that N & Z’s workings do not provide proof of anything for the technical reasons he stated but that does not mean they are wrong.
There is substantial evidence of a relationship between atmospheric masses and the surface temperature enhancement for various planets and moons in the solar system
The presentation by this chap provides several worked examples which are hard to ignore:
I prefer to get basic concepts and physical laws right before worrying about equations and it seems clear to me that the proposals I have put forward based on conduction and convection do provide an arguable alternative to the radiative hypothesis.
Is anyone else going to suggest a means of choosing which is more likely to be correct?
“provides several worked examples”
Here this “chap” curve fits an ideal equation with all measured data similar to N&Z (V&R) fit. There is no law against doing so but claiming curve fitting invalidates the GHE does not follow simple logic. All exoplanets can be curve fitted to the same eqn. once the data is measured but the point is to acquire enough knowledge to estimate exoplanet nature prior to having that measured data.
I don’t see that ‘chap’ as having engaged in curve fitting. He simply points out facts and gives an equation that actually works.
What no one has done other than me is to provide a conceptual description that is capable of explaining those particular facts better than the radiative theory can.
It comes down to this:
If the formation of an atmosphere causes a temporary drop in outgoing radiation then the pressure based theory is correct.
If that formation causes a temporary drop in surface temperature then the radiative theory is correct.
Which is it and why?
”I don’t see that ‘chap’ as having engaged in curve fitting.”
Then Stephen doesn’t understand curve fitting. What I mean here is the guy consulted papers (did you read them?) for information on measured planet median P(z), density(z) and he/they used the IGL to compute median T(z). So when he uses two of the parameters to back calculate the third, he is not adding anything new from the papers. And he just plots the various planets examined with the parameters chosen to show a good fit on the IGL curve like N&Z (V&R) did. No one typically measures avg. atm. density(z) in situ on a planet, it is typically found from probe (radio signal modulation) and T(z) calculated using IGL.
”Which is it and why?”
Consider the tank of water in the experiment above that should be teaching Stephen something about convection in a fluid warmed from the bottom in a gravity field.
Stephen now imagines the time of “formation of the atmosphere.” Let’s use that tank to add some climate formation realism. Imagine someone dropping in 4” diameter stones randomly different s[peeds,directions, 3” diameter iceballs even more randomly, someone sloshing the tank around, the tank glass walls and bottom still being poured from molten glass, then someone sloshes in very hot water & about 40% splashes out, all the time random temperature hot water cups are placed under the fluid on and off. THEN at some point while all this happening, the cup of hot water is placed as in the video and we are asked to answer 2 questions about the resulting convection.
Does this lab test exhibit a temporary drop in outgoing radiation or a temporary drop in surface temperature?
Alas, this experiment exhibits both answers to these questions at different times. To determine your answer, would require examination at a certain time & that is past history. No data is available. Thus, if you want to believe that there is drop in outgoing radiation or a drop in surface temperature, you may do so. No one can prove you wrong. But you cannot prove you are right. When faced with an undecidable proposition, you may believe whatever you wish. As you often do.
i) So you are suggesting that since the atmospheric parameters were calculated from probes by applying the gas laws the outcomes can safely be ignored? I don’t think so.
ii) With that tank there is no rotating sphere, no gas subject to the gas laws and no density gradient with height so that model is just silly. It is simple logic that if a flow of energy is split into two components then the first flow will drop by an amount equivalent to the second flow. Calling that ‘imagination’ is silly. One can easily resolve the answer to my rhetorical question by noting that conduction and convection work more slowly than radiation so introducing such processes can never reduce the surface to a temperature below S-B and the pressure hypothesis must be correct but if you can demonstrate otherwise then please do so.
”the outcomes can safely be ignored?”
No. What makes you write that? That the video invalidates the GHE does not follow simple logic.
The tank is on a rotating oblate spheroid. The tank contains a fluid. There is a density gradient with height. Find one for fluid called air. With particles used and find it demonstrates the same physics, no surface warming, particles return at ambient.
”..introducing such processes can never reduce the surface to a temperature below S-B..”
True all objects radiate at S-B not at a temperature below S-B. The pressure hypothesis must be correct only if you can demonstrate such in a test or observe such in nature. Until then all you have is imagination.
Until you have something new Stephen, I may not reply.
I didn’t say the outcomes from probes and the application of the gas laws could be safely ignored. I was pointing out that that was what you seemed to be saying.
As for the tank experiment it contains a liquid that does not observe the gas laws. Can you link to one that does use a gas ?
To study/test/learn about convective physics, all Stephen needs is a fluid warmed from below in a gravity field. The IGL is immaterial. Water and air are both fluids.
Willis, it is unfair to comment “nonsense” on anyone’s writing until you have read their exact words viz. “When you comment, please QUOTE THE EXACT WORDS THAT YOU ARE DISCUSSING, so that we can all understand the nature of your objections.” My memory also is not perfect and I prefer no abbreviations.
“do you agree with N&Z”…?
Those following are Willis’ words. Quote N&Z (V&R) exact words and I will keep it civil (top post also) and agree or disagree with them based on experiment and observation of nature. I’ve written my own words based on Willis’ proposition of Ar earth atm. above.
The N&Z name reversal was to demonstrate the nakedness of those who imagine their superiority. Sure worked. There have been other works showing up the same ritualistic beliefs in other ways, but also quite purposeful. The targets are expected to be the last to see the joke. Therein lies the effect.
ATE has been closely debated and tested for around a decade, maybe more, in its current debate. My own database concerning it is around 1GB. It took me on a journey into Quantum Theory and very much else. It will not go away, but nor do I mistake it for any ‘revealed truth’ that all must follow. Popper learnt his trade in my country, NZ, and I would be a fool to forget what he learnt and taught us.
heh- popper proved that nobody can prove anything – obviously a divinely revealed troof.
(i suppose i need to add an iron tag, but i was tempted to let it fly over you head again as it did the first time when popper zoomed ya.)
Trick, I linked to an entire post that I wrote about Nikolov and Zeller, which discussed their theory in detail. If that didn’t satisfy you, nothing will. I also invited you to find any flaws in my post. You have given me nothing.
Still waiting …
w.
Brett Keane December 1, 2017 at 10:26 am
If what you mean by “worked” was that it got their paper pulled from publication, got them mentioned in Retraction Watch, and exposed them to ridicule around the web, well yeah, it “sure worked” all right …
w.
Willis, your post was 2012. The 2017 paper didn’t exist back then. Quote N&Z (V&R) words from the 2017 paper being discussed per your note: “When you comment, please QUOTE THE EXACT WORDS THAT YOU ARE DISCUSSING, so that we can all understand the nature of your objections.” Also your be civil applies.
Still civilly waiting…
Trick December 1, 2017 at 7:24 pm Edit
Trick, obviously, somewhere I said the word “nonsense”. I’m sure I’ve said it lots of times. But since you have not quoted what I said, I fear you are making no sense. AFAIK I did NOT say that their 2017 paper was nonsense, because I have not read their 2017 paper. Sorry, not interested, they are puerile curve-fitters who’ve burnt their bridges with me. So … please quote whatever it was I said about “nonsense” that you are objecting to, and I’ll be happy to take a look at what I said.
In any case, it is totally unclear why you think I’m obliged to discuss a paper I haven’t read and won’t read …
w.
Willis, just yesterday you wrote specifically that a 2017 paper you haven’t read is nonsense. Then asked me a question about what they supposedly wrote using your words not theirs. Despite your top post request to always use the author’s words. The entire Willis quote is nearby:
https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1/#comment-2681969
” if you are impressed by their equation “
Which 2017 eqn.?
”do you agree with N&Z that a non-GHG atmosphere, say argon, can permanently raise the temperature of a planet?”
Where specifically do they write this in the 2017 paper? These are Willis’ words, I’ll respond to their words, point them out.
”In any case, it is totally unclear why you think I’m obliged to discuss a paper..”
You are not; you commented on the 2017 paper in your own block quote as “..such nonsense….just as meaningless as theirs..” and then asked me a question “do you agree” to something in it – so point out the something so I can answer. Simple re-re-request. Waiting.
Trick December 2, 2017 at 5:28 am
Trick, thanks for (finally) linking to what you are referring to. HOWEVER, what I actually said in the comment you linked to was as follows. YOU brought up the 2017 paper, viz:
I answered:
Please read that again, and point out to us where ANYTHING in what I said refers to the 2017 paper. I specifically said “five years ago”, which on my planet is NOT 2017. I also SPECIFICALLY SAID I DID NOT READ THEIR PAPER!!! So please, stop accusing me of claiming their 2017 paper was nonsense. I said no such thing.
Who said anything about a 2017 equation? Not me. I’m discussing their so-called “MIRACLE” equation.
Again, what is it with the 2017 paper? I have been discussing their claim that they can predict the temperature of a planet by knowing only the pressure and the solar input. If that is so, then it must work whether there are GHGs in the atmosphere or not. It’s called “a logical deduction from their claims”.
When you quote a man, don’t strip out the context. Here is what I said:
Now … am I talking about their 2017 paper there? NO. I’m talking about the paper discussed in “The Mystery of Equation Eight“.
If you had actually read “The Mystery of Equation Eight”, perhaps you might have noticed that it was written in 2012 … so your claim that I was referring to 2017 is just more of your endless BS.
RE-READ WHAT YOU QUOTED OF MY WORDS, and point out to me where it says anything about their 2017 paper. Simple re-re-request. Waiting.
w.
Willis you are being temperamental, it must be a mirthy day for you. I had written a longer response point by point but it isn’t worth posting it. Made me feel better though.
”Finally, do you agree with N&Z that a non-GHG atmosphere, say argon, can permanently raise the temperature of a planet?”
You will have to point me to where 2012 N&Z write this if you want to discuss further. My first impression is they did not write this.
It’s okay, gnomish does it fine. Some heads are close to the ground.
A C Osborn December 1, 2017 at 12:45 pm
Back to the Original Post.
Who has heard of Marc-Auguste Pictet?
Who has seen his experiment?
Yes to both.
Do you still believe Cold Objects can make Warmer objects warmer without the warm object being heated some other way?
Of course the Pictet experiment proves it.
Do you believe that the Cold Radiation from a colder object can be focused using a Mirror, just like Heat can?
Of course, I liked Rumford’s explanation of Pictet’s experiment where he used the analogy with sound:
“According to this hypothesis, cold can with no more propriety by considered the absence of heat than a low or grave sound can be considered as the absence of a higher or more acute pitch; and the admission of rays which generate cold involves no absurdity and creates no confusion of ideas.”
Comports well with our current understanding.
So based on that logic if a cold object CAN warm a continually heated warm object, CO2 at -80C can only warm the Earth’s Surface while the Sun is also warming it.
-80ºC has nothing to do with anything!
Once the Sun goes down it will not warm it at all. So at the moment no CO2 induced Warming in the UK for 16 hours per day.
Actually it will, because in its absence the incoming radiation would be from a much colder source. Be thankful for the warming 24hrs/day.
It was established at the beginning of this post that Cold Object cannot warm a warmer object UNLESS THE WARMER OBJECT IS BEING HEATED, otherwise it can only slow the cooling.
At Night the Earth is COOLING, so the radiation from a -80C CO2 Molecule is a Cooler Object.
Whic part do you not understand?
How did the Pictet experiment prove that a cold object makes a warm object colder?
It made the Thermometer colder from 16ft away by focusing the Ice’s Radiant Energy on it.
Sorry this “How did the Pictet experiment prove that a cold object makes a warm object colder?” should have been
How did the Pictet experiment prove that a cold object makes a warm object warmer?
It made the Thermometer colder from 16ft away by focusing the Ice’s Radiant Energy on it.
AC says: “How did the Pictet experiment prove that a cold object makes a warm object colder?”
The analogy here would be if Pictet started with dry ice at the focus of the mirror, then switched to regular ice @ur momisugly 0C. With the dry ice, the object at the focus of the other mirror would cool significantly (until some balance was achieved between heat in from the room and heat out to the dry ice). Perhaps that object would cool to 10C in a 20 C room. If the dry ice was replaced with regular ice, the object would warm (perhaps from 10 C to 15 C) — even though the object was already above the temperature of the ice! ‘
The radiation from the cold object (ice) makes the warm object (at the focus of the second mirror) warmer than it would have been with the radiation from the even colder object (dry ice).
The radiation from the cold object (the atmosphere) makes the warm object (the surface) warmer than it would have been with the radiation from the even colder object (the 3 K background radiation).
Tim Folkerts December 1, 2017 at 4:44 pm
AC says: “How did the Pictet experiment prove that a cold object makes a warm object
colderwarmer?”The analogy here would be if Pictet started with dry ice at the focus of the mirror, then switched to regular ice @ur momisugly 0C. With the dry ice, the object at the focus of the other mirror would cool significantly (until some balance was achieved between heat in from the room and heat out to the dry ice). Perhaps that object would cool to 10C in a 20 C room. If the dry ice was replaced with regular ice, the object would warm (perhaps from 10 C to 15 C) — even though the object was already above the temperature of the ice!
Exactly, Pictet’s variation on the experiment did that. When he started with snow at the one focus to which Nitric acid had been added (colder than 0ºC) he achieved a temperature about 5ºC colder than if snow itself was used. Thus if the cold temperature is increased the thermometer recorded a higher temperature, this is exactly what happens when the cold background of space is replaced by the warmer atmosphere.
The radiation from the cold object (ice) makes the warm object (at the focus of the second mirror) warmer than it would have been with the radiation from the even colder object (dry ice).
The radiation from the cold object (the atmosphere) makes the warm object (the surface) warmer than it would have been with the radiation from the even colder object (the 3 K background radiation).
A C Osborn December 1, 2017 at 3:30 pm
It was established at the beginning of this post that Cold Object cannot warm a warmer object UNLESS THE WARMER OBJECT IS BEING HEATED, otherwise it can only slow the cooling.
At Night the Earth is COOLING, so the radiation from a -80C CO2 Molecule is a Cooler Object.
Whic part do you not understand?
Where you get the -80ºC CO2 from!
If you need to ask you do not even understand the Composition of the Atmosphere and where the CO2 window is that can absorb LWIR.
AC, if you are implying (as you indeed seem to be!) that the 15 um wavelength of radiation of CO2 implies that the radiation is “-80 C” based on Wein’s Law, then you have only a rudimentary understanding of thermal radiation! Wein’s law only gives the peak radiation for a blackbody.
Tim, I am merely quoting others, so please provide me with the actual temperature, if it is below Zero, or even below the Temperature of the Earth’s Surface then the same rule applies.
Back to the Original Post.
Who has heard of Marc-Auguste Pictet?
Who has seen his experiment?
Do you still believe Cold Objects can make Warmer objects warmer without the warm object being heated some other way?
Do you believe that the Cold Radiation from a colder object can be focused using a Mirror, just like Heat can?
So based on that logic if a cold object CAN warm a continually heated warm object, CO2 at -80C can only warm the Earth’s Surface while the Sun is also warming it.
Once the Sun goes down it will not warm it at all. So at the moment no CO2 induced Warming in the UK for 16 hours per day.
A C Osborn wrote, “Do you believe that the Cold Radiation from a colder object can be focused using a Mirror, just like Heat can?”
There’s no such thing as “cold radiation.” Radiation does not have temperature.
Perhaps you are confused because your light bulbs are specified with a “color temperature.” That doesn’t mean the light from the light bulb is any particular temperature, it is just a rough indication of the spectral distribution of the light which that bulb produces. It means that the light is supposed to LOOK similar to the human eye to light which WOULD be emitted by a black body of that temperature. Here’s a pretty good article:
http://www.soundandvision.com/content/led-vs-cfl-bulbs-color-temp-light-spectrum-and-more
Note that LED light bulbs have actual temperatures much lower than incandescent bulbs, but usually have a higher “color temperature.” That is possible because LED bulbs (unlike incandescent bulbs) are not at all like back-body emitters.
GHGs are nothing like black-bodies, either.
GHGs are colorants, because they change the absorption spectrum of the atmosphere. That is, they change its “color,” albeit in the far infrared part of the EM spectrum, rather than in the visible part. If you don’t think a color change can affect temperature then take your shoes off while standing on a light-colored sidewalk on a hot summer day, and then step, in bare feet, onto a black asphalt parking lot. That will certainly clear up your misconception!
A C Osborn wrote, “Do you still believe Cold Objects can make Warmer objects warmer without the warm object being heated some other way?”
Practically speaking, warmer objects are always being heated “in some other way.” Otherwise they would cool continually, and eventually become very, very cold, indeed.
But even it such a hypothetical case, if you were to wrap the object in a Space Blanket it would cool more slowly.
 
A C Osborn wrote, “…CO2 at -80C can only warm the Earth’s Surface while the Sun is also warming it.
Once the Sun goes down it will not warm it at all. So at the moment no CO2 induced Warming in the UK for 16 hours per day.”
Wrong. Most of the radiation which CO2 absorbs comes from the Earth. So it does, indeed, help make the surface a bit warmer even at night.
The first question I asked you is have you heard of “Marc-Auguste Pictet?”
And the second question was have you seen his experiment?
I suggest you take a look.
As to your answers, you obviously did NOT read what I said.
The whole point of this post is that if photons hit an object that is being constantly heated it will make it warmer (unless it is Argon Gas according to our Poster).
The Earth’s Surface is the object receiving sunlight, so photons from a colder object can warm it.
When the sunlight stops the heating stops, now the COOLING Earth’s surface can only have it’s Cooling Slowed, which is not Heating. This point was established at the begining of this post.
If the sun goes out will the Earth not ” cool continually, and eventually become very, very cold, indeed.”
Dave says: “Radiation does not have temperature.”
Thermal radiation DOES have a temperature — the temperature of the object that emitted it. This is most obvious with blackbody radiation but applies to non-black-bodies as well. Among other things, this mean that thermal radiation inside a 3000 K furnace cannot be focused to warm anything inside the furnace above 3000K, just like the thermal IR in a 300 K room cannot be focused to warm anything inside the room above 300 K. No fancy mirrors or lenses or filters can get around these limits — or the 2nd Law of Thermodynamics could be violated!
Tim, that’s wrong.
“Thermal radiation” is simply radiation emitted from a hot object or substance. It is emitted at a spectrum of wavelengths, which are determined by a combination of several factors, including temperature, and the physical characteristics of the object. Analysis of that emission spectrum can give clues to the temperature, which is how those no-touch IR thermometers work. It can also give clues about the object’s composition, which is how emission spectroscopy works.
However, the radiation, itself, has no temperature. There is no such thing as “20 C radiation.”
If you filter radiation from a hot source, to eliminate or attenuate some of the wavelengths, you’ll get a different spectrum, which might “fool” an IR thermometer, by mimicking the emissions of a blackbody at a different temperature.
That radiation will warm anything that absorbs it, regardless of the temperature of the emitter or the temperature of the absorber. There’s no little tag attached to a 15µ photon which says what the temperature was of the source that emitted it. That source could have been at 20°C or 20,000°C, there’s no way to tell, and it simply doesn’t matter.
Inference of an emitting object’s temperature from the spectrum of its emissions is basically a statistical exercise, like repeating an experiment many times to discovery the likelihood of various outcomes, but with a very large number of experiments (photons). But if you filter the radiation, e.g., though a tinted glass, or a prism, or a diffraction grating, it’s like throwing out undesired experimental outcomes, so you’ll draw incorrect inferences from it.
Here’s an example of that thought-experiment, with an object of unknown temperature emitting light, which is filtered by wavelength, and two sensors, sampling different slivers of the object’s emission spectrum
http://sealevel.info/temperature_through_prism.png
Do you think Sensor A can correctly determine the temperature of the light source?
Do you think Sensor B can correctly determine the temperature of the light source?
Do you think that if the two sensors tried to deduce a temperature they would deduce the same temperature?
Do you think that varying the temperature of the object would change the temperatures deduced by the two sensors?
Mr Burton, I am shocked do have no concept of Power & Energy?
Let me ask you
1 Would you be prepared to sit in a room being bathed in White Light Radiation from a Bulb?
2 Would you be prepared to sit in a room being bathed in Radiation from an IR Heater?
3 Would you be prepared to sit in a room being bathed in Ultra Violet Radiation?
4 Would you be prepared to sit in a room being bathed in X Ray Radiation?
5 Would you be prepared to sit in a room being bathed in Gamma Radiation?
6 Would you be prepared to sit in a room being bathed in Beta Radiation?
7 Would you be prepared to sit in a room being bathed in Neutron Radiation?
Dave, I think we are looking at two sides of the same coin.
Thermal radiation from a blackbody (or cavity) does indeed have a temperature. There is a thing called a “photon gas” with temperature and pressure and entropy, etc. Thermodynamics and statistical mechanics still applies to photons. https://en.wikipedia.org/wiki/Photon_gas.
“That radiation will warm anything that absorbs it, regardless of the temperature of the emitter or the temperature of the absorber. There’s no little tag attached to a 15µ photon which says what the temperature was of the source that emitted it. That source could have been at 20°C or 20,000°C, there’s no way to tell, and it simply doesn’t matter.”
Planck’s Law gives an upper limit on the intensity of radiation at any frequency (the spectral irradiance). No amount of filtering or focusing can make the thermal radiation from a cooler source as bright as the radiation from a warmer blackbody source. At 1 um, 5000K blackbody thermal radiation has a spectral radiance of ~ 7 kW / (sr*m^2*nm). 4000 K radiation is about 3.5 kW / (sr*m^2*nm). 3000 K radiation is about 1 kW / (sr*m^2*nm). No filter or focusing or different radiating surface can generate 7 kW / (sr*m^2*nm) from a 3000 K sources. If the intensity is 7 kW / (sr*m^2*nm), the surface must be (at least) 5000 K. Filtering and focusing cannot fool this sort of measurement into think bright light came from a cool source.
Thus the NET effect of ALL 15 um photons does have some limitations. No matter the sizes or shapes, no matter the materials or filters, no matter the mirrors or lenses — the net flow of 15um photons will always be from warmer to cooler. So there is no way (without other energy sources) to use the radiation from a 20 C object to raise the temperature of a 30 C object.
PS Astronomers routinely use a system like your Sensor A/Sensor B to find the temperature of stars. https://en.wikipedia.org/wiki/Color_index
Time’s up. The answers to the four questions are: No, No, No, No.
Tim Folkerts wrote, ” So there is no way (without other energy sources) to use the radiation from a 20 C object to raise the temperature of a 30 C object.”
There are always other energy sources. Otherwise it wouldn’t stay at 30°C.
Think about a 30°C “object” floating in the dark in deep, deep interstellar space, with no incoming radiation, but with enough of a radioisotopic heat source to maintain its temperature at a steady 30°C. (Of course the output of real radioisotopic heaters very slowly drops, as the radioisotope decays, but let’s pretend ours is steady.)
Now, if, instead of putting your object in the dark in interstellar space, you put the same object, with the same radioisotopic heat source, in a 20°C vacuum bottle container (with some sort of thermostatic control to keep the container at exactly 20°C), the “30°C” object will heat up very dramatically. In fact, it will become an 81.something°C object.
See if you can figure out how to calculate the new equilibrium temperature of the formerly-30°C object. I’ve given you the first two digits of the answer (81). See if you can figure out the next two digits.
Dave, I think we are talking past each other. You are focusing on certain details, while I am focusing on others.
“There are always other energy sources. Otherwise it wouldn’t stay at 30°C.”
First, I never claimed the object was staying at 30 C — only that it happened to be at 30 C at that particular moment. Second, it is quite possible for thermal radiation to be the only source of energy. Put the object inside an evacuated box where the walls are 20C.
My point is that any object inside the box will equilibrate to 20 C because it equilibrates with the *radiation* emitted by the walls. The radiation has a temperature of 20 C. There is no way around this. There is also no way for the object inside to equilibrate to any other temperature using passive devices to focus/reflect/filter the 20 C thermal IR inside the chamber. Fancy filters designed to ‘trick’ an IR thermometer cannot make the object warmer or cooler than 20 C once equilibrium is reached.
Also, measuring the intensity of the radiation inside the box (or the radiation coming from a blackbody surface) at any single wavelength is sufficient to determine the temperature. This radiation will follow the curve predicted by Planck’s Law. Here measuring the spectral irradiance at a single wavelength with “Sensor A” will indeed tell you the temperature.
“Now, if, instead of putting your object in the dark in interstellar space, you put the same object, with the same radioisotopic heat source, in a 20°C vacuum bottle container (with some sort of thermostatic control to keep the container at exactly 20°C), the “30°C” object will heat up very dramatically. In fact, it will become an 81.something°C object.”
Dave, it is pretty easy to do those calculations (assuming blackbody surfaces, as you seem to have done). It’s 81.56 C. Of course, any typical ‘vacuum bottle’ will have reflective walls with an emissivity of ~ 0 rather than ~ 1, so the heated object inside would get MUCH warmer than 81.56 C.
I wrote, “There are always other energy sources. Otherwise it wouldn’t stay at 30°C.”
Tim replied, “First, I never claimed the object was staying at 30 C — only that it happened to be at 30 C at that particular moment…”
Oh, come on. If there were no source of energy at all, it would equilibrate (well, asymtopically approach) 0 K. In the real world there are always other energy sources. How did your object get to 30°C?
I also wrote, “Now, if, instead of putting your object in the dark in interstellar space, you put the same object, with the same radioisotopic heat source, in a 20°C vacuum bottle container (with some sort of thermostatic control to keep the container at exactly 20°C), the “30°C” object will heat up very dramatically. In fact, it will become an 81.something°C object.”
Tim did the calculation correctly, and replied, “…(assuming blackbody surfaces, as you seem to have done). It’s 81.56 C. Of course, any typical ‘vacuum bottle’ will have reflective walls… so the heated object inside would get MUCH warmer than 81.56 C.”
Correct! Without reflective walls it would be 81.56°C, and if some of the object’s own radiation is reflected back to it then its temperature would rise even higher.
And so you have seen that the presence of a nearby (surrounding) 20°C object would raise the temperature of the formerly-stable-at-30°C object by more than 50 degrees!
In just the same way, warming the Earth’s chilly atmosphere can warm the already-warmer ground. Please explain this to A C Osborn et al!
Tim also wrote, “Put the object inside an evacuated box where the walls are 20C. … any object inside the box will equilibrate to 20 C because it equilibrates with the *radiation* emitted by the walls.”
Only if it has no other energy source. Such an object (with no other source of energy) would drop to near 0 K if it were in deep, deep interstellar space instead of in your 20°C box.
But restore the radioisotopic heater to the object that I described (the radioisotopic heater which puts out enough energy to keep the object at 30 °C), and put that object into your 20 °C box, and the object’s temperature will rise to 81.56 °C (or, as you correctly noted, even higher, if the box has reflective walls).
But then Tim wrote, The radiation has a temperature of 20 C. There is no way around this.”
Nooooo! Wrong! And you were doing so well. 😞
The radiation from the 20 °C box has no temperature. It is as readily absorbed by a 30 °C object or an 81°C object as it is by a 15°C object, and it has exactly the same effect: it warms them.
And Tim wrote, “Also, measuring the intensity of the radiation inside the box (or the radiation coming from a blackbody surface) at any single wavelength is sufficient to determine the temperature. This radiation will follow the curve predicted by Planck’s Law. Here measuring the spectral irradiance at a single wavelength with “Sensor A” will indeed tell you the temperature.”
Only you know the distance to the object, and only if the object is a true blackbody… and there are no true blackbodies, only approximations.
I wrote, “Only you know the distance to the object, and only if the object is a true blackbody…”
I should have added, “, and only if you don’t do something tricky, like put the object at the focal point of a parabolic reflector.”
Dave, we are getting pretty far afield. And I really do agree with much of what you are writing. My my difficulty is hwo you keep insisting on specific conditions that are different from what I am discussing – and then saying that disproves my point.
For example, I say ” it is quite possible for thermal radiation to be the only source of energy. ” You reply “If there were no source of energy at all, it would equilibrate (well, asymtopically approach) 0 K. ” You get rid of the thermal radiation and then claim a different temperature — which of course is possible without the thermal radiation that was posited.
Or my 30 C object was simply any old object that happened to be 30 C. Your 30 C object became a blackbody with a built-in heater that would maintain 30 C when in deep space far from any other heat sources.
———————–
As for the radiation within the evacuated chamber, that cavity radiation is basically perfect blackbody radiation. The intensity of the thermal radiation is independent of the size of the cavity or the distance from the walls to the detector. Furthermore, your fancy parabolic reflector will STILL not change this intensity (assuming the reflector has also been allowed to come to the temperature of the walls of hte chamber). That parabolic reflector will NOT change the temperature of the photon gas and will NOT change the temperture of any object placed at the focus.
———————–
But yes, I agree that “In just the same way, warming the Earth’s chilly atmosphere can warm the already-warmer ground. ” When there is an independently heated object with some relatively steady power supply (like the earth or your radioactive 30 block), then raising the temperature of hte surroundings will raise the temperature of the original, warmer object.
THIS IS THE IMPORTANT POINT as far as global warming goes. All the rest of what we are discussing is more esoteric phsyics. 🙂
Tom, what if the “surroundings” is a non-linear, chaotic system with huge forcings varying at all time scales? What if the minuscule increase in the temperature of the dynamic system caused directly by a minor gas is unmeasurable?
What if temperature measurements of the “surroundings” do not comport with assumed forced warming?
I think the above questions are your: “THIS IS THE IMPORTANT POINT as far as global warming goes.”
Tim Folkerts wrote, But yes, I agree that “In just the same way, warming the Earth’s chilly atmosphere can warm the already-warmer ground. ” When there is an independently heated object with some relatively steady power supply (like the earth or your radioactive 30 block), then raising the temperature of hte surroundings will raise the temperature of the original, warmer object. / THIS IS THE IMPORTANT POINT as far as global warming goes. All the rest of what we are discussing is more esoteric phsyics. :-)”
True, and of course it means the Slayers are all wrong.
Tim also wrote, “As for the radiation within the evacuated chamber, that cavity radiation is basically perfect blackbody radiation. The intensity of the thermal radiation is independent of the size of the cavity or the distance from the walls to the detector. Furthermore, your fancy parabolic reflector will STILL not change this intensity (assuming the reflector has also been allowed to come to the temperature of the walls of hte chamber). That parabolic reflector will NOT change the temperature of the photon gas and will NOT change the temperture of any object placed at the focus.”
I think there’s a misunderstanding about what I was talking about. The question is, can Sensor A tell what temperature the source object is?
http://sealevel.info/temperature_through_prism_with_parabolic_reflector.png
The answer is No, it cannot. Sensor A has only one clue about the temperature of the source: the intensity of the narrow band of orange light which the sensor receives. That intensity is affected by several factors, including the temperature of the source, the composition of the source, the size of the source, the distance of the source from the prism & sensor, and the existence of the parabolic reflector. All of those factors affect the intensity of the orange light which the sensor receives, and the sensor cannot tell them apart.
Adding the parabolic reflector increases the intensity of the orange light. Reducing the object’s temperature decreases the intensity of the orange light. Doing both could either increase or decrease the intensity, or leave the intensity unchanged. There’s no way for Sensor A to tell what has happened to the temperature merely from the intensity of the orange light, which is the only clue it has.
And there’s no such thing as “photon gas,” and photons have no temperature.
Dave a couple quick comments.
1) 88,000 hits for “photon gas” say you are wrong about the existence of photon gases. Some of the first links are to places like Nature magazine, Wikipedia, and several major research universities. If you dislike the idea of a “photon gas”, take it up with them.
2) Suppose I have a narrow band pass filter on my camera and take a picture of the sun. The pixels for the image of the sun will show some intensity — related to the brightness and temperature of the sun. A telephoto lens or a mirror will make MORE pixels bright, but will not make a specific pixel brighter. As such, the brightness of any pixel that is aimed at the sun can provide an estimate of the temperature of the sun. Certainly measure multiple wavelengths gives a better temperature estimate. Certainly anything between the sensor and the source can make the results inaccurate. Certainly tiny sources like stars that are too small to fully illuminate a pixel won’t measure accurately. But in principle, a single wavelength can tell you the temperature of a blackbody surface.
Tim wrote, “88,000 hits for “photon gas” say you are wrong about the existence of photon gases.”
I stand corrected, and thank you.
Tim also wrote, “2) Suppose I have a narrow band pass filter on my camera and take a picture of the sun. The pixels for the image of the sun will show some intensity — related to the brightness and temperature of the sun. A telephoto lens or a mirror will make MORE pixels bright, but will not make a specific pixel brighter.”
Assuming that your telephoto lens is the same aperture as the regular lens, it will make the originally lit pixels dimmer. But what’s your point?
Tim continued, “As such, the brightness of any pixel that is aimed at the sun can provide an estimate of the temperature of the sun.”
No, it can’t. It tells you only the intensity of the light which is falling on that camera pixel, which depends on many factors, including your distance from the Sun, the aperture of your lens, the focal length / size of the focused solar image, the clarity of you filter & lens, etc.
Tim continued, “But in principle, a single wavelength can tell you the temperature of a blackbody surface.”
No, it cannot.
Of course the object doesn’t necessarily resemble a blackbody, and certainly is not a perfect blackbody. But that actually doesn’t matter, in this case, because we’ve filtered out all but the orange light. Sensor A cannot possibly deduce the temperature of the emitting object merely from the intensity of the orange light which it emits.
“Do you believe that the Cold Radiation from a colder object can be focused using a Mirror, just like Heat can?”
Well any sort of EM radiation can be reflected and focused – lasers and radio waves and x-rays and thermal IR. So yes, “cold radiation” (thermal radiation from cold objects) can indeed be reflected and focused.
However, some care needs to be taken with your phrase “just like”. There are limits to how well thermal IR can be focused. For example, in a room at 20 C, a flat surface would receive about 400 W/m^2 of incoming thermal IR. With mirrors and lenses, you could focus this thermal IR to provide a maximum of …. still 400 W/m^2 of thermal IR! But 400 W/m^2 of sunlight could be focused to provide 1000’s of W/m^2 of incoming power.
Conversely, if your room was mostly at 20 C but a small portion was @ur momisugly 0 C and radiating @ur momisugly 315 W/m^2, then you could uses mirrors to block out the 20 C radiation and “focus” just on the 0C region so that only 315 W/m^2 radiation arrived at a surface (allowing the surface to cool well below 20 C). This is basically what Pictet’s experiment did.
Tim, Mr Eshenbach says it can’t be done for diffuse Radiation, would yor small portion be diffuse radiation or not?
I have just re-read the crap that comes from Mr Burton to make his point.
First of all he replaces a Heat sink at 3.5K with another heat sink at 293.15K and not only that but this new heat sink is not allowed to re-radiate at the same rate as the original Object it is receiving heat from and has to maintain a temperature 293.15K.
So he has introduced some kind of new Energy source in to the equation to keep the bottle cold.
And has the nerve to say “In just the same way, warming the Earth’s chilly atmosphere can warm the already-warmer ground. Please explain this to A C Osborn et al!”
Absolutely Typical.
We now have a few CO2 Molecules that do not re-radiate out to space, there is nothing between them and the Earth to absorb their energy before it gets to the Earth and they are no longer surrounded by any gases removing their Excess energy by collisions.
A C Osborn, I have no idea what you’re talking about. I said nothing about a “Heat sink at 3.5K.” In fact, I didn’t say anything about any heatsinks, nor about anything at 3.5K. None of what you wrote makes any sense.
If there’s something I wrote that doesn’t make sense to you, please quote it, or link to it, so I’ll know what you’re talking about.
And why don’t you see if you can do the calculation that Tim did, and figure out the next digit of the equilibrium temperature of the formerly-30°C object (the digit after “81.56”), assuming that the 20°C surrounding container reflects none of the formerly-30°C object’s radiation back to the formerly-30°C object.
I didn’t say the outcomes from probes and the application of the gas laws could be safely ignored. I was pointing out that that was what you seemed to be saying.
As for the tank experiment it contains a liquid that does not observe the gas laws. Can you link to one that does use a gas ?
Mr Eschenbach, would you like me to list the many many Scientific pieces of literature and courses that say exactly the same thing as that Wiki Quoute?
All matter above absolute Zero emit Thermal Radiation and it has nothing to to do with Molecules and all to do with Excited Atoms.
So do want the list or not, if so how many before you are convinced?
Here is just the Dictionary version
thermal radiation
See more synonyms on Thesaurus.com
noun, Thermodynamics.
1.
electromagnetic radiation emitted by all matter above a temperature of absolute zero because of the thermal motion of atomic particles.
A C Osborn December 2, 2017 at 4:34 am Edit
Someone here linked to the actual NIST spectrum for argon. Go study it.
Yes, there are lots of claims that “every object” radiates, which is 100% true for the common case in which people are referring to SOLID objects. And you can indeed find that statement all over the web.
However, it is not true for monatomic gases, AS THE LINK TO THE ACTUAL MEASURED SPECTRUM OF ARGON DEMONSTRATES CONCLUSIVELY.
And if you don’t see that in the NIST results, then you don’t know what you’re talking about, and I’ll bow out.
w.
A C Osborn December 1, 2017 at 3:10 pm Edit
No wonder you have trouble with scientific concepts, you’re getting your information from Wikipedia. You do realize that ANYONE IN THE WORLD could have written that sentence on Wikipedia, right? Including a 16-year old high school student? Because that is where I first heard that claim, in high school … and like you, I believed it. Until I found out how IR absorption happens and learned about monatomic gases.
Yes, it is widely believed, as you believe, that IR emission applies to everything. You can find that claim lots of places … but not in the spectroscopic lab.
Read the link to lab results that Phil gave above. It clearly shows that there are NO absorption bands for argon in the thermal IR range. NONE.
Sheesh … yes, A C, all SOLIDS emit thermal radiation, because all of them have molecular bonds between the atoms that can bend and flex. That’s how thermal IR is absorbed. The energy in the photon is converted into mechanical energy of the flexing of the bonds.
But monatomic gases (helium, neon, argon, etc.) have no molecular bonds. None. So they cannot either absorb or emit IR.
w.
Willis: “It clearly shows that there are NO absorption bands for argon in the thermal IR range. NONE.”
Click on the “strong lines” tab to find the absorption bands for argon in the thermal IR range.
Trick, that is not the same thing. These emission are for Argon that has been excited to higher levels (for example, in a gas discharge tube) and then falls back to some intermediate level. These lines are not available to cool Argon in the ground state.
It is much like the Pfund series or Humphreys Series for atomic hydrogen. Only very hot hydrogen (or hydrogen that has undergone an energetic collision) will emit such radiation. https://en.wikipedia.org/wiki/Hydrogen_spectral_series
Ar ref. please. The wiki page is only for hydrogen. The only ref. NIST has is G. Norlén, Phys. Scr. 8, 249 (1973). I’ll challenge my local college librarian to track it down. I have learned they like these, get a chance to practice their trade. Again, the only way that steady state equilibrium can be obtained in Willis’ challenge is for Ar to shift its emission spectrum into regions of nonzero emissivity. It doesn’t seem like nature should abhor a pure Ar atm., should be able to handle it.
Trick December 1, 2017 at 4:08 pm
Willis: “It clearly shows that there are NO absorption bands for argon in the thermal IR range. NONE.”
Click on the “strong lines” tab to find the absorption bands for argon in the thermal IR range.
Those are emission lines when you pass current through the Ar thereby electronically exciting it up that ladder of levels that I gave you, then it decays from one level to another emitting those wavelengths. Heat up the Ar so it’s not in the ground state then you could get some absorption in those bands but it would take ~2000ºC which is not what we’re talking about here.
Here’s the spectrum from an electronic discharge tube containing Argon.
Phil. 6:27pm, your NIST link does not have this detail. It merely references a 1973 paper. Perhaps you have a better link to an experiment that determines the spectral emissivity of Ar at earth atm. STP. Until then all I can do is look up the 1973 paper. Perhaps there are no tests for spectral emissivity of Ar at earth STP so that no one can ref. an experiment to form a reasoned response to Willis’ questions.
Trick December 1, 2017 at 7:31 pm
Phil. 6:27pm, your NIST link does not have this detail. It merely references a 1973 paper. Perhaps you have a better link to an experiment that determines the spectral emissivity of Ar at earth atm. STP. Until then all I can do is look up the 1973 paper. Perhaps there are no tests for spectral emissivity of Ar at earth STP so that no one can ref. an experiment to form a reasoned response to Willis’ questions.
There are no experiments for spectral emissivity of Ar at earth STP because there is no emission so there’s nothing that can be measured, it’s all in the ground state. When I used my FTIR spectrometer in my lab I had to take account of any background CO2 in the air, I didn’t have to do that for Argon because no lines exist.
“There are no experiments for spectral emissivity of Ar at earth STP..”
Every object radiates at all temperatures, all frequencies all the time, there are no exceptions. If Ar STP emissivity can’t be or hasn’t been measured by today’s instruments who knows about tomorrow’s. If there are no experiments, then any answer to Willis Ar atm. question is speculative at best, not borne out by test.
The cooling of a 1atm., 255K Ar transparent container in deep space would occur whether we can measure the Ar emissivity or not just like the atm. around Willis’ proposition.
Trick December 2, 2017 at 5:36 am
“There are no experiments for spectral emissivity of Ar at earth STP..”
Every object radiates at all temperatures, all frequencies all the time, there are no exceptions.
That’s your fundamental mistake, it’s not true, no gases do that for example.
If Ar STP emissivity can’t be or hasn’t been measured by today’s instruments who knows about tomorrow’s. If there are no experiments, then any answer to Willis Ar atm. question is speculative at best, not borne out by test.
There are experiments all the time, anytime anyone measures the emissions from air at STP, the emissions from Ar don’t exist. This is well known and the reason for it is well known, the measured energy levels for the Argon atoms have been measured and I gave you the data above. The lowest energy level is so far above the ground state that there is no chance of it being populated at STP. Take the Boltzmann distribution, the fraction of a gas that will occupy that excited state (or above) when at a temperature of 300K is ~2×10^-185, (there are only about 10^80 atoms in the universe)!
The cooling of a 1atm., 255K Ar transparent container in deep space would occur whether we can measure the Ar emissivity or not just like the atm. around Willis’ proposition.
Yes it will cool down because the container will emit to space, not the contents.
“There are experiments all the time, anytime anyone measures the emissions from air at STP, the emissions from Ar don’t exist.”
I’ve asked repeatedly for a cite to all these experiments Phil. So far nothing, no response except NIST which shows plenty of lines in Ar in air in IR band from 1973 testing. Just post up, cite your best published experiment confirming your comment, I’ll dig it up at college library if can’t find it online. I found the NIST ref. 73 today.
What is the energy needed to go from atomic Ar rotational base energy to the first excited quantized level for instance? Commonly for spinning constituents of air in meteorology applications this is about 1/3 kT with quantized spacing order of kT.
Ok, dispense with the container. Let the 255K Ar sample float free in space, large enough sample held together long enough by gravity say as large as Earth atm. Are you saying Ar won’t radiatively cool for the life of the universe? Universe heat death occurs and the Ar will still be 255K?
“That’s your fundamental mistake, it’s not true, no gases do that for example.”
Phil. there is a miscommunication going on here, the atm. radiation text I use Bohren 2006 does not agree with you (and his earlier one 1998). Possibly if we ask enough questions eventually get down to why you so vehemently disagree with the text.
All solids, all liquids, all gases, all plasmas emit at all frequencies at all temperatures all the time. The Planck radiance formula is never zero at any temperature and at any frequency. Whatever primoridial Ar existed at the temperatures of the big bang, it must still be at that temperature (or maybe a lower one above some cutoff) if Ar at STP cannot radiate or absorb according to you. My first impression was you really meant instruments aren’t sensitive enough, not sure now.
There is a big disconnect here and we ought to be able to find it. My interpretation is for the Ar atm. of Willis emissivity just shifts to regions for which the gas can emit.
Phil. this comment of yours might be a starting point to ask questions to iterate to reason for disconnect.
https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1/#comment-2682654
Tim writes: “These emission are for Argon that has been excited to higher levels..”
The NIST ref. 73 informs that the testing pressure for the lines they report was at 0.2torr. This is LOT less than STP. It also says a current of 10mA was used, at 120v (which it doesn’t say) that is little over 1 watt. At STP, the global energy balances all show order of 100x this watt level. This seems to conflict with your 2000C electronic discharge tube example wattage, how does 1W get to a need for 2000C?
You write “Heat up the Ar so it’s not in the ground state…2000C “ yet Bohren writes to go up from base to 1st rotational quantum level for typical STP atomic constituent is 1/3kT (1998 p. 119), this is 1.1*10^-21 J. Hardly 2000C. Why?
Does this create a route to start to connect?
Trick, the Boltzmann constant is k = 8.6e-5 eV/K, or conversely that would be 11,600 K/eV.
* Since accelerating an electron through 120 V would give an electron up 120 eV of energy, that corresponds to about 1,400,000 K!
* the binding energy for most valence electrons is on the order of 10 eV, so 120 V is more than enough to ionize them!
* room temperature is about 0.025 eV, so thermal energies are typically insufficient to excite electrons to higher level.
However you look at it, 120 V is a LOT of energy and corresponds to very high temperatures! This can easily excite electrons to higher orbitals or ionize atoms completely — in ways that 300 K thermal vibrations never can.
Tim, what causes you to bring up electronic base level to 1st excited level energy? The 2000C electronic discharge tube? Yes, that is a diversion, way more energy ~100kT level jump not order of ~1/3kT typical for atomic rotationals (spinning), similar the collisional energy available in troposphere. Agree, no or very rare electronic transitions in Earth troposphere, don’t need to discuss those, atm. is not an electronic discharge tube. Trying to connect the dots on the atm. constituent rotational quantum jumps as in most common emission/absoption of photons in the troposphere.
Trick December 2, 2017 at 3:51 pm
“There are experiments all the time, anytime anyone measures the emissions from air at STP, the emissions from Ar don’t exist.”
I’ve asked repeatedly for a cite to all these experiments Phil. So far nothing, no response except NIST which shows plenty of lines in Ar in air in IR band from 1973 testing. Just post up, cite your best published experiment confirming your comment, I’ll dig it up at college library if can’t find it online. I found the NIST ref. 73 today.
The NIST report shows no emission lines from Ar in air at STP, it shows lines emitted from Ar in an electronic discharge tube. Such lines do not exist in the absence of such excitation.
What is the energy needed to go from atomic Ar rotational base energy to the first excited quantized level for instance? Commonly for spinning constituents of air in meteorology applications this is about 1/3 kT with quantized spacing order of kT.
You appear to have some severe misconceptions about the physical chemistry involved.
Molecules have three different modes of excitation:
Electronic, in which electrons occupy different energy levels, these are separated by the most energy and require UV light to excite them.
Vibrational, in which the molecular bonds vibrate and if the bond has a dipole can be excited by light. These exist as substructure to the electronic levels and are much more closely separated and so are excited by IR radiation.
Rotational, in which the molecule rotates and if the molecule has a dipole can be excited by light. These exist as substructure to the vibrational levels and are much more closely separated and so are excited by microwave radiation.
In all of these cases absorption will only occur if the incident light energy exactly matches the separation of two of the states. When emission occurs from an excited state to another lower state then the wavelength of the emitted light will exactly correspond to the separation between the states. Consequently there will be a discrete set of wavelengths emitted, the gas will not “radiate at all temperatures, all frequencies all the time”.
Argon is an atom not a molecule so it only has electronic states, no vibrational or rotational states.
Consequently your question: “What is the energy needed to go from atomic Ar rotational base energy to the first excited quantized level for instance?”, makes no sense.
As the NIST site I showed you says the electronic ground state is separated from the first excited state by 93143.7653 cm-1 (in the far UV, separation over 1000kJ/mole).
Your statement: “Commonly for spinning constituents of air in meteorology applications this is about 1/3 kT with quantized spacing order of kT”, refers to rotational levels which Argon does not have.
Phil.: “Argon is an atom not a molecule so it only has electronic states, no vibrational or rotational states.”
Ok, this would be where you and Bohren differ, he writes ALL atoms have rotational states that are quantized without exception. So, thanks for that I have some reading to do. He is very good at citing source material.
The NIST ref. uses so much specialist wording I would have to spend time learning the lingo if it confirms this or not. You are then writing the spinning argon atoms do so as smoothly as they translate, that there is no quantum separation in rotational energy states as found in (all?) other atoms and molecules. Photons are not absorbed/emitted as rotational quanta in atomic Ar. Interesting. This strikes me as odd and not covered in Bohren’s work that I have found as Ar is not that important in meteorology. I may be interested enough to dig into this further.
“…it shows lines emitted from Ar in an electronic discharge tube.”
The NIST report uses an etalon to excite Ar in air at collisional energy levels MUCH lower than those found in the atmosphere. If it is only Ar electronic states being excited above base state by light quanta then possibly those electronic states for Ar can also be energized with the energy available in the lower atm. unlike the more prevalent constituents. Since you say these have not been found then they could be too feeble for modern instruments (which Bohren mentions) or the researchers weren’t looking for them.
Willis’ question excites more profound discussions than (possibly) expected.
Trick December 2, 2017 at 6:16 pm
Tim, what causes you to bring up electronic base level to 1st excited level energy? The 2000C electronic discharge tube? Yes, that is a diversion, way more energy ~100kT level jump not order of ~1/3kT typical for atomic rotationals (spinning), similar the collisional energy available in troposphere. Agree, no or very rare electronic transitions in Earth troposphere, don’t need to discuss those, atm. is not an electronic discharge tube. Trying to connect the dots on the atm. constituent rotational quantum jumps as in most common emission/absoption of photons in the troposphere.
It is not a diversion, Ar atoms don’t have rotational levels the 93,000 cm-1 ‘jump’ is the smallest one possible.
Trick December 2, 2017 at 4:07 pm
“That’s your fundamental mistake, it’s not true, no gases do that for example.”
Phil. there is a miscommunication going on here, the atm. radiation text I use Bohren 2006 does not agree with you (and his earlier one 1998). Possibly if we ask enough questions eventually get down to why you so vehemently disagree with the text.
All solids, all liquids, all gases, all plasmas emit at all frequencies at all temperatures all the time.
Not true, gases don’t do that.
Trick December 3, 2017 at 6:48 am
Phil.: “Argon is an atom not a molecule so it only has electronic states, no vibrational or rotational states.”
Ok, this would be where you and Bohren differ, he writes ALL atoms have rotational states that are quantized without exception. So, thanks for that I have some reading to do. He is very good at citing source material.
Then Bohren’s wrong, try reading any college level physical chemistry text.
The NIST ref. uses so much specialist wording I would have to spend time learning the lingo if it confirms this or not. You are then writing the spinning argon atoms do so as smoothly as they translate, that there is no quantum separation in rotational energy states as found in (all?) other atoms and molecules. Photons are not absorbed/emitted as rotational quanta in atomic Ar. Interesting. This strikes me as odd and not covered in Bohren’s work that I have found as Ar is not that important in meteorology. I may be interested enough to dig into this further.
A monatomic gas has three degrees of freedom which are the three translational modes, no monatomic gas has any rotational or vibrational modes. Neither do homonuclear diatomic such as N2 and O2 have IR active rotational or vibrational modes due to the lack of a dipole.
“…it shows lines emitted from Ar in an electronic discharge tube.”
The NIST report uses an etalon to excite Ar in air at collisional energy levels MUCH lower than those found in the atmosphere.
No, they’re much higher than this in the atmosphere see my post above.
If it is only Ar electronic states being excited above base state by light quanta then possibly those electronic states for Ar can also be energized with the energy available in the lower atm. unlike the more prevalent constituents. Since you say these have not been found then they could be too feeble for modern instruments (which Bohren mentions) or the researchers weren’t looking for them.
No they just do not exist.
“..try reading any college level physical chemistry text.”
I have one on order. On line texts (the 3 or 4 I found) tend to all agree with Bohren so far and not Phil. but nothing I can relate to test just words so not worth following up until I can find one with test data. I have at least a related source explaining misconceptions about this stuff that he cites but haven’t been to the library yet. Perhaps you could recommend a text that traces source material from actual testing. I’ve learned that atomic rotational quanta are so uninteresting they have been given comparatively little attention. Vibrationals having to do with chemical bonds have more significance.
“No, they’re much higher than this in the atmosphere see my post above.”
They? The etalon testing of NIST Ref. was done at much lower collisional energies than exist in the atm. so any rotational relaxation lines produced in those tests will also find enough energy in lower atm. collisions. The trouble is the ref. does not label them as rotational OR electronic unless the specialist lingo they use is not communicating to me (yet).
Trick December 3, 2017 at 6:30 pm
They? The etalon testing of NIST Ref. was done at much lower collisional energies than exist in the atm. so any rotational relaxation lines produced in those tests will also find enough energy in lower atm. collisions. The trouble is the ref. does not label them as rotational OR electronic unless the specialist lingo they use is not communicating to me (yet).
Look at the NIST report I cited, in the L column it shows you the electronic configuration, in the ground state it shows all 6 electrons in the 3P level, in the excited states one of the electrons has been raised to a higher level so 5 3P electrons plus one 4S electron and so on so they’re explicitly electronic.
[Italics dropped, second paragraph. .mod]
“Look at the NIST report I cited…”
I did, thanks for the decoding. If you look up the 1973 source material though there are way more pages of lines than the NIST link provides. It is at least possible those can be decoded into rotationals. It is also possible the rotational relaxation energies occur below the sensitivity of the etalon apparatus. The lingo is specialist and decoding will take some time.
Also, these listed electronic relaxations would statistically occur in a lower Earthian argon atm. as it would have more collisional energy available than in the collisional energy available in the etalon for the 1973 testing.
Trick December 4, 2017 at 1:17 am
“Look at the NIST report I cited…”
I did, thanks for the decoding. If you look up the 1973 source material though there are way more pages of lines than the NIST link provides. It is at least possible those can be decoded into rotationals. It is also possible the rotational relaxation energies occur below the sensitivity of the etalon apparatus. The lingo is specialist and decoding will take some time.
No it is not possible , Argon (or helium, neon etc.) does not have rotational states, you are confusing the emission lines from a discharge tube (which has many lines over a wide range of wavelengths) with the energy levels of the Argon atom.
Also, these listed electronic relaxations would statistically occur in a lower Earthian argon atm. as it would have more collisional energy available than in the collisional energy available in the etalon for the 1973 testing.
No they won’t, as I pointed out above the fraction of Ar atoms at 300K existing in the first excited state is about 10^-185. By the way an etalon is an interferometric devise used to measure the wavelengths with high accuracy, it is not a device for generating emissions.
The NIST data is from an etalon per the 1973 paper. So far that is the only test data you have cited supporting your position. The other authors I’ve found do not support your conclusions but I am still digging into their source material, too early to draw a conclusion.
Phil.: “Argon is an atom not a molecule so it only has electronic states, no vibrational or rotational states.”
https://www.bing.com/images/search?q=Rotational+states+of+Argon&qpvt=Rotational+states+of+Argon&FORM=IGRE
More non existent states
https://www.bing.com/images/search?q=Electronic+states+of+Argon&qpvt=Electronic+states+of+Argon&FORM=IGRE
A C Osborn December 4, 2017 at 11:09 am
More non existent states
https://www.bing.com/images/search?q=Electronic+states+of+Argon&qpvt=Electronic+states+of+Argon&FORM=IGRE
That shows the same electronic states given by the NIST report I linked to:
http://raptor.physics.wisc.edu/overview/over4.gif
Mr Eschenbach, would you like me to list the many many Scientific pieces of literature and courses that say exactly the same thing as that Wiki Quoute?
All matter above absolute Zero emit Thermal Radiation and it has nothing to to do with Molecules and all to do with Excited Atoms.
So do want the list or not, if so how many before you are convinced?
Here is just the Dictionary version
thermal radiation
See more synonyms on Thesaurus.com
noun, Thermodynamics.
1.
electromagnetic radiation emitted by all matter above a temperature of absolute zero because of the thermal motion of atomic particles.
A C Osborn December 2, 2017 at 4:38 am
Mr Eschenbach, would you like me to list the many many Scientific pieces of literature and courses that say exactly the same thing as that Wiki Quoute?
All matter above absolute Zero emit Thermal Radiation and it has nothing to to do with Molecules and all to do with Excited Atoms.
So do want the list or not, if so how many before you are convinced?
Since the statement is untrue I don’t care how many places you can find it, it’s still wrong.
Here is just the Dictionary version
thermal radiation
See more synonyms on Thesaurus.com
noun, Thermodynamics.
1.
electromagnetic radiation emitted by all matter above a temperature of absolute zero because of the thermal motion of atomic particles.
That is poorly written and misunderstood by you. For example Argon will emit thermal radiation if you heat it to about 2000K, it’s not in the case of Argon due to the motion of atomic particles but sub-atomic particles (electrons). In the case of CO2 it’s due to the motion of the atoms (vibration and rotation of the molecular bonds) and occurs at lower temperatures.
Phil, great, thank you.
Can you now explain how the the texts are all wrong.
They all say that any radiation from an object above Zero K is called Thermal Radiation, they do not call it anything else.
So can you provide me with the definitive Reference that says either that radiation is no emitted or it is called something else.
When you do that can you also provide a reference that says that that “something else” doesn’t involve photons being emitted?
More non existent states
https://www.bing.com/images/search?q=Electronic+states+of+Argon&qpvt=Electronic+states+of+Argon&FORM=IGRE
Phil, so I provide images of many examples of what you say does not exist and you refer to the NIST experiment once again. Which appears to prove – nothing.
All you have shown is how limited the NIST experiment is when it is contradicted by so many seperate sources.
Actually you haven’t in fact judging by what you’ve linked to it seems unlikely you’ve even looked at them, here’s one of them: http://pedalmag.com/wp-content/uploads/2016/05/Electron_Pro_Danish_FrameSet_cote_20160425.png
You haven’t showed a single example of a rotational transition but you have confirmed my reference to the electronic states of Argon: https://www.nature.com/articles/srep15254/figures/5
There are no images in the many you linked that had anything to do with rotational states of Argon, a few which show the same information about the electronic states as I linked to in the NIST report, and a lot of irrelevant images from various sources such as lasers, bicycles, helmets etc.
A C Osborn commented:
First, A C, you seem to think that I am the Editor of the journal. I am not. I pointed out to the Editor what was going on. He pulled the paper, as he should have, to protect the reputation of his journal—it was an attempt at underhanded deception. But notice … HE pulled the paper, not me. If you have a problem with that, I suggest you take it up with the Editor. I have no problems or regrets about my own actions, I’d notify the Editor again if I had it to do again.
It also appears that you are confused. The post you pointed to contains the “gory details”, but not the gory details of their “theory”. It has the gory details of N&Z getting their paper yanked for scientific malfeasance.
However, as you point out, that post doesn’t “do anything to destroy the theory” … for a simple reason. It wasn’t about the theory. It was about their attempt to scam the journal.
However, this post, which I’ve linked to three times now, shows that their theory is a scientific joke. If you find a real problem with my deconstruction of their curve-fitting, I’m all ears. But to date … nothing.
w.
“But to date…nothing”
Apart from:
https://tallbloke.wordpress.com/2012/02/09/nikolov-zeller-reply-eschenbach/
Tony December 2, 2017 at 12:04
Sorry for my lack of clarity. By “nothing”, I meant “nothing of substance”. If you think their defense of their own work holds water, more fool you. But as I proved above, there is no possible way that their MAGICAL! equation works without GHGs …
And N&Z still don’t understand that when you have 5 tunable parameters and only 8 data points, that it is very easy to fit them. In my linked post, The Mystery of Equation 8, I did a better job of fitting them with only 3 parameters.
So … how come you are not patting me on the back and congratulating me for improving on the N&Z equation?
Heck, Tony, believe what you want. Everyone here who does understand the issues knows that you don’t. You’ve completely trashed your own reputation … oh, wait, you don’t have a reputation, because you don’t even have the balls to sign your own name to your own words. So you can walk away, take up a new alias, and continue your nonsense.
Sadly,
w.
The post I linked to is a reply to “The Mystery of Equation 8”, by Nikolov and Zeller themselves. I recommend readers have a look at both and decide for themselves.
I note that you do not refer to my other reference which shows that many scientists and mathematicians ha very acceptably done exactly the same thing as N&Z.
Why is that?
A C Osborn December 2, 2017 at 3:41 am Edit
Because:
a) Science is not a democracy settled by votes, and
b) You haven’t even done yourself the favor of QUOTING YOUR OWN WORDS so that I could find the link you are mumbling about, and
c) If as you say they did “exactly the same thing as N&Z”, what N&Z did was a trivial and ludicrous curve fitting exercise involving 5 free parameters. Why would I want more of that?
I note you do not refer to my other post which shows that I have very acceptably done exactly the same thing as N&Z, and even better. I fitted their data with two less parameters. How come you are not congratulating me???
If that’s not enough reasons, let me know, I’m sure there are more ..
w.
when u burn ants with a magnifying glass, you are not concentrating temperature and the lens doesn’t get hot because it is transparent.
maybe some basic definitions are in order.
maybe start with temperature.
gnomish, thanks for the comment, but it’s not at all clear who or what you are referring to here. Please quote what you’re responding to, because as it stands, it’s too vague to be meaningful.
w.
well, there are some describing radiant energy as cold or hot – as if a thermometer could take its temperature, for one example.
that’s not going to get anybody to logic town.
then there’s a ‘thermal radiation’ meme. if IR is meant, then use that term, no? radiation is EM energy no matter if it’s IR or UV or X and any absorbed radiation of any frequency becomes kinetic energy, no?
if one wishes to refer to ‘color temperature’ of a black body- that’s a spectrum but color can not be measured with a thermometer.
imo, this language is but a baby step from phlogiston – but at least there’s a definition for that so i don’t have to parse idiosyncratic language of every person just to see if i can find any sense in a statement. i just can’t keep track of the babble. it is babel.
temperature is not an attribute of a 445nm blue laser’s monochromatic radiation but boy does it burn stuff.
gnomish December 1, 2017 at 10:34 pm
I repeat my request. Quote what you are talking about.
The infrared spectrum is roughly divided into two parts, generally called “near infrared” and “far infrared”. Near infrared is “near” to visible light in wavelength. We get a lot of that from the sun. Far infrared, on the other hand, comes from objects at earthlike temperatures. As a result it is often called “thermal IR”. And just to confuse matters, far infrared is also often called “LWIR”, or long-wave infrared. Sorry you don’t like it, but those are the terms in common use.
The temperature of a black-body for a given emitted radiation is often called “brightness temperature”, or the “S-B temperature”, for the Stefan-Boltzmann equation used to relate temperature and thermal radiation.
Again, however, your lack of quoting what you are referring to is biting you. The only reference in this thread to “color temperature” refers to light bulbs … and again, this is common usage, viz:
This relates to the fact that as you heat something, say a bar of steel, you can tell its temperature by the color of the light that it emits. See Westinghouse on the subject, they know about light bulbs.
I’m sorry, but it’s not babble in the slightest to anyone familiar with the field. These are all called “terms of art”, meaning words that have a special meaning in a particular field. Every specialty has them. You are obviously unfamiliar with them … but that’s no reason to get all huffy with people who are familiar with them.
Regards,
w.
dude, you want to be deliberately obtuse, fine.
off you feak, then.
Great word that Gnomish Obtuse.
Mr Eschenbach, what colour is an Argon Lamp?
So based on the colour what Temperature is it at?
You know, Argon, that stuff that cannot emit photons.
A C Osborn December 2, 2017 at 4:27 am
Great word that Gnomish Obtuse.
Mr Eschenbach, what colour is an Argon Lamp?
So based on the colour what Temperature is it at?
You know, Argon, that stuff that cannot emit photons.
Let’s get it straight as you’ve been told above Argon can’t absorb IR and therefore can’t emit at the temperatures experienced in the atmosphere.
Color temperature is applied to blackbody emitters which emit a continuum of light the peak of which shifts with temperature (Wien’s Law) so as temperature changes our perception of its color changes. In common parlance this is frequently used for incandescent light bulbs, but pyrometers used to measure furnace temperatures use the light spectrum to do so. Since incandescent bulbs have a hot filament which emits as a blackbody this was straight forward, however it’s not so with LEDs which only emit in the visible, but the manufacturers have an equivalent scale for them so that consumers can chose the bulb that best fits their application. Discharge lamps such as argon, helium etc are different again because they emit discrete line spectra (such as I showed you above for Argon), the color balance will change with temperature but it’s not as straightforward as in a continuous distribution. The temperature of an Argon lamp will depend on how much current you run through it and the perceived color will change as well.
No you get it straight.
Mr eschenbach incorrectly stated that there are “things” ( he did not state what things other than they are not solid objects) that CANNOT emit Thermal Radiation, no qualification that it is Infra Red Radiation.
All Radiation from any matter above Zero K is classed as Thermal Radiation and Argon is above Zero K.
The Atmosphere does not come in to this part of the conversation.
Up post you stated “No, the energy levels for Ar are well known and will require excitation in the far UV”.
Is there any far UV in the Radiation coming from the Sun?
.
Strictly speaking there is no such thing as thermal radiation. It’s a term that has (In common parlance) come to characterise electromagnetic radiation emitted by matter above absolute zero that we encounter in common environments.
It is all electromagnetic radiation and there is no diff between 10um Co2 photon from an atmospheric molecule and one that pops out of a Co2 laser.
It’s also misused a lot. Argon can radiate but not in the normal conditions encountered in this thread.
A C Osborn December 2, 2017 at 6:40 am
No you get it straight.
Mr eschenbach incorrectly stated that there are “things” ( he did not state what things other than they are not solid objects) that CANNOT emit Thermal Radiation, no qualification that it is Infra Red Radiation.
All Radiation from any matter above Zero K is classed as Thermal Radiation and Argon is above Zero K.
But it won’t emit until it gets to a couple of thousand degrees!
The Atmosphere does not come in to this part of the conversation.
Up post you stated “No, the energy levels for Ar are well known and will require excitation in the far UV”.
Is there any far UV in the Radiation coming from the Sun?
Not anywhere where you’ll find argon on this planet.
Again, you completely miss the point of the question.
If the whole Atmosphere is Argon as Mr Eshenbach suggests (probably an impossible situation considering where Argon comes from), would the Sun provide Far UV to warm it up?
A C Osborn December 2, 2017 at 4:27 am
Are you claiming that argon “cannot emit photons”? Because I never said that. I said that it does not emit in the range of “thermal IR”.
And no, “thermal IR” doesn’t mean “any temperature”. It is a term used interchangeably with “longwave IR” to indicate the part of the IR spectrum that occurs at earthlike temperatures.
w.
OK, Mr Eshenbach, I can see that you can never be pinned down as you change what you mean every time you are challenged on EXACTLY what you said.
What you originally said and I quote
Mr Eschenbach says “Is there some law of nature I don’t know about, that says “All things must radiate thermal energy”?”
Now you are saying that they DO radiate Thermal Energy, just not Thermal IR Energy.
There is a word for that.
A C Osborn December 2, 2017 at 1:25 pm Edit
Yeah, there is. It’s regarding you not noticing the context of the discussion, in which we’ve been talking about thermal IR. The word for that is unobservant.
And the word for your comment is “nitpicking”. If that’s all you can find wrong in my writings, I’m a happy man.
I do notice that nobody has found any problems with my proof that a non-GHG atmosphere cannot permanently warm the surface … including you. So you decide to nitpick instead.
Your choice … me I prefer science, but if you want to pick nits, be my guest.
w.
A C Osborn December 2, 2017 at 10:18 am
Again, you completely miss the point of the question.
If the whole Atmosphere is Argon as Mr Eshenbach suggests (probably an impossible situation considering where Argon comes from), would the Sun provide Far UV to warm it up?
Only far up in the thermosphere.
You may recall that in an attempt to simplify and clarify this matter, I asked the following question:
Willis Eschenbach November 30, 2017 at 3:48 pm
We’ve had responses of yes, it’s possible, and no, no way. Hans Jelbring has a whole theory about how a non-GHG atmospher can keep the surface warmer. Nikolov and Zeller say the same, with a different mechanism.
Now it’s time to settle the question. I say I can prove, not assert but prove, that not only is Jelbring proposed mechanism wrong—I can prove that there is no possible mechanism by which a non-GHG atmosphere can maintain a warmer surface.
(As an aside before we begin, before you claim that “nothing can be proven in science”, consider: you certainly can’t prove that there are no black swans … but it’s easy to prove that there are white swans …)
The type of proof that I will use is “proof by contradiction”. This involves assuming that what is under investigation is true, and then seeing if that assumption leads to any fatal contradictions.
SO: let us assume we have an airless black-body planet that has a non-GHG atmosphere. Say it is being evenly radiated from all sides at a rate of 340 W/m2, the strength of the sun. It is also emitting 340 W/m2, so it is in steady-state. And Stefan-Boltzmann gives that planet a temperature of about 5°C.
Now assume that we introduce an atmosphere of argon, which despite nay-sayers neither radiates nor emits in the thermal IR range. We let the transients settle down until it is once again at steady-state. (There may be transient heating or cooling from the expansion or compression of the atmosphere, but these are one-time events.)
Now, according to Hans Jelbring (and others), gravity acts on the atmosphere to make the surface permanently warmer. So let us ASSUME that he is right, and see what that gets us. Suppose his mechanism adds 10°C to the temperature. So the temperature of the planet settles down at a new steady-state, with a temperature of 15°C, maintained by the atmosphere. What’s not to like?
What’s not to like is that IF that is the case, then Stefan-Boltzmann says the planet must be radiating at the rate of about 390 W/2. But if the planet is at steady-state, this would violate the First Law of Thermodynamics. The planet at steady-state would constantly be radiating MORE than it is receiving, which is impossible.
Therefore, not only Jelbring’s hypothesis but any mechanism claiming to warm the planet by atmospheric action alone without GHGs is impossible. It leads to a violation of the First Law.
Q.E.D.
Comments invited.
w.
PS–As I said above, I believed Jelbring’s hypothesis when I first read it, and I said so publicly … but after I’d studied and learned more, I changed my mind. I’m curious if those of you who said yes, it is possible for a non-GHG atmosphere to permanently warm the surface, will have the courage to do the same …
For the longer version of this same proof, see A Matter Of Some Gravity
SB is a model that has validity in a defined context and not outside that context.
some soup is 95C but the thermos that contains it does not radiate the SB spectrum corresponding to its temperature cuz the heat can’t move by conduction or radiation (ok, the mirror returns only 99.9%, fine)
the thermos mirror is the radiant surface that is observable and it’s just not hot. but the soup is.
and this is a planet covered deep in soup.
so the surface of the thermos will be warmer than without the soup but it is heated by conduction, mostly.
the liquid storing the energy heats the surface and SB has no say in the matter.
http://www.westinghouselighting.com/images/pageassets/color-temperature-kelvin-light-appearance-ambience.jpg
and SB radiation calcs don’t confine radiation to long wave end of the spectrum.
http://hyperphysics.phy-astr.gsu.edu/hbase/wien.html big duh, right? when we have a stellar fusion generator so handy to look at any day…
lots of radiation has nothing to do with messrs s & b or wein or planck.
just cuz u got a hammafor…
Willis,
I like to read the longer version, as the gravity argument is used here too… But the reference points back to this discussion…
Thanks, Ferdinand, I fixed it.
w.
“..which despite nay-sayers neither radiates nor emits in the thermal IR range.”
A Willis’ speculation until proven by experiment. Hence any conclusion is likewise speculative. Willis cannot be proven wrong, nor can Willis be proven right.
Trick December 2, 2017 at 6:05 am
As mentioned repeatedly, you’ve been given the actual NIST spectrographic results that show exactly that. They are not theory. They are the results of an EXPERIMENT.
The fact that you don’t seem to understand them does not invalidate them.
w.
Willis: “..that show exactly that.”
What is it you mean by “that” Willis?
The NIST experimental data show plenty of lines emitted from atomic Ar in the IR in air. You must not have clicked on the right tab. Your Ar atm. will have an emissivity to multiply a nonzero Planck radiance at 255K and IR wavelength. Thus it will have a surface warming effect above an ideal transparent atm. The question is how much.
I guessed for an Ar emissivity of 0.01 (vs. normal atm. at 0.8 global median) then calculated around +0.3K (255.3K above 255K for transparency) assuming common Earth pressure, insolation, surface parameters.
As Phil. points out no one (so far as is yet discovered anyway) has run an emissivity test on Ar at typical atm. pressure (1atm.). Thus no one can prove your answer wrong or right for an argon atm. If you choose to change the proposition to an atm. consisting of a gas with a known emissivity (1atm) then experiment can support your proposition answer.
An immediate failure at the first point.
You cannot have a planet evenly radiated from all sides with 340 W/m2.
Is it totally surrounded by suns?
Therefore it cannot be in a steady state or thermal equilibrium, the Earth is not and has never been in equilibrium.
Seriously?
Nobody every claimed that 340 was an instantaneous value. Do you think you’ve made a new discovery with that rotation thingy?
Are you denying that the Earth’s Surface and Atmosphere is NOT in Equilibrium?
A C Osborn December 2, 2017 at 3:31 am
Dude, it’s a THOUGHT EXPERIMENT … I can absolutely have it surrounded by thousands of suns. Did you read “A Matter Of Some Gravity” as I suggested? Obviously not …
w.
Yes I know it is make believe and not any kind of Science.
A C Osborn December 2, 2017 at 1:02 pm
You’re really going to object to thought experiments all over again? Didn’t you learn from the first time? We went through that upthread, were you not paying attention? Maxwell’s Demon is a thought experiment. Schrodingers Cat is a thought experiment. Are you going to sit there and claim that neither of those are “any kind of science” just because they are thought experiments?
Sheesh … been there, done that …
w.
AC, if you don’t like the term “thought experiment” then try “homework problem”. That is essentially what these scenarios are. The scenarios focus in on a few concepts to see what textbook science would predict as the answer.
Another perspective on “A Matter of Some Gravity”:
http://theendofthemystery.blogspot.co.uk/2012/02/incompetent-skeptics-ii-willis.html
Tony, your “Other Perspective” is interesting and makes some good points. But it also makes some very fundamental mistakes about thermodynamics.
Tony December 2, 2017 at 4:52 pm
Tony, even you should be able to see the errors in this statement from “Another perspective” …
His claim is that when a warmer object absorbs radiated energy from a cooler object, the warmer object cannot then radiate the energy that it has received. Why? He doesn’t mention that part.
Now if you believe that, Tony, then you are as innocent of thermodynamic knowledge as is Harry Dale Huffman.
And if you don’t believe it, then why are you cluttering up this site with that kind of garbage?
Finally, there are lots of “competent scientists defending the greenhouse effect”, including a number on this very thread. Another point against Mr. Huffman’s screed.
w.
PS—I don’t trust men who use all three of their names. And Harry Dale Huffman is one more data point in favor of that rule of thumb.
Willis posts a link to a five year old article he’s written as if to say, “if you can’t find a fault with this, it’s right”. My point is that these discussions have already been had. If you don’t like (or can’t explain without misrepresenting) what Huffman is saying, there are blog posts and papers by others with a different take (or the same take written in a different way). There are also over one thousand comments in response to the original article, in the first place, many of which are not in agreement with Willis. I’m just making sure it’s realised that in each case (each time Willis links to one of his own articles) it’s known that this wasn’t the final word on the matter.
Willis:
i) Kinetic energy at the surface can either be conducted away or radiated away. Both cannot occur simultaneously. If one rises the other drops. So, if you have kinetic energy sufficient to produce a temperature of 288k as per the Earth’s measured surface temperature there is no physics that prevents 255k radiating to space and the other 33k being conducted to the atmosphere.
ii) If that 33k goes upwards from the surface in convective ascent (effectively disappearing from the radiative energy budget as PE) then it comes down in convective descent (reappearing within the radiative energy budget as KE at the surface) but at a later time. It is well established physics that adiabatic cooling in ascent and warming in descent is fully reversible.
iii) The original 33k taken from the surface by conduction in the formation of the atmosphere came from energy that would otherwise have radiated to space. That being the case it cannot have reduced surface temperature below 255k which was the surface temperature pre atmosphere as per S-B.
iv) If the surface temperature remained at 255k whilst the atmosphere was being formed then it MUST rise by 33k when the adiabatic loop closes.
v) Although the surface temperature is then 288k you still have ongoing conduction of 33k upward to maintain convective overturning PLUS 255k of radiation going out to space which keeps the combined radiative and adiabatic loops stable and the atmosphere in hydrostatic equilibrium indefinitely.
Your previous objection to such a scenario missed the issue of the process of creating the atmosphere in the first place when the first convective overturning cycle formed. That is where the non zero consequence of conduction and convection comes from so that there is no breach of the first law.
You can only counter that by suggesting that a surface at 288k can both radiate at 288k AND conduct at 33k simultaneously and THAT would be the true breach of the first law.
“Kinetic energy at the surface can either be conducted away or radiated away. Both cannot occur simultaneously.”
Yes! Please remember this! (also, it would be much more productive in general to talk about energy and power, rather than temperature.
“So, if you have kinetic energy sufficient to produce a temperature of 288k as per the Earth’s measured surface temperature there is no physics that prevents 255k radiating to space and the other 33k being conducted to the atmosphere.”
Actually, if you have kinetic energy sufficient that will be producing 390 W/m^2 of thermal IR, there IS physics that allows it to radiate away only 240 W/m^2 of those 390 W/m^2 to space . It’s called “Green house gases”. Without these gases, it would indeed radiate 390 W/m^2 if the surface were 288 K, causing very rapid cooling.
It could only be GHGs as a cause if the surface temperature drops whilst the atmosphere is forming.
If the surface temperature remains at 255 and radiation to space drops instead the cause is atmospheric mass within aa gravitational field.
Please explain how the surface temperature could drop during the formation of the atmosphere without a breach of S-B and the first law.
“You can only counter that by suggesting that a surface at 288k can both radiate at 288k AND conduct at 33k simultaneously and THAT would be the true breach of the first law.”
Objects routinely conduct and radiate as they independent processes, no breach Stephen. Planck’s radiation law was developed at room temperature and ~1atm. Conduction and convection were simply minimized.
The total of radiation and conduction cannot exceed the energy required to fulfil the S -B equation. If the Earth’s surface at 255k were to both radiate 255 to space and at the same time conduct 33k to the newly forming atmosphere the consequent surface temperature fall would be a breach of S-B.
Your proposal is that the same parcel of kinetic energy can both radiate photons and conduct the same energy that was released by the photons simultaneously. That is a breach of the first law.
Tim Folkerts, are you prepared to discuss the “Mechanics” of how those GHGs do there job?
AC, there are innumerable details, but the basic mechanism is pretty straightforward. The earth absorbs ~ 240 W/m^2 of power from the sun on average. To balance out (and in the long-term, earth does come quite close to balancing out), the earth must also radiate ~ 240 W/m^2 of thermal IR to space. This would correspond to an average temperature of ~ 255 K. GHGs in the cold upper atmosphere radiate at a temperature well below 255 K, and hence radiate weakly to space. To balance, some region MUST radiate at a higher temperature than 255 K, radiating more strongly to space to compensate. That region would be the surface.
Tim, that is what they keep saying and the diagrams show, it is “how it actually works” that I am interested in.
Let’s start with if CO2 is suplying the W/m2 where in the Atmosphere does this happen, ie how high and at what Temperature are they?
Next Moisture does the same, but presumably lower in the Atmosphere.
If that is correct why do dry Deserts get both Hotter & Colder than the tropics, aren’t all High temp records not registered in “dry” areas?
I completely understand that moisture (and the Seas) keep the Temperature more stable, but not Hotter.
“The total of radiation and conduction cannot exceed the energy required to fulfil the S -B equation.”
Sure it can Stephen. This is no breach of 1LOT. A molecule can emit a photon and bang into another slower molecule all at the same time, losing energy by a transfer of energy by conduction and radiation simultaneously. Possibly your wording is not conveying the math you are trying to verbalize, but I know you can’t read math so trying to get you to link or show what you really mean in math is probably futile.
You need a faster process of energy shedding than radiation to reduce the surface temperature when the atmosphere is forming. Since conduction and convection work slower than radiation the energy must come from a reduction in outgoing.
If a molecule bumps into another so as to reduce its energy content by conduction then the reduction in energy content is replaced by fresh incoming energy instead of that fresh incoming energy radiating out.
Conduction is an interruption in the free flow of radiation not an accelerant.
The atmosphere formed 4 billion years ago. Your insistence in continuing to bring it up is bright shining evidence that your theory can not support a simple energy flow accounting that demonstrates energy balance.
The planet is in dynamic equilibrium! Fit your theory to that unpleasant truth..
I thought you had given up when I used your own style of analogy to show you why you were wrong.
The events at the formation of the atmosphere when there was an interruption of the free flow of radiation through the system are critical to the surface temperature enhancement and will remain so for as long as the sun shines and thereby keeps the mass of the atmosphere suspended off the surface.
Stephen I didn’t go away. When I started our discussion I was trying to get you to demonstrate a simple flux accounting for our current, dynamically equilibrated atmosphere. Since you insisted on bringing up 4 billion year old developing atmosphere I engaged that argument.
We disagree on that accounting. I gave my best shot at trying to change your thinking on that formation and failed. Like you, i know when the horse is dead. I won’t go there anymore. I will even go one step further and grant that it all worked back then exactly as you say.
It is irrelevant to our initial discourse. Any ‘bump’ that is 4 billion years old is long gone from the system, right? By your own prior statements, you agree that today’s atmosphere has a net zero interchange of energy with the surface. The basis of your theory is that the atmosphere has no radiant energy interchange.
So we both agree that we are left with an atmosphere, by your reckoning that, from an energy consideration, can be completely eliminated from condideration. This leaves us with a surface at 288 k with an input of 240 watts/m^2. There is no reconciling that. It is not in balance. You need to explain the imbalance as some hitherto absent energy source
It is certainly not reconciled as an artifact of a 4 billion year old atmosphere formation under a vastly different sun.
For this question, I am still engaged and waiting for an answer devoid of the formation arguments.
You can’t ignore the imbalance created during the formation of the atmosphere.
As per the figures I gave you there were 10 cheese balls on the table at the start but by delaying the deliveries you are then actually holding 10 on the table and 5 in the bucket forever unless something else changes.
The effect of that initial temporary energy imbalance remains until the sun stops shining.
If one adds mass to the atmosphere the imbalance increases. If one removes mass the imbalance decreases.
Atmospheric mass used conduction and convection to delay radiative throughput and the surface temperature rose as a result.
That is why the temperature at 1 bar pressure on Venus is similar to that on Earth after adjusting for distance from the sun and why the surface beneath such a massive atmosphere is so hot.
More mass in the atmosphere created a bigger radiative imbalance during the formation of the Venusian atmosphere so the surface is much hotter after accounting for the distance from the sun.
So, in a nutshell, you claim that these 4 4 billion year old energies are still with us?
I respectfully repeat my request to show me the accounting.
I gave you the accounting in the cheese ball analogy. You start with 10 balls but by manipulating timing on one occasion you end up with 15 in stock forever.
I can see you never did stock control in retail.
No fair again. No cheese balls this time. That was a different problem for a different time. All those old balls got eaten in the cretacious. And, phase delays are not allowed in steady state accounting. All phase delays are moot after 1 cycle and the cyclical nature of our atmosphere has a frequency much much smaller than 4 billion years.
Once again I respectfully request the energy accounting for today’s dynamic equilibrium. BTW, that means no net energy exchange for the system, the atmosphere, and the surface. Please remember that this means you can’t include 4 billion year old bump in today’s accounting. That would be another no fair.
You are correct on the retail. Never did it.
Did however do microwave engineering, software engineering and physical oceanography, modeling boundary free ocean circulations, antenna field distributions, and network topology and capacity modeling.
Those bits have some accounting. Hope that is the last snark where we share the size of our d***s.
BTW. No fair bringing up Venus. No fair bringing up mass. No fair bringing up convection.
1. Net energy exchange at surface atmosphere Boundary is zero
2. No radiant interchange with atmosphere
3. 240 w/m^2 In
4. Surface is 288 K
Those are your claims and the only thing on the table.
Show the accounting!
You seem incapable of the mathematical agility required for retail stock control (multiple moving parts and variable timing) so no point engaging with you further since I’m sure other readers will get the idea.
Do you not realise that the radiatively inert atmosphere has a conductive energy exchange with the surface and the surface has a radiative energy exchange with space?
Adiabatically warmed air descending along the lapse rate slope places air at 255k adjacent to the surface and warms colder surfaces directly or causes warmer surfaces to cool less quickly.
By that means conduction and convection achieve exactly the same effect as that proposed for GHGs but it can’t be both or you get a surface temperature of 321k which we don’t.
So you have 33k up and down in the adiabatic loop with 255k in and out for the radiative loop.
But first you have to get the atmosphere up off the ground to create the adiabatic loop.
Paul Bahlin December 3, 2017 at 5:15 am
The planet is in dynamic equilibrium!
I assume that you mean “Thermodynamic equilibrium”
You do actually understand the concept of “Thermodynamic equilibrium”
What is Thermodynamic Equilibrium? Part-1
written by: Haresh Khemani • edited by: Swagatam • updated: 6/2/2011
Thermodynamic Equilibrium Defined
In an isolated system when there is no change in the macroscopic property of the system like entropy, internal energy etc, it is said to be in thermodynamic equilibrium. The state of the system which is in thermodynamic equilibrium is determined by intensive properties such as temperature, pressure, volume etc.
Whenever the system is in thermodynamic equilibrium, it tends to remain in this state infinitely and will not change spontaneously. Thus when the system is in thermodynamic equilibrium there won’t be any spontaneous change in its macroscopic properties.
Conditions for Thermodynamic Equilibrium
The system is said to be in thermodynamic equilibrium if the conditions for following three equilibrium is satisfied:
1) Mechanical equilibrium
2) Chemical equilibrium
3) Thermal equilibrium
1) Mechanical equilibrium: When there are no unbalanced forces within the system and between the system and the surrounding, the system is said to be under mechanical equilibrium. The system is also said to be in mechanical equilibrium when the pressure throughout the system and between the system and surrounding is same. Whenever some unbalance forces exist within the system, they will get neutralized to attain the condition of equilibrium. Two systems are said to be in mechanical equilibrium with each other when their pressures are same.
2) Chemical equilibrium: The system is said to be in chemical equilibrium when there are no chemical reactions going on within the system or there is no transfer of matter from one part of the system to other due to diffusion. Two systems are said to be in chemical equilibrium with each other when their chemical potentials are same.
2) Thermal equilibrium: When the system is in mechanical and chemical equilibrium and there is no spontaneous change in any of its properties, the system is said to be in thermal equilibrium. When the temperature of the system is uniform and not changing throughout the system and also in the surroundings, the system is said to be thermal equilibrium.
So let’s just take the last one.
When was the Earth’s Temperature Unchanging and when was the Temperature within the Earth System (Which includes the Core, the Crust, the Surface and the Atmopshere Unchanging?
Nice cut and paste diversion. I especially admire the part where you remind me that earth has parts. Never knew that.
Oh wait…..those other parts aren’t in what we agreed were the boundary parameters of the discussion. So go back , read the agreed upon parameters and answer the question.
BTW, your assumption is wrong. I specifically said dynamic equilibrium, no net change to energy flux. There could we’ll be millions of people jumping up and down trying to change our orbit at this very moment. They won’t change our radiation budget.
I am well aware that there are mouse farts that will cause momentary out of balance conditions. Fortunately for all of us, there aren’t enough mice on the planet now, or even 4 billion years ago to give that the slightest relevance to the discussion
Stephen Wilde December 3, 2017 at 7:27 am
I do get the idea. You think that by accusing your opponent in the debate of stupidity, that you get to walk away without answering his question.
You also think that some story about cheese balls is a proper accounting of the change in today’s energy flows that you claim exist because of a one-off event four billion years ago
So yes, Steven, “other readers will get the idea” … but you might not like the idea we get.
ANSWER THE MAN’S QUESTION! Other readers are getting the idea you don’t have an answer
And while you are thinking about how to answer it, consider. There are transient, one-off events all the time. Volcanoe erupt. Meteors crash. Hurricanes move immense amounts of energy around. These cause huge one-time changes to the energy content of the atmosphere and the earth.
Now if your theory is true, and the effect of one-time events persist forever as you claim … then where are the signatures of the other one-off events?
In fact, in a dynamic system like the climate, replete with homeostatic mechanisms, such events sink without a trace, and the status quo ante is quickly restored.
Best regards,
w.
Cheeseballs were introduced by the other guy.
Stock control is a perfect example of my point.
Energy used to raise an atmosphere off the ground is indubitably present for the life of the atmosphere.
Energy is not flux until it flows through a surface. You can have all the PE you wish up there but it will always be kg • m^2/s^2. Same for all the KE.
We are counting flux with units kg/m^3.
The fact that it sits on The surface and has been there for a long time has no bearing on the discussion.
It doesn’t ‘sit’ on the surface. It moves up and down in vast columns spread around the world and, being additional energy to that involved in the S-B equation it heats cooler surfaces and reduces the cooling rate of hotter surfaces.
Could not disagree with this more strongly. The atmospheric mass does just sit there. It has a PE that is function of mass and height. Parcels move up. Parcels move down. The mass of the atmosphere is not changing is it? Gravity? Height?
The PE is not a flux. You have already agreed that there is no net energy exchange with the surface from the day to day ups and downs of atmosphere.
Are you claiming that is the PE that creates your surface temperature enhancement?
Beneath downward columns PE is the only source of the ‘extra’ KE that arrives at the surface.
It is true that it is matched by the removal of KE beneath rising columns but there is a one cycle delay built in from the formation of the atmosphere so that they are out of phase and do not cancel at the surface provided fresh insolation to the surface continues.
You should have a closer look at my stock management example.
There is an ‘extra’ 33k at the surface which cannot be radiated to space because as fast as it appears beneath falling air it is taken up again in rising air.
Wow, just checked back on this thread and it is still going strong!
Willis,
Thanks for responding to my first post on Nov 26 2017 10:05 pm but you did not respond to my second post on Nov 28 2017 12:21 pm. I assume you lost it in all the other responses you are tracking so I thought I would ask the key question again in the hope that you are still tracking this thread.
It seems to me there is an elephant in the room. In my opinion, there is little doubt that, with bodies radiating back and forth to each other, you can work out the net energy flux using an ‘energy budget’ concept. However, this does not seem to work with temperatures.
For example, consider a white hot bar of steel weighing 1 Kg with a temperature of 1500 c. If an identical bar, also at 1500 c is placed next to it, the temperature of the first bar will not increase. If an infinite number of new bars, all at 1500 c, are added, the temperature of the first bar would still not increase.
Thus, despite an infinite increase in energy, there would be no temperature increase.
Yet, you argue that if an ice cube was put next to the white hot bar, the temperature of the bar would go up because the ice is blocking an even colder source (outer space). Somehow, this trivial change in energy would increase the temperature of the white hot bar when, in the other example, an infinite amount of new energy was not able to.
This does not seem to make sense.
It seems that , under certain circumstances, the energy budget approach works with joules but not with temperatures. Do you agree?
your argument is akin to galileo’s logic.
can something heavier fall faster? if you tie 2 lighter objects together do they suddenly fall faster than each other?
so i think you’ve shown that ‘warmer’ is not . but rate of cooling to some lower temperature than the original has changed for the bars that are surrounded.
What does get much warmer if we are talking about atmosphere is the air gap between the two or more steel plates. Which changes other parameters as well as radiation flow.
Yes you have.
But that does not meet the requirements of Cooler Warming Hotter for at least half the time on a day to day basis.
So as LWIR has little affect on water ie 70% of the Earth and it only warms the other 30% half the time from many kilometres away through a much thicker lower Atmosphere it is truly a mighty molecule.
so if you cut the hot steel bar in 2 pieces, each piece will make the other warmer than if they didn’t.
if you send one of them on a trip at near light speed,when it gets back, they will each be older than the other, as well.
the last ingredient to unlock the dream of infinite free energy is to put schrodinger’s cat in a transparent box and schrodinger’s dog in another transparent box where they can see each other.
by jove, i think i’ve got it! 😀
Every diversion to another hypothetical is evidence that you’ve abandoned the current one because you couldn’t make it stick.
Doesn’t make you wrong. Could be a tactical move, of course. But, if that’s ALL you ever do, It’s called dancing.
Paul Bahlin
December 3, 2017 at 9:25 am
Every diversion to another hypothetical is evidence that you’ve abandoned the current one because you couldn’t make it stick.
Doesn’t make you wrong. Could be a tactical move, of course. But, if that’s ALL you ever do, It’s called dancing.
Mr Bahlin, let me ask you thi as you are an avid believer in Cold making hot hotter.
Her is the ACTUAL situation
Two object in normal room temperature Air.
Object A at 25C and being heated by a constant energy source
Object B at 0C.
When I introduce Object B to within a few millimetres of Object A WHAT HAPPENS TO THE TEMPERATURE OF OBJECT A?
Before you answer bear in mind that I did this experiment 3 hours ago.
Don’t know or care. In my house object B would be a beer and I would drink it before A radiation was allowed to touch it.
Why do you keep dancing?
Bernard, thanks for the reminder.
With regards to the earth, we are talking about a steady-state situation. The earth’s temperature only changed by 0.2% over the twentieth century.
So to consider an equivalent situation, say we have a hot bar of steel weighing 1 Kg with a temperature of 50° c. Let’s assume for the moment that it is a blackbody floating in outer space. If so, it is radiating at about 620 W/m2. If it is in steady-state, that means it is also receiving 620 W/m2 from somewhere, say a radioactive element in the core of the bar.
So … it will sit there, just like the earth, in steady-state, for centuries. It gets 620 W/m2 from its radioactive core, basically 0 W/m2 from the surroundings, and it radiates the same total. Steady state, input = output.
Now, we plop down another bar of steel next to it. Let’s say that it is heated internally to 50°C as well, and radiates at the same rate as the other bar, 620 W/m2.
As a result, in the new situation, the first bar of steel is no longer receiving 0 W/m2 from its surroundings as before. Instead, it is receiving some of the energy that is being radiated by the second bar. Not all of it, obviously, most of the radiation of the second bar is going out to space. But some of the radiation from the second bar is hitting the first bar.
And when that happens, the first bar is no longer in steady-state. Now it is getting more energy than it is radiating … and when that happens, the first bar will warm up until the radiation going out matches the radiation coming in.
If this is not clear, I encourage you to play with the online calculator I linked to above …
Further questions or objections?
Bring’em on … that’s what this site is for, to try to find holes in or support for the claims of other scientists.
My best to you,
w.
(PS—Note that in this thought experiment, both bars will heat up until they reach a stable steady-state situation where they both are radiating what they absorb plus what comes from the radioactive core. Note also that this is the same thing that happens in my Steel Greenhouse. If you add a steel shell around an object heated by a constant energy flow, both the shell and the object will heat up.
For example: if we put a light bulb inside a thermos bottle, both the light bulb and the thermos bottle will get warmer …)
Mr Eschenbach, but the Earth’s surface is NOT in a heated steady state is it?
A heated steady state is NOT the same as an Average Radiation input if the input continuously varies from zero to 1300 W/m2 is it?
Every night the surface on the night side is no longer being heated, it is in a cooling state, do you agree that the surface cools at night?
A C Osborn December 2, 2017 at 3:24 am
Mr Eschenbach, but the Earth’s surface is NOT in a heated steady state is it?
A heated steady state is NOT the same as an Average Radiation input if the input continuously varies from zero to 1300 W/m2 is it?
Every night the surface on the night side is no longer being heated, it is in a cooling state, do you agree that the surface cools at night?
The Earth’s surface is continually emitting the same amount of radiation to space, if the surface is uniform, at any given time it will emit the same from the dark side as it did some hours previously. That’s a steady state.
At any given spot on the earth the budget will change in a periodic way, the amplitude of the fluctuation will depend on the rotation rate. Hence the amplitude of the fluctuation on the moon is higher than the earth is greater because of it’s very slow rotation rate. The amplitude will also depend on the presence of an atmosphere which will tend to smooth out the fluctuation, the extremely dense atmosphere of Venus leads to a very low amplitude.
Thanks Willis,
No fair! You replaced my thought experiment with your thought experiment and then defended your thought experiment. You did not address my elephant!
My point is that if you add infinity amounts of 1500 c white hot steel to a 1 Kg bar of 1500 c white hot steel, the 1 Kg bar will not increase its temperature. A split second later, everything will start to cool but that does not change the fact that the temperature of the bar did not go up despite the addition of an infinite amount of new energy! Thus, adding new energy does not always increase temperature!
You claim that if you place an ice cube next to my 1500 c white hot bar the temperature of the bar will actually increase because the ice cube is blocking an even colder source (outer space).
So, an ice cube increases the temperature of the white hot bar but a massive amount of white hot steel added does not?!
Doesn’t this drive a coach and horses through the ‘energy budget’ approach to predicting temperature?
Best regards
Sorry, Willis, I did not note which of your replies I was replying to.
My reply
Bernard Lodge December 2, 2017 at 8:55 am
was in response to your reply of:
Willis Eschenbach December 2, 2017 at 1:13 am
That is not the point Phil, only when an Object or Surface is being heated will the photons from a cooler object increase it’s temp.
So at night how can CO2 at well below Zero increase the Surface Temp as shown by the Trenberth diagram?
It can only very temporarily slow the cooling.
As I said below I conducted the test yesterday and again just now, 2 objects of equal tem cannot even warm each other let alone a colder one.
Seeing as you are answering my questions can I ask you why after around 30 years and Billions of $s the Climate Scientists cannot give a categorical answer of the affect of Water Vapour and Clouds?
Because I have a series of questions around the Mechanics of the Physics of GHG.
A C Osborn December 2, 2017 at 3:24 am
In the short term, no. In the long term, yes … and generally we have been talking here about the long-term steady-state average surface temperature.
w.
Yes you have.
But that does not meet the requirements of Cooler Warming Hotter for at least half the time on a day to day basis.
So as LWIR has little affect on water ie 70% of the Earth and it only warms the other 30% half the time from many kilometres away through a much thicker lower Atmosphere it is truly a mighty molecule.
Willis Eschenbach December 2, 2017 at 12:18 pm
Bernard Lodge December 2, 2017 at 9:05 am
Phil. December 2, 2017 at 6:00 am
Thanks Phil,
I think you would agree that the average temperature of both bars would be the same. My base point is that the the temperature of the first bar did not go up despite the addition of a new energy source.
Bernard, I did NOT say the temperature would go up.
I said, and very clearly I thought, that the temperature would be WARMER THAT IT WOULD BE WITHOUT THE SECOND BAR.
There is a very large difference between the two.
+++++++++++++++++++++++++++++++++++++++++++++++++
Willis,
Thanks for adding to my reply to Phil, but you did not actually reply to my original question to you which I will ask again:
Consider a white hot bar of steel weighing 1 Kg with a temperature of 1500 c. If an identical bar, also at 1500 c is placed next to it, the temperature of the first bar will not increase. In fact, if an infinite number of new bars, all at 1500 c, are added, the temperature of the first bar would still not increase.
Thus, despite an infinite increase in energy, there would be no temperature increase. This seems to contradict the energy budget approach to deriving temperatures in that no matter how much extra energy was added, the temperature would not increase.
I’m wondering how the Trenberth Energy Balance chart would cope with this scenario of extra energy with no temperature effect?
Best regards
Consider that you are using energy, as a term, a bit too sloppily. In all of these discussions we dwell way too much on temperature. Temperature is tertiary effect and we all talk about it too much because we live with it in our every day lives and we are very familiar with it.
Energy is what matters and even there that’s a bit over used. What is really really important is flux, the flow of energy, or in this case the exchange of energy. Flow causes energy variance which causes temp variance.
So we talk of our familiar, temperature, which is really tertiary effect of the most important physics which is flow, the most mysterious to our senses. If you asked 100 people about how their body senses heat they would all be confident that their skin is covered with little temperature sensors that somehow communicate temperature to our brains. Not true though.
Our skin senses flow. When you ‘feel’ cold It’s because you are losing energy and your skin sensors are telling you the direction of flow. You are NOT feeling cold. This is why you can walk into a 74 F room in winter and feel nice and cozy. Walk into the same room during a Florida summer and you’ll think you walked into a meat locker.
In the case of the iron bars this is how I approach it….
Each bar contains identical joules, therefore each bar is radiating identical flux. Any two bars that are exchanging flux with each other, must (with identical geometries) be having a net zero energy exchange and will therefore be incapable of changing each other’s energy content. The only thing left is that energy radiates away from both bars in the directions that are not facing each other.
The net loss of energy can only occur now over less geometry than would occur if they were not near each other. So you have two bars that will have identical temperature decay. It will take longer for them to stabilize (net zero exchange with environment)than either would alone.
You might even agree with me that temperature is not very relevant to the discussion and we figured it out without ever mentioning it.
Bernard Lodge December 1, 2017 at 9:52 pm
It seems to me there is an elephant in the room. In my opinion, there is little doubt that, with bodies radiating back and forth to each other, you can work out the net energy flux using an ‘energy budget’ concept. However, this does not seem to work with temperatures.
For example, consider a white hot bar of steel weighing 1 Kg with a temperature of 1500 c. If an identical bar, also at 1500 c is placed next to it, the temperature of the first bar will not increase.
Actually the adjacent sides of the two bars will increase. Assuming the bars are in a furnace then the first bar is at steady state with the furnace lining which will be cooler than the bar itself, when the second bar is placed next to the first it replaces the radiation from the cooler wall and therefore the bar will be hotter. This is the same principle as the heat shield used with thermocouples. In that case the temperature measured by a thermocouple is lower than the flame temperature because it is radiating heat to the surroundings, if a cylindrical shield is placed around the ThC that is much hotter than the surroundings and the measurement error is reduced. NACA produced reports on different designs of shields to be used in gas turbine engines back in the late 40s. Read any college text on radiation heat transfer and you’ll find many examples of these applications with worked calculations.
Phil. December 2, 2017 at 6:00 am
Thanks Phil,
I think you would agree that the average temperature of both bars would be the same. My base point is that the the temperature of the first bar did not go up despite the addition of a new energy source. A split second later they may start to cool but the second bar was added and the temperature of the first bar did not change. In fact, there could be an infinite amount of hot new bars added and the temperature of the first bar would still not go up.
Thus, not all new energy sources cause an object’s temperature to increase.
I conducted that experiment Bernard suggested yesterday with 2 objects at 60C and a few mms apart with cork between the two. Both were cooling, how much hotter did they both get when the cork was removed?
Zero, that is correct 0 not even 0.1C difference.
So you specify the steel plates to be in a furnace, where did Bernard Lodge mention a furnace?
Why would you change the parameters totally to make a point?
Bernard Lodge December 2, 2017 at 9:05 am
Bernard, I did NOT say the temperature would go up.
I said, and very clearly I thought, that the temperature would be WARMER THAT IT WOULD BE WITHOUT THE SECOND BAR.
There is a very large difference between the two.
Since you seemed to want a temperature that goes up, I gave the example of a bar heated by a constant energy source. Given your example, no, the temperature of the bar will NOT go up. But it will be warmer than it would have been without the second bar.
Regards,
w.
MMMMMMMMMM, “Given your example, no, the temperature of the bar will NOT go up. But it will be warmer than it would have been without the second bar.”
Temperature will NOT go Up = Warmer than it would have been.
The First Bar would be “Warmer”?
I think Mr Eschenbach must be getting tired.
I think he must mean that it would cool more slowly, so hold it’s heat longer.
A C Osborn December 2, 2017 at 9:06 am
I conducted that experiment Bernard suggested yesterday with 2 objects at 60C and a few mms apart with cork between the two. Both were cooling, how much hotter did they both get when the cork was removed?
Zero, that is correct 0 not even 0.1C difference.
So you specify the steel plates to be in a furnace, where did Bernard Lodge mention a furnace?
Why would you change the parameters totally to make a point?
The important parameter of the background temperature was not specified, without that the question was impossible, I chose a perfectly reasonable value.
Phil, as I was the one doing the experiment in my house the The Background Temperature was ambient in my kitchen, ie 20 C.
Now perhaps you will tell why this flood of photons being “Absorbed” by the objects did not raise the temperature of either at 60 C.
Because we keep hearing that Photons are photons and their Energy must make an [
obsorbingabsorbing] object hotter, even if they are coming from a colder object.Sorry Absorbing Object.
After all the second object at 60 C is obscuring an area of the kitchen that is at 20 C when the cork is removed.
Surely it must have [obscured] it from view just like a CO2 particle does with Space?
Damn, I wish there was an Edit facility on here.
I meant obscure.
Sounds like a really poorly thought out experiment, it appears that you think that the cork is at 20ºC? How are you measuring your objects’ temperature?
Willis Eschenbach
December 1, 2017 at 9:21 pm:
Just like on Earth, what is to stop conduction from the ground (heated by the sun) from warming the gas, and getting convection going? Also as here. If the gas is not warmable by radiation, only conduction could unfreeze it…..
So what, the conduction and convection don’t cause loss to space in the case of a non GHG atmosphere so the surface temperature will remain the same.
The surface temperature stays the same while the atmosphere is forming via convective ascent of atmospheric mass but as soon as that mass gets back to the surface in convective descent then the surface temperature rises above S-B.
…the surface temperature rises above S-B.”
No, the surface emits AT S-B. 288K T, emission is from an object at 288K. If at 255K, radiation is from an object at 255K.
Not if conduction and convection intervene it doesn’t.
Your require the same unit of kinetic energy to carry out two jobs simultaneously. Doesn’t operate like that.
“Your require the same unit of kinetic energy to carry out two jobs simultaneously.”
There is no such requirement. The energy transfer processes conductive, convective and radiative are all independent.
Stephen Wilde December 2, 2017 at 8:12 am
The surface temperature stays the same while the atmosphere is forming via convective ascent of atmospheric mass but as soon as that mass gets back to the surface in convective descent then the surface temperature rises above S-B.
And the emission increases which will drop the temperature back……
Air reaching the surface beneath falling columns produces KE which raises surface temperature but it is immediately taken up beneath the adjacent rising column and so cannot radiate to space.
Furthermore there is a phase delay of one cycle left over from the formation of the atmosphere which causes a net warming effect of 33k at the surface.
“Air reaching the surface beneath falling columns produces KE which raises surface temperature”
No. There are no falling columns. When surface air is warmed and rises, ambient higher pressure air at the surface moves in to replace the lower pressure surface air. Ambient surface air. As shown in the convection experiment above. Testing (and observation) is your imagination’s downfall.
Higher pressure IS a falling column.
if your falling air is forever supplying energy, What happens when It’s all down?
“Higher pressure IS a falling column.”
Again, simple convection testing shows there are no falling columns Stephen, the high pressure fluid moves in laterally at the surface to the low pressure fluid areas. The rising convective columns are mixed at upper levels. But i know Stephen will refuse to learn from testing and continue to imagine nature of meteorology.