Guest Post by Willis Eschenbach
Once again, the crazy idea that downwelling longwave radiation (DLR, also called infra-red or IR, or “greenhouse radiation”) can’t heat the ocean has raised its ugly head on one of my threads.
Figure 1. The question in question.
There are lots of good arguments against the AGW consensus, but this one is just silly. Here are four entirely separate and distinct lines of reasoning showing that DLR does in fact heat the oceans.
Argument 1. People claim that because the DLR is absorbed in the first mm of water, it can’t heat the mass of the ocean. But the same is true of the land. DLR is absorbed in the first mm of rock or soil. Yet the same people who claim that DLR can’t heat the ocean (because it’s absorbed in the first mm) still believe that DLR can heat the land (despite the fact that it’s absorbed in the first mm).
And this is in spite of the fact that the ocean can circulate the heat downwards through turbulence, while there is no such circulation in the land … but still people claim the ocean can’t heat from DLR but the land can. Logical contradiction, no cookies.
Argument 2. If the DLR isn’t heating the water, where is it going? It can’t be heating the air, because the atmosphere has far too little thermal mass. If DLR were heating the air we’d all be on fire.
Nor can it be going to evaporation as many claim, because the numbers are way too large. Evaporation is known to be on the order of 70 w/m2, while average downwelling longwave radiation is more than four times that amount … and some of the evaporation is surely coming from the heating from the visible light.
So if the DLR is not heating the ocean, and we know that a maximum of less than a quarter of the energy of the DLR might be going into evaporation, and the DLR is not heating the air … then where is it going?
Rumor has it that energy can’t be created or destroyed, so where is the energy from the DLR going after it is absorbed by the ocean, and what is it heating?
Argument 3. The claim is often made that warming the top millimetre can’t affect the heat of the bulk ocean. But in addition to the wind-driven turbulence of the topmost layer mixing the DLR energy downwards into lower layers, heating the surface affects the entire upper bulk temperature of the ocean every night when the ocean is overturning. At night the top layer of the ocean naturally overturns, driven by the temperature differences between surface and deeper waters (see the diagrams here). DLR heating of the top mm of the ocean reduces those differences and thus delays the onset of that oceanic overturning by slowing the night-time cooling of the topmost layer, and it also slows the speed of the overturning once it is established. This reduces the heat flow from the body of the upper ocean, and leaves the entire mass warmer than it would have been had the DLR not slowed the overturning.
Argument 4. Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean. The DLR is about two-thirds of the total downwelling radiation (solar plus DLR). Given the known heat losses of the ocean, it would be an ice-cube if it weren’t being warmed by the DLR. We know the radiative losses of the ocean, which depend only on its temperature, and are about 390 w/m2. In addition there are losses of sensible heat (~ 30 w/m2) and evaporative losses (~ 70 w/m2). That’s a total loss of 390 + 30 + 70 = 490 w/m2.
But the average solar input to the surface is only about 170 watts/square metre.
So if the DLR isn’t heating the ocean, with heat gains of only the solar 170 w/m2 and losses of 390 w/m2 … then why isn’t the ocean an ice-cube?
Note that each of these arguments against the idea that DLR can’t warm the ocean stands on its own. None of them depends on any of the others to be valid. So if you still think DLR can’t warm the ocean, you have to refute not one, but all four of those arguments.
Look, folks, there’s lot’s of good, valid scientific objections against the AGW claims, but the idea that DLR can’t heat the ocean is nonsense. Go buy an infrared lamp, put it over a pan of water, and see what happens. It only hurts the general skeptical arguments when people believe and espouse impossible things …
w.
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JAE states: The “bottom line” is that THERE IS ABSOLUTELY NO EMPIRICAL EVIDENCE OF ANY RADIATIVE GREENHOUSE EFFECT.”
How do you define “radiative greenhouse effect”? It is difficult for me to try to defend something without knowing precisely what you mean.
And what would constitute “empirical evidence” to you?
I am fully convinced that there is readily available empirical evidence (and theoretical support) that
1) IR radiation from the atmosphere shines down toward the surface
2) this IR is due to “greenhouse gases” (and clouds and aerosols to a lesser extent)
3) the surface absorbs much of that IR
4) conservation of energy requires that additional energy will result in a higher temperature
5) the observed average surface temperature is higher than would be with only sunlight heating it.
Which of these do you disagree with? Then I could try providing specific support and/or empirical evidence. Or is this not what you mean by the “radiative greenhouse effect” ?
PS I can be a bit obstinate myself , and I can love a good discussion too. 🙂
Tim Folkerts @ur momisugly September 5, 2011 at 9:00 pm
Putting aside your obfuscations to JAE, and I hope as one old fart to another that JAE does not mind me interjecting, it seems to me that you totally ignore that he asked you to explain to him your view on “surface integration”.
My understanding of what you describe is that it is mathematically applicable to integrate the normal (vertical) components of the surface hemispherical emissions. I have no problem with that, but the point is that it is only part of the total hemispherical surface emission. There is sideways stuff (horizontal vectors) that is part of the whole that you neglect. What adds to the confusion is that the sideways stuff should initially conceptually be considered in an environment of a transparent atmosphere, in which case it would just escape directly to space. However, the real-world situation is that it is less simple when the atmosphere is partially opaque. Would you care to discuss this trifling complexity?
4) conservation of energy requires that additional energy will result in a higher temperature
This is not correct. The radiation is from a colder body so at best can only very slightly slow the cooling rate. So saying
4) conservation of energy requires that the additional energy might result in slower surface cooling
would be more accurate
In that if 100 balls are in a box and 20 leave and one return. The box now contains 81 balls not 80. But I’m unsure if the one ball does anything more than returns (if it does in fact return) and immediately leaves again without having any impact.
Bob,
Trying to type – let alone explain – surface integrals is not really effective in this forum. You will have to pursue that on your own, I fear. (Besides, I am a bit rusty and don’t really want to take the hours to get back up to speed on it all before spending even more hours explaining it. I have done surface integrals before as an undergrad and as a grad student, and I know it will work here)
But the net result in this case is not that tough to understand. One last analogy. You have a very large warehouse with fire sprinklers on the ceiling. The sprinklers are spaced 1 m apart on a grid, and each sprinkler sprays 390 liters/minute. Once the sprinklers have been on for more than a few seconds, then an average of 390 liters/min will arrive at each square meter of the floor of the warehouse, INDEPENDENT OF the direction the water is spraying. The water could be in a stream straight down; the water could be spraying uniformly in all directions; the water could be spraying in a cone @ur momisugly 45 degrees. The “vector components” of the water shooting out of the sprinklers is not important; only the net flow really matters.
Similarly, the vector components of the photons don’t really matter (for net transfer of energy in a transparent atmosphere). If 390 W/m^2 leave a 1mx1m square of the surface, then 390 W/m^2 will reach a 1mx1m square of a surface drawn around the earth at any height (again ignoring the absorption for the time being). We don’t need to worry the the 390 W are spreading out over some “sphere” or that some of the photons are going mostly sideways.
[Since the atmosphere is not perfectly transparent, then the directions would matter a bit, but that is a next level deeper than what I think we are discussing here.
Gravity is a slight inconvenience since gravity is not significant for photons, but the analogy works just fine in the absence of gravity, since the water is spraying generally toward the floor and will eventually get to the floor with or without gravity.
One final challenge in the analogy is that water drops splatter off one another, while photons tend not to interact, but again that is a minor distraction that doesn’t affect the result.]
RJ says:
I like your analogy. The box clearly represents a section of ground. The balls represent energy. Since no balls are arriving, it must be night. (During the day, “balls” of sunlight would be arriving.)
So we have a section of ground with some energy (and hence some temperature). During the night it loses 20 “balls of energy” — which must represent IR radiation leaving the ground. The returning “balls” would be IR coming from elsewhere (in this case the atmosphere) down to the ground.
We now consider two difference locations. In one location (call it “Phoenix), less energy heads down to the ground, so the ground has less energy (80 “balls” worth) at the end of the night and is cooler at dawn. In another location (call it “Atlanta”) more energy returns, so it has more energy (81 “balls” worth) and is a little warmer at dawn.
Your model has perfectly shown how “back-radiation” affects diurnal ranges. Areas where more energy shines down (humid, cloudy areas like Atlanta), the diurnal variation is less. Areas where less energy shines down (dry, clear areas like Phoenix), the diurnal variation is greater.
[The next step would be to note that the NEXT day when 20 “balls of sunshine” arrive, the area with more back-radiation will end up with more energy and hence a higher temperature (101 balls vs 100 balls). At this rate, the area with back-radiation would gain 1 “ball of energy” each day, warming up a bit more every day. At some point the area with back-radiation will warm up enough to emit 21 balls of energy (perhaps 110 balls at the start of the night, losing 21, gaining 20 sunshine and 1 “back-radiation” so it holds steady at 110 every morning, as opposed to the other area which will only be at 100 units of energy. This models the “radiative GHE” quite, although the relative values are not quite right. ]
“[The next step would be to note that the NEXT day when 20 “balls of sunshine” arrive, the area with more back-radiation will end up with more energy and hence a higher temperature (101 balls vs 100 balls).”
But it does not work this way. The highest maximum temperature in a day will never increase due to backradiation (unless the atmosphere is warmer than the surface)
All backradiation might do is increase the average surface temperature over a period due to a reduce cooling rate. The sun and only the sun warms (except where the atmosphere is warmer than the surface)
Its like a jacket can not increase the body’s heat above the heat from the body. But if the jersey had its own heat source then it might.
RJ says: “All backradiation might do is increase the average surface temperature over a period due to a reduce cooling rate. ”
YES! The cooling rate of the surface would be greater if it were surrounded by the 3K temperature of outer space (ie well below 0 C on average).
The current GHG’s and clouds “reduce the cooling” (ie prevent the earth from getting so close to 3K) by surrounding the earth with a ~270 K jacket. This allows the sun to warm the earth well above the temperature it could warm the earth without the help of the “reduced cooling”.
If we add more GHGs and reduce the cooling even more, the obvious conclusion is that the surface could be warmed even more by the sun. (At THIS point people can start arguing about how feedbacks might affect the amount of “reduced cooling”.)
Tim Folkerts @ur momisugly September 6, 2011 at 7:31 am
Tim, re your warehouse sprinkler analogy:
If you remove the walls, the outer (Omni directional) sprinklers would wet the ground ballistically outside the warehouse. The inner ones would not do this because of interference. Remove gravity and air drag and the outer sprinklers would spray all the way out to space.
A better analogy would be to replace the sprinklers with lamps that shine equally in all directions. (with no interference, and remove the walls). If you stand some distance from the warehouse you will see light coming at you horizontally. The more lamps there are, or the greater the power, the brighter will be this non-downward light. You could consider just having an opening in the wall at eye level, and painting everything inside black if you are worried about reflection.
I have no problem with integrating the downward light, but it is not all of the light.
I like Tim’s analogies with sprinklers and invisible mirrors to try and explain the integration over angles, but at the end of the day they are just analogies and no substitute for the maths. What you need to do is to look at the derivation of the Stefan-Boltzmann equation. The total power emitted from a ‘back body’ is obtained by taking Planck’s formula for the radiation density and integrating it
1) over all frequencies
2) over all solid angles in a hemisphere
3) over the area of the surface.
Now if you do 1) and 2) but not 3) you get the Stefan-Boltzmann formula i.e it is the power emitted by an infinitesimal area (a point) integrated over all angles. So the angles have already been taken care of. All that remains is to integrate over the area, which is simple – it just means take the power per unit area and multiply by the area. In other words ,you do not have to worry about angles , that has already been included.
Now you may not believe that, but if so you should get a good physics book and consult it. Or, this happens to be one of the cases where wikipedia gives the right answer, so you could just search for ‘Stefan-Boltzmann derivation’ and see what you get.
jimmi_the_dalek @ur momisugly September 6, 2011 at 5:35 pm
Thanks for that Jimmi,
I’ve had a quick look at Wiki’s S-B derivation, and have some difficulty because I’ve not done maths like that for over 50years. (I’m an ol’ fart just like JAE and maybe Tim whom seems to admit recently the same difficulty). However, from a quick look it seems to me that Wiki’ describes an integration of hemispherical emission from a small flat surface emitting into a female hemispherical receptor. (= the original Stefan experiment?). Could you please describe how that integration becomes normal to the emitter (as you imply) rather than the total that it emits in all directions as received by the female hemisphere?
Oh and BTW, it has been mentioned somewhere that in various engineering problems such as in heat exchangers, that field of view calculations in two-body situations are not easy. However, if S-B is normal to the transmitter, it would be fairly straightforward, yes, no?
If for instance two bodies are inclined in modestly different planes and S-B emission is normal to those different planes…… Uh? ……. One of the bodies might receive nothing from the other! (which is obviously not the case)
Oh, and convex or spherical surfaces?
Tim Folkerts says:
September 6, 2011 at 1:05 pm
“the obvious conclusion is that the surface could be warmed even more by the sun.”
NO. It is not the obvious conclusion. In fact it is a flawed conclusion
CO2 and backradiation just at best slows the rate of cooling. (and therefore increases the AVERAGE temperature). It never warms (unless the atmosphere is warmer than the surface).
If the highest surface temperature possible on a given day in a given location is 20 degrees. Backradiation will never rise this to above 20 degrees. This is the mistake many so called climate scientists make.
The sun warms. Backradiation might slow the cooling rate after the sun has done the warming.
Further to my September 7, 2011 at 12:11 am to Jimmi, I’ve had another thought on Jimmi’s recommendation to review the Wiki’ article on S-B law, and particularly its derivation. Here is an extract:
http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
Clearly, this is still talking of total energy radiated as having hemispherical distribution, and the sort of integration of surface emission talked of by Tim is not a simple summation of individual small flat surfaces emitting hemispherically. If we look at the Trenberth et al cartoon the field of view of the 390 W/m^2 is normal to the surface. However, with hemispherical radiation, the horizontal stuff is part of the 390 and in a transparent atmosphere would escape to space outside the field of view.
G Karst,
Don’t you love the voodoo of modern particle physics?? If you are in the FRAME of the particel it will have one mass, BUT, if you are in another frame observing the particle it will have another mass. In my ignorant opinion this is the area where relativity falls apart!!!
On another note, I have read where even photons have mass. This of course ties in with solar sails etc. I am just wondering when this will be built into the GCM’s. If the earth is emitting 390w/m2 maybe someone can compute what the “lift” for the atmosphere will be from all this energy going UP!!! Then there are the issues of the individual particles being struck by all these photons. Is it inertia or heat or both that is being transferred?? I am still waiting for someone to explain all this to us!!
Tim Folkerts,
“The current GHG’s and clouds “reduce the cooling””
Unfortunately you have stated only part of the issue. The water vapor and other specie absorb incoming solar radiation lowering the maximum the sun can raise the surface temp.
GHG’s are MODERATORS!!! They prevent the hot peaks AND the cold valleys!!!!! Both are necessary for the liveability of the earth!!
The following diagram shows that even in a transparent atmosphere the surface emission vectors are not constant at all altitudes as claimed by Tim
http://bobfjones.files.wordpress.com/2011/09/surface-integration.png?w=588&h=568
Kuhnkat,
I think you asked above somewhere how to post drawings.
I use MS Paint ,which is now fairly easy and more versatile to use in Windows 7. Older versions are cunningly designed to drive you mad and are harder to learn.
I used to store drawings/photos at the Flickr website to access URL’s, but Yahoo have taken over, and they have foiled my attempts so far to access my previous work.
Instead, I have a FREE WordPress website for posting stuff on the web, such as several articles here at WUWT and Jo Nova‘s. So in the example above, I just posted it onto my website and grabbed the URL of the image.
It is easy to get a free WordPress website
Bob_FJ,
thanks for the suggestions.
Bob_FJ says: September 6, 2011 at 4:03 pm
Bob, the trouble here (and with your diagram in the later post) is that there is no “outside” the warehouse”. You never get to the “end of the earth” so there are always more “sprinklers (or lights) on the other side”.
The lights themselves ARE a good analogy. Let me leave you with this riddle. Suppose you have a large building (large enough that the walls are far away from you compared to the height of the ceiling — or you could paint the walls white and get pretty much the same result in a smaller room). The ceiling is covered with banks of lights (eg lights like this everywhere http://www.kulekat.com/articles/wp-content/themes/tuned-100/images/LED/led-grow-light-panel.jpg). Now suppose you make the roof twice as high. The lights, being farther away, will look dimmer. But being farther away, they will look closer together. Will the floor be illuminated better, worse, or the same as before? (I submit the room will look just a bright — again, assuming the wall are very far away from you, like a large warehouse or big-box store.)
kuhnkat says:
Blockquote>Tim Folkerts says, “The current GHG’s and clouds “reduce the cooling””
Unfortunately you have stated only part of the issue. The water vapor and other specie absorb incoming solar radiation lowering the maximum the sun can raise the surface temp.
GHG’s are MODERATORS!!! They prevent the hot peaks AND the cold valleys!!!!! Both are necessary for the liveability of the earth!!
I don’t really disagree with what you said. I was pointing out that the main affect of GHGs on surface temperature is to emit “thermal” IR (> 4 um) toward the ground. Since sunlight has very little energy in the thermal IR spectrum, GHGs themselves absorb very little incoming solar radiation. This was the issue that was being discussed. The net effect of gaseous GHGs (which is a bit redundant since the second “G” = “gas”) like CO2, CH4, and H2O vapor is to keep the earth from being a much colder place. (H2O vapor DOES absorb a bit of near IR, so it would keep a BIT of sunlight out, but the effect on outgoing IR would be greater it seems, so it should still be a net warming effect.)
Clouds (liquid H2O), as you rightly point out, have several competing effects. They reflect away incoming sunlight, leading to a cooling affect (especially during the day). They emit thermal IR, leading to a warming affect (especially at night). This is a very handy moderating effect. They have a second moderating effect too: warmer weather -> more evaporation ->; more clouds ->; less sunlight –>; cooler weather.
* GHGs are “WARMERS” (but have minimal moderating effects that I can imagine).
* water droplets are “MODERATORS” (in a variety of ways)
It is the dual nature of water (gas and liquid) that makes it unique (and challenging) with its affects on climate.
Tim_Folkerts,
“I don’t really disagree with what you said. I was pointing out that the main affect of GHGs on surface temperature is to emit “thermal” IR (> 4 um) toward the ground.”
Even that is debatable. The tired cartoon shows half up and half down when some part of that energy is being transferred to the local atmosphere speeding convection. We know the conductivity of the ground and the atmosphere is very poor so the transfer speeding convection each morning would not be very fast based on conductivity only.
It would be really nice if they someday gave us a better cartoon to play with. My WAG is that in the lower atmosphere the transfer of energy through collision is actually as, if not more, important than the DLR.
Tim Folkerts,
Doesn’t water vapor also have very wide absorption characteristics similar to water droplets?? I’ve looked it up before but may have only looked at liquid water spectra.
Tim Folkerts @ur momisugly September 9, 2011 at 10:33 am
Nice try Tim, but you overlook the fundamental issue. Unlike your water sprinkler analogy, there is no interference between light waves, so that even those in the middle of a very large array will keep going until there is some other interference, such as your warehouse walls. In a transparent atmosphere, the horizontal stuff will head out to space, but cannot be seen in the normal field of view. (as depicted in the Trenberth cartoon). That “missing” stuff is part of the Trenberth S-B 390 W/m^2, that he illustrates as entirely normal to the surface, so thus there is a paradox
BTW, did you notice that your hero Travesty Trenberth has been very naughty again, this time in the fracas over the Spencer and Braswell paper.
Dear Moderator,
Sorry, but in my September 8, 2011 at 5:31 pm , I made a wee mistake causing later posts to be all in italics. In line 2, I inserted the italics closure slash the wrong side of the I. Sorry for the chaos. Would it help if I Email the whole thing?
jimmi_the_dalek,
On a day of awful weather, I’ve just been browsing up through this humungous thread, and found yours of August 31, 2011 at 7:19 pm
Nope! Unless you can validly contradict what Wicki’ says about it, as I repeat next:
Further to my September 7, 2011 at 12:11 am to Jimmi, I’ve had another thought on Jimmi’s recommendation to review the Wiki’ article on S-B law, and particularly its derivation. Here is an extract:
http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
Clearly, this is still talking of total energy radiated as having hemispherical distribution, and the sort of integration of surface emission talked of by Tim is not a simple summation of individual small flat surfaces emitting hemispherically. If we look at the Trenberth et al cartoon the field of view of the 390 W/m^2 is normal to the surface. However, with hemispherical radiation, the horizontal stuff is part of the 390 and in a transparent atmosphere would escape to space outside the field of view.
kuhnkat says:
>Doesn’t water vapor also have very wide absorption characteristics similar
>to water droplets?? I’ve looked it up before but may have only looked at
>liquid water spectra.
here are copies of spectra that represent what I understand of the spectra for liquid and gaseous water.
http://www.randombio.com/spectra.png
http://www.physics.umd.edu/grt/taj/104a/watopt.gif
The scales are rather different, but they show distinct sorts of behavior. H2O gas has several moderately broad bands of absorption in the IR range. H2O liquid is clearest near 400 nm, absorbing progressively more effectively on either side of that minimum.