# Sun-Cloud-Ocean Update

Guest essay by Mike Jonas

so much to say…so little time.” – Roy Spencer

I have at last found the time for the next step in my Sun-Cloud-Ocean calculations. But first, I would like to thank everyone who commented on my previous article. Some are addressed directly below, and all comments (well, most) were useful. You found, or helped me to find, a number of important errors and new lines of thought. As I said last time “If I’ve stuffed up, I want to know that right away, so please get a critical comment in asap.“. The same applies this time!

[For those not familiar with some of the abbreviations used, there is a list of abbreviations at the end of this article, along with data and code files].

1. Preamble

2. Quick Summary

3. Summary

3.1 Energy balance

3.2 Evaporation

4. Method

4.1 Input Data

4.2 The Matching Process

5. Absorption changes from last time

Abbreviations

1. Preamble

In an earlier article, I expressed the opinion that Infra-Red radiation (IR), eg. as from Greenhouse Gases (GHGs), did not warm the ocean as effectively as the wavelengths of direct solar radiation which penetrated into the ocean (ITO): “The GHG process involves only IR, which cannot penetrate the ocean more than a fraction of a millimetre, where its energy goes mainly into evaporation. ie, the energy goes straight back into the atmosphere.” and “The ITO warms the ocean well below the surface with little direct effect on the atmosphere.”.

Some time after my last article was published, I realised that Nick Stokes’ (NS) diagram (Figure 2) provided an opportunity to test my statement. If I could reproduce the diagram from first principles – ie. reverse engineer it – then I would have the means to calculate whether IR and direct solar radiation did indeed differ in their effect, and if so by how much.

It took me a long time, but I have completed (to my satisfaction) the reverse engineering, using a notional “average” patch of ocean over one 24-hour day without upwelling. The comparisons for IR and direct solar radiation were then very simple. The results are summarised below, and then the whole process is described in more detail.

There is one important caveat. The results can only be as good as the assumptions that went into them. The ocean surface is a pretty volatile place, and everything is necessarily some kind of approximation. It is possible that changed assumptions could give significantly different results.

2. Quick Summary

· With no upwelling, nearly a third of all input radiation remains in the ocean at the end of the day. It is not all lost on the same day.

· Results do support the idea that a proportion of inward IR is immediately lost in evaporation, but the proportion comes out at about 17% rather than “mainly”.

· IR and solar radiation do differ in their ability to warm the ocean. A Watt of direct solar radiation is nearly 50% more effective at warming the ocean than a Watt of IR.

· Retained energy can build up for later upwelling, eg. with El Niño, AMO or PDO, or for transport towards the poles.

· Results suggest that from 1983-2009, cloud changes were responsible for a bit over 90% (90.6%) of global warming, man-made CO2 for less than 10% (9.4%).

My take: Changes to direct solar radiation as caused by changes in cloud cover are much more important than changes to back radiation as caused by man-made GHGs. Solar energy is always being stored in the ocean, and it is reasonable to suppose that this energy is the key to Earth’s climate, as also evidenced by the global climate (= atmospheric temperature) changes that result from warming/cooling phases of ENSO, AMO, PDO, etc. Upwelling would seem to be a (or the?) major mechanism by which the stored energy is delivered from the ocean to the atmosphere. We have to understand the ocean if we want to understand climate.

3. Summary

3.1 Energy balance

The Kiehl and Trenberth (K&T) energy balance (Figure 1) that I use for some of the input information is useful and informative, but it conceals as much information as it reveals. It shows a perfect energy balance, but it shows nothing of the different timescales involved. Looking at K&T it would be easy to suppose that the energy coming in during a day also goes out – as is shown, for example, in the diagrams that Nick Stokes presented (Figure 2).

In fact, much of the energy from solar radiation remains in the ocean at the end of the day. This is the heat which builds up and is transported poleward or is later released by an El Niño or by the AMO or PDO, etc. In the “average” patch of ocean, 168 Wm-2 of direct solar radiation and 324 Wm-2 of back radiation enter the ocean, and if there is no upwelling then 160 Wm-2 stays there – only 332 Wm-2 escapes to the atmosphere. The actual numbers may be surprising, but the concept should not be. We all know that the ocean transports heat from the tropics towards the poles, and that the ocean oscillations release heat that has built up in the ocean over a period of time. Heat cannot build up if it is not being retained in the first place. An implication is that much of the energy shown by K&T escaping from the ocean has been there for quite a long time – it does not all escape on the day it came in, it does not escape at a steady rate, and it is not released uniformly across the oceans. By implication, over short periods or even up to a few decades, atmospheric temperature may bear little or no relationship to the global temperature.

3.2 Evaporation

In the calculations, I test the possibility that some of the inward IR gets lost in evaporation by, for example, exciting water molecules at the surface so that they escape into the atmosphere. There is an obvious limit to how much can get lost in this way, because less than half of the excited molecules would go in the right direction. I don’t put any restriction on the parameterisation for this, so the reverse engineering is free to settle on any percentage.

For the balance of thermals and evaporation, I assume that the rate varies linearly with temperature. In the real world, other factors such as wind speed are important, so there is an implicit assumption that these other factors remain unchanged. Given that we are working with a notional “average” patch of ocean over a single day, that should not be an issue.

Results suggest that about 17% of the energy from inward radiation that does not get past the top 10µm goes straight into evaporation (see parameter ‘v’ in worksheet parms). This evaporation is not temperature dependent. Results also indicate that my earlier assertion that IR’s energy “goes mainly into evaporation” is incorrect : partly, yes, but “mainly”, no (Roy Spencer will be pleased, I think).

IR and solar radiation do indeed differ in their ability to warm the ocean. Looked at in isolation, a Watt of direct solar radiation is nearly 50% more effective at warming the ocean than a Watt of IR.

I had expected a difference, but I had thought that the ratio would be higher. The result also shows the reason for the lower ratio: the mechanism is not what I had expected. The major factor is the temperature gradient near the ocean surface – IR isn’t fully effective at slowing the flow of energy from the ocean to the atmosphere.

In the previous article, I estimated that man-made CO2 contributed only 9% or less of the global warming over the 1983-2009 period. There were some errors in that article, addressed below. The corrected figures for the 19983-2009 period are +0.65 Wm-2 for man-made CO2, and +4.5 Wm-2 for direct solar radiation (all direct solar radiation, not just ITO). The CO2 figure is much higher than before, as explained below, and the direct solar figure is a bit lower. With the results from the NS reverse engineering, the man-made CO2 contribution to 1983-2009 global warming came out at about 9.4%, but the calculation has changed as described below.

If Dr. Antero Ollila is correct, then the figure for CO2 1983-2009 would be +0.38 Wm-2, not +0.65 Wm-2, giving a lower contribution (only 5.7%) from CO2.

3.4. Energy Accumulation

The results for different SSTs (see worksheet parms in spreadsheet OceanDiurnal.xlsx) are so similar, that the notion that absorbed solar energy at depth can build up over a long period is supported. [NB. Just ‘supported’, not proven. In interpretation of the figures, be aware of how they were calculated.]. The fact that ocean temperature just below the surface is higher than at depth means that the only way that energy at depth can escape to the atmosphere is by convection, ie. by mixing or by upwelling.

The ~118 Wm-2 retained from 10m to 100m depth is enough to warm that ocean band by nearly 10 deg C in a year (see the Heat content calculator in worksheet FluxDescr in spreadsheet OceanDiurnal.xlsx). Obviously the heat wouldn’t necessarily be retained for a whole year, but this shows that significant heat build-up is possible.

Note: After writing everything up, I have noticed that the ‘Thermals and evapotranspiration’, at 104 Wm-2, is higher than K&T’s 102 Wm-2. It should if anything be a bit lower, which suggests that I should have used a slightly higher SST. Maybe 19 deg C instead of the 18 deg C that I used. The results would change slightly, but the overall pattern and conclusions would remain unchanged. Percentage of inward IR lost to immediate evaporation would come down from 18% to 17%. I have changed the text above to use the lower number.

4. Method

The aim was to reproduce the Day & Night temperature profiles in the NS diagrams using the K&T energy budget figures, and using solar and absorption data as presented in the previous article.

The spreadsheet, OceanDiurnal.xlsx, models the upper ocean bands of a notional “average” patch of ocean in 20-second steps over one 24-hour day. Data for all inputs of energy is used unchanged, but data for outputs is used as a guide only with variable parameters. The parameters were then optimised to find the combination of inputs and outputs, together with the energy flows within the ocean, which matched both the Day and the Night NS temperature profiles in a single daily cycle.

It is all explained in spreadsheet OceanDiurnal.xlsx, worksheet FluxDescr, so I won’t repeat the details here. You can play with the figures in the spreadsheet, of course, but to run new optimisations you will need an external optimiser. You can verify that the result is correctly optimised by changing the ‘optimised’ parameters in worksheet parms.

4.1 Input Data

K&T data is taken from:

Figure 1. Global annual average energy budget, from here).

The data that the reverse engineering is trying to match is taken from the NS diagram:

Figure 2. The diurnal (day-night) cycle in the top few metres of the ocean. From Nick Stokes’ blog Moyhu. NB. The two panels have different scales on the x-axes (that’s not an issue at all, just be careful to see the panels correctly).

The part of the NS diagrams that I am trying to match is the top part, ie. below 1m. If you look closely at the diagrams you will see that the vertical axis is vague (“5-10m”), and that it is not accurately to scale. On a true log scale, using 10m for the last point, it looks like this:

Figure 3. Data from the NS ‘Day’ diagram. Y axis is ocean band number. “-1” is surface, 0 is to 1µm, 1 is to 10µm, then increase by a factor of 10 per band to: band 7 is to 10m. Bands 8 to 100m and 9 >100m are not covered in the NS diagrams. In the spreadsheet, band 1 is surface to 10µm.

I also use absorption data as reported on last time, but with corrections (see 5. below) and with the IR wavelengths that are missing from SORCE data estimated to match the K&T data. See spreadsheet OceanDiurnal.xlsx worksheet Absorption.

4.2 The MatchingProcess

Bands 1, 2 (10µm, 0.1mm) are very thin, and a large amount of energy goes into and out of them with very small residuals, so calculating their temperature accurately is not practical. I therefore tie bands 1 and 2 to band 3 (1mm) for optimising purposes, using temperature differences from band 3 to match the NS diagram. The adjustments needed are small, averaging a lot less than 0.01 Wm-2. I then optimise for just band 3.

The resulting match looked like this:

Figure 4. The match to NS Data obtained by the reverse engineering process.

The match at band 3 is accurate, but any attempt to match the deeper bands exactly failed because the entire profile is effectively dictated by band 3. Put simply, if energy flows more between bands – conduction radiation or mixing – then band 3 cannot get up to its daytime temperature in the NS diagram. If they flow less then the heat can’t get out fast enough at night.

5. Absorption changes from last time

This para refers to assumptions and calculations in the previous post. The changes listed here were made in the absorption spreadsheet from last time. The results as used are shown in worksheet Absorption of spreadsheet OceanDiurnal.xlsx.

Changes were:

· Previously, I effectively ignored energy entering the ocean at depths beyond 10m. This energy is undoubtedly added to the system, so this time I account for it. Note that this energy cannot be released into the atmosphere by conduction or radiation because the higher ocean layers are warmer, so it is actually likely to accumulate in the system for longer than energy from other wavelengths. The ocean thermocline is typically well below 100m, so is not an issue.

· A bad arithmetic error in a RF calculation, pointed out by commenter Donald L. Klipstein was corrected. Thanks, Donald, much appreciated.

· Error corrected: Different units were used for CO2 (Wm-2 actual) and ITO (ocean Wm-2 global equivalent).

· A more subtle logical error was corrected: Last time, I left out non-ITO wavelengths when estimating the proportions of warming from CO2 and ITO, because I argued that it’s the ITO wavelengths that drive multi-decadal global temperature. But CO2 operates via non-ITO wavelengths, so I should have included those wavelengths from the sun too, for correct comparison.

· For 1983-2009, I previously used solar and cloud data averaged over all the ocean. This time, I calculated them in 5-degree latitude bands in order to get a more accurate weighted trend 1983-2009. The end result was a slightly smaller trend in cloud effect over the period.

The latest results show that IR is not as effective, Watt for Watt, as direct solar radiation at warming the ocean. This is now taken into account, too.

One of my statements (“I use SORCE data for 2003. All years are almost identical.”) was challenged by Bob Weber (“All years are not ‘almost identical’ in solar activity …”). I do agree that over extended periods all years are not ‘almost identical’, but the years covered in the SORCE data, 2003-2016, were almost identical:

Figure 5. Composition of solar radiation by wavelength, from SORCE. 14 separate curves are plotted, for the 14 years 2003-2016. They are all almost exactly the same, apart from gaps where data is missing..

Bob also asked ‘A practical question’: “how long does it take for varying solar energy deposited at depth to resurface?”. The question goes to climate’s absolute core. The results reported here show that a lot of energy is deposited. I argue that the upwelling timescale is variable. In an earlier post, I said the time taken “could be days or months (eg, it might up-well quite quickly), it could be years (eg, waiting to be scooped up in an El Nino), it could be decades (eg, accumulating until an ocean oscillation such as the AMO or PDO brings it to the surface), or it could even be many centuries (eg, taken down into the deep ocean by the THC).”.

I will try to reply to Leif Svalgaard’s comments (eg. here, here) in a later post.

Abbreviations

CO2 – Carbon Dioxide

GHG – GreenHouse Gas

ITO – Into The Ocean [Band of Wavelengths approx 200nm to 1000nm]

K&T – Kiehl and Trenberth

NS – Nick Stokes

SORCE – Solar Radiation and Climate Experiment

SST – Sea Surface Temperature

THC – ThermoHaline Circulation

Wm-2 or W/m2 – Watts per square metre

Attachments (data and code)

· The calculations reported here are in spreadsheet OceanDiurnal.xlsx. The spreadsheet also contains a guide to the calculations, see worksheet FluxDescr.

· Spreadsheet DifferenceSummary.xlsx shows the differences referenced in 3.3 below.

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April 30, 2017 7:52 am

Well, [Mike], it does look like you have indeed “stuffed up”. I am afraid I only got a couple of sentences into your preamble before this became apparent so regret it looks pointless to read the rest when you have made such a fundamental error. You appear to think that downwelling IR from GHGs in the atmosphere can transfer heat to the ocean. Downwelling IR from a COLD gas cannot transfer heat to a WARMER ocean, or indeed anything that is warmer. It’s pretty basic physics/thermodynamics really. If you disagree I challenge you to prove it with an experimental set up. If you are right and I am wrong I think you will be famous and deserving of at least one Nobel prize !

April 30, 2017 7:57 am

What is the experiment that shows this? Assertions without references are just opinion.

April 30, 2017 8:07 am

It is called, The Second Law of Thermodynamics, and if there is one thing we know with dead certainty, this is it.

April 30, 2017 8:26 am

Mark – I wish I had known of the phrase “Assertions without references are just opinions” much earlier in my life but I fear I would have gotten even more school detentions than usual. It would indeed be good to see some experimental evidence of radiative heat transfer between hot and cold things being only in one direction. I am looking for it on t’internet right now but suspect the older books in the library may be more helpful. Anyone else got a link/reference ?
I just take it as fundamentally given basic physics as taught to me and seen taught to others over many decades that although everything emits EM based on how hot it is the heat can only flow in one direction, hot to cold. It it ever went the other way it would be trivial to construct a perpetual motion/free energy machine with a few thermo-electic generator cells, wires, resistor, battery,etc.

April 30, 2017 9:44 am

The 2nd law of thermodynamics is a statistical thing. Stefan-Boltzmann law can explain why – statistically – heat moves from hotter to cooler. https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law It does not say that every photon knows where to go. A blackbody radiating LWIR at the surface continues net emission of LWIR. All that the OP proposes is that some LWIR can be absorbed – if it’s heading in the right direction.

April 30, 2017 8:08 pm

That is a mistake. You ought to stop right there.

Frank
April 30, 2017 8:32 pm

Anyone looking for references showing that photons (radiant energy) can travel from cold to hot can find a half dozen at: https://scienceofdoom.com/2010/10/07/amazing-things-we-find-in-textbooks-the-real-second-law-of-thermodynamics/
Individual molecules have kinetic energy, but not a temperature – so they are neither hot or cold. Therefore, the 2LoT doesn’t constrain the behavior of individual molecules and photons. Collisions change the kinetic energy of individual molecules millions of times per second. A Boltzmann distribution of molecular speed wouldn’t exist if collisions transferred energy only from faster-moving to slower-moving molecules.
To address this problem, thermodynamic temperature is defined as being proportional to the MEAN kinetic energy of a large GROUP of rapidly colliding molecules. The temperature of such a group is doesn’t change erratically millions of times every second. Heat is defined as the NET flux of energy (including radiation) between two groups of rapidly colliding molecules. It always blows from hot to cold. Temperature, heat and thermodynamics are concepts applied to the macroscopic world.
The KT energy balance diagram shows the fluxes of energy between the surface, atmosphere, sun, and space – not the net flux of energy – heat. If you calculate the net fluxes, they are always from hot to cold.
An individual molecule has no way of “knowing” the mean kinetic energy of its neighbors. A photon carries no “information” about the molecule that emitted it. Any molecule a photon encounters has no way of knowing the local temperature and deciding whether to absorb or scatter an incoming photon.
Many people have difficulty navigating the interface between macroscopic and microscopic behavior Large groups of molecules follow the laws of the laws of thermodynamics and Newtonian mechanics, even though individual molecules, photons and other particles are not constrained by these laws. They follow the laws of quantum mechanics. The field of statistical mechanics explains how the behavior of large groups of molecules following the laws of quantum mechanics produces the laws of thermodynamics. Microscopic disorder become macroscopic entropy.

gymnosperm
April 30, 2017 9:23 pm

“An individual molecule has no way of “knowing” the mean kinetic energy of its neighbors. A photon carries no “information” about the molecule that emitted it.”
A photon certainly does carry information about the molecule that emitted it. The type of emission we are talking about here is when an electron regresses to a lower state. The photon emitted carries the energy (and information) of the quantum separating the states.
Quanta are exact. If the quantum of the emitted photon does not match exactly the quanta between available electron transitions (and electrons) in the receiving molecule, the receiving molecule is transparent and the photon sails right through. This can happen between two otherwise identical molecules if the receiving molecule already has all its electrons at higher states than the emitted photon’s quantum.

Kristian
May 1, 2017 4:04 am

Frank says, April 30, 2017 at 8:32 pm:

Anyone looking for references showing that photons (radiant energy) can travel from cold to hot can find a half dozen at: https://scienceofdoom.com/2010/10/07/amazing-things-we-find-in-textbooks-the-real-second-law-of-thermodynamics/

As you’ll notice, Frank, The Badger is NOT saying that individual photons aren’t able (or “allowed”) to travel from cold to hot. What he’s saying is simply that the atmosphere – being cooler than the surface – isn’t able (or “allowed”) to HEAT (as in ‘directly raise the temperature of’) the warmer surface. It doesn’t matter how far down into the ocean surface each IR photon from the atmosphere is able to penetrate, and it doesn’t matter if each individual IR photon is indeed able to bring energy from a cooler atmosphere to a warmer surface. The flow of HEAT is all that matters. And it goes UP and only up. The surface – being on average warmer than the atmosphere above – does not gain ANY “internal energy” [U] in its thermal exchange with the atmosphere. Nothing at all. It LOSES internal energy in its thermal exchange with the atmosphere. Which means it COOLS to the atmosphere. The heat moves from surface to atmosphere, from warmer to cooler.
And so, what is simply being pointed out is that IR photons from the atmosphere down to the surface can NOT directly raise the temperature of the surface (and therefore also cannot directly cause an increase in evaporation). This is NOT the way “insulation” (of any kind) works …

Frank
May 1, 2017 5:40 am

Gymnosperm wrote: “A photon certainly does carry information about the molecule that emitted it. The type of emission we are talking about here is when an electron regresses to a lower state. The photon emitted carries the energy (and information) of the quantum separating the states.”
The wavelength of the photon does tell you about the difference in energy between to state in the emitting molecule. It doesn’t tell you for sure what molecule emitted any particular photon or anything about its local temperature. No transition is totally forbidden in QM; there is some probability that any thermal infrared photon arriving at the surface came from CO2, H2O, or CH4, etc. Even O2 or N2 or H2. When you look at MANY PHOTONS – for example – from the sun, you can deduce its temperature from the peak emission, However, a SINGLE photon arriving at the surface (to be absorbed or reflected) could come from the sun or the atmosphere. The sun puts out plenty of thermal infrared photons, but most of its POWER arrives as visible and near IR photons.
You know what large numbers of photons will do. You don’t know what a single photon does.

May 1, 2017 6:06 am

Michael Moon is correct -the fourth proposal of Thermodynamics , also called the 2nd law of Thermodynamics. A proof is set out in the book “The Theory of Heat Radiation” by Dr Max Planck 1914. My copy is from Project Gutenberg ebook #40030. You will need to understand mathematics to follow it. Somewhere I also have a proof by Dr. Boltzmann, I think it is written in German and also has considerable maths. In Plank’s book he talks about energy waves and entropy. There is no mention of so called photons. Dr Willis Lamb Jr who won a Nobel prize for Physics in 1955 wrote a peer reviewed published article entitle Antiphoton in 1955 in which he clearly states that there are no photons and all energy can be described in terms of waves. Climate alarmists like to use the term photons as emitting in all direction. Planck shows that is not so. When someone makes comments around photons one can know they have no understanding of heat transfer.

May 1, 2017 7:06 am

What he’s saying is simply that the atmosphere – being cooler than the surface – isn’t able (or “allowed”) to HEAT (as in ‘directly raise the temperature of’) the warmer surface.

At least in the atm, that is sort of what it’s doing. As the air tries to cool, it has to convert more gaseous water vapor to a liquid (doing work), the latent heat that is released, is supplementing the energy available at the surface limiting the rate of heat loss. If you add co2, it just doesn’t have to supplement it as much to maintain the temps defined by dew points.

MarkW
May 1, 2017 8:06 am

Kristian: Doesn’t matter. When the atmosphere warms, it radiates more. Some of that extra radiation will impact the oceans, making them warmer than they would have been under a colder atmosphere.

Frank
May 1, 2017 11:58 am

Cementafriend writes: “Michael Moon is correct -the fourth proposal of Thermodynamics , also called the 2nd law of Thermodynamics. A proof is set out in the book “The Theory of Heat Radiation” by Dr Max Planck 1914. My copy is from Project Gutenberg ebook #40030. You will need to understand mathematics to follow it. Somewhere I also have a proof by Dr. Boltzmann, I think it is written in German and also has considerable maths. In Plank’s book he talks about energy waves and entropy. There is no mention of so called photons. Dr Willis Lamb Jr who won a Nobel prize for Physics in 1955 wrote a peer reviewed published article entitle Antiphoton in 1955 in which he clearly states that there are no photons and all energy can be described in terms of waves. Climate alarmists like to use the term photons as emitting in all direction. Planck shows that is not so. When someone makes comments around photons one can know they have no understanding of heat transfer.
Cementafriend, Badger, Mark, Moon, et al: 1914 physics written for physicists in German. Dr. Lamb (a skeptic of quantum mechanics) writing in 1955. 2Lot as stated by Clausis. How do we integrate all of these old statements of into what we know about physics today.
In 1979, Feynman gave a series of lectures about the interaction of light and matter and the theory of quantum mechanics meant for a general audience. The lectures were published as a book “QED” and are available on youtube at: https://www.youtube.com/watch?v=eLQ2atfqk2c
The discussion of whether light is a particle or a wave is discussed beginning at 36:00-40:30 (or better 34:30-40:30). The rules for adding arrows (vectors) that follow (and in the next lecture) explain why photons produce wave-like behavior. When discussing the accuracy of QM at 10:00 minutes, I believe 1.00112 comes from the Lamb shift – measured by QM-skeptic WIllis Lamb. You should know that hundreds of physicists were unable to adapt to the reality of quantum mechanics. Einstein famously said that God doesn’t dice with the universe. In this lecture, Feynman spends much time explaining why “rationalizing the behavior of nature” is a waste of time, because many of his older colleagues failed to abandoned old rationals. The only ideas that are of value are the ones that make predictions in accord with observations – and quantum mechanics seemed highly irrational compare with electromagnetic waves. And wave mechanics does explain a large fraction the behavior of light – but not all of it. Anyone who really wishes to understand QM and modern physics should invest the time to listen to the first two lectures in this series. Or as Feynman says around 24:30, if you don’t like the predictions of QM, “go to another universe” where the rules are simpler, philosophically more pleasing ….
Good luck

cinaed
May 1, 2017 12:53 pm

I agree with The Badger April 30, 2017 at 7:52 am.
A few comments of my own.
I don’t understand what Mike Jones means by “man-made GHGs”.
The first thing he should notice about a greenhouse is that it’s a physical structure which prevents air movement and the convection of heat when the doors are closed.
And the gases inside the greenhouse are same as the gases outside the greenhouse – except for possibly some “man-made GHGs”.
A car is an example of a small greenhouse.
If one sits in a car with windows rolled up and the air conditioner turned off around noon on a hot summer day, you can statistically feel the kinetic energy of the gas molecules slamming into your face.
The air feels hot. The velocities of the molecules will typically have a statistical distribution approximated by Boltzmann distribution – the peak of the distribution provides an estimate of the most probable temperature.
However, if you place your hand on the part of the car where the Sun isn’t or wasn’t shining recently, you’ll notice the surface is cool.
In fact, if you place your cheek on the glass where the Sun light isn’t shining it feels cold.
How can this be if the carbon dioxide causes heating?
But I digress.
My question is related to the phrase “man-made GHGS”.
What evidence does he have that only men pass gas in a greenhouse? Why would the women have their mouths and noses taped shut?

Frank
May 1, 2017 1:43 pm

Kristian wrote: And so, what is simply being pointed out is that IR photons from the atmosphere down to the surface can NOT directly raise the temperature of the surface (and therefore also cannot directly cause an increase in evaporation). This is NOT the way “insulation” (of any kind) works …
The water molecules on the surface of the ocean do not “know” whether DLR is arriving from the atmosphere to “heat” or “not heat” the surface of the ocean. Collisions cause some water molecules at the top of the surface to break free from surface tension and enter the gas phase. The temperature of the water -not incoming radiation – determines how many collisions allow water molecules to evaporate.
The same thing is true for emission of thermal radiation by water molecules in the skin layer of the ocean. The molecules do what they do because of the local temperature – no anything to do with what is happening around them.
The local temperature CHANGE is determined by the net flux of all sources of energy gain or loss via radiation, conduction or convection (including the latent heat of evaporation). The BIG picture is all that matters; not the source or destination of various energy fluxes. Molecules and photons are not constrained by the 2LoT, but the laws of QM “conspire” to produce bulk behavior that is consistent with the 2LoT.
Heat is the net energy flux between two objects. Heat has nothing to do with any one-way flux. Temperature change requires heat – the net result of all incoming and outgoing fluxes. When you believe that one source of energy is less important that another, you are violating conservation of energy. All W/m2 are created equal. None can raise or lower temperature more effectively than another.

Kristian
May 1, 2017 2:12 pm

MarkW says, May 1, 2017 at 8:06 am:

Kristian: Doesn’t matter. When the atmosphere warms, it radiates more. Some of that extra radiation will impact the oceans, making them warmer than they would have been under a colder atmosphere.

The atmosphere doesn’t warm if the surface doesn’t warm first, MarkW. And even if it HAD somehow warmed before the surface, then the reduced temperature gradient between the surface and the atmosphere is what would’ve forced the surface to become warmer, because the total (not just the radiative) surface heat loss rate would’ve gone down as a result. “Back radiation” from a cooler atmosphere CANNOT directly raise the temperature of the already warmer surface.

Kristian
May 1, 2017 3:43 pm

Frank says, May 1, 2017 at 1:43 pm:

Temperature change requires heat – the net result of all incoming and outgoing fluxes.

No. “Heat” [Q] is simply the spontaneous transfer of energy from a hot place to a cold one during a thermal exchange. It goes ONE way only. What you’re talking about here is the NET heat [Q_net] of a system, which is Q_in minus Q_out. Net heat for a system in thermal equilibrium is ZERO. That doesn’t mean that there are no heat transfers going on. Both Q_in and Q_out can very well be large, only EQUALLY large. For the surface of the Earth, Q_in is essentially equal to the solar heat flux (ASR_sfc, net SW), while Q_out is the sum of the radiative, conductive and evaporative heat losses:
Q_in = Q_rad(SW) = 165 W/m^2
Q_out = Q_rad(LW) + Q_cond + Q_evap = [398-345=] 53 + 24 + 88 = 165 W/m^2
We’re ONLY discussing the Q_rad(LW) process here, Frank. That’s a surface heat LOSS. Which means it acts to COOL the surface.

When you believe that one source of energy is less important that another, you are violating conservation of energy. All W/m2 are created equal. None can raise or lower temperature more effectively than another.

One more time: It’s not about whether one W/m^2 is equal to another or not. It’s all about the DIRECTION they move. The Q_rad(SW) moves INTO the surface and thus heats is. The Q_rad(LW), on the other hand, moves OUT of the surface and thus can and does not heat it; it rather COOLS it.
This is what the “conservation of energy” principle is all about, Frank.

Kristian
May 1, 2017 3:52 pm

Sorry, that should be “dynamic equilibrium” (steady state) rather than “thermal equilibrium”.

May 1, 2017 5:42 pm

Kristian,
“’Back radiation’ from a cooler atmosphere CANNOT directly raise the temperature of the already warmer surface.”
This statement arises because of Trenberth’s improper conflation of the energy transported by photons and the energy transported by matter where he calls their sum returned to the surface ‘Back radiation’. Gray body radiation from clouds and re-emissions from GHG’s are in the form of photons and these can be added to the arriving solar photons in order to establish how much photon energy the surface must emit in LTE and its consequential average temperature. Energy transported by matter must be returned to the surface while photons can be returned to the surface or emitted into space. Whatever effect the energy transported by matter has is already accounted for by the net LTE surface emissions which when converted to a temperature by SB is a close approximation to the average temperatures measured by thermometers close to the surface.
The energy balance is really as simple as considering only the photons where the energy transported by matter has a zero sum influence on the net photon emissions by the surface. Some fraction of the surface emissions absorbed by the atmosphere are returned to the surface and the remaining fraction is emitted out into space. The fraction returned to the surface is added to the photons arriving from the Sun to establish the required LTE emissions by the surface. The remaining fraction is added to the energy passing through the transparent windows in the atmosphere whose LTE sum is equal to the photon energy arriving from the Sun. Geometrically, the fraction is about 1/2 since photons arrive over half the area across which the atmosphere emits them and the data confirms this.

Dr Deanster
May 1, 2017 6:31 pm

Kristen …. I’m no physics guy, … but I seem to recall reading somewhere that the “skin” temperature of the ocean surface is actually cooler than the measured SST, due to the loss of heat with evaporation. Given that IR only impacts the skin and cannot penetrate beyond impact, I wonder f it is possible that IR down welling could in such circumstances transfer energy to the SS skin, and contribute to latent heat transfer associated with evaporation
Granted, this wouldn’t heat the ocean, but it would supply some of the heat of evaporation leading to a conservation of energy in the immediate layers below.

Frank
May 1, 2017 9:57 pm

Kristian wrote:
Q_in = Q_rad(SW) = 165 W/m^2
Q_out = Q_rad(LW) + Q_cond + Q_evap = [398-345=] 53 + 24 + 88 = 165 W/m^2
Frank replies: This is all screwed up. Q is usually symbol for heat transfer between two objects, but Kristian has written an equation of surface ENERGY BALANCE. That is why both quantities equal 165 W/m2. Most of these quantities are energy transfer between the surface and the atmosphere, but Q_rad(SW) is from the sun and about 40 of 396 W/m2 is from the surface to space.
Now let’s properly calculate the HEAT flow from the surface to the atmosphere. It should obey the 2LoT – be a positive number as written.
Q_sur_to_atm = Q_rad(OLR) – Q_rad(DLR) + Q_evap + Q_cond = 358 – 345 + 88 + 24 = 125 W/m2
Kristian writes: “One more time: It’s not about whether one W/m^2 is equal to another or not. It’s all about the DIRECTION they move.”
Both are important. By tradition, incoming energy fluxes are given a positive sign and outgoing fluxes a negative sign. When the sum is zero, the temperature isn’t changing. When the sum is positive (negative), the temperature/internal energy is rising (falling). Your don’t ignore DLR, or say 1/3 of DLR “doesn’t stick”. If you correct for reflection/scattering of incoming radiation, absorptivity = emissivity means you apply the same correction factor to outgoing radiation (of the same wavelength). If you want to discuss net LWR cooling (OLR-DLR), use the word NET. Those DLR photons are real. The GHGs that emit them certainly don’t know up from down. However, net flux usually refers to heat transfer. The idea that DLR somehow “doesn’t count” or is cancelled by some type of interference (which only occurs with coherent light sources) with OLR is bogus. So is the idea that DLR boils off the skin layer. The skin layer is cooler than the bulk of the ocean – as it must be for convection to return heat from SWR to the skin layer (from which it can reach the atmosphere and then space).

Kristian
May 2, 2017 3:04 am

Frank says, May 1, 2017 at 9:57 pm:

Kristian wrote:
Q_in = Q_rad(SW) = 165 W/m^2
Q_out = Q_rad(LW) + Q_cond + Q_evap = [398-345=] 53 + 24 + 88 = 165 W/m^2
Frank replies: This is all screwed up. Q is usually symbol for heat transfer between two objects, but Kristian has written an equation of surface ENERGY BALANCE. That is why both quantities equal 165 W/m2. Most of these quantities are energy transfer between the surface and the atmosphere, but Q_rad(SW) is from the sun and about 40 of 396 W/m2 is from the surface to space.

Uhm, I don’t even know where to begin here. If you recall, Frank, just prior to the heat budget expressed above I wrote the following:
“Frank says, May 1, 2017 at 1:43 pm:
Temperature change requires heat – the net result of all incoming and outgoing fluxes.
No. “Heat” [Q] is simply the spontaneous transfer of energy from a hot place to a cold one during a thermal exchange. It goes ONE way only. What you’re talking about here is the NET heat [Q_net] of a system, which is Q_in minus Q_out. Net heat for a system in thermal equilibrium is ZERO. That doesn’t mean that there are no heat transfers going on. Both Q_in and Q_out can very well be large, only EQUALLY large. For the surface of the Earth, Q_in is essentially equal to the solar heat flux (ASR_sfc, net SW), while Q_out is the sum of the radiative, conductive and evaporative heat losses:”
What I have set up above, then, is the HEAT BUDGET of the global surface of the Earth, where Q_in = Q_out = Q_net. That shouldn’t be so hard to grasp.
In turn, while the surface Q_in is basically just the solar heat (Q_rad(SW), “net SW”, ASR), the surface Q_out is the total of three separate components: 1) the radiative, 2) the conductive, and 3) the evaporative heat loss.
I denote the surface radiative heat loss by Q_rad(LW). That’s the “net LW”, UWLWIR minus DWLWIR: [398-345=] 53 W/m^2 (numbers from Stephens et al., 2012).
Further, I denote the conductive heat loss by Q_cond (24 W/m^2) and the evaporative heat loss by Q_evap (88 W/m^2).
All in all, the sum of these three components make up the TOTAL surface heat loss:
Q_rad(LW) + Q_cond + Q_evap = Q_out.
Which balances the surface heat INPUT (“gain”) from the Sun: Q_rad(SW) = Q_in.
None of the terms in my budget above are mere “flows of energy”. They’re ALL actual HEAT fluxes.
The “temperature change” requiring “heat” that you’re talking about specifically depends on the NET HEAT (Q_in – Q_out = Q_net) of the system in question. Looking at ONE heat transfer out of two only gets you half-way.
Still, you go on:

Now let’s properly calculate the HEAT flow from the surface to the atmosphere. It should obey the 2LoT – be a positive number as written.
Q_sur_to_atm = Q_rad(OLR) – Q_rad(DLR) + Q_evap + Q_cond = 358 – 345 + 88 + 24 = 125 W/m2

Dear me! Frank, you have put the two conceptual component “flows of energy” making up the ONE radiative heat flux between surface and atmosphere up as discrete heat fluxes in their own right: Q_rad(OLR) and Q_rad(DLR). Big no-no! Read carefully now: The UWLWIR and DWLWIR are NOT themselves radiative HEAT fluxes! They MAKE UP the unidirectional radiative heat flux from sfc to atm between them! Q is heat is thermodynamics. And so only the net is a Q, not the UWLWIR and the DWLWIR individually.
Here’s the CORRECT set-up for the sfc-atm heat transfer:
Q_sfc→atm = Q_rad(LW), sfc→atm + Q_cond + Q_evap = [398-345-20=] 33 W/m^2 + 24 W/m^2 + 88 W/m^2 = 145 W/m^2.
However, the surface of the Earth is involved in TWO separate heat transfers at the same time:
1) Sun → Sfc (Q_in), and
2) Sfc → Atm/Space (Q_out).
1) HEATS the surface, while 2) COOLS the surface.
The photons from the atmosphere to the surface are part of 2), not 1). It’s that simple. So all talk of photons from the atmosphere somehow heating/warming the surface directly and/or directly causing an increase in evaporation rates is thoroughly misguided.

By tradition, incoming energy fluxes are given a positive sign and outgoing fluxes a negative sign. When the sum is zero, the temperature isn’t changing. When the sum is positive (negative), the temperature/internal energy is rising (falling).

Indeed.

Your don’t ignore DLR, or say 1/3 of DLR “doesn’t stick”.

It’s not me making this claim, so this particular point you will have to discuss elsewhere.

If you want to discuss net LWR cooling (OLR-DLR), use the word NET.

I am. However, there’s also another name for the “net LW”, Frank, and that’s “radiant HEAT” (or “the radiative HEAT flux”), since “heat” is – by definition – the NET exchange/transfer of energy between two systems. If you simply say “heat”, then you can drop the “net” modifier, because everyone SHOULD know what the heat term actually represents.

Those DLR photons are real.

As are the UWLWIR photons, Frank. And the exchange is instantaneous, continuous and simultaneous. So macroscopically (thermodynamically) you will only ever see the NET exchange. It is basically all there is. The NET exchange is what affects the U, not each photon coming in or going out. Why? Because an individual photon absorption or emission event is distinctly a QUANTUM REALM phenomenon, NOT a thermodynamic one. We need statistics to get from the quantum (micro) to the thermo (macro) realm.
Think of it this way (a simple analogy):
Imagine you have your hand stretched out with the open palm facing up. In your palm lies, say, two dimes. Two people are standing on either side of your hand, one holding a single dime, the other one nothing at all. (The two people are really just one; they simply represent the dual nature of the exchange at hand. You’re the surface, the two people are the atmosphere, and the dimes are photons.)
Here’s what happens: The person holding the single dime places it in your hand, at the very same moment as the other one grabs the two that were there already, removing them from your hand. That is, these two actions/operations happen simultaneously.
The question then becomes: Did you ever have THREE dimes in your hand during this exchange?
The answer is of course “no”. First you had TWO. Then you had ONE. And that’s it. The first of the original two was simply exchanged with another one, while the second was lost.
People here have a tendency to look at and interpret ONE event at a time. Theoretically. Mathematically. And that’s where they get confused. They analyse its effect IN ISOLATION from everything else. They estimate its effect AS IF the other (opposing) one didn’t happen at the exact same time. They only look at the photon absorption and “forget” or “ignore” the simultaneous (and larger) photon emission. Such a narrow scope doesn’t work if you want to discuss THERMODYNAMIC effects. Then you will only fool yourself into thinking that there are (two) SEPARATE thermodynamic effects (an actual change in U and T) being caused by (two) SEPARATE thermodynamic processes. There aren’t. There is just the one.
The NET effect – the THERMODYNAMIC effect – of the thermal radiative exchange between sfc and atm is that the atmosphere doesn’t give the surface ANY energy at all (zero dimes), while the surface gives IT some energy (one dime), but LESS energy than it would’ve handed to space in the same situation (two dimes).

The idea that DLR somehow “doesn’t count” or is cancelled by some type of interference (which only occurs with coherent light sources) with OLR is bogus. So is the idea that DLR boils off the skin layer. The skin layer is cooler than the bulk of the ocean – as it must be for convection to return heat from SWR to the skin layer (from which it can reach the atmosphere and then space).

This I agree with.

Kristian
May 2, 2017 9:08 am

co2isnotevil says, May 1, 2017 at 5:42 pm:

This statement arises because of Trenberth’s improper conflation of the energy transported by photons and the energy transported by matter where he calls their sum returned to the surface ‘Back radiation’.

He most certainly doesn’t. This is confusion on YOUR part only, I’m afraid.

Gray body radiation from clouds and re-emissions from GHG’s are in the form of photons and these can be added to the arriving solar photons in order to establish how much photon energy the surface must emit in LTE and its consequential average temperature.

Absolutely not! Photons from the atmosphere do NOT cause photons to be emitted from the surface. That’s basically what you’re saying here. This is precisely the flawed budget logic of “Mainstream Climate Science”.
In a thermodynamic budget you specifically do NOT add together solar radiation and atmospheric radiation into ONE incoming energy transfer to the surface. The two are NOT equivalent entities AT ALL! They’re part of completely separate heat transfers; the former of the Sun Sfc (heat IN), the latter of the Sfc Atm/Space (heat OUT).
If you jumble it all together in the same bowl and stir it around, you’ll completely lose sight of “cause and effect”. You will fool yourself into thinking that atmospheric DWLWIR is somehow equivalent to the solar heat flux and that it therefore has the exact same direct, discrete thermodynamic effect on the surface. It’s not. And it doesn’t have. This is elementary stuff …!
Photons from the sky do NOT make up an independent (macroscopic) flux of energy to the surface like the solar heat flux. They are integrated into the continuous thermal radiative EXCHANGE between the surface and the atmosphere, which nets out to a radiant heat flux moving UP. They are part of the surface heat LOSS, not the surface heat GAIN.
In order to fully appreciate that these two propositions aren’t really mutually exclusive at all, even though they might appear to be, one has to be able to keep two thoughts in one’s head at the same time, and distinguish between the MICROscopic (quantum) and the MACROscopic (thermodynamic) realms. One also has to have an understanding of what radiation REALLY IS. It isn’t anything like a straight arrow on a piece of paper. And a thermal radiative transfer between two warm, radiating objects is likewise NOT like two opposing such arrows. Those are just conceptual, mental constructs, highly simplified mathematical models of reality. The true nature of radiation is MUCH more complex than that, and we basically need statistics (probabilistic averages) to be able to describe it as a macroscopic phenomenon.

Energy transported by matter must be returned to the surface while photons can be returned to the surface or emitted into space.

I’m sorry, but it very much seems you do not understand how the climate system works.
The energy (“heat”, really) brought up from the surface towards the troposphere by convection (the bulk movement of air) is never returned to the surface. It is released to space via radiation/emission from the tropospheric column. This is a fundamental principle.
The AIR itself naturally returns to the surface. The HEAT it contained, however, doesn’t. It escaped to space on the way.

The energy balance is really as simple as considering only the photons where the energy transported by matter has a zero sum influence on the net photon emissions by the surface.

You need to reconsider. Because this is plain wrong. This will get you nowhere.

wildeco2014
May 2, 2017 3:03 pm

I’ve considered all these issues in fine detail and came up with a simple and elegant solution several years ago. The
Divide the relevant energy fluxes into two discrete loops which remain separate for as long as an atmosphere remains supported off the surface against gravity in hydrostatic equilibrium
Loop 1 is radiative energy out equalling radiative solar energy in as observed. Basically, solar energy passes straight through at hydrostatic equilibrium.
Loop 2 is conducted energy adiabatically lifted up in the ascent phase of convection equal to conducted energy adiabatically returned to the surface in the descent phase of convection. You don’t need the atmosphere to actually conduct to the solid surface because the energy involved just goes straight up again in the next convective overturning cycle.
If you think it through, that cuts through all the detail and accords with observations.
That Trenberth conflates the two loops is obvious but understandable given that the temperature of atmospheric GHG molecules in contact with the surface represents the sum of both loops so such molecules do indeed radiate at the total of both loops but what is being missed is that the vast majority, if not all, of the energy carried by those GHG molecules is derived from Loop 2 which is the mass induced greenhouse effect.
If radiative imbalances do arise from the radiative characteristics of GHGs then convection adjusts to neutralise the effect of any such imbalances.

Kristian
May 2, 2017 9:12 am

Dr Deanster says, May 1, 2017 at 6:31 pm:

Kristen …. I’m no physics guy, … but I seem to recall reading somewhere that the “skin” temperature of the ocean surface is actually cooler than the measured SST, due to the loss of heat with evaporation. Given that IR only impacts the skin and cannot penetrate beyond impact, I wonder f it is possible that IR down welling could in such circumstances transfer energy to the SS skin, and contribute to latent heat transfer associated with evaporation
Granted, this wouldn’t heat the ocean, but it would supply some of the heat of evaporation leading to a conservation of energy in the immediate layers below.

No. The skin is cooler than the bulk of the surface layer below. But it is STILL warmer than the air above. If it weren’t, then heat would constantly be transferred DOWN from the atmosphere to the ocean. Which is ridiculous.
BTW, my name is “Kristian”.

george e. smith
May 2, 2017 2:14 pm

“””””….. Frank
April 30, 2017 at 8:32 pm …..”””””
Finally someone talking sensibly.
“””””….. An individual molecule has no way of “knowing” the mean kinetic energy of its neighbors. …..”””””
Consider a group of molecules, at some thermodynamic Temperature. Why not say Avogadro’s number of molecules; each with its own serial number (which we can’t read) numbered from 1, up to 6.023E+23
So any time we look at that sample it has a Maxwell/Boltzmann distribution of energies, and on average, every pixel on that graph is occupied by precisely one molecule, always filling the whole pattern, that is representative of the sample Temperature.
Now as Frank says, these molecules are in constant collisions with each other, and exchanging energy among themselves, but for some reason, we always see the same pattern fully occupied, because the Temperature has not changed.
But now Mother Gaia; the Queen of all Maxwell’s Demons, SHE CAN read those serial numbers, so she can see exactly which molecule is in which pixel.
So just like lobsters hopping into and out of lobster pots (for shelter), our molecules are hopping from one pixel to another as they collide with somebody else, So Mother Gaia can see what they all are doing, but we see nothing going on, because we can’t read the serial numbers.
So over a long period of time, with the Temperature staying fixed, it is likely that any one molecule; how about number 314159, might be located could eventually have occupied every single pixel location. we can’t see that but MG knows.
So we can opine, that ANY one molecule, over a long period of time, can have an instantaneous kinetic energy, that has exactly the same energy distribution, as the Maxwell-Boltzmann distribution that characterizes the thermo-dynamic Temperature of the whole sample.
So I would argue, that although as Frank has said, a single molecule does ot know what its neighbors are doing, and therefore does not know instantaneously what its Temperature is, over a long period of time, it does have the exact same MB energy distribution as the whole assemblage, so we are justified in saying it has the same Temperature as the whole sample, but we cannot at any one instant observe its Temperature.
Mother Gaia is never going to show us that serial number 314159, so we can watch it hop around the MB distribution just like she can.
And as for photons which have no mass; they know nothing of the original source from which they came.
Only some photons have a frequency or wavelength, which is characteristic of a particular electron energy level transition in a particular atom or molecule.
The EM radiation that is emitted by EVERY massive body by virtue of its Temperature, has no related electron energy levels at all, it is a CONTINUUM radiation, and ALL possible photon energies are possible, limited only by the Planck distribution.
Such THERMAL EM radiation needs only an accelerating electric charge to radiate, and when molecules collide, the whole charge distribution about the colliding bodies gets distorted, so you have accelerating electric charge, and hence EM radiation during the collision process, which is an eternity of time, as far as plenty of time to radiate.
The particle in free flight may have zero electric dipole moment, and hence no dipole antenna to radiate, but during a collision even the most symmetrical electric charge distribution is distorted, and you get a non-zero electric dipole moment. There also are other higher order polar moments, such as quadrupole, which also can radiate.
Individual photons do NOT know anything about Temperature.
G

April 30, 2017 8:01 am

Sorry, Mike not Roy, my bad.

April 30, 2017 8:08 am

You need to differentiate between the transfer of heat by convection and the transfer of heat by photons. Photons don’t have a ‘temperature’, but are energy that reflects the temperature of a distant body. How else does the Earth warm when the space it’s in contact with is so cold?
The Trenberth diagram adds confusion by conflating the energy transported by photons (solar, surface emissions, GHG effect, etc.) with the energy transported by matter (thermals, latent heat, etc.). What effect do thermals, latent heat and other non radiative transports have, other than the effect they have on the temperature of the surface which is already accounted for by the 390 W/m^2 of surface radiation? Can you see how non radiative transport between the surface and atmosphere has a zero sum effect relative to surface emissions?

April 30, 2017 8:33 am

Photons certainly do have a temperature! What an absurd comment. They also have a frequency/wavelength, corresponding to their color. What do you think you are seeing when something glows red-hot, or even white-hot?

April 30, 2017 8:23 pm

Michael,
Temperature is a property of mass. A photon has no mass and is a bundle of energy quantified in Joules. When matter absorbs Joules, it’s temperature can increase, but the amount of temperature increase depends on the heat capacity of that matter.
When we see something red hot, we are observing BB emissions which is a flux of photons, which if its an ideal Planck spectrum, the peak frequency/wavelength is given by Wein’s Displacement Law and we can infer the temperature of the matter emitting the photons, but this is not the temperature of the photons.

April 30, 2017 8:39 am

2. EM radiation can be considered either as a wave or as photons. These are just constructs of the human mind to aid our understanding. The fact that we can use either depending on what we (or our cat) is doing illustrates the point that neither of them are “real”. It’s just a way of thinking.
3. The Trenberth diagram is shite in so many ways I tend to ignore it completely. It bears no relation to the real world not least as it is a fudged average of night and day. About as useful as calculating how long my roast chicken dinner will take to cook based on the average annual temperature of my oven. If you can show me 2 Trenberth style diagrams, one for day and one for night, which are consistent in themselves AND consistent with Kenneth’s original average I might chat about them further. BTW I have thrown out the challenge to produce the Day & Night Trenberths for months now and NOBODY has even attempted it.
Correction: they probably have attempted it, realised the impossibility of dividing a crock of shit into 2 crocks and decided to keep very quiet about it.

Duane
April 30, 2017 8:59 am

You are correct.
Michael Moon is incorrect.
Photons are particles of light with discrete energy that travel through the cold void of space from the sun to our planet, where they interact with matter and transfer their energy to said matter, without regard to any measurable “temperature”, which is a measurement of heat energy that only applies to matter, i.e., a substance that has mass. Photons have no mass, unlike molecules or atoms.

Leopoldo
April 30, 2017 10:04 am

photons when absorbed by atoms increase the energy of those atoms. Anyway I think that a photon has some amount of energy and this is basic physics since QM was accepted. The formula is E=h*c/lambda and h is the Planck constant. Lambda is the photon’s wavelength and c the speed of light

April 30, 2017 10:08 am

Duane,
Is radiation a particle or a wave? Temperature is the average kinetic energy of the molecules of the mass in question. Photons have no mass of course. They “have” a temperature in the same way that 470nm radiation “has” a color. Spare me your parsing of my language.

May 1, 2017 1:27 am

If using the SB equation for a real world atmospheric calculation it needs to be used with RIGOUR. This means it must include the emissivity term as there are no spherical black cows in the troposphere (yet).
Radiation FROM CO2 must be calculated using the correct emissivity of the gas. Not difficult. Been measured under many differing conditions. Loads of graphs of it in my old physics and heat engineering book, yes it does vary. PLEASE go and look it up.
And include it in ALL the SB equations relating to CO2 outgoing downward LWIR.

May 1, 2017 10:05 am

The temperatures of the atmosphere and the clouds it contains are irrelevant when establishing the surface temperature, but instead are slaved to it. Co2 does little to warm the atmosphere directly, as shown by the lack of a hot spot, but instead, returns some of the absorbed photons back to the surface which are then added to the photons arriving from the Sun making the surface a little warmer than it would be otherwise. The GHG’s and clouds in the atmosphere turn a black body radiator like the Moon into a gray body radiator like the Earth, where the effective emissivity relative to the surface temperature is about 0.62.
The big mistake made by consensus climate science is to assume that the energy of photons absorbed by GHG’s is ‘thermalized’ and almost immediately converted into the kinetic energy of atmospheric gas molecules in motion and that these faster gas molecules (not photons) then warm the surface. This is clearly incorrect as the atmosphere is cooler than the surface and gas molecules in motion do not radiate photons, however; energized Co2 molecules do when they return to the ground state.
From space, we see only about a 50% reduction in the emitted power in absorption bands and the only way to explain this is as re-emissions from GHG molecules high in the atmosphere that are not absorbed on the way into space. Nothing else that high in the atmosphere has the mass and thermal capacity to emit this much energy in the absorption bands. The remaining 50% of the energy is what gets returned to the surface making it warmer than the Sun can do on its own.
This plot of monthly average planet emissions vs. monthly average surface temperature is the Achilles of consensus climate science which confirms that the average relationship between the planets emissions and the surface temperature is given almost exactly by, P = 0.62*o*T^4, where P is the emissions of the planet, T is the surface temperature and o is the SB constant and 0.62 is the effective emissivity of the planet. The sensitivity factor, dT/dP is the slope of this relationship.

Duane
May 1, 2017 5:50 am

Michael – no, light energy (photons) does not “have temperature”, since photons are massless. The notion of “color temperature” does not refer to a non-existent “temperature of light”, but rather refers to the wavelength band emitted by an “ideal black body radiator” that has mass and emits light energy in the form of photons. The temperature of the black body mass determines the wavelength of light emitted by that mass, while the photons themselves have no temperature.
And no, I am not parsing your words – I am stating your words are not correct.

May 1, 2017 10:16 am

Duane,
The color temperature refers to the wavelength of maximum radiation per Wein’s Displacement. For example, the color temperature of the planet is about 288K since the peak radiation tracks the peak emissions of the surface at 288K. However; the radiant temperature is only about 255K. In general, the color temperature is always higher than the radiant temperature.
The color temperature of the emissions from a true color LCD panel is over 6000K. Obviously, if the LCD panel was emitting the power corresponding to 6000K, it would burn your face from across the room.

Frank
May 1, 2017 1:09 pm

Michael Moon wrote: “Photons certainly do have a temperature! What an absurd comment. They also have a frequency/wavelength, corresponding to their color. What do you think you are seeing when something glows red-hot, or even white-hot?”
Many hot materials give off blackbody radiation. The hotter the material, the shorter the wavelength at which MAXIMUM power is emitted – which determines the color you see. However, a material that is red-hot still emits photons at many wavelengths from microwaves to ultraviolet. As temperature rises, the material emits more microwaves, not less. The problem is that the spectrum of power vs wavelength is re-scaled and power is rising with T^4.
The wavelength of a single photon can’t tell you the temperature of the material that emitted it. The blackbody spectrum created by large numbers of photons tells us the temperature of the surface of the sun is about 5700 K. A single photon is emitted or absorbed totally by chance: Einstein coefficients for the emission or absorption of a photon. After a lot of mathematical processing, the coefficients produce cross-sections used for both absorption and emission.

May 1, 2017 2:54 pm

The wavelength of a single photon can’t tell you the temperature of the material that emitted it

Line emitters have the some color temperatures from radio to gamma, and it matters in lasers. The my not use it as a term, and use J’s instead, but lasers makeup for whatever energy is, by emitting more of them, or it doesn’t get very hot.
BTW : gamma’s come out of the nuclei, everything with a longer wavelength than that, is from electrons in one way or another.
BTW II : Why doesn’t your hand push through solid materials?

george e. smith
May 2, 2017 10:51 am

Heat is the random mechanical energy of massive particulate matter. The absolute thermodynamic scale of Temperature (kelvin) is defined in terms of that ean energy per degree of freedom of the various mechanical motions those particles can have, such as translational in xyz, and rotation about three mutually perpendicular axes.
So without physical matter that can be found somewhere on the periodic table of the elements, there can be NO HEAT !
So HEAT (energy) is NOT transported by electro-magnetic radiation, such as we get from the sun.
As far as the ENERGY that earth receives from the sun is concerned; it is for all intents and purposes almost pure BLACK BODY thermal radiation in the electromagnetic spectrum , so you can consider it as Hertz / Maxwell waves, of as photons in chunks of h.nu.
We make ALL of the heat right here on earth by absorbing most of that solar radiant energy in lossy processes, such as absorbing it in the deep oceans. Botanical processes convert some of it into bio mass, etc., but most of it is immediately converted to WASTE HEAT; but we get NO HEAT from the sun. we get almost pure Black body radiation energy; NOT HEAT. There is NO massive material to transport mechanical energy of HEAT (noun) either by conduction or convection.
YES I do know the sun hurls a thimble full of charged particles at the earth each year.
If you want to compute the possible conductive heat transfer to the earth along a long (guard ringed) rod (ignoring that the sun will evaporate much of it) with copper, you will get a few microwatts per square METER.
The sun is powered by the force of gravity. That force alone would collapse the sun into nothingness, if it wasn’t for thermonuclear fusion coming along and screwing things up to stop the collapse. That raises inner Temperatures to multi million kelvin temperatures, and somehow some of the thermal energy makes it to the cold surface which is somehow only 6,000 kelvin, and the surface neutral plasma is the resultant black body radiator that send EMR to earth.
Dr S can correct all of my approximations and mistakes but that is basically what is happening.
I don’t care what professor YOU had at whatever UNIVERSITY who told YOU that “HEAT” (noun) is transported by electro-magnetic radiation.
(S)He’s just plain wrong. Heat requires the presence of matter, and without that matter, there is no such thing as TEMPERATURE.
G

April 30, 2017 8:28 am

I agree 100%.

April 30, 2017 8:46 am

This flat earth model with the surface of the earth receiving three times as much energy as the sun provides should have been consigned to the dustbin years ago.

commieBob
April 30, 2017 8:55 am

The formula for radiative heat transfer between an object (the ocean) and its colder surroundings (the atmosphere) is:

q = ε σ (Th^4 – Tc^4) Ac
where
q = heat transfer per unit time (W)
ε = emissivity coefficient of the object (one – 1 – for a black body)
σ = 5.6703 10-8 (W/m^2 K^4) – The Stefan-Boltzmann Constant
Th = hot body absolute temperature (K)
Tc = cold surroundings absolute temperature (K)
Ac = area of the object (m^2)

This isn’t speculative science, it’s engineering.
Clearly a colder object influences how fast a warmer object loses heat by radiation. The warmer object doesn’t cool as fast … it stays warmer longer … however you want to put it.
If you don’t like the concept that downwelling long wave infrared from greenhouse gasses (the main, by far, being water vapor) can cause the temperature at the top of the ocean to be warmer than it otherwise would, you can fall back on the above equation. The net result is the same.

Ian W
April 30, 2017 10:03 am

I note that you do not include evaporation and the loss of latent heat – the loss of latent heat through evaporation which is increased by any infrared photons exciting the surface molecules and greatly increased by wind over the surface and is considerably greater than the loss due to radiation alone.

Steve Case
April 30, 2017 10:04 am

commieBob – 8:55 am

Clearly a colder object influences how fast a warmer object loses heat by radiation. The warmer object doesn’t cool as fast … it stays warmer longer … however you want to put it.

B I N G O !
Reasonably curious people should be able to come up with all sorts of thought experiments that lead to that conclusion.

Retired Engineer John
April 30, 2017 10:50 am

This discussion simplifies some of the factors in radiative heat transfer too much. The formula you show is classic thermodynamics and works in most circumstances. The ε = emissivity coefficient must be held the same for both surfaces. On Skylab electronics boxes, we used a paint, S13-G, that had an emissivity that varied with frequency. It reflected radiation at the Sun’s frequency and allowed the electronics to radiate and cool since the emissivity was different at that temperature. Radiation, both radiated and absorbed, comes in discrete units called photons and when a photon is emitted, it does not know where it is going, it simply goes. It is possible for special setups over limited temperature ranges to transfer energy from a cold surface to a hot surface: the two temperatures must separated some amount and an optical filter placed between the two that transmits the photons from the lower temperature source and blocks the photons from the high temperature source.

commieBob
April 30, 2017 11:30 am

Retired Engineer John April 30, 2017 at 10:50 am
… It is possible for special setups over limited temperature ranges to transfer energy from a cold surface to a hot surface …

Once you can separate the frequencies involved you’re getting into something akin to antenna systems. You can use the same antenna to transmit on one frequency and receive on the same frequency at the same time!

May 1, 2017 12:06 am

>>
This isn’t speculative science, it’s engineering.
<<
I’ve done the math, and there’s nothing in thermodynamics that prevents a colder object from transferring heat to a warmer object as long as the warmer object transfers more heat to the colder object. I’d post my example, but it’s long and boring and no one would read it.
Jim

May 1, 2017 1:30 am

and the emissivity of CO2 is ?? So the calculation of the intensity of DWIR then becomes ???

commieBob
May 1, 2017 5:25 pm

The Badger May 1, 2017 at 1:30 am
and the emissivity of CO2 is …

According to Kirchoff’s law of thermal radiation:

For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity. link

So it’s emissivity is the same as its absorption coefficient.
The thing is, it’s mostly irrelevant. Water vapor is, by far, the main greenhouse gas. CO2 doesn’t matter that much. It absorbs way less energy than water vapor.
The other thing that makes CO2’s emissivity irrelevant is that it doesn’t emit most of the energy that it absorbs. It loses most of its energy through collisions with other molecules. So, for downwelling longwave IR, the emissivity that matters is that of water vapor. link

daveburton
May 1, 2017 6:00 pm

commieBob, thank you to you, Frank, Jim, Ed, etc., for trying to educate the folks who don’t want to learn. (I don’t have the patience for it today.)
However, you wrote one thing that I disagree with: “The other thing that makes CO2’s emissivity irrelevant is that it doesn’t emit most of the energy that it absorbs. It loses most of its energy through collisions with other molecules…”
Although it is true that the vast, vast majority of the LWIR which CO2 absorbs in the atmosphere is transferred away in collisions with other air molecules, rather than remitted, that does not mean its LWIR emissions are negligible or irrelevant. Don’t forget that, although the time it takes for a CO2 molecule to lose energy by collision is only nanoseconds, the same is true for the time it takes for a CO2 molecule to gain energy by collision. The CO2 molecules are continually excited by collisions with other air molecules, so even though the probability of any one such excitation causing a photon emission is exceedingly small, the CO2 can still end up emitting a lot of LWIR.
The extremely high probability of collisional energy transfer compared to photon emission simply means that:
1. the CO2 molecules stay at the same temperature as the rest of the air, regardless of how much or how little LWIR they absorb; and,
2. the amount of 15 µm LWIR which the CO2 in the air emits depends on only two things: the air temperature, and the partial pressure of the CO2.

commieBob
May 1, 2017 7:04 pm

daveburton May 1, 2017 at 6:00 pm
1. the CO2 molecules stay at the same temperature as the rest of the air, regardless of how much or how little LWIR they absorb; and,
2. the amount of 15 µm LWIR which the CO2 in the air emits depends on only two things: the air temperature, and the partial pressure of the CO2.

I think both statements are true. The other thing to bear in mind is that the blackbody temperature of 15 um is around -80°C. In that light we have this 2011 comment by davidmhoffer.

What ever CO2 does or does not do, well over 90% of it exists in the atmosphere at temperatures low enough that water vapour is also low enough that CO2 becomes significant by comparison. link

I’m dimly remembering, or maybe misremembering, a paper I read a while ago and can’t find again. Given the choice between believing me or you, I would probably be wise to choose you.

Ed Bo
April 30, 2017 9:02 am

You need to understand the difference between gross radiative energy flows and the resulting net flow. All that the 2nd Law requires is that the net energy flow be from the hotter body to the colder body.
This has been well understood since the initial formulation of the 2nd Law. In Clausius’ own words:
“the principle [ that heat cannot by itself pass from a colder to a wamer body] implies that in the immediate interchange of heat between two bodies by conduction and radiation, the warmer body never receives more heat from the colder one than it imparts to it…we find that in one and the same process heat may be carried from a colder to a warmer body and another quantity of heat transferred from a warmer to a colder body without any other permanent change occurring. In this case we have not a simple transmission of heat from a colder to a warmer body, or an ascending transmission of heat, as it may be called, but two connected transmissions of opposite characters, one ascending and the other descending, which compensate each other.”
(Originally from Clausius’ Fourth Memoir, published in Philosophical Magazine, 1854)
So Clausius, who did more than anyone else to formulate the 2nd Law, has a specific term for the transmission of thermal energy from a colder to a warmer body: the “ascending transmission of heat”.

April 30, 2017 3:40 pm

And thus the Earth doesn’t warm the sun.

daveburton
May 1, 2017 6:17 pm

Tom Trevor wrote, “And thus the Earth doesn’t warm the sun.”
Correction: the Earth doesn’t warm the Sun very much.
The Earth does warm the Sun a little bit.
Only about one photon in 200,000 emitted by the Earth happens to strike the Sun, but those which do are absorbed by it, and they warm it every so slightly.
Not significantly, or course. But not quite zero, either.

Ed Bo
May 1, 2017 10:38 pm

William:
What we are talking about here, and what Clausius was talking about, is radiative emissions purely from thermal energy. In modern terms, from substances in “local thermodynamic equilibrium” (which I point out is different from broader thermodynamic equilibrium).
This rules out the microwave klystron tubes you refer to, as well as the more recent lasers and LEDs, none of which are in LTE. (And I have worked professionally in these fields.)
And you completely misunderstand the nature of thermal emission and absorption. Objects of certain temperature can emit EM radiation of many different frequencies, and EM radiation of a given frequency can be emitted by objects of many different temperatures.
Let’s use 15-micron radiation (long-wave infrared) as an example, because that is emitted significantly by gaseous CO2 at many different temperatures.
Consider an ocean surface at a typical temperature of 15C (288K). It is constantly subjected to downwelling 15-um radiation from the CO2 in the atmosphere, as shown by countless measurements. Sometimes this comes from CO2 that is warmer than 288K, and sometimes from CO2 that is colder than 288K.
But there is NOTHING in that 15-um radiation that contains any information as to whether it was emitted from a warmer or colder body. Liquid water’s propensity to absorb or reflect it depends only on the material properties of the water.
And we know, again from countless measurements, that liquid water absorbs 98-99% of EMR of this and similar LWIR wavelengths/frequencies, regardless of the temperature of the emitting body.
Now, Kirchhoff’s law shows us that emissivity equals absorptivity for any given wavelength/frerquency (again, for objects in LTE), so the water surface is emitting 98-99% of the possible blackbody radiative flux density in these LWIR wavelengths.
And if the water is warmer than the atmosphere radiating towards it, the water will radiate at a higher intensity than the intensity that it receives, so the NET radiative flux is from warmer to colder, as the 2nd Law demands.
Many posting here believe there is NO radiative flux from colder to warmer. Clausius anticipated their confusion with the passage I quoted. His viewpoint is still valid today. Every E&M physics text and engineering heat transfer text I have ever seen explains radiative heat transfer as a two-way radiative exchange.

April 30, 2017 9:09 am

Perhaps I can offer a bit of help on this. I teach this to chemical engineering students at UCLA (University of California at Los Angeles), and other universities.
The common misconception is that “heat cannot flow from cold to hot.” Or, “heat always flows from hot to cold.”
The correct statement is “net heat always flows from hot to cold.” This is shown by the Stefan-Boltzmann equation
Q = k (Th^4 – Tc^4),
where Q is net heat transferred, Th is hot object’s absolute temperature, Tc is cold object’s absolute temperature, and k is an appropriate constant for the heat transfer units chosen.
The crucial part of the SB equation is that little minus sign. That is there to show that the heat flowing from the cold object to the hot object does indeed exist.
One need not trust my writings on this; the fact is that every engineer who works in the heat transfer field knows this to be absolutely true. We engineers use this on a daily basis to design, monitor, optimize, and operate literally millions of radiant heaters, or fired heaters, around the world. This also is used in kilns, heat treatment systems, annealing ovens, to name but a few.
If anyone doubts the statement that heat can and does flow from cold to hot, please do this simple and safe experiment to prove it to yourself.
Make a cup of hot tea or coffee, or just place hot water into the cup.
Place the cup of hot liquid on a table top for a minute or two. just long enough for the cup to slightly warm the table surface just below the cup.
Lift the cup a few inches, and insert your bare hand between the cup and the table.
The bare hand will be able to feel warmth flowing from both directions, above it and below it.
The table is, obviously, a bit cooler than the cup above it. Yet, heat is flowing from the table toward the cup. The bare hand interrupts the heat flow and can easily detect this. .

April 30, 2017 9:38 am

Roger Sowell:
I don’t doubt that the way you teach this, the way you understand it and the usefulness of such method do indeed “work” for the students, engineers, designers, or whatever. However I dispute that it represents the reality of the phenomena. This hinges on the definition of heat and the definition of radiative flux. Th^4 is radiative flux, Tc^4 is radiative flux, they have a magnitude. They are always flowing out of the body, it’s radiation based on the temperature of the body. They only have a sign when you have 2 bodies interacting, the sign shows which is hotter and which is cooler. When you do the SB equation you are summing radiative fluxes in order to calculate the heat flow from hot to cold. You are not calculating net heat transfer you are calculating net radiative flux in order to calculate heat transfer.
You probably think they are the same thing, I don’t. Maybe the difference is the way we were taught but I suspect there may well be more to it than that.

Butch
April 30, 2017 9:52 am

The heat from the table is flowing to your hand, NOT to the cup !! Very dumb anology…

SkepticGoneWild
April 30, 2017 11:05 am

Roger,
I have to disagree, strongly. There is no “net heat” or two-way heat transfer. Per the Second Law of Thermodynamics, heat does not spontaneously transfer from cold bodies to warm ones. These are thermodynamic “laws”, not “suggestions”. Your unique definition of “heat” would also cause violations of the First Law as well, since the self-warming cycle would continue repeating itself over and over, with both objects ending up in a higher energy state than initial conditions.
The two terms on the right side of the heat flow equation are independent variables. Q is the dependent variable. The two terms on the right side of the equation are not individual “heat” terms. If you were to plot the Planck curves for both objects, Q is the difference, or area between the two curves. And if you have T1 = T2, Q is zero, even though each body is emitting energy towards each other.
Your thought experiment is incorrect as well. Your hand is not interrupting heat flow from the cooler table to the hot cup. The table has been warmed up from the hot cup of water and is now warmer than your hand, so the warmer table from below is transferring heat to your cooler hand, and the even hotter cup above is transferring heat to the top of your cooler hand. This is NOT two way heat transfer as your imagine. Once the table cools down below the temperature of your hand, you will no longer feel any heat transfer from the table below.

April 30, 2017 11:28 am

As Roger Sowell’s biggest and most aggressive detractor on this site, I want to take this moment to say that on this matter he is 100% correct (though the analogy he chose sucks, so I shall present my own).
Consider a lake, out of which flows water over a dam. The water fall goes over a turbine, causing it to spin. This in turn spins a generator which in turn produces electricity. The electricity is carried by wires to your home where it is available for use on demand from an electrical outlet. You plug a kettle into the outlet, and in short order, the water in the kettle boils.
Is the lake hotter than boiling water? How about the water fall? The turbine? What about the generator? The electric wires? The outlet? WHAT? NONE OF THOSE THINGS ARE HOTTER THAN THE BOILING WATER IN THE KETTLE?
Well then badger, moon and the rest of the crowd, how did the water in the kettle get hotter than the lake, the water fall, the turbine, the generator, the electric wires and the outlet?
You’re trying to take a simplistic view of “heat” and apply it to all forms of energy. As the example above shows, energy can be transmitted in many different forms with no regard for the temperature of the final destination.

April 30, 2017 11:46 am

Roger,
I have to disagree. There is no “net heat” or two-way heat transfer. Per the Second Law of Thermodynamics, heat does not spontaneously transfer from cold bodies to warm ones. These are thermodynamic “laws”, not “suggestions”. Your unique definition of “heat” would also cause violations of the First Law as well, since the self-warming cycle would continue repeating itself over and over, with both objects ending up in a higher energy state than initial conditions.
The two terms on the right side of the heat flow equation are independent variables. Q is the dependent variable. The two terms on the right side of the equation are not individual “heat” terms. If you were to plot the Planck curves for both objects, Q is the difference, or area between the two curves. And if you have T1 = T2, Q is zero, even though each body is emitting energy towards each other.
Your thought experiment is incorrect as well. Your hand is not interrupting heat flow from the cooler table to the hot cup. The table has been warmed up from the hot cup of water and is now warmer than your hand, so the warmer table from below is transferring heat to your cooler hand, and the even hotter cup above is transferring heat to the top of your cooler hand. This is NOT two way heat transfer as your imagine. Once the table cools down below the temperature of your hand, you will no longer feel any heat transfer from the table below.

April 30, 2017 1:13 pm

For Davidmhoffer,
“As Roger Sowell’s biggest and most aggressive detractor on this site,”
Perhaps true, but you have some fierce competition for that title.
“. . . I want to take this moment to say that on this matter he is 100% correct.”
Thank you, Mr. Hoffer. Quite a surprise, and shows you had excellent schooling in thermodynamics.

April 30, 2017 1:16 pm

skepticgonewild April 30, 2017 at 11:46 am
Roger,
I have to disagree. There is no “net heat” or two-way heat transfer. Per the Second Law of Thermodynamics, heat does not spontaneously transfer from cold bodies to warm ones.

The laws of thermodynamics REQUIRE that all bodies be emitting and absorbing at all times, and that their measured temperature be the net of these. Gawd, Read the d@mn laws, learn them. Then come back and apologize for getting it so blatantly wrong and dragging what should have been an interesting thread into the mud.

April 30, 2017 1:26 pm

For Butch,
“The heat from the table is flowing to your hand, NOT to the cup !! Very dumb anology…”
It’s very easy to move the hand out of the path of the two IR flows. Then back in between again. It is absurd to believe that the cooler area on the table top “knows” when the hand is there, and when it isn’t.
One can perform the experiment on two-way radiative transfer in any number of other ways.
A cup of hot coffee in the hand, and a hot campfire nearby, for example.
A candle flame for the hotter object, and a cup of hot beverage as another example.
A warm wall after sunset and any hotter object, such as a candle flame.
My engineer colleagues and I used to goof around at work around the big heat exchangers in the oil refineries, where we knew for a fact how hot each heat exchanger was. One pair had an exchanger surface at 500 degrees F, and its neighbor was 350 degrees F. Simply standing between them, one could feel the heat impinging on either side of the face, or a bare hand held up between them. The cooler exchanger certainly radiated heat at all times.
For all the naysayers on this one, you are entitled of course to your own opinion, but the cooler objects in the world do not care one bit. They radiate away, now and as they always have. Will do so in the future, too.
That little minus sign is incredibly important.

April 30, 2017 1:40 pm

Roger Sowell
and shows you had excellent schooling in thermodynamics.
Roger, as you know, I take every opportunity to say when you are wrong 😉
I have no schooling in thermo. When I became interested in climate, I sat down and started reading. The laws are not hard to understand, you don’t even need calculus to learn how to apply SB Law, it isn’t that hard. Yet threads like this are regularly bombed by people demonstrating that they haven’t bothered to learn the basics yet lecture everyone else as to what they are.

April 30, 2017 1:41 pm

Your argument is stupid beyond belief. The subject is radiative heat transfer. The operation of a hydroelectric dam has nothing to do with radiative heat transfer. The dam water could be hot or cold. OMG you need to take some physics and thermo courses.

April 30, 2017 1:57 pm

Roger,
I truly feel sorry for your students.
Per the heat flow equation, heat (Q) is only transferred when you have a positive difference in temperatures. And heat is only transferred from warm bodies to cool ones per the Second Law of Thermodynamics. This is not subject to interpretation.
Stand next to a large block of ice. The ice is emitting thermal radiation in your direction. Does the ice warm your skin up? LMAO. No. Go into an ice cave. You may THINK that you will be warmed up with all the IR hitting you from the ice. But you will get hypothermia and die. Go experiment this if you want, and report back on the results.

April 30, 2017 1:59 pm

Your argument is stupid beyond belief. The subject is radiative heat transfer.
Which is a two way street with more traffic in one direction than the other ACCORDING TO THE LAWS OF THERMODYNAMICS.
My example was to demonstrate that cold things can indeed make other things warmer. You want to argue that the energy in a cold lake cannot make water in a kettle boil? Be my guest. If you don’t understand that energy transfer from one body to another is in the same way comprised of energy flows that don’t care what the temperature of the absorbing body is… well, then I’m not the one who needs to study thermo. All over the world, thousands of engineers design and build all manner of things based on the exact math and understanding of thermo that Roger explained. If they were wrong, the stuff they designed wouldn’t work. But like the water in the kettle that boils because of cold water in a lake, it does.

April 30, 2017 2:02 pm

Stand next to a large block of ice. The ice is emitting thermal radiation in your direction. Does the ice warm your skin up? LMAO. No.
Stand next to nothing at all. Nothing at all is -273 degrees C. Which is warmer, nothing at all or the block of ice at say 0 deg C?
heat is only transferred from warm bodies to cool ones per the Second Law of Thermodynamics.
You misunderstand the difference between heat and radiated energy. Go find out. Then come back and apologize.

John Harmsworth
April 30, 2017 4:45 pm

I present no academic credentials on this question as I am more or less self taught, but I have little doubt that skeptic is incorrect in his understanding. If he were correct then the cooler body in any two body relationship would behave as if it were at absolute zero (zero radiation). This is clearly wrong. It is just a simplistic understanding that matters at very close scrutiny .I expect many engineers understand it to be true. Perhaps most climate scientists as well, though 97% don’t seem to understand anything except funding. More interesting to me when considering heat transfer at the ocean surface is sublimation, wind effect and vapour pressures. I never see the experts discuss this despite the huge role latent heat plays in Earth’s energy balance.

April 30, 2017 10:37 pm

The water from the lake making electricity to boil water example has nothing to do with any temperature difference. The potential energy of the lake water isn’t related to its temperature, whether hot or cold (except perhaps as it effects water density). The energy for the production of electricity comes from the fall of water down the gravity well, its potential energy (gained when it was moved to the lake) converted to kinetic in the fall. That kinetic energy is captured by the turbine to be converted to electricity.

charles nelson
April 30, 2017 11:54 pm

According to Boltzman, if you have a sealed container with one corner partitioned off containing a gas and you remove the partition, the gas will spread into the rest of that container and occupy the space available. According to Boltzman it is ‘statistically possible’ at some point in the future for that same volume of gas to re occupy the partitioned corner it was originally released from.
You gotta watch Boltzman.

May 1, 2017 5:29 am

Roger Sowell April 30, 2017 at 9:09 am

Lift the cup a few inches, and insert your bare hand between the cup and the table.
The bare hand will be able to feel warmth flowing from both directions, above it and below it.
The table is, obviously, a bit cooler than the cup above it. Yet, heat is flowing from the table toward the cup. The bare hand interrupts the heat flow and can easily detect this. .

The bare hand between the cup and the table is the third object (at 98.4 F) which is colder than the other two radiating objects (hot coffee & warm table top ). The bear hand facing towards coffee would fill warmer than the hand facing towards table top. All objects radiate (net) towards colder surface and in this case towards the hand between other two object.
Roger Sowell April 30, 2017 at 1:26 pm

My engineer colleagues and I used to goof around at work around the big heat exchangers in the oil refineries, where we knew for a fact how hot each heat exchanger was. One pair had an exchanger surface at 500 degrees F, and its neighbor was 350 degrees F. Simply standing between them, one could feel the heat impinging on either side of the face, or a bare hand held up between them. The cooler exchanger certainly radiated heat at all times.

Yes you could know which heat exchanger was hotter because your body was the colder object near two radiating objects which were hotter than your body. All objects radiate (net) towards colder surface and in this case towards you standing between two heat exchangers of varying warmth. If you move away the exchanger surface at 500 degrees F will radiate (net) towards the exchanger at 350 degrees F.

May 1, 2017 8:14 am

andyhce
That kinetic energy is captured by the turbine to be converted to electricity.
It is amusing when people dispute my point while making it for me. The point was that energy moves in many ways. In this example, several different ways, before making water boil. Moon and Badger keep yelping about “heat” not being able to move from colder to warmer. They ignore the fact that energy can be transmitted by many mechanisms other than heat, as you just showed in my quote of your words. They ignore that radiated energy is completely different from heat, and like electricity, doesn’t know the temperature of the thing it is heating. Consequently they come to a wrong conclusion. Radiated energy is no more heat than is electricity. When they learn to apply the actual laws as written, then they’ll come to the same conclusion. If they can prove their warped application of the laws is correct, they’ll win Nobel prizes.

old construction worker
May 1, 2017 3:46 pm

A Patel; “All objects radiate (net) towards colder surface and in this case towards you standing between two heat exchangers of varying warmth. If you move away the exchanger surface at 500 degrees F will radiate (net) towards the exchanger at 350 degrees F.” The only way to know for sure if “heat” from the 350 degrees F exchanger is reaching the 500 degrees F exchanger is to monitor the energy needed to operate the 500 degrees F exchanger. If the energy needed decreases then “heat” from the cooler exchanger is adding “heat” to the hotter exchanger.

commieBob
May 1, 2017 6:06 pm

John Harmsworth April 30, 2017 at 4:45 pm
… I never see the experts discuss this despite the huge role latent heat plays in Earth’s energy balance.

Your instinct is good. Latent heat is huge. If you look at Figure 1 above you will note that latent heat is responsible for removing 78 w/m^2 from the planet’s surface. The net radiation is 168 – 390 + 324 = 102 w/m^2. So, when you net out radiation, latent heat is responsible for removing three quarters of the heat from the surface of the planet.

Leo Smith
April 30, 2017 9:20 am

However radiative heat transfer to a warmer surface is less than to a colder one.
You can treat that as a negative transfer, for the purposes of mathematics.
Or its simpler to say that all objects above absolute zero exchange radiation, the net amount being proportional to the temperature difference.
After all, where radiation is concerned the colder object doesn’t know. in advance how hot the hotter its. It will radiate energy. It just receives more in exchange.
If we took your position at face value, cloudy nights would be just as cold as clear ones. Cloud being often at or below freezing point.

Tom in Florida
April 30, 2017 1:14 pm

Questions:
If two objects similar in makeup are touching one another, with one very warm and the other very cold, will the colder object radiate any energy into (or onto) the warmer object?
Will this energy be overwhelmed by the radiated energy from the warm object so that the radiated energy from the colder object never reaches the warm object?
Or is it still just a simple matter that the radiated energy from the warm object, being so much greater, “hides” any energy coming from the colder object so it cannot be detected?

Samuel C Cogar
April 30, 2017 4:37 pm

Will this energy be overwhelmed by the radiated energy from the warm object so that the radiated energy from the colder object never reaches the warm object?

Tom in Florida, instead of “visualizing” radiated energy ……… kinda “visualize” it’s a dark night and you are outdoors and you are holding a 100W “spotlight” and your friend is 100 feet away and holding a 40W “spotlight” and you both are pointing your “spotlights” directly at one another.
Now Tom, do you suppose that your friend will be able to see the “radiated energy” from your “spotlight” or that you will be able to see the “radiated energy” from his “spotlight”?
Or you can visualize the above noted “radiated energy” …… by visualizing throwing a large rock and a small rock into a pool of water and observe what happens when the water waves or ripples created by the big rock …… and the water waves or ripples created by the small rock ….. come in contact with each other.

April 30, 2017 5:29 pm

This is good:
“After all, where radiation is concerned the colder object doesn’t know. in advance how hot the hotter its. It will radiate energy. It just receives more in exchange.”
Sea ice is colder than what it sits on most of the time. It emits warmth back at that the ocean keeping it warmer.

Loren C. Wilson
April 30, 2017 9:31 am

To the Badger: I think you are looking at the net process, while this paper breaks the net process down into both the incoming radiative heat transfer and the outgoing. Basically, a photon can be absorbed by an electron in an atom as long as its energy is equal to or greater than the energy difference for the next step up for the quantized energy levels of the electron. If all the electrons are in energy levels where the next step exceeds the energy of the incoming photon, then the photon is not absorbed. The photon can be from a “cooler” object and can be absorbed by a “hotter” object as long as the condition above is met. Very hot objects do not absorb photons from cold objects since all of the electrons in the hot object are already in energy levels that the photon from the cold object with energy (E=hv) is too low to promote. If the two atoms emitting photons are close in energy, then there is radiative heat transfer between them. What both we and the warmists believe is that electrons in atoms in the atmosphere absorb either incoming or outgoing emr and re-emit it, some of which comes back toward the surface of the earth. If this layer weren’t there, the radiative heat transfer from the earth would be greater, and the surface of the earth cooler than it is. An example of this is the difference between the temperature at night with a humid sky versus a dry sky. When the air is dry, we lose heat faster to space.

May 1, 2017 12:48 am

Ok, here we go again.
Reminder-Note to self; Don’t mention the Gravity Induced Temperature Gradient.
Look, I have said it before, photons do not really exist, they are just a construct of the human mind to try to explain what we observe in the real world and in lab experiments. We can also use the wave explanation. Sometimes both work, sometimes only one. It is self evident a real thing cannot be both only a wave and only a massless particle at the same time. It must be something else but we have not got our heads round it yet.
An EM wave and/or a stream of photons are not in of themselves energy. They are a transport mechanism which can move energy from real world objects A to B . You can only transfer energy via an interaction between EM/photon and real matter. IF it were the case that EM/photon is energy in of itself then it would be possible to transfer energy from one EM beam to another WITHOUT involving real matter. EM waves do not interact like that they all go about their business AS A TRANSPORT MECHANISM entirely independently.
The radiative flux is a measurement of the intensity of an EM wave.Expressed as an intensity per square metre (and if you do it correctly per steradian too). It gives, in the SB equations, the MAXIMUM energy that MAY be transferred, it’s a measurement of the POTENTIAL to transfer energy. NOT energy IN the beam of radiation in of itself. The analogy of voltage in an electrical circuit is useful here. Radiative flux intensity is Voltage, Heat flow is current. Consider a circuit with 3 batteries, each of 2 volts, a series resistor of 1 ohm is connected to the positive of each battery (I hope you are drawing it out in your mind or on paper), the end points of the 3 one ohm resistors are connected together and returned to the 3 negatives of the batteries via a single additional 1 ohm resistor. What is the current flowing in the additional 1 ohm resistor ? I’ll give you a clue, I lied about the number of batteries, there are actually 6 but I hid 3 of them under the bench.
Have we got it ? Voltage is the measurement of the POTENTIAL to transfer energy. You cannot just add voltages in a circuit, you have to consider the current which is driven by the voltage at the relevant point feeding the resistor. Radiative fluxes are the POTENTIAL to transfer energy (heat) in a thermal system. You cannot just add radiative fluxes where there are more than one sources of it (more than one object at a temperature). Hence the repeated example of the 16 electric fires warming your face.
I think if you carefully study the various so called energy balance diagrams (Trenberth and derivatives) produced by the IPCC and how they have changed over time it does become obvious that the clever people at the top do in fact know all this and they have VERY carefully crafted said diagrams, all the little notes, etc to deliberately fool you (and fool some of their own minions too). The killer con as I call it is the use of “radiative flux” and showing W/m^2 as the vast majority of people do actually think that is real energy (my kettle has a label on it which say a number of WATTS) and erroneously think the arrows represent real energy transfer in the arrow direction. It IS very clever as a con but it is NOT real science or reality.

May 1, 2017 10:24 am

>>
What is the current flowing in the additional 1 ohm resistor ? I’ll give you a clue, I lied about the number of batteries, there are actually 6 but I hid 3 of them under the bench.
<<
Before you lied about the other three batteries, the current through that additional 1 ohm resistor should be 1.5 amperes. Each of the other resistors experience a 0.5 ampere current (1.5 amperes – 1.0 amperes). The voltage across those resistors is 0.5 volts. The voltage across the additional resistor is 1.5 volts. That is, if you’re assuming Kirchhoff’s voltage and current laws, Ohm’s law, and the rules governing power supplies.
It’s been a long time since I did circuit analysis. Did I get it right?
Jim

old construction worker
May 1, 2017 4:45 pm

“An example of this is the difference between the temperature at night with a humid sky versus a dry sky. When the air is dry, we lose heat faster to space.” I some what disagree with you. At night it takes longer for a humid sky to cool down that a dry sky but both will have close to the same overnight low temperature. Day time temperatures will be higher in the dry sky than the humid sky. example: Shreveport, La vs. Yuma, Az. both are about 90 miles from a large body of water, both have about the same population and both are along the same longitude.

May 2, 2017 5:48 am

“An example of this is the difference between the temperature at night with a humid sky versus a dry sky. When the air is dry, we lose heat faster to space.” I some what disagree with you. At night it takes longer for a humid sky to cool down that a dry sky but both will have close to the same overnight low temperature. Day time temperatures will be higher in the dry sky than the humid sky. example: Shreveport, La vs. Yuma, Az. both are about 90 miles from a large body of water, both have about the same population and both are along the same longitude.

It does cool slower with a higher dew point, the air in the tropics, has twice the energy per cubic meter than a deserts does (75 kJ/m^-3 vs 35kJ/m^-3), but deserts cool twice as much at night. (~9kJ/m^-3 vs 18 kJ/m^-3). But it also doesn’t necessarily cool to the same temp, as cooling is regulated to dew point by adaptive cooling at night (and dew points are higher on humid days).

Leopoldo
April 30, 2017 9:46 am

I am not an expert in physics, but IR radiation from CO2 has only a few photons of some concrete frequency. Those photons are no able to get absorbed by other molecules. While the IR radiation coming from water molecules can get absorbed by other water molecules. Our own boy that is reach in water, by example. While the light can be absorbed easily by water, I suppose. Nearly a third of the solar radiation is in near infrared, and some bands are absorbed by water molecules. The question is the concrete frequency bands of radiation.
But perhaps visible light in general can be absorbed by any atoms in general, Of this part I am not sure.
Observing this graphic we can see that the atmosphere absorbs some ratio of visible light. The same can be said of the water that do not have a high albedo. Then, with the same argument that the atmosphere absorbs visible light, the ocean can absorb this light and all Near IR and IR radiation that is compatible with water molecules. So it makes sense the oceans are absorbing solar energy from the visible light spectrum.

Robert of Ottawa
April 30, 2017 10:30 am

Radiation does indeed transfer in both directions, that’s why the equation is sigma(T1**4 – T2**4)

April 30, 2017 1:09 pm

You, sir, are awarded an A in the class.

April 30, 2017 2:00 pm

But heat only transfers in one direction. What happens when T1 = T2? Q is zero. Are the bodies still emitting radiation?

April 30, 2017 3:51 pm

For skepticgonewild,
“But heat only transfers in one direction. What happens when T1 = T2? Q is zero. Are the bodies still emitting radiation?”
Almost true, but not quite. As I wrote above, the correct statement is “Net heat only transfers in one direction.”
When T1 = T2, or Th = Tc in my nomenclature, Q is indeed zero. Both bodies radiate according to their individual temperatures, at the exact same rate. Each plate is sending out radiative energy, but it is also receiving the same amount as it sends. Thus, there is zero net heat transfer. Because there is zero net heat transfer, neither object warms or cools.
It’s a bit like two jugglers, tossing bowling pins to each other. The same number of pins fly back and forth.

Mike Flynn
April 30, 2017 5:42 pm

Roger Sowell,
You wrote –
“Because there is zero net heat transfer, neither object warms or cools.”
Two objects, at the same temperature, surrounded by vacuum containing no other source of heat, will cool.
As an example of reality, even two blocks of stone emitting radiation such that both they and the surface on which they sit have zero net heat transfer, will cool after the Sun sets.
In any case, the Earth has demonstrated cooling over the last four and a half billion years.
No energy accumulation, no heating, no GHE. Just a lot of misleading nonsense masquerading as science, by the look of it.
Not even a semblance of a testable hypothesis regarding the GHE, GHGs, and all the rest of the claptrap spouted by those who should (but obviously don’t) know better.
Please quote anything with which you disagree, if you wish to discuss.
Cheers.

April 30, 2017 5:54 pm

I am going to use the wave analogy here, not photons. As I have said before both are a construct of the human mind to help explain what we observe in reality but it is unlikely EM radiation is actually a real massless particle AND a wave, it’s probably something else entirely we have not yet got our head around. Wave thinking or photon thinking seems to work, usually. Here is some wave thinking (and the person at the back who whispered “hand waving” can STFU);
A spherical hot body (or cow) in a vacuum in infinite space with no other objects anywhere at all radiates EM as per SB (T^4).
Does it lose heat ? No. Does it get colder? No. Does it emit energy? No. The radiative flux intensity (of the EM wave) can be calculated, we know it falls off as 1/r^2. So if the radiation leaving the cow is not “energy” why do we call it such when we look at energy transfer diagrams. Because energy transfer only happens between 2 or more cows. The radiation , the EM wave is not “energy” in itself but is a transport mechanism for potentially transferring energy. However we don’t call it “potential” energy as it confuses us with the other one, the P.E. you get with your hydroelectric dam for example. An EM wave (of any frequency) can only result in the transfer of energy when it interacts with some physical matter. I know this because the air above my house is teeming with thousands of different EM waves all happily going about their business without interfering with one another. But if I stick a bit of wire up and connect it to my Ham Radio Transceiver we can get some tiny fractions of real energy (microvolts). Back to the 2 cows.
Hot cow is radiating, Cold cow is radiating. The energy transfer ONLY takes place when the outgoing radiation from a cow “hits” the other cow. Now the actual process as to how this happens is not fully understood but we do have the equations which give the correct result. And we often do define the heat transfer as being from hot to cold. Whether you say there is a 2 way heat flow and therefore a net heat flow from hot to cold or whether you say there is only a one way heat flow may not matter at all when it comes to designing stuff, etc because the equations we use always seem to work. The interesting thing is that whichever you want to go with as a real explanation you would derive the same equation. It also works if (I say erroneously) you call radiative flux “energy” because it will work if there is more than a Uni-cow universe i.e. where there is the potential to transfer energy from one body to another.
Now if I look at the Uni-cow universe I can see that no energy is being emitted, and to me that looks like the starting condition for the 2 cow situation where they are a distance apart and “switched ON” at the same time. So what we need to resolve is that when the radiation from cold cow hits hot cow radiating surface does an energy transfer take place from cold to hot OR does no energy transfer take place but it only happens at the surface of the cold cow i.e. from hot to cold. If the outgoing radiative flux is only dependent upon the absolute temperature of the object i.e. the outgoing flux is the same irrespective of anything else happening then the outgoing flux of the hot cow does not change depending on whether there is a cold cow in its universe or not.
If we put an existence chopper (that’s a motorised wheel with slots cut in it but operating in the 4th dimension) in front of the cold cow such that it is alternately there and not there we do not find that the outgoing flux of the hot cow is modulated. Hot cow obeys SB irrespective. So the ability of the hot cow to potentially transfer actual energy out over the EM transport mechanism is fixed, it does not vary based upon incoming EM. How much energy is actually transferred is decided at the surface when incoming EM hits the surface. So it must be what goes on at the surface that determines the amount of energy transferred over the EM pathway. i.e what the EM waves meet, or “see” . So if the incoming wave sees a hotter surface then no energy is transferred but if the incoming wave sees a cooler surface then the amount of energy is determined by the difference in the temperatures.

Chimp
April 30, 2017 6:20 pm

The blood, milk, urine, brain water and other liquids in the space vacuum-dwelling cows would instantly boil, explode and spatter both with energetic and messy but frozen matter.
OK, I’ll shut up now.

Nick Stokes
April 30, 2017 7:41 pm

“A spherical hot body (or cow) in a vacuum in infinite space with no other objects anywhere at all radiates EM as per SB (T^4).
Does it lose heat ? No. Does it get colder? No. “

Really? So a warm body in space will stay warm forever?
In fact such a body will cool by the elementary calculation – heat loss as per T^4 translated to T change via specific heat. Or if there is a heat source, cooling until the T^4 loss balances that.

April 30, 2017 8:19 pm

skepticgonewild: What happens when T1 = T2? Q is zero. Are the bodies still emitting radiation?
Yes.

tonyM
April 30, 2017 8:51 pm

Roger Sowell:
I think Leopoldo was far closer to the mark than has been described so far.
The net difference transfer equation presupposes similar emissivity for the objects in radiative contact by his use of a common constant “k.” Water has a far higher emissivity than CO2 at surface temperatures which puts a real dampener on net energy transfer.
CO2 is even more constrained by the limited frequency emission.
As Leopoldo says CO2 emits in certain bands only. It would seem strange that specific frequency quantum absorption would continue beyond saturation of quantum energy levels for a given T of the ocean surface unless the CO2 was at a higher T. Surplus photons would not be absorbed or alternatively any absorption is accompanied by simultaneous emission of equivalent photons.
Further, hard to see how CO2 partially blocking limited bands of emission from the ocean prevents partial, if not full, compensation by other heat transfer mechanisms.
No one has ever shown that a colder body can ever INCREASE the T of a hotter body. While on this subject I have a query for an observation made by TonyB about high cirrus clouds passing by and “heating” icy roads at night in winter. Anyone have an explanation (other than NASA)? Is there any reflection or latent heat by cloud formation involoved?

wildeco2014
April 30, 2017 11:43 pm

Any increase in atmospheric opacity reduces the rate of radiative loss to space. If the rate of radiative loss to space drops below the rate at which warmth is conducted upwards from the ground beneath then the icy roads become less cold than they otherwise would be.

April 30, 2017 10:59 pm

Is it not the case that the rate of radiation emitted from any source goes up as its temperature is raised? When a warmer object receives radiation from a colder object, does that not result in the warmer object radiating at a still higher rate?
How many photons from the colder object must be absorbed to assure the emission of one more photon from the warmer object than the warmer object would have emitted without there being a colder radiating object near enough to interact with it?
For how long a time can the warmer object experience a net increase in energy before it losses that gained photon? Is it long enough for human instruments to register an increased temperature?
Take it down to the one water molecule receives one IR photon. Does that not make the water molecule more likely to radiate a photon sooner than it would have otherwise?
While when it will radiate (lose) the additional photon is not exactly defined, surely it is possible to calculate the average time between absorption and emission. If that is very short, as it is with CO2 in the atmosphere, is there really time for the radiation from the colder air to measurably increase the water temperature?

May 1, 2017 12:57 am

Radiation is not heat. It represents the potential to transfer heat. SB gives the MAXIMUM possible heat transfer. Maximum theoretical possible only occurs in fictional spherical black cows. Real world SB equations include the term “emissivity”.
What is the emissivity of CO2. Go look it up.

May 1, 2017 9:27 am

This for tonyM, re
“no one has ever shown that a colder body can ever INCREASE the T of a hotter body.”
Actually, this is done routinely. Consider a car engine’s exhaust header – fairly hot – that radiates heat onto the nearby starter motor. The starter motor fails quickly. The solution is a sheet of metal, a heat shield, bolted so that radiant heat is blocked from hitting the starter motor.
The result is the exhaust header gets hotter, the skin temperature increases. The much cooler sheet metal of the heat shield radiates heat back to the exhaust header. Literally hundreds of thousands of cars were built with such heat shields.
[It is the catalytic converter that gets much hotter than the exhaust pipes and muffler, but same principle holds. .mod]

Rick C PE
May 1, 2017 11:40 am

Got to go with Nick Stokes on Badger’s cow in an empty universe example. The cow would radiate energy off into infinite space per S-B law. It does not matter if there is no other matter to receive the energy. It would simply become entropy (second law) and the poor cow would eventually reach absolute zero temperature (3rd law).

May 1, 2017 6:28 pm

For Mike Flynn,
“Two objects, at the same temperature, surrounded by vacuum containing no other source of heat, will cool.”
That is true. However, it is not the same situation for the climate GHG discussion. Also, not the same situation as my example above. Even though the Earth is a sphere, it is very large (8000 miles diameter approximately) with a fairly thin atmosphere (most of the air is 20 miles or less, and it grows much less dense with higher altitude). For that situation, a relatively good approximation is a flat surface, Earth, perhaps 100 miles by 100 miles, and the atmosphere above is also flat. The point is, a given zone is not surrounded by the vacuum of space. Only one side has the space-vacuum, the far side of the atmosphere.
“As an example of reality, even two blocks of stone emitting radiation such that both they and the surface on which they sit have zero net heat transfer, will cool after the Sun sets.”
True again, but the stones’ cooling occurs by convection, conduction, and also radiant heat. I wrote a piece about skyscrapers cooling in this manner. The fact is, the sides that face each other radiate heat toward each other, thereby minimum heat leaves or exits from those two sides. The radiant heat exits from the sides that do not face each other.
“In any case, the Earth has demonstrated cooling over the last four and a half billion years.”
I would agree in the most general terms with that, but of course there have been many warming periods also. Notable warming periods are the end of glaciation periods that last for approximately 100,000 years. There is speculation but zero proof of how exactly the 100,000 year glaciations ended.
“No energy accumulation, no heating, no GHE. Just a lot of misleading nonsense masquerading as science, by the look of it.”
As just above, somehow an awful lot of heat energy accumulated in a short time to melt the ice caps and continental glaciers; not just once but multiple times. The precise source of all that heat energy is still not proven, at least not to many people’s satisfaction; including my own.
It would be nice to know if a series of large volcanoes erupted in a short period, with the dark ash falling on the ice sheets. Or, underwater volcanic activity heated one of the key ocean currents. Or, a meteor smacked the Earth hard enough to splash dark material onto the ice sheet. Or, perhaps the Sun blasted away at abnormal intensity. No one knows. I’ve read some peer-reviewed material on the proposed causes of warming; none are convincing.
“Not even a semblance of a testable hypothesis regarding the GHE, GHGs, and all the rest of the claptrap spouted by those who should (but obviously don’t) know better.”
Disagree on this one. My knowledge squares with the statement by Professor Richard Lindzen (MIT), that GHG warming is trivially true but numerically insignificant.
As proof of GHE, or better expressed, radiant heat absorbed and re-radiated by CO2 and water vapor, my chemical engineering colleagues and I refer to the well-known properties of luminous gases in fired heaters. It turns out that furnace design (for industrial fired heaters that burn coal, oil, natural gas, or other such carbonaceous fuels) requires an adjustment for the combustion products’ composition. Otherwise, the furnace does not work as expected.
The basic textbooks on chemical engineering, and heat transfer, all have a comprehensive section on this. The key parameters are gas composition, gas pressure (and hence the CO2 and water vapor partial pressures), flame temperature, and mean beam length. Here, beam length is the radiating distance. It is not far in a furnace, typically measured in inches. see e.g. Perry’s Chemical Engineers’ Handbook, 8th Edition, (2008) pg 5-15 et seq. The same material is in the same Handbook, 5th Edition (1973) (that I used in undergrad) starting on pg 10-48 et seq.
The upper atmosphere where the CO2 and water vapor radiate energy back to Earth, aka TOA or top of atmosphere, has far different properties compared to a fiery furnace’s interior. It turns out that heat radiated is far less for colder temperatures, lower partial pressures, and longer beam length. TOA is at approximately -50 deg C, CO2 partial pressure is very, very low at 400 ppm and (probably) one-tenth bar, and the beam length is measured in miles, not inches. In contrast, a furnace’s radiant heat section is at approximately 2600 degrees F, partial pressures are much, much higher at one atmosphere and several percent (not ppm), and the beam length is a few inches.
And that is why (I am certain) that Dr. Lindzen says the GHE is trivially true (we know for a fact that CO2 and water vapor can absorb and re-radiate), but numerically insignificant. The values are so small as to be meaningless.
“Please quote anything with which you disagree, if you wish to discuss.”
Not all disagreement, but happy to discuss any of this.
The absorption and re-radiation of IR by CO2 and water vapor absolutely is true. It’s all a matter of degree.
And, for Mike Jonas, the current piece’s author, I suspect you are fairly close to being on the right track. The answer is that atmospheric CO2 has a trivially unmeasurable effect on the Earth’s energy balance. It may take a while for all the calculations to prove that out, but that is indeed the correct answer.
The chemical engineers have known this for decades. It is why our fired furnaces work, in the countless millions, around the globe, 24/7.

April 30, 2017 10:36 am

As a net effect, by the laws of physics, you are quite right.
As for being correct in the real world, you have completely failed. Migawd, I thought that I was ignorant of the oceans, growing up in the desert as I did. But it didn’t take a nice dip in the summer ocean for me to know that the vast majority of the ocean is COLDER than the atmosphere above it.

April 30, 2017 4:53 pm

That was actually a reply to the one that was skirting dangerously on the edge of ad-hominem argument, Mike.
I saw that point – the key word is net. Downwelling IR affects the net heat accumulation in that layer, which in turn affects the heat that is transferred out of it, by whatever physical route, to cooler places. Which is why you take it into account, and try to get the numbers right. The (insert self-moderated descriptive here) original commenter has an English language comprehension problem, apparently.

John Harmsworth
April 30, 2017 5:13 pm

We forget that we live on a water world. The ocean surface doesn’t need to warm above atmospheric temperature to warm us up. We live in a constant interaction between sea and sky. If upwelling sea water displaces water which is a degree or two warmer or cooler, it can have considerable effect on air temperatures by virtue of the change in surface heat transfer. Massive quantities of ocean water descend and ascend constantly. Where, when , how much and to what effect are almost completely unknown. The idea that “the science is settled” is utterly laughable when we consider that our atmosphere interacts with oceans that contain a thousand times more heat and the cycling of that heat is not understood.

Rick C PE
May 1, 2017 11:21 am

Mike Jonas:
“What you say is true, but I am addressing the issue that at 1mm depth (one millimetre), the ocean is warmer than both the atmosphere above and the ocean below.”
It seems to me your model describes a static flat ocean surface. You say that heat transfer from the upper 1 mm to water below can only occur by convection. Convection is driven by temperature related density difference. But in the real ocean the upper surface layers are surely substantially and continuously mixed by wind induced wave action. I have rarely seen a waveless ocean surface. I would think that this would tend to result in the water temperature “seen” by the air above to be colder than the air and result in a net radiant transfer from the atmosphere to the water over much of the ocean. It also seems to me the LWR heating of the upper 1 mm would be immediately distributed to deeper levels by wave action mixing.

Hans-Georg
April 30, 2017 10:54 am

Why should not a warmer atmosphere give energy to a colder ocean? Since the temperatures of the atmosphere vary much more quickly than the temperature of the water, the interaction in many parts of the earth occurs even in the daytime, in the weather of some days and also in the season. This requires no experimentation, but logical consideration. It is not the case that the heat storage capacity of the atmosphere would be exhausted in the heat retention capacities of the CO2, also the water vapor stored in the air heats up and can give energy to the water if it is warmer than the water surface under the atmosphere. I am by no means an AGW supporter, rather the opposite. But such fundamental errors as in this post are provoking laughter.

April 30, 2017 2:29 pm

“The GHG process involves only IR, which cannot penetrate the ocean more than a fraction of a millimetre, where its energy goes mainly into evaporation. ie, the energy goes straight back into the atmosphere.” and “The ITO warms the ocean well below the surface with little direct effect on the atmosphere.”
That quote from Mike Jonas about the physical affects of LWIR is accurate and represents good science – the opposite being Trenberth’s position that some of the AGW is hiding in the ocean.
The studies I have read show that LWIR does penetrate the ocean to a depth of a millimeter, not a fraction thereof; so, I would only disagree with Jonas on that minor issue.

April 30, 2017 8:05 pm

The Badger: Downwelling IR from a COLD gas cannot transfer heat to a WARMER ocean, or indeed anything that is warmer.
You need to consider the stochastics. The temp is proportional to the mean kinetic energy of the molecules, but there is a distribution of energy states. The molecules at the lower energy states can absorb the IR, even as the molecules at the higher energy states can’t. Then take the evaporation and advection/convection into account — those are transferring energy from the ocean surface to the atmosphere. The second law is not violated by these energy flows.

May 1, 2017 1:35 am

Show me your SB equation for calculating the magnitude of the downwelling IR from CO2 using the correct value for the emissivity of CO2 please.

May 1, 2017 2:48 pm

The Badger: Show me your SB equation for calculating the magnitude of the downwelling IR from CO2 using the correct value for the emissivity of CO2 please.
you shifted your ground from this statement of yours, which is false : Downwelling IR from a COLD gas cannot transfer heat to a WARMER ocean, or indeed anything that is warmer.

daveburton
May 1, 2017 7:23 am

Oh, good grief. The Badger wrote, “you… “stuffed up” … [& made] a fundamental error. You appear to think that downwelling IR from GHGs in the atmosphere can transfer heat to the ocean. Downwelling IR from a COLD gas cannot transfer heat to a WARMER ocean, or indeed anything that is warmer. It’s pretty basic physics/thermodynamics…”
And then it was off to the muck races with well over 100 comments in the thread.
I am so, so</i tired of obstinately ignorant Sky Dragon Slayers like The Badger polluting WUWT with such gibberish.
Frank, mark4asp, Duane, commieBob, Steve Case, Jim Masterson, Ed Bo, Roger Sowell, davidmhoffer, John Harmsworth, and probably some others all had a go at educating Badger and his Slayer friends, to no avail. It’s like arguing with D**g C*tton. (In fact it is possible we are arguing with him — he apparently has many sock puppets.)
The Slayers have no wish to be educated. Logic and evidence are irrelevant to them.
Mods, can no one rid us of their turbulent tripe?

May 1, 2017 4:03 pm

Badgers don’t wear socks. Anthony knows my real name and can confirm I am not the WUWT Anti-Christ.
Why not consider my point made earlier that we take 1 (or 3) quality physics/thermodynamics text books written by knowledgeable PhDs and agree the specific explanations relevant here as a “given”. The only thing that interests me really now is that we all have a correct understanding of what is happening in the real world. If you disagree with something I say why not just point it out politely. If you start being rude (as per above) it could confuse readers as they might think they are on a AGW blog.
A little humour, a little sarcasm and a very small measure of good natured insults are fine, I quite like the feel of this blog generally where these things can be seen but going over the top with downright insults is not really going to help the debate. And you will have to admit the 100 comments are not all of the same flavour or even in agreement on the science !
Calling someone a Slayer is the same type of insult as Denier. Let’s leave the insults out of it (I don’t think I did any, maybe a bit of sarc humour) and TRY to get back to a nice polite debate about the science. Once we have got that sorted , and can agree of the physics/thermodynamics here in this particular example (DWIR/heat transfer) then, AND ONLY THEN can we move on to the most important bit…
Subject to Anthony’s approval, of course.
[nope, I do not approve. It’s crap theory, no better than chemtrails IMHO, so don’t take the discussion any further. – Anthony]

daveburton
May 1, 2017 7:34 pm

Sorry, Badger, I’m being a jerk. My excuse — not a good reason, just an excuse — is that I just had exactly the same argument with D.C., over on Amazon, and it left me with zero patience for such nonsense. He stubbornly insisted, in disregard of all evidence, that, “Atmospheric radiation from a colder atmosphere cannot heat a surface that is already warmer.”
That’s crap theory, no better than chemtrails.
(Digression: Amazon has deleted all of D.C.’s comments, but before they did I saved a copy of the entire main thread. It got weirder and weirder, as it progressed. By the end D.C. was insisting that when diesel engines compress air in their cylinders the compression doesn’t heat the air, except a little bit by “friction,” because the laws of thermodynamics prohibit it. If anyone wants to read a whole lot of crazy nonsense, together with repeated fruitless attempts by several different people, including me, to help him understand it, then email me, and I’ll send it to you.)
Dr. Roy Spencer has made many patient attempts to explain what is wrong with the Slayers’ claims, including their claim that radiation from cooler sources cannot heat warmer destinations:
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/
Here is my weaker attempt to explain of how radiation from CO2 in the cooler atmosphere can warm the already warmer earth below:
Speaking of crap theory, no better than chemtrails, recall this wisdom from Peter Drucker:
“If you can’t quantify it, you don’t understand it.”
So if you want to understand why “gravity induced temperature gradient” theory is also crap theory, no better than chemtrails, you could start by reading my attempt to work through the numbers, last Fall, here:

daveburton
May 2, 2017 8:01 am

I wonder where the gaps went, between my paragraphs? Anyone know how I managed to mess that up?

Samuel C Cogar
May 3, 2017 5:28 am

I wonder where the gaps went, between my paragraphs? Anyone know how I managed to mess that up?

If you are using MS Word as your “text editor” ….. me thinks one can set the “paragraph spacing” for each different text file that you create on or copy to your PC.
So, normally, iffen you finish a paragraph it requires two (2) depressions of the “ENTER” key to get the ‘gap’ between paragraphs.
But iffen your “paragraph spacing” is set different then only one (1) depression of the “ENTER” key is required to get the ‘gap’ between paragraphs ….. and when that text/data is transmitted to the WUWT Server it only “sees” the one (1) CR or “ENTER” code …… and thus no “gap” between your paragraphs.

daveburton
May 3, 2017 7:34 am

Thanks, Samuel. No, I’m not using MS Word as the text editor.
For short comments (like this one) I generally just type into the Reply box (and paste a copy to my logs before clicking “Post Comment”).
For longer comments, like the one which lost its gaps between paragraphs, and for comments in which I need to enter a special character (like the degree symbol) via alt-codes, I usually compose in a simple text editor (Windows Notepad or Notepad++), typically with some manually entered HTML markup (which I often botch).
In both cases, I customarily use a double linebreak between paragraphs. But somehow, in this case, the extra linebreaks went away. I don’t know why.
I keep copies of almost everything. Here’s the raw text for the comment which lost its gaps between paragraphs:
http://www.sealevel.info/WUWT_comment-2491667.txt

Samuel C Cogar
May 4, 2017 6:37 am

Here’s the raw text for the comment which lost its gaps between paragraphs:

Dave B, the contents of the “raw text” doesn‘t mean anything. It’s the data character’s “editing parameters” that determines what you are seeing/reading.
Each and every character “position” of your per se “raw text” has 1 or 2 or 10 “editing parameters” that may or may not be assigned to it. So, iffen you have a 1K text file, ….. it is really a 2K or 3K text file with the aforesaid “editing parameters” occupying the other 1K or 2K or even 3K of data storage (file size).
And once created with a “new” file, those parameters stay with that file for ever n’ ever, unless you change them. And when you copy data from that file ….. you can either keep or change those parameters when you paste it.

daveburton
May 4, 2017 9:15 am

Samuel C Cogar wrote, “Each and every character “position” of your per se “raw text” has 1 or 2 or 10 “editing parameters” that may or may not be assigned to it.”
Samuel, that’s only true when the text is stored in word processor document (or in the Windows clipboard). When it is stored in a plain, ASCII text file, or even a UTF8 Unicode text file, all that formatting information is lost.
Sometimes keeping that formatting info around is useful, sometimes not. So in Chrome instead of just pasting (which pastes the text with the formatting info) you can right click and “Paste as plain text” to just paste the text, with the formatting info removed.
In Mozilla Firefox, and in many programs, there is no “paste as plain text” option. So, as an alternative, you can simply paste your text into Windows Notepad (which discards all formatting info), and then select/copy in Notepad (which puts it back into the Windows clipboard, but without the formatting info), and then paste it where you want it.
There is one small potential “gotcha” with Windows notepad: “Word wrap.” If “word wrap” is enabled, then the “soft” line breaks do get copied to the Windows clipboard. That’s usually not what you want, so if your purpose iis to remove ALL formatting info, be sure to disable “word wrap” before copying the text to the Windows clipboard. (This doesn’t seem to be an issue with Notepad++.)
Anyhow, when I compose blog comments I use plain text, only, and editors which only support plain text (like Windows Notepad). So I don’t think that’s the problem.
I do use (and completely depend upon) the indispensable Ditto clipboard manager. If you’re not using it, then take my advice: get it now. You won’t know how you ever got along without it. (My suggestion: just refuse it network access when it asks; it only wants that so it can share your clipboard history across your multiple machines, which most people don’t need or want.)
Hmmm… I notice that in my raw text one of the lines was indented. I wonder if that could have caused an issue?

daveburton
May 4, 2017 9:18 am

THIS IS A TEST. This SHOULD have doubled lines between the paragraphs. I wonder whether it will?
This one of my favorite quotes, from Peter Drucker:
“If you can’t quantify it, you don’t understand it.”
So if you think you understand something, but you haven’t quantified it, you don’t really understand it.

daveburton
May 4, 2017 9:19 am

Crud. That wasn’t it.

Samuel C Cogar
May 5, 2017 4:32 am

Dave B, sorry I couldn’t help. But I did find out that you know far, far, far more about the current status of “text-editing” programs than I do. And I am now far too old to be worrying about learning such details and particulars.

MarkW
May 1, 2017 8:03 am

From your first statement you have proven that you have no idea how radiation works.
While it is true that conduction of heat always goes from warm to cold, radiation works differently.
Every object radiates. The amount that it radiates depends on it’s temperature. Warmer objects radiate more than colder objects.
This means that colder objects are transferring energy to warmer objects all the time.
When the atmosphere warms, it transfers more energy than it did when it was colder, the heat of other objects isn’t relevant.

May 1, 2017 9:00 am

When the atmosphere warms, it transfers more energy than it did when it was colder

All else being equal. They are not.

Mark L Gilbert
May 1, 2017 9:49 am

At the risk of confirming my own ignorance, i have a question. Why does heat not work like electricity?
Perhaps a simple analogy to explain what I mean would suffice. Electricity only flows (actually) from Negative to positive… transfer of electrons. HOWEVER, there are P and N junction transistors. Sometimes electrons flow by adding themselves to the outer valence shell, but strangely, it works backwards as well, but through the flow of “holes” (empty spots in the outer valence shell) although considerably slower. The electrons still flow in the correct direction, but the effect of the circuit is reversed.
In the same way, if a colder object of two objects is made warmer, it will conduct heat from the warmer object less effectively and quickly.
You (it seems to me) are just being pedantic and insisting on something physically impossible. The heat will still flow the same direction, however a static measurement would reveal that the colder object can actually warm the warmer object, by removing less heat. Of course I may be entirely wrong, in which case I apologize profusely.

commieBob
May 1, 2017 7:13 pm

Heat flow by conduction is a direct analog of Ohm’s law. Heat flow = (T1 – T2) x conductance.
Heat flow by radiation is more akin to antenna theory.

Samuel C Cogar
May 3, 2017 8:10 am

OH MY MY, after all these many, many years of believing what I was taught, …. that thermal (heat) energy could be transferred or propagated vis three (3) methods, …. radiation, convection and conduction.
Now “radiation” means, infers, implies that the thermal (heat) energy is transferred via a wave or particle.
And “convection” means, infers, implies that the thermal (heat) energy is transferred via the movement of a “heated” mass or entity to a different location.
And “conduction” means, infers, implies that ….., well, this is what it means, to wit:

conduction — 2. Physics. a. the transfer of heat between two parts of a stationary system, caused by a temperature difference between the parts..
http://www.dictionary.com/browse/conduction

But, but, but, ….. technically, there is no such thing as a stationary system at the atom or molecule level unless it’s per se “temperature” is at Absolute Zero, ….. meaning completely void of any thermal (heat) energy. And iffen there is no such thing as a stationary system ….. then there is no such thing as the “conduction of thermal (heat) energy”, ….. RIGHT?
So, my question is, …….. is not “radiation” (at the molecular level) also the method of transferring thermal (heat) energy within or through a per se “stationary system”?

Stephen Wilde
May 3, 2017 8:25 am

No radiation is involved in conduction.
The vibrating molecules ‘knock against’ each other and thereby redistribute energy of motion (heat) between themselves.
That is why one cannot have the surface radiating a parcel of energy at the same time as it is conducting the same parcel of energy to the atmosphere.
The same parcel of energy cannot be in two places or involved in two processes simultaneously.
Thus an object that is in the process of conducting its energy away will show a level of radiation lower than would be expected from its temperature.
For the Earth’s surface it follows that it shows a temperature of 288K which is made up of 255K engaged in radiation and 33K engaged in conduction.
The S-B equation does not apply to an object that is conducting and convecting at the same time as it is radiating.
S-B is a radiation only equation which should only be used for surfaces radiating directly to a vacuum.

Trick
May 3, 2017 9:49 am

Stephen stumbles yet again over atm. radiative physics: “Thus an object that is in the process of conducting its energy away will show a level of radiation lower than would be expected from its temperature.”
Physically Stephen, testing shows any object in an atm. will always show a level of radiation equal to what is expected from its (~stable) temperature.
“For the Earth’s surface it follows that it shows a temperature of 288K which is made up of 255K engaged in radiation and 33K engaged in conduction.”
Physically Stephen, calibrated instrumentation shows a global median near surface temperature of ~288K which is made up of ~288K engaged in radiation. Testing shows any object will always show a level of radiation equal to what is expected from its temperature whether in near vacuum or at any atm. pressure.
“The S-B equation does not apply to an object that is conducting and convecting at the same time as it is radiating. S-B is a radiation only equation which should only be used for surfaces radiating directly to a vacuum.”
Physically Stephen, testing for S-B development was at 1bar from objects conducting and convecting (but minimized) at the same time as they were radiating. S-B is a radiation only equation which can be used for all surfaces radiating directly to a vacuum or any atm. pressure as demonstrated by extensive testing. Room temperature and 1bar was used for the original testing to develop S-B theory. Conduction and convection are independent energy transfer processes (as you correctly imply) which can be correctly superposed as in Earth energy budget diagrams such as the one in the top post.

Stephen Wilde
May 3, 2017 10:09 am

Should have said that a surface at 288K doesn’t radiate to space at 288K if conduction of 33K is going on in the atmosphere between the surface and space.
It will be radiating at 288k at the surface but due to conduction the atmosphere is opaque to 33K of that.

Stephen Wilde
May 3, 2017 10:19 am

The truth of what I say is in Trick’s reference to pressure. Higher pressure results in molecules packed closer together so that more conduction and convection can occur at a given level of incoming radiation. Thus higher pressure reduces the proportion of surface radiation that makes it past the atmosphere and out to space. At 1 bar pressure, 33K goes into conduction and convection rather than radiation to space. For a higher pressure it would be more than 33K.
Note that this is a mass induced phenomenon. Not radiative at all.
The greater the pressure, the lower the proportion of surface radiation that escapes to space past the mass of an atmosphere because more conduction and convection is occurring.
Pressure is simply a product of atmospheric mass subjected to a gravitational field so the basic variables for the greenhouse effect are simply atmospheric mass, the strength of the gravitational field and the level of incoming solar radiation.
By referring to pressure, Trick concedes the issue.

Trick
May 3, 2017 10:16 am

Much better physics Stephen 10:09am since that is what is measured as reported in various long term observed energy budgets such as the one in the top post. Improved wording: It will be radiating at 288k at the surface but due to optical depth theory the atmosphere is opaque to (radiation amounting to) 33K up to the TOA (a height defined by observing satellite orbit).

Trick
May 3, 2017 10:57 am

The truth is found from testing and calibrated instrumental observation Stephen, not assertion. Changing pressure on an object was found in testing to have no effect on the amount of radiation from that object, the object’s temperature was the driver. S-B added the object’s measured emissivity/absorptivity, reflectivity, transmissivity.
”Thus higher pressure reduces the proportion of surface radiation that makes it past the atmosphere and out to space.”
Yes, the optical depth is a function of the total pressure et. al.. Tests find increasing atm. opacity with increasing total local pressure.
”At 1 bar pressure, 33K goes into conduction and convection rather than radiation to space.”
At 1 bar pressure, an amount of radiation of about 33K that doesn’t make it thru the window goes into conduction and convection rather than radiation to space.
”For a higher (total) pressure it would be more than 33K.”
Yes, the atm opacity would increase with higher total pressure.
”Note that this is a mass induced phenomenon.”
The mass then induces the radiation field.
Physically more mass, then more radiation field, higher total surface pressure, more atm. opacity.
”The greater the pressure, the lower the proportion of surface radiation that escapes to space past the mass of an atmosphere because more conduction and convection is occurring.”
Yes, the window from the surface closes as the total pressure increases. Conduction and convection distribute the added energy at the surface. The atm. opacity also increases as the increased mixing ratio of increasingly higher extinction coefficient gas.
.
Concede (Total) Pressure is simply a product of (radiating) atmospheric mass subjected to a gravitational field so the basic variables for the greenhouse effect are simply atmospheric mass (effect on total pressure, various well mixed gas extinction coefficients and wv, colloid liquid/frozen water on atm. optical depth), the strength of the gravitational field (effect on total local pressure) and the level of (absorbed) incoming solar radiation.

Stephen Wilde
May 3, 2017 11:17 am

Well, we seem to be in agreement at last.
A molecule with a temperature of 288K at the surface radiates at 288K but 33K fails to get out to space. The failure to get out to space being a consequence of conduction and convection rather than the presence of GHGs.
The greenhouse effect being mass induced rather than radiation induced.
Since the mass of the entire atmosphere is involved it follows that any effect from GHGs is magnitudes too small to measure.

Trick
May 3, 2017 12:01 pm

A molecule with a temperature of 288K at the surface radiates at 288K but amount of radiation in 33K fails to get out to space. The failure to get out to space being a consequence of atm. opacity which is increased by increased presence of GHGs & increased total atm. pressure.
The greenhouse effect being radiation from atm. mass & which is increased by increased atm. opacity.
The mass of the entire atmosphere is involved; the effect from GHGs has been measured in the lab and in the wild.

Stephen Wilde
May 3, 2017 12:17 pm

Radiative gases are not needed. Conduction and convection occur without them. Since KE becomes PE during uplift so that rising air cools you still get the adiabatic lapse rate even without GHGs.

Stephen Wilde
May 3, 2017 12:21 pm

Thus, for Earth, the surface would still be 33K warmer than S-B with no GHGs present.
The atmosphere could never become isothermal because of the conversion of KE to PE with height.
The experiments with bounded columns don’t apply because the adiabatic lapse rate relies on an exponential fall in pressure and density with height and any experiment with bounded sides removes the exponential element.

Trick
May 3, 2017 12:55 pm

Stephen 12:17, radiative gases are indeed needed or the sun could not irradiate the L&O surfaces. The atm. also irradiates the L&O surfaces. Yes, energy transfer by conduction and convection are independent processes from the process of radiative transfer. DALR is -g/Cp so you still get the adiabatic lapse rate even without non-condensing GHGs (IR active gas, really) except for their minuscule effect on Cp.
Stephen, 12:21pm: Just when you almost had it “a temperature of 288K at the surface radiates at 288K” you backslide: “the surface would still be 33K warmer than S-B” no surface has ever tested warmer than S-B, all tests have shown temperature consistent with S-B. After all, S-B was developed FROM testing.
”The atmosphere could never become isothermal because of the conversion of KE to PE with height.”
Never? Earth standard atm. IS measured on avg. isothermal (295.07K) for ~9-10km of height in the mid-latitude tropics.
”any experiment with bounded sides removes the exponential element.”
Why?

talldave2
May 1, 2017 10:02 am

That seems to be a common point of confusion, especially on the “cloudless nights” assertions. The IR from the atmosphere doesn’t so much “warm the ground” as it “prevents the ground from being even colder by radiating a little, because space radiates even less.”
You have to remember the Second Law is only true on average. While a colder object generally doesn’t transfer net warmth to a warmer one, (as others have pointed out) it is absolutely not true that a photon can never carry energy from a colder object to a warmer object. A cold blanket can keep you warmer than no blanket.

Matt S
May 2, 2017 6:33 am

Regardless of heat ocean heat retention triggered by IR, the heat retained comes from SW. IR can only decrease the skin gradient and cause the ocean to retain SW energy. IR energy does not penetrate, and can not conduct up a positive gradient to depths.
Thus the oceans can not absorb IR energy and can not delay the effect IR has on the atmosphere, thus the supposed ‘ocean heat uptake’ as an excuse for poor CO2 response is not valid. TCR and ECS are one and the same and they are low.

Mike Flynn
April 30, 2017 8:03 am

The ocean does not accumulate heat. The hottest water remains on top, being less dense than that below. At night, it radiates energy to outer space, becomes cooler, sinks and displaces less dense water to the top . . .
The following day, the cycle repeats . . .
The oceans sit on top of a large molten blob known as the Earth. Crustal thickness is at a minimum below the deep oceans. At a depth of 10 kms, the water will be around 3 C. The rock surrounding the water will be very roughly 250 C at the same depth.
Anybody still surprised why deep water doesn’t freeze? Don’t forget the unknown quantity of heat from below, in the form of thermal vents, volcanoes, and the globe encircling mid-ocean ridges, continuously releasing magma directly into the oceans.
No GHE.
Cheers.

ferdberple
April 30, 2017 8:26 am

The deep oceans don’t freeze because the density curve reverses below 4C which keeps the deep ocean at 4C.

Nick Stokes
April 30, 2017 10:24 am

” the density curve reverses below 4C”
Not for salt water.

Owen in GA
April 30, 2017 2:53 pm

Salt water is pretty linear in density right down to its freezing point of -1 to -2 C (depending on specific salinity).

Retired Engineer John
April 30, 2017 3:56 pm

Don’t forget about the 4K joules of energy per mole of sodium chloride necessary to hydrate the sodium chloride. This energy must be removed before the water will freeze and causes the curve of lower temperature verses energy removed to be nonlinear near freezing

george e. smith
April 30, 2017 8:24 pm

The density of sea water increases monotonically all the way down to the freezing point. There is a curve of the Density versus Temperature as a function of salinity, and any sea water is plenty salty enough; I thin 18% dissolved salts is typical of most sea water. The 4 deg. C maximum density is strictly a property of PURE water containing no dissolved anything.
G

Leopoldo
April 30, 2017 9:57 am

even if only on the top layers the ocean accumulates heat. Other question is that it is also loosing a part of this heat by evaporation. How much it accumulates heat and how fast it looses heat can be observed when a current like the Gulf Stream is getting slowly colder as it moves to the northern latitudes. In higher latitudes, the water is cooling because it gets less insolation.
While in lower latitudes the rate of heat is maximum, but not so much evaporation. For the rate of evaporation must be related to the relative humidity of the air. Then a small parcel of ocean near the equator can be storing more energy than it looses. This is a question of mere common sense.

Robert Stewart
April 30, 2017 12:58 pm

The tiny layer that absorbs the incoming radiation is well mixed into the upper ocean under most oceanic conditions. Flat calms are the exception, although they are not infrequent in equatorial waters. The heated water may be slightly more buoyant than the water beneath, but the difference is not enough to cause it to stay at the surface with the turbulence created by moderate wind waves. I recall an analysis of the dispersion of tiny oil droplets in a moderate sea state, and the conclusion was that the droplets could reach depths on the order of 100 m. This was confirmed with water samples taken at that depth in the vicinity of an oil spill. These oil droplets had much lower densities than any water heated at the surface. But conditions at the surface in that tiny layer must be quite interesting, since any water vapor that enters a parcel of air will cause that air to be of lower density, and its buoyancy will tend to cause it to rise, replacing the surface air with a parcel that doesn’t have as much water vapor. The transport of the heat water at the surface water must be episodic with the greatest transfer occurring under very windy conditions. It would be a mistake to model this as some sort of simple diffusion.

Gary Pearse
April 30, 2017 3:59 pm

Mike, careful to note that warm water with more salt dissolved in it can sink below cooler water with less salt. Evaporation accumulates salt for one thing in the surface layer and cooler rainfall on top will float on top of warmer salt water below. And of course freezing of Arctic ice on the sea rejects salt, enriching cold water beneath the ice and increasing it’s density and it sinks. If I have learned a bit of climate physics from all the fine contributors here, I believe this is the generator of the mighty deep water cold current that moves cold water along the bottom of the world’s oceans and draws warm surface waters into the Arctic to be cooled.

April 30, 2017 8:04 am

324 Absorbed by Surface. No, no such thing happens, the sky does not heat the Earth. The SUN heats the Earth, alone. Trenberth’s diagram is an embarrassment to my Alma Mater, U of Michigan. I had to look it up, as it was not covered in my amazingly difficult Transport of Heat and Mass class senior year, but, when radiation from a cooler body falls on a warmer body it is simply reflected and no Heat or Energy is transferred. Second Law, not just a good idea but…

Grant
April 30, 2017 8:09 am

Isn’t a great deal of the ocean, say about the equator cooler than the air above it?

April 30, 2017 8:26 am

The surface of the ocean is in the process of evaporating, difficult to actually say what its temperature is. Half a micron below the surface we cannot measure the temperature. But, we know it is not evaporating and so must be slightly warmer than the surface. Trenberth’s diagram does not differentiate between land and ocean, shows the mistaken 324 W/m-squared Everywhere. And, the layer of air “above” the ocean is in the process of receiving the evaporated water, also difficult to know its temperature. One meter above, yes, might be a little warmer in the Tropics, but radiating in all directions all the time. This analysis is far more complex than can be diagrammed as Fig. 1.
[“mistaken 324 W/m^2” ? 342 W/m^2, right? .mod]

RACookPE1978
Editor
April 30, 2017 8:31 am

It is not just the equatorial zone.
All areas of the ocean between the equator and (about) 60-65 north are cooler than the air above the ocean during “average” conditions. (A transient storm or cold front (a weather “event”) is different.) Further north, it is only in the very brief few days of the Arctic summer than the arctic average air temperature of 2-5 degrees C approaches and (sometimes briefly exceeds) the steady 2-4 deg C arctic ocean surface temperature.

April 30, 2017 8:42 am

No, Mods, 324 Back Radiation is the fundamental error in the diagram. 342 is wrong too, as you cannot take 1366 W/m-squared and divide it by four, as this would imply a much cooler Sun, but, no, I meant 324 Back Radiation.

April 30, 2017 8:54 am

Michael Moon:
Mi casa, su casa. Tell me, Michael, based on your understanding of what happens on Earth, what do you think is the explanation for the surface temperature of Venus

April 30, 2017 9:56 am

What, 90 times the atmospheric pressure of Earth? 96% CO2? 0.71 AU from the sun receiving almost twice the incident radiation? What could be different?

April 30, 2017 9:57 am

Oh, and Venus day is longer than its year.

Leo Smith
April 30, 2017 9:21 am

The sky can either be said to heat the earth or to stop the earth from cooling as much, at night.
Take your pick. The effect is real.
Cloudy nights are warmer.

April 30, 2017 4:44 pm

Leo Smith,
You confuse the Earth with the Earth’s atmosphere. Yes CO2 impedes cooling at the TOA. To say that this heats the surface of the Earth implies a vast misunderstanding of heat transport. The Earth is a solid object, its atmosphere is gaseous.

Kristian
May 1, 2017 8:31 am

Leo Smith says, April 30, 2017 at 9:21 am:

The sky can either be said to heat the earth or to stop the earth from cooling as much, at night.

Er, no. It can ONLY be said to “stop the earth from cooling as much”. Heating is the OPPOSITE thing to insulation.

Take your pick. The effect is real.

Only the latter effect is real. The atmosphere INSULATES the surface, it doesn’t HEAT it. The Sun heats the surface. Because it is hotter than the surface. The surface heats the atmosphere. Because it is warmer than it.

Cloudy nights are warmer.

Yes. And cloudy days are cooler. And guess what, the overall (net) effect of clouds on earth’s surface T is … cooling. Significant cooling.

April 30, 2017 1:40 pm

“Trenberth’s diagram is an embarrassment to my Alma Mater, U of Michigan.”
I think Trengerth’s diagram is an embarrassment to all of science. Let us not blame UofM. The fact that the “consensus” has let this [expletive self censored], anti-science diagram be used to justify the CO2 delusion is a crime against mankind and science.
By the way, in case some here have missed it, the sun shines on a 3-D sphere and the intensity of the light at the equator in the daytime is much greater than the average that Trenberth wants to use with his diagram. Have you ever went barefoot on sand at the beach in the tropics? Hot as heck ain’t it?
Trenberth’s diagram is meant to deceive.

Chimp
April 30, 2017 1:44 pm

Trenberth is an embarrassment to New Zealand, just as Gavin Schmidt and Phil Jones are to the UK, and Michael Mann and a host of fellow unindicted coconspirators are to the USA. They have nearly succeeded in destroying science, while also costing the world trillions in squandered treasure and millions of human lives lost. And on the order of 100 million birds and bats.

george e. smith
May 1, 2017 11:25 am

Kristian has it backwards: “””””….. “Cloudy nights are warmer. ……””””””
Actually, …… warmer nights are cloudy …… Because it was MUCH warmer; and humid, during the previous day. But warmer nights are still colder that the previous day before sunset, and it WILL continue to cool over night; clouds notwithstanding.
G

April 30, 2017 8:10 am

“from 1983-2009, cloud changes were responsible for a bit over 90% (90.6%) of global warming, man-made CO2 for less than 10% (9.4%)”
Are you suggesting all atmospheric CO2 is man made or that of all greenhouse gasses, including water vapor, ONLY anthropogenic CO2 has a warming effect on a global scale?

Hans-Georg
April 30, 2017 11:09 am

With the varying cloud cover, we come to the natural factors for climate change. However, there is still an interaction between a warmer atmosphere and a cooler ocean. Just as between warmer ocean and cooler atmosphere. This is called the inertial balance of the climate system. The difference between atmosphere and ocean consists only in the different absorption capacity of energy, before a temperature change occurs in the medium itself. Therefore, temperatures in the atmosphere vary much more than in the ocean. In the real ocean, however, there is no inertia as under laboratory conditions. In the real ocean, there are different pressure ratios, different salt contents and different irradiation states, which ensure movement in otherwise sluggish water.

Scott
April 30, 2017 11:49 pm

Bert I think you have hit the nail on the head here the comment that “· Results suggest that from 1983-2009, cloud changes were responsible for a bit over 90% (90.6%) of global warming, man-made CO2 for less than 10% (9.4%).”
The more correct statement is “· Results suggest that from 1983-2009, cloud changes were responsible for a bit over 90% (90.6%) of global warming, greenhouse gases account for less than 10% (9.4%). CO2’s contribution is 4% of 9.4% or 0.384% and mans contribution to that is 4% of 0.384% or 0.02%.” in other words negligible.

Scott
May 1, 2017 5:46 pm

Hi Mike how did you separate our water vapour impact from CO2? or are you saying the clouds are the only water vapour source in the atmosphere?
in addition if you believe 90% of the increase in CO2 is man-made how did you measure this?

commieBob
April 30, 2017 8:29 am

The term ‘upwelling’ can have more than one meaning:
It can mean water rising to the top of the ocean. link
It can also refer to long wave infrared energy being emitted in an upward direction. link

Clyde Spencer
April 30, 2017 8:39 am

Mike,
I believe that the 30 Wm-2 shown as being reflected from the surface of the Earth in the Kiel/Trenberth diagram is a lower-bound. As I argued in https://wattsupwiththat.com/2016/09/12/why-albedo-is-the-wrong-measure-of-reflectivity-for-modeling-climate/ I don’t believe that specular reflection is adequately addressed in estimates of surface reflection.

Kevin Terry
April 30, 2017 8:41 am

Take a large body of water that is sited in a near tropical zone where the high altitude precludes extensive cloud cover. Measure the amount heat/temperature gained by ITO radiation. Lake Titicaca is a good example (16S Lat average temperature <14C ref Wikipedia) Another example is Lake Victoria which is on the equator but is on average colder than the lakes Tanganyika & Malawi which are both located at higher latitudes. The dominant factor appears to be altitude. Another interesting study is the average temperature of the Dead Sea which is much hotter than all of the above but is not located in the tropics.

April 30, 2017 8:44 am

Trenberth et al 2011jcli24 Figure 10
This popular balance graphic and assorted variations are based on a power flux, W/m^2. A W is not energy, but energy over time, i.e. 3.4 Btu/eng h or 3.6 kJ/SI h. The 342 W/m^2 ISR is determined by spreading the average discular 1,368 W/m^2 solar irradiance/constant over the spherical ToA surface area. (1,368/4 =342) There is no consideration of the elliptical orbit (perihelion = 1,415 W/m^2 to aphelion = 1,323 W/m^2) or day or night or seasons or tropospheric thickness or energy diffusion due to oblique incidence, etc. This popular balance models the earth as a ball suspended in a hot fluid with heat/energy/power entering evenly over the entire ToA spherical surface. This is not even close to how the real earth energy balance works. Everybody uses it. Everybody should know better.
An example of a real heat balance based on Btu/h is as follows. Basically (Incoming Solar Radiation spread over the earth’s cross sectional area, Btu/h) = (U*A*dT et. al. leaving the lit side perpendicular to the spherical surface ToA, Btu/h) + (U*A*dT et. al. leaving the dark side perpendicular to spherical surface area ToA, Btu/h) The atmosphere is just a simple HVAC/heat flow/balance/insulation problem.
So the K-T diagram is thermodynamic rubbish, earth as a ball in a bucket of hot mush is physical rubbish, the Δ 33 C w/ atmosphere is obvious rubbish, the layered models are unrelated to reality rubbish.
http://writerbeat.com/articles/15582-To-be-33C-or-not-to-be-33C
http://writerbeat.com/articles/16255-Atmospheric-Layers-and-Thermodynamic-Ping-Pong

April 30, 2017 8:47 am

Spot on.

April 30, 2017 8:58 am

Proposal: AutoMod word filter update; add “Trenberth” to “Na2i” and “Den1er”

April 30, 2017 8:50 am

Exactly right. How come the K-T diagram has been around so long when it is complete nonsense?

April 30, 2017 9:02 am

How come the K-T diagram has been around so long when it is complete nonsense?
Because you keep referring to it & using it in your arguments you mad fools !! Just STOP IT !

tetris
April 30, 2017 9:21 am

Why do people still say that the sun is “rising” and “setting” – based on the erroneous but RC church promoted earth-centered diagram of our solar system? The KT diagram is CAGW church promoted and habits, once set, die hard.

Chimp
April 30, 2017 10:10 am

What was once thought literally true is now just a convenient expression. But of course you’re free to say that “your point on earth has now come into view of the sun” if you want, rather than “the sun has risen”.

Nick Stokes
April 30, 2017 11:00 am

“How come the K-T diagram has been around so long when it is complete nonsense?”
A good question. The most logical reason is that it isn’t nonsense.

RACookPE1978
Editor
April 30, 2017 1:02 pm

Nick Stokes

“How come the K-T diagram has been around so long when it is complete nonsense?”

A good question. The most logical reason is that it isn’t nonsense.

Inaccurate, misleading, and only “almost good” for “average temperate climates” in an average temperate latitude around 41 north. Dead wrong anywhere else though.
Too low for the “average tropics.” Too high for even the mid-summer arctic (or antarctic) latitudes.
Good for average, simplistic propaganda though.

April 30, 2017 11:44 am

Nick,
Why does the K-T diagram indicate 324 W/m2 of backradiation going down from the atmosphere, yet only 165 W/m2 going up from the atmosphere? I thought GHG’s from the atmosphere returned half back to the earth’s surface and half back up to space?

Nick Stokes
April 30, 2017 11:51 am

” I thought GHG’s from the atmosphere returned half back to the earth’s surface and half back up to space?”
No, it doesn’t work like that. There are multiple absorption end emission events.

April 30, 2017 1:01 pm

Cheap answer Nick. Not good enough.

April 30, 2017 1:30 pm

How logical is the K-T diagram from a thermodynamic standpoint to have 342 w/m2 striking the TOA from the sun (which happens to be the only energy source that can heat the earth/atmosphere) then be multiplied to 492 W/m2 at the surface. It’s thermodynamic hocus-pocus.
Why not use the same principle and engineer a device where the energy output is greater than the input?

Nick Stokes
April 30, 2017 3:51 pm

“Not good enough.”
Well, you should do a little more to substantiate your claim. I guess it is based on the notion that any one emission event has no preferred direction. But what then happens to the energy? Near TOA, of that which goes up, a substantial part escapes. Of that which goes down, most is absorbed again, and that heat is then the basis for further bi-directional emission, with a further chance to escape to space. This doesn’t add up to an even split.

Bill Illis
April 30, 2017 4:42 pm

CO2 does NOT emit any energy downward until you get to 3 kms high or so in the atmosphere. Even then it is very small downward flux at this height and those emitted photons only travel 100 metres or less before some other CO2 molecule intercepts them.
There is actually ZERO back-radiation from CO2 when you are standing at the surface. None.
If a CO2 molecule absorbs photons in its frequency within 3 kms of the surface, it transfers that energy to another atmospheric molecule before it can emit it. This is described as “thermalized” into the atmosphere by climate scientists when they have to talk about it in an audience that understands this, but they never say it in public when their supporters are around.
There is NO backradiation from CO2 at the surface. There is lots from water vapor but none from CO2.
Now go up in the atmosphere 10 kms high, well now there is CO2 backradiation but it never gets all the way down to the surface. At this height, CO2 is now the primary molecule which is emitted Earth’s energy back to space. A completely different story than climate science ever tells you.

Nick Stokes
April 30, 2017 4:58 pm

“If a CO2 molecule absorbs photons in its frequency within 3 kms of the surface, it transfers that energy to another atmospheric molecule before it can emit it.”
True. But so is the converse. The energy needed for emission comes from the surrounding gas (not from a previous absorption). If air with GHGs is IR absorptivity, it must have equal emissivity (Kirchhoff).
And again, it’s observed. Here is Grant Petty’s Alaska spectrum view is surface looking up
Peak CO2 is at about 667 cm^-1. The spectrum shows that is emitted on average at about 267K. It’s above an ice sheet at around 273K.

April 30, 2017 10:00 pm

Forrest
“The calculations and assumptions behind it are so complex, however, that it is almost certainly wrong.”
1,368 / 4 = 342
Hardly complex, not wrong, but useless since it does not slightly resemble the earth’s actual heating.

May 1, 2017 1:09 am

>>
The most logical reason is that it [KT 97] isn’t nonsense.
<<
It has problems. KT 97’s figure 7 is obviously meant for non-scientists or they would show error bars (some of those error bars are 100% or more which is another reason they don’t show them).
KT 97 gets the IR window wrong too. There are three main windows: the radio window (used by radio astronomers) from 1 cm to 11 meters; the visible window from 300 nm to 1100 nm (which is handy since our eyes are sensitive from 390nm to 700 nm); and the IR window from 8 microns to 14 microns. KT 97 say it’s from 8 microns to 12 microns–a mistake?
KT 97 also say this about the IR window:

The estimate of the amount leaving via the atmospheric window is somewhat ad hoc. In the clear sky case, the radiation in the window amounts to 99 W/m^2, while in the cloudy case the amount decreases to 80 W/m^2, showing that there is considerable absorption and re-emission at wavelengths in the so-called window by clouds. The value assigned in Fig. 7 of 40 W/m^2 is simply 38% of the clear sky case, corresponding to the observed cloudiness of about 62%.

Notice the problem with interpolation here. The clear case of 99 W/m^2 is straightforward, but what’s that 80 W/m^2? Is that 80 W/m^2 at 100% cloudiness, 62% cloudiness, or is it just nonsense? If it’s at 100% cloudiness, then they should be interpolating between 99 W/m^2 and 80 W/m^2 or about 87 W/m^2. If it’s 62% cloudiness, then 80 W/m^2 should be in the diagram. Instead, they interpolate between 99 W/m^2 and 0 W/m^2. They don’t even do that right. It should be slightly less than 38 W/m^2, but the diagram shows 40 W/m^2. They rounded by more than 2 W/m^2. If you’re worried about a missing 0.5 W/m^2, then rounding by more than 2 W/m^2 may be a solution to your missing heat.
There is a great deal of IR leaving the planet through the IR window. Some of the hand-wringing over CO2 is because CO2 has an absorption band at around 15 microns (well outside of KT 97’s window range). The argument use to be about how much CO2’s 15 micron absorption was affecting IR escaping through the window. We don’t seem to argue about that any more–apparently people are assuming it’s automatically a problem.
Jim

Tom Dayton
April 30, 2017 8:49 am

One easily understandable explanation of the long-understood mechanism by which IR from the atmosphere’s heating of the top very thin layer of ocean, insulates the underlying layers thereby reducing loss of the energy that the oceans absorb from non-IR: https://scienceofdoom.com/2010/10/06/does-back-radiation-heat-the-ocean-part-one/

Tom Dayton
April 30, 2017 8:54 am
george e. smith
May 1, 2017 12:08 pm

Only Chemists plot EM Radiation Spectra on a ” wave number ” (AKA a NON-SI frequency scale)horizontal axis. It creates the ILLUSION that CO2 is absorbing at the very peak. What they DON’T tell you; and they put it in small print so you can’t read it, is that the VERTICAL scale is in watt.m^-2.cm^-1.
Howzat for a bastardized system of units: meters and centi-meters all in one sentence ??
Physicists always use a ” wavelength ” scale as the horizontal axis. Often in microns, and usually so for spectra including both solar and LWIR surface thermal radiation.
AND the vertical axis is in watt.meter^-2.micron.
In BOTH cases what is being plotted is ….. SPECTRAL RADIANT EMITTANCE …..
Radiant emittance is ….. watt per square meter (SI unit) ; and the SPECTRAL part adds per increment of spectrum width; so per micron if the horizontal axis is in wavelength units of microns, or in the Chemist’s case in ….. per wave number (cm^-1) …..
So what ?? Well the CO2 band occurs at a lowish wavenumber around 667, which is 10,000 microns divided by 15, and the CO2 band is not that many wavenumbers wide, so the per wavenumber spectral increment is much larger, exaggerating the height of the spectrum.
The atmospheric window of course is at about 10 microns wavelength, which is around 1,000 wavenumbers.
And don’t forget that those CO2 band photons have a lower photon energy; about 78 meV.
Peak solar photons are about 2.7 electron volts.
That is milli electron volts.
So the Chemists spectrum is deceptive. Nobody I know of ever plotted the solar spectrum on a wavenumber horizontal scale. NOTE: I said (I know of).
Now I can sort of understand where chemists are coming from. Photon energies are proportional to frequency, and different spectral emission or absorption lines do relate to electron energy level transitions. But why use CGS units instead of SI.
G

Tom Dayton
April 30, 2017 8:56 am
Grant
April 30, 2017 9:40 am

The experiment aboard the Tangaroa showed that with increased clouds the temperature of the skin of the water was heated and therefore acted as an insulator for the water below it, and so proves increased CO2 heats the oceans.
Seems to me you’d need to sort out the relative effects of cloud variation and CO2. Wouldn’t CO2 be inconsequential compared to clouds?

Tom Dayton
April 30, 2017 8:58 am

There are lots more places, across many years, including standard textbooks, where this is explained. So I don’t understand why Mike Jonas did not take that well accepted explanation into account, let alone explain why he thinks it is wrong.

wildeco2014
April 30, 2017 9:03 am

Atmospheric pressure determines the amount of energy taken up by the latent heat of evaporation
I can accept that an atmosphere in hydrostatic equilibrium with a surface pressure of 1 bar as for Earth might result in 17% of downward IR being used in evaporation
The issue though is whether only 17% of additional IR from GHGs is taken up by evaporation.
I would submit that in fact 100% of such additional IR must be taken up by increased evaporation otherwise the hydrostatic equilibrium created by atmospheric mass conducting and convecting would be disrupted and eventually the atmosphere would be lost
I cannot see that Mike’s work deals adequately with that issue unless I have missed something

April 30, 2017 9:37 am

In the troposphere strictly apply hydrostatic equilibrium. This applies to atmospheres up to a pressure of less than 100 hPa.
http://ds.data.jma.go.jp/tcc/tcc/products/clisys/STRAT/gif/zt_nh.gif
http://ds.data.jma.go.jp/tcc/tcc/products/clisys/STRAT/gif/zt_sh.gif

Nick Stokes
April 30, 2017 11:53 am

“I would submit that in fact 100% of such additional IR must be taken up by increased evaporation”
So what happens to all that water vapor?

wildeco2014
April 30, 2017 12:05 pm

More rain. But so little more in relation to the natural base level that one could never measure it.
Likewise an unmeasurable shift in the global air circulation pattern.
Sun and oceans create the larger changes that we are able to observe.

Nick Stokes
April 30, 2017 12:19 pm

“More rain. But so little more in relation to the natural base level that one could never measure it.”
For 100% of 324 W/m2, at LH 2.23e+6 J/kg, that is about 4.8 m/year. A lot of rain!
[????? .mod]

wildeco2014
April 30, 2017 12:45 pm

Most of the ‘extra’ heat at the surface comes from kinetic energy derived from the descent of adiabatically warmed air around the globe and not from GHGs at all
As another commenter says above, Trenberth mistakenly adds together the radiative and non radiative downward energy flows to arrive at the stated number as if both were radiative
In relation to the non radiative portion which is the bulk of it atmospheric pressure limits the evaporative response to 17%
The radiative portion from GHGs is trivial but the evaporative response to the radiative portion is 100% and 100% of very little is still very little
So your figures are entirely wrong

Nick Stokes
April 30, 2017 3:53 pm

“The radiative portion from GHGs is trivial”
This is just arm-waving. Trenberth’s figure is based on measurement.

wildeco2014
April 30, 2017 5:49 pm

His measurements do not distinguish between kinetic energy derived from the radiative properties of GHGs and kinetic energy derived from the location of those GHGs along the lapse rate slope
The latter is by far the greater and comes from non radiative processes of conduction and convection

April 30, 2017 8:27 pm

Nick Stokes: For 100% of 324 W/m2,
You earlier wrote “additional”: 4/342 = 1.17% additional rainfall. You’d be hard pressed to notice that in the rainfall records.

Nick Stokes
April 30, 2017 9:16 pm

Really? Where?
Here is the calc in detail
342 W/m2, if turned into LH by evap, =342/2.23e+6 (LH J/kg)
=145e-6 kg/s/m2
=145e-6 * 33e+6 (s/year) kg/yr/m2
=4785 kg/yr/m2
= 4.8 m^3/m2/yr
=4.8 m/yr rainfall over ocean
~ 3m/year globally

Nick Stokes
April 30, 2017 9:21 pm

Oops – I see I used 342 instead of the back IR 324. So that is 4.6m/year.

Nick Stokes
April 30, 2017 9:23 pm

Oops again – I was right first time (mis-typed 2.33 for LH evap)

May 1, 2017 2:42 pm

Nick Stokes: “I would submit that in fact 100% of such additional IR must be taken up by increased evaporation”
So what happens to all that water vapor?

The first sentence is a quote that started you on your way to calculations. You went from the “such additional IR” to “all that water vapor”, missing the point that the topic that had been introduced (and then returned to) was an increment.

Nick Stokes
May 1, 2017 7:27 pm

“was an increment”
OK, how much?
In the head post, it is unclear. But in the comment that I responded to, it seems to be all additional IR due to GHG, which is pretty much the 324 W/m2 – maybe a bit less for clouds.

Duane
April 30, 2017 9:07 am

Mr. Jonas – this is all very interesting, as far as mathematical models go. What is the experimental evidence of how different wavelengths of light penetrate and transfer energy from photons to varying depths of typical oceanic salt water? I understand there has been quite a bit of observational work done on that with respect to calculating depths of water for various wavelengths of visible light … this explains why there is a maximum depth at which all visible sunlight does not exceed (near 490 feet) and why various colors (wavelengths of visible light) tend to “wash” out at varying depths, leading to essentially monochromatic visible light below about 33 feet or so. The former number is a far greater depth than the numbers you use for IR and visible sunlight for purposes of energy transmission to the water column.
Where are your sources for the penetration depths you cite for purposes of your model? I’m not disputing – I just want to understand the sources of the assumptions you plugged into your model.

Chimp
April 30, 2017 10:16 am

The sea appears blue because red and yellow wavelengths of visible light penetrate only to shallow depths, while green and blue go deeper. UV gets deepest, down to 600 feet, IIRC, which I might not.

Chimp
April 30, 2017 12:04 pm

You’re right. Blue light is scattered off the surface of the ocean, but that which does get in goes deeper.
http://oceanexplorer.noaa.gov/explorations/04deepscope/background/deeplight/diagram3_220.jpg
This shows UV shallower than I recalled.
http://www.thekeytoislam.com/theme/images/g-ocean-light.jpg
But this says it goes down through the photosynthesis zone, ie 200 meters:
http://www.whoi.edu/oceanus/feature/shedding-light-on-light-in-the-ocean
Maybe it depends upon wavelength. All UVC is absorbed by the atmosphere, making and breaking ozone. So is most UVB, maybe 90%. About half of UVA gets through.

Gary Pearse
April 30, 2017 4:20 pm

Actually water’s natural color is very light blue. A glass of water looks colorless, but…

April 30, 2017 10:21 am

Science of Doom has all that sufficiently laid out and explained. There is a link to it upthread.

Duane
May 1, 2017 5:51 am

Thank you, Mike – you answered my question.

george e. smith
May 1, 2017 12:35 pm

Duane, the penetration of solar radiation in sea water, is very well documented. Numerous good graphs can be found referenced in …. The Infra-Red Handbook …. Which is a standard reference text book. And the references are to literally thousands of original peer reviewed papers.
But for the peak penetration in the ocean, which is around 480 nanometers in the blue-green region, the absorption coefficient, is about 10^-4 cm^-1, and this is also close to the solar spectrum peak near 500 nm.
So if alpha is 10^-4, then the 1/e absorption depth is 10^4 cm or 100 meters, which is about 328 feet.
So 36.8% of blue green solar radiation survives down to 328 feet. So 5% will survive to three times that or 984 feet, and 1% still remains at five times or 1640 feet.
There is a (very) deep sea large squid called the ” cockeyed squid ” which has two asymmetrical eyes. One of them is a big dinner plate sized eye, and the other is a much smaller wide angle eye.
The squid swims on its side, with the dinner plate looking up at the surface, and residual solar energy, while the smaller wide angle eye looks at the whole ocean depth below it to detect bio-luminescence from potential prey creatures.
So solar energy is deposited much deeper in the ocean than most people think.
By comparison, the maximum spectral absorption of sea water occurs at a wavelength of 3 microns, which is near tail end near IR solar, and much too short for LWIR surface thermal emissions. Alpha is about 8,000 cm^-1 at 3.0 microns, so the 1/e absorption depth is 1.25 microns, and 99% is absorbed in just 6.25 microns of sea water.
But a more typical thermal radiation absorption in sea water is about ten times lower so 99% is absorbed in perhaps 50-60 microns; and certainly by 500 microns (1/2 mm) 99% of anything related to climate is gone.
G

April 30, 2017 9:15 am

If you build any scientific argument on false foundations you are wasting all your time, it will all be garbage. Any apparent truths will be coincidence.
If you build any scientific argument on acceptable foundations but then incorporate a falsity in it you are still wasting ALL your time even if some of the conclusions are correct. You are also wasting the time of others.
I want RIGOUR (Defn: The quality of being extremely thorough and careful ) in ALL OF IT.
This means NOT using any of the obvious crap which has been produced by the CAGW camp. So for my project this week I am going to make a list of all the stuff the Warmists say that is NOT crap and is sound science. You can use that.
I may not need a week.

Leo Smith
April 30, 2017 9:24 am

Be careful. You yourself have already shown a woeful lack of understanding of heat transfer by radiation.

April 30, 2017 4:27 pm

I shall be exceedingly careful in future….when boiling my kettle. Careful where I put my hands that is, lest I be fooled into thinking that some part of my own flesh is a suitable device for measuring electro-magnetic radiation (or counting photons) if I stick it in the way of a hot thing and a much hotter thing.
Unfortunately it appears that being careful on this blog is not what is required but perhaps being clear is. We do all need to be clear about the difference between EM radiation (visible & IR), Heat, Temperature, and specifically Heat Transfer. Having read everyone’s comments though I think that if I ask for all contributors definitions the number of different answers obtained will exceed the number of fingers on both hands whether I have them on the kettle or not.
Could we not all agree on 1 (or 3) quality physics/thermodynamics text books which actually spell all this out and effectively “sticky” it as a “given and accepted truth” ? I mean the PhDs are all in agreement, aren’t they, and we are not going to argue with those experts (unless of course it is about the gravity induced thermal gradient which they have apparently all forgotten).
NB. Remember Basil Fawlty – don’t mention the Gravity Induced Thermal Gradient.

Leo Smith
May 1, 2017 1:00 am

Badger: Its not that the Phd’s are not in agreement, its that you have singularly failed to understand what they are actually saying.
You are the sort of guy who tells me that ‘electrician flows from the positive terminal of the battery to the negative’.
And thereby claims electrons dont exist, let alone holes, and that transistors therefore cannot work.
On reflection I suspect your confusion arises from the total inability to separate science from religion. You revel in the certainty that both bring you.
And yet, as Korzybski reminds us, “The map is not the territory”. Such models as we construct to represent the worlds both in our minds are in our theories are only as good as they are insofar as they accord with reality.
At the micro level, heat moves between cold and hot objects freely. Cold objects can occasionally heat hot objects, its just that statistically it’s a very rare event. You could plug your kettle into the wall and have it instantly turn to a block of ice. It’s so unlikely that classical science doesn’t deal with the possibility.
Maxwell’s demon is of course the classic thought experiment.
Heat is just the kinetic energy of molecules. That kinetic energy can at any time result in photons being emitted as radiant energy. At almost any wavelength. Those photons will heat anything they get absorbed by. Including very fast moving molecules in a much hotter object.
Heat does move from colder objects to hotter ones, but the overwhelming majority of it moves in the other direction. The laws of superposition apply, and if you want to consider only macro flows, its convenient to treat the process as a single flow from hot to cold.
But that isn’t the only valid approach.
The laws of thermodynamics do not preclude this.
You should learn to distinguish between broadly useful and approximately correct physics, taught as truth, from the actual truth as understood by those who investigate and lay down those laws.
“2>1” does not preclude “2+4 > 1+4”
The actual truth is that the truth is unattainable and science does not help us attain it.
Science merely eliminates what is actually definitely completely wrong, like a significant effect on climate from man made CO2.
http://vps.templar.co.uk/Cartoons%20and%20Politics/no-stupid-people-warning-sign.png
We have no way to tell what is true, in terms of human maps applied to Reality™. Instead we must settle for eliminating what is definitely false. When we have eliminated teh impossible, what is left is not the Truth, its merely a set of ideas we haven’t been able to eliminate (yet).Science is merely a refined and formalized approach to sucking it and seeing what happens. And trying to derive general principles from the particular cases that Time presents us with. (And of course Time itself is only an artefact of memory. We move through time continually, we are all time travellers, it’s just that we can’t remember doing it. Each time has its own set of memories :-))
Disprove that if you can….;-)
And that is the old think of a number trick that we used to play as kids, writ large over popular science and dialectic materialism.
The way to control peoples’ minds is to offer them what seem to be self evident truths, and then base a completely faux superstructure on those propositions that are in the end either faux themselves or irrelevant.,
And the reason this is possible is because of the fundamental misconception that Truth is attainable, in terms of ideas about the world, rather than the world-in-itself (as Kant might have said).
What marks the fanatic, is that they consider themselves in possession of the One Truth.
Well Science ain’t it. And Climb it Cyans certainly ain’t it. And though the second law of thermodynamics is often dumbed down to ‘you can’t pass heat from a colder to a hotter’ this is true only in a limited sense of overall macro flows.
And for the purposes of mathematics 10 watts going in one direction can just as easily be expressed as 30 watts going in one direction and 20 watts going in the other.
Mutatis mutandis these are saying exactly the same thing. To a scientist or mathematician or engineer.
But not, it seems, to badgers.