<p>Check out www.arm.gov an ongoing set of Atmospheric Radiation Measurement experiments at many locations around the globe that measures many aspects of the atmosphere including atmospheric back radiation using infra-red spectroscopy.  This work started in the mid 90's.  Some of the cleanest data comes from the <a href="https://www.arm.gov/tour/north-slope.html" rel="nofollow noreferrer">https://www.arm.gov/tour/north-slope.html</a> North Slope of Alaska ARM research station.</p>

<p>Here are some articles on the research.</p>

<p><a href="https://www.nature.com/articles/nature14240" rel="nofollow noreferrer">https://www.nature.com/articles/nature14240</a></p>

<p><a href="https://newscenter.lbl.gov/2015/02/25/co2-greenhouse-effect-increase/" rel="nofollow noreferrer">https://newscenter.lbl.gov/2015/02/25/co2-greenhouse-effect-increase/</a></p>

CC BY-SA 4.0
2018-11-27T15:48:56.740Z
2018-11-27T15:48:56.740Z
Kevin Gardiner
<p>The thing is, we don´t have any instruments(to my knowledge) to measure the intensity of that radiation. But, we measure the intensity of atmospheric emission from space with satellites, which is seen as the "effective temperature". Logically this is the same as the net radiation downwards. And heat transfer then confirms that no net radiation arrives at the surface. Which is equal to "no radiation at all". Because "net" radiation is the only radiation that can heat anything.</p>

<p>There are lots of people claiming that measurements done with IR sensors and pyrgeometers confirm backradiation. This is not true and it is based on a misunderstanding of the function of the device, as well as misapplication of the stefan-boltzmann equation. Both IR-thermometers and pyrgeometers use the s-b equation  in combination with a thermocouple or thermopile(pyrgeometer). They are both a kind of more advanced and more sensitive thermometers. The thermopile consist of a sensor measuring a gradient across itself indirectly by changes in resistance acting on a current through the thermopile. So, by determining the gradient in the thermopile, it can sense the amount of heat transferred to the surroundings using the s-b equation. When someone claims measurements show "back-radiation", they refer to measurements done with these devices. As most interested people understand, a measurement of heat transfer away <strong>from</strong> the device, cannot show incoming radiation from a colder atmosphere. If you measure something hotter than the temperature of the thermopile, sure, then you have high accuracy. But a measurement of "back-radiation" from the atmosphere is per definition a measurement of something colder than the device. Because the atmosphere is almost always colder(exception is inversion, but that is not relevant). So any measurement of backradiation made from the surface, is a measurement of nothing. It is simply what the device <strong>doesn´t</strong> measure coming from the atmosphere. This is what I mean with "misapplication" of the s-b equation. It can not, and should not, be used that way. That is, to add flux densities that are not calculated as "net" transfer.</p>

<p>In regards to your suggested experiment. Unless the gas is hotter than the instruments(if using IR-instruments), you will measure a heat sink, not any backradiation. And, yes, that is what greenhouse gases really are. Heat sinks. The earth emit its heat to the ultimate bottomless heat sink in space, the only observed infinity, and greenhouse gases are added heat sinks along the highway for heat, that leads to 3 Kelvin space.</p>

CC BY-SA 3.0
2017-05-14T22:59:14.213Z
I am aware of the limitations on using pyrgeometers as indeed anyone will be if they bothered to actually read the manufacturers manuals and understood how they work and what they can and cannot do. It amazes me that even well educated professors think they can just grab some instrument out of the suppliers catalogue and use it in a lab experiment without even thinking about how said instrument works.(or RTFM). Some clever people sure are dumb ! Nice to see you are not one of them.
user7733
CC BY-SA 3.0
2017-05-05T19:03:23.933Z
2017-05-05T19:03:23.933Z
Emil Junvik
<p>Backradek wrote:
&quot;No, this is not the case. The atmosphere is almost black and opaque at certain wavelengths such as 15um (CO2). Then only the external layers radiate, mostly.&quot;</p>
<p>Absolutely correct.</p>
<p>Also remember that there exists a mean free path length / altitude / air density relation. Mean free path length increases exponentially with an increase in altitude.</p>
<p>Thus the net mean free path length for upwelling radiation is longer than for downwelling radiation, and given the potentially millions upon millions of times re-emitted radiation could be re-absorbed / re-emitted in random directions, this means that a very great majority of downwelling radiation is redirected to upwelling... especially so with the ~10.4 m surface-level extinction depth at 14.98352 µm wavelength.</p>
<p>Given that ~10.4 m extinction depth and the nature of absorption (50% being absorbed within the first 10% of that depth,50% of the remainder being absorbed in the next 10% of that depth, 50% of the remainder being absorbed in the next 10% of that depth, etc., etc., etc.) that would mean that fully 50% of the so-called 'backradiation' from CO2 would have to come from within ~1.04 m of the surface... where there would be practically no energy density gradient between surface and atmosphere.</p>

CC BY-SA 4.0
2024-02-07T10:30:17.817Z
2024-02-07T10:30:17.817Z
Mr_Magoo
<p>Gerrit wrote:
&quot;Stefan-Beltzmann and Kirchhoff laws apply always, everything radiates, including convecting gases&quot;</p>
<p>Incorrect.</p>
<p><a href="https://i.imgur.com/QErszYW.gif" rel="nofollow noreferrer">https://i.imgur.com/QErszYW.gif</a></p>
<p>Idealized blackbody objects &gt; 0 K emit at all times, whereas graybody objects &gt; 0 K above their ambient emit. That's why the S-B equation in graybody form includes the temperature of the cooler object.</p>
<p>In reality, what is happening with the S-B equation (the form meant to be used with graybody objects) is a subtraction of the energy density of the cooler object from the energy density of the warmer object to arrive at the energy density gradient; the radiant exitance of the warmer object predicated upon the energy density gradient.</p>
<p>Idealized Blackbody Object (assumes emission to 0 K and ε = 1 by definition):</p>
<p>q_bb = ε σ (T_h^4 - T_c^4) A_h</p>
<pre><code> = 1 σ (T_h^4 - 0 K) 1 m^2

 =    σ  T^4
</code></pre>
<p>Graybody Object (assumes emission to &gt; 0 K and ε &lt; 1):</p>
<p>q_gb = ε σ (T_h^4 - T_c^4) A_h</p>
<p>The 'A_h' term is merely a multiplier, used if one is calculating for an area larger than unity [for instance: &gt;1 m^2], which converts the result from radiant exitance (W m-2, radiant flux per unit area) to radiant flux (W).</p>
<p>Temperature is equal to the fourth root of radiation energy density divided by Stefan's Constant (ie: the radiation constant).</p>
<p>e = T^4 a</p>
<p>a = 4σ/c</p>
<p>e = T^4 4σ/c</p>
<p>T^4 = e/(4σ/c)</p>
<p>T = 4^√(e/(4σ/c))</p>
<p>T = 4^√(e/a)</p>
<p>q = ε σ (T_h^4 – T_c^4)</p>
<p>∴ q = ε σ ((e_h / (4σ / c)) – (e_c / (4σ / c))) A_h</p>
<p>Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).</p>
<p>W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))</p>
<p>∴ q = (ε c (e_h - e_c)) / 4</p>
<p>Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).</p>
<p>W m-2 = (m sec-1 (ΔJ m-3)) / 4</p>
<p>One can see from the immediately-above equation that the Stefan-Boltzmann equation is all about subtracting the radiation energy density of the cooler object from the radiation energy density of the warmer object.</p>
<p>∴ q = σ / a * Δe</p>
<p>Canceling units, we get W m-2.</p>
<p>W m-2 = (W m-2 K-4 / J m-3 K-4) * ΔJ m-3</p>
<p>For graybody objects, it is the radiation energy density differential between warmer object and cooler object which determines warmer object radiant exitance.</p>
<p>Remember that a warmer object will have higher energy density at all wavelengths than a cooler object:</p>
<p><a href="https://i.sstatic.net/qPJ94.png" rel="nofollow noreferrer">https://i.sstatic.net/qPJ94.png</a></p>
<p>This is why 2LoT in the Clausius Statement sense implies that system energy cannot ascend an energy density gradient,&quot;without some other change, connected therewith, occurring at the same time&quot;... that &quot;some other change&quot; typically being external energy pumping that system energy up the energy density gradient. That's what happens in, for instance, AC units and refrigerators.</p>
<p>Most people cannot think in terms of energy, energy density and energy density gradient. We need to analogize to something they’re familiar with. Thus, just as, for instance, water only spontaneously flows down a pressure gradient, energy only spontaneously flows down an energy density gradient. That’s 2LoT in the Clausius Statement sense, in a nutshell. So one tack to take is to ask people if water can ever spontaneously flow uphill. Of course they’ll say, “No, water cannot flow uphill on its own.” Then show them dimensional analysis.</p>
<p>mass (M), length (L), time (T), absolute temperature (K), amount of substance (N), electric charge (Q), luminous intensity (C)</p>
<p>We denote the dimensions like this: [Mx, Lx, Tx, Kx, Nx, Qx, Cx] where x = the number of that dimension. We typically remove dimensions that are not used.</p>
<p>Force:                      [M1 L1 T-2]     /</p>
<p>Area:                       [M0 L2 T0]      =</p>
<p><strong>Pressure:                 [M1 L-1 T-2]</strong>  /</p>
<p>Length:                     [M0 L1 T0]      =</p>
<p><strong>Pressure Gradient:            [M1 L-2 T-2]</strong></p>
<p>Explain to them that Pressure is Force / Area, and that Pressure Gradient is Pressure / Length. Remind them that water only spontaneously flows down a pressure gradient (ie: downhill). Then introduce energy. Tell them that energy is much like water. It requires an impetus to flow, just as water requires an impetus (pressure gradient) to flow. In the case of radiative energy, that impetus is a radiation energy density gradient, which is analogous to (and in fact, literally is) a radiation pressure gradient.</p>
<p>Energy:                     [M1 L2 T−2] /</p>
<p>Volume:                     [M0 L3 T0]      =</p>
<p><strong>Energy Density:               [M1 L-1 T-2]</strong>  /</p>
<p>Length:                     [M0 L1 T0]      =</p>
<p><strong>Energy Density Gradient:  [M1 L-2 T-2]</strong></p>
<p>Explain to them that Energy Density is Energy / Volume, and Energy Density Gradient is Energy Density / Length. Highlight the fact that Pressure and Energy Density have the same units (<strong>bolded above</strong>). Also highlight the fact that Pressure Gradient and Energy Density Gradient have the same units (<strong>bolded above</strong>).</p>
<p>So we’re talking about the same concept as water only spontaneously flowing down a pressure gradient (ie: downhill) when we talk of energy (of any form) only spontaneously flowing down an energy density gradient. Energy density <strong>is</strong> pressure, an energy density gradient <strong>is</strong> a pressure gradient… for energy.</p>
<p>It’s a bit more complicated for gases because they can convert that energy density to a change in volume (1 J m-3 = 1 Pa), for constant-pressure processes, which means the unconstrained volume of a gas will change such that its energy density (in J m-3) will tend toward being equal to pressure (in Pa). This is the underlying mechanism for convection.</p>

CC BY-SA 4.0
2024-02-07T10:07:52.623Z
2024-02-07T10:17:09.757Z
2024-02-07T10:17:09.757Z
Mr_Magoo
<p>It seems like you use a tube of sufficient volume that the effects of the interior walls are de minimus on the air in the tube, and long enough for there to be some temperature gradient top to bottom.  Then you need a "piston" made of wire mess of a material with high emissivity.   The wire mesh is so that the piston has de minimus effect on convection.  Finally you need a hot plate at the bottom that produces a known amount of radiation and  thermometer close to the plate to monitor temperature.   Then you measure the temperate with "no piston" in the tube, and compare with the piston in the tube at various distances from the plate.</p>

<p>It seems this would enable one to prove/disprove back-radition.</p>

CC BY-SA 4.0
2018-11-20T23:50:53.143Z
2018-11-20T23:50:53.143Z
david russell
<p>"Logically this is the same as the net radiation downwards".</p>

<p>No, this is not the case. The atmosphere is almost black and opaque at certain wavelengths such as 15um (CO2). Then only the external layers radiate, mostly. The rest of the atmosphere transport heat by convection.</p>

<p>The atmosphere is black because an absorbed IR photon energy will excite a CO2 molecule, for example, for a time which is 1000 x bigger than mean time between collisions in the case of low altitude. The molecule will deexcite through collision first.</p>

<p>Due to the gravity gradient, the top of the atmosohere is low pressure and mean time between collisions is much longer. We can see CO2 radiation from a thicker layer, this is the little glitch in the middle of the 15um band dip.</p>

<p>But at low altitude, only 10m of atmosphere will backradiate and the energy density is low because the temperature difference with earth is small.</p>

<p>The backradiation diagram from Nasa in ipcc AR4 is simply wrong. This happens (remember Mann hockeystick scandal).</p>

CC BY-SA 4.0
2020-02-10T14:08:12.777Z
Hi Backradek, welcome to Earth Science. I think you have mispreresented [radiative-convective equilibrium](http://glossary.ametsoc.org/wiki/Radiative-convective_equilibrium), at least your point goes against established scientific consensus. All atmospheric layers radiate ([Stefan-Boltzmann Law](https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law)), which transports heat when layer temperatures differ, greenhouse effect is (more or less) because TOA radiation is from colder layers than the surface. Can you cite sources supporting your claim that IPCC AR4 quoted energy diagram is wrong?
gerrit
CC BY-SA 4.0
2020-02-12T14:27:03.080Z
@gerrit "Established scientific consensus" means nothing in sciences. Science is not based on consensus and not on authority sophisms. Climate science is new, polluted by politics, and very far from any kind of convergence even if this had some meaning. The SB and Kirchhoff laws apply to a black body facing space, not to convecting gasses. Basic physics tell us that the pressure gradient in the atmosphere leads to a dissymetry in the CO2 reemission (see my other post), leading to an IR flux towards high altitudes. This flux of course has a cooling effect.
Backradek
CC BY-SA 4.0
2020-02-12T16:07:29.660Z
This flux has a cooling effect on the atmosphere and a heating effect on the surface. You are right that scientific consensus can be wrong, if you claim it to be wrong you need to provide strong (peer reviewed) evidence. Stefan-Beltzmann and Kirchhoff laws apply always, **everything** radiates, including convecting gases (but radiation is not the only means of heat transport, hence the radiative-*convective* equilibrium). Your other answer didn't answer the question and was converted to a comment, unfortunately the second half of the post was lost in the conversion (flag that post for access).
gerrit
CC BY-SA 4.0
2020-02-13T18:26:54.383Z
How could this flux from the surface cool the atmosphere and heat the surface ? Violation of the laws of thermodynamics. Energy goes from hot to cold. What heats the atmosphere is for example the dip around 15um in the emission spectrum of Earth, given the fact that the total average emitted power must be equal to the sun average input. Yes everything radiates. If you use SB law for a thing which is not a blackbody and not in equilibrium and not facing space, then you have a little problem, no ? This is empirical at best. In the case of Earth radiating to space, then I have less of a problem.
Backradek
CC BY-SA 4.0
2020-02-13T19:29:18.300Z
You misunderstood what I meant by "cool the atmosphere and heat the surface".  Of course, directly speaking, radiation cools the surface and heats the atmosphere; I mean relative to the absence of those greenhouse gases. The presence of greenhouse gases heat the surface, the presence of greenhouse gases cools the upper trophosphere. Does not violate laws of thermodynamics.
gerrit
CC BY-SA 4.0
2020-02-13T19:31:00.223Z
The total emitted power must equal the average input (apart from a small imbalance which is climate change), but that doesn't tell us at what temperature that happens. Greenhouse effect significantly increases that temperature. SB assumes blackbody, does not assume facing space. Everything radiates, you stated the opposite  Will you add some sources to your post, please?
gerrit
CC BY-SA 4.0
2020-02-09T10:14:03.640Z
2020-02-09T10:14:03.640Z
Backradek
<p>Based on our current knowledge of CO$_2$ and current instrumental techniques, how big would a laboratory sample of air with 400ppm of CO$_2$ need to be in order to quantify the back radiation from CO$_2$? </p>

<p>I am envisioning a large container with many cubic metres of gas, probably thermally isolated longitudinally and with a radiative IR input at one end (<em>A</em>) plus measuring apparatus at the other end (<em>B</em>) and some distributed thermistors throughout. We know the thermal input at <em>A</em>, measure the thermal output at the <em>B</em> and measure the trapped thermal energy via the thermistors. </p>

<p>I would like to get a result with something like +/- 1% accuracy. Would this be possible? Assume I have a budget of at least $10,000,000.</p>

CC BY-SA 3.0
2017-04-17T05:32:21.757Z
I suspect this can be done with less gas volume and much less money
Gimelist
CC BY-SA 3.0
2017-04-21T11:05:43.030Z
this question might be among the top ten of hardest to answer,there is so many things that needs to be solved to get a correct measurment.
trond hansen
<meteorology><atmosphere><climate><climate-change><temperature>
How to design an experiment to quantify back radiation from atmospheric carbon dioxide?
user7733
# How to design an experiment to quantify back radiation from atmospheric carbon dioxide?

(Asked by: username_0 on 2017-04-16)

Based on our current knowledge of CO$\_2$ and current instrumental techniques, how big would a laboratory sample of air with 400ppm of CO$\_2$ need to be in order to quantify the back radiation from CO$\_2$?

I am envisioning a large container with many cubic metres of gas, probably thermally isolated longitudinally and with a radiative IR input at one end (*A*) plus measuring apparatus at the other end (*B*) and some distributed thermistors throughout. We know the thermal input at *A*, measure the thermal output at the *B* and measure the trapped thermal energy via the thermistors.

I would like to get a result with something like +/- 1% accuracy. Would this be possible? Assume I have a budget of at least $10,000,000.

**Tags:** meteorology, atmosphere, climate, climate-change, temperature

> **Comments:**
> **username_1**: I suspect this can be done with less gas volume and much less money  
> **username_2**: this question might be among the top ten of hardest to answer,there is so many things that needs to be solved to get a correct measurment.  

---

## 6 answers

### Answer by username_4 on 2017-05-05. Score: 5

The thing is, we don´t have any instruments(to my knowledge) to measure the intensity of that radiation. But, we measure the intensity of atmospheric emission from space with satellites, which is seen as the "effective temperature". Logically this is the same as the net radiation downwards. And heat transfer then confirms that no net radiation arrives at the surface. Which is equal to "no radiation at all". Because "net" radiation is the only radiation that can heat anything.

There are lots of people claiming that measurements done with IR sensors and pyrgeometers confirm backradiation. This is not true and it is based on a misunderstanding of the function of the device, as well as misapplication of the stefan-boltzmann equation. Both IR-thermometers and pyrgeometers use the s-b equation in combination with a thermocouple or thermopile(pyrgeometer). They are both a kind of more advanced and more sensitive thermometers. The thermopile consist of a sensor measuring a gradient across itself indirectly by changes in resistance acting on a current through the thermopile. So, by determining the gradient in the thermopile, it can sense the amount of heat transferred to the surroundings using the s-b equation. When someone claims measurements show "back-radiation", they refer to measurements done with these devices. As most interested people understand, a measurement of heat transfer away **from** the device, cannot show incoming radiation from a colder atmosphere. If you measure something hotter than the temperature of the thermopile, sure, then you have high accuracy. But a measurement of "back-radiation" from the atmosphere is per definition a measurement of something colder than the device. Because the atmosphere is almost always colder(exception is inversion, but that is not relevant). So any measurement of backradiation made from the surface, is a measurement of nothing. It is simply what the device **doesn´t** measure coming from the atmosphere. This is what I mean with "misapplication" of the s-b equation. It can not, and should not, be used that way. That is, to add flux densities that are not calculated as "net" transfer.

In regards to your suggested experiment. Unless the gas is hotter than the instruments(if using IR-instruments), you will measure a heat sink, not any backradiation. And, yes, that is what greenhouse gases really are. Heat sinks. The earth emit its heat to the ultimate bottomless heat sink in space, the only observed infinity, and greenhouse gases are added heat sinks along the highway for heat, that leads to 3 Kelvin space.

> **Comments:**
> **username_0**: I am aware of the limitations on using pyrgeometers as indeed anyone will be if they bothered to actually read the manufacturers manuals and understood how they work and what they can and cannot do. It amazes me that even well educated professors think they can just grab some instrument out of the suppliers catalogue and use it in a lab experiment without even thinking about how said instrument works.(or RTFM). Some clever people sure are dumb ! Nice to see you are not one of them.  

---

### Answer by username_3 on 2018-11-27. Score: 3

Check out www.arm.gov an ongoing set of Atmospheric Radiation Measurement experiments at many locations around the globe that measures many aspects of the atmosphere including atmospheric back radiation using infra-red spectroscopy. This work started in the mid 90's. Some of the cleanest data comes from the <https://www.arm.gov/tour/north-slope.html> North Slope of Alaska ARM research station.

Here are some articles on the research.

<https://www.nature.com/articles/nature14240>

<https://newscenter.lbl.gov/2015/02/25/co2-greenhouse-effect-increase/>


---

### Answer by username_5 on 2024-02-07. Score: 0

username_7 wrote:
"No, this is not the case. The atmosphere is almost black and opaque at certain wavelengths such as 15um (CO2). Then only the external layers radiate, mostly."

Absolutely correct.

Also remember that there exists a mean free path length / altitude / air density relation. Mean free path length increases exponentially with an increase in altitude.

Thus the net mean free path length for upwelling radiation is longer than for downwelling radiation, and given the potentially millions upon millions of times re-emitted radiation could be re-absorbed / re-emitted in random directions, this means that a very great majority of downwelling radiation is redirected to upwelling... especially so with the ~10.4 m surface-level extinction depth at 14.98352 µm wavelength.

Given that ~10.4 m extinction depth and the nature of absorption (50% being absorbed within the first 10% of that depth,50% of the remainder being absorbed in the next 10% of that depth, 50% of the remainder being absorbed in the next 10% of that depth, etc., etc., etc.) that would mean that fully 50% of the so-called 'backradiation' from CO2 would have to come from within ~1.04 m of the surface... where there would be practically no energy density gradient between surface and atmosphere.


---

### Answer by username_5 on 2024-02-07. Score: 0

username_8 wrote:
"Stefan-Beltzmann and Kirchhoff laws apply always, everything radiates, including convecting gases"

Incorrect.

<https://i.imgur.com/QErszYW.gif>

Idealized blackbody objects > 0 K emit at all times, whereas graybody objects > 0 K above their ambient emit. That's why the S-B equation in graybody form includes the temperature of the cooler object.

In reality, what is happening with the S-B equation (the form meant to be used with graybody objects) is a subtraction of the energy density of the cooler object from the energy density of the warmer object to arrive at the energy density gradient; the radiant exitance of the warmer object predicated upon the energy density gradient.

Idealized Blackbody Object (assumes emission to 0 K and ε = 1 by definition):

q\_bb = ε σ (T\_h^4 - T\_c^4) A\_h

```
 = 1 σ (T_h^4 - 0 K) 1 m^2

 =    σ  T^4

```

Graybody Object (assumes emission to > 0 K and ε < 1):

q\_gb = ε σ (T\_h^4 - T\_c^4) A\_h

The 'A\_h' term is merely a multiplier, used if one is calculating for an area larger than unity [for instance: >1 m^2], which converts the result from radiant exitance (W m-2, radiant flux per unit area) to radiant flux (W).

Temperature is equal to the fourth root of radiation energy density divided by Stefan's Constant (ie: the radiation constant).

e = T^4 a

a = 4σ/c

e = T^4 4σ/c

T^4 = e/(4σ/c)

T = 4^√(e/(4σ/c))

T = 4^√(e/a)

q = ε σ (T\_h^4 – T\_c^4)

∴ q = ε σ ((e\_h / (4σ / c)) – (e\_c / (4σ / c))) A\_h

Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).

W m-2 = W m-2 K-4 \* (Δ(J m-3 / (W m-2 K-4 / m sec-1)))

∴ q = (ε c (e\_h - e\_c)) / 4

Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).

W m-2 = (m sec-1 (ΔJ m-3)) / 4

One can see from the immediately-above equation that the Stefan-Boltzmann equation is all about subtracting the radiation energy density of the cooler object from the radiation energy density of the warmer object.

∴ q = σ / a \* Δe

Canceling units, we get W m-2.

W m-2 = (W m-2 K-4 / J m-3 K-4) \* ΔJ m-3

For graybody objects, it is the radiation energy density differential between warmer object and cooler object which determines warmer object radiant exitance.

Remember that a warmer object will have higher energy density at all wavelengths than a cooler object:

<https://i.sstatic.net/qPJ94.png>

This is why 2LoT in the Clausius Statement sense implies that system energy cannot ascend an energy density gradient,"without some other change, connected therewith, occurring at the same time"... that "some other change" typically being external energy pumping that system energy up the energy density gradient. That's what happens in, for instance, AC units and refrigerators.

Most people cannot think in terms of energy, energy density and energy density gradient. We need to analogize to something they’re familiar with. Thus, just as, for instance, water only spontaneously flows down a pressure gradient, energy only spontaneously flows down an energy density gradient. That’s 2LoT in the Clausius Statement sense, in a nutshell. So one tack to take is to ask people if water can ever spontaneously flow uphill. Of course they’ll say, “No, water cannot flow uphill on its own.” Then show them dimensional analysis.

mass (M), length (L), time (T), absolute temperature (K), amount of substance (N), electric charge (Q), luminous intensity (C)

We denote the dimensions like this: [Mx, Lx, Tx, Kx, Nx, Qx, Cx] where x = the number of that dimension. We typically remove dimensions that are not used.

Force: [M1 L1 T-2] /

Area: [M0 L2 T0] =

**Pressure: [M1 L-1 T-2]** /

Length: [M0 L1 T0] =

**Pressure Gradient: [M1 L-2 T-2]**

Explain to them that Pressure is Force / Area, and that Pressure Gradient is Pressure / Length. Remind them that water only spontaneously flows down a pressure gradient (ie: downhill). Then introduce energy. Tell them that energy is much like water. It requires an impetus to flow, just as water requires an impetus (pressure gradient) to flow. In the case of radiative energy, that impetus is a radiation energy density gradient, which is analogous to (and in fact, literally is) a radiation pressure gradient.

Energy: [M1 L2 T−2] /

Volume: [M0 L3 T0] =

**Energy Density: [M1 L-1 T-2]** /

Length: [M0 L1 T0] =

**Energy Density Gradient: [M1 L-2 T-2]**

Explain to them that Energy Density is Energy / Volume, and Energy Density Gradient is Energy Density / Length. Highlight the fact that Pressure and Energy Density have the same units (**bolded above**). Also highlight the fact that Pressure Gradient and Energy Density Gradient have the same units (**bolded above**).

So we’re talking about the same concept as water only spontaneously flowing down a pressure gradient (ie: downhill) when we talk of energy (of any form) only spontaneously flowing down an energy density gradient. Energy density **is** pressure, an energy density gradient **is** a pressure gradient… for energy.

It’s a bit more complicated for gases because they can convert that energy density to a change in volume (1 J m-3 = 1 Pa), for constant-pressure processes, which means the unconstrained volume of a gas will change such that its energy density (in J m-3) will tend toward being equal to pressure (in Pa). This is the underlying mechanism for convection.


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### Answer by username_6 on 2018-11-20. Score: 0

It seems like you use a tube of sufficient volume that the effects of the interior walls are de minimus on the air in the tube, and long enough for there to be some temperature gradient top to bottom. Then you need a "piston" made of wire mess of a material with high emissivity. The wire mesh is so that the piston has de minimus effect on convection. Finally you need a hot plate at the bottom that produces a known amount of radiation and thermometer close to the plate to monitor temperature. Then you measure the temperate with "no piston" in the tube, and compare with the piston in the tube at various distances from the plate.

It seems this would enable one to prove/disprove back-radition.


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### Answer by username_7 on 2020-02-09. Score: -2

"Logically this is the same as the net radiation downwards".

No, this is not the case. The atmosphere is almost black and opaque at certain wavelengths such as 15um (CO2). Then only the external layers radiate, mostly. The rest of the atmosphere transport heat by convection.

The atmosphere is black because an absorbed IR photon energy will excite a CO2 molecule, for example, for a time which is 1000 x bigger than mean time between collisions in the case of low altitude. The molecule will deexcite through collision first.

Due to the gravity gradient, the top of the atmosohere is low pressure and mean time between collisions is much longer. We can see CO2 radiation from a thicker layer, this is the little glitch in the middle of the 15um band dip.

But at low altitude, only 10m of atmosphere will backradiate and the energy density is low because the temperature difference with earth is small.

The backradiation diagram from Nasa in ipcc AR4 is simply wrong. This happens (remember Mann hockeystick scandal).

> **Comments:**
> **username_8**: Hi username_7, welcome to Earth Science. I think you have mispreresented [radiative-convective equilibrium](http://glossary.ametsoc.org/wiki/Radiative-convective_equilibrium), at least your point goes against established scientific consensus. All atmospheric layers radiate ([Stefan-Boltzmann Law](https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law)), which transports heat when layer temperatures differ, greenhouse effect is (more or less) because TOA radiation is from colder layers than the surface. Can you cite sources supporting your claim that IPCC AR4 quoted energy diagram is wrong?  
> **username_7**: @username_8 "Established scientific consensus" means nothing in sciences. Science is not based on consensus and not on authority sophisms. Climate science is new, polluted by politics, and very far from any kind of convergence even if this had some meaning. The SB and Kirchhoff laws apply to a black body facing space, not to convecting gasses. Basic physics tell us that the pressure gradient in the atmosphere leads to a dissymetry in the CO2 reemission (see my other post), leading to an IR flux towards high altitudes. This flux of course has a cooling effect.  
> **username_8**: This flux has a cooling effect on the atmosphere and a heating effect on the surface. You are right that scientific consensus can be wrong, if you claim it to be wrong you need to provide strong (peer reviewed) evidence. Stefan-Beltzmann and Kirchhoff laws apply always, **everything** radiates, including convecting gases (but radiation is not the only means of heat transport, hence the radiative-*convective* equilibrium). Your other answer didn't answer the question and was converted to a comment, unfortunately the second half of the post was lost in the conversion (flag that post for access).  
> **username_7**: How could this flux from the surface cool the atmosphere and heat the surface ? Violation of the laws of thermodynamics. Energy goes from hot to cold. What heats the atmosphere is for example the dip around 15um in the emission spectrum of Earth, given the fact that the total average emitted power must be equal to the sun average input. Yes everything radiates. If you use SB law for a thing which is not a blackbody and not in equilibrium and not facing space, then you have a little problem, no ? This is empirical at best. In the case of Earth radiating to space, then I have less of a problem.  
> **username_8**: You misunderstood what I meant by "cool the atmosphere and heat the surface".  Of course, directly speaking, radiation cools the surface and heats the atmosphere; I mean relative to the absence of those greenhouse gases. The presence of greenhouse gases heat the surface, the presence of greenhouse gases cools the upper trophosphere. Does not violate laws of thermodynamics.  
> **username_8**: The total emitted power must equal the average input (apart from a small imbalance which is climate change), but that doesn't tell us at what temperature that happens. Greenhouse effect significantly increases that temperature. SB assumes blackbody, does not assume facing space. Everything radiates, you stated the opposite  Will you add some sources to your post, please?  

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