I am a layman with a little beginners knowledge.

This question is an effort to prove or disprove the Greenhouse Effect (GHE) theory as this is the basis of "back radiation" as described by the GHE theory, i.e. Some of the Infrared (IR) radiation coming from the surface of the earth is absorbed by "greenhouse gasses" (GHGs) and about 50% is re-radiated back towards the earth, heating the surface more than it would be without GHGs.

As all objects warmer than absolute zero radiate electromagnetic radiation at frequencies according to their temperature, according to Max Planck, I proposed the following experiment:

Hang an object in the center of a box with perfectly reflective surfaces facing inward on the inside so as to reflect the radiation from the object back on itself. Mirrors would be good enough as it is said about 50% of earth's radiation that is absorbed by GHGs is re-radiated back to earth. Make sure the box and the object are stabilized at room temperature before placing the object in the box. Measure the temperature of the object at the beginning and after some time period to compare.

Will the object get warmer from re-absorbing its own radiation? I don't think so, but maybe somebody with a better understanding of the physics of radiation could help. I have not had time to perform this experiment, yet. But, I will as I can get to it.

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    $\begingroup$ Welcome AlexJ. I don't understand how that experiment has a bearing on the claim that GHGs inhibit the emission of radiation back into outer space. In your experiment, there is no external source of radiation like the sun which is constantly adding more radiation to the object... the claim as I understand it, is that the rate of emission decreases due to GHGs, while the rate of absorption from the sun stays more or less the same, both effects combined then lead to the overall warming effect. $\endgroup$
    – Amit
    Jun 7, 2023 at 19:40
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    $\begingroup$ @Amit check my answer. But, yes, it is a combination effect. $\endgroup$ Jun 7, 2023 at 20:39
  • $\begingroup$ What is the analog for the sun in your “experiment”? $\endgroup$
    – Dale
    Jun 7, 2023 at 21:09
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    $\begingroup$ I vote to reopen because the question is not about unproven theories but very specifically about what happens when an object reabsorbs the energy in the setup described by the OP. In addition, it has generated interest and discussion among users. $\endgroup$
    – Themis
    Jun 11, 2023 at 20:12

3 Answers 3


Yes, the presence of GHGs results in an increase in temperature. As they serve to trap heat within the atmosphere, this reduces the radiative flux from the planet, meaning there is no longer an equilibrium state. To restore said equilibrium, the temperature of the Earth must rise such that it can radiate more.

It's important to also remember that there's an external source of energy (the sun) adding to the system, which just makes matters worse.

  • $\begingroup$ You are just wrong. Check my answer. $\endgroup$ Jun 7, 2023 at 20:38
  • $\begingroup$ I've amended my answer. It seems I've forgotten more from undergrad than I thought. Thank you for correcting me. $\endgroup$ Jun 7, 2023 at 20:49

Actually, it does.

The total energy radiated per unit surface area, over all frequencies, per unit time, of a body, is the radiant emittance, and it is dictated by Stefan-Boltzmann law to be $$j=\frac{\mathrm d^3E}{\mathrm d^2A\,\mathrm dt}=\varepsilon\sigma T^4$$ where $T$ is the absolute thermodynamic temperature, $\sigma$ is a universal constant, and $0<\varepsilon\leqslant1$ is the emissivity, a measure of how good something absorbs radiation, or reflects it.

By an ingeniuous argument due to Kirchhoff (the same Kirchhoff who was lucky to have given us the trivial looking laws of circuit analysis), the rate at which a body emit light is related to the rate at which it absorbs light. The argument is that $j$ must be equally fast outwards of body as inwards to body at thermal equilibrium.

This same argument is now going to cause the trouble. The ideal blackbody has $\varepsilon=1$, and at whatever $T$ you have, that is some fixed value of $j.$ Now, if you have that fixed value of $j$ and you reflect some light that was originally about to leave, then that makes $\varepsilon<1$, and then the only way to achieve thermal equilibrium is to raise $T.$ i.e. For a given rate of energy radiation, in or out, the thermal equilibrium temperature of a blackbody is the least of all possible bodies.

I was set the question of using these values to derive the average temperature of the Earth during an exam, and exam conditions are not the correct time to be mind blown over how precise everything has to be! By considering the approximations even a tiny bit wrongly, the temperature of the Earth will easily be inhospitable for life, and so we alter $\varepsilon$ of the Earth to humanity's peril.

  • $\begingroup$ I have a hard time understanding this intuitively. Putting an object inside a mirrored box will obviously not result in its temperature increasing without bound. The object we're considering is a closed system, where does the extra energy come from? $\endgroup$ Jun 7, 2023 at 20:47
  • $\begingroup$ Everybody else covered---the box is absorbing sunlight and also needs to radiate at the rate at which sunlight is coming in, in order to achieve thermal equilibrium. $\endgroup$ Jun 7, 2023 at 20:51
  • $\begingroup$ Yes, the box will get warmer absorbing the sun's radiation. That's not what the question is - the experimental apparatus is a mirrored box with an object inside and nothing else. $\endgroup$ Jun 7, 2023 at 20:57
  • $\begingroup$ It is more important to answer to the context that the person is asking, rather than to the literal text of the question. The OP is obviously interested in understanding how global warming works, and not at all interested in the idealised reflective box doing unrepresentative things. Note that all the answerers have come to the same stance; your insistence upon a pedantic treatment is too weird; if you want an answer to that, ask a new question for that. $\endgroup$ Jun 7, 2023 at 21:00
  • $\begingroup$ I've come across this sort of situation before since I joined the PSE, and I never know if it justifies editing the question. Since we can directly address what they were really trying to ask, does it even make sense to leave the original question as it is? I don't know. Maybe a question better suited for meta. $\endgroup$ Jun 7, 2023 at 21:24

If the object is not thermally isolated from the rest of the universe, then the mirrors or thermal insulators can prevent the object from losing heat as fast as it gains heat from somewhere else (e.g. the sun), which will cause its temperature to rise. We could do this by shining light in with a bright spotlight through a hole in the mirrors or insulators, or by shining light in at a frequency for which the insulators are transparent, and then re-radiating light away in the infrared spectrum, for which the insulators are more opaque. This is the greenhouse effect, which is your everyday experience of getting into a car that has been exposed to sunlight for a few hours and finding it is much warmer than the outside world.

If the object is not inert, then the mirrors or thermal insulators can prevent the object from losing heat as fast as it converts other forms of energy to heat (e.g. nuclear potential energy to heat via radioactive decay, or chemical potential energy to heat via respiration), which will cause its temperature to rise. This is your everyday experience of warming up when you cover yourself with blankets.

If the object and the reflectors are thermally isolated from the rest of the universe, and if the object is inert, the answer is no. To heat something at constant pressure, you must add energy, so any temperature increase must have energy transported across the boundary of the system. Furthermore, to increase the temperature of something without doing mechanical work, you must put it in thermal contact with something at a higher temperature. The 2nd Law prohibits any system of mirrors or lenses from increasing light intensity at any point above the intensity at which it was emitted, which is a function of the temperature of the thing that emitted it. If the object in question has uniform temperature and uniform thermal emissivity, then the mirrors cannot increase the temperature of any part of the object. This is conservation of etendue. The common experience of this is the function of a thermos, which indeed surrounds the object (you hot or cold beverage) with a mirror (the metal interior flask) and does its best to isolate it from the exterior world with an insulator (an empty layer between the interior and exterior flask without much matter in it to conduct heat).


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