2
$\begingroup$

Say I have a 1 $m^2$ source of visible light emitting 1,000W worth of light. This is shining down upon 1 $m^2$ of ground.

If we treat the ground as a blackbody, my understanding is that eventually it'll heat up until it reaches a temperature at which it's emitting 1,000W back in the form of thermal radiation. From the Stefan-Boltzmann equation, $1000 \frac{W}{m^2} = \sigma T^4$, therefore $T = 364.4K $.

And likewise this matches the temperature of the source (?): since the source is also emitting 1000 $\frac{W}{m^2}$ over 1 $m^2$, the same formula applies. All is well -- the two bodies have reached thermal equilibrium and are at the same temperature.

Now say I add a pane of glass between the source and the ground. The glass allows visible light through, but absorbs & reflects back the infrared thermal radiation the ground emits. Say it absorbs 100% and emits 50% out in each direction. (EDIT: This would be equivalent to say the glass heats up and emits its own IR emission from its own temperature.)

Eventually thermodynamic equilibrium is reached. The 1000W from the source gets absorbed & re-emitted up from the ground. The 1000W of IR hits the glass, is absorbed, and 500W is emitted upwards and 500W downwards. Now the ground is receiving 1500W (1000W visible light + 500W infrared radiation), which it absorbs & emits upwards as IR, and the pane absorbs this and reflects 750W up and 750W down. This repeats until we reach this equilibrium:

  • at the entrance to the system: 1000W comes in from the source and 1000W exits the system
  • at the pane of glass: 2000W of thermal radiation from the ground is absorbed, 1000W is emitted upwards, 1000W is emitted downwards
  • at the ground: 1000W from source + 1000W reflected radiation comes in, and 2000W thermal radiation emitted upwards

None of this violates the first law of thermodynamics -- energy isn't being created or destroyed, just accumulated at a higher density via the reflections in the system.

But in this situation the ground is emitting 2000 $\frac{W}{m^2}$ which gives it a temperature of 433.4K, which is more than the source. And note that if the glass instead reflected 90% of the radiation downward and only 10% upward, the equilibrium would be with the ground emitting 10,000 $\frac{W}{m^2}$ = 648 K.

(EDIT: To put it differently, if the glass absorbed say only 10% of the IR radiation which it used to heat up, but reflected 90% back where it came from -- then the multiplier would be higher. It wouldn't be 'selectively reflecting' since it treats IR the same whatever direction it comes from -- its just that the source is visible light so these properties don't affect it.)

Does this violate the second law? If so, what about the setup makes it impossible? If not, then how is it possible?

A further question: say the source is hotter but the energy is dispersed. e.g. say it emits 2000 $W/m^2$ but it disperses such that only 1000 $W/m^2$ reaches the system above. Now the source itself is 433.4K instead of 364.4K. Or it can be emitting 20,000 $W/m^2$ at the source but it's even further and disperses more. Does it change the situation at all since still, only 1000 $W/m^2$ is coming in?

(For context: this question arose from a comment on an answer to this question here: Why can't fire heat something hotter than itself, but electromagnetic waves can heat something indefinitely?)

$\endgroup$
7
  • 2
    $\begingroup$ Your special glass that transmits radiation in one direction but not the other is an equivalent of Maxwell's demon - once you take into account what actually happens in the glass, the paradox would go away. $\endgroup$
    – Roger V.
    Commented Mar 6, 2023 at 10:59
  • 1
    $\begingroup$ Hmm interesting. Ok in that case the potentially infinite multiplier effect goes away, but the 2x effect is still possible for the glass that emits 50% in both directions? $\endgroup$
    – Cloudyman
    Commented Mar 6, 2023 at 11:12
  • 1
    $\begingroup$ But, can’t there exist a material that reflects (instead of absorbs) infrared light while allowing visible light through? $\endgroup$
    – Cloudyman
    Commented Mar 6, 2023 at 11:42
  • 1
    $\begingroup$ Suppose we have such a material: as it absorbs the infrared light it becomes heated, so it starts re-emitting this light, till the absorption and emission equilibrate. $\endgroup$
    – Roger V.
    Commented Mar 6, 2023 at 11:48
  • 1
    $\begingroup$ Indeed that’s the point of the question — it equilibrates but with a higher wattage under the glass than above it. Say it allows 100% of visible light through, and reflects 90% IR and allows 10% IR through (in any direction). The equilibrium will be when it’s receiving 10,000W of IR from the ground, allowing 1,000W to pass (ie equilibrium with net source) and 9,000W reflected. $\endgroup$
    – Cloudyman
    Commented Mar 6, 2023 at 12:07

2 Answers 2

1
$\begingroup$

Your source has to cool itself (because infrared radiation impinges upon it) and also generate low-entropy visible light. The mechanism that achieves this shunts entropy somewhere else, which compensates for the out-of-equilibrium steady-state condition in which the cool visible light faces the hot plate.

Once this mechanism is incorporated into the thought experiment, the Second Law will be satisfied.

$\endgroup$
1
  • $\begingroup$ Hmm interesting. So the answer is, it's possible, and can multiply to the extent a real barrier could have these properties, and the reason is that the generation of the visible light needs to be considered in the whole system. So like the other Q, again a 'cooler' source leads to a 'hotter' object, but only because of the whole system. Another thermodynamic 'good news' :D $\endgroup$
    – Cloudyman
    Commented Mar 6, 2023 at 16:53
1
$\begingroup$

The question is a little ambiguous because the nature of the source is not clear. A laser, for example, can heat up an object far hotter than the temperature of the laser itself, which is how things like laser welders and cutters work. That is because the laser light has very low entropy and the light is not generated thermally but through electromagnetic work.

From the context of the rest of the question it does not seem that a laser is what you have in mind. It seems that you have more in mind a gray-body source thermally interacting with a black-body object. For clarity, I believe that you intend a source that has an emissivity of 1.0 in the visible range and is transparent outside the visible range. The ground is, of course, a black body with emissivity 1.0 at all wavelengths. The glass seems to be the reverse of the source, with an emissivity of 1.0 outside of the visible range and transparent within the visible range.

Because we are using gray bodies the Stefan Boltzmann law is not applicable. You would need to use Planck's law instead. The visible range is a rather narrow range of wavelengths. To achieve the same 1 kW output over this narrow band of wavelengths will require a much higher source temperature (1709 K).

Without the glass, the ground will absorb all of the incident visible radiation. This will increase the temperature of the ground and therefore increase the amount of radiation emitted both in and outside of the visible range. As the visible range is small, most of the radiated energy will escape the system, with equilibrium achieved when the ground is radiating 1 kW outside of the visible range, which will be a much lower temperature than the source temperature (364 K). At that temperature the ground will radiate a negligible amount in the visible range (9.3 fW) which will be absorbed by the source and will negligibly increase the source temperature so that the source radiates the 1 kW plus the negligible amount of the visible back-radiation from the ground.

With the glass, the visible radiation passes through the glass and is completely absorbed by the ground. As before, this requires a very high temperature for the source since the visible range is so narrow. The ground absorbs it and heats up. As it does, it emits black body radiation and the source absorbs the visible radiation while the glass absorbs the non-visible radiation. This increases the temperature of the source negligibly and the temperature of the glass a lot. The glass then radiates half of its power to escape the system and half back to the ground. Equilibrium is achieved when the glass is radiating 2 kW outside of the visible range, 1 kW of which escapes and 1 kW is re-emitted back to the ground. This occurs when the glass reaches the temperature of the ground from the previous case (364 K). The ground, in turn, will be radiating 2 kW outside the visible range plus a negligible amount in the visible range. So the ground will be at a higher temperature than in the previous case (433 K), but not much higher since most of the radiation is still outside the visible range.

$\endgroup$
7
  • $\begingroup$ I see ok. So the breakdown was in considering the source as the same temperature at the ground — due to the nature of the source it’s actually much hotter. So the ground heats to the 2kW thermal radiative equivalent but that’s still a lot lower temperature than the source. $\endgroup$
    – Cloudyman
    Commented Mar 6, 2023 at 21:22
  • $\begingroup$ @Claudiu yes. And the reason the source is much hotter is because it is a gray body (if I understood your intention). The gray body has to be hotter to radiate the 1 kW over the visible band. $\endgroup$
    – Dale
    Commented Mar 6, 2023 at 21:32
  • $\begingroup$ Minor note, wouldn't the glass also be the temperature of the ground (433 K) since it's emitting 2kW total? (1 kW that escapes and 1 kW towards the ground) $\endgroup$
    – Cloudyman
    Commented Mar 6, 2023 at 22:24
  • 1
    $\begingroup$ Ah never mind, it has double the surface area (2m^2), so it's still at 1 kW/m^2 . $\endgroup$
    – Cloudyman
    Commented Mar 6, 2023 at 22:32
  • 1
    $\begingroup$ @Claudiu you will need to post a new question. As you can see, such questions take a bit of analysis to answer. I cannot do that in comments. Let me know here when you do post the question. It will help if you specify details in your question regarding the various objects. Black bodies are easy, but for everything else please specify the wavelength ranges where it has a given emissivity or transparency. $\endgroup$
    – Dale
    Commented Mar 7, 2023 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.