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This is a conceptual question I can't quite wrap my head around.

Take two blackbodies with temperatures $T_{hot} > T_{cold}$. Both should have a spectral intensity described by Planck's law $$I(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \frac{1}{\exp{(hc/\lambda k_B T)} - 1}$$

That seems to imply that the spectral intensity of a hot object at each wavelength is higher than that of a colder object.

Now, what I don't exactly understand is whether something would change if both objects interact with each other in the following way:

  • the hot object has an internal source and maintains its temperature while radiating
  • all the radiated power from the hot object is received by the cold object (which should absorb all of it, since it's a blackbody)

As I understand it, the colder object heats up and starts radiating at shorter and shorter wavelengths (Wien's law) until it reaches the same temperature as the hot object (then the incoming and outgoing power should be the same and the temperature doesn't change anymore).

During the time the colder object heats up - what exactly happens to the energy/power difference between what the hot object emitted, and what the colder object re-radiates? If the intensity at each wavelength is lower for the colder object than the hotter object, then there is some left over. Is that just exactly the energy that is changing the temperature of the colder object?

Maybe even more to the point, can there be a scenario where the colder object surpasses the hot object in intensity only in a (longer) wavelength region?

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  • $\begingroup$ You seem to understand the process. I do not know what else you need to know about the first part of the question. As for the question if the colder object can get hotter than the hot object, the answer is no. $\endgroup$
    – Mauricio
    Mar 1 at 10:49
  • $\begingroup$ Maybe that last sentence is actually what is not clear to me: does being colder always imply that the intensity at every wavelength must be lower than for a hotter object? I understand that a hot object can't get hotter; I guess the confusion is more about whether the spectrum of the cold object can shift in such a way that the intensity (only) at longer wavelengths can be higher than those from a hot object, while the overall intensity is still lower. Kind of like a down-conversion maybe. $\endgroup$
    – DK2AX
    Mar 1 at 11:10
  • $\begingroup$ The requirement is that each object absorbs all emission from the other. $\endgroup$
    – my2cts
    Mar 1 at 12:16
  • $\begingroup$ Can you just simply plot Placnk law for two different temperatures? Even if for some wavelength, the cold object emits more, what matter is the total flux (integral over all frequencies) from one body to the other, the flux of the hotter object will be higher because it is peaked at higher frequencies and thus higher energies. $\endgroup$
    – Mauricio
    Mar 1 at 12:16

1 Answer 1

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Let's split the answer in few steps:

  • recall Stefan-Boltzmann laws to write emitted power flux
  • write energy balance of the systems involved in the problem
  • set a power source to keep constant energy of system 1
  • assume negligible radiance from the infinity

In what follows, I'll assume emittivity = 1 i.e. black bodies.

Stefan-Boltzmann law

Following Stefan-Boltzmann law, each body emits a power per unit-surface equal to

$$M_i = \sigma T_i^4 \ ,$$

being $\sigma$ Stefan-Boltzmann constant.

Emitted power fluxes

Power flux emitted from the body 1 can be written as the sum of the fraction of power reaching body 2 and the fraction that does not reach it, as

$$P_{1} = \underbrace{S_{21} \sigma T_1^4}_{P_{21}} + \underbrace{S_{\infty1} \sigma T^4_1}_{P_{\infty 1}} \ ,$$

while the power flux emitted from body 2 can be written as $$P_{2} = \underbrace{S_{12} \sigma T_2^4}_{P_{12}} + \underbrace{S_{\infty2} \sigma T^4_2}_{=P_{\infty2}} \ ,$$

and the net power (since many parts of the infinity look at other parts of the infinity) emitted from the infinity reads

$$P_{\infty} = \underbrace{S_{1\infty} \sigma T_{\infty}^4}_{=P_{1\infty}} + \underbrace{S_{2\infty} \sigma T^4_\infty}_{=P_{2\infty}} \ .$$

Energy balance

By geometry, $S_{ij} = S_{ji}$, and thus the total energy balance of the bodies reads $$\begin{aligned} \dot{E}_1 & = -P_1 + P_{12} + P_{1\infty} + R = \sigma \left[ S_{12}(T_2^4 - T_1^4) + S_{1\infty}(T_\infty^4 - T_1^4) \right] + R \\ \dot{E}_2 & = -P_1 + P_{12} + P_{1\infty} \qquad = \sigma \left[ S_{12}(T_1^4 - T_2^4) + S_{2\infty}(T_\infty^4 - T_2^4) \right] \\ \end{aligned}$$

Now, assuming that $T_\infty$ is constant, and the source $R$ keeps the energy of system 1 constant, $\dot{E}_1$, $$R = -\sigma \left[ S_{12}(T_2^4 - T_1^4) + S_{1\infty}(T_\infty^4 - T_1^4) \right] \ ,$$

and assuming that a constitutive equation relating the total energy of a system with its temperature, $E(T)$,

  • temperature $T_1$ is constant, as long as $E_1$ is constant
  • you can solve the dynamical equation for temperature evolution of system 2 as

$$\dot{E}(T_2) = - \sigma \left( S_{12} + S_{2\infty} \right) T^4_2 + \sigma \left( S_{12} \overline{T}_1^4 + S_{2\infty} \overline{T}_{\infty}^4 \right) \ . $$

Negligible contribution of the infinity

If you can neglect the contribution of the infinity, as an example if it's in thermal equilibrium with system 2 (or temperature difference is not so high, if compared with $T_1 - T_2$), you get the following equation

$$\dot{E}(T_2) = - \sigma S_{12} T^4_2 + \sigma S_{12} \overline{T}_1^4 \ , $$

from which, considering a monotonically increasing function $E(T)$, it should be easy to realize that system 2 approaches the constant temperature $\overline{T}_1$, reaching equilibrium $\dot{E}_2 = 0$ when $T_2 = T_1$.

Remarks.

  • This conclusion agrees with principles of thermodynamics stating that net heat flux occurs from hot to cold bodies.
  • If you're interested in the spectrum of the emission, you can recover it once the temperature of the black-body is known.
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