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For the purposes of this question, "lighter" and "darker" refer to the absorptive qualities of the objects. Darker objects absorb more light, and therefore appear darker.

I'm trying to understand the emissivity of lighter versus darker objects. I get why dark objects absorb more light than lighter ones (available energy transitions, probability of interaction with photons, etc). But why do they also emit more thermal radiation?

I'm familiar with the thought experiment with a light and dark object in an enclosed box reaching equilibrium (much better explained here), but that is more of a proof and not a physical explanation. Do dark objects emit more for the same reasons they absorb more, just in reverse?

Community
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Dave
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    Yes, exactly. Same reason in reverse. – garyp Dec 30 '15 at 18:24
  • I always thought that it is that "reverse" explanation: an energy transition is available both to absorb a photon (go from lower to higher energy) or to emit a photon (from higher to lower), the frequency of the photon is the same. – rodrigo Dec 30 '15 at 18:27
  • Dave, what is the difference between your "proof" and a "physical explanation"? And if there is a difference, isn't a proof always the "better, safer thing" to have than just a "physical explanation"? – Luboš Motl Dec 30 '15 at 18:27
  • I always had trouble understanding why a ruby is red but also it emits red light when used in a laser. So I would love to see a good answer on this question. – Gyro Gearloose Dec 30 '15 at 18:30
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    Beware being too simplistic here. Black objects absorb visible light better than a white object does. That does not mean it absorbs or emits more thermal radiation than a white one does. – BowlOfRed Dec 30 '15 at 18:37
  • @Lubos A proof shows that it is true without explaining the physical interactions that make it true. Neither is better or worse -- they're just different. A different example: A ball bounces. Proof: conservation of energy. Explanation: the elasticity the ball allows it to briefly absorb the kinetic energy of its fall before turning it right back to kinetic energy. – Dave Dec 30 '15 at 18:56
  • @BowlOfRed Yeeeah, definitely cutting a few corners to keep the question concise but I hope my point comes across. Happy to make edits for clarity if you have a suggestion. – Dave Dec 30 '15 at 18:58
  • Dear @Dave, so you don't mean a "proof vs physical explanation". You mean an "argument using overall macroscopic quantities" vs "argument using microscopic, constructive concepts". You can't really do the latter too well here because "black" and "white" are concepts describing the overall macroscopic appearance of an object, not its microscopic composition, so one must unavoidably use some of the macroscopic arguments that, in your logic, turn an explanation into a "proof". – Luboš Motl Dec 30 '15 at 19:49
  • Kevin Ye's or a basically equivalent argument is really the only sensible answer one can get. One could discuss what is happening microscopically. But one would still need to translate the concepts of "black" and "white" to properties of the microscopic building blocks which is rather complicated. And at the end, what one really wants to deduce are just statistical properties referring to absorption and emission of large amounts of light quanta. So one has to use overall thermodynamic properties like the temperature of both bodies, emissivity, absorptivity etc. somewhere in the explanation – Luboš Motl Dec 30 '15 at 19:54

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The microscopic laws of physics all (excepting the weak interaction) have the property of being invariant on time-reversal. In a classical context that means that if I show you a very short film clip of pool balls colliding, you'll have a hard time knowing if I have shown it to you forward or in reverse. In quantum mechanics the meaning of "invariant" is a little different, but it implies that cross-section for the forward and reversed process are identical.

Now consider the emission and absorption of light. All possible modes are quantum mechanical interactions of the electro-magetic variety, which means that they are time-reversal invariant. So, any mechanism that emits photons of a particular wavelength efficiently from some high-energy also absorbs photons of the same wavelength efficiently from the lower-energy state. And those low energy states will be available because the thing is emitting all the time.

And the reversed argument also applies. Wavelengths that aren't emitted efficiently have no efficient mechanism for being absorbed either.

This is basically why a single quantity---the (wavelength dependent) emissivity serves to parameterize both emission and absorption.

This also points out one of the big puzzles of physics: how do microscopically reversible laws result in macroscopic irreversibility ala the 2nd Law of Thermodynamics. (And one answer comes from Statistical Mechanics.)

dmckee --- ex-moderator kitten
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Emitting without absorbing, or vice versa, is forbidden by the Second Law of Thermodynamics.

Suppose we have two objects facing each other, both at temperature $T$. Let their emissivities be $\alpha_1$ and $\alpha_2$, and let their absorbances be $\beta_1$ and $\beta_2$. Then the rate at which object 1 gains energy from object 2 is proportional to $\beta_1 \alpha_2$, and the rate at which object 2 gains energy from object 1 is proportional to $\beta_2 \alpha_1$.

These two rates must be equal, or else you could use one to spontaneously heat the other, and run a perpetual motion machine off the result. Therefore $\beta_1 \alpha_2 = \beta_2 \alpha_1$, i.e. $\beta_1/\beta_2 = \alpha_1/\alpha_2$. An object that emits a lot of radiation must also absorb a lot.

In general, absorbance/emissivity depend on frequency. This same argument can be used to show $\alpha(\omega)$ must be proportional to $\beta(\omega)$.

The general thermodynamic principle is detailed balance, i.e. a forward process must be balanced by its reverse process. On a deeper level, this comes from time reversal symmetry (i.e. if you take a semiclassical model of the radiation, it follows because the coefficients of the raising and lower operators must be equal, which follows because the Hamiltonian is Hermitian, which is a manifestation of T symmetry).

If you want to try it out, you can also use this exact same reasoning to show that (1) one-way mirrors don't exist, and (2) you can't use lenses to focus the Sun's light to make something hotter than the Sun.

knzhou
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The thermal radiation, for an ideal blackbody, is determined by its absolute temperature and Stefan-Boltzmann law:

\begin{equation} P = \sigma T^4 \end{equation} where $\sigma$ is known as Stefan-Boltzmann constant. For a white object, it reflects electromagnetic wave (assuming otherwise materials used are the same for two objects) in the visible region while the black one absorbs more, therefore its temperature in equilibrium is smaller than that of a black object in your thought experiment. This means

\begin{equation} T_{white} < T_{black} \end{equation} so $P_{white} < P_{black}$.

Kevin Ye
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  • This is exactly the proof I wasn't looking for... But maybe good to have here as a reference. – Dave Dec 30 '15 at 18:59
  • @Dave This is a physical explanation already. Just as Lubos said in the previous comment, your "proof" is actually a physical explanation (and mine is not rigorous proof either). – Kevin Ye Dec 30 '15 at 23:11
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    This isn't the answer, OP is asking for the case where a white and black object have the same temperature. – knzhou Dec 31 '15 at 04:18
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It depends rather on what you call "more" . The blackbody laws are pretty clear about what percent of photons at a given wavelength are absorbed (for a black body or a "grey body", etc) and how the re-emitted energy is distributed across wavelengths. It is important, however, to be aware that the curve you're most likely to find in books and on web pages is the power distribution. If you want to calculate how many photons are emitted at each wavelength, the curve shifts to the right (longer wavelengths) due to the variation in energy per photon at different wavelengths. Thus my comment on what "more" means.

Carl Witthoft
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