I understand why a black body absorbs every frequency(it is the definition of a black body!) but i do not understand why it also radiates at all frequency spectrum.
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asked Jan 18, 2016 at 17:57
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$\begingroup$ Is your problem that you think the material the body is made up should only have distinct transitions and therefore only emit light at these frequencies? Is this compatible with assumption that it absorbs every frequency? If you do not like to think about the body at all, than imagine (as sometimes is done) the emission coming from a small hole in the body on one side $\endgroup$– BortCommented Jan 18, 2016 at 18:07
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$\begingroup$ @Bort my problem is that when someone says that a black body absorbs all frequencies it means that it also radiates at all frequencies, so why does one also mean the other? $\endgroup$– TheQuantumManCommented Jan 18, 2016 at 18:16
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$\begingroup$ have a look at this hyperphysics.phy-astr.gsu.edu/hbase/mod6.html $\endgroup$– anna vCommented Jan 18, 2016 at 19:55
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4 Answers
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Absorption necessitates emission (and vice-versa). If you can absorb some energy (say by absorbing a photon) then you can also emit that energy (by emitting a photon with the same frequency). Therefore if a black body can absorb all frequencies, it must also be able to emit all frequencies.
Edit: Thanks to @Bort!
Absorption and emission necessitate each other because of T-symmetry, i.e. time-reversed absorption is equivalent to emission.
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$\begingroup$ Thanks for the answer, but why should what you tell hold truth? By what law one thing means also the other(emmision at all frequencies means absorption at all frequencies and vice versa)? $\endgroup$ Commented Jan 18, 2016 at 18:15
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$\begingroup$ Good question, and I'm sorry but I don't have a good answer as to why the exchange of energy is fundamentally reversible. Every absorption/emission event I can think of can also be reversed, although I'm not aware of any law dictating this. $\endgroup$– JudgeCommented Jan 18, 2016 at 18:22
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1$\begingroup$ Symmetry under time reversal or (by say Noether-Theorem in a classical view) conservation of energy $\endgroup$– BortCommented Jan 18, 2016 at 18:24
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$\begingroup$ @Bort i am not talking about reversing the procedure in time. I am talking about emmision and radiation at all frequencies? Why does one automatically mean the other? $\endgroup$ Commented Jan 18, 2016 at 18:27
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$\begingroup$ @Bort, excellent point! Absorption is the same as emission if time is reversed :) Genius. Quick question: doesn't the conservation of energy just demands that the amount emitted and absorbed be the same, not that both can fundamentally happen? $\endgroup$– JudgeCommented Jan 18, 2016 at 18:29
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The definition of a blackbody is not just that it should absorb light perfectly at all wavelengths; it should also be in thermal equilibrium at some temperature.
If this is the case, then at equilibrium, all absorption processes must be balanced by emission processes. If that were not the case then the populations of energy levels would change, or the average speed of particles would change, leading to a changing temperature as the object heated up or cooled down.
Thus the reason that a blackbody emits as much radiation as it absorbs is really by definition. An object that did not fulfil this condition would not be a blackbody.
As to how this can happen, the usual illustration is the "two-level atom". At equilibrium there is a detailed balance between absorption of photons by atoms in level 1 with emission (both spontaneous and stimulated) by atoms in level 2. At equilibrium this implies a relationship between the Einstein A and B coefficients that determine the rates of these processes. The quantum mechanical root of this relationship is that the Hamiltonian perturbation caused by the electromagnetic field is a Hermitian operator, such that $|D_{if}|=|D_{fi}|$, where $i,f$ are the initial and final states and the transition probabilities are proportional to $D^2$.
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First at all, really black bodies do not exist. We simulate them, making a hole in a cavity with thick walls and painting the walls black.
As you know, colour surfaces emit the radiation they get from the surrounding space, in a narrow range. Black surfaces absorb the incoming electromagnetic radiation and this of course has to heat them up. Otherwise a body that is hotter (warmer) the surrounding space, emit EM radiation, but this time in the infrared spectrum. Than more radiation hits the body than more the emission spectrum shifts to the visible light.
That the black body emits light in a wide spectrum has to do with the black painting and the cavity. As long as the wavelength of the light can oscillate in integer units between the walls of the cavity the light gets shifted and shifted to longer wavelengths, the theoretical longest wavelength is the distance between the walls. But meanwhile the wave gets dissipated.
answered Jan 18, 2016 at 18:49
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I think a black body radiating in a certain range of spectrum would cause some serious problems, since we wouldn't be able to explain why it preferred that particular range.
Therefore, a black body must radiate in all frequencies, because one interval is not physically preferable than the other.
