Why is the black body spectrum bell shaped if it's absorptance is equal to 1?

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• carter7gindenv
carter7gindenv
TL;DR Summary
Why does a black body spectral intensity is bellshaped if it's absorptance equal 1?
This bother me too much.
A black body is said to absorb every incoming radiation.
Looking at absorptance we have :
So if a Black body absorbs everything we have A=1
Φ is a flux meaning object/time.
This means that whatever the wavelength a black body will absorb the maximal flux.
However the spectre of a black body intensity is bellshaped meaning that the emitted flux is higher at some wavelength.
Therefor it would mean that the absorptance should not be A=1 whatever the wavelength thus contradicting the fact that a black body absorbs every incoming radiation.

I believe I'm thinking in circle there. Save me please.

EDIT: Maybe I'm confusing Absorptance/absorptivity and Emittence/emissivity.
Half the question about that says it is the same thing and the other half that it isn't. I really can't tell and I'm hungry.

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Homework Helper
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Are you assuming that after a blackbody in thermal equilibrium absorbs a certain amount of energy at a given wavelength, it will emit the same amount of energy at the same wavelength? I can't tell from your post if you think that's the case.

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carter7gindenv
Gold Member
I believe I'm thinking in circle there. Save me please.
Are you making some unjustified conclusions here? The 'receiving body ' is not a tuned circuit. The 'Bell curve" refers to the radiated energy spectrum for a particular temperature. If you put a warm black body next to a 'hotter' black body, the two will have different spectra until they reach equilibrium, at which point the spectra will be the same and heat one way will equal the heat transferred the other way.

The short wavelength peak of the hotter body will mean that the cooler body is receiving a lot more short wavelengths than it's radiating etc. etc. (initially)

carter7gindenv and vanhees71
carter7gindenv
Are you assuming that after a blackbody in thermal equilibrium absorbs a certain amount of energy at a given wavelength, it will emit the same amount of energy at the same wavelength?

I do. I'm visualizing it in a similar way as photon absorption by electron which rises their energy state until they release that photon at the same wavelength (except for when they do it with a step)

Are you making some unjustified conclusions here? The 'receiving body ' is not a tuned circuit. The 'Bell curve" refers to the radiated energy spectrum for a particular temperature. If you put a warm black body next to a 'hotter' black body, the two will have different spectra until they reach equilibrium, at which point the spectra will be the same and heat one way will equal the heat transferred the other way.

The short wavelength peak of the hotter body will mean that the cooler body is receiving a lot more short wavelengths than it's radiating etc. etc. (initially)

I admit that it is hard to put in text what I have in my head.
Maybe I can state it otherwise.

For a black body we have A=Φabsreceived =1
With Kirchhoff law we have αλλ
Because of those two equations I read it as "If a flux constant over the whole spectrum is received by the black body then it has to emit a flux constant over the whole spectrum and not a bellshaped."

Thank you again for your patience and time.

Gold Member
until they release that photon at the same wavelength
This is the problem with your idea. A black body always consists of highly dense matter and not the sort of isolated atoms you get told about in the very first Quantum Theory lessons. You can't expect photon in equals photon out. The incoming energy goes 'into the pool' of thermal energy and the probability of the energy of any particular emitted photon follows the Planck radiation law.

carter7gindenv
carter7gindenv
ooOOOOooh
So it means that the output flux is not dependent on the input flux? Doesn't that contradict Kirchhoff law of thermal radiation? I made another post days ago here where my problem resided in my confusion of emissivity/emission/emittence and Absorbptivity/absorption/absorbance.
I think here again it is the case. I'm trying to find a correct differentiation of those terms online but it is hard and everywhere it seems that there is some degree of confusion.

I know that usually books>webpages but one of the paper theory on the subject I have make that exact confusion so I've become paranoid

PS: I'd really like to praise the kindness of people on this forum dealing with convoluted question with patience. Thank you.

For a black body we have A=Φabsreceived =1
With Kirchhoff law we have αλλ
Because of those two equations I read it as "If a flux constant over the whole spectrum is received by the black body then it has to emit a flux constant over the whole spectrum and not a bellshaped."
I'm trying to think of a good analogy to explain the difference between absorption and the BB spectrum. Okay, here goes. Let's say a rock concert is held in a stadium, and all the stadium exit doors are the same. They all have the same efficiency of letting people go through. The concert ends and people want to leave. However, the sociological situation is that people bought seats in different areas and they parked in different areas, so they don't distribute themselves evenly across the doors. Different fluxes of people come out the doors, even though all the doors have the same ability to let people out.

With an ideal black body, even though all the wavelengths have perfect emissivity not all wavelengths emit the same amount of energy.

carter7gindenv
Gold Member
Output flux is only the same as input flux when the temperatures are the same. Imagine two black bodies at the two foci of an ellipsoid. All the energy radiated by one will enter the other. BUT more energy will leave the hotter body; the spectral peak of the hotter body will be at the higher energy.
This is only repeating what an been said already

carter7gindenv
ooOOOOooh
So it means that the output flux is not dependent on the input flux? Doesn't that contradict Kirchhoff law of thermal radiation?
What SophieCentaur is trying to tell you in post #5 is that in some particular window of wavelengths, a black body in equilbrium doesn't need to have the same number of photons in as out. In this small window, the energy flux out will not equal the energy flux in. However, if you sum across all the wavelengths, if the system is in equlibrium, you will find the total energy flux inward equals the the total energy flux outward.

carter7gindenv and sophiecentaur
Gold Member
@Dr_Nate yes. The OP is confusing the (Domed) emission spectrum with the (Flat) absorptivity of a black body. The cooler black body will absorb EVERY photon it receives and is not at all frequency sensitive. Having accepted that fact about the path from A to B then the same thing applies for the reverse path B to A. When the temperatures are the same, there will be equal Energy flow in each direction i.e. equilibrium.

carter7gindenv
Gold Member
@carter7gindenv Just imagine a laser shining on a black surface. Would you expect the surface to 'glow' with re-emitted laser light (it's BLACK' so there can be no reflection)? The surface, of course, will warm up and produce its own (low temperature) thermal emission spectrum.

Now replace the laser with a white hot black body emitter. Each element of its spectrum will behave like the laser beam and warm up the surface.

carter7gindenv
Homework Helper
What SophieCentaur is trying to tell you in post #5 is that in some particular window of wavelengths, a black body in equilbrium doesn't need to have the same number of photons in as out.
In this context, we are talking about a black body that is in thermal equilibrium with an environment that does not have the characteristic black body spectrum, right?

In an equilibrium environment where all of the participants are black bodies, all of them would emit and absorb identically at any given frequency, surely.

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carter7gindenv and Dr_Nate
In this context, we are talking about a black body that is in thermal equilibrium with an environment does not have the characteristic black body spectrum, right?

In an equilibrium environment where all of the participants are black bodies, all of them would emit and absorb identically at any given frequency, surely.
Thanks for pointing that out. I moved the goal post from what the OP was saying. OP was talking about an absorptivity of 1. I was imagining the ideal black body of interest and some other source that isn't an ideal black body, which I did not at all explain.

Now that I think about it for a little bit, if I detail the system, I think it might be the same number of photons in and out at each wavelength. Imagine a perfectly reflective and insulated box with the ideal black body and a less ideal black body at the same temperature. I think each body will individually have the same number of photons in and out at each wavelength.

jbriggs444
Homework Helper
I think each body will individually have the same number of photons in and out at each wavelength.
I had very much the same idea in mind.

hutchphd
Gold Member
an environment that does not have the characteristic black body spectrum, right?
Aren't most environments like that? There are objects at different temperatures all around us so the received spectrum could (and does) have many different and very distinct peaks. Predicting what happens in that case would be much harder.
I had very much the same idea in mind.
That is too simplistic, I think. Bodies with different areas would emit (and absorb) different numbers of photons at a given frequency even is they are all at the same temperature.
I think this thread is getting a bit too convoluted now that the OP's basic misconception has been sorted out. If you have a black object inside a large black box then most of the photons will be ones that leave the box and arrive at another point on the box. The object inside would account for a small amount of the photons flying around., corresponding to its (smaller) surface area.

Homework Helper
That is too simplistic, I think. Bodies with different areas would emit (and absorb) different numbers of photons at a given frequency even is they are all at the same temperature.
Yes, but for every individual frequency, assuming equilibrium with a black body spectrum, the claim is that the energy absorbed and emitted by a particular body at that frequency will match.

If it emits heavily in the spectral red band then it will also absorb heavily in the spectral red band.

Gold Member
If it emits heavily in the spectral red band then it will also absorb heavily in the spectral red band.
I can't help interpreting that statement in a way that it seems to be wrong - or at least incomplete. Of course, at equilibrium with just one other body or surface, what you state is correct but things just aren't in equilibrium in most instances.
Your statement seems to imply that there's some selectivity about absorption. Can you really mean that?

Homework Helper
I can't help interpreting that statement in a way that it seems to be wrong - or at least incomplete. Of course, at equilibrium with just one other body or surface, what you state is correct but things just aren't in equilibrium in most instances.
Certainly. That is why I specified equilibrium in this case.

Yes, I agree that in our everyday environment (cold object, hot small sun), red objects look red.

Gold Member
If it emits heavily in the spectral red band then it will also absorb heavily in the spectral red band.
It will only "absorb heavily" the red wavelengths if they are there in high levels. The statement still implies some selectivity and that misconception has already been made and put to bed in the thread. This is true whether or not there is equilibrium.

Also, if the object is painted Red, it will not absorb red wavelengths - but then we have introduced some selectivity.

Homework Helper
It will only "absorb heavily" the red wavelengths if they are there in high levels. The statement still implies some selectivity and that misconception has already been made and put to bed in the thread.
Let us place our red absorber in a bath of radiation with a black body spectrum centered in the green. Assume that it has attained thermal equilibrium.

As you point out, it cannot absorb more red than is present. Nor can it absorb red more heavily than a black body sitting to one side. The black body is a perfect absorber, after all. But it can absorb red more heavily than it will absorb green, for instance. Naturally, since the incident radiation must go somewhere, our red absorber must reflect red poorly while it reflects green well.

The claim is that our red absorber re-emits red well so that, under these conditions, it re-emits as much red as it absorbs. And it re-emits green poorly, it also emits as much green as it absorbs. Of course at the same time, by definition, the red and green that it is reflecting will both match incoming and outgoing.

The net (counting reflection, absorption and re-emission) incoming and outgoing spectra for this non-black object should be identical and match the spectrum of the bath with which it is at equilibrium.

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hutchphd
carter7gindenv
Thank you all for your participation it really helped me to weed out some of the bad starting hypothesis I had!
I've read a bit your extended discussion and I'm really happy that my interrogation sparked other curiosity!

Cheers!

Gold Member
The net (counting reflection, absorption and re-emission) incoming and outgoing spectra for this non-black object should be identical and match the spectrum of the bath with which it is at equilibrium.
Most statements that we can find connecting absorption, emission and temperature are based on black body behaviour. I think we are actually arguing from different standpoints because that's the most likely situation; neither of us is 'ignorant'. Hashtag Dialogue of the Deaf.
At some point there has been confusion between equilibrium and non-equilibrium situations and, even in the original post, there are assumptions made (cart before horse) about the outcome of a thought experiment. There seems to be even an implication that higher energy photons can be produced by low energy received photons. I think that can be discounted; the processes can surely be regarded as linear (?).

Let's forget my 'red painted' object and stick to a black object - it is a bit of a red (not black) herring. Coloured objects will still have the same Absorption and Emissivity at any particular frequency, Painting an object can't alter its final equilibrium temperature when in a box of a different colour. The argument would be slightly more convoluted but it has to be the case.

The only objection I have to what has been said is to do with the implication that a black body is somehow frequency selective. By definition of a black body, it cannot be. There has also been some implication that the same photon enters and leaves a body, which is definitely misguided (that pesky Hydrogen Atom model).
A cool body (say 200K) will be emitting the majority of its photons in the long infra red region. If it's next to a yellow hot body (say 2000K) it will be absorbing most photons which are in the visible spectrum and its absorption spectrum will totally match the spectrum of the incident light, irrespective of its own temperature.

carter7gindenv