I have a question

1. Jan 27, 2008

zdream8

I was wondering how to prove the problem about an expanding black body.
There is a black body at a given temperature. All lengths are expanded by a factor of 2. Then it should still be a black body, but at a lower temperature.
I understand why this should happen, but I was wondering if anyone could show me how the proof works.
I found the equation for energy density

I(\lambda,T) =\frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda kT}}-1}
.
(sorry, I just copied it and that looks bad, but it's easy to find online)

but I wasn't really sure what to do with it. The wavelengths are obviously going to increase and the photon density is going to go down by appropriate factors...I'm just not sure how it all fits together.
Thanks. :)

Last edited by a moderator: Jan 28, 2008
2. Jan 27, 2008

Staff: Mentor

When one says, expand, does one infer work. Only if a hot body expands and exerts a force over a distance, i.e. does work, would it cool.

If the blackbody simply expands, without doing work, it remains at temperature, but the heat flux per unit surface area decreases, i.e. the number of photons per unit area decreases.

3. Jan 27, 2008

zdream8

This was brought up in context of the expanding universe, I forgot to mention.
And like I said, I realize that it makes sense, I just need help manipulating the equation and mathematically representing the concepts to do a semi-formal proof.
Thanks. :)

4. Jan 28, 2008

marcus

Just to get a bit more definite, since it about an expanding universe, suppose we picture something concrete like a volume V of space with a lot of photons in it, with blackbody temp T

so now suppose distances double

the new volume is 8V

and it contains the same number of photons as before but their wavelengths have all doubled so they represent only half as much energy

so the new energy density is 1/16 of the old.

that means the temperature is now T/2 (use the energy density form of the fourthpower Stefan Boltzmann Law)

IS THIS WHAT YOU HAD IN MIND? because if so it is very easy to write down the equations that go along with it

5. Jan 28, 2008

zdream8

Yes, that's basically what I'm talking about, thanks. But in writing the final equation, then the energy density would equal (1/16)*the original equation? And the T in the final equation would be (T/2)? (Is that correct?) Would that be all the changes? This is where I get confused...because then shouldn't the lambdas be 2*lambda? But this makes the end factor different. Could you show me how the equations work? Sorry, it seems really simple, but I'm stuck on something.
And also, based on it being in the same form, is it implied that it remains a black body? I understand physically that nothing in expansion alone would change any proportions, so it would remain a black body, but I don't know if that's "good enough" with the math.
Thanks again.

Last edited: Jan 29, 2008