Black Body Energy Spectrum

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Homework Statement



Write the volume element of d3p as a function of "nu". Assume spherical symmetry in doing this change of variables so write d3p = 4[tex]\pi[/tex]p2dp.

Homework Equations



[tex]n(\nu)=\frac{1}{e^{\frac{h \nu}{kT}} -1}[/tex]

[tex]\epsilon=\frac{2}{h^3}\int h \nu \cdot n(\nu)d^3p[/tex]

The Attempt at a Solution



I have zero idea of where to even start with this. As stupid as this is, I don't even understand "d3p = 4[tex]\pi[/tex]p2dp" or what the d3p even is. I don't think I've ever come across an integral that has used this type of notation before.

Any help to even get me started would be appreciated. :redface:
 

Answers and Replies

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The p in the equation above is simply the momentum of photons. There is a relationship between momentum and frequency for photons:

p = [tex]\frac{h\nu}{c}[/tex]

With this you can figure out p^2 ( simply square it) and a relationship between dp and dv. With both of these pieces of information you should be able to replace d3p with an expression that is entirely in terms of dv , v and constants. Your new integration range will simply be over all of the frequencies so from 0 to infinity.
 
  • #3
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The p in the equation above is simply the momentum of photons. There is a relationship between momentum and frequency for photons:

p = [tex]\frac{h\nu}{c}[/tex]

With this you can figure out p^2 ( simply square it) and a relationship between dp and dv. With both of these pieces of information you should be able to replace d3p with an expression that is entirely in terms of dv , v and constants. Your new integration range will simply be over all of the frequencies so from 0 to infinity.

Ah got it. That helps out a ton. Thank you.

I couldn't figure out what that d^3p term meant.
 
  • #4
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So I am pretty sure I stumbled through this and got it mostly right, but I would I really like to understand this for future problems. Here is what I did...at least from what I remember. I already turned in my assignment a like I said...I stumbled through it.

[tex]n(\nu)=\frac{1}{e^{\frac{h \nu}{kT}} -1}[/tex]

[tex]\epsilon=\frac{2}{h^3}\int h \nu \cdot n(\nu)d^3p[/tex]

Combining the two.

[tex]\epsilon=\frac{2}{h^3}\int h \nu \cdot \frac{1}{e^{\frac{h \nu}{kT}} -1} d^3p[/tex]

Part 1 I don't get.

d3p = 4Π2dp Where does this change occur? i.e. how would one know to do this?

Using that and subbing into my integral so far.

[tex]\epsilon=\frac{2}{h^3}\int h \nu \cdot \frac{1}{e^{\frac{h \nu}{kT}} -1} 4\pi p^2 dp[/tex]

Here is where I probably messed up as I don't think I ever actually learned how to do this. My answer did match another integral in book. So that's a plus. :rofl:

[tex]p=\frac{h\nu}{c}[/tex]

Using my masterful, and most likely wrong subbing skills I come up with this.

[tex]p=\frac{h\nu}{c}[/tex]

[tex]dp=\frac{h}{c}d\nu[/tex]

[tex]\frac{c}{h}dp=d\nu[/tex]

Then substitute again....

[tex]\epsilon=\frac{2}{h^3}\int h \nu \cdot \frac{1}{e^{\frac{h \nu}{kT}} -1} 4\pi \left(\frac{h \nu}{c}\right)^2 \frac{c}{h} d\nu[/tex]

This eventually simplifies to...

[tex]\frac{8\pi}{hc}\int \frac{\nu^3}{e^{\frac{h \nu}{kT}} -1} d\nu[/tex]


Now...I am quite certain this isn't right. I really had no idea what to do here so I just tried to make it match the final integral I found in my text using u-sub type rules. I don't think I've ever had to find relationships between dp and dv or any other sort of integral. So you can see my confusion.

Any pointers or tips would be appreciated. I tried searching google for videos/lectures on this subject but couldn't find anything.
 

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