How Do You Convert d3p to a Function of Frequency Nu in Spherical Coordinates?

[erok81]

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$$\pi$$p²dp.

Homework Equations

$$n(\nu)=\frac{1}{e^{\frac{h \nu}{kT}} -1}$$

$$\epsilon=\frac{2}{h^3}\int h \nu \cdot n(\nu)d^3p$$

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$$\pi$$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.

 

[shreepy]

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

p = $$\frac{h\nu}{c}$$

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.

 

[erok81]

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

p = $$\frac{h\nu}{c}$$

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.

 

[erok81]

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.

$$n(\nu)=\frac{1}{e^{\frac{h \nu}{kT}} -1}$$

$$\epsilon=\frac{2}{h^3}\int h \nu \cdot n(\nu)d^3p$$

Combining the two.

$$\epsilon=\frac{2}{h^3}\int h \nu \cdot \frac{1}{e^{\frac{h \nu}{kT}} -1} d^3p$$

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.

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

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.

$$p=\frac{h\nu}{c}$$

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

$$p=\frac{h\nu}{c}$$

$$dp=\frac{h}{c}d\nu$$

$$\frac{c}{h}dp=d\nu$$

Then substitute again...

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

This eventually simplifies to...

$$\frac{8\pi}{hc}\int \frac{\nu^3}{e^{\frac{h \nu}{kT}} -1} d\nu$$


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.