# Shifting integration variable when determing population densities

1. Mar 26, 2014

### "Don't panic!"

Hi,

I'm hoping someone can enlighten me on this as I'm a little bit fuzzy on the reasoning:

Say I have a space-time dependent field $B_{a}$ that interacts with fermions such that it affects their energy dispersion. It appears in the energies in the form

$$E\sim\sqrt{\left(\vec{p}+\vec{B}\right)-m^{2}}+B_{0}$$

Why is it, that when I then calculate the number density of fermions in such a scenario, i.e.

$$n\sim\int^{+\infty}_{-\infty}\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{\exp{\left(E/k_{_{B}}T\right)}+1}$$

(where in this case the chemical potential is negligible) that I can only shift the integration variable, such that $\vec{p}\rightarrow \vec{p}+\vec{B}$ (thus "absorbing" the 3-vector components of $B_{a}$), if I consider $B_{a}$ to be constant?

2. Mar 26, 2014

### "Don't panic!"

Apologies for the spelling mistake in the title of the thread by the way, should be "determining" , but don't know how to retroactively edit it!

3. Mar 26, 2014

### dauto

What do you think would happen to d3p if B is not constant?

4. Mar 26, 2014

### "Don't panic!"

Would it be that it becomes time dependent and thus coupled to the fluctuations in B over time?

5. Mar 26, 2014

### "Don't panic!"

or more explicitly, that you would also introduce an additional integral over $d^{3}B$?

6. Mar 26, 2014

### dauto

Slow down with the questions and answer my question in post #3

Last edited: Mar 26, 2014
7. Mar 26, 2014

### "Don't panic!"

sorry, they were my attempts at a possible answer (shouldn't have included the question marks)!

I assume that you would have $d^{3}p\rightarrow d^{3}\left(p+B\right)=d^{3}p'$ and so, as B is not constant, one could not talk of set momentum states for the fermions as they would fluctuate in time depending on the fluctuations in B.

8. Mar 26, 2014

### dauto

Correct. If you have a explicit form for B then you might attempt a solution. You can't go much further with the general expression, I don't think

9. Mar 26, 2014

### "Don't panic!"

ok, that's cleared things up a bit. Thanks for your time.