1
$\begingroup$

I am reading this article and I am having trouble understanding a calculation there.

In it, this following equation is obtained:

$$ \frac{\partial}{\partial t} \int p_i n d \tau + \int p_i \left( \frac{\partial n}{\partial \textbf{x}} \frac{\partial \epsilon}{\partial \textbf{p}} - \frac{\partial n}{\partial \textbf{p}} \frac{\partial \epsilon}{\partial \textbf{x}}\right) d \tau = 0 .$$

where $\epsilon$ and p are the energy and momentum of the quasiparticles, and,

$$ d \tau = g \frac{d^3 p}{(2 \pi)^3} .$$

with g the degeneracy of each state.

This equation was next rewritten, by algebraic manipulation and integration by parts, in the form,

$$\frac{\partial}{\partial t} \int p_i n d \tau + \frac{\partial}{\partial \textbf{x}} \int p_i \frac{\partial \epsilon}{\partial \textbf{p}} n d \tau + \frac{\partial}{\partial x_i}\int n \epsilon d \tau - \int \epsilon \frac{\partial n }{\partial x_i} d \tau =0 .$$

My question is how was this second form obtained? I can't follow the calculation.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

It's just a plain double integration by parts. Recall the bold vectors dot each other, and since there is no integration over x, its divergence/gradient term survives. $$\left( \frac{\partial n}{\partial \textbf{x}} \frac{\partial \epsilon}{\partial \textbf{p}} - \frac{\partial n}{\partial \textbf{p}} \frac{\partial \epsilon}{\partial \textbf{x}}\right) =\frac{\partial }{\partial \textbf{x}}\left( n \frac{\partial \epsilon}{\partial \textbf{p}}\right ) - \frac{\partial }{\partial \textbf{p}}\left ( n \frac{\partial \epsilon}{\partial \textbf{x}}\right) . $$

Your second starting integral, thus, reduces to $$ \int\!\! d\tau ~~ p_i \left ( \frac{\partial }{\partial \textbf{x}}\left( n \frac{\partial \epsilon}{\partial \textbf{p}}\right ) - \frac{\partial }{\partial \textbf{p}}\left ( n \frac{\partial \epsilon}{\partial \textbf{x}}\right) \right ) \\ = \frac{\partial }{\partial \textbf{x}}\int\!\! d\tau ~~ p_i n \frac{\partial \epsilon}{\partial \textbf{p}} - \int\!\! d\tau ~~ p_i \frac{\partial }{\partial \textbf{p}}\left ( n \frac{\partial \epsilon}{\partial \textbf{x}}\right) \\ = \frac{\partial }{\partial \textbf{x}}\int\!\! d\tau ~~ p_i n \frac{\partial \epsilon}{\partial \textbf{p}} + \int\!\! d\tau ~~ \frac{\partial p_i}{\partial \textbf{p}} n\frac{\partial \epsilon}{\partial \textbf{x}}\\ = \frac{\partial }{\partial \textbf{x}}\int\!\! d\tau ~~ p_i n \frac{\partial \epsilon}{\partial \textbf{p}} + \int\!\! d\tau ~~ n\frac{\partial \epsilon}{\partial x_i} \\ = \frac{\partial }{\partial \textbf{x}}\int\!\! d\tau ~~ p_i n \frac{\partial \epsilon}{\partial \textbf{p}} + \frac{\partial}{\partial x_i}\int n \epsilon d \tau - \int \epsilon \frac{\partial n }{\partial x_i} d \tau . $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.