# Does Ashcroft and Mermin chapter 13 problem 4 have a misprint?

I've been working through Ashcroft and Mermin to review some aspects of solid state physics and I think problem 4 in chapter 13 might have a misprint. Since there don't seem to be errata available I thought I would post about it here to see if others have encountered this. I'll show my work to make clear my concerns. To start the total derivative in equation (13.8) needs the following term added to it to account for the dependence of $\mu$ on time:

$$\frac{\partial g^{0}}{\partial\mu}\frac{\partial\mu}{\partial t}$$

We can neglect any term involving the gradient of temperature since it is constant, and we have

$$g(t) = g^{0}-\int_{-\infty}^{t}dt'\exp\left[-\frac{t-t'}{\tau}\right]\left(-\frac{\partial f}{\partial \epsilon}\right) \left[e\vec{E}\cdot\vec{v}+\vec{v}\cdot\nabla\mu + \frac{\partial\mu}{\partial t} \right]$$

I've assumed an energy-independent relaxation time, which seems reasonable in light of the result quoted in A&M. Now we put everything in terms of the Fourier modes given in the book, leading to (in what follows $\vec{E}$ and $\delta\mu$ are assumed to be in Fourier space):

$$g^{0}-\mathcal{R}\left[\int_{-\infty}^{t}dt'\left(-\frac{\partial f}{\partial\epsilon}\right)\exp\left[-\frac{t-t'}{\tau} +i\left(\vec{q}\cdot\vec{r}-\omega t\right) \right] \left(e\vec{v}\cdot\vec{E}+i\left(\vec{v}\cdot\vec{q}-\omega\right)\delta\mu\right)\right]$$

where $\mathcal{R}$ indicates the real part of the subsequent expression. The time dependence is completely contained in the exponential, so the integral is straightforward and I find

$$g^{0} -\mathcal{R}\left[\frac{\exp\left[i\left(\vec{q}\cdot\vec{r}-\omega t\right)\right]}{1/\tau -i\omega}\left(-\frac{\partial f}{\partial\epsilon}\right)\left(e\vec{E}\cdot\vec{v}-i\left(\omega-\vec{v}\cdot\vec{q}\right)\delta\mu\right) \right]$$

This looks something like what Ashcroft and Mermin have, but is sufficiently different that I can't tell if I am making an error or A&M has a misprint. Does anybody know? I've discussed this with a few other physicists and no one has been able to identify an error on my part yet.

No, there is no misprint in A&M. (1) The $r$ vector in the exp factor is $r(t')$, which needs to be differentiated as well. That will give $(1/\tau) - i(\omega - q \dot v)$ in the denominator. (2) $g_0$ is not $f(E(k))$. Note that chemical potential is now $\mu + d\mu(r,t)$. so you need to expand $g_0$ to linear order in $d\mu(r,t)$ which gives another term containing $d\mu(q,w)$.