Consider the continuity equation for an electron gas: $$\tag{1} \nabla \cdot\left[n(\boldsymbol{r}, t) \frac{\partial}{\partial t} \tilde{\boldsymbol{r}}(\boldsymbol{r}, t)\right]=-\frac{\partial}{\partial t} n(\boldsymbol{r}, t) $$ where $\tilde{\boldsymbol{r}}=\boldsymbol{r}_{0} e^{i \boldsymbol{k} r-i \omega t}$ describes the electron position and $n(\boldsymbol{r}, t) \approx n_{0}+n_1(\boldsymbol{r}, t)$ the density of conduction electrons. My professor wrote in his notes that with the following assumptions: $$\tag{2} \begin{array}{l} \left|n_{1}(\boldsymbol{r}, t)\right| \ll n_{0} \\ \left|\frac{\partial}{\partial t} n_{1}(\boldsymbol{r}, t)\right| \ll\left|\frac{\partial^{2}}{\partial t^{2}} \tilde{\boldsymbol{r}}(\boldsymbol{r}, t)\right| \\ \left|\nabla n_{1}(\boldsymbol{r}, t)\right| \ll\left|\nabla \frac{\partial}{\partial t} \tilde{\boldsymbol{r}}(\boldsymbol{r}, t)\right| \end{array} $$ The following relation can be obtained from the continuity equation: $$\tag{3} \nabla n \approx-n_{0} \nabla\left(\nabla \cdot \boldsymbol{\tilde{r}}\right) $$ However, I really cannot see how that comes about. I have tried to find the connections that show that eq. $(3)$ can be obtained from eq. $(1)$ and $(2)$, but I do not see it.
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$\begingroup$ In the last equation in $(2)$, do you mean $\nabla n_1(\mathbf{r},t)$ instead of $\nabla \cdot n_1(\mathbf{r},t)$? $\endgroup$– secavaraCommented Mar 7, 2021 at 12:28
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$\begingroup$ You might be right. The problem is that my teacher used the symbol $\nabla$ for both gradient and divergence operator. However, I know that in the continuity equation it should be a divergence ($\nabla\cdot$) and I know that on the right side of eq. $(3)$ it should be $-n_{0} \nabla(\nabla \cdot \tilde{\boldsymbol{r}})$ but he used the symbol $\nabla$ in all equations. $\endgroup$– LogiCommented Mar 7, 2021 at 12:33
1 Answer
I'm not fully satisfied with every step in this answer, I think I am missing a couple of details, but hopefully it is still helpful to you.
Starting from (1), using $\nabla\cdot(\phi\mathbf{A})=\phi\nabla\cdot \mathbf{A}+\nabla\phi\cdot \mathbf{A}$, we get
$$ n \nabla\cdot\frac{\partial \bar{r}}{\partial t}+ \nabla n\cdot\frac{\partial \bar{r}}{\partial t}=-\frac{\partial n}{\partial t}. $$
We can then use the first assumption, that $|n_1|\ll n_0$, to replace $n$ in the first term of the equation above with $n_0$. So far, I'm happy it's correct. I think we now need to use the third assumption to argue that we can neglect the second term in the above equation, but as far as I can see it doesn't quite follow. Regardless, if we neglect the second term we have
$$ n_0 \nabla\cdot\frac{\partial \bar{r}}{\partial t}=-\frac{\partial n}{\partial t}. $$
Taking the gradient of this equation we find
$$ n_0 \nabla\left(\nabla\cdot\frac{\partial \bar{r}}{\partial t}\right)=-\nabla\left(\frac{\partial n}{\partial t}\right), $$
which, on changing the order of the differentials leads to
$$ n_0 \frac{\partial}{\partial t}(\nabla(\nabla\cdot\bar{r}))= -\frac{\partial }{\partial t}\nabla n. $$
Integrating both sides w.r.t time, and assuming the constant of integration is zero, we get
$$ \nabla n = -n_0 \nabla(\nabla\cdot\bar{r}), $$
which is what you wanted to show.
I see from the comments that you've already edited the third assumption once, but I'd go back and check the second and third ones again. The second isn't dimensionally consistent, and in the third you take the gradient of a vector quantity.
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$\begingroup$ Thank you for your contribution. I eventually talked with my professor and sorted it out. The confusion came down to the fact that he had been a bit lazy and sloppy with the definitions and notation. $\endgroup$– LogiCommented Mar 15, 2021 at 19:43