Why is the mean free time of electrons independent of applied electric field?

Question

In the drude model we can easily get that the mean free time $\tau$ is $$\tau =\frac{m_e}{\rho n e^2}$$

So from above, we can see that the mean free time depends on properties of the material alone (i.e. the resistivity $\rho$ etc) and not on the applied field. We further see this independence from $\vec{E}$ when we use the Drude theory to calculate that $$\frac{d\vec{p}(t)}{dt}=-\frac{\vec{p}(t)}{\tau}+\vec{f}(t)$$ where $\vec{f}(t)$ is the electric force per electron. That is, $\vec{f}(t)$ gives a measure of $\vec{E}$ and from above, we can freely adjust $\vec{E}$ and hence increase or decrease the drift velocity all the while maintaining a constant $\tau$. The fact that $\tau$ remains constant even if we increase $\vec{E}$ does not sit well with me. If we apply a larger field, then the acceleration felt by each particle will increase and hence they should cross the mean free distance in a shorter amount of time. I have 3 ways of potentially fixing this issue although I'm not sure whether they are correct or not and would love some help on this issue.

My first solution would simply be to assume that $\tau$ is in fact dependent on $\vec{E}$ however it is extremely weakly dependent on $\vec{E}$ because at standard temperatures, the average velocity of the electrons is very high ($\pm 500 m/s$). Thus the additional speed gained due to the presence of a field of normal strength is very small in comparison to the speed without the field present and so we can approximate this as a constant $\tau$ provided the field is not absurdly strong. I'm not very confident in this explanation at all though.

The second solution is to observe that when an electron collides with an ion, the directionality of its velocity is random. Those electrons who attain velocities in the direction of the acceleration caused by the $\vec{E}$ field (the direction of electron flow in the field) will have their mean free times reduced as they speed up due to the field. However, for each electron that attains a forward velocity, there is on average one that attains a backward velocity (opposite the direction of electron flow in the field) and these electrons decelerate/ slow down due to the field and hence the time taken to travel their mean free paths increases thus increasing $\tau$ for them. Overall then, we get a constant $\tau$ on average. I am more confident in this solution but still not totally confident.

The final solution I have is to assume that the mean free distance changes in the presence of an electric field. A stronger $\vec{E}$ means faster electrons which could potentially mean that the collision cross section of the ions decreases. Thus the electrons accelerate in the presence of the field and hence travel faster but now they have to travel a larger mean free distance on average since they are less likely to collide. I am also not very confident in this.

Are any of these explanations correct or is it some combination of the above? Any help on this problem would be greatly appreciated as this issue is giving me nightmares!

Answer 1

Your first conjecture is the correct one: the drift speed predicted by the simple equation with friction, given in the OP, is much smaller than the thermal velocity of electrons, and hence the collision times are only weakly dependent on the electric field $\sim\frac{v_F}{v_d}$.

Indeed, the drift equation is actually written for average quantities (average momentum) - it can be obtained by averaging of the kinetic equation, which gives the full distribution function for electron velocities, e.g., something like $$ \propto \exp\left[-\frac{m(v-v_d)^2}{2k_B T}\right] $$

That is, the drift equation should be really written as $$ \frac{d\langle\mathbf{p}\rangle}{dt}=-\frac{\langle\mathbf{p}\rangle}{\tau}+f(t) $$ This averaged nature of the equation with friction becomes problematic when we deal with a non-parabolic band, i.e., when we consider a full band. Say (taking for simplicity 1D case) $$ \epsilon(p)=-\Delta\cos\left(\frac{pa}{\hbar}\right)\rightarrow v(p)=\frac{d\epsilon(p)}{dp}=\frac{\Delta a}{\hbar}\sin\left(\frac{pa}{\hbar}\right) $$ and we have $$ \frac{d\langle p\rangle}{dt}=-\frac{\langle v(p)\rangle}{\tau'}+f(t), $$ since the frequency of collisions is still controlled by velocity, while the rest of the equation follows from Hamiltonian/Heisenberg equations of motion for momentum.

A simple method of dealing with such a situation was employed, e.g., by Esaki and Tsu in their seminal paper, where they essentially reset the momentum to zero after every collision, whereas the collision times are expected to be random, distributed according to the exponential law: $P(t)=\tau^{-1}e^{-t/\tau}$.

See also:
Is current density independent of applied fields for Bloch electrons?
Is the electrical conductivity of a perfectly periodic crystal really infinite?

Answer 2

For a second if we let that $\tau$ depends on another parameter than that of the material itself, then $$\frac{d\mathbf{p}}{dt}=\mathbf{F}-\frac{\mathbf{p}}{\tau}$$ Consider only the electric field, We get

$$\frac{d\mathbf{p}}{dt}=-e\mathbf{E}-\frac{\mathbf{p}}{\tau}$$ $$\frac{d\mathbf{p}}{dt}=0=-e\mathbf{E}-\frac{\mathbf{p}}{\tau}\ \ \text{(in steady state)}$$ $$e\mathbf{E}=-\frac{m\mathbf{v}}{\tau}$$ Further for conductors $$\mathbf{J}=\sigma \mathbf{E}=-n e\mathbf{v}\Rightarrow e\mathbf{E}=-\frac{ne^2}{m\sigma}m\mathbf{v}$$ Comparing the two equation we get $$\boxed{\tau=\frac{m\sigma}{ne^2}}$$

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