Relaxation time and average electron velocity in Drude model

[Sunny Singh]

If τ is the relaxation time, τ means, on average the time between two collisions for an electron moving under a constant electric field inside a metal. Now according to the assumptions of drude model, the electron acquires an additional velocity of $$\frac{-eEt}{m}$$where t is the time elapsed since its last collision. The average velocity of the electron between time t=0 and t=τ (since an average electron spends time τ under acceleration) must be $$\frac{-eE}{m}\frac{1}{τ} \int_{0}^{τ} t dt=\frac{-eEτ}{2m}$$ but in all the books, they simply write <t>=τ and hence the average electron velocity becomes $$\frac{-eEτ}{m}$$I don't understand how they can write this if they really mean τ to be the time it takes, on average for an electron to collide with another ion. Doesn't this make their definition of average velocity to be maximum velocity acquired by electrons? And if so how can this maximum velocity of an electron be used to find the current density? I think I'm confused here at something very basic about averages. Please explain me the point where I'm messing up the logic.

 

[vanhees71]

The Drude model assumes non-interacting electrons moving in the positive background with some friction. The EoM thus is
$$m \dot{\vec{v}}=q \vec{E} -m \gamma \vec{v}.$$
In the DC case you have
$$\vec{v}=\frac{q}{m \gamma} \vec{E}.$$
Further
$$\vec{j} = q n \vec{v}=\frac{q^2 n}{m \gamma} \vec{E},$$
where $n$ is the conduction-electron-number density. From this the el. conductivity is
$$\sigma=\frac{q^3 n}{m \gamma}.$$
The relaxation time is $\tau=1/\gamma$. It's the average free time. The distribution of the free time is
$$P(t)=N \exp(-\gamma t).$$
Normalization
$$\int_0^{\infty} \mathrm{d} t P(t)=\frac{N}{\gamma}=1 \;\Rightarrow \; N=\gamma.$$
From this
$$\langle t \rangle = \int_0^{\infty} \mathrm{d} t P(t)=-\gamma \partial_{\gamma} \int_0^{\infty} \exp(-\gamma t)=-\gamma \partial_{\gamma} \frac{1}{\gamma}=\frac{1}{\gamma}=\tau.$$