Why is the resistivity $\rho_{xy}$ independent of scattering time in the Drude model for the Hall effect

Question

I was reading about the Hall Effect, and how it can be explained through the Drude model of conductivity. I was looking at the 2D model, as I'm mainly interested in 2 dimensional electron gasses. You take electrons in a magnetic and electric field, use the Lorentz force, and add in some scattering time $\tau$ heuristically in order to find current densities. These are then related to the conductivity tensor, which by inversion gives you the resistivity tensor.

This has two unique components: $\rho_{xx} = \frac{m^*}{n_se^2\tau}$ and $\rho_{xy} = \frac{B}{e n_s}$. Here, $m^*$ is the effective mass and $n_s$ the electron density. What strikes me is that the $\rho_{xy}$ component is independent of $\tau$. I don't really understand this, because isn't it the scattering mechanism that plays a crucial role, by giving a mean drift velocity? The cyclotron orbits caused by the magnetic field aren't always completed due to $\tau$, so shouldn't this play a role in the resistivity?

Answer 1

$\rho_{xy}$ is related to the ratio of the $x$ component of current density to the voltage in the $y$ direction, when the $y$ component of the current is $0$. Decreasing the scattering time will increase the $x$ component of the drift velocity and so will increase mean force on a charge carrier, leading to a larger Hall voltage. However the current is also proportional to the drift velocity, so this is already taken into account in Hall resistivity. The $y$ component of the drift velocity is fixed at 0, an so is clearly not affected by the scattering time.