# Derive the drift velocity from drifting Maxwellian distribution

Having a drifting Maxwellian (Maxwell-Boltzmann) distribution: $$f(\vec v) = n \left(\dfrac{m}{2\pi k_B T}\right)^\frac{3}{2}\exp\left({-\dfrac{m[(v_x-a)^2+v_y^2 + v_z^2]}{2k_BT}}\right)$$ where $$a$$ is the drift velocity, is it possible to derive this drift from the moment of the distribution $$\langle \vec v \rangle = 1/n \int_{- \infty}^\infty \vec v f(\vec v) d^3v$$

I assume that both $$\langle v_y \rangle=0$$ and $$\langle v_z \rangle=0$$ (from the Maxwellian being an even function in these directions), but does $$\langle v_x \rangle=a$$? If so, please help me derive it, I couldn't do it myself neither could I find the anwer in textbooks.

• Yes, the even integrals (y and z directions) will be zero... Commented Aug 9, 2023 at 20:36

According to your formulas, $$\langle v_x \rangle = \left( \frac{m}{2\pi k_B T} \right)^{3/2} \int_{-\infty}^\infty v_x e^{-\frac{m[(v_x - a)^2 + v_y^2 +v_z^2]}{2k_B T}} d v_x d v_y dv_z$$
Shifting $$v_x$$ by defining $$v'_x = v_x - a$$ we have
$$\langle v_x \rangle = \left(\frac{m}{2\pi k_B T}\right)^{3/2} \int_{-\infty}^\infty (v'_x +a) e^{-\frac{m[v_x'^2 + v_y^2 +v_z^2]}{2k_B T}} d v'_x d v_y dv_z = a \left( \left(\frac{m}{2\pi k_B T}\right)^{3/2} \int_{-\infty}^\infty e^{-\frac{m[v_x'^2 + v_y^2 +v_z^2]}{2k_B T}} d v'_x d v_y dv_z\right) = a$$ I used that the $$v'_x e^{-(...)}$$ integral vanishes because it is odd in $$v'_x$$, and the remaining $$a e^{-(...)}$$ integral simplifies with the prefactor to give simply $$a$$ as expected.
This result just follows from identifying this distribution with a gaussian distribution of mean $$(a,0,0)$$ and standard deviation $$\sigma = \sqrt{\frac{k_B T}{m}}$$