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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.

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

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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 $\sigma = \sqrt{\frac{k_B T}{m}}$

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