# Speed distribution in 1 dimension

In 3D, the maxwell velocity distribution is:

$$f = \left(\frac{\alpha}{\pi} \right)^{\frac{3}{2}} e^{-\alpha v^2} d^3 \vec v$$

To get the speed distribution in 3D, we simply expand $d^3\vec v = 4\pi v^2 dv$

Thus in 3D, the maxwell speed distribution is:

$$w = 4\pi^2 \left(\frac{\alpha}{\pi} \right)^{\frac{3}{2}}v^2 \space e^{-\alpha v^2} d v$$

In 2D, the maxwell velocity distribution is:

$$f = \left(\frac{\alpha}{\pi} \right) e^{-\alpha v^2} d^2 \vec v$$

To get the speed distribution in 2D, we simply expand $d^2\vec v = 2\pi v\space dv$

Thus in 2D, the maxwell speed distribution is:

$$f = 2\pi\left(\frac{\alpha}{\pi} \right) v \space e^{-\alpha v^2} d \vec v$$

In 1D, the maxwell velocity distribution is:

$$f = \left(\frac{\alpha}{\pi} \right)^{\frac{1}{2}} e^{-\alpha v^2} d \vec v$$

Following the same line of thought, how do I get the 1D speed distribution?

You should simply multiply by two, to get $$f =2 \left(\frac{\alpha}{\pi} \right)^{\frac{1}{2}} e^{-\alpha v^2} \text d v.$$ This is because your integral over speed will now be from zero to infinity, and you need to 'fold over' the integral from minus infinity to zero. Otherwise, there are no further geometrical factors - the integral $\int\text d\vec v=\int_{-\infty}^\infty \text dv$ is alreadyin the form you need it, and you just need to figure out what to do with negative velocities.