Maxwell-Boltzmann velocity distribution graph

Question

The graph looks like a bell curve, but why is the bell curve shifted? The actual gaussian distribution is symmetric about $x$ axis but MB distribution curve is not symmetric. Why is that?

Answer 1

If you consider a MB distribution along one axis, say for the velocity along the $x$-direction you indeed get a gaussian. Because, by indicating the probability of measuring $v_x$ as $P(v_x)$, you have $$P(v_x)=P(-v_x)$$ you also get that the distribution is symmetric.

However, if instead of $v_x$ you are interested in the absolute value of the velocity $P(|v_x|)$ or, in 3D, $$P(v=\sqrt{v_x^2+v_y^2+v_z^2})$$ then of course this distribution can not be symmetric becuase $v>0$ always. This is why the MB distribution for the magnitude of the speed is not symmetric, because it only deals with the magnitude of the speed i.e. the positive part of the velocity.

On the other hand, the MB distribution for individual components (i.e., $v_x$ or $v_y$ or $v_z$) is a symmetric gaussian.

In mathematical terms, for each component $i$ you get

$$P(v_i) \propto e^{-v_i^2/2 \sigma^2}$$ (symmetric gaussian, between $-\infty$ and $\infty$), whereas for the magnitude $v$ you get

$$P(v)\propto v^2 e^{-v^2/2T}$$ i.e. the "shifted" bell curve, defined between $0$ and $\infty$.

Answer 2

The PDF:

$$f(v)dv_xdv_ydv_z$$

Isn't the same as:

$$f(v)dv$$

Measured results are the probability of a state associated with a particular velocity magnitude. You need a factor of $4\pi v^2$ in front.

One way of thinking about this is to imagine a probability "sphere", since the probability is a function of each of the $v_x$, $v_x$, and $v_z$ coordinates. The probability of a state with a particular velocity magnitude will be proportional to the surface area of that sphere, and not just any of the individual components, since,

$$v=\sqrt{v_x^2+v_y^2+v_z^2}$$