# Statistical independence of $x,y,z$ dimensions for Maxwell velocity distribution function

I have been looking into the derivation of the Maxwell speed distribution function as for instance given in https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution#Distribution_for_the_speed or elsewhere.

In all these treatments it is claimed that the velocity distribution functions in the $$v_x, v_y, v_z$$ directions are statistically independent of each other, which justifies writing the combined probability as the product of the individual probabilities, i.e.

$$f_\mathbf{v} \left(v_x, v_y, v_z\right) = f_v (v_x)f_v (v_y)f_v (v_z)$$

where in this case

$$f_v (v_i) = \sqrt{\frac{m}{2 \pi kT}} \exp \left(-\frac{mv_i^2}{2kT}\right)$$.

However, assume for instance a particular particle has a velocity along the $$x$$-axis, so that its speed

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

In this case it follows obviously that the probability $$f_v (v_y) = f_v (v_z) =0$$ for any values $$v_y, v_z >0$$, so how can one claim statistical independence here when clearly the value of $$v_x$$ limits the possible values of $$v_y$$ and $$v_z$$ depending on the overall speed/energy of the particle?

So generally speaking, my point is that the functional relationship

$$f_\mathbf{v} \left(v_x, v_y, v_z\right) = f_v (v_x)f_v (v_y)f_v (v_y)$$

(which is used for instance to derive the Maxwell-Boltzmann speed distribution) implies that the velocity components are independent variables. However, when reducing the problem to just one independent variable by using the speed $$v$$ instead of $$v_x, v_y, v_z$$, this is not the case anymore as each component can be expressed in terms of the other two via $$v$$.

• Why are $f(v_y)=f(v_z)=0$? Commented Mar 29 at 18:43
• @KyleKanos $f_v (v_y) = f_v (v_z) =0$ only for $v_y, v_z >0$ (because $v_x =v$, the particle can not have any non-zero velocity components in the $y$ and $z$ directions) Commented Mar 29 at 18:50
• How can $v_y$ or $v_z$ have non-zero velocity if $\mathbf{v}=v_x\mathrm{e}_x$? If they are non-zero then you have an ill-defined starting point. Commented Mar 29 at 19:05
• @KyleKanos I said they don't have any non-zero velocities. Commented Mar 29 at 19:09
• According to your reasoning, if you know that a particle has a specific speed, its distribution function ceases to be Maxwell-Boltzmann. This is not the way probability distributions work. By the way, math says that $\sqrt{a^2}=|a|$, not $a$. Commented Mar 30 at 6:36

1. the assertion that $$v_x = v$$ doesn't mean much the other velocities, other than they are unlikely to be large.
2. There is no preferred direction and physics doesn't care about your coordinates. Try to write a probability distribution with $$xy$$, $$xz$$ or $$yz$$ correlations that is invariant under rotations. I suspect it would be hard, or, you'd have to confine them to a spherical surface in velocity space of radius $$v$$, and that would just be weird. Very low entropy.
• My point was that the functional relationship $f_\mathbf{v} \left(v_x, v_y, v_z\right) = f_v (v_x)f_v (v_y)f_v (v_y)$ (which is used to derive the Maxwell-Boltzmann speed distribution) implies that the velocity components are independent variables. However, when reducing the problem to just one independent variable by using the speed $v$ instead of $v_x, v_y, v_z$, this is not the case anymore as each component can be expressed in terms of the other two via $v$. Commented Mar 30 at 16:11
• Right, it goes as $f(|\vec v|)u(\cos\theta)2\pi u(\phi)$ where $u(x)$ is uniform on $[0, 1]$