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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 $v$.

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  • $\begingroup$ Why are $f(v_y)=f(v_z)=0$? $\endgroup$
    – Kyle Kanos
    Commented Mar 29 at 18:43
  • $\begingroup$ @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) $\endgroup$
    – Thomas
    Commented Mar 29 at 18:50
  • $\begingroup$ 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. $\endgroup$
    – Kyle Kanos
    Commented Mar 29 at 19:05
  • $\begingroup$ @KyleKanos I said they don't have any non-zero velocities. $\endgroup$
    – Thomas
    Commented Mar 29 at 19:09
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    $\begingroup$ 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 $a$. $\endgroup$ Commented Mar 30 at 6:36

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  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.

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  • $\begingroup$ 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 $v$. $\endgroup$
    – Thomas
    Commented Mar 30 at 16:11
  • $\begingroup$ Right, it goes as $f(|\vec v|)u(\cos\theta)2\pi u(\phi)$ where $u(x)$ is uniform on $[0, 1]$ $\endgroup$
    – JEB
    Commented Mar 30 at 17:00

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