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In the book of Prigogine, Modern Thermodynamics at page 29, it is given that

$$P(\mathbf{v})\,\text dv_x\,\text dv_y\,\text dv_z=\left(\frac{m}{2\pi k_bT}\right)^{3/2}e^{-mv^2/2k_bT}\,\text dv_y\,\text dv_y\,\text dv_z.$$

The main argument is that since by Boltzmann's principle, $P(E) \propto e^{-E/(k_BT)}$, by direct substitution, $P(\mathbf{v})$ should be given as above.

However, for a particle to have a velocity $\mathbf{v}$, it hast to have a kinetic energy $E = 0.5 mv^2$, and since the probability of latter is $\propto e^{-E/(k_BT)}$, when we have a energy $E$, there are infinitely many possible velocities a particle could have, so when we calculate $P(\mathbf{v})$, we should account for this degeneracy by dividing $e^{-E/(k_BT)}$ by the surface area of the sphere $v^2 = const$.

However, the book does not mention any such degeneracy; why ? what am I missing in here?

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$dv_xdv_ydv_z \ne v^2dv$

where the righthand side $v\equiv ||\vec v||$. So that:

$$ P({\bf v})d^3{\bf v} \ne P(v)dv$$

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  • $\begingroup$ $dv_xdv_ydv_z \not= v^2dv$ $\endgroup$
    – Our
    Commented Mar 17, 2020 at 19:32
  • $\begingroup$ $v^2 dv = v_x^2 dv + v_y^2dv + v_z^2 dv = \frac{dv}{dv_x} v_x^2 + ... = \frac{v_x}{v}dv_x v_x^2 + ....$ $\endgroup$
    – Our
    Commented Mar 17, 2020 at 19:34
  • $\begingroup$ @onurcanbektas thx. An infinitesimal cube is never the same as a shell of infinitesimal thickness...hence the 2 P's are different P's. $\endgroup$
    – JEB
    Commented Mar 18, 2020 at 21:15
  • $\begingroup$ JEB, thanks for your answer, but your "answer" is not answering to my question; $E$ is a function of the speed $v$ and there is a whole surface of velocity vectors having the same speed, so if the probability for $E$ is given by the Boltzman's principle, shouldn't we account for the aforementioned degeneracy when we calculate $P(\vec{v})$? $\endgroup$
    – Our
    Commented Mar 18, 2020 at 21:19
  • $\begingroup$ @onurcanbektas Each degree of freedom is independent, so $E_x$, $E_y$, $E_z$ are all populated independently. $\endgroup$
    – JEB
    Commented Mar 19, 2020 at 0:14

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