# Maxwell's velocity distribution does not account for degeneracy in direction

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?

$$dv_xdv_ydv_z \ne v^2dv$$

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

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

• $dv_xdv_ydv_z \not= v^2dv$
– Our
Mar 17, 2020 at 19:32
• $v^2 dv = v_x^2 dv + v_y^2dv + v_z^2 dv = \frac{dv}{dv_x} v_x^2 + ... = \frac{v_x}{v}dv_x v_x^2 + ....$
– Our
Mar 17, 2020 at 19:34
• @onurcanbektas thx. An infinitesimal cube is never the same as a shell of infinitesimal thickness...hence the 2 P's are different P's.
– JEB
Mar 18, 2020 at 21:15
• 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 $P(\vec{v})$?
– Our
Mar 18, 2020 at 21:19
• @onurcanbektas Each degree of freedom is independent, so $E_x$, $E_y$, $E_z$ are all populated independently.
– JEB
Mar 19, 2020 at 0:14