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The Maxwell-Boltzmann distribution for the kinetic energies of particles is given as ( https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution#Distribution_for_the_energy )

$$f(E) = 2 \sqrt{\frac{E}{\pi}} \left[\frac{1}{kT}\right]^\frac{3}{2} \exp\left(-\frac{E}{kT} \right)$$

This implies obviously $f(0)=0$. However, when I consider the detailed balance condition for a collision event (which should hold for all collision events in a closed system in equilibrium) this does not appear to be valid:

if we consider the collision rate of two particles with energy $E1$, $E2$ to energies $E1'$, $E2'$ this must be equal to the inverse collision rate $E1'$, $E2'$ -> $E1$, $E2$ in equilibrium as the distribution $f$ can not change ( https://en.wikipedia.org/wiki/Detailed_balance#Microscopic_background ). The collision rate is generally given by the product of the densities times the collision cross section times the relative speed of the two particles. Now collision cross sections are invariant under time reversal, so we can ignore these here. This means we are left with the following detailed balance equation (using just the distribution functions i.e. the normalized densities)

$$f(E_1')\cdot f(E_2')\cdot \overline{|\vec{v_1'}-\vec{v_2'}|} = f(E_1)\cdot f(E_2)\cdot \overline{|\vec{v_1}-\vec{v_2}|} $$

where we have averaged over the relative speed of the two particles corresponding to the (isotropic) orientation of the velocity vectors. However, for elastic collisions the relative speed is unchanged by the collision i.e.

$$|\vec{v_1'}-\vec{v_2'}| = |\vec{v_1}-\vec{v_2}|$$

so we can simplify the above equation to

$$f(E_1')\cdot f(E_2')= f(E_1)\cdot f(E_2) $$

Now for elastic collisions we also have because of energy conservation

$$E_1'+E_2' = E_1+E_2$$

If we now consider the special case $E_2' =0$, the detailed balance equation thus becomes

$$f(E_1+E_2)\cdot f(0) = f(E_1)\cdot f(E_2) $$

and therefore

$$f(0) = \frac{f(E_1)\cdot f(E_2)}{f(E_1+E_2)}$$

However, as $E_1$ and $E_2$ are arbitrary, and assuming $f(E)$ is not identically $0$, all terms on the right hand side of this equation must be $>0$, from which follows

$$f(0)>0$$

which obviously contradicts the well know formula for the Maxwell-Boltzmann distribution of kinetic energies.

How can we explain/resolve this contradiction?

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    $\begingroup$ This calculation seems to be dropping way too many factors. What happens in a collision depends on both the momenta (not just the energies) and the interaction potential. At the very least, there is a measure factor converting $d^3p$ to $dE$ which is dropped, which ought to set $f(0) = 0$. $\endgroup$
    – knzhou
    Commented Apr 21 at 18:02
  • $\begingroup$ @knzhou I only dropped the factors that can be dropped because of the symmetries involved for elastic collisions under equilibrium conditions. My equation $$f(E_1')\cdot f(E_2')= f(E_1)\cdot f(E_2) $$ is indeed identical to the well known detailed balance condition $f_1'\cdot f_2'-f_1\cdot f_2 =0$ that comes out of the Boltzmann equation under equilibrium conditions, and the solution for this contradicts $f(0)=0$ as well (see onlinelibrary.wiley.com/doi/pdf/10.1002/0471724254.app2 ) $\endgroup$
    – Thomas
    Commented Apr 21 at 18:38
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    $\begingroup$ It's not the same thing, because the usual statement of detailed balance is in momentum-space (or equivalently, in this case, in velocity-space). The energy-space distribution is related to the momentum-space distribution by a Jacobian factor, which just vanishes at $E = 0$. To say it another way, your definition of $f$ is related to the $f$ in the source you cite by a scaling factor, which turns your equation into $0 = 0$. $\endgroup$
    – knzhou
    Commented Apr 21 at 18:46
  • $\begingroup$ @knzhou The scaling factor in the reference I linked to above ist just a constant factor, not velocity dependent. A distribution function $f\propto v^2 \cdot exp(-v^2)$ would not fulfil the detailed equilibrium condition as in general $$v_1'^2\cdot v_2'^2\cdot e^{-(v_1'^2+v_2'^2)} \neq v_1^2\cdot v_2^2\cdot e^{-(v_1^2+v_2^2)} $$ At this point it is only about energies. The momentum terms have dropped out earlier already due to the symmetries of the problem under (isotropic) equilibrium conditions. $\endgroup$
    – Thomas
    Commented Apr 21 at 19:56
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    $\begingroup$ Again, you are mixing up the speed distribution $f_v(v)$ with the energy distribution $f_E(E)$. They aren't the same function; they're related by $f_v(v) = f_E(E(v)) \, (dE/dv)$. The speed distribution is indeed zero at $v = 0$. If you just write "$f$" for a generic distribution it's easy to mix the two up. $\endgroup$
    – knzhou
    Commented Apr 21 at 20:04

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