# Maxwell-Boltzmann distribution - find error in derivation

I have a derivation of the Maxwell-Boltzmann distribution:

Consider a gas consisting of only one type of molecules, which is in an equilibrium with a heat reservoire of temperature T.

Since all three room directions $x,y,z$ are equivalent, the probabilty distribution has to be the same for every room direction. The system is also invariant under arbitrary reflections. Thus the probability distribution only depends on the absolute value of the velocity. Hence $p_i(v_i)=f(v_i^2)$, $i=x,y,z$ and a yet to be determined function $f$.

Since the velocities in each room direction are independet, we have $p(v_x,v_y,v_z)=f(v_x^2)f(v_y^2)f(v_z^2)$ and since the system is invariant under rotations, the velocity distributions have to be independent, too. Now we can rotate the velocity vector onto the $z$-axis and obtain $f(v_x^2)f(v_y^2)f(v_z^2)=f(0)f(0)f(v_x^2+v_y^2+v_z^2)$. But this property only holds for an exponential function, $f(x)=Ae^{-\alpha x}$, which yields the parametric form of the Maxwell-distribution $p(v_i)=Ae^{-\alpha v_i^2}$.

This derivation does not make direct use of the Hamiltonian of the system and I am supposed to find out at what part it is faulty. I.e. the derivation uses an assumption that does not hold in general.

Now I think that we cannot assume that the physical system is invariant under arbitrary reflections as for example the spins of fermions are not invariant under reflection, but I am not sure with this.

Can anyone help me with this? Thanks!

• How to understand your words "the physical system is invariant under arbitrary reflections" ? Does it mean that the direction x is equivalent with -x, i.e. the distribution you seek depends on (v_x)^2 but not on v_x? – Sofia Nov 14 '14 at 17:36
• (continuation) And why do you find a problem in that the Hamiltonian wasn't used in the considerations. Now, what I see is that the particular assumption you do, i.e. which is not generally true, is isotropy, i.e. all the directions in space are equivalent. If there were some field in the room to which the particles were sensitive, the isotropy were gone. – Sofia Nov 14 '14 at 17:40
• With fermions the considerations are totally different, they obey the exclusion principle, s.t. the statistics is different. You can find a simple description in Wikipedia. – Sofia Nov 14 '14 at 17:49
• Yes, I think that is what is meant by arbitrary reflections. But whether my system is isotropic or not I can determine from the Hamiltonian? – dinosaur Nov 14 '14 at 17:54
• Of course, isotropy you see in the Hamiltonian, but I am skeptical about the need of it here. If in the Hamiltonian you make a rotation around an axis that passes through the origin, and the Hamiltonian doesn't change, it is isotropic. For instance, a spherically symmetrical Hamiltonian is isotropic. – Sofia Nov 14 '14 at 18:04

Since the velocities in each room direction are independent, we have $p(v_x,v_y,v_z)=f(v_x^2)f(v_y^2)f(v_z^2)$
In other words, it is assumed that the distribution function of three variables $v_x,v_y,v_z$ factorizes into product of three distribution functions, each being a function of one argument only.