# Maxwell-Boltzmann distribution -- Maxwell's argument

pellman
The attached image shows the text I am following. I get that the 3-D pdf F can only depend on the speed v. I also understand that if f_x , f_y, f_z are the pdfs of the individual components of velocity, then rotational invariance requires them to all be the same function f_x = f_y = f_z = ϕ. What seems to be missing is that F must be separable, that is F must satisfy

F = f_x f_y f_z

Why must F be separable?

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## Answers and Replies

Gold Member
2021 Award
That's the assumption that the three velocity components are stochastically independent. Thus you have the ansatz
$$F(v)=f(v_x) f(v_y) f(v_z).$$
To get the functions ##F## and ##f## we take the logarithmic derivative with respect to ##v_x##, which leads to
$$\frac{1}{F(v)} \partial_{v_x} F(v)=\frac{1}{F(v)} \frac{v_x}{v} F'(v)=\frac{f'(v_x)}{f(v_x)}$$
or
$$\frac{F'(v)}{v F(v)}=\frac{f'(v_x)}{v_x f(v_x)}.$$
Now we define
$$\Phi(v)=\frac{F'(v)}{v F(v)}, \quad \phi(v_x)=\frac{f'(v_x)}{v_x f(v_x)}.$$
Then we have
$$\Phi(v)=\phi(v_x).$$
Then we take the derivative of this equation with, say, ##v_y##. The right-hand side doesn't depend on ##v_y## and thus
$$\frac{v_y}{v} \Phi'(v)=0 \; \Rightarrow \; \Phi'(v)=0 \; \Rightarrow \; \Phi(v)=C=\text{const}.$$
Thus we have
$$\Phi(v)=\frac{F'(v)}{v F(v)}=C \; \Rightarrow \; \frac{F'(v)}{F(v)}=C v \; \Rightarrow \; \ln \left (\frac{F(v)}{F(0)} \right) = \frac{C}{2} v^2 \; \Rightarrow \; F(v)=F(0) \exp \left (\frac{C}{2} v^2 \right).$$
For this to be properly normalizable distribution function, we must have ##C<0##, and thus the distribution is Gaussian. Now writing ##C=-2 \gamma## with ##\gamma>0##
$$F(v)=F(0) \exp (-\gamma \vec{v}^2)$$
and
$$f(v_x)=[F(0)]^{1/3} \exp(-\gamma v_x^2).$$

pellman
Thank you for that. I follow all the steps but the first one. Why do we assume that the three velocity components are stochastically independent?

Gold Member
That's an assumption. You can not derive it otherwise it would not be an assumption. It is quite intuitive though.

vanhees71
pellman
Thanks but I don't see it. There must be some argument in favor of the assumption. Right?

Gold Member
2021 Award
You are right. It's not a convincing argument since it's not correct in the relativistic case though it has the same symmetries argued with here.

The correct argument for the Maxwell-Boltzmann distribution is kinetic theory, which leads to the characterization of the equilibrium phase-space distributions as those of maximum entropy given the constraints imposed by conservation laws. This leads, for the canonical ensemble of the non-interacting gas, to
$$f(t,\vec{x},\vec{p})=\frac{1}{Z} \exp(-\beta H), \quad \beta=\frac{1}{k_{\text{B}} T}, \quad Z=\int_{V} \mathrm{d}^3 x \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \exp(-\beta H).$$

Gold Member
That's an assumption. You can not derive it otherwise it would not be an assumption. It is quite intuitive though.

I am not sure whether this is merely an assumption. In his "Lectures on Kinetic Theory of Gases and Statistical Physics“ [1], Alexander A. Schekochihin writes in section “2.1. Maxwell’s Distribution” (page 16):

Maxwell (1860) argued (or conjectured) that the three components of the velocity vector must be independent random variables.10

10It is possible to prove this for classical ideal gas either from Statistical Mechanics (see x11.10) or by analysing elastic binary collisions (Boltzmann 1995; Chapman & Cowling 1991), but here we will simply assume that this is true.

[1]
Lectures on Kinetic Theory of Gases and Statistical Physics

pellman
There are two strangely mirrored but opposite assumptions here. On the one hand, stochastic independence amounts to saying "If we made a series of measurements to get the distribution of velocity components along one axis, it gives us no information about the distribution along orthogonal axes." However, rotational invariance means that if we know the distribution of the velocity component along any one axis, we know it for ALL axes.

$$f(\vec{x},\vec{p})=\frac{g}{(2 \pi \hbar)^3} \exp(-\beta \sqrt{c^2 \vec{p}^2+m^2 c^4}),$$