# Maxwell-Boltzmann distribution -- Maxwell's argument

• pellman
In summary, the conversation discusses the assumption of stochastic independence for the three velocity components in the Maxwell-Boltzmann distribution. This assumption is intuitively reasonable but not necessarily true for all cases, as it is not valid in the relativistic case. The correct derivation of the distribution is through Boltzmann's kinetic/transport equation and his H theorem.

#### 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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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
$$f(v_x)=[F(0)]^{1/3} \exp(-\gamma v_x^2).$$

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?

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

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

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
$$f(t,\vec{x},\vec{p})=\frac{1}{Z} \exp(-\beta H), \quad \beta=\frac{1}{k_{\text{B}} T}, \quad Z=\int_{V} \mathrm{d}^3 x \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \exp(-\beta H).$$

dRic2 said:
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

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 anyone axis, we know it for ALL axes.

Well, it's an assumption which by chance is true for the non-relativistic gas in classical approximation. The correct answer is Boltzmann's derivation from his kinetic/transport equation and his H theorem, which identifies the state of maximum entropy (with entropy defined as the Shannon-Jaynes entropy of information theory, as anticipated by Landauer and Szilard before) as the equilibrium state.

Already for the relativistic ideal gas it's not true but, in the rest frame of the fluid, reads
$$f(\vec{x},\vec{p})=\frac{g}{(2 \pi \hbar)^3} \exp(-\beta \sqrt{c^2 \vec{p}^2+m^2 c^4}),$$
which does not factorize in the manner assumed by Maxwell.

pellman

## 1. What is the Maxwell-Boltzmann distribution?

The Maxwell-Boltzmann distribution is a statistical distribution that describes the distribution of particle speeds in a gas at a given temperature. It is named after James Clerk Maxwell and Ludwig Boltzmann, who developed the theory of kinetic theory of gases in the 19th century.

## 2. How is the Maxwell-Boltzmann distribution derived?

The Maxwell-Boltzmann distribution is derived from the kinetic theory of gases, which states that the average kinetic energy of gas particles is directly proportional to the temperature of the gas. The distribution is then derived using statistical mechanics principles, taking into account the different possible velocities and the probability of each velocity occurring.

## 3. What is the significance of the Maxwell-Boltzmann distribution?

The Maxwell-Boltzmann distribution is important in understanding the behavior of gases, as it allows us to predict the distribution of particle speeds in a gas at a given temperature. It also helps us understand other properties of gases, such as pressure and diffusion.

## 4. How does the Maxwell-Boltzmann distribution relate to the ideal gas law?

The Maxwell-Boltzmann distribution is closely related to the ideal gas law, which describes the relationship between pressure, volume, temperature, and the number of moles of an ideal gas. The distribution helps us understand the behavior of individual gas particles, while the ideal gas law describes the overall behavior of a gas as a whole.

## 5. Is the Maxwell-Boltzmann distribution applicable to all gases?

The Maxwell-Boltzmann distribution is applicable to ideal gases, which are gases that follow the kinetic theory of gases and do not interact with each other. Real gases, on the other hand, may deviate from the distribution at high pressures or low temperatures due to intermolecular interactions. However, the distribution is still a useful approximation for real gases in many cases.