# How does one regain the Maxwell-Boltzmann distribution using relative velocity and extensivity?

Let's say I have $$2$$ gases $$A$$ with $$N_A$$ particles and $$B$$ with $$N_B$$ particles in a universe with only one space and one time dimension. Both at the same temperature. Let the distribution of relative velocity between $$2$$ gas particles be:

$$P(k) = \text{Probability of finding a particle where the relative velocity with respect to another particle is k}$$

and

$$n_a = \text{number of pairs with one of gas molecule's of A with relative velocity k}$$

Thus the probability of finding (from gas $$A$$) $$n_a$$ particles with relative velocity $$k$$ is:

$$n_a = N_A^2 P(k)$$

The probability of finding (from gas $$B$$) particles $$n_b$$ with relative velocity $$k$$ is:

$$n_b = N_B^2 P(k)$$

Lets say I combine gases $$A$$ and $$B$$. Then there will be particles of gas $$A$$ which have relative velocity with respect to particles of gas $$B$$ ($$*$$). To account for them, we know:

$$v_{AC} + v_{CB} = v_{AB}$$

Where $$v_{AC}$$ is the velocity of $$C$$ with respect to $$A$$. Hence, if $$v_{AB} = k$$ then the possible relative velocities of are given by:

$$v_{AC} = z$$ and $$v_{CB} = k-z$$

where $$z$$ can be any real number. Hence, to account for ($$*$$) the cross term velocities $$P_2(k)$$ are given by:

$$P_2(k) = \int_{-\infty}^\infty P(k-z)P(z) dz$$

Since both gases are in thermal equilibrium the relative velocity distribution should be an extensive property:

$$(n_A + n_B) =(N_A + N_B)^2 P(k)$$

But we should also be able to use our previous calculations:

$$(n_A + n_B) = N_A^2 P(k) + N_B^2 P(k) + 2 N_A N_B P_2(k)$$

On comparing coefficients to ensure the extensive property:

$$P_2(k) = P(k)$$

Or:

$$P(k) = \int_{-\infty}^\infty P(k-z)P(z) dz$$

## Question

Is this correct? How does one start from this and regain the Maxwell Boltzmann distribution?

• Commented Sep 2, 2021 at 11:52

While this is not fully satisfying. Let us use the Maxwell distribution as an ansatz:

The Maxwellian distribution function for particles moving in only one direction, if the direction is $$x$$

$$f(v_x) dv_x= (\frac{A}{\pi})^{1/2} e^{-A v_x^2}$$

with a paramter $$A$$ and normalization constant $$\frac{A}{\pi}$$. Now if we want a $$2$$'nd particle with velocity $$v_x+k$$:

$$f(v_x+k) dv_x= (\frac{A}{\pi})^{1/2} e^{-A (v_x+k)^2}$$

The probability of seeing $$2$$ particles each the other with relative velocity $$k$$ is given by:

$$P(k) dk= \int_{-\infty}^\infty f(v_x) f(v_x+k) dv_x = (A/\pi) \int_{-\infty}^\infty e^{-A (v_x^2 + (v_x+k)^2)} dv_x = (\frac{A}{2 \pi})^{1/2} e^{-\frac{A}{2} k^2} dk$$

And indeed:

$$\int_{-\infty}^{\infty} P(k-z) P(z) dz = P(k)$$

Two thoughts:

1.

$$n_a^2 = N_A^2 P(k)$$

Where is this from. This is not N choose 2. From $$N_A$$ atoms there are $$\tfrac{1}{2} N_A^2$$ pairs and I guess probabilities are independent. So the number of matches $$n_a$$ is $$n_a=\tfrac{1}{2} P ~N_A^2$$ Not $$n_a = \sqrt{P~N_A^2}$$. Even if $$n_A$$ was the number of molecules participating in a match, that would be $$n_a= 2~ \cdot \tfrac{1}{2} P ~N_A^2$$ not $$n_a^2=...$$

1. Frankly if:

$$E(\vec{v_a}) - E(\vec{v_b}) =0$$

Then, because we have the same temperature, we can just have one gas with $$N=N_A+N_B$$ particles there will never any reason or meaning in saying which gas a particle is from.

• oh ok. Yeah I might have to disappear suddenly we’ll see. I think there are only $\tfrac{1}{2} N^2$ relative v’s. But as mentioned if defn of $n_a$ is pairs, it wouldnt be $n_a^2=...$ for a single k. Are we talking about a single k for that line? Is that the defn of $n_a$? Commented Sep 2, 2021 at 10:28
• "there will never any reason or meaning in saying which gas a particle is from." I don't believe I wrote anything that contradicts that Commented Sep 2, 2021 at 10:45