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:

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


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

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:

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


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


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


2 Answers 2


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:


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

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.

  • $\begingroup$ 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 $n_a$? $\endgroup$
    – Al Brown
    Commented Sep 2, 2021 at 10:28
  • 1
    $\begingroup$ "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 $\endgroup$ Commented Sep 2, 2021 at 10:45

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