2
$\begingroup$

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?

$\endgroup$
1

2 Answers 2

0
$\begingroup$

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:

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
    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 $\endgroup$ Sep 2, 2021 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.