1)The expression for $Mean$ $Free$ $Path$ (ie) ,
$$λ=\dfrac{1}{πd²\dfrac{n}{V}}$$ is derived under the assumption that the particles except the one( let's consider it as particle $A$) you're considering are at rest .Now this assumption is wrong for obvious reasons .
So the correction involves finding the relative velocity of any two particles (ie) the relative velocity between particle $A$ and the particle that is going to collide at the moment with $A$ ,let it be $B$ .
Now $\vec v_{BA}$=$\vec v_{B}$-$\vec v_{A}$.
Velocity of B with respect to A is $\vec v_{BA}$ .
Velocity of B and A with respect to the earth frame is $v_{B}$ & $v_{A}$ .
Now initially we have considered $\vec v_B$ to be 0.
So $\vec v_{BA}$ becomes -$\vec v_{A}$.
Now if $\vec v_{B} \ne 0$.
Then $\vec v_{BA}$=$v_b$-$v_a$
Squaring on both sides we get ,
$$\vec v_{BA}^2=(\vec v_{B}-\vec v_{A})(\vec v_{B}-\vec v_{A})$$.
$$\vec v_{BA}^2=\vec v_{A}^2 +\vec v_{B}^2-2\vec v_A.\vec v_B$$.
Now $\vec v_A.\vec v_B$ vansihes because there are a large number of particles and the velocity distribution of particles look like a bell curve ,so for every possible $\vec v_A.\vec v_B$ ,there exists - $\vec v_A.\vec v_B$,therefore on average $\vec v_A.\vec v_B$ has to be zero .
$\implies$ $\vec v_{BA}^2=\vec v_{A}^2 +\vec v_{B}^2-2\vec v_A.\vec v_B$.
becomes ,
$\vec v_{BA}^2=\vec v_{A}^2 +\vec v_{B}^2$.
Now $\vec v_{A(avg)}^2$=$\vec v_{B(avg)}^2=\vec v_{avg}^2$,as they have the same velocity distributions,so they should have same average values .
Therefore ,
$\vec v_{relative}^2$=2$\vec v_{(avg)}^2$.
$\vec v_{relative}$=√2$\vec v_{avg}$.
My questions are
Is my reasoning for $\vec v_{A}.\vec v_{B}$ being 0 correct ?
Is my reasoning for $\vec v_{A(avg)}^2$=$\vec v_{B(avg)}^2=\vec v_{avg}^2$ correct ?
Why does a factor of √2 in average Velocity decrease the mean free path by a factor of √2 ? I can kind of understand it ,but I don't know how to put it in proper words
