 #1
rogdal
 10
 2
 Homework Statement:

Given a box of dimensions ##V=L \times L \times L## in the ##xyz## space which has ##N \gg1## particles of a certain gas, the velocity of a random particle is given by the MaxwellBoltzmann distribution:
$$\rho_v(\vec{v})=\frac{1}{\sqrt{(2\pi \beta)^3}}e^{\frac{v^2}{2\beta}},$$
where ##\beta = k_BT/m##
With this data, prove that the probability distribution function with which a particle chosen at random collides against the wall located in the plane ##z = L## is:
$$\rho_{collision}(\vec{v})=2\rho_v(\vec{v})\theta(v_z),$$
where ##\theta(v_z)## is the Heaviside theta function.
 Relevant Equations:
 $$\rho_v(\vec{v})=\frac{1}{\sqrt{(2\pi \beta)^3}}e^{\frac{v^2}{2\beta}}$$
I have considered two scenarios:
1) A particle that has just collided with the wall at ##z=L## is moving with a velocity ##v_z<0## moving away from the wall. Hence, the probability that this particle has of colliding again is ##0##, so its distribution is also ##0##.
2) A particle moving with positive ##v_z##. Since the particles are constrained to move in that box, it will necessarily collide against the wall, and assuming a completely elastic collision, the velocity distribution in this case would be ##2\rho_v(\vec{v})## because after the collision the particle emerges with the same velocity it had before.
Would these arguments be correct?
Thanks in advance!
1) A particle that has just collided with the wall at ##z=L## is moving with a velocity ##v_z<0## moving away from the wall. Hence, the probability that this particle has of colliding again is ##0##, so its distribution is also ##0##.
2) A particle moving with positive ##v_z##. Since the particles are constrained to move in that box, it will necessarily collide against the wall, and assuming a completely elastic collision, the velocity distribution in this case would be ##2\rho_v(\vec{v})## because after the collision the particle emerges with the same velocity it had before.
Would these arguments be correct?
Thanks in advance!