Gas in a box with Maxwell-Boltzmann distribution

[rogdal]

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 Maxwell-Boltzmann 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!

 

[TSny]

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.

Hello, @rogdal. I don't understand your argument. The particle does not emerge from the collision with the same velocity it had just before the collision. Since $v_z$ changes in the collision, the final velocity vector is not the same as the initial velocity vector. If $v_z$ just changes sign in the collision with no change in $v_x$ or $v_y$, then the initial and final speeds would be the same. But I don't see why this would give the factor of 2 in the answer.

I'm not completely sure of the interpretation of the statement of the problem:

prove that the probability distribution function with which a particle chosen at random collides against the wall located in the plane $z = L$ is:
$$\rho_{collision}(\vec{v})=2\rho_v(\vec{v})\theta(v_z),$$

Two possible interpretations come to my mind:

(1) Consider some arbitrary interval of time that is long enough for many particles to strike the wall. Imagine making a list of the incoming velocity $\vec v$ for each particle that strikes the wall during the time interval. What is the probability distribution $\rho_{\rm collision}(\vec v)$ associated with this list?

(2) Pick an arbitrary particle of the gas. Wait until it collides with the wall. (It could be a long wait, but that's ok.) Write down the incoming velocity $\vec v$ for this collision. Repeat this many times. What is the probability distribution $\rho_{\rm collision}(\vec v)$ associated with this list?

I favor the second interpretation. It seems to be more in line with the way I would literally interpret the wording of the problem statement. Also, I think the second interpretation will give the answer that's stated in the problem; whereas, I don't believe the first interpretation would.

 

[rogdal]

Hello @TSny. Many thanks for your answer. Yes, obviously, after thinking it a bit, the argument I used in my 2nd scenario for justifying the origin of the 2 extra factor makes no sense.

Let me try to think about the probability distributions from your interpretations:

(1) In this case, we would obtain exactly the Maxwell-Boltzmann distribution $\rho_{collision}(\vec{v}) = \rho_v(\vec{v})$. However, since we are only considering positive $v_z$ from, let's say, 0 to infinity, in this particular case, $\rho_v(\vec{v})$ wouldn't be normalized. Integrating, instead of 1 one would obtain 0.5, which is not correct. But if we normalize it by adding a factor of 2, we would get exactly $\rho_{collision}(\vec{v}) = 2\rho_v(\vec{v})\theta(v_z)$. Does this make sense?

(2) Instead the Maxwell-Boltzmann distribution for a group of particles, this would be the Maxwell-Boltzmann distribution just for a single particle, right? This distribution would be given by:

$$\rho_{collision}(\vec{v}) = \left(\frac{m}{2kT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$
The factor of 4πv^2 would arise from the fact that the velocity distribution is for a single particle rather than for a group of particles.The velocity would be proportional to the surface area of a sphere of radius v around the particle. How would this expression be linked to the final answer that I'm told to prove?

Thanks!

 

[TSny]

Hello @TSny. Many thanks for your answer. Yes, obviously, after thinking it a bit, the argument I used in my 2nd scenario for justifying the origin of the 2 extra factor makes no sense.

Let me try to think about the probability distributions from your interpretations:

(1) In this case, we would obtain exactly the Maxwell-Boltzmann distribution $\rho_{collision}(\vec{v}) = \rho_v(\vec{v})$. However, since we are only considering positive $v_z$ from, let's say, 0 to infinity, in this particular case, $\rho_v(\vec{v})$ wouldn't be normalized. Integrating, instead of 1 one would obtain 0.5, which is not correct. But if we normalize it by adding a factor of 2, we would get exactly $\rho_{collision}(\vec{v}) = 2\rho_v(\vec{v})\theta(v_z)$. Does this make sense?

(2) Instead the Maxwell-Boltzmann distribution for a group of particles, this would be the Maxwell-Boltzmann distribution just for a single particle, right? This distribution would be given by:

$$\rho_{collision}(\vec{v}) = \left(\frac{m}{2kT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$
The factor of 4πv^2 would arise from the fact that the velocity distribution is for a single particle rather than for a group of particles.The velocity would be proportional to the surface area of a sphere of radius v around the particle. How would this expression be linked to the final answer that I'm told to prove?

Thanks!

I think your argument for the first interpretation is actually the one that should be used for the second interpretation.

We can consider a simple model to illustrate the difference between the first and second interpretations. Imagine a box of gas for which the molecules are uniformly spread in space and the molecules move only parallel to the z-axis. The molecules do not collide with themselves. So, each molecule just bounces back and forth between two walls and never changes its speed. Suppose half the molecules have a speed of 500 m/s and the other half a speed of 100 m/s. Then the velocity distribution would be 25% with velocity +500 m/s, 25% with - 500 m/s, 25% +100 m/s, and 25% -100 m/s. Or, if you consider molecules with $v_z >0$, then 50% of those will have velocity +500 m/s and 50% +100 m/s.

(1) If you watch the wall at $z = L$, then you will see molecules of velocity + 500 m/s and velocity +100 m/s strike the wall. But the faster molecules hit the wall more often than the slow molecules (just because they are moving faster). So, the velocity distribution for the molecules that hit the wall during some arbitrary interval of time will be more than 50% +500 m/s. That is, this velocity distribution $\rho_{\rm collision}(\vec v)$ of the molecules striking the wall is different from the velocity distribution of the molecules in the box that have $v_z > 0$.

(2) However, if you pick a molecule at random in the box that has $v_z > 0$, there is a 50% chance that it has a velocity of +500 m/s and a 50% chance of +100 m/s. Thus, a molecule picked at random will have a 50% chance of hitting the wall at +500 m/s and a 50% chance of hitting the wall at +100 m/s. So, for particles picked randomly, the velocity distribution $\rho_{\rm collision}(\vec v)$ is the same as the velocity distribution of the molecules in the box that have $v_z >0$.

I think similar arguments hold for the actual gas which has the Maxwell-Boltzmann distribution.

In discussing interpretation (2), you wrote

This distribution would be given by:

$$\rho_{collision}(\vec{v}) = \left(\frac{m}{2kT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$

The right-hand side is the Maxwell-Boltzmann probability distribution for speeds rather than velocity. But, the question is asking for the velocity probability distribution.

 

[vela]

(2) Instead the Maxwell-Boltzmann distribution for a group of particles, this would be the Maxwell-Boltzmann distribution just for a single particle, right? This distribution would be given by:

$$\rho_{collision}(\vec{v}) = \left(\frac{m}{2kT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$
The factor of 4πv^2 would arise from the fact that the velocity distribution is for a single particle rather than for a group of particles.The velocity would be proportional to the surface area of a sphere of radius v around the particle. How would this expression be linked to the final answer that I'm told to prove?

The different distributions aren't for single particle vs. many particles. It's for Cartesian vs. spherical. You have $dv_x\,dv_y\,dv_z = v^2 \sin\theta\,dv\,d\theta\,d\phi$. That's where the $v^2$ comes from. When you integrate out the angles to get a distribution in terms of the speed only, you get the factor of $4\pi$.

 

[rogdal]

Oh, I see... Many thanks @TSny and @vela for your help. I understand it now!