# Angle of particles hitting a wall - Kinetic theory of gases

In Blundel's "Concepts of thermal physics" it says that the number of particles hitting a wall with speeds between $$v$$ and $$v+dv$$ and angle between $$\theta$$ and $$\theta +d\theta$$ to the normal of the surface, per unit area per second is: $$v\cos\theta\ nf(v)\mathrm{d}v\frac{1}{2}\sin\theta\ \mathrm{d}\theta$$ where $$n$$ is the number of particles per unit volume, $$f(v)$$ is the Maxwell-Boltzman speed distribution function.

Regarding this I have three questions:

1. They call it "the number of particles hitting a wall" but isn't that in reality the probability distribution of the number of particles hitting a wall with speeds between $$v$$ and $$v+dv$$ and angle between $$\theta$$ and $$\theta +d\theta$$ to the normal of the surface, per unit area per second?

2. As the angle is measured from the normal of the surface, then it varies from 0 to $$\pi$$/2, and it is easy to see that it can never be $$\pi/2$$ therefore it should have a null probability at that angle, but what about 0? if a particle has an angle $$0$$ with the normal surely it will hit the wall.

3. How come that the average cos$$\theta$$ for these particles is $$\frac{2}{3}$$ and not $$\langle cos\theta \rangle=\frac{1}{\frac{\pi}{2}-0}\int^{\pi/2}_0cos\theta d \theta$$ ? And how do I derive it?

I don't know if the derivation for that expression is in the book or not, but for clarity and completeness I'll briefly write it here:

For a single particle, the probability of it having velocity $$v$$ to $$v+dv$$ and angle $$\theta$$ to $$\theta + d\theta$$ is given by

$$\int_\theta^{\theta+d\theta}g(\theta')d\theta'\int_v^{v+d\theta}f(v')dv' = f(v)dv\;g(\theta)d\theta.$$

$$g(\theta)$$ is the probability distribution of $$\theta$$ for a particle. Crucially, if the particle has equal probability of moving in any direction, $$g(\theta)$$ isn't constant (if this isn't clear, consider that there's only one possible particle direction with $$\theta=0$$, but a particle with $$\theta=\pi/2$$ could be moving in any direction orthogonal to the wall).

The correct expression is $$g(\theta) = \frac{1}{2} \sin(\theta)$$, meaning the probability of a particle having speed/angle in the given constraints is

$$f(v)dv\frac{1}{2}\sin(\theta)d\theta.$$

Now consider that the particle will only collide with the wall in one second if its in the volume of distance $$v \cos(\theta)$$ away from the wall (and $$\theta<\frac{\pi}{2}$$). For a single particle, this gives the probability of the particle colliding with the wall with the given speed/velocity as

$$v\cos(\theta) \frac{1}{V}f(v)dv\frac{1}{2}\sin(\theta)d\theta.$$

Q1. Since we're assuming the particles don't interact, the way to go from the 1-particle expression to the many-particle expression is the central limit theorem. You're right in that strictly we'd need a distribution, but for any macroscopic number of particles the CLT will give such a sharp peak that it may as well be a single value. This is probably discussed in your textbook somewhere, or googling the central limit theorem might help if you're not familiar with it and this isn't clear.

Q2. Basically, this isn't a problem. Remember that $$\theta$$ is a continuous variable, so the probability of being at any specific angle is 0. If you integrate from 0 to something you get a non-0 answer.

Q3. Your mistake is that you're averaging over every particle with $$\theta < \frac{\pi}{2}$$, not just the ones that collide with the wall. To get the right answer, you need to use the expression above as the probability distribution for all particles that hit the wall. Hence, you get

$$\langle \cos(\theta)\rangle = \frac{\int_0^{\pi/2} p(\theta,v)\cos(\theta)}{\int_0^{\pi/2} p(\theta,v)} ,$$

where $$p(\theta,v)$$ is your expression. The speed parts cancel so you get a straightforward trig integral to solve.