Mean Collision Time in a Gas: Deriving the Probability of Collision

[fayled]

Other ways of looking at it are that the probability of a collision in time dt is
dt/τ
or
fdt
where τ is the mean collision time and f is the mean collision rate.

 

[fayled]

This is from Feynman's lectures:

We may often wish to ask the following question: “What is the chance that a molecule will experience a collision during the next small interval of time dt?” The answer, we may intuitively understand, is dt/τ. But let us try to make a more convincing argument. Suppose that there were a very large number N of molecules. How many will have collisions in the next interval of time dt? If there is equilibrium, nothing is changing on the average with time. So N molecules waiting the time dt will have the same number of collisions as one molecule waiting for the time Ndt. That number we know is Ndt/τ. So the number of hits of N molecules is Ndt/τ in a time dt, and the chance, or probability, of a hit for anyone molecule is just 1/N as large, or (1/N)(Ndt/τ)=dt/τ, as we guessed above. That is to say, the fraction of the molecules which will suffer a collision in the time dt is dt/τ. To take an example, if τ is one minute, then in one second the fraction of particles which will suffer collisions is 1/60. What this means, of course, is that 1/60 of the molecules happen to be close enough to what they are going to hit next that their collisions will occur in the next second.

So basically this is what I want to understand, but I still don't...

So basically he says that the number of hits for N molecules in time dt is Ndt/τ, but then he says the probability of a hit for one molecule is 1/N as large, which is dt/τ. How is this? Surely 1/N multiplied by the number of hits for N molecules is average number of hits for one molecule in time dt, not the probability of a single hit.

 

[haruspex]

Is there a way to see that nσvdt will have to always be between 0 and 1, even in the limit then?

n, σ, and v are all fixed. As dt tends to zero, nσvdt tends to zero, so it is not merely less than 1, it is as close to zero (without actually being zero) as you care to make it.

 

[fayled]

n, σ, and v are all fixed. As dt tends to zero, nσvdt tends to zero, so it is not merely less than 1, it is as close to zero (without actually being zero) as you care to make it.

Ok, thanks, I think I get it now - dt is obviously tiny and so dt/mean collision time will be some fraction, telling us the number of collision a given molecule will have. As this is a fraction we can say it is the probability of the given molecule colliding with another molecule in time dt.

The numerical example above helped me a lot (so I seemingly still depend on numbers for intuition :().

For my derivation in the OP, nσvdt is just a fraction and so works as a probability too.