# Mean collision time in a gas

• fayled
The probability of a molecule not colliding up to time t is given by P(t)=-nσv.The probability of a molecule not colliding up to time t multiplied by not colliding in time dt is given by P(t+dt)=P(t)(1-nσvdt)Therefore P(t+dt)=-nσv.Which is a normalized probability distribution with mean time 1/nσv, which is the required result.f

## Homework Statement

Derive an expression for the mean collision time in a gas where the collision cross-section is σ and the number density is n.

## The Attempt at a Solution

I'm just using my book to do this, and I can't get past the first bit...

It says
Consider a particular molecule moving at speed v with all other molecules in the gas stationary. In a time dt, the molecule sweeps out a volume σvdt, and if another molecule lies inside this volume, there will be a collision. With n molecules per unit volume, the probability of a collision in time dt is therefore nσvdt.

I'm sure the quantity nσvdt is just the number of molecules in the volume it sweeps out in time dt, so how does this give a probability? And it can't be normalized can it?

The remainder of the derivation is
define P(t) as the probability of a molecule not colliding up to time t.
Then P(t+dt)=P(t)+(dP/dt)dt
However P(t+dt) is the probability of a molecule not colliding up to time t multiplied by not colliding in time dt, i.e
P(t+dt)=P(t)(1-nσvdt)
Then
(1/P)dP/dt=-nσv
P(t)=exp(-nσvt)
The probability of not colliding up to time t, then colliding in the next dt is
P(t)nσvdt=exp(-nσvt)nσvdt
which is a normalized probability distribution with mean time 1/nσv which is the required result.

This all makes sense assuming the first bit, but I just can't see why it is right, can anyone help please?

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This all makes sense assuming the first bit, but I just can't see why it is right, can anyone help please?
I'm sure the quantity nσvdt is just the number of molecules in the volume it sweeps out in time dt,
You see everything but the following, so far, so read it again, and think about what it says. And think about the definition of "number density."
the number density is n.

You see everything but the following, so far, so read it again, and think about what it says. And think about the definition of "number density."

The number density is the number per unit volume, so we could write it as
n=N/V where N is the total number and V is the volume, so the probability (according to my book) that a molecule has a collision in time dt is given by
Nσvdt/V.

Unfortunately that doesn't enlighten me in any way :(

the number per unit volume,

Yup, but I'm multiplying it by a volume, namely σvdt, so it just gives a number, not a probability, right?

σvdt, so it just gives a number

Hmm, the number that could collide in an infinitesimal amount of time? I don't think I can see what you're getting at, sorry...

I can't get past the first bit...
I couldn't either. I didn't realize textbooks had become this bad. Sucked me under right along with you.
nσvdt is just the number of molecules in the volume it sweeps out in time dt
Turn this expression into an equation; I'll do it since the book had me confused enough to try to make you follow its "mountain from a molehill" approach;
dn = nσvdt .

I couldn't either. I didn't realize textbooks had become this bad. Sucked me under right along with you.

Turn this expression into an equation; I'll do it since the book had me confused enough to try to make you follow its "mountain from a molehill" approach;
dn = nσvdt .

So nσvdt represents a differential number of molecule that our molecule moving with speed v could collide with in time dt, right? How does this translate into the probability of a molecule colliding in time dt?

dn = nσvdt .
Integrate this expression and see if it makes sense as far as probabilities go.

Integrate this expression and see if it makes sense as far as probabilities go.

Wait, is your n in dn the same as the n on the right hand side, because n is the number density, whilst dn is a differential number that will colide with our chosen molecule at speed v...

Should we use dN=nσvdt?

Works for me. You do see where this is going?

Works for me. You do see where this is going?

Nope, not sure where we're going.

See if I can get my brains unscrambled enough to make this work. Start over from the point where the book lost both of us.
probability of a collision in time dt is therefore nσvdt.
In time dt, the molecule will collide, or not collide with another molecule; the sum of probabilities for both outcomes is unity, 1, one. The number density n is the total number of molecules randomly distributed in the total volume. In some fraction of the total volume, say σvdt, the probability of finding a gas molecule is proportional to n times that volume fraction.
Does this turn into a probability for you now?

See if I can get my brains unscrambled enough to make this work. Start over from the point where the book lost both of us.

In time dt, the molecule will collide, or not collide with another molecule; the sum of probabilities for both outcomes is unity, 1, one. The number density n is the total number of molecules randomly distributed in the total volume. In some fraction of the total volume, say σvdt, the probability of finding a gas molecule is proportional to n times that volume fraction.
Does this turn into a probability for you now?

Well I suppose saying that
nσvdt
is proportional to the probability of a collision makes sense, however, in the step

'however P(t+dt) is the probability of a molecule not colliding up to time t multiplied by not colliding in time dt, i.e
P(t+dt)=P(t)(1-nσvdt)'

we seem to require that it actually equals (i.e isn't just proportional to) this in order to say 1-nσvdt is the probability of not colliding in time dt.

Page 9 here http://www.damtp.cam.ac.uk/user/tong/kintheory/kt.pdf derives the same thing but just without finding the form for the probability of collision in time dt, which might be useful in terms of understanding what's going on (although I haven't got anywhere with it).

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I thought I had deleted "proportional" before I posted that last reply. Does it make sense to you without using the word "proportional?"

I thought I had deleted "proportional" before I posted that last reply. Does it make sense to you without using the word "proportional?"

No, sorry.

So we're saying the probability of a chosen molecule colliding with other molecules in time dt = volume fraction in time dt * total number of molecules in volume?

I can't even see why this can't exceed 1 for a start... The fraction of the volume is cleary between 0 and 1, but N is going to be huge because its the number of molecules...

probability of a chosen molecule colliding with other molecules in time dt = volume fraction in time dt * total number of molecules in volume?
Volume swept by one molecule (σvdt) over total volume, that fraction times total number of molecules in the total volume.

Volume swept by one molecule (σvdt) over total volume, that fraction times total number of molecules in the total volume.
Yep, bit ambiguous sorry, but I did mean total volume, so

probability of a chosen molecule colliding with other molecules in time dt = volume fraction in time dt * total number of molecules in total volume

So why must this be between zero and one?

So why must this be between zero and one?
You are absolutely correct --- it could hit every other molecule in the box Avogadro's number of times squared.
You, take a break. I'm going to take a break. My preference is to make your book work for you rather than to subvert the course synopsis. Just need a couple hours to twist my mind around their way of thinking.

You are absolutely correct --- it could hit every other molecule in the box Avogadro's number of times squared.
You, take a break. I'm going to take a break. My preference is to make your book work for you rather than to subvert the course synopsis. Just need a couple hours to twist my mind around their way of thinking.

Ha, thanks for persisting.

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You are absolutely correct --- it could hit every other molecule in the box Avogadro's number of times squared.
You, take a break. I'm going to take a break. My preference is to make your book work for you rather than to subvert the course synopsis. Just need a couple hours to twist my mind around their way of thinking.

So if dt and so vdt is small enough that our chosen molecule could only possibly collide with a maximum of one other molecule in this time, nσvdt would give us the probability of collision in time dt then. However, we can't just assume this about dt can we.

To continue, there's no obvious statement that the "scale" of "number density" is changing back and forth as the discussion proceeds, and this may be causing the difficulty. When number density is given as "n" for the problem, it's referring to a large scale, number of molecules in a syringe barrel, room, breath of air, party balloon, bubble rising in an aquarium, and is not measurably different for random samples of similar sizes. When number density is used to count collisions of one molecule moving through a volume of space σvdt, the variations in number density from one volume to another of equal size can be enormous; one volume could be completely unoccupied (nlocal,1 = 0), and another double or triple the large scale or macroscopic average (nlocal,2 = 3n). All that can be said is that the sum of the local number densities times their populations divided by the total number of local sample volumes will average to be "n" the large scale density. Then when looking at the number of collisions of a single molecule in the volume σvdt, all that is known is the overall probability of the actual number density the molecule experiences in that volume, or n.

To continue, there's no obvious statement that the "scale" of "number density" is changing back and forth as the discussion proceeds, and this may be causing the difficulty. When number density is given as "n" for the problem, it's referring to a large scale, number of molecules in a syringe barrel, room, breath of air, party balloon, bubble rising in an aquarium, and is not measurably different for random samples of similar sizes. When number density is used to count collisions of one molecule moving through a volume of space σvdt, the variations in number density from one volume to another of equal size can be enormous; one volume could be completely unoccupied (nlocal,1 = 0), and another double or triple the large scale or macroscopic average (nlocal,2 = 3n). All that can be said is that the sum of the local number densities times their populations divided by the total number of local sample volumes will average to be "n" the large scale density. Then when looking at the number of collisions of a single molecule in the volume σvdt, all that is known is the overall probability of the actual number density the molecule experiences in that volume, or n.

Yup, I see this. So I guess we just assume our macroscopic number density is valid on microscopic scales here.

However, we're still not really any closer to understanding why nσvdt is the probability of a collision in time dt are we?

However, we're still not really any closer to understanding why nσvdt is the probability of a collision in time dt are we?
Toss a coin a million times, and it should come up half and half heads and tails. Probability of either for a single toss is half and half. You're not going to call it more than half the time.
Sweep a volume of nσvdt a million times and you'll get one million times nσvdt collisions; sweep it once, and maybe you get nσvdt collisions, maybe you get more, maybe you get fewer.

Toss a coin a million times, and it should come up half and half heads and tails. Probability of either for a single toss is half and half. You're not going to call it more than half the time.
Sweep a volume of nσvdt a million times and you'll get one million times nσvdt collisions; sweep it once, and maybe you get nσvdt collisions, maybe you get more, maybe you get fewer.

Yeah, but we never established why nσvdt has to be between zero and one did we?

Equating the average number of hits to the probability of a hit is just an approximation that works for very small values. Treat the hits as a Poisson process. If the probability of k hits is ##\frac {\lambda^k e^{-\lambda}}{k!}## then the average number of hits is λ. The probability of no hits is ##e^{-\lambda} ≈ 1 - \lambda##, so the probability of at least one hit is also λ.

Equating the average number of hits to the probability of a hit is just an approximation that works for very small values. Treat the hits as a Poisson process. If the probability of k hits is ##\frac {\lambda^k e^{-\lambda}}{k!}## then the average number of hits is λ. The probability of no hits is ##e^{-\lambda} ≈ 1 - \lambda##, so the probability of at least one hit is also λ.

But it seems to me that the average number of hits here in time dt, nσvdt, can exceed one and so how does it make sense as a probability?

But it seems to me that the average number of hits here in time dt, nσvdt, can exceed one and so how does it make sense as a probability?
dt is an arbitrarily short time. In the calculus step you will take the limit as dt tends to zero.

dt is an arbitrarily short time. In the calculus step you will take the limit as dt tends to zero.

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

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.

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.

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.

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.