# Volume of velocity-space in Maxwell-Boltzmann distribution

• I
I'm having trouble understanding the role of the volume of velocity space when deriving the Maxwell-Boltzmann speed distribution.

1. If we wish to compute the speed distribution from the velocity distribution we work with, say, $$p(v) dv \propto v^2e^{-av^2} dv$$ where the ##v^2 dv## comes from considering the volume of velocity space available to the states with speeds in ##[v, v+dv]##. This makes sense, at least mathematically. However ..

2. This distribution seems to say that, if we consider the states close to ##v=0##, we will find no particles with speed 0 (and not many close by), as there is no volume space available to them there (and this appears to be independent of the underlying Boltzmann distribution itself, since ##v^2 f(v) \to 0## as ##v \to 0## for any physically realistic ##f(v)##)

3. But this seems to clash with what the Boltzmann distribution predicts: the probability of finding a particle in a microstate of 0 energy (hence 0 velocity) is greater than finding it with any other energy since the probability of it being in a microstate ##r##, energy ##E_r## is ##p(r) \propto e^{E_r/kT}##.

So I'm confused. The Boltzmann distribution seems to predict that we will often find gas particles with 0 velocity. But the Maxwell-Boltzmann speed distribution seems to say that we will find *no* particles with 0 speed (##\Leftrightarrow## 0 velocity), since there is no velocity space available to them.

Can anyone clear up my confusion?

Khashishi
The Boltzmann distribution tells you the relative probability of each state being occupied, but it doesn't tell you how many states are available. There are more states with higher speed than lower speed, so even though the relative probability of any of those states being filled is lower, the most likely speed is higher than 0. The number of available states with speed ##|v|## is proportional to ##v^2##.

If you had weighted dice which had twice as much chance of rolling 1 as the other sides, the most likely roll of 5 dice is obviously going to be ##(1,1,1,1,1)##, but the most likely sum is not 5.

• nasu
The Boltzmann distribution tells you the relative probability of each state being occupied, but it doesn't tell you how many states are available. There are more states with higher speed than lower speed, so even though the relative probability of any of those states being filled is lower, the most likely speed is higher than 0. The number of available states with speed ##|v|## is proportional to ##v^2##.

Thanks. However, I don't think that you are doing any more than restating in words what I described more mathematically in the OP. And I'm not really confused about it from the POV of the probability argument - it's the physical intuition that I'm lacking.

Having thought about the question some more, however, I think part of my problem is that:

a) I'm not precisely sure where the "density of states" arguments really come from. (I'm not entirely sure if it's a classical or QM based thing)

b) I'm not clear to what extent the "density of states" term i.e. the ##v^2 dv## term is independent of the Boltzmann exponential term. AFAICS the argument that gives the M-B speed distribution would rule out *any* physical system having a large number of states with low energy since the volume of available phase space ##\to 0## as ##E \to 0## - that surely can't be right though?

In short, I need to go away and think some more about precisely what I don't understand, as I can't really articulate it to myself yet.

Homework Helper
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One item that you might find of use is that the density of states, i.e. the number of states ## \Delta N ## in an interval (of momentum or velocity space), is usually expressed in momentum space or k-space. In momentum space, the density is given by ## \Delta N=V \Delta^3 p/h^3 ##. This is derived by setting up periodic boundaries so that ## \exp^{ik_x x}=\exp^{ik_x (x+L_x)} ##. The result is ## L_x=(2 \pi) n_x ## where ## n_x ## is an integer, and similarly for ## L_y ## and ## L_z ##. ## \Delta^3 p ## with spherical symmetry in p-space can be written as ## 4 \pi p^2 \, dp ## where the ## p ## is now the magnitude of the momentum. This magnitude of momentum is proportional to the speed ## v ## in the Maxwell-Boltzmann distribution, so that the ## \Delta N ## can also be expressed as a function of ## v^2 \, dv ##.

One item that you might find of use is that the density of states, i.e. the number of states ## \Delta N ## in an interval (of momentum or velocity space), is usually expressed in momentum space or k-space. In momentum space, the density is given by ## \Delta N=V \Delta^3 p/h^3 ##. This is derived by setting up periodic boundaries so that ## \exp^{ik_x x}=\exp^{ik_x (x+L_x)} ##. The result is ## L_x=(2 \pi) n_x ## where ## n_x ## is an integer, and similarly for ## L_y ## and ## L_z ##. ## \Delta^3 p ## with spherical symmetry in p-space can be written as ## 4 \pi p^2 \, dp ## where the ## p ## is now the magnitude of the momentum. This magnitude of momentum is proportional to the speed ## v ## in the Maxwell-Boltzmann distribution, so that the ## \Delta N ## can also be expressed as a function of ## v^2 \, dv ##.

Thanks. It's not clear from your post but I'm guessing that the periodicity that you mention is essentially the QM "states of particle in a box" argument. If so, I'm familiar with it, and have seen derivations (e.g. in Mandl) but part of my confusion is that many treatments seem to play fast and loose with some classical stat. mech. "volume of phase space" arguments here, when really the whole thing should be quantum, AFAICS - and they don't really make it clear what they're doing (to my mind, at least). Mandl does in fact discuss precisely this matter, but I've yet to get a chance to read it properly. I'm leaving this question for now anyway, till I can pursue it in more detail.

• I think that this is clearer in my mind now:

For any physical system, we need to do three things:

1. Decide if it is described by the canonical ensemble. To do that, we need to identify a constant T heat bath, and the "system" in contact with said heat bath. (This may be subtle e.g. the "system" may be one molecule in a gas at constant T). Once we've done that, we know that for any state/energy of state ##r, E_r##, we have ##p(r) \propto e^{-\beta E_r}## but there may be *many* states with this energy, say ##n(E_r)## for a discrete set of states.

2. We now need to analyse the system from a different POV and we have to find the number of available states per ##E_r## (This is ##n(E_r)##). This depends entirely on the physical characteristics of the system.

3. If we then want to find p(system has energy ##E_r##), we bear in mind that each state is disjoint i.e. the system can only be in one state at a time. Hence, we have ##p(E_r) \propto n(E_r)e^{-\beta E_r}## (from basic probability theory). (Modify in the obvious way for continuous states).

So, any decent book, in analysing systems, should make these first two steps clear. However, it seems (to my mind) that many don't, and often, if there is only one state per energy (e.g. system of paramagnetic dipoles) then it is not mentioned at all. I find this confusing and, when a proper analysis isn't done, then assumptions, maybe obvious to the author but not to the reader, can be hidden.

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