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Deriving the Maxwell-Boltzmann speed distribution

  1. Dec 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Using the Boltzmann distribution for a small system in contact with a heat reservoir at temperature T, find the Maxwell-Boltzmann distribution.

    2. Relevant equations
    The Boltzmann distribution states that the probability density of a system in contact with a heat reservoir at temperature T having energy ε is proportional to e-ε/kT where k is the Boltzmann constant.

    3. The attempt at a solution
    So my book basically does this for me, but I'm having trouble understanding a few things. So say we treat an individual molecule in the gas as our small system, and the reservoir is the rest of the gas at fixed T, then the probability density of having kinetic energy in the x direction of 0.5mvx2 is proportional to
    e-0.5mvx2/kT
    Then if this is the probability density for the KE in the x direction, it must be the same proportionality for velocity in the x direction so the probability density for vx is proportional to
    e-0.5mvx2/kT.
    The same logic follows for y and z so we get the same factors for these
    e-0.5mvy2/kT and e-0.5mvz2/kT.

    Then the probability of having velocity in the interval [vx,vx+dvx]x[vy,vy+dvy]x[vz,vz+dvz] is the product of the individual probability densities so we have
    e-0.5mv2/kTdvxdvydvz

    We can get the speed from changing to spherical polars, integrating over theta and phi to get the probability of being in [v,v+dv] and this gives something proportional to
    v2e-0.5mv2/kTdv.

    So assuming that is all the correct logic, why couldn't I have started by saying that if our molecule is the small system, and the gas our reservoir, then the probability density of having KE 0.5mv2 is proportional to e-0.5mv2/kT? This disagrees with the actual result but follows the same initial step, just considering the overall KE and not the x,y,z components.
     
    Last edited: Dec 18, 2014
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  3. Dec 18, 2014 #2
    To attempt to answer this myself:

    The derivation of the Boltzmann factor using a canonical ensemble (system plus heat reservoir) assumes that each possible energy of the system (which is much less than the total combined energy of the system plus reservoir) corresponds to a single microstate of the system. Here, my system is the particle, with the reservoir the rest of the gas, and I need to choose it's energy so that each energy corresponds to a single microstate. vx will suffice in that case, because each x direction kinetic energy is associated with one vx microstate. v will not suffice, because each total kinetic energy from the speed corresponds to an infinite number of microstates (namely, in velocity space, a sphere of radius v centred on the origin). So we can only say that the probability density for the velocity components, and not the speed, are proportional to the Boltzmann factor.

    Can anybody confirm this?
     
  4. Dec 18, 2014 #3

    TSny

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    Yes, I think you have the right idea. But I would quibble with your first statement. The derivation of the Boltzmann factor does not assume that there is only one (micro) state corresponding to a given energy. The probability of a molecule being in a particular state is proportional to the Boltzmann factor even if there are many other states with the same energy. You should think of the Boltzmann factor as associated with the probability of a particular state rather than a particular energy.

    Otherwise, your argument is good. If you want the probability that a molecule has a speed between v and v + dv, then you are not asking for the probability of a state. Rather, you are asking for the sum of the probabilities of all of the states of the molecule that have a speed between v and v + dv. Since all of these states have essentially the same energy, the probability for the speed to be between v and v + dv will be proportional to the product of the Boltzmann factor with this energy and the ”number of states with speeds between v and v+dv” (i.e., the volume of a spherical shell of radius v and thickness dv).
     
  5. Dec 18, 2014 #4
    Ah, ok then. The derivation I am using only seems to do it for one energy having one microstate which is why I thought that. Thanks!
     
  6. Dec 18, 2014 #5
    Having thought about this, how does this make sense? The probability of a particle being in some microstate is surely uniform (i.e the microstates are equally probable), but because the microstates occur in different numbers, the probability of a given macrostate is not uniform and is instead described by the BD.
     
  7. Dec 18, 2014 #6

    TSny

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    The microstates of a molecule interacting with the other molecules in the gas are not equally probable. For example, the probability that a molecule in the gas is in a state where the translational kinetic energy is 100kT is much smaller than the probability that it is in a state with energy kT. That's what the BD is telling you.

    Maybe you are thinking of the microcanonical ensemble for an isolated thermodynamic system of fixed energy E, where all the microstates of the system (with energy E) are equally probable.
     
  8. Dec 18, 2014 #7
    The way you use the word state makes me think your state refers to the particle having a certain energy. When I use microstate I mean for a given energy, there is only one possible microscopic configuration that can create this energy. Is this what you mean?

    Pages 38-40 here http://books.google.co.uk/books?id=T0luBAAAQBAJ&pg=PA47&source=gbs_toc_r&cad=3#v=onepage&q&f=false give the derivation of the Boltzmann factor I'm considering...

    The derivation assumes there is one microstate for each energy - how can I see from this that it applies when there are multiple microstates for a given energy?
     
  9. Dec 18, 2014 #8

    TSny

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    No. For the canonical ensemble you have a system that is in thermal equilibrium with a heat bath at temperature T. In general there can be a large number of different states of the system that all have the same energy. For the gas, you can take a molecule as the "system" and the rest of the gas as the heat bath. If I specify the energy of the molecule (say E = 2kT) then there will be many states of the molecule that have that energy. For example, if I assume a monatomic gas, where the only type of molecular energy is translational KE, then there will be many states of a molecule with the same energy. The states will differ in the direction of the velocity of the molecule.

    The BD gives the probability of the system being in a particular state. See for example, http://en.wikipedia.org/wiki/Boltzmann_distribution.

    Yes, that particular text does make the assumption of only one microstate per energy value in deriving the BD. But, that assumption is not necessary. I think the derivation given there can essentially be used with minor modification to show that the BD applies to a state of the system (even if there are many other states with the same energy). Since the result of the BD shows that the probability of a state depends only on the energy of the state (for a given T), all microstates of the system with the same energy will have the same probability. But, states of different energy will have different probabilities.
     
  10. Dec 18, 2014 #9
    I think I get it now.

    Going back to my original answer to my own problem, I see it is just aswell that the probability of a particular state with some energy is proportional to the Boltzmann factor for that energy (i.e not just a particular energy being proportional to that factor), because I said having a kinetic energy along x of 0.5mvx2 corresponded to one microstate (velocity vx in the x direction), but it corresponds to two doesn't it (we can have -vx as well). So if the derivation only worked for single microstate energies, I wouldn't be able to get anywhere...

    I find this weird though - the book has derived it only for this special case so I would expect this special case to be able to be applied to the velocity distribution (as on page 48, it claims to have got the velocity distribution from the BD)...
     
    Last edited: Dec 18, 2014
  11. Dec 18, 2014 #10

    TSny

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    Yes, for 1D motion of a molecule along the x-axis, there would be two states of energy 0.5mvx2 corresponding to opposite directions of velocity. So, the probability that the molecule has that energy would be twice the probability that the molecule is in a state with that energy and also traveling in the positive x direction. In this case, the probability that the molecule is in a particular state (which has energy E) and the probability that the molecules has the energy E differ by a factor of 2. Both probabilities would be proportional to the Boltzmann factor ##e^{- \frac{E}{kT}}##.

    More generally, the probability that the system has energy E turns out to be proportional to g(E)##e^{- \frac{E}{kT}}## where g(E) is the number of states of the system that have the energy E. g(E) is the "degeneracy" of the energy E.

    For your problem of the gas molecule, you can think of g(E) as being proportional to the volume of the spherical shell of radius v and thickness dv. So, the probability that the molecule has speed between v and v+dv is proportional to v2##e^{- \frac{E}{kT}}##.
     
  12. Dec 18, 2014 #11
    Oh I see now, it's pretty obvious to prove for a general number of microstates of the system just by multiplying the initial probability by g(E) instead of 1 (so I don't get why my book didn't bother doing it). So g(E)=2 for E=0.5mvi2.

    I'm a little confused about the logic of knowing that different microstates for the same energy have equal probability of occuring, whilst compared to those of another energy, they are different. Why is this? (not just in the context of a gas, but in terms of the general example using the canonical ensemble).

    Then once I know this, I can just divide the probability of having energy E, by the number of it's microstates, to give the probability of a single microstate with energy E, which is simply the BD again...
     
    Last edited: Dec 18, 2014
  13. Dec 18, 2014 #12

    TSny

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    Yes, g(E) = 2 for the case of a molecule restricted to move in one dimension and having only translational KE.
    For an isolated system the total energy of the system will be constant. A basic postulate of statistical mechanics is that all of the microstates of the isolated system corresponding to the same total energy are equally probable. See the top of page 37 of your book. This corresponds to the “microcanonical ensemble”.

    The canonical ensemble deals with a system, S1, that is in thermal contact with a large heat reservoir, S2, at temperature T. The total system, Stot = S1 + S2, can be thought of as an isolated system for which the microcanonical ensemble applies. That is, different states of the total system corresponding to the same total energy Etot = E1 + E2 are equally probable. You then use this idea to derive the BD for the probability of occurrence of states of the system S1 while it is in thermal contact with S2. The BD then tells you that the probability of occurrence of a state ##\small a## of S1 (while it is in contact with S2) depends only on the energy E##\small a## of the state of system S1 and the temperature of S2: Prob(state ##\small a##) ##\propto e^{-\frac{E_a}{kT}}##. States of S1 with different energies will therefore have different probabilities of occurring.
     
  14. Dec 19, 2014 #13
    Ok, so
    system + reservoir -> make up the microcanonical ensemble, in which each microstate is equally likely
    system -> makes up the canonical ensemble

    In the derivation of the BD, the first step is to say the probability of the system having energy e is given by
    P(e)∝g(e)Ω(E-e)
    where E is the total energy of the system. This proportionality uses the fact that all microstates are equiprobable in the microcanonical ensemble.

    This leads to
    P(e)∝g(e)exp(-e/kT)
    which gives the probability distribution of the system having an energy e.

    I still can't see why I'm allowed to divide by g(e) to give the probability of one of the microstates with energy e unfortunately (Edit 3: I see why the equiprobability of the overall system microstates allow this now. I used this fact to be able to multiply by g(e) in the first place, but somehow my brain finds it harder to connect the logic to division).

    Edit: Am I allowed to think about the derivation in the following way:
    Let's consider the probability of the system being a specified single microstate. Then we expect
    P(any microstate with energy e)∝Ω(E-e)
    giving
    P(any microstate with energy e)∝exp(-e/kT)
    Is this perhaps what my book was trying to get across by saying each energy has a single microstate (I've just basically changed the wording a bit, which makes it clearer to me)?
    If so then that partially solves my issue, it's just it doesn't intuitively make much sense to me.

    In my head it just feels like a big contradiction - the overall system microstates are equally likely, yet the system microstates are not equally likely depending on energy...

    Edit 2: I suppose it makes sense - the overall system microstates are equally likely, but the probability of having a system microstate depends on the number of overall system microstates that exist that allow the system to be in that microstate...
     
    Last edited: Dec 19, 2014
  15. Dec 19, 2014 #14

    TSny

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    Yes! Think of the BD as giving the probability of a state rather than an energy.
    Yes!
     
  16. Dec 19, 2014 #15
    Great, thankyou very much, I understand this all a lot better now :)
     
  17. Dec 19, 2014 #16
    Really sorry to bring this up again, but I'm struggling to understand the example on pages 40 and 41 of the link I previously sent. If you would be able to have a quick look at it I would be very grateful, but obviously feel free to give up on me after all this time...

    So the basic idea is
    microstates: the individual equally likely combinations of energy quanta in the lattice
    macrostate: the frequency (i.e probability) distribution of these quanta
    If you randomly juggle the quanta around, as you do more and more iterations, you get macrostates that correspond to greater numbers of microstates, and eventually we get something that looks like a BD - so the BD seems to give us a distribution for having the maximum number of microstates for our macrostate. We want to get as many microstates for our macrostate as we can, because this is what happens when we are in thermal equilibrium, as our system we derived the BD for was, so this seems to agree with what we have done before.

    However, I'm trying to make this run in analogy with the system-reservoir thing we had going on, but I can't really make physical links between what everything represents.

    What actually do the quanta and the 400 spaces in the lattice represent?
     
  18. Dec 19, 2014 #17

    TSny

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    OK
    The lattice as a whole is an isolated system with a fixed total energy. As you said, each microstate of the whole lattice is equally probable as the total energy shuffles around the lattice. To relate this to the idea of a system in contact with a heat reservoir you can pick one lattice site S and treat this single site as a “system”. The rest of the lattice can then be thought of as a heat reservoir in thermal contact with S. The “state” of the system S is defined by how many energy quanta it contains. (This is a case where there is one-to-one correspondence between energy of the system S and the state of system S.)

    The lattice starts out with the energy distributed among the lattice sites in a very improbable way; namely, exactly one energy quantum per lattice site. But once the energy has had time to shuffle randomly among the sites, the system reaches “thermal equilibrium” and you can then consider the equilibrium distribution of energy over the lattice sites. Figure 4.9 shows the equilibrium distribution by plotting the number N(n) of lattice sites that have n quanta. The figure shows that N(n) follows a Boltzmann type distribution where N(n) is proportional ##e^{-bn}## for some positive constant ##\small b##. Or, equivalently, N(n) is proportional to ##e^{-\beta E(n)}## where ##\small E(n)## is the energy of a site that contains n quanta and ##\small \beta## is a constant.

    Since all the lattice sites are on equal footing, the probability PS(n) that the specific site S has n quanta equals the total number of sites that have n quanta, N(n), divided by the total number of sites in the lattice, Ntot. That is, PS(n) = N(n)/Ntot. Since N(n) is a Boltzmann distribution, so is PS(n). Thus, the probability to find system S in state n is given by a Boltzmann distribution.
     
  19. Dec 28, 2014 #18
    Yes, you could've started out that way. But you have to keep in mind that there can be more than one microstate for each speed so that the probability might be proportional to more factors that contain v. For each infinitesimal interval of speed (v, v + dv), there is an entire spherical shell of infinitesimal thickness in [itex]v_xv_yv_z[/itex]-space. The volume of this shell is [itex]4\pi v^2dv[/itex]. The probability density for speed is not only proportional to the boltzmann factor, but also to this [itex]4\pi v^2[/itex] factor. You can think of it this way: each point in that spherical shell represents a unique microstate (unique velocity) but when you calculate the SPEED for each point (on this shell), you will get the same value. So to get the number of unique microstates that correspond to a speed value within the interval (v, v+dv), you have to sum up the boltzmann factors of each point on this shell. But the boltzmann factor is the same on this shell so this sum turns into a multiplication: [itex] 4\pi v^2\exp{\frac{-mv^2}{2kT}}dv[/itex]. This is proportional to the number of microstates within the infinitesimal shell. All you have to do now is find the normalizing factor (just some constant Z) because when you integrate a probabliity density function over all possible states, you should get a value of 1.
     
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