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Probability distribution for a rotating gas

  1. Feb 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A cylinder contains an ideal gas and rotates at angular speed w. Find the probability that a molecule is at radial position r from the axis of the cylinder.

    2. Relevant equations
    Boltzmann distribution, P(E)∝Ω(E)exp(-E/kT)
    where Ω(E) describes the degeneracy of the energy level E.

    3. The attempt at a solution
    E=mr2w2/2 with molecules having mass m. Ω(E)∝r because the area goes like 2πrl. Then P(E)∝rexp(-mr2w2/2kT). However apparently this is wrong and there should be no r present in the probability. I don't understand this - it's like for the Maxwell Bolztmann distribution the probability of having speed v has a degeneracy ∝v2. Also, apparently I should be using E=-mr2w2/2 - why is this?

    This is seen in 4.1 on page 1 here
    http://www.physics.ohio-state.edu/~jay/7602/Feb ho2.pdf

    Thanks for any help!
     
  2. jcsd
  3. Feb 19, 2015 #2

    TSny

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    In the link, the quantity n(r) is the number density; i.e.,number of particles per unit volume. Thus, n(r)dV would be the number of particles in a volume dV at radial distance r.

    Can you see how to relate that to the probability that a particle lies between r and r+dr?
     
  4. Feb 19, 2015 #3
    Do you mean I should have rn(r)∝P(E) instead of n(r)∝P(E) and so the factor r cancels meaning my P(E) is correct?
     
  5. Feb 19, 2015 #4

    TSny

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    Yes. You want to make sure it's clear to you why this is so.
     
  6. Feb 19, 2015 #5

    TSny

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    Do you have the sign correct for E and for the argument of the exp?
     
  7. Feb 19, 2015 #6
    Hmm I don't really understand it - say we're considering the x component of kinetic energy... I believe P(E)∝exp(mvx2/2kT) here, and my book then says that P(vx)∝exp(mvx2/2kT), i.e that probability of having some vx2 directly transforms into the probability of having vx. This seems to be fine so why isn't the above?
     
  8. Feb 19, 2015 #7
    Nope, I'm aware my E has the wrong sign - I'm not sure why though, surely KE is 0.5Iw2. The Boltzmann factor has the negative in...
     
  9. Feb 19, 2015 #8

    TSny

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    You're not dealing with KE here. KE has relevance to the probability distribution of velocities. But here you want the probability distribution for radial position. So, the energy you are dealing with is an effective potential energy associated with the "centrifigal force". This energy depends on radial position.
     
  10. Feb 19, 2015 #9
    Ok, so if the centrifugal force is F=mw2r then the potential is found by integration as -mw2r. However, I'm not entirely happy - why do I have to consider the centrifugal force (i.e the force observed in the rotating frame), and not the centripetal force (the force as observed outside of the rotating frame) - I'm guessing its something to do with us wanting things from the perspective of a molecule but I can't quite pinpoint it. If we're in the non-rotating the frame then the sign reverses and we're back to where we began (albeit by being less wrong).
     
  11. Feb 19, 2015 #10

    TSny

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    It's in the rotating frame that you have a stationary gas. In the inertial lab frame, the gas is not stationary and the elements of gas undergo acceleration (centripetal). So, it's conceptually simpler to go to the rotating frame where the fictitious centrifugal force is easily handled by introducing the effective potential energy.

    I have never studied thermodynamics of accelerating systems, so I would not feel comfortable trying to analyze the system from the point of view of the non-rotating reference frame.
     
  12. Feb 19, 2015 #11
    Ok, I'm fairly happy with that.

    Going back to the first issue I was having, why is this the case:
    say we're considering the x component of kinetic energy... I believe P(E)∝exp(mvx2/2kT) here, and my book then says that P(vx)∝exp(mvx2/2kT), i.e that probability of having some vx2 directly transforms into the probability of having vx

    Whereas here, the probability of having some r2 doesn't directly transform into the probability of having r, because we said rn(r)∝rP(r)∝P(E)
     
  13. Feb 19, 2015 #12

    TSny

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    exp(-mvx2/2kT)dvx is proportional to the probability of finding a particle with an x-component of velocity between vx and vx + dvx. Likewise for the other components. So, the probability of finding a particle that simultaneously has x-component of velocity between vx and vx + dvx, y-component of velocity between vy and vy + dvy, and z-component of velocity between vz and vz + dvz, is proportional to exp[-m(vx2+vy2+vz2) /(2kT)]dvx dvy dvz.

    However, if you want the probability P(E)dE that a particle has a kinetic energy between E and dE, then you must take into account the degeneracy of E. This degeneracy is proportional to the area (4 π v2) of a sphere of radius equal to the speed v. Thus P(E) dE ∝ v2exp[-mv2/(2kT)].

    Similarly, in the rotating gas, n(r) as used in the link is proportional to the probability of finding a particle in a small volume dxdydz located a distance r from the center. The "degeneracy" or number of volume elements located a distance r from the center is proportional to r for a cylinder. So, the probability of a particle "being a distance r from the center" is the probability that it is in any one of these volume elements. Thus, the probability of a particle lying anywhere between r and r + dr is P(r)dr ∝ rn(r)dr.
     
  14. Feb 19, 2015 #13
    Ah so it is all consistent, thanks a lot :)
     
  15. Feb 19, 2015 #14
    Sorry, I know I said I was happy with everything, but I've tried to do this problem via another method, using the grand partition function and I seem to be misunderstanding something. I think for simplicity we can switch to the example of molecules in an isothermal atmosphere now - as on page 174 here (section 13.2.1.1):
    http://www.google.co.uk/url?sa=t&rc...DcqdHU2eepXvXRRlKgNj7qA&bvm=bv.86475890,d.d2s

    I understand that when writing down the grand partition function, you can sum over N and split it into a product of:
    exp(BNu)
    the N particle partition function (i.e not the grand one).

    Note B=beta, u=mu.

    Now the usual partition function should just be a product of the N particle partition function for the translational modes of an ideal gas (I'm fine with this) and that due to the gravitational field itself (which I'm confused about). So my choice for this would be exp(-Bmgz), and then I would want to integrate over all states (partition function is sum over states) and then raise it to the power N. However the integration isn't done for some reason (sort of makes sense - it would diverge! However why? Because if we go back to my original problem with the rotating gas, these notes do the same thing - they don't integrate/sum over all states, and in that case the integral won't diverge).
     
  16. Feb 20, 2015 #15

    TSny

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    If I'm understanding your question, you are asking why exp(-βmgz) wasn't included in the N-particle partition function and integrated over z.

    I believe it is due to the fact that the author is constructing a grand partition function for a particular height z. That is, the "system" can be considered to be a thin horizontal layer of gas at a height z. The number of molecules in this layer is not constant due to diffusion in and out of the layer. So, z is a parameter which is fixed for a particular layer. You could put the exp(-βmgz) into the partition function, but z would not be integrated. So, the partition function for a single particle would have an overall factor of exp(-βmgz). Then, the partition function for the layer when it has N particles would have a factor [exp(-βmgz)]N = exp(-βmgzN). When constructing the grand partition function you sum the N_particle partition funciton over all N with an additional factor of exp(βμN) in the summation. Thus, you end up summing exp[β(μ-mgz)N] times the N-particle partition function. This gives the author's expression for the grand partition function.

    This is my understanding, anyway.
     
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