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I Is an integral over phase space reasonable?

  1. Sep 1, 2016 #1
    In statistical physics, the partition function should be calculated in the whole phase space.This is finally an integral over the phase space, like ∫d3Nqd3Np...

    The problem is that the integral covers some cases that different particles have the same generalized coordinates and momenta. So, is this reasonable?
     
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  3. Sep 1, 2016 #2

    Orodruin

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    It is not the same generalised coordinate. You have N copies of the same single particle system and therefore need N different copies of the configuration space for one particle in order to describe the many particle configuration space. Each copy is different, even though you may be using the same way to describe each single particle state. (You could use different parametrisations of the single particle states, but this really seems like an overcomplication.)

    The phase space integral is well defined.
     
  4. Sep 1, 2016 #3

    Orodruin

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    Also, the points with same coordinates and momentum have measure zero and do not contribute to the integral.
     
  5. Sep 2, 2016 #4
    Thank you!

    Why should we separate the N particles into single particles and do integrals in different copies of generalized coordinates. I mean, considering the N particle in a same container, I cannot figure it out.

    Your second reply is very interesting. is there any mathematical evidence or physical explanation?
     
  6. Sep 2, 2016 #5

    Orodruin

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    The configuration space is just a collection of coordinates that tell you the configuration of the system. With N particles and no additional constraints, these can be taken to be the spatial coordinates of the particles. There is nothing strange going on here.
    There is nothing strange here either. Just that the integral over a single point of any finite function is zero.
     
  7. Sep 2, 2016 #6
    Thank you. I think I may take the calculation of partition function as an example. Just simplify the case. Three spatial coordinates (three energy levels), two particle, no inter-particle interaction. There are 6 combinations
    1 2 3
    x x 0 E1
    x 0 x E2
    0 x x E3
    xx 0 0 E4
    0 xx 0 E5
    0 0 xx E6
    The partition function is [exp(-\beta*E1) + exp(-\beta*E2) + ...+ exp(-\beta*E6)], but the last three terms in the sum do have contribution to the calculation. Anywhere wrong with it?
     
  8. Sep 2, 2016 #7

    Orodruin

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    It is utterly unclear what you mean by this.
     
  9. Sep 2, 2016 #8
    Just transform the integral into summation considering only two particles and three discrete allowed spatial coordinates. "1-3" are three coordinates. "x" means one particle occupies one coordinate. "xx" means two particle occupy the same coordinate. "0" is null.
     
  10. Sep 2, 2016 #9

    Orodruin

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    Classically, this would be fine unless you have some reason to assume that two particles cannot be in the same state. Depending on whether your particles are distinguishable or not you may be missing states. The discrete case is also manifestly different from the continuous case. You can no longer assume that a single given state has measure zero so it does not really relate to your original question.
     
  11. Sep 2, 2016 #10
    Yes, if the particles are distinguishable, I missed the states of exchanging the particles. This is not the topic here. The integral should be a generalized form which can be reduced to any special case like what I just used.
     
  12. Sep 2, 2016 #11

    Orodruin

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    I think you are focusing way too much on this issue, which is not really an issue. Any state in your total phase space can be determined by giving the single particle states of each particle. If there are any states that are forbidden, you can simply remove them from your total phase space and then you will not be adding those to the partition function. Whether or not this (a) should be done or (b) actually matters might differ from situation to situation.
     
  13. Sep 2, 2016 #12
    Dear Orodruin, I know this cannot be an issue because it is quite fundamental. It is where I get confused when I read the textbook. I know my understanding most probably went wrong, but I don't know where. I understand what you mean, but this does not seem to have made it clear. I am reading the book <<Thermodynamics and Statistical Mechanics>> by W. Greiner (1995). The part of dealing with the noninteracting particles is to derive a partition function in the form of single-particle one using the integral over the phase space (page 170). If like what you said the deduction should differ from situation to situation and there will be no general results.
     
  14. Sep 2, 2016 #13

    Orodruin

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    Of course there are general results, but they might depend on what kind of assumptions you make. For example, when you deal with quantum statistical mechanics, it will make a large difference whether you consider bosons or fermions because the phase spaces will be different.

    When you consider classical, distinguishable, non-interacting particles, the phase space is just the N copies of the single particle phase space.

    I do not have the book by Greiner so I cannot check explicitly what you are referring to.
     
  15. Sep 2, 2016 #14

    mfb

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    If the particles do not interact, and if they are distinguishable, there is nothing wrong with two particles at the same location in space and with the same momentum. In the integral that is as interesting as a particle at the arbitrary coordinate origin: so what. You can even shift the coordinate system for one of the particles, and suddenly they will have different coordinates while nothing physical changed.
     
  16. Sep 2, 2016 #15
    The P170 of this book deduce the partition function of n noninteracting particles in a canonical ensemble. The result is
    Z_n = (Z_1)^n/n! ----nondistinguishable
    Z_n = (Z_1)^n -- distinguishable
    Z_1 is the one-particle partition function. There is no other assumption. So this is why I get confused, because points with multiple particles at the same location are covered in the integral and this is hard to understand from physics. BTW, mfb's post seems to be similar to your second reply. I think you are right, single points do not contribute to the integral mathematically. But how about a discrete case like I mentioned.
     
  17. Sep 2, 2016 #16
    Thank you, mfb. Even they are not distinguishable, the n-particle partition function (canonical) can still be expressed in single-particle partition function: Z_n = (Z_1)^n/n!

    I can figure it out mathematically that the single points (which correspond to multiple particles at same locations) do not contribute to the integral, as Orodruin also said. I just get confused in the discrete case where these points contribute to the summation (see above). Maybe I missed something when I compared a discrete case with a continuous case?
     
  18. Sep 2, 2016 #17

    mfb

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    If you have fermions and discrete states, then I think you have to take into account that some states are impossible. That is some sort of interaction then, because the state of one particle depends on the state of the other particle. For bosons there is no strict exclusion but you get different statistics then.
     
  19. Sep 2, 2016 #18
    I think Z_n = (Z_1)^n/n! should be a general result deduced from the integral. There is no other constraints in the deduction. If we consider interactions, this is not a noninteracting system any more. A general result should also be applicable to discrete cases. But if we exclude the points with multiple particles at same locations in a discrete case, we cannot get the same expression of Z_n. This is the difference.
     
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