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I don't understand partition function

  1. Nov 29, 2013 #1
    is the boltzmann factor the probability of a particular state of a system?

    can someone explain the partition function to me (qualitatively please!), we've been using it in class and i don't get it. we derived what the partition function was for a general system. usually when learning physics and i don't understand something it's because i'm looking at the math and i don't understand the physical interpretation of something, what something "is" physically. but in this case, we're rather studying physical phenomena using statistics, and so i have to be able to look at things and understand what's going on statistically (statistics is a class which i haven't taken yet). So, statistically, i don't understand what the partition function "is". What does it tell me?
    all i know is what i can do with it, kind of... i can convert between the boltzmann factor of a particular state and the probability of that state, and that's all i know. could someone give me a better feel for what the partition function is?

    i've looked on wikipedia and google searched it but i didn't really get much..
  2. jcsd
  3. Nov 29, 2013 #2


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    The partition function in particular, and the Boltzmann statistics in general come straight out of probability theory; once I figured out just how this worked, it made a lot more sense for me.

    The situation is that you have two objects, a "system" and a "reservoir", which together form a larger system, which we'll call the kaboodle.

    The system and reservoir each can exchange energy with each other, and do so back and forth, the net effect one way or the other being zero when the system and reservoir are in thermal equailibrium (when they have the same temperature). The kaboodle we treat as a closed system, so that energy can flow in between each part of the kaboodle, but no energy can flow in or out of the kaboodle.

    Probability comes into play when we say that the energy of the system can have different values so long as the total energy of the kaboodle remains constant. The question we ask then, is:
    " What is the probability that the system has energy U, given that the system plus reservoir has total energy E?"
    Working it out, you can show that the probability that the system has total energy U is equal to its Boltzmann factor divided by the partition function.

    The partition function conceptually is just a normalization constant so all the probabilities add up to one.
  4. Nov 29, 2013 #3


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    Yep. The partition function is the normalization constant for the probabilities. (or that's how I like to think of it, anyway). For the canonical ensemble, we know the probability of observing the system having energy ##E_i## must be proportional to
    [tex]\exp(-\beta E_i)[/tex]
    So, if we say ##1/Z## is the constant of proportionality, then we have
    [tex]P_i = 1/Z \exp(-\beta E_i)[/tex]
    And when we sum over all states ##i##, we get:
    [tex]\sum_i P_i = 1/Z \sum_i \exp(-\beta E_i)[/tex]
    and we know that (since probabilities must add to 1)
    [tex]\sum_i P_i = 1 [/tex]
    so therefore:
    [tex]Z = \sum_i \exp(-\beta E_i)[/tex]
    Last edited: Nov 29, 2013
  5. Nov 29, 2013 #4



    here is an image of my notes about the derivation of the partition function. all the terms circled in yellow are terms that are all equal to each other (but not all terms that are equal to each other have been circled)

    the bottom most yellow circle: above that is P(S2) which i forgot to circle, but those two terms are equal to one another, and are the terms on the right hand side; ie on the RHS the numerator and demoninator are both equal to one another, so could i say that Z is ultimately equal to 1?
  6. Nov 29, 2013 #5


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    Not all the stuff inside the yellow circles is equal. ##S_R(S_2) - S_R(S_1) = -\frac{1}{T}((U_{sys}(S_2)-U_{sys}(S_1))## does not imply that ##S_R(S_2)=-\frac{1}{T} U_{sys}(S_2)##
  7. Nov 29, 2013 #6
    how not? for the case where we ignore any volumetric changes due to energy exchange, and if there are no particles being exchanged, how is what you wrote not true? if you'd like you could replace the equal sign to a ≈, but aside from this, in this context, why does it not imply that

    the SR(S2) corresponds to the -(1/T)Usys(S2) and the the SR(S1) corresponds to the -(1/T)Usys(S1) i thought.
  8. Nov 29, 2013 #7


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    nope. If you think about it, mathematically, you can add whatever you want to both ##S_R(S_2)## and ##S_R(S_1)## and the left hand side of the equation doesn't change. So mathematically, that equation does not imply that ##S_R(S_2)=-\frac{1}{T} U_{sys}(S_2)##.
  9. Nov 29, 2013 #8


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    Also, it is true that
    [tex]\frac{P(S_2)}{P(S_1)} = \exp((S_R(S_2)-S_R(S_1))/k)[/tex]
    But this does not mean that
    [tex]P(S_2) = \exp(S_R(S_2)/k)[/tex]
    for similar reasons.
  10. Nov 30, 2013 #9


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    I wrote a series of notes on Boltzmann statistics, but the conversion to pdf shredded my handwriting for some reason.


    In it, I show how to get the Boltzmann factors, and how they relate to the total probabilities, along with a lot of other general knowledge. If you find it's useful, then that's great.
    You don't actually have to use the partition function to get the probabilities to come out right. It's just a very convenient way of doing things.

    Also, it covers how to deal with multiple states of the same energy (degenerate states)
  11. Nov 30, 2013 #10
    i found this pdf very helpful

    Attached Files:

  12. Nov 30, 2013 #11


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    This probably repeats what everybody else says, but it's short.

    Assume that you have a small system that can exist in a number of discrete states of energy [itex]\mathcal{E}_j[/itex]. This system is in thermal contact with a much, much huger system, say, a reservoir of water, which has vastly more possible states.

    Let [itex]W(E)[/itex] be the number of states of the reservoir with energy [itex]E[/itex]. Now, suppose we place the small system into the reservoir, and allow it to exchange energy with the reservoir. The basic assumption is that all states (of the composite small system + reservoir) having the same total energy [itex]E[/itex] are equally likely. So the probability of the small system being in state [itex]j[/itex] is just the fraction of all states of small system + reservoir such that the reservoir has energy [itex]E - \mathcal{E}_j[/itex]. This is given by:

    [itex]P_j = \dfrac{W(E - \mathcal{E}_j)}{\sum_{i} W(E - \mathcal{E}_{i})}[/itex]

    The final steps are to define the entropy of the reservoir, [itex]S(E)[/itex], to be

    [itex]S(E) = k ln(W(E))[/itex]

    or [itex]W(E) = e^\frac{S(E)}{k}[/itex] where [itex]k[/itex] is Boltzmann's constant.

    and we define temperature via:

    [itex]\frac{dS}{dE} = \frac{1}{T}[/itex]

    With these definitions,

    [itex]W(E - \mathcal{E}_{i}) = e^\frac{S(E - \mathcal{E}_i)}{k} \approx e^\frac{S(E) - \mathcal{E}_i \frac{dS}{dE}}{k} = e^\frac{S(E) T - \mathcal{E}_i }{kT}[/itex]

    where we used a Taylor series to approximate [itex]S[/itex]

    So the probability becomes:

    [itex]P_j = \dfrac{e^\frac{S(E) T - \mathcal{E}_j }{kT}}{\sum_i e^\frac{S(E) T - \mathcal{E}_i }{kT}}[/itex]

    The factor of [itex]e^\frac{S(E) T}{kT}[/itex] cancels from numerator and denominator, leaving:

    [itex]P_j = \dfrac{e^\frac{ - \mathcal{E}_j }{kT}}{\sum_i e^\frac{ - \mathcal{E}_i }{kT}}[/itex]

    Okay, so it wasn't that short. It seemed simpler when I started.
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