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I am trying to understand why the Boltzmann distribution is of the specific form:

$$ f(E) = A e^{-E/B} $$ for some constants $A$ and $B$. I am following the discussion here:

https://courses.physics.ucsd.edu/2017/Spring/physics4e/boltzmann.pdf

The author gives an example of a six particle system with eight discrete energy levels. I am happy with the argument that:

$$ n(E_i) = \sum_j n_{i,j} \, p_j $$

where $p_j$ is the probability of the system being in the $j$-th microstate, $n_{i,j}$ is the number of particles with energy $E_i$ in the $j$-th microstate and $n(E_i)$ is the expected number of particles to be found with energy $E_i$.

I have checked numerically that the distribution does have an approximately negative exponential shape in this small example.

I get confused however at the next step of the argument. Let $f(E_i)$ denote the probability that a given particle has energy $E_i$. Then, since there are six particles in our system we have:

$$ f(E_i) = \frac{n(E_i)}{6} $$

Clearly $f(E)$ has the same shaped distribution as $n(E)$, since they are proportional. Consider now the probability $h(E_0 + E_2)$ that a pair of particles has combined energy $E_0 + E_2$.

By my calculation: $$ h(E_0 + E_2) = 2f(E_2)f(E_0) + f(E_1)^2 $$ Why? Because either one particle has energy $E_0$ while the other has energy $E_2$ or both particles have energy $E_1$.

This is now, however, what the author has written. He claims that: $$ h(E_0 + E_2) = f(E_0)f(E_2) $$ I agree that if we could show that: $$ f(E_i + E_j) = f(E_i)f(E_j)$$ Then the exponential form of the distribution would follow immediately. However I fail to see the relation between the probability $f(E_i + E_j)$ that a single particle has energy $E_i + E_j$ and the probability $h(E_i + E_j)$ that a pair of particles has combined energy $E_i + E_j$.

Obviously I am doing something wrong. Any clarification would be greatly appreciated.

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  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Jul 9 '18 at 7:31
  • $\begingroup$ Jorge Hirsch "Classical Concept Review 7: Derivation of the Boltzmann Distribution". Physics 4E, Quantum Physics, Spring 2017, University of California, San Diego. $\endgroup$ – Andrew Davidson Jul 9 '18 at 10:51
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You seem to be slightly misunderstanding the text. $h(E_1+E_2)$ is not the probability that two particles have a total energy $E_1+E_2$.

The argument goes like this: Imagine that you have a bag of particles with a discrete, equally-spaced set of allowed energies. The probability of reaching into your bag and pulling out a particle with energy $E_i$ is $f(E_i)$, which means that the probability of reaching into your bag and pulling out two particles with respective energies $E_0$ and $E_2$ is $f(E_0)\times f(E_2)$.

Notice that this is not the probability of pulling out two particles with total energy $E_0+E_2$. To get the latter, we would need to consider all of the possible ways (read: microstates) in which we might get that total. Neglecting subtle arguments about particle distinguishability, we could have

  • Particle A has energy $E_0$ while particle B has energy $E_2$
  • Particle A has energy $E_2$ while particle B has energy $E_0$
  • Particle A and particle B both have energies $E_1$

However, on the bottom of page 2:

If we now make the reasonable assumption that all microstates occur with the same probability [...]

This is the crucial assumption, and is actually a very deep one. If we make it, however, it says that each of the three possibilities enumerated above occurs with the same probability. It is this probability which we call $h(E_0+E_2)$.

If this is true, then in particular, we have that $h(E_0+E_2)=f(E_0)\times f(E_2)^\dagger$, which is the first of the three possible microstates. The only $f$ which could satisfy such a relationship is an exponential function, as stated in the text.

Of course, this logic holds if the allowed energies are not evenly spaced; I only included that part to mirror your question. The fundamental assumption is that each energetically allowed microstate occurs with equal probability.


$^\dagger$ Your linked text has a typo - on the top of page 23, that should be $f(E_1)\times f(E_2)$, not $f(E_1)+f(E_2)$.

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  • $\begingroup$ Thankyou for your answer. I will have to think about it some more. Does what you are saying about the assumption of equally likely microstates imply that $f(E_1)^2 = f(E_2)f(E_0)$ ? (and I agree about the typo). $\endgroup$ – Andrew Davidson Jul 9 '18 at 10:39
  • $\begingroup$ I think I get it. For the full 6 particles, and fixed energy, all possible ways that the energy could be divided up amongst them occur with equal probability. Restricting now to the two particle system we have the same assumption, so indeed $f(E_1)^2 = f(E_2)f(E_0)$. $\endgroup$ – Andrew Davidson Jul 9 '18 at 10:45
  • $\begingroup$ Yes, that's exactly right. Based on that assumption, it follows that the likelihood of a particular macrostate is equal to $h$ times the number of microstates that correspond to it, and that the most probable macrostate is the one which corresponds to the largest number of microstates. For technical reasons (including that this number is absurdly large for systems of ~$10^{23}$ particles) it is more useful to talk about the natural log of this number; if $W$ is the number of microstates corresponding to a particular macrostate of the system, then we call $\ln(W)$ its entropy. $\endgroup$ – J. Murray Jul 9 '18 at 11:15
  • $\begingroup$ (In standard thermodynamic units, we define the entropy with a factor of the Boltzmann constant, so $S=k\ln(W)$, but this changes precisely nothing and should not obscure the clear meaning of $S$) $\endgroup$ – J. Murray Jul 9 '18 at 11:17
  • $\begingroup$ Let me check if I understand. There is a normalization constant $A = 1/Z$. where $Z = \sum_i e^{-E_i/B}$. We need to show that $f(E_1 + E_2) = Z f(E_1)f(E_2)$. Note firstly that $f(0) = 1/Z$. If $E = E_1 + E_2 + E_3$ is the total energy of the system and there are $n$ particles, then by the equiprobability of microstates we have: $f(E_1+E_2) f(E_3) f_(0)^{n-2} = f(E_1)f(E_2)f(E_3)f(0)^{n-3}$ and so f(E_1 + E_2) = f(E_1)f(E_2)/f(0) = Z f(E_1)f(E_2)$ as desired. $\endgroup$ – Andrew Davidson Jul 10 '18 at 0:19

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