An average number, $N$, of bosons of spin $S = 0$ is conned to a two-dimensional domain with surface $A$. The gas is ultrarelativistic with a single particle energy $\epsilon = cp$, where $c$ is the speed of light in vacuum and $p$ is the absolute value of the momentum. Compute the pressure, P, of this system as a function of z, A and T. While still being in the high-temperature regime, use the result for $N(z, A, T)$ to find $P(N, A, T)$ (keep up to quadratic terms in N). Discuss your results and the relation to the ideal gas law.

$N(z, A, T)$ in function of $z$ has been previously calculated:

$$N = \frac{2\pi A}{(h \beta c)^2}(z + \frac{z^2}{4})$$


$$\beta = \frac{1}{K_B T}$$

$$z = e^{\beta \mu}$$

The following equation also holds:

$$\frac{P}{K_B T} = \frac{2\pi }{(h \beta c)^2}\sum_{n=1}^{\infty} \frac{z^n}{n^3}$$

With this information we should be able to get $P(N,A,T)$ (keeping up to quadratic terms in N), but I am not getting the stated pressure.

The provided solution is:

$$P \approx \frac{NK_BT}{A} (1 - \frac{N(h \beta c)^2}{16 \pi A})$$

  • 1
    $\begingroup$ I can see one or two areas where clarification might help people to answer your question. Firstly, the $n=2$ term in your sum is $z^2/8$, not $z^2/4$, so this is not consistent with your first equation. Secondly, maybe it would help to define some quantities such as $z$, and explain what system you are considering, and where the first two equations come from. Can you edit your question to explain these points a bit more? $\endgroup$
    – user64968
    Dec 2, 2018 at 16:59
  • $\begingroup$ @LonelyProf I have added more details. If there is more information you think would be helpful, please let me know. $\endgroup$
    – JD_PM
    Dec 3, 2018 at 19:28
  • $\begingroup$ Thanks. This clarifies what the series in $z$ applies to, and gives some useful background. I'm puzzled by the appearance of $V$ in that equation, though, when it does not feature in the final solution, or in any other equation. Also, do take a moment to review the policy on homework and related questions which may apply here. $\endgroup$
    – user64968
    Dec 3, 2018 at 19:50
  • $\begingroup$ @LonelyProf My bad, actually it has to be $A$ because we are dealing with a surface. $\endgroup$
    – JD_PM
    Dec 3, 2018 at 19:58

1 Answer 1


Since this is clearly an exercise, I don't think it's appropriate to give the full solution, but I see no harm in pointing you in the right direction.

You are aiming to convert an expansion in $z$ into an expansion in the number density $\rho=N/A$. These are often called virial expansions. Your first equation can be written $$ \rho = q\left[z+\frac{z^2}{4}\right] \qquad\text{where}\qquad q = \frac{2\pi}{(h\beta c)^2} . $$ I have defined $q$ just to cut down on needless clutter. The brute force approach to this would be to solve this equation, to give $z$ in terms of $\rho$, and then substitute this expression into your series formula for $P/K_BT$ as a function of $z$. You can do this; it's a bit fiddly, since it is a quadratic equation in $z$, but it could be done.

For your purposes, a slightly more straightforward approach is to assume that you will get a result of the form $$ z = c_1 \rho + c_2 \rho^2 + \ldots $$ where you need to determine the coefficients $c_1$, $c_2$ etc. Hopefully you can see why there is no need for a $\rho$-independent term $c_0$, when you look at the equation for $\rho$. Also, it should be obvious that you don't need to consider more terms like $c_3$ etc, if you are only interested in the final result having the first couple of powers of $\rho$.

So you determine the coefficients $c_1$, $c_2$ by substituting that expression for $z$ into the right hand side of the equation for $\rho$, and equating the coefficients of corresponding powers of $\rho$ on each side. The term linear in $\rho$ will give the coefficient $c_1$ immediately: $$ [\rho^1]:\qquad 1 = q \, c_1 . $$ The quadratic term in $\rho$ will give an equation for $c_2$, which also involves $c_1$, $$ [\rho^2]:\qquad 0 = \text{I'll leave you to work this out.} $$ Since you just determined $c_1$, this will be easy to solve to get $c_2$.

So now you have $z\approx c_1 \rho + c_2 \rho^2$ and you can insert this expression into the given series expansion for $P/K_BT$, only keeping the terms in $\rho$ and $\rho^2$. If you do all this carefully, you will get the answer provided.

  • $\begingroup$ Thank you, your explanation is really helpful! I got $z\approx \rho -1/4 \rho^2$ which led me to the provided answer :) By the way, for the sake of curiosity, this is the complete exercise and the provided answer: imgur.com/a/dnILMPR imgur.com/a/hrCNdoK imgur.com/a/4i5tei1 What I did not know was how to get EQ 26. $\endgroup$
    – JD_PM
    Dec 4, 2018 at 19:36

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