# Getting the pressure of a system by series expansion

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:

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

Where:

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

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

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:

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

• 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?
– user64968
Dec 2 '18 at 16:59
• 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.
– user64968
Dec 3 '18 at 19:50
• @LonelyProf My bad, actually it has to be $A$ because we are dealing with a surface. Dec 3 '18 at 19:58

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.
• 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. Dec 4 '18 at 19:36