# Exploring Ultrarelativistic Particles in 3-D Cubic Boxes

• PerUlven

#### PerUlven

The given problem:
The permitted energy values for a massless (or ultrarelativistic) particle (kinetic energy much larger than rest energy) in a 3-dimensional cubic box of volume V = L^3, can be expressed in terms of quantum numbers $n_{x}$, $n_{y}$ and $n_{z}$:

$\epsilon = \frac{hc\sqrt{n_x^2 + n_y^2 + n_z^2}}{2L}$,
where n$_{x}$, n$_{y}$ and n$_{z}$ must be positive integers.

a) What are the lowest two energy levels for this system and their degeneracy?

b) Write down an expression for the canonical partition function $Z_1[\itex] for 1 particle at low temperature c) Determine the energy U and heat capacity [itex]C_V[\itex] in the limit of low T. Relevant equations b) This is the equation I've been trying to use for Z: [itex]Z = \sum_i e^{-\frac{\epsilon_i}{kT}}$

c) The "shortcut formula" $U = -\frac{\partial}{\partial \beta}\ln Z$, where $\beta = 1/kT$,
and $C_V = \left ( \frac{\partial U}{\partial T} \right )_{N,V}$

Attempt at a solution
a) $\epsilon_1 = \frac{hc\sqrt{3}}{2L}$, not degenerate ($d = 1$)
$\epsilon_2 = \frac{hc\sqrt{6}}{2L}$, thrice degenerate $d = 3$

b) $Z_1 = e^{-\frac{\epsilon_1}{kT}} = e^{-\frac{\sqrt{3}hc}{2LkT}}$

Could anyone tell me if is correct? b) doesn't seem correct to me, since they're asking us to sketch U(T) and Cv(T) at low T and comment on the temperature dependence later in the task, seing as neither of them depend on T...

I tried another approach after some hint from another student, setting $\epsilon_1' = 0$ and $\epsilon_2' = \epsilon_2 - \epsilon_1$. Then the temperature dependence doesn't disappear when I derivate to find U and Cv, but the expressions doesn't look very nice. This is part 1 of a problem, so I really think the answers should "look better".

This is what I get then:
$U = \frac{\sqrt{6}-\sqrt{3}}{\exp\left(\beta\frac{(\sqrt{6}-\sqrt{3})hc}{2L}\right) + 1}\left(\frac{hc}{2L}\right)$
$C_V = \left(\sqrt{6}-\sqrt{3}\right)^2\left(\frac{hc}{2L}\right)^2 \frac{\exp\left(\frac{(\sqrt{6}-\sqrt{3})hc}{2LkT}\right)} {kT^3\left(\exp\left(\frac{(\sqrt{6}-\sqrt{3})hc}{2LkT}\right) + 1\right)^2}$