How do ultrarelativistic particles behave in 3-D cubic boxes?

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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<br /> <br /> c) Determine the energy U and heat capacity C_V[\itex] in the limit of low T.&lt;br /&gt; &lt;br /&gt; &lt;b&gt;Relevant equations&lt;/b&gt;&lt;br /&gt; b) This is the equation I&amp;#039;ve been trying to use for Z:&lt;br /&gt; Z = \sum_i e^{-\frac{\epsilon_i}{kT}}&lt;br /&gt; &lt;br /&gt; c) The &amp;quot;shortcut formula&amp;quot; U = -\frac{\partial}{\partial \beta}\ln Z, where \beta = 1/kT, &lt;br /&gt; and C_V = \left ( \frac{\partial U}{\partial T} \right )_{N,V}&lt;br /&gt; &lt;br /&gt; &lt;b&gt;Attempt at a solution&lt;/b&gt;&lt;br /&gt; a) \epsilon_1 = \frac{hc\sqrt{3}}{2L}, not degenerate (d = 1)&lt;br /&gt; \epsilon_2 = \frac{hc\sqrt{6}}{2L}, thrice degenerate d = 3&lt;br /&gt; &lt;br /&gt; b) Z_1 = e^{-\frac{\epsilon_1}{kT}} = e^{-\frac{\sqrt{3}hc}{2LkT}}&lt;br /&gt; &lt;br /&gt; Could anyone tell me if is correct? b) doesn&amp;#039;t seem correct to me, since they&amp;#039;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...
 
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I tried another approach after some hint from another student, setting \epsilon_1&#039; = 0 and \epsilon_2&#039; = \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:
<br /> U = \frac{\sqrt{6}-\sqrt{3}}{\exp\left(\beta\frac{(\sqrt{6}-\sqrt{3})hc}{2L}\right) + 1}\left(\frac{hc}{2L}\right)<br />
<br /> 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}<br />
 
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