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Homework Help: Finding canonical partition function and U when permitted E-levels are given (3d box)

  1. Nov 14, 2011 #1
    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 [itex]n_{x}[/itex], [itex]n_{y}[/itex] and [itex]n_{z}[/itex]:

    [itex]\epsilon = \frac{hc\sqrt{n_x^2 + n_y^2 + n_z^2}}{2L}[/itex],
    where n[itex]_{x}[/itex], n[itex]_{y}[/itex] and n[itex]_{z}[/itex] 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 [itex]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}}[/itex]

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

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

    b) [itex]Z_1 = e^{-\frac{\epsilon_1}{kT}} = e^{-\frac{\sqrt{3}hc}{2LkT}}[/itex]

    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...
  2. jcsd
  3. Nov 17, 2011 #2
    Re: Finding canonical partition function and U when permitted E-levels are given (3d

    I tried another approach after some hint from another student, setting [itex]\epsilon_1' = 0[/itex] and [itex]\epsilon_2' = \epsilon_2 - \epsilon_1[/itex]. 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}
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