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Specific heat for a 1-D lattice

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data
    A 1-D lattice consists of a linear array of N particles (N>>1) interacting via spring-like nearest neighbor forces. The normal node frequencies are given by
    [tex]\omega_n=\omega_0\sqrt{\,\,2-2\cos\left(2\pi n/N\right)}[/tex]
    where [itex]\omega_0[/itex] is a constant and n an integer ranging from -N/2 and +N/2. The system is in thermal equilibrium at temperature T. Let cv be the constant length specific heat.

    a) Compute cv for the T -> infinity.

    b) For T -> 0, [itex]c_v\rightarrow A \omega^{-\alpha}T^{\gamma}[/itex] where A is a constant you need not compute. Compute the exponents [itex]\alpha\text{ and }\gamma[/itex].


    2. Relevant equations
    [tex]U=(1/2+n)\hbar\omega\\\\c_v=\left.\frac{d U}{d T}\right|_V[/tex]



    3. The attempt at a solution
    a) Using the energy of a harmonic oscillator listed above, we have
    [tex]U=N(1/2+n)\hbar\omega_0\sqrt{\,\,2-2\cos\left(2\pi n/N\right)}[/tex]
    which does not have T in it explicitly. It seems n should depend on T, that in order to cause it to go to the next excited mode, one would have to add enough energy to increment n but I am not sure how to express that.

    b) For T->0, the lowest energy state is when n=0 -> wn=0, no motion/energy. That does not really tell me anything useful though. I feel its the same question, how does the energy (or n) depend on T?
     
  2. jcsd
  3. Mar 23, 2012 #2
    You are confusing your formulas here. [tex] U_1 = \hbar\omega_n = \hbar \omega_0 \sqrt{2-2 \cos\left( \frac{2 \pi n}{N} \right) } [/tex]

    is the energy for a single oscillator

    What you would usually do here is to calculate the partition function, and then find the thermodynamics from that. In this case, you cannot really calculate it explicitly, but you can find T->0 and T-> infinity limits.
     
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