Solid State Physics 2013
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Re: The Debye calculation

Posted by Steve Simon on March 15, 2013, 7:44 pm, in reply to "Re: The Debye calculation"

 The counting of degrees of freedom used for equipartition theorem is not the same as the counting used here. For a single oscillator mode, the energy is always hbar omega (n+1/2) ** If there are two directions of motion, there are two modes etc. If you derive specific heat from this (by using the bose factor for n) then you take the high temperature limit, you will get Cv = kb. In order to get this from equipartition theorem you must then count momentum and position as two degrees of freedom. So the two agree. [ You can also check using classical stat mech (homework problem 1.1) that you get Cv=kb and in essence you are counting x and p as different degrees of freedom. ]

 Message Thread The Debye calculation - Sean March 15, 2013, 12:29 am Re: The Debye calculation - Steve Simon March 15, 2013, 11:16 am Re: The Debye calculation - Sean March 15, 2013, 7:05 pm Re: The Debye calculation - Steve Simon March 15, 2013, 7:44 pm « Back to index