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  • $\begingroup$ Thank you very much for your response. If I am understanding correctly, taking the integral over momentum essentially allows us to account for the $g_i$'s (since those were supposed to account for degeneracy). If I did the math right, this means $\int{d^3p}=4\pi \int{dE \ m\sqrt{2mE}}$ (assuming spherical symmetry). It seems almost too nice for it to work out like this in general... If you have a reference I can consult for further clarification that would be extremely helpful... as a student self learning its been quite difficult to clarify these seemingly minor inconsistencies $\endgroup$
    – user62783
    Commented Jul 5 at 5:45
  • $\begingroup$ I went through the calculations and ended up at your answer. To clarify, this only works in the case of 0 interactions between particles in a box with the reservoir having "infinite"/very large energy? $\endgroup$
    – user62783
    Commented Jul 5 at 7:20
  • $\begingroup$ The point is, if you have a box of finite volume, strictly speaking you get a sum and not an integral. And you always get the extra V when you approximate the sum as an integral. I think your $g_i$ only accounts for stuff like spin degeneracy, but not the $\sqrt{E}$ density of state you write down here. As for reference, any solid state book will talk about this in chapter 2 (chapter 1 is always general handwaving, you know.) Piers Coleman's "Intro to many-body physics" is quite accessible. Or try Ashcroft & Mermin if you prefer old classics. $\endgroup$
    – T.P. Ho
    Commented Jul 5 at 7:28
  • $\begingroup$ I was getting the wrong answer because I didn't know how to account for the density of states--looks like this can be evaluated directly via integrating energy, same as your method but interesting to see that this works: if $\Gamma(E)=\sum_{e_i <E} \sim L_x L_y L_z \frac{4\pi}{3}\frac{2mE}{(\bar{h} \pi)^2}^{3/2}*2^{-3}$ (1/8 since x,y,z>0), $\sum_{e_i} f(e_i)=\int{f(e_i)d\Gamma}=\int{f(e_i)\frac{d\Gamma}{dE}dE}$. changing variables gives $\int{f(e(\vec{p}))\frac{d^3 p}{(2\pi \bar{h})^3}}$. cool to see it work using this method as well (even though its essentially the same as yours) $\endgroup$
    – user62783
    Commented Jul 5 at 7:30
  • $\begingroup$ sorry for the messy latex--ran out of space. but thanks a lot for clarifying. makes much more sense now. I assumed this was a result of some much more complicated logic but failed to see there were (very) simplifying assumptions put In place. this kind of upsets me since when I began learning stat mech one of the first things I did was derive that $\Gamma$... would've saved me a massive headache if I tried to use it earlier $\endgroup$
    – user62783
    Commented Jul 5 at 7:33