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Finding total energy as a function of the Fermi Energy

  • Thread starter hb1547
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  • #1
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Homework Statement


"The numerator of this fraction:

[tex]\overline{E}=\frac{\int \! E N(E)D(E)dE}{\int \! N(E)D(E)dE}[/tex]
(N(E) is the number of particles in an energy state, D(E) is the density of states)

is the total (as opposed to the average particle) energy, which we'll call [tex]U_{total}[/tex] here (In other words, the total system energy U is the average particle energy [tex]\overline{E}[/tex] times the total number of particles N.) Calculate [tex]U_{total}[/tex] as a function of [tex]E_{Fermi}[/tex] and use this to show that the minimum (T=0) energy of a gas of spin-1/2 fermions may be written as:

[tex]U_{total}=\frac{3}{10} (\frac{3\pi^2 \hbar^3}{m^{3/2}V})^{2/3} N^{5/3}[/tex]

Homework Equations


- The above.
[tex]D(E) = \frac{(2s+1)m^{3/2}V}{\pi^2 \hbar^3 \sqrt{2}}E^{1/2}[/tex]
[tex]N(E)_{FD} = \frac{1}{e^{(E-E_{f})/k_{B}T}+1}[/tex]
[tex]N(E_{F})_{FD} = \frac{1}{2}[/tex]
[tex]\overline{E} = k_{B}T[/tex]

The Attempt at a Solution


I'm having a hard time deciding what should be multiplied together, and in which ways.

I thought it would be [tex]N(E_{F})_{FD} * \overline{E}[/tex] (the ones above), yet that obviously isn't a function of [tex]E_{Fermi}[/tex], it's just a function of temperature.

I guess I'm having a hard time deciding which equations I should be using.
 

Answers and Replies

  • #2
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Just finished the same problem. I calculated the total energy using the numerator term in the average energy equation. You can integrate from 0 to E Fermi because T = 0. Then plug in the Fermi energy equation from the book (i'm assuming this is from Harris). It's eqn 9-42. After a lot of cancelation you get the expression given in the book.
 
  • #3
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Yeah this is the Harris problem haha.

So I plug in the Fermi energy equation for E in the top, as well as [tex]N(E)_{FD}[/tex] and [tex]D(E)[/tex]? Sounds good, I'll try it and see where it gets me. Thanks, I was just leaving the term as E and it wasn't working too well
 
  • #4
35
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Hmm actually I'm still having some trouble with this one. I get why you can integrate from 0 to [tex]E_{Fermi}[/tex], yet I'm not following when you plug in the equation for [tex]E_{Fermi}[/tex]. Did you do it before or after integrating?

If before, I keep getting to an integral I can't solve:
[tex]\int_{0}^{E_{F}} \! \frac{E^{1/2}}{e^{(E-E_{F})/k_{b}T}+1} dE[/tex]
(there are a bunch of constants on the outside as well)

If after, not a lot seems to cancel out -- the integral just becomes more complicated
 
  • #5
7
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Remember that in
[tex]
\overline{E}=\frac{\int \! E N(E)D(E)dE}{\int \! N(E)D(E)dE}
[/tex]
The numerator represents the total energy. This is what you want to use to calculate it. Your limits are from 0 to Fermi energy. The occupational number (N(E)) turns to 1, because at T = 0 it is a step function equaling to 1 up to the Fermi energy. The density function (D(E)) is given in the book Eqn 9-39. Do not forget to include the E that is not part of the density function. After integrating, you should have total energy in terms of Fermi energy. Now plug in the expression for Fermi energy Eqn 9-42, and simplify.
 
  • #6
35
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OH I get it -- that clears it up a lot. Thanks! I was able to get it from there, I really appreciate the help!
 

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