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Degenerate matter total energy

  1. Jul 1, 2008 #1


    Fermi energy:
    [tex]E_f =\frac{\hbar^2 \pi^2}{2m L^2} n_f^2[/tex]

    The number of states with energy less than [tex]E_f[/tex] is equal to the number of states that lie within a sphere of radius [tex]|\vec{n}_f|[/tex] in the region of n-space where nx, ny, nz are positive. In the ground state this number equals the number of fermions in the system:
    [tex]N = 2\times\frac{1}{8}\times\frac{4}{3} \pi n_f^3[/tex]

    The factor of two is because there are two spin states, the factor of 1/8 is because only 1/8 of the sphere lies in the region where all n are positive, the octant of positive quantum numbers.

    [tex]n_f=\left(\frac{3 N}{\pi}\right)^{1/3}[/tex]

    According to Wikipedia (ref. 1), the degenerate matter total energy is obtained by integrating the Fermi energy of each state over the allowed states:
    [tex]E_t = {\int_0}^{N_0} E_f(N) dN = {3\over 5} N_0 E_f[/tex]

    However, according to ref. 2, total energy is:
    [tex]E_t = \frac{1}{8} \int_0 ^{N_0} E_f(N) dN[/tex]

    1/8 is from the octant of positive quantum numbers in a sphere region where all n are positive.

    My question is, why is the octant of positive quantum numbers included in the ref. 2 integration, but not included in the Wikipedia integration?

    If the octant of positive quantum numbers is included in the integration, then why is the number of spin states also not included?

    Also, I noticed that the Fermi energy equation in ref. 1 does not match the equation in ref. 2. Does this mean that the ref. 2 solution is incorrect?

    Reference:
    Fermi energy - Wikipedia
    PHYS 385 Lec. 18 - Lecture 18 - Degenerate matter - Professor Boal
     
    Last edited: Jul 1, 2008
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  3. Jul 1, 2008 #2

    malawi_glenn

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    Well it doesent matter where you put your 1/8, as long as you do it and know why you do it. Athor of wiki is making this from the begining, the author of ref 2 does it afterwards, since you deal with quadratic entities, and they are symmetric with respect to sign interchange.

    Also regarding reference 2, the energy for each state is [tex] E_n = n^2 p^2 h^2 \pi^2 / mL^2 [/tex]

    So the author of ref 2 has already included the factor 2 from the number of spin states there ;-) But he has not explicity said that he has, so one has to be careful.

    You can see if you get the true result using both references conventions, as a test for yourself

    Anymore questions regarding this?
     
  4. Jul 1, 2008 #3

    [tex]E_t = {\int_0}^{N_0} E_f(N) dN = {\int_0}^{N_0} \frac{\hbar^2 \pi^2}{2m L^2} \left( \frac{3 N}{\pi} \right)^{2/3} dN[/tex]

    [tex]E_t = \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2}{2m L^2} {\int_0}^{N_0} N^{\frac{2}{3}} dN = \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2}{2m L^2} \left( \frac{3}{5} N^{\frac{5}{3}} \right) = \frac{3^{\frac{5}{3}} \pi^{\frac{4}{3}} \hbar^2}{10 m L^2} N^{\frac{5}{3}}[/tex]

    Degenerate matter total Fermi energy:
    [tex]\boxed{E_t = \frac{3^{\frac{5}{3}} \pi^{\frac{4}{3}} \hbar^2}{10 m L^2} N^{\frac{5}{3}}}[/tex]
     
  5. Jul 1, 2008 #4

    malawi_glenn

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    looks ok, good job!
     
  6. Jul 1, 2008 #5
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    Last edited: Jul 1, 2008
  7. Jul 1, 2008 #6

    malawi_glenn

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    Wait 3sec and I'll check your calculations more carefully
     
  8. Jul 1, 2008 #7

    malawi_glenn

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    You intepretation is wrong.

    the subscript 0 is perhaps confusing you...

    You have entierly correct in post #3, but you must keep subscript 0:

    [tex]E_t = \frac{3^{\frac{5}{3}} \pi^{\frac{4}{3}} \hbar^2}{10 m L^2} N_0^{\frac{5}{3}}[/tex]

    That is the total energy of a system of N_0 identical fermions at T=0K.

    The Fermi energy is defined as the maximum energy, i.e at particle # N_0
    [tex]E_f =\frac{\hbar^2 \pi^2}{2m L^2} \left(\frac{3 N_0}{\pi}\right)^{2/3}[/tex]

    Then it is a trivial thing to rewrite the total energy as something times E_f
     
  9. Jul 1, 2008 #8

    [tex]E_t = {\int_0}^{N_0} E_f(N) dN = {\int_0}^{N_0} \frac{\hbar^2 \pi^2}{2m L^2} \left( \frac{3 N}{\pi} \right)^{\frac{2}{3}} dN[/tex]

    [tex]E_t = \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2}{2m L^2} {\int_0}^{N_0} N^{\frac{2}{3}} dN = \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2}{2m L^2} \left( \frac{3}{5} N_0^{\frac{5}{3}} \right) = \frac{3^{\frac{5}{3}} \pi^{\frac{4}{3}} \hbar^2}{10 m L^2} N_0^{\frac{5}{3}}[/tex]

    Degenerate matter total Fermi energy:
    [tex]\boxed{E_t = \frac{3^{\frac{5}{3}} \pi^{\frac{4}{3}} \hbar^2}{10 m L^2} N_0^{\frac{5}{3}}}[/tex]
     
  10. Jul 1, 2008 #9

    malawi_glenn

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    yes, and by using:

    [tex]E_f =\frac{\hbar^2 \pi^2}{2m L^2} \left(\frac{3 N_0}{\pi}\right)^{2/3}[/tex]

    You'll get the famous result:

    [tex] E_t = \frac{3}{5}N_0E_f [/tex]
     
  11. Jul 1, 2008 #10

    Integration by substitution:
    [tex]E_t = \frac{3}{5} N_0 E_f = \frac{3}{5} N_0 \left[ \frac{\hbar^2 \pi^2}{2m L^2} \left(\frac{3 N_0}{\pi}\right)^{2/3} \right] = \frac{3^{\frac{5}{3}} \pi^{\frac{4}{3}} \hbar^2}{10 m L^2} N_0^{\frac{5}{3}}[/tex]

    [tex]\boxed{E_t = \frac{3}{5} N_0 E_f}[/tex]
     
  12. Jul 2, 2008 #11

    Degenerate total Fermi energy:
    [tex]E_t = \frac{3^{\frac{5}{3}} \pi^{\frac{4}{3}} \hbar^2}{10 m L^2} N_0^{\frac{5}{3}}[/tex]

    The elimination of L in favor of V:
    [tex]L^2 = V^{\frac{2}{3}}[/tex]

    Degenerate total Fermi energy:
    [tex]E_t = \frac{3^{\frac{5}{3}} \pi^{\frac{4}{3}} \hbar^2 N_0^{\frac{5}{3}}}{10 m V_0^{\frac{2}{3}}}[/tex]

    Degenerate Fermi pressure:
    [tex]P_f = - \frac{\partial E}{\partial V}[/tex]

    Integration by differentiation substitution:
    [tex]P_f = - \frac{3^{\frac{5}{3}} \pi^{\frac{4}{3}} \hbar^2 N_0^{\frac{5}{3}}}{10 m} \left( \frac{V_0^{- \frac{2}{3}}}{dV} \right)[/tex]

    Differentiation identity:
    [tex]\frac{V^{-n}}{dV} = -n V^{-n - 1} = \left(-\frac{2}{3} \right) V^{-\frac{2}{3} - \frac{3}{3}} = \left(- \frac{2}{3} \right) V^{-\frac{5}{3}}[/tex]

    [tex]N_0 = \frac{M_0}{m} \; \; \; \rho_0 = \frac{M_0}{V_0}[/tex]

    Integration by substitution:
    [tex]P_f = - \left(- \frac{2}{3} \right) \frac{3^{\frac{5}{3}} \pi^{\frac{4}{3}} \hbar^2}{10 m} \left( \frac{N_0}{V_0} \right)^{\frac{5}{3}} = \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2}{5 m} \left( \frac{M_0}{m V_0} \right)^{\frac{5}{3}} = \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2 \rho_0^{\frac{5}{3}}}{5 m^{\frac{8}{3}}}[/tex]

    Degenerate Fermi pressure:
    [tex]\boxed{P_f = \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2 \rho_0^{\frac{5}{3}}}{5 m^{\frac{8}{3}}}}[/tex]

    Is my equation solution for degenerate Fermi pressure incorrect?

    Reference:
    PHYS 390 - Lecture 18
    PHYS 390 - Lecture 19
     
    Last edited: Jul 2, 2008
  13. Jul 2, 2008 #12

    malawi_glenn

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    You have the answer in eq.3 in the souce 'lecture 19', is it difficult to see if the answers match??
     
  14. Jul 2, 2008 #13


    That degenerate Fermi pressure solution depends on if this solution for Fermion number is valid for a degenerate neutron star?

    Fermion number:
    [tex]\boxed{N_0 = \frac{M_0}{m_n}}[/tex]

    key:
    [tex]M_0[/tex] - total stellar mass
    [tex]m_n[/tex] - neutron mass
     
  15. Jul 2, 2008 #14

    malawi_glenn

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    What?

    How to go from a fermi pressure in theory to a Neutron Star are some more steps dude.

    What is your aim by doing all these calculations on degenerate matter?
     
  16. Jul 2, 2008 #15

    To validate my current model of a neutron star.

    Reference:
    Orion1 - neutron star model
     
  17. Jul 2, 2008 #16

    malawi_glenn

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    hmm one validates models by doing observation, but anyways, was just curious.
     
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