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[PoM] Electron gas pressure

  1. Nov 26, 2016 #1

    BRN

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    Hi guys! I have a problem with this exercise:

    1. The problem statement, all variables and given/known data

    The stars called white dwarfs may have inside them a density in order of 1011 kg m-3. For semplicity, we assume:
    • these stars are made with non interacting protons and electrons at the same quantity and with uniform density;
    • Inside, the electrons are relatevistic with energy ε=c|p|=cħ|k|;
    • Temperature is nothing.
    In these hypotheses is evaluated the pressure exerted by this electron gas.

    3. The attempt at a solution
    I'm at T=0, but electrons are relatevistic, so P=-∂U/∂V.

    Internal energy is:

    U=∫0εFg(ε)ε dε

    where:

    states density g(ε)=(2me3/2V√ε)/(√2π2ħ3)

    ε=ħck ; k2=p22=me2v22=2εme2 ⇒ k=√(2εme2)

    Then:

    U=∫0εF (2me3/2V√ε)/(√2π2ħ3) √(2εme2) ħc dε =

    = (me3/2V c εF3/2 √(εFme))/(π2ħ2)

    So:

    P=-∂U/∂V=-(me3/2 c εF3/2 √(εFme))/(π2ħ2)

    Fermi energy is:

    εF2/(2me)(3π2N/V)2/3

    Then:

    P=-3/4ħ2c(3π2)1/3(N/V)4/3

    This solution is wrong. The correct solution is:

    P=(ħ c(3π2)1/3(N/V)4/3)/4

    Could someone tell me where I'm wrong?
     
  2. jcsd
  3. Nov 26, 2016 #2

    TSny

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    This is not the correct density of states for this problem. You will need to derive the expression for g(ε) based on the assumption that ε = c|p|.
     
  4. Nov 28, 2016 #3

    BRN

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    I checked on my book and for fermions and low temperature, the density of states is:

    $$ g(\epsilon)=\frac{(2m_{e})^{3/2}V2}{4\pi^{2}\hbar^{3}}\epsilon^{1/2} $$

    but replacing ε with $$ \hbar c \sqrt{\frac{2m_{e}\epsilon}{\hbar^{2}}} $$ solution is wrong again...

    How to calculate the relativistic density of states?
     
  5. Nov 28, 2016 #4

    TSny

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    This is for the nonrelativistic case. You need the highly relativistic case.

    You can just repeat the derivation for the nonrelativistic case, but use the highly relativistic formula E = cp for the energy of an electron. Here is a link that derives the nonrelativistic case.
    http://web.eecs.umich.edu/~fredty/public_html/EECS320_SP12/DOS_Derivation.pdf

    The derivation for the relativistic case is the same up through equation 4b. You will need to modify the rest of the derivation.
     
  6. Nov 29, 2016 #5

    BRN

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    I don't understand... :(

    I start with:
    $$ g(k)dk=\frac{Vk^{2}}{\pi^{2}}dk $$
    and:
    $$ \epsilon=c\hbar k \Rightarrow k=\frac{(\epsilon-\epsilon_{c})}{c\hbar} \Rightarrow dk=\frac{d\epsilon}{c\hbar} $$
    then:
    $$ g(k)dk=\frac{V(\frac{\epsilon-\epsilon_{c}}{c\hbar})^2}{\pi^2}\frac{d\epsilon}{c\hbar}\Rightarrow g(\epsilon)d\epsilon=\frac{(\epsilon-\epsilon_{c})^2}{c^3\pi^2\hbar^3}d\epsilon $$
    the density of states is:
    $$ \frac{(\epsilon-\epsilon_{c})^2}{c^3\pi^2\hbar^3} $$
    but ## \epsilon_{c}=?? ##
     
  7. Nov 29, 2016 #6

    TSny

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    Just set ##\epsilon_c## to zero. ##\epsilon_c## comes from dealing with electrons in the conduction band of a solid.

    From ##\epsilon = cp = c \hbar k##, you can see that ##k = \frac{\epsilon}{c \hbar}##. So, use that rather than ##k = \frac{\epsilon - \epsilon_c}{c \hbar}##.

    Your work looks good, but did you drop a factor of ##V## near the end?
     
    Last edited: Nov 29, 2016
  8. Nov 30, 2016 #7

    BRN

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    I'm lost... :(

    whith ##\epsilon_{c}=0##
    $$ U=\int_{0}^{\epsilon_{F}} g(\epsilon)\epsilon d\epsilon=\int_{0}^{\epsilon_{F}} \frac{\epsilon^2}{c^3\pi^2\hbar^3} c \hbar \frac{\epsilon}{c \hbar}d\epsilon=\frac{\epsilon_{F}^4}{4c^3\pi^2\hbar^3}$$
    $$ \epsilon_{F}=\frac{\hbar^2}{2m_{e}}(3 \pi^2\frac{N}{V})^{2/3} $$
    then
    $$ U=\frac{\hbar^5}{4c^32^4m_{e}^4}3^{8/3}\pi^{10/3}(\frac{N}{V})^{8/3} $$
    $$ P=-\frac{\partial U}{\partial V}=\frac{\hbar^5}{8c^3m_{e}^4V}(3\pi^2)^{5/3}(\frac{N}{V})^{8/3} $$

    a disaster...


    in order to pass from g(k) to g(ε) i need to divide for V...

    please do not hate me!!!
     
  9. Nov 30, 2016 #8

    TSny

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    This looks good. But I don't understand why there isn't a factor of ##V##, unless ##U## represents the total energy per unit volume. In your first post, your expression for ##U## for the nonrelativistic case does have a factor of ##V##.

    This is the Fermi energy for the nonrelativistic case. You will need to derive it for the extreme relativistic case. Hint: What does ##\int_0 ^ {\epsilon_F}g(\epsilon) d\epsilon ## represent?

    Why is that?

    Of course not. Hang in there. :woot:
     
  10. Dec 1, 2016 #9

    BRN

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    Ok, here we go!
    I took possession of my mental faculties.:smile:

    The number of k states within in spherical shell is:
    $$ g(k)dk=\frac{Vk^{2}}{\pi^{2}}dk $$
    so
    $$ \epsilon=c\hbar k \Rightarrow k=\frac{\epsilon}{c\hbar} \Rightarrow dk=\frac{d\epsilon}{c\hbar} $$
    Now, the relativistic density of the states come from:
    $$ g(k)dk=\frac{V(\frac{\epsilon}{\hbar c})^2}{\pi^2}\frac{d\epsilon}{\hbar c}=\frac{V\epsilon^2}{\pi^2\hbar^3 c^3}d\epsilon $$
    With the number of particles within the sphere of radius ## \epsilon_{F} ##, is possible to calculate the Fermi energy:
    $$ N=\int_{0}^{\epsilon_{F}} g(\epsilon) \epsilon d\epsilon=\int_{0}^{\epsilon_{F}}\frac{V\epsilon^2}{\pi^2\hbar^3 c^3}d\epsilon=\frac{V\epsilon_{F}^3}{3\pi^2\hbar^3 c^3}\Rightarrow\epsilon_{F}=\hbar c\sqrt[3]{\frac{3N\pi^2}{V}} $$
    Then, the internal energy is:
    $$ U=\int_{0}^{\epsilon_{F}} g(\epsilon) \epsilon d\epsilon=\int_{0}^{\epsilon_{F}}\frac{V\epsilon^2}{\pi^2\hbar^3 c^3}\hbar c \frac{\epsilon}{\hbar c}d\epsilon=\frac{V\epsilon_{F}^4}{4\pi^2\hbar^3 c^3}=\hbar c\frac{(3N)^{4/3}\pi^{2/3}}{4V^{1/3}} $$
    In the end, the pressure is
    $$ P=-\frac{\partial U}{\partial V}=\frac{\hbar c}{4}(3\pi^2)^{1/3}(\frac{N}{V})^{4/3} $$
    The star is made of by protons and electrons in same quantity, that is made of by idrogen. Then:
    $$ \frac{N}{V}=N\frac{\rho N_{A}}{A10^{-3}}=6.0221*10^{37}[m^{-3}]\Rightarrow P=5.7713*10^{24}[Pa] $$

    It was not difficult this exercise!
    I have to stop studying at night...:frown:

    Tanks very much, your help was crucial!:wink:
     
  11. Dec 1, 2016 #10

    TSny

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    OK. Good work!
     
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