What is the pressure exerted by an electron gas in white dwarf stars?

[BRN]

Hi guys! I have a problem with this exercise:

1. Homework Statement
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.

The Attempt at a Solution

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

Internal energy is:

U=∫₀^ε_Fg(ε)ε dε

where:

states density g(ε)=(2m_e^3/2V√ε)/(√2π²ħ³)

ε=ħck ; k²=p²/ħ²=m_e²v²/ħ²=2εm_e/ħ² ⇒ k=√(2εm_e/ħ²)

Then:

U=∫₀^ε_F (2m_e^3/2V√ε)/(√2π²ħ³) √(2εm_e/ħ²) ħc dε =

= (m_e^3/2V c ε_F^3/2 √(ε_Fm_e))/(π²ħ²)

So:

P=-∂U/∂V=-(m_e^3/2 c ε_F^3/2 √(ε_Fm_e))/(π²ħ²)

Fermi energy is:

ε_F=ħ²/(2m_e)(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?

 

[TSny]

states density g(ε)=(2m_e^3/2V√ε)/(√2π²ħ³)

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|.

 

[BRN]

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?

 

[TSny]

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

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

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

How to calculate the relativistic density of states?

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.

 

[BRN]

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}=?? ##

 

[TSny]

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?

 

[BRN]

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...

but did you drop a factor of VV near the end?

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

please do not hate me!

 

[TSny]

whith ##\epsilon_{c}=0##
$$ U=\int_{0}^{\epsilon_{F}} g(\epsilon)\epsilon d\epsilon=\int_{0}^{\epsilon_{F}} \frac{\epsilon^2}{c^3\pi^2\hbar^3} c \hbar \frac{\epsilon}{c \hbar}d\epsilon=\frac{\epsilon_{F}^4}{4c^3\pi^2\hbar^3}$$

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##.

$$ \epsilon_{F}=\frac{\hbar^2}{2m_{e}}(3 \pi^2\frac{N}{V})^{2/3} $$

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?

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

Why is that?

please do not hate me!

Of course not. Hang in there.

 

[BRN]

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

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...

Tanks very much, your help was crucial!

 

[TSny]

OK. Good work!