# Homework Help: [PoM] Electron gas pressure

Tags:
1. Nov 26, 2016

### BRN

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. Nov 26, 2016

### TSny

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|.

3. Nov 28, 2016

### 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?

4. Nov 28, 2016

### TSny

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.

5. Nov 29, 2016

### BRN

I don't understand... :(

$$g(k)dk=\frac{Vk^{2}}{\pi^{2}}dk$$
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 $\epsilon_{c}=??$

6. Nov 29, 2016

### 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?

Last edited: Nov 29, 2016
7. Nov 30, 2016

### 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...

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

8. Nov 30, 2016

### TSny

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.

9. Dec 1, 2016

### 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!

10. Dec 1, 2016

### TSny

OK. Good work!