We can nowrelate the Fermi energy to the number of particles in the box.

EF=

π^2 ̄h^2

2 mL^2

rn^2 =

π^2 ̄h^2

2 mL^2

(

3 N

π

)^23

=

π^2 ̄h^2

2 m

(

3 N

πL^3

)^23

=

π^2 ̄h^2

2 m

(

3 n

π

)^23

We can also integrate to get thetotal energyof all the fermions.

Etot= 2

1

8

∫rn

0

4 πr^2

r^2 π^2 ̄h^2

2 mL^2

dr=

π^3 ̄h^2

2 mL^2

rn^5

5

=

π^3 ̄h^2

10 mL^2

(

3 N

π

) (^53)

π^3 ̄h^2
10 m

(

3 n

π

) (^53)
L^3
where the last step shows how the total energy depends on the number of particles per unit volume
n. It makes sense that this energy is proportional to the volume.
The step in whichEFandEtotis related toNis often useful.

EF=

π^2 ̄h^2

2 m

(

3 N

πL^3

)^23

Etot=

π^3 ̄h^2

10 mL^2

(

3 N

π

)^53

13.1.2 Degeneracy Pressure in Stars

The pressure exerted by fermions squeezed into a small box is whatkeeps cold stars from collapsing.
White Dwarfs are held up by electrons and Neutron Stars are held upby neutrons in a much smaller
box.

We cancompute the pressurefrom the dependence of the energy on the volume for a fixed
number of fermions.

dE = F~·d~s=PAds=PdV

P = −

∂Etot

∂V

Etot =

π^3 ̄h^2

10 m

(

3 N

π

)^53

V−

(^23)

P =

π^3 ̄h^2

15 m

(

3 N

π

)^53

V−

(^53)

P =

π^3 ̄h^2

15 m

(

3 N

π

) (^53)
L−^5 =
π^3 ̄h^2
15 m

(

3 n

π

) (^53)
The last step verifies that the pressure only depends on the density, not the volume and theN
independently, as it should. We will use.

P=

π^3 ̄h^2

15 m

(

3 N

π

) (^53)
L−^5 =
π^3 ̄h^2
15 m

(

3 N

π

) (^53)
V
− 5
3