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256

 

 

 

6

Stellar Equilibrium

 

R

9

= 2.25

(6.3.67)

 

 

4

GM/c2

What does this mean? It means that in order for a star to sustain its own weight when its radius becomes smaller than 2.25 times the Schwarzschild radius, one needs an infinity pressure, namely it is impossible. The star necessarily collapses, all matter falling within the surface r = GMc2 that, as we are going to see in later chapters, is an event horizon. The final equilibrium state reached by the star is that of a black-hole.

6.4 The Chandrasekhar Mass-Limit

The conclusion of the last section raises an important question: is the collapse into a black-hole the necessary end point in the evolution of any star, when all of its energy is exhausted? The answer would be yes if General Relativity had been the only novelty of XXth century Physics. There was, however, another quite relevant novelty: Quantum Mechanics. Together with Quantum Mechanics came also Pauli Exclusion Principle. Precisely this latter is responsible for another source of pressure which, quite unexpectedly, offers a star the last chance to survive in an equilibrium state when all other hopes are already lost. The same principle and the same mathematical modeling actually describes two quite different equilibrium states that correspond to the end-point in the evolution of stars of medium initial size and of much larger size, respectively. The first of this two equilibrium states is that of White Dwarf, while the second is that of Neutron Star. The difference is provided by the nature of the fermions that compose the degenerate Fermi gas filling the star, electrons in the case of white dwarfs neutrons in the second case. The main difference resides in the mass of such fermions, which determines the actual size of the equilibrium radius. Apart from that, the sustaining mechanism is essentially the same in both cases and it can be understood by studying the rather astonishing properties of a completely degenerate gas of an extremely large number of fermionic particles.

6.4.1The Degenerate Fermi Gas of Very Many Spin One-Half Particles

Let us consider a system composed by a very large number N of spin s = 12 free particles. If this system is deprived of energy to allow for their excitations, all particles will fall into the lowest available energy states. Yet, since Pauli exclusion principle forbids that two fermions occupy the same level, they will pile up at increasing energy levels. Taking into account the degeneracy 2 of each level, due to spin 12 , we can write the following equation:

N = d3n (6.4.1)

2

6.4 The Chandrasekhar Mass-Limit

257

where n is the 3-vector of integer wave-numbers labeling a quantum state in a cubic box of size L × L × L. The momentum of such a state is:

p

2π

n

(6.4.2)

 

−→ =

L −→

 

Combining (6.4.1) with (6.4.2) we can write:

N

=

 

V

 

 

d

3p

=

V

4π pF p2 dp

2

(2π )3

 

(2π )3

 

|p|≤pF

 

 

0

 

 

 

 

 

 

 

 

 

 

=

 

V

 

4π

pF3

 

 

 

 

(6.4.3)

 

 

 

 

 

 

 

 

 

 

 

(2π )3

3

 

 

 

 

where V = L3 denotes the volume of the box surrounding the fermion system. If we introduce the fermion density:

N

 

ρf = V

(6.4.4)

from (6.4.3) we obtain the following expression for the Fermi momentum pF :

pF = 3π 2ρf 1/3

(6.4.5)

Next we calculate the ground state energy of a system of relativistic spin one-half particles near the Fermi temperature. Naming mf the mass of these particles, by definition we obtain:

 

 

 

 

 

 

 

 

 

c

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E0

=

2

 

 

 

−→

 

+

mf

 

 

 

 

 

 

 

 

 

 

 

 

 

 

p

p

 

 

p 2

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

|−→|≤ F

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

8π cV

pF p2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

 

p2

+

m2 c2 dp

 

 

 

 

 

 

 

 

 

(2π )3

0

 

 

 

 

 

 

 

 

 

f

 

 

 

 

 

 

 

 

=

 

cV

mf c pF p2

5

1

+

 

 

p

2 dp

(6.4.6)

 

 

π 2 3

mf c

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let us now define the new variable x =

 

 

p

. Using this notation we obtain the

mf c

 

following expression for the ground state energy:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V

c5m4

xF x2

 

 

 

 

 

 

 

 

E0

=

 

1

+

x2 dx

(6.4.7)

 

 

π 2 3

 

 

 

 

 

 

f

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

Hence:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E0

=

c5mf4 1

 

 

 

 

 

 

 

 

 

 

 

(6.4.8)

 

 

 

 

 

 

 

 

 

 

 

f (xF )

 

 

 

 

 

 

 

 

 

 

 

 

N

π 2 3

ρ

 

 

 

 

 

 

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