Download C02-TOPIC- Equation of state of degenerate gas

The equation of state of a degenerate gas
REFERENCE: Shu
We begin with three important equations: Two from classical physics,
and one from quantum physics.
[Eq. 1]
PV = NkT
3
[Eq. 2]
K =
kT
2
(p)(r) > h
[Eq. 3]
Eq. 1 is the perfect gas law of chemistry, where P is the pressure the
gas exerts on its surroundings, V his the volume to which the gas is
constrained to occupy, N is the number of gas particles, T is the
temperature of the gas in Kelvin and k is Boltzmann's constant and has
the value
k = 1.3810-16 ergK-1, where 1 erg = gcms-2.
Eq. 2 relates the kinetic energy of an ideal gas to its temperature.
Finally, Eq. 3 is Heisenberg's uncertainty principle, which states
one cannot measure both position and momentum exactly.
The
precisely the position is known, (i.e.: r is very small) the
precisely we know the exact momentum p (i.e.: p must exceed h/r
Eq. 3). h is Planck's constant and has the value
that
more
less
from
h = 6.6310-27 ergs.
Each of the N particles of the gas has a kinetic energy given by
K = 12 <pv>,
[Eq. 4]
Substitution of Eq. 4 into Eq. 2, and solving for kT we have
3
2
3
<pv> = 2 kT
1
3
<pv> = kT
which we then substitute out of Eq. 1:
PV =
1
3
N<pv>.
[Eq. 5]
We now define the number of particles per unit volume n  N/V and
rewrite <pv> as (1/m)<p(mv)> = (1/m)<p>2 so that Eq. 5 becomes
P =
n<p>2
.
3m
[Eq. 6]
Eq. 6 is valid for any perfect gas and gives the pressure in terms of
the momentum of each particle instead of the temperature of the gas.
For stars of the size and mass of the sun Eq. 6 describes the pressure
well enough for our purposes.
But for white dwarfs and neutron stars
(and black holes) it does not. As a star burns its fuel in fusion, the
radiation pressure opposes the self-gravitational pull of the star's
gases, and for many billions of years the radiation pressure and the
self-gravity are in equilibrium. Eventually, nuclear reactants are no
longer available in sufficient quantity to maintain the fusion rates
needed to counter gravity, and the star begins to collapse.
As it
collapses its density increases and its volume decreases to the point
that the average space r each molecule has to move about in is very
miniscule.
At the same time, its momentum change p is also
constrained.
Each mass m cannot move far.
Therefore it must
constantly be changing direction as it "reflects" off of nearby
particles. It cannot have a large change. By Heisenberg's uncertainty
principle (Eq. 3) we know that (p)(r) > h.
Thus
p > h/r.
Eventually, the density becomes so large that p  h/r.
When this
point is reached, we say the gas has become degenerate. Eq. 6 becomes
P 
nh2
.
.
3m(r)2
Note that n varies inversely with the volume:
n2/3 
[Eq. 7]
Thus n  1/(r)3 so that
1
.
.
(r)2
Substitution of this result into Eq. 7, dispensing with the 3 in the
denominator, and ignoring the approximations, we finally arrive at the
equation of state of a degenerate gas:
P =
n5/3h2
m .
[Eq. 8]
We will look more closely at the consequences of Eq. 8 when we study
the white dwarf and the neutron star.