Statistical Physics, Second Revised and Enlarged Edition

178 Statistics under extreme conditions

includerelativitywithratherfew problems. Step (a)isidenticalto theabove, sinceit
involvesk-space geometry only. The expression forkkkFis thus unchanged. However,
in steps (b)and(c), we must allowfor appropriate modifications to thedispersion
relation. For aparticleofrest massmandmomentumk,the energy–momentum
relationship isε^2 =(mc^2 )^2 +(ck)^2. In the extreme relativistic limit, this simplifies
toε=cp,the samedispersion relation asfor photons. Hence,inthe extremelimit,
step (b)givesμ=ckkkF,leadingtoU=^34 Nμ.Step(c)becomesP=^13 U/V,since

nowε∝V−^1 /^3 .Puttingthese results together to calculate the pressureinthe extreme
relativistic limit,we obtain

P=

1

4

c( 3 π^2 )^1 /^3 (N/V)^4 /^3 ( 15 .3)

When thegas is in an intermediate regime, the full dispersion relation must be used,
and some unpleasant integrals must be computed. What happens is that (1 5 .2) and
(15.3)join up smoothly.

The stability requirement. This is difficult to work out exactly, but quite easy to get
the roughidea. Solet usjust concentrate on the roughidea! Considerfirst a starin
theabsence ofgravitation. We treat the star as a numberNofspin-^12 gas particles,
confined within a ‘box’ of volumeV=^43 πR^3 whereRis the radius of the star.In the
absence of gravity, the density of the gas will be uniform. If the star were now to be
collapsedbyafurther amountδR,an amount ofexternalworkwouldneedtobedone
equal toP 4 πR^2 δR, i.e. to the pressure times the volume change.
In practice, ofcourse, this energy mustbe suppliedbytheloss ofgravitational
potentialenergyofthe star. We needto turngravityon! Thisiswhere the trouble
starts for our calculation, since the density of the star is no longer uniform. If it were
to remainuniform, a straightforwardintegration givesfor the gravitationalpotential
energyofasphericalmass the expression –αGM^2 /R,withthe constantα= 3 /5.
HereMis the total mass of the star. For a non-uniform star it is a rough and reasonable
assumption that this expression willremainvalid(dimensionalanalysis!),but that the
true value ofαmight differ somewhat from 3/5. Thus thegain in potential energyfor
areductioninradius ofδRis equal toαGM^2 δR/R^2.
Our staristhus stablewhen thereisbalancebetween thegravitationalpotential
energychange and the work required to combat the pressure, i.e. when

P=α

G

4 π

M^2 /R^4 (15.4)

Let us suppose that the non-relativistic approximation for thepressure (15.2) is valid.
This turns out to be the case for neutron stars and for light enough white dwarf stars.
Putting together (15.2) and (15.4) gives us the relationship between mass and size for
astable star. WritingVas^43 πR^3 ,wehave

KN^5 /^3 /R^5 =M^2 /R^4 ( 15 .5)