Mathematical Principles of Theoretical Physics

(Rick Simeone) #1

428 CHAPTER 7. ASTROPHYSICS AND COSMOLOGY


2.Fluid dynamic model.The fluid dynamic model takes the momentum representation
equations coupling the heat equation. Under the condition (7.2.45) and the radial symmetry,
they are given as follows:


∂Pr
∂ τ

+


1


ρ
Pr

∂Pr
∂r

+


1


2


∂v
∂r
(7.2.51) Pr^2


=νe−v

[


1


r^2


∂r

(


r^2

∂Pr
∂r

)



2


r^2

Pr+

1


2



∂r

(


∂v
∂r

Pr

)]


+γe−v


∂r

[


e−v/^2
r^2


∂r

(r^2 ev/^2 Pr)

]


−e−v

[


∂p
∂r


ρ
2

( 1 −βT)

∂eu
∂r

]


,


∂T


∂ τ

+


1


ρ

Pr

∂T


∂r

=


κe−v
r^2


∂r

(


r^2

∂T


∂r

)


(7.2.52) +Q(r),


∂ ρ
∂ τ

+


e−v/^2
r^2


∂r

(7.2.53) (r^2 ev/^2 Pr) = 0.


3.Equation of state.We know that the gravitational field equations represent thelaw of
gravity, which essentially dictates the gravity related unknowns:eu,ev,R=ek/^2 ,φ.
The laws for describing the matter field are the motion equation (7.2.51), the heat equation
(7.2.52), and the energy conservation equation (7.2.53). To close the system, one needs
to supplement an equation of state given by thermal dynamics, which provides a relation
between temperatureT, pressurep, and energy densityρ:


(7.2.54) f(T,p,ρ) = 0 ,


which depends on the underlying physical system.
In summary, we have derived a consistent model coupling the gravitational field equa-
tions, the fluid dynamic equations and the equation of state consists of eight equations solving
for eight unknowns:ψ=eu,α=ev,R=ek/^2 ,φ,Pr,T,pandρ.


4.Energy conservation formula.From the energy conservation equation (7.2.53), we can
deduce energy conservation in the following form


(7.2.55) R^3 r^2 ev/^2 Pr+


1


4 π

d
dt

Er= 0 for 0<r< 1 ,

wherer=1 stands for the boundaryR=ek/^2 of the star,Eris the total energy in the ballBr
with radiusr.
To see this, we first note that the volume differential element of the Riemannian manifold
is given by
dV=



g 11 g 22 g 33 drdθdφ=e^3 k/^2 r^2 ev/^2 sinθdrdθdφ.

Taking volume integral for (7.2.53) onBrimplies that


d
dt


Br

ρdV+R^3

∫r

0


∂r

(


r^2 ev/^2 Pr

)∫π

0

∫ 2 π

0

sinθdθdφ= 0 ,

which leads to
dMr
dt


+ 4 πR^3 r^2 ev/^2 Pr= 0 ,
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