Schwarzschild metric: a static spherically symmetric
solution of vacuum Einstein equations
A general spherically symmetric static metric can be written as
ds2=Adt2−Bdr2−r2(dθ2 +sin2θdφ2),
(1)
where A and B are functions of only the radius r. Calculating the
Christoffel symbols1
and the Ricci tensor2 and then
solving the vacuum Einstein equations Rab=0 gives3
the famous Schwarzschild metric
ds2 =
1−
R
r
dt2−
dr2
1−
R
r
−r2(dθ2+sin2θdφ2).
(2)
The integration constant R is determined from the Newtonian limit,
R=2M where M is the mass of the central body4. It is called gravitaional or Schwarzschild radius.
Motion in the Schwarzschild metric
In the Schwarzschild metric the geodesic equations,
[(d)/(ds)](gabub)=[1/2]gbc,aubuc, for
a=t,θ,φ are:
d
ds
1−
2M
r
dt
ds
=0 ;
(3)
d
ds
r2
dθ
ds
= r2sinθcosθ
dφ
ds
2
;
(4)
d
ds
r2sin2θ
dφ
ds
=0 .
(5)
Instead of the r-geodesic we shall divide the expression for
the Schwarzschild metric (2) by ds2:
1=
1−
2M
r
dt
ds
2
−
1−
2M
r
−1
dr
ds
2
−r2
dθ
ds
2
+sin2θ
dφ
ds
2
(6)
The first three equations can be integrated as
θ =
π
2
, r2
dφ
ds
=J ,
1−
2M
r
dt
ds
=E ,
(7)
where J and E
are constants. The fourth equation then becomes
1=
E2
1−
2M
r
−
J2
r4
r′2
1−
2M
r
−
J2
r2
,
(8)
where r′ ≡ [(dr)/(dφ)]. Traditionally one makes a variable
substitution r=1/u,
(1−2Mu)=E2−J2u′2−J2u2(1−2Mu),
(9)
and then differentiates the equation once. Assuming u′ ≠ 0 this gives
u"+u=
M
J2
+3Mu2 .
(10)
In this form the last term is a relativistic correction to the othewise
non-relativistic equation.
The light rays travel along the null-geodesics where
ds2=0. Consequently instead of ds one needs to use some parameter
dλ in the geodesic equations [(Dka)/(dλ)]=0, where
ka=[(dxa)/(dλ)] and also the unity in the left-hand side of
equation (6) has to be substituted with zero. This apparently
leads to the equation
u"+u=3Mu2 ,
(11)
which describes the trajectory of a ray of light in the Schwarzschild metric.
Exercises
Consider a non-relativistic equatorial (θ = π/2) motion of a
planet with mass m around a star with mass M discribed by a
Lagrangian5
L=
1
2
m(
⋅
r
2
+r2
⋅
φ
2
)+
mM
r
.
Write down the Euler-Lagrange equations
∂
∂t
∂L
∂
⋅
q
=
∂L
∂q
, q=r,φ .
Using the first integral r2[(φ)\dot]=J rewrite the r-equation as an
equation for the function u(φ), where u=1/r, and compare with
(10).
Show that a light ray can travel around a massive star in a circular
orbit much like a planet. Calculate the radius (in Schwarzschild
coordinates) of this orbit. Answer: r=[3/2](2M).
Footnotes:
1
Γrrr=[1/2] [(B′)/(B)], Γttr=[1/2] [(A′)/(A)], Γrtt=[1/2] [(A′)/(B)], Γθθr=[1/(r)], Γrθθ=−[(r)/(B)], Γφφr=[1/(r)], Γrφφ=−[(r sin2θ)/(B)], Γφφθ=cotθ, Γθφφ=−sinθcosθ.
2Rtt = [(A")/(2B)]+[(A′)/(B)]([1/(r)]−[(B′)/(4B)]−[(A′)/(4A)]), Rθθ = 1−([(r)/(B)])′−[1/2]([(A′)/(A)]+[(B′)/(B)])[(r)/(B)], Rrr = −[(A")/(2A)]+[(A′B′)/(4AB)]+[(A′2)/(4A2)]+[(B′)/(rB)].
3
Making a linear
combination BRtt+ARrr=0 gives A′B+AB′=0 ⇒ [(A′)/(A)]+[(B′)/(B)]=0 ⇒ AB=1.
Then Rθθ=0 gives
B=[1/(1−[(R)/(r)])], A=1−[(R)/(r)],
where R is an integration constant.
4 in the units
where G=1
5 [(A)\dot] ≡ [(∂A)/(∂t)]
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