PHZ 6607 Notes
Steven Hochman and Shawn Mitryk
10/21/2008
Contents
1 Book recommendations
1.1 Cosmology: . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 GR Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
1
2 Homework Question #28
2
3 Homework Question #4
2
4 Schwarzschild Star
5
5
6
1
1.1
Homework
Book recommendations
Cosmology:
1) Modern Cosmology - Dadelson
2) Cosmology - Weinberg *
1.2
GR Topics
1) Relativist’s Toolkit - Poisson *
2) Advanced GR - Stewart
* Tentatively scheduled to be used next semester
1
2
Homework Question #28
ds2 = −[1 −
2M 2
dr2
]dt +
+ r2 dΩ2
2M
r
[1 − r ]
(1)
Find t̂(t, r) such that ds2 = −a(r)dt̂2 + 2b(r)dt̂dr + r2 dΩ a(r) and b(r) are
not both arbitrary.
Could use dt̂ =
dt̂
dt
dt
+
dt̂
dr
dr
We can define:
r̂ = r̂(r) = r =⇒ dr̂ = 0dt + dr
for clarification:
q 2M
1−
dt̂
= ± a(r)r
dt
dt̂
dr
= f ([1 − 2M
], a(r), b(r))
r
Must Also have:
d dt̂
dt̂
= dtd dr
dr dt
d
f ([1
dt
−
2M
], a(r), b(r))
r
= 0 =⇒
d
[±
dr
q
1− 2M
r
a(r)
]=0
Thus: t̂(t, r) = Kt + g(r)
Equaling Components of ds2
1
1− 2M
r
dt̂ 2
dt̂
− a( dr
)
= 1 + 2b dr
Now we can solve for a(r) and b(r). Along t̂(r, t) surfaces we have flat space
geometry
q
Solutions: b(r) = 2M
r
√ √
√
2M
√
√
g(r) = 2 2M r + 2M ln[ r−
]
r+ 2M
3
Homework Question #4
Divide Parallel of latitude into ”n” sections
2 Methods: Geometric, Geodesic Equation
2
Figure 1: Prescription for calculating the curvature of a sphere.
Make Great circle symmetric in φo
x = R sin θo cos φo
y = ±R sin θo sin φo
z = R cos θo
(2)
(3)
(4)
Using Cross product we can define the plane of the great circle and thus
any point on the circle.
This results in: tan θ cos φ = constant
Then, to find the angle we must calculate the slope of the line at the intersection point of great circle and parallel of latitude.
dθ
dθ dφ
slope: dφ
or some dλ
, dλ |φ=φo
then: cos ∆ = u+ · u−
where u+ , u− are tangent vectors in direction of great circle at the intersection
point.
Thus the accumulated angle = n · ∆
Given the solution: sin ∆ =
2 sin φo cos φo cos θo
1−sin2 φo sin2 θo
3
Figure 2: Summation of exterior angles.
Limited by: 2φo =
2π
N
In the limit φo → 0
∆ = 2φo cos θo
N ∆ = N 2π
cos θo =⇒ N ∆ = 2π cos θo
N
limθo →0 2π cos θo = 2π
P
∆
δ = 2π
P
= 2π(1 − cos θo )
A = R2 2π(1 − cos θo ) (spherical cap)
curvature =
P
∆
A
=
1
R2
It is also possible to find constants of motion via geodesics on the sphere
We would get dλ = √ sin2θdθ
sin θ−J 2
Substituting cos α =
Thus dφ =
Jdλ
sin2 θ
√cos θ
1−J 2
=⇒ dλ = dα
and tan β =
We get tan θ cos φ =
√ J
1−J 2
tan λ
J
=⇒ dφ = dβ
(same constant as geometric equation)
4
4
Schwarzschild Star
ds2 = −e2φ dt2 + e2Λ dr2 + r2 dΩ2
Solutions:
1
) 2 − 12 (1 −
e2φ = [ 32 (1 − 2M
R
e2Λ = [1 − 2M
r2 ]
R3
1
p=
2M 2 21 2
r ) ]
R3
1
2M 2
2 2
ρo [(1− 2M
3 r ) −(1− R ) ]
R
[3(1− 2M
R
1
1
) 2 −(1− 2M
r2 ) 2 ]
R3
In the limit: M = 94 R
p(0) → ∞
e2φ (0) → 0
p
T µν = diag(ρe−2φ , pe−2Λ , rp2 , r2 sin
2 θ)
Consider
Tνµ = [−ρ, p, p, p]
Computing
Tνµ Tµν = ρ2 + 3p2
Gµν Gνµ ∝ Rνµ Rµν
=⇒ r = 0 for (M = 49 R) is singular
Light curves in this geometry:
ds2 = −e2φ dt2 + e2Λ dr2 + r2 dΩ2
E = −pt = −e2φ ṫ and J = pφ = r2 φ̇
2
2
=⇒ 0 = −E
+ e2Λ ṙ2 + Jr2
e2φ
2
ṙ2 = e−2Λ [e−2φ E 2 − Jr2 ]
dr
is maximized when J = 0 (outward orientation)
dλ
=⇒ ṙ2 = e−2(φ+Λ) E 2
5
(5)
Outside of the star: e2Λ =
1
1− 2M
r
and e2φ = [1 −
2M
]
r
=⇒ ṙ2 = E 2 =⇒ r = Eλ + ro
Thus r is an affine variable.
Inside the star:
E2
ṙ2 = e2(φ+Λ)
=⇒ E 2 as r → R
2
ṙ is smooth as light leaves the star
Coming back to E = −pt = e2φ ṫ
dt
= eE2φ ) · dλ
=⇒ ( dλ
dr
√
ṙ = e−2(φ+Λ) E 2
Λ
dt
= eeφ (holds inside and outside)
dr
Outside:
dt
r
= [1−12M ] =⇒ t = r + 2M ln [ 2M
− 1] + constant
dr
dt2
dr2
r
=
−2M r−2
[1− 2M
]2
r
r
r∗ = r + 2M ln [ 2M
− 1]
Inside:
dt
=
dr
1
1
2 2
[1− 2M
3 r ]
R
5
·
1
1
1
2 2
[ 32 (1− 2M
) 2 − 21 (1− 2M
3 r ) ]
R
R
Homework
Analyze outward radial light ray inside the Schwartzchild star. (ρ = ρo )
dt2
M
4
M
4
dt
absolutely (+) or (-) of dr
or dr
2 Consider R ≤ 9 and R > 9
6