PHZ 6607 Notes
Steven Hochman
9/9/2008
1. Stress tensor
Stress tensor for scalar field.
1
T µν = g µλ g νσ ∂λ Φ∂σ Φ − g µν [ g λσ ∂λ Φ∂σ Φ + V (Φ)]
2
Stress tensor for E.M.
(1)
1
µν
TEM
= F µλ Fλ ν − g µν F λσ Fλσ
(2)
4
R
Suppose we have S= Ld4 x, we can vary S with respect to the metric to
get T µν
δS
→ T µν
δgµν
1
µν
∇µ TEM
= ∇µ (F µλ Fλν − g µν F λσ Fλσ ) = 0
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This conservation law is fundamental.
∇µ T µν = 0, ∀matter
This conservation of T
µν
(3)
(4)
(5)
implies
∇µ [Gµν = 8πGN Tµν ] = 0
(6)
1
T µν = g µλ g νσ ∂λ Φ∂σ Φ − g µν [ g λσ ∂λ Φ∂σ Φ + V (Φ)]
2
(7)
∇µ gνσ = 0
(8)
1
∇µ T µν = (∇µ g µλ ∂λ Φ)g νσ ∂σ Φ + g µλ ∂λ Φ∇µ (g νσ ∂σ Φ)
1 µν
− 2 g ∂σ Φ∇µ (g λσ ∂λ Φ) − 21 g µν ∂λ Φ∇µ (g λσ ∂σ Φ)
−g µν ∇µ V (Φ)
where
g ∂λ Φ∇µ (g νσ ∂σ Φ) = ∇µ Φ(∇µ ∇ν Φ)
g µν ∂σ Φ∇µ (g λσ ∂λ Φ) = ∇σ Φ∇ν · ∇σ Φ
g µν ∂λ Φ∇µ (g λσ ∂σ Φ) = ∇λ Φ∇ν · ∇λ Φ
(∇µ g µλ ∂λ Φ)g νσ ∂σ Φ − g µν ∇µ V (Φ) = ∇ν [∇2 Φ −
µλ
=⇒ ∇µ T µν = ∇ν Φ[∇2 Φ −
∂V
]
∂Φ
∂V
] + ∇µ Φ(∇µ ∇ν Φ) − ∇σ Φ∇ν · ∇σ Φ
∂Φ
(9)
2. Timelike and Spacelike Vectors
We take the metric to be ds2 = −c2 dt2 + dx2 + dy 2 + dz 2 When in the
curved space, we can still have curves which are timelike, spacelike or lightlike.
Suppose we have a vector X µ
X µ = [αλ, R cos(βλ), R sin(βλ), 0]
µ
= [α, −βR sin(βλ), βR cos(βλ), 0]
U µ = dX
dλ
2
U = −α2 + β 2 R2 (sin2 (βλ) + cos2 (βλ))
If we choose α = ±βR, then U 2 = 0 We then want to check if it is a
geodesic.
aµ = [0, −β 2 R cos(βλ), −β 2 R sin(βλ), 0]
a2 = β 4 R2 6= 0, it’s not a geodesic
(When aµ = f (λ)U µ , it’s geodesic)
When choosing α = ±βR , it’s a lightlike curve if choosing α =
2
U = −1, but it’s not a geodesic curve
p
1 + β 2 R2 =⇒
3. Homework Question #6
metric : ds2 = −(1 −
rs 2
dr2
2
2
2
2
2
)dt +
rs + r dθ + r sin θdφ
r
1− r
2
(10)
X µ = [αλ, R, π2 , βλ]
Here we pick θ to be a constant π2 , and r = R
U µ = [α, 0, 0, β]
+ r2 β 2 = −(1 −
U = −(1 −
rs
)α2
r
choose β such that −(1 −
rs
)α2
R
2
rs
)α2
R
+ R2 β 2
+ R2 β 2 = −1
p
β = R1 (1 − rRs )α2 − 1
for vector X µ = (t, r, θ, φ)
dt dr dθ dφ
U µ = ( dλ
, dλ , dλ , dλ )
dθ
We can make dλ
= 0 because it’s a geodesic in an equilateral plane and
=0
Pt here = −E is a constant for a geodesic
Pθ2
Pφ = J = constant
g µν Pµ Pν = −m2
1
rs
E2
2
) + 2 J2 =
−
rs + Pr (1 −
1− r
r
r
dr
Pr = mgrr dλ
, use Ẽ =
=⇒
Ẽ 2
1 − rrs
dr
dλ
J˜ =
0
lightlike
−m2 timelike
rs
)(1
r
= 0 and
+
J˜2
)
r2
d2 r
dλ2
(12)
timelike proper time parametrized
=0
˜2
2
˜2
dr
d r
rs
2J
3rs J
dr
2( dλ
) dλ
)( dλ
2 ) = (− r 2 + r 3 −
r4
d2 r
1
rs
2J˜2
3rs J˜2
( dλ
)=0
2 ) = 2 (− r 2 + r 3 −
r4
rs r
rs R
2
˜
=⇒ J = (2−3 rs ) = (2−3 rs )
r
R
dt
dt
t = αλ, dλ
= α, Pt = gtt dλ
= −Ẽ
rs
=⇒ Ẽ = (1 − R )α
˜ J˜ = r2 β
Pφ = gφφ dφ
= J,
dλ
β 2 = Rrs3 ( 2−13rs )
R
3
(11)
J
m
(for timelike m 6= 0)
dr 2
( dλ
)
1 ˜2
0
lightlike
+
rs + 2 J =
−1 timelike
1− r
r
dr 2
=⇒ ( dλ
) = Ẽ 2 − (1 −
For circular motion:
E
,
m
Timelike circular motion (geodesic) occurs only for R > 32 rs
(∀R > 23 rs , it’s timelike)
For lightlike,
˜2
d2 r
dλ2
dr 2
( dλ
) = Ẽ 2 − (1 − rrs ) Jr2
˜2
= ( Jr3 [2 − 3rrs )] = 0, for R = 32 rs ∃ lightlike geodesic
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