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Lecture 2
Classical Fields in Curved Spacetime
1 Conventions and useful expressions
ημν = diag(+1, − 1 , − 1 , −1) = η
μν , (1)
x
μ = (x
0 , x
1 , x
2 , x
3 )
T = (ct, x
i )
T = (ct,⃗x )
T = (t,⃗x )
T , (2)
xμ = ημν x
ν = (t, −x⃗ )
T , (3)
xμ = gμν (x)xν^ , (4)
ds
2 = gμν (x)dx
μ dx
ν = dx
μ dxμ, (5)
σ μν =
g
σρ (∂μgνρ + ∂ν gρμ − ∂ρgμν ) , (6)
∇μϕ = ∂μϕ , ϕ = ϕ(x), (7)
∇μAν = ∂μAν − Γ
λ μν Aλ^ , Aμ^ =^ Aμ(x),^ (8)
∇μA
ν = ∂μA
ν
ν μλA
λ , (9)
∇μA
μ
−g
∂μ
−gA
μ
, f our − divergence, g = det(ˆg), (10)
∇ρTμν = ∂ρTμν − Γ
λ ρμTλν^ −^ Γ
λ ρν Tμλ,^ (11)
Fμν = ∇μAν − ∇ν Aμ = ∂μAν − ∂ν Aμ (due to Γ
λ μν = Γ
λ νμ),^ (12)
Dμ := ∇μ + iAμ , gauge covariant derivative, (13)
R
μ ναβ =^ ∂β^ Γ
μ να −^ ∂αΓ
μ νβ + Γ
μ σβ Γ
σ να −^ Γ
μ σαΓ
σ νβ ,^ (14)
Rμν = R
α μαν =^ ∂ν^ Γ
α μα −^ ∂αΓ
α μν + Γ
β μαΓ
α νβ −^ Γ
α μν Γ
β αβ ,^ (15)
R = R
μ μ =^ G
μν Rμν , (16)
δgμν = −gμρgνσδg
ρσ , (17)
δg = gT r(ˆg
− 1 .δˆg) = g(g
μν δgμν ) = −ggμν δg
μν , (18)
δ
−g = (
−g)
′ gδg^ =^ −^
δg
2
−g
−ggμν δg
μν , (19)
δR
δgαβ^
= Rαβ , (20)
Tμν =
−g
δ(
−gLM )
δgμν^
(only in signature (+, −, −, −)). (21)
2 Scalar Field in Curved Spacetime
Let us first consider scalar fields in curved spacetime:
S [ϕ, ∂μϕ] =
Z
−ηd
4 x
η
μν ∂μϕ∂ν ϕ − V (ϕ)
Z
d
4 x
−ηLϕ, (22)
δS = 0 ⇒ ∂α
∂Lϕ
∂(∂αϕ)
∂Lϕ
∂ϕ
∂Lϕ
∂ϕ
dV
dϕ
∂α
∂Lϕ
∂(∂αϕ)
η
μν ∂α
∂(∂μϕ)
∂(∂αϕ)
∂ν ϕ + ∂μϕ
∂(∂ν ϕ)
∂(∂αϕ)
∂α
η
μν δ
α μ ∂ν^ ϕ^ +^ η
μν ∂μϕδ
α ν
∂α
δ
α μ ∂
μ ϕ + δ
α ν ∂
ν ϕ
= ∂α∂
α ϕ ⇒
□ϕ = −
dV
dϕ
, where ∂α∂
α ϕ = □, Klein − Gordon (26)
Gauge invariance and coupling to the electromagnetic field:
LF = −
Fμν F μν^ − AμJμ, (27)
∂α
∂LF
∂(∂αAβ )
∂LF
∂Aβ
= 0 ⇒ ∂αF
αβ = J
β (homework). (28)
Global U (1) symmetry: ϕ(x) −→ ϕ
′ (x) = eiαϕ(x), α = const ⇒
Lϕ = Lϕ′^ , ϕ(x) = ϕ 1 (x) + iϕ 2 (x).
Local U (1) gauge symmetry: ϕ
′ (x) = e
iα(x) ϕ(x) ⇒ Lϕ ̸= Lϕ′^.
To restore gauge invariance −→ minimal coupling −→ Dμ = ∂μ + iAμ ⇒
Tαβ =
−g
δ(
−gLϕ)
δgαβ^
−g
−ggαβ
g
μν ∂μϕ∂ν ϕ −
m
2 ϕ
2 − V (ϕ)
−g∂αϕ∂β ϕ
∂αϕ∂β ϕ − gαβ
g
μν ∂μϕ∂ν ϕ −
m
2 ϕ
2 − V (ϕ)
δS
δϕ
δSϕ
δϕ
= 0 ⇒ δSϕ = 0 ⇒
Z
d
4 x
−g
g
μν (δ(∂μϕ)∂ν ϕ + ∂μϕδ(∂ν ϕ)) −
m
2 δ(ϕ)
2 − δV (ϕ)
Z
d
4 x
−gg
μν ∂ν ϕ∂μ(δϕ)+
Z
d
4 x
−gg
μν ∂μϕ∂ν (δϕ)−
Z
d
4 x
−g
m
2 ϕ +
dV
dϕ
δϕ =
Z
d
4 xδϕ
−∂μ
g
μν √ −g∂ν ϕ
− ∂ν
g
μν √ −gg
μν ∂μϕ
−g
m
2 ϕ +
dV
dϕ
Z
d
4 xδϕ
−∂μ
−gg
μν ∂ν ϕ
−g
m
2 ϕ +
dV
dϕ
−g
∂μ
−gg
μν ∂ν ϕ
2 ϕ = −
dV
dϕ
⇒ (□ + m
2 )ϕ = −
dV
dϕ
□ϕ = ∇μ∇
μ ϕ =
−g
∂μ(
−gg
μν ∂ν ϕ), Laplace − Beltrami. (38)
Short summary:
Gμν + Λgμν = −κTμν (39)
∇μG
μν = 0, Bianchi identity, (40)
∇μT
μν = 0, conservation of energy − momentum, (41)
(∇μ∇
μ
2 )ϕ = −
dV
dϕ
, □ϕ = ∇μ∇
μ ϕ =
−g
−gg
μν ∂μϕ
Homework:
SF = −
Z
d
4 x
−gFμν F
μν ,
a) T
(F ) αβ =^
−g
δ(
−gLF )
δgαβ^
b)
δSF
δAα
4 Dirac Equation in Curved Spacetime
(iγ
μ Dμ − m)Ψ = 0, (43)
γ
μ (x) = Γ
a e
μ a (x),^ spacetime gamma matrices,^ (44)
Dμ = ∂μ +
i
4
ωabμ σab, (45)
σab =
i
2
[Γa, Γb], Γa − the constant gamma matrices, (46)
ηab = e
μ a e
ν b gμν^ , M inkowski metric,^ (47)
gμν (x) = η
ab e
a μ(x)e
b ν (x),^ e
a μ(x)^ −^ tetrads,^ (48)
g
μν (x) = η
ab e
μ a (x)e
ν b (x),^ e
μ a (x)^ −^ inverse tetrads,^ (49)
e
μ a eμb^ =^ ηab,^ (50)
e
a μe
μ b =^ δ
a b ,^ (51)
e
a μe
ν a =^ δ
ν μ,^ (52)
e
μ a =^ g
μν e
b ν ηba,^ (53)
ω
a μb =^ e
a λe
σ b Γ
λ σμ +^ e
a λ∂μe
λ b , spin connection.^ (54)
Homework: Write the Dirac equation in the Schwarzschield background.
ds
2
Rg
r
dt
2 −
Rg
r
dr
2 − r
2
dθ
2
2 θdϕ
2
Rg = 2M. (56)
Hints:
ds
2 = e
a ⊗ e
b ηab = ηabe
a μe
b ν dx
μ dx
ν = gμν dx
μ dx
ν , (57)
e
a = e
a μdx
μ , tetrad f orms (cof rame) (58)
ηabe
a μe
b ν =^ gμν^ →^ solve the system.^ (59)
ds^2 = f (r)dt^2 −
f (r)
dr^2 + r^2 (dθ^2 + sin^2 θdϕ^2 ) = (e^0 )^2 − (e^1 )^2 − (e^2 )^2 − (e^3 )^2.
e
0 = e
0 μdx
μ = e
0 0 dx
0
0 1 dx
1
0 2 dx
2
0 3 dx
3 , (61)
e
1 = e
1 μdx
μ = ... (62)
e
2 = e
2 μdx
μ = ... (63)
e
3 = e
3 μdx
μ = ... (64)
Note 1: For charged particles under the influence of external Lorentz force
one has
x ¨
λ
λ μν x˙
μ x˙
ν = −
q
m
F
λα x˙
β gαβ , (71)
where F λα^ is the Maxwell electromagnetic tensor.
Eq. (71) is useful when interpreting antiparticles (solutions with „negative
energy”). The problem in QFT with antiparticles are two as far as interpretation
goes: negative energy or travelling backwards in time. The negative energy
problem can be cured if we transfer the minus sign in −Ept to the time. Thus,
one gets travelling backwards in time. The latter can be cured by considering
equation (71) under the change τ −→ −τ :
d
2 x
λ
d(−τ )^2
λ μν
dx
μ
d(−τ )
dx
ν
d(−τ )
q
m
F
λα dx
β
d(−τ )
gαβ ⇒
x ¨
λ
λ μν x˙
μ x˙
ν = +
q
m
F
λα x˙
β gαβ. (72)
One notes that the only change is the sign of the charge q. Therefore,
the interpretation of antiparticles is particles with positive energy , moving
forward in time , but with opposite charge to the charge of the particles.
Note 2:
δϕ
2 = 2ϕδϕ, δV (ϕ) =
dV
dϕ
δϕ. (73)
Note 3:
Z
V 4
d
4 x∂μ(δϕ)
−gg
μν ∂ν ϕ =
Z
V 4
d(δϕ)
−gg
μν ∂ν ϕ =
δϕ
−gg
μν ∂ν ϕ
∂V 4
Z
d(
−gg
μν ∂ν ϕ)δϕ =
Z
d
4 x∂μ
−gg
μν ∂ν ϕ
δϕ.
Note 4: Gμν + Λgμν = −κTμν /∇
μ ⇒
μ Gμν | {z } 0 (Bianchi identity)
μ gμν | {z } 0 (metric compatibility)
= −κ∇
μ Tμν | {z } 0 (energy−momentum conservation)
Note 5:
l =
Z
P
ds =
Z
P
q
ϵgμν (x)dxμdxν^ , the f unctional of the lenght of the curve P.
ϵ = ±1(timelike/spacelike curves).
l =
Z
P
v u u tϵgμν^ (x)^
dxμ
dτ |{z} x ˙μ
dxν
dτ |{z} x ˙ν
dτ dτ | {z } dτ 2
τ is the proper time (an affine parameter along the curve P).
Note 6: Relativistic dispersion relation
In Minkowski spacetime p
μ = (E,⃗p )
T ⇒ p
2 = p
μ pμ = ϵm
2 ⇒ ημν p
μ p
ν
ϵm^2 ⇒ E^2 −p⃗ 2 = ϵm^2.
In general curved spacetime pμpμ ̸= E^2 −p⃗ 2 , but pμpμ = ϵm^2 is OK, because
p
μ pμ = gμν p
μ p
ν ⇒
p
μ pμ = gμν p
μ p
ν = g
μν pμpν = ϵm
2 , relativistic dispersion relation in curved spacetime.
On the other hand U μ^ := ˙xμ^ ⇒ L^2 = ϵgμν x˙μ^ x˙ν^ = ϵ x˙μ^ x˙μ ⇒ ϵU μUμ = 1 /ϵ ⇒
ϵ
2 |{z} 1
U
μ Uμ = ϵ ⇒ U
μ Uμ = ϵ. Multiplying by m
2 : mU
μ | {z } pμ
mUμ |{z} pμ
= ϵm
2 ⇒
p
μ pμ = ϵm
2 .
Hence, we have three „equivalent” statements for the relativistic dispersion
relation:
L =
q
ϵgμν (x) ˙xμ^ x˙ν^ = 1, (74)
U
μ Uμ = gμν (x)U
μ U
ν = ϵ, (75)
p
μ pμ = gμν (x)p
μ p
ν
ϵm^2 , for massive particles
0 for massless particles
where U
μ = ˙x
μ and p
μ = mU
μ .
