General Relativity Notes
September 15th, 2014
Peisen Ma
We may start by writing the Maxwell Eqs. as the following form:
∇ i
B i
∇ i
E i
= 4 πρ ijk
∇ j
E k
= −
˙ i ijk
∇ j
B k
= 0
= i
+ 4 πJ i d where ∇ i
T i j is a derivative operator, and then the electromagnetic stress tensor may be written as:
4 πT i j
= − E i
E j
− B i
B j
+
1
2
δ j i
E k
E k
+ B k
B k
(5)
In vacuum we have ρ = 0 and J i
= 0, thus we can reach the conclusion that
4 πP i
+ ∇ j
T i j
= 0 (6)
(1)
(2)
(3)
(4) in which the energy flux is defined by the following equation
4 πP i
= ijk
E j
B k
.
(7)
Now if we act of 4 πP i
∇ j on the electromagnetic stress tensor T i j and add a term to the result, we would obtain an equation which has exactly the same left hand terms as eqn(6), while the right hand side would be some other terms. To show that the right hand terms add up to exactly 0, we can introduce a symmetric tensor g ij called metric which allows us to relate the covector to the corresponding contravariant vector:
E = g ij
E j
(8)
However the introduction of the metric is not enough to complete the proof as we would see some ∇ k g ij terms. Therefore we have to command the metric to be in a certain form to be compatible with Maxwell’s equations.
1

The action for a free and massless scalar field can be written as
S =
Z d
4 x
√
− g
1
2
∇
µ
φ ∇ µ
φ , and the variation of the action is
δS =
Z d
4 x
√
− g ( ∇
µ
δφ ) ∇ µ
φ .
(9)
(10)
Integrating by part gives us
δS = −
Z d
4 x ∇
µ
√
− g ∇ µ
φ δφ .
(11)
In class we have
δS =
=
Z d
4 x
δS
δφ
Z d
4 x
√
− g
δφ
√
1
− g
δS
δφ
δφ .
(12)
(13)
Since the stationary point of the action requires the functional derivation to be 0, thus if we compare eqn (13) with eqn (11), we should arrive at
√
1
− g
δS
δφ ( x )
≡ √
1
− g
∇
µ
√
− g ∇ µ
φ = 0 , (14) which is exactly the covariant equation of motion for the free massless scalar field. Moreover, we can define the conserved current density as
√
1
− g
∇
µ
J
µ
= 0 .
(15)
2

The invariant separation square ds
2 as follows ds
2 and proper time square
≡ g
µν dx
µ dx
ν ≡ − c
2 dτ
2
, dτ
2 are defined
(16) where the metric g
µν has the space like signature (-+++). For can make the following choice
λ in dx
µ
, we dλ spacelike: dλ → ds lightlike: dλ timelike: dλ → dτ
Consider especially the timelike case, we can define the 4-velocity as,
U
µ
≡ dx µ
, dτ then the eqn (16) can be rewritten as,
− c
2
= g
µν dx µ dτ dx ν dτ
≡ g
µν
U
µ
U
ν
, which is known to be the normalization of the 4-velocity vector.
(17)
(18)
3