Chong-Sun Chu
Solution
Mid Examination
Introduction to Relativity I (PHYS431000)
1. (Baisc concepts)
(a) State the condition for a spacetime transformation
xµ → x0µ = Λµ ν xν
to be a Lorentz transformation.
Soln: Lorentz transformation preserves the proper distance
dss = −ηµν dxµ dxν .
Therefore the transformation matrix Λµ ν satisfies
ηµν Λµ α Λµ β = ηαβ .
[4 pts]
µν
(b) State the condition for a quantity T λ to be a Lorentz tensor.
Soln: Under a Lorentz transformation, T µν λ has to transform as
T µν λ → T 0µν λ =
∂x0µ ∂x0ν ∂xγ αβ
T γ
∂xα ∂xβ ∂x0λ
[4pts]
(c) A two index object X µν is defined as the sum of two vectors
X µν = Aµ + B ν
Is X µν a tensor?
Soln: Not a tensor as X µν does not transform as a tensor. or one can consider the product with
another two vectors
X µν Wµ Vν = (Aµ Wm )Vν + (V ν Bν )Wµ
Since it is not a scalar, so X µν cannot be a tensor.
[4pts]
i
(d) One may model the universe as a collection of dust with a rest mass density ρ(x , t) (as measured
in their comoving frame) and a local velocity ~v (xi , t). Derive the energy momentum tensor for
this universe.
Soln: Consider a fluid element at position xi and at time t. In the comoving frame of the fluid
element, the energy momentum tensor is given by
00
T(rest)
= ρ,
otherwise zero
[3pts]
In the labratory frame, the energy momentum tensor is given by
αβ
T µν = Λµ α Λν β T(rest)
= ρΛµ 0 Λν 0 ,
where Λµ α is the transformation matrix for the Lorentz transformation which bring the comoving
frame to the frame moving with velocity −v i . Explicitly, Λµ α is given by
Λ0 0 = γ,
1
Λ0 i = Λi 0 = γv i ,
√
where γ = 1/ 1 − v 2 .
We get
[5pts]
T 00 = ργ 2 ,
T i0 = T 0i = ργ 2 v i ,
T ij = ργ 2 v i v j .
Or, in terms of the 4-velocity uµ = γ(1, v i ), we have
T µν = ρuµ uν .
[5pts]
2. (Energy momentum tensor)
(a) Write down the equation for the local conservation law for the energy-momentum tensor. From
it derives that the energy and momentum of a closed system is conserved.
Soln:
∂µ T µν = 0.
[2pts]
Energy E = P 0 and momentum P i are defined by
Z
P µ = d3 xT µ0 .
Therefore
dP µ
=
dt
Z
d3 x∂0 T µ0 = −
Z
d3 x∂i T µi = 0
since the integral is over a closed system and so there is no boundary contribution.
[5pts]
(b) Write down the energy-momentum tensor for a perfect fluid with mass density ρ and pressure P .
Soln:
T µν = pη µν + (p + ρ)uα uβ ,
where ua is the 4-velocity of the fluid element.
[3pts]
(c) Show that the energy-momentum tensor for a closed system obeys the virial theorem
Z
Z
1 d2
T ij d3 x =
T 00 xi xj d3 x.
2 dt2
Soln: Using the conservation law, we have
Z
Z
d2
d
00 i j 3
T xx d x = −
∂k T k0 xi xj d3 x
dt2
dt
Z
d
=
(T i0 xj + T j0 xi )d3 x
dt
Z
= − (∂k T ik xj + ∂k T jk xi )d3 x
Z
= 2 T ij d3 x,
where we have used integration by parts and drops the boundary terms.
(1)
(2)
(3)
(4)
[15pts]
3. (Perfect fluid)
(a) Using the conservation of energy-momentum, show that the flow of a perfect fluid is adiabatic,
i.e.
dσ
= 0,
dτ
where τ is the proper time for a fluid element.
2
Soln: We have the conservation of energy-momentum:
∂p
∂
+
((p + ρ)uα uβ )) = 0.
∂xα
∂xβ
[2pts]
and conservation of particle number:
∂
(nuα ) = 0.
∂xα
[2pts]
Contract the first eqn with uα and use
uα ∂β uα = 0,
and the conservation of particle number, we get the desired result.
[6pts]
(b) Consider a fluid with density ρ, temperature T , pressure p, entropy per particle σ and number
density n. Write down the first law of thermodynamics for the fluid in terms of ρ, T, p, σ and n.
Soln: The first law of thermodynamics for a system reads
T dS = dE + pdV
Dividing it by the number of particle, and expressing the result in terms of σ, ρ, P and n, we get
ρ
1
T dσ = pd( ) + d( )
n
n
[5pts]
(c) For a perfect fluid with equation of state ρ = ρ(n) where n is the particle number density, show
that T µ µ is negative if and only if
d log ρ
4
< .
d log n
3
Soln:
T µ µ = (p + ρ)uµ uµ + pgµµ = 3p − ρ.
Therefore T µ µ < 0 iff
p/ρ < 1/3.
Now from part (a), we have dσ = 0 and hence
dρ
p + ρ dn
=
.
ρ
ρ n
Therefore
T µ µ < 0,
⇐⇒
d log ρ
4
< .
d log n
3
[10pts]
4. (Affine connection)
(a) Starting from its definition
Γλµν =
∂ 2 ξ α ∂xλ
∂xµ ∂xν ∂ξ α
derive the transformation law of the affine connection Γλµν
Soln: use chain rule and simple differentiation, we get
Γ0λ
µν =
∂x0λ ∂xβ ∂xγ α
∂x0λ ∂ 2 xρ
Γ
+
βγ
∂xα ∂x0µ ∂x0ν
∂xρ ∂x0µ ∂x0ν
[5pts]
3
(b) Use the affine connection to construct the covariant derivative V µ ; ν for a 4-vector V µ . Show that
it transforms covariantly.
Soln:
∂V µ
V µ; ν =
+ Γµ να V α
∂xν
Direct computation shows that
V 0µ ; ν =
∂x0µ ∂xβ α
V β
∂xα ∂x0ν
[5pts]
(c) Show that
1
∂ √
V µ; µ = √
( −gV µ )
−g ∂xµ
and derive the integral form of Gauss law in general relativity.
hint: You may use
1
∂gρλ
∂gνλ
∂gρν
−
).
Γµ νλ = g µρ ( λ +
ν
2
∂x
∂x
∂xρ
Soln:
V µ; µ =
∂V µ
+ Γµ µλ V λ .
∂xµ
Now
Γµ µλ
=
=
=
1 µρ ∂gρµ
∂gρλ
∂gµλ
1
∂gρµ
g ( λ +
−
) = g µρ λ
µ
ρ
2
∂x
∂x
∂x
2
∂x
1 ∂
log(−g)
2 ∂xλ
1
∂ √
√
−g.
−g ∂xλ
(5)
(6)
(7)
[10pts]
Integrate over a 4-volume V with boundary Σ, we have
Z
Z
Z
√
√
∂ √
d4 x −gV µ ; µ =
d4 x µ ( −gV µ ) =
dAµ Vµ −g
∂x
V
V
Σ
This is the desired Gauss law.
[5pts]
4