Exercise Sheet No. 2
Maneenate Wechakama, Jochen Klar
Cosmology and early Universe
Exercise Sheet No. 2
7.4.2008, to be handed in on 14.4.2008 (in the lecture)
1. Tensor Operations [4 points]
a) Let space be two dimensional, with coordinate (x1 , x2 ). Suppose the tensors
V a , Wa , P ab , Qab and M a b are measured to have the values
V 1 = 2,
V 2 = 3;
2 −1
ab
P =
;
3 6
ab
W1 = 4,
W2 = 5
0 2
4 3
a
Qab =
; M b=
.
4 7 ab
2 1 ab
(1)
(2)
Calculate the following tensors:
(1)
(2)
(3)
(4)
α = V a Wa
T b = P ab Wa
F a c = P ab Qbc
Gab = M c b Qca
b) Translate the following 3-vector identities into index notation, and prove them:
~ · (B
~ × C)
~ =B
~ · (C
~ × A)
~ =C
~ · (A
~ × B)
~
(1) A
~ =A
~ · ∇f + f ∇ · A
~
(2) ∇ · (f A)
~ × B)
~ =B
~ ·∇×A
~−A
~·∇×B
~
(3) ∇ · (A
The cross product can be calculated using the three-dimensional total-antisymetric
Levi-Civita symbol, which is defined as:


if αβγ is an even permutation of (1, 2, 3)
1
(3)
αβγ = −1 if αβγ is an odd permutation of (1, 2, 3)


0
otherwise (two indices are equal)
Using matrix notation it can be displayed



0 0 0
0 0



0 0 1 ; αβ2 =
0 0
αβ1 =
0 −1 0
1 0
in three dimensions as:



−1
0 1 0
0  ; αβ3 =  −1 0 0  (4)
0
0 0 0
or in two dimensions as:
0 1
αβ =
−1 0
(5)
In index notation the cross product can now be expressed as:
(A × B)α = αβγ Aβ Bγ
(6)
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Exercise Sheet No. 2
Maneenate Wechakama, Jochen Klar
c) Let Aab be an anti-symmetric contravariant tensor. Show that Aab is antisymmetric as well. (Remember that the metric tensor is defined to be symmetric.) Next let S ab = S ba be a symmetric tensor. Show that Aab S ab = 0
2. Maxwell’s equations in special and gereral relativity [10 points]
In classical electrodymics the properties of the electromagnetic fields are decribed
by Maxwell’s equations. For a linear medium and in cgs units these are written:
∇ · E = 4πρ
1
∇ × E = − ∂t B
c
∇·B=0
4π
1
∇×B=
j + ∂t E
c
c
(7)
where E denotes the electric field, B the magnetic field, ρ the charge density, j the
current density, and c the speed of light.
a) In four dimensional notation, E and B
day tensor:

0
E1
E2
E3
 −E1
0
−B
B
3
2
Fab = 
 −E2 B3
0
−B1
−E3 −B2 B1
0
are represented by the covariant Fara



(8)
Find the contravariant version of the Faraday tensor F ab = η ac η bd Fcd . Use the
Minkowki metric ηab = diag (1, −1, −1, −1)
b) Show that equations (7) can be written as:
∂a F ab =
4π b
j
c
and εabcd ∂ b F cd = 0
(9)
where j b = (cρ, j) ist the four-current and εabcd the four-dimensional LeviCivita-symbol.
Hint: Calculate the time component and the space components seperately.
c) The relation between the vector potential Aa and the covariant Faraday tensor
is given by:
Fab = ∂a Ab − ∂b Aa
(10)
The corresponding relation for the contravariant tensor is:
F ab = ∂ a Ab − ∂ b Aa
(11)
Show that, using the Lorenz gauge ∂a Aa = 0, equations (9) can be written as:
Aa =
4π a
j
c
(12)
where Aa ≡ ∂b ∂ b Aa is the d’Alembertian.
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Exercise Sheet No. 2
Maneenate Wechakama, Jochen Klar
d) In general relativity, the covariant form of equation (10) is given by:
Fab = ∇a Ab − ∇b Aa
(13)
Write out the right hand in terms of Christoffel symbols. Show that the the
Christoffel symbols cancel, leaving equation (10).
3. Metric connection [6 points]
It was shown in class that the covariant derivative of a vector can be expressed as:
(∇b V )a = ∂b V a + Γabc V c
(14)
Given a 2nd rank tensor Mcd , we wish to calculate its covariant derivative (∇b M )cd .
Let V c and W d be two arbitrary vectors and let f = Mcd V c W d be a scalar function.
a) Using the product rule, find ∂b f
b) Using the product rule, find ∇b f
c) For a scalar function (a) and (b) are the same. Thus by setting ∂b = ∇b and
using equation (14), obtain
(∇b M )cd = ∂b Mcd − Γabc Mad − Γabd Mca
(15)
Hint: Remember that dummy indices can easily be swapped!
d) Now suppose that the covariant derivative of the metric vanishes, ie
(∇b g)cd = 0
(16)
and that the christoffel symbols are torsion free, that is
Γabc = Γacb
(17)
Using the above equations (15) - (17), and keeping in mind that the metric
is symmetric (gab = gba ) prove that the christoffel symbols may be calculated
from the metric:
1
Γabc = g ad (∂b gcd + ∂c gbd − ∂d gbc )
2
(18)
Hint: Once you obtain a single expression for ∂a gbc you can cylce through
indices and get two additional equations, one for ∂b gca and one for ∂c gab . These
can be added and subtracted in order to obtain equation (18).
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