1st Set of Examples 1
1. If F αβ is antisymmetric on its two indices. Show that
Fµ α,ν Fαν = −Fµα,β F αβ
SOLUTION
We have to use the metric tensor ηµν in order to lower the indices. So we get:
βσ
ησα )
Fµ α,β F βα = (Fµγ η γα ),β (F βσ ησα ) = (Fµγ ),β F βσ (η γα ησα ) + Fµγ η γα
,β (F
(1)
Since ηµν is constant, ηµν,β = 0 and we get:
Fµγ,β F βσ (η γα ησα ) = Fµγ,β F βσ δσγ = Fµγ,β F βγ = −Fµγ,β F γβ
(2)
where in the last passage we have used the antisymmetry of the tensor Fµν . Considering
that γ is a dummy index, we can relabel it so that:
Fµ,βα F βα = −Fµα,β F αβ
1
(3)
The problems have been chosen by the Problem Book in Relativity and Gravitation By Alan P. Lightman,
Richard H. Price, William H. Press, Saul A. Teukolsky, Published by Princeton University Press, 1975
1
2. In a coordinate system with coordinates xµ , the invariant line element is ds2 =
ηαβ dxα dxβ . If the coordinates are transformed xµ → x̄µ , show that the line
element is ds2 = ḡµν dx̄µ dx̄ν , and express ḡµν in terms of the partial derivatives
∂xµ /∂ x̄ν . For two arbitrary vectors U and V, show that
U · V = U α V β ηαβ = Ū α V̄ β ḡαβ
SOLUTION
We start by writing the differentials in terms of the new coordinate:
ds2 =ηαβ dxα dxβ
∂xα µ ∂xβ ν
dx¯
dx̄
∂ x̄µ
∂ x̄ν
∂xα ∂xβ µ ν
= ηαβ µ ν dx̄ dx̄
∂ x̄ ∂ x̄
=ηαβ
(4)
If we write the line element as ds2 = ḡµν dx̄µ dx̄ν and compare this expression with the
last passage of the equation above, we can identify:
ḡµν = ηαβ
∂xα ∂xβ ∂ x̄µ ∂ x̄ν
(5)
Now consider a vector U α . The transformation of coordinate xµ → x̄µ for U α is:
Uα =
∂xα µ
Ū
∂ x̄µ
(6)
So for the scalar product U · V we get:
U · V = U α V β ηαβ
∂xα ∂xβ = Ū µ µ V̄ ν ν ηαβ
∂ x̄
∂ x̄ α
β
∂x
∂x
= Ū µ V̄ ν
ηαβ
∂ x̄µ ∂ x̄ν
= Ū µ V̄ ν ḡµν
2
(7)
3. If Λαβ and Λ̄αβ are two matrices which transform the components of at ensor
from one coordinate basis to another, show that the matrix Λαγ Λ̄γβ is also a
coordinate transformation.
SOLUTION
Take two coordinate transformations as:
µ
µ
ν
∂ x̄α
=
∂xβ
(8)
∂ x̃α
∂xβ
(9)
x̄ = x̄ (x )
Λαβ
x̃µ = x̃µ (xν )
Λ̃αβ =
and
The product matrix is given by:
Λαγ Λ̃γβ =
∂ x̃α ∂ x̃γ
∂xγ ∂xβ
(10)
The expression above can seem to differ from a usual coordinate transformation.We will
show that this is not the case. Take another transformation of coordinate such as:
x̄µ = x̄µ (xν ) = x̄µ [x̃γ (xν )]
then
(11)
∂ x̄α ∂ x̃γ
∂ x̄α
=
(12)
∂xβ
∂ x̃γ ∂xβ
This is the same expression of (10) also if it differs for the x̃γ in the denominator: the
partial derivative are always taken respect on the argument variable, and it is possible
to call these variables with different ’names’ without that the meaning of the operation
changes. So the expression (12) is a transformation of coordinate as well as (10).
Λ̄αβ =
3
4. Show that the second rank tensor F which is antisymmetric in one coordinate
frame (Fµν = −Fνµ ) is antisymmetric in all frames. Show that the contravariant
components are also antisymmetric (F µν = −F νµ ). Show that symmetry is also
coordinate invariant.
SOLUTION
We start by trasforming the tensor Fµν :
F̄µν = Λαµ Λβν Fαβ = −Λαµ Λβν Fβα .
(13)
where in the last passage we have use the antisymmetry propriety of Fµν . Now if we
consider that α and β are two dummy indices, we can relabel them, for example naming
α → β and β → α so that:
F̄µν = −Λαµ Λβν Fβα = −Λβµ Λαν Fαβ = −F̄νµ
(14)
So if a tensor is antisymmetric in one coordinate frame, it is antisymmetric in all the
coordinate frame. The same applied to an antisymmetric contravariant tensor:
F µν = g αµ g βν Fβα = −g αµ g βν Fαβ = −g βµ g αν Fβα = −F µν
(15)
where we have used the antisymmetry propriety of F µν and the fact that α and β are
dummy indices and, consequently, we can relabel it.
4
5. Let Aµν be an antisymmetric tensor so that Aµν = −Aνµ and let S µν be a
symmetric tensor so that S µν = S νµ . Show that
Aµν S µν = 0 .
For any arbitrary tensor Vµν establish the following two identities:
V µν Aµν =
1 µν
1
V − V νµ Aµν V µν Sµν = V µν + V νµ Sµν
2
2
If Aµν is antisymmetric, then Aµν S µν = −Aνµ S µν = −Aνµ S νµ . Because µ and ν
are dummy indices, we can relabel it and obtain:
Aµν S µν = −Aνµ S νµ = −Aµν S µν
so that Aµν S µν = 0, i.e. the product of a symmetric tensor times an antisymmetric one is equal to zero.
SOLUTION
Since the µ and ν are dummy indexes can be interchanged, so that
Aµν S µν = −Aνµ S µν = −Aνµ S νµ = −Aµν S µν ≡ 0 .
Each tensor can be written like the sum of a symmetric part V̄µν = 12 Vµν + Vνµ and an
antisymmetric
part Ṽµν = 12 Vµν − Vνµ so that a Vµν = V̄µν + Ṽµν = 12 Vµν + Vνµ + Vµν −
Vνµ = Vµν . So we have:
V µν Aµν =
1
1
1
V̄µν Aµν + Ṽµν Aµν == Ṽµν Aµν = Vµν − Vνµ Aµν
2
2
2
(16)
since we have already show that the scalar product of a symmetric tensor with an antisymmetric one is equal to zero. In the same way we can show that:
V µν Sµν =
1
1
1
V̄µν Sµν + Ṽµν Sµν == V̄µν Sµν = Vµν + Vνµ Sµν
2
2
2
5
(17)
6. (a) In a n-dimensional metric space, how many independent components are
there for an r-rank tensor T αβ... with no symmetries?
(b) How many independents component are there if the tensor is symmetric
on s of its indices?
(c) How many independents component are there if the tensor is antisymmetric on a of its indices?
SOLUTION
(a) In the case of no symmetries the number of independent components is nr (where r
is the rank of the tensor).
(b) If we have s symmetric indices, then we have to calculate how many inequivalent way
we have to choose them (including the repetitions) in a set n. The number is given by:
(n + s − 1)!
(n − 1)!s!
while the remaining r − s indices can be chosen in nr−s ways so that the number of
independent component is:
(n + s − 1)!
nr−s
(n − 1)!s!
(c) If we have a antisymmetric indices, then we have to calculate how many inequivalent
way we have to choose them (including the repetitions) in a set n. The number is given
by:
n!
(n − a)!a!
while the remaining r − a indices can be chosen in nr−a ways so that the number of
independent component is:
n!
nr−a
(n − a)!a!
Note that for a = n we have just a possibility to choose the a indices, while for a > n
there is no possibility: this means that all the components must be zero.
6
7. If F is antisymmetric, T is symmetric and V is an arbitrary tensor, give
explicit formulation for the following :
(a) V[µν] , F[µν] , F(muν) , T[µν] , T(µν) , V[µνγ] , T(µν,γ) , F[µν,γ] .
(b) Show that Fµν = A[ν,µ] − A[µ,ν] implies
Fµν,γ + Fγµ,ν + Fνγ,µ = 0
SOLUTION
(a)
1
Vµν − Vνµ
2
F[µν] = Fµν
V[µν] =
F(µν) = 0
(Fµν is antisymmetric)
T(µν) = Tµν
T[µν] = 0
(Tµν is symmetric)
V[µνγ] =
1
Vµνγ − Vµγν + Vγµν − Vγνµ + Vνγµ − Vνµγ
6
1
Fµν,γ + Fγµ,ν + Fνγ,µ
3
1
= Tµν,γ + Tγµ,ν + Tνγ,µ
3
F[µν,γ] =
T[µν,γ]
(b) Fµν is an antisymmetric tensor so:
Fµν,γ + Fγµ,ν + Fνγ,µ = 3F[µν,γ]
but Fµν = −A[µ,ν] and consequently F[µν,γ] = −A[[µ,ν],γ]
But we have:
A[[µ,ν],γ] =
−
=
=
1
1 1
(A[µ,ν,γ] − A[ν,µ,γ] ) =
(Aµ,ν,γ − Aν,µ,γ + Aγ,µ,ν − Aµ,γ,ν + Aν,γ,µ − Aγ,ν,µ )
2
2 6
1
(Aν,µ,γ − Aµ,ν,γ + Aγ,ν,µ − Aν,γ,µ + Aµ,γ,ν − Aγ,µ,ν )
6
1 2
(Aµ,ν,γ − Aν,µ,γ + Aγ,µ,ν − Aµ,γ,ν + Aν,γ,µ − Aγ,ν,µ )
2 6
A[µ,ν,γ]
(18)
7
and since Aµ,ν,γ = Aµ,γ,ν , it follows
F[µν,γ] = −A[[µ,ν],γ] = −A[µ,ν,γ] = 0
and so:
Fµν,γ + Fγµ,ν + Fνγ,µ = 3F[µν,γ] = 0
8
8. Show that the Kronecker delta δνµ is a tensor.
SOLUTION
We have just to show that δνµ transform like a tensor. We have for a transformation of
coordinate xµ → x̄γ (xµ ):
δ̄νµ =
∂ x̄µ ∂xσ γ ∂ x̄µ ∂xσ
δ =
= δ̄νµ
∂xγ ∂ x̄ν σ ∂xσ ∂ x̄ν
where the last equality follows from the fact that
each other. So δνµ transforms like a tensor.
9
∂xσ
∂ x̄µ
and
∂ x̄µ
∂xσ
(19)
are the matrix inverse of
9. Prove that, except for scaling by a constant, there is an unique tensor αβγδ
which is totally antisymmetric on all its 4 indices. The usual choice is to take
0123 = 1 in Minkowsky coordinates. What are the components of in a generic
coordinate system, with metric gµν ?
SOLUTION
For a totally antisymmetric vector with rank r and a antisymmetric components in a
n-folds, we have already shown that the number of independent components is given by:
nr−a
n!
(n − a)!a!
For αβγδ we have n = a = 4 so that there is just one possibility to choose the component,
i.e. once time that 0123 is given, the tensor is fixed in an unique way. In Minkowski
coordinates, we have:
0123 = −1023 = 1032 = ... = −1
In a generic frame:
¯αβγδ =
∂xa ∂xα ∂xβ ∂xγ ∂xδ
=
det
αβγδ
µνλσ
∂ x̄µ ∂ x̄ν ∂ x̄λ ∂ x̄σ
∂ x̄a
(20)
Now we have to find the relation between the expression above and the metric tensor gµν .
We have already shown that:
∂xα ∂xβ
ηαβ
(21)
ḡµν =
∂ x̄µ ∂ x̄ν
and so:
∂xa 2
∂xa 2
det[ḡµν ] = |det
|
det[η
]
=
−|det
|
(22)
αβ
∂ x̄a
∂ x̄a
It folllows from the expression above that:
det
∂xa = [−det(ḡµν )]1/2
∂ x̄a
(23)
Substituting this last expression in (20) we get:
¯αβγδ = [−det(ḡµν )]1/2 αβγδ
10
(24)
10. In an orthonormal frame, show that
αβγδ = −αβγδ
What is the analogous relation in a general frame, with metric tensor gµν ?
SOLUTION
In an orthonormal frame we get:
αβγδ = η αµ η βν η γλ η δσ µνλσ
(25)
Since that in this frame only the elements on the diagonal are different from zero we get:
0123 = η 00 η 11 η 22 η 33 0123 = −0123
(26)
because η 00 = −1 and η ii = 1. So in this frame:
αβγδ = −αβγδ
(27)
In order to find the expression in a generic frame, we use the equation (24) and so we get:
αβγδ = [− det(ḡµν )]1/2 αβγδ = −[− det(ḡµν )]1/2 αβγδ
11
(28)
11. Prove that:
a)
g
µν
1
Rµν − gµν R = g µν Sµν = 0
4
b)
g λρ Cλµνρ = 0
SOLUTION
a) Proof that Sµν is traceless
µν
S = g Sµν = g
µν
1
1
1
Rµν − gµν R = g µν Tµν − g µν gµν R = R − · 4 · R = 0
4
4
4
b)
g λρ Cλµνρ = g λρ Rλµνρ
1 λρ
−
g (gλρ Sµν + gµν Sλρ − gλν Sµρ − gµρ Sλν )
2
1
−
R g λρ gλρ gµν − g λρ gλν gµρ
12
1
1
= Rµν − (4Sµν + 0 − Sµν − Sµν ) − R (4gµν − gµν )
2
12
1 µν
= Rµν − Sµν − Rg
4
= 0
12
(29)
12. Prove that:
a)
gαβ g µβ ,γ = −g µβ gαµ,γ
b)
g,α = −ggβγ g βγ ,α = gg βγ gβγ,α
c) In a coordinate frame
Γα αβ = log |g|1/2
,β
d) In a coordinate frame
g µν Γα µν = −
1
g αν |g|1/2 ,ν
1/2
|g|
e) In a coordinate frame
Aα ;α =
1
|g|1/2 Aα ,α
1/2
|g|
f) In a coordinate frame
Aβα;β =
1
1/2 β
− Γλ αµ Aµλ
|g|
A
α
,β
1/2
|g|
g) In a coordinate frame if Aαβ is antisymmetric
Aαβ ;β =
1
1/2 αβ
|g|
A
,β
|g|1/2
h) In a coordinate frame
S = S;α ;α =
1
|g|1/2 g αβ S,β ,α
1/2
|g|
SOLUTION
a) Since gαµ g µβ = δαβ we get gαµ,γ g µβ + gαµ g µβ ,γ = 0
b) g λν = g −1 Gλν → g λν ,γ = −g −2 g,γ Gλν = −g −2 g,γ g λν g and thus
g,γ = −gλν g λν ,γ g
c) In a coordinate frame Γµ αβ = 21 g µν (gνα,β + gνβ,α − gαβ,ν ) thus Γα αβ = 12 g αν gνα,β and
thus by b)
1
1
Γα αβ = g,β = (log |g|),β = log |g|1/2 ,β
2g
2
13
d)
g µν Γα µν = −g αβ ,β − Γβ λβ g λα = −g αβ ,β − log |g|1/2
g λα = −g αν ,ν − log |g|1/2
,λ
αν
−1/2
= −g αν ,ν − |g|1/2
,ν g |g|
,ν
g αν
(30)
e)
Aα ;α = Aα ,α + Γα βα Aβ = Aα ,α +
1
1
|g|1/2 ,β Aβ = 1/2 |g|1/2 Aα ,α
1/2
|g|
|g|
f)
Aβα;β = Aβα,β + Γβµβ Aµα − Γλαβ Aβλ = Aβα,β +
=
1
1/2
|g|
Aµα − Γλαµ Aµλ
,µ
1/2
|g|
1
1/2 β
|g|
A
− Γλαµ Aµλ
α
,β
1/2
|g|
g)
Aαβ ;β = Aαβ ,β + Γαµβ Aµβ + Γβµβ Aαµ = Aαβ ,β + Γαµβ Aµβ +
=
1
1/2
|g|
Aαµ
,µ
|g|1/2
1
1/2 αβ
|g|
A
+ Γαµβ Aµβ
,β
1/2
|g|
But Γαµβ = Γα(µβ) , so if Aµβ = A[µβ] , then the last term vanishes
h)
1
S = S,α g αβ ;β = 1/2 |g|1/2 S,α g αβ ,β
|g|
14
(31)