1St Set of Examples 1

Home , Antisymmetric tensor

1st Set of Examples 1

1. If F αβ is antisymmetric on its two indices. Show that

α ν αβ Fµ ,νFα = −Fµα,βF

SOLUTION

We have to use the metric tensor ηµν in order to lower the indices. So we get:

α β γα βσ βσ γα γα βσ Fµ ,βF α = (Fµγη ),β(F ησα) = (Fµγ),βF (η ησα) + Fµγη ,β(F ησα) (1)

Since ηµν is constant, ηµν,β = 0 and we get:

βσ γα βσ γ βγ γβ Fµγ,βF (η ησα) = Fµγ,βF δσ = Fµγ,βF = −Fµγ,βF (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:

α β αβ Fµ,βF α = −Fµα,βF (3)

1The 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 = α β µ µ ηαβdx dx . If the coordinates are transformed x → x¯ , show that the line 2 µ ν element is ds =g ¯µνdx¯ dx¯ , and express g¯µν in terms of the partial derivatives ∂xµ/∂x¯ν. For two arbitrary vectors U and V, show that

α β ¯ α ¯ β U · V = U V ηαβ = U V g¯αβ

SOLUTION

We start by writing the diﬀerentials in terms of the new coordinate:

2 α β ds =ηαβdx dx ∂xα ∂xβ =η dx¯µ dx¯ν αβ ∂x¯µ ∂x¯ν (4) ∂xα ∂xβ =η dx¯µdx¯ν αβ ∂x¯µ ∂x¯ν

2 µ ν If we write the line element as ds =g ¯µνdx¯ dx¯ and compare this expression with the last passage of the equation above, we can identify:

∂xα ∂xβ g¯ = η (5) µν αβ ∂x¯µ ∂x¯ν Now consider a vector U α. The transformation of coordinate xµ → x¯µ for U α is: ∂xα U α = U¯ µ (6) ∂x¯µ So for the scalar product U · V we get:

α β U · V = U V ηαβ ∂xα ∂xβ = U¯ µ V¯ ν η ∂x¯µ ∂x¯ν αβ (7) ∂xα ∂xβ = U¯ µV¯ ν η ∂x¯µ ∂x¯ν αβ ¯ µ ¯ ν = U V g¯µν

2 α ¯ α 3. If Λβ and Λβ are two matrices which transform the components of at ensor α ¯ γ from one coordinate basis to another, show that the matrix Λγ Λβ is also a coordinate transformation.

SOLUTION

Take two coordinate transformations as: ∂x¯α x¯µ =x ¯µ(xν)Λα = (8) β ∂xβ and ∂x˜α x˜µ =x ˜µ(xν) Λ˜ α = (9) β ∂xβ The product matrix is given by: ∂x˜α ∂x˜γ ΛαΛ˜ γ = (10) γ β ∂xγ ∂xβ 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:

x¯µ =x ¯µ(xν) =x ¯µ[˜xγ(xν)] (11)

then ∂x¯α ∂x¯α ∂x˜γ Λ¯ α = = (12) β ∂xβ ∂x˜γ ∂xβ This is the same expression of (10) also if it diﬀers for thex ˜γ 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).

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:

¯ α β α β Fµν = ΛµΛν Fαβ = −ΛµΛν 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:

¯ α β β α ¯ Fµν = −ΛµΛν Fβα = −ΛµΛν Fαβ = −Fνµ (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:

µν αµ βν αµ βν βµ αν µν F = g g Fβα = −g g Fαβ = −g g Fβα = −F (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 µν 5. Let Aµν be an antisymmetric tensor so that Aµν = −Aνµ and let S be a symmetric tensor so that Sµν = Sνµ. Show that

µν AµνS = 0 .

For any arbitrary tensor Vµν establish the following two identities: 1 1 V µνA = V µν − V νµA V µνS = V µν + V νµS µν 2 µν µν 2 µν

µν µν νµ If Aµν is antisymmetric, then AµνS = −AνµS = −AνµS . Because µ and ν are dummy indices, we can relabel it and obtain:

µν νµ µν AµνS = −AνµS = −AµνS

µν so that AµνS = 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

µν µν νµ µν AµνS = −AνµS = −AνµS = −AµνS ≡ 0 .

¯ 1 Each tensor can be written like the sum of a symmetric part Vµν = 2 Vµν + Vνµ and an ˜ 1 ¯ ˜ 1 antisymmetric part Vµν = Vµν − Vνµ so that a Vµν = Vµν + Vµν = Vµν + Vνµ + Vµν − 2 2 Vνµ = Vµν. So we have: 1 1 1 V µνA = V¯ A + V˜ A == V˜ A = V − V A (16) µν 2 µν µν µν µν 2 µν µν 2 µν νµ µν 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: 1 1 1 V µνS = V¯ S + V˜ S == V¯ S = V + V S (17) µν 2 µν µν µν µν 2 µν µν 2 µν νµ µν

5 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 V = V − V [µν] 2 µν νµ

F[µν] = Fµν

F(µν) = 0

(Fµν is antisymmetric) T(µν) = Tµν

T[µν] = 0

(Tµν is symmetric) 1 V = V − V + V − V + V − V [µνγ] 6 µνγ µγν γµν γνµ νγµ νµγ 1 F = F + F + F [µν,γ] 3 µν,γ γµ,ν νγ,µ 1 T = T + T + T [µν,γ] 3 µν,γ γµ,ν νγ,µ

(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: 1 11 A = (A − A ) = (A − A + A − A + A − A ) [[µ,ν],γ] 2 [µ,ν,γ] [ν,µ,γ] 2 6 µ,ν,γ ν,µ,γ γ,µ,ν µ,γ,ν ν,γ,µ γ,ν,µ 1 − (A − A + A − A + A − A ) 6 ν,µ,γ µ,ν,γ γ,ν,µ ν,γ,µ µ,γ,ν γ,µ,ν 12 = (A − A + A − A + A − A ) 2 6 µ,ν,γ ν,µ,γ γ,µ,ν µ,γ,ν ν,γ,µ γ,ν,µ

= 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 µ 8. Show that the Kronecker delta δν is a tensor. SOLUTION

µ We have just to show that δν transform like a tensor. We have for a transformation of coordinate xµ → x¯γ(xµ):

∂x¯µ ∂xσ ∂x¯µ ∂xσ δ¯µ = δγ = = δ¯µ (19) ν ∂xγ ∂x¯ν σ ∂xσ ∂x¯ν ν

∂xσ ∂x¯µ where the last equality follows from the fact that ∂x¯µ and ∂xσ are the matrix inverse of µ each other. So δν transforms like a tensor.

9 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 Minkowski 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: n! nr−a (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:

∂xα ∂xβ ∂xγ ∂xδ ∂xa ¯ = = det (20) αβγδ ∂x¯µ ∂x¯ν ∂x¯λ ∂x¯σ αβγδ ∂x¯a µνλσ

Now we have to ﬁnd the relation between the expression above and the metric tensor gµν. We have already shown that: ∂xα ∂xβ g¯ = η (21) µν ∂x¯µ ∂x¯ν αβ and so: ∂xa ∂xa det[¯g ] = | det |2 det[η ] = −| det |2 (22) µν ∂x¯a αβ ∂x¯a It folllows from the expression above that: ∂xa det = [− det(¯g )]1/2 (23) ∂x¯a µν Substituting this last expression in (20) we get:

1/2 ¯αβγδ = [− det(¯gµν)] αβγδ (24)

10 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 αµ βν γλ δσ = η η η η µνλσ (25)

Since that in this frame only the elements on the diagonal are diﬀerent from zero we get:

0123 00 11 22 33 = η η η η 0123 = −0123 (26)

because η00 = −1 and ηii = 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:

1/2 1/2 αβγδ αβγδ = [− det(¯gµν)] αβγδ = −[− det(¯gµν)] (28)

11 11. Prove that: a) 1 gµν R − g R = gµνS = 0 µν 4 µν µν

b) λρ g Cλµνρ = 0

SOLUTION

a) Proof that Sµν is traceless

1 1 1 S = gµνS = gµν R − g R = gµνR − gµνg R = R − · 4 · R = 0 µν µν 4 µν µν 4 µν 4

λρ λρ g Cλµνρ = g Rλµνρ 1 − gλρ (g S + g S − g S − g S ) 2 λρ µν µν λρ λν µρ µρ λν 1 − Rgλρ (g g − g g ) 12 λρ µν λν µρ 1 1 = R − (4S + 0 − S − S ) − R (4g − g ) µν 2 µν µν µν 12 µν µν 1 = R − S − Rgµν µν µν 4 = 0 (29)

12 12. Prove that: a) µβ µβ gαβg ,γ = −g gαµ,γ b) βγ βγ g,α = −ggβγg ,α = gg gβγ,α c) In a coordinate frame α 1/2 Γ αβ = log |g| ,β d) In a coordinate frame 1 gµνΓα = − gαν|g|1/2 µν |g|1/2 ,ν

e) In a coordinate frame 1 Aα = |g|1/2Aα ;α |g|1/2 ,α f) In a coordinate frame 1 Aβ = |g|1/2Aβ − Γλ Aµ α;β |g|1/2 α ,β αµ λ

g) In a coordinate frame if Aαβ is antisymmetric 1 Aαβ = |g|1/2Aαβ ;β |g|1/2 ,β

h) In a coordinate frame 1 S = S ;α = |g|1/2gαβS ;α |g|1/2 ,β ,α

SOLUTION

µβ β µβ µβ a) Since gαµg = δα we get gαµ,γg + gαµg ,γ = 0

λν −1 λν λν −2 λν −2 λν b) g = g G → g ,γ = −g g,γG = −g g,γg g and thus

λν g,γ = −gλνg ,γg

µ 1 µν α 1 αν c) In a coordinate frame Γ αβ = 2 g (gνα,β + gνβ,α − gαβ,ν) thus Γ αβ = 2 g gνα,β and thus by b) 1 1 Γα = g = (log |g|) = log |g|1/2 αβ 2g ,β 2 ,β ,β

13 d) First we prove that αβ α µβ β µα g ,γ = −Γ µγg − Γ µγg (30)

αβ µβ λα µβ λα α µβ β λα g ,γ = −gλµ,γg g = ··· = − (Γλµγ + Γµλγ) g g = −Γ µγg − Γ λγg here we also used that gλµ,γ = Γλµγ + Γµλγ Then by using eq (30) and putting β = γ we can write:

µν α αβ β λα αβ 1/2 λα αν 1/2 αν g Γ µν = −g ,β − Γ λβg = −g ,β − log |g| ,λ g = −g ,ν − log |g| ,ν g αν 1/2 αν −1/2 = −g ,ν − |g|,ν g |g| (31) e) 1 Aα = Aα + Γα Aβ = Aα + |g|1/2 Aβ ;α ,α βα ,α |g|1/2 ,β 1 = |g|1/2Aα (32) |g|1/2 ,α f)

β β β µ λ β A α;β = A α,β + Γ µβA α − Γ αβA λ 1 = Aβ + |g|1/2 Aµ − Γλ Aµ α,β |g|1/2 ,µ α αµ λ 1 = |g|1/2Aβ − Γλ Aµ |g|1/2 α ,β αµ λ g)

αβ αβ α µβ β αµ A ;β = A ,β + Γ µβA + Γ µβA 1 = Aαβ + Γα Aµβ + |g|1/2 Aαµ ,β µβ |g|1/2 ,µ 1 = |g|1/2Aαβ + Γα Aµβ (33) |g|1/2 ,β µβ

α α µβ [µβ] But Γ µβ = Γ (µβ), so if A = A , then the last term vanishes β αβ h) By using equation (32) (assume for example that A = S,αg ) we get: 1 S = S gαβ = |g|1/2S gαβ (34) ,α ;β |g|1/2 ,α ,β