Tensor Analysis and Geometry

Tensor Analysis & Geometry Spherical Coordinates x1 = r sin θ cos φ , x 2 = r sin θ sin φ , x 3 = r cosθ , x 4 = ct 3-dimensional line element:

(

ds 2 = dr 2 + r 2 dθ 2 + sin 2 θdφ 2

)

Christoffel Symbols Γrmn = [mn, r ] =

Christoffel Symbol of the first kind:

1  ∂g rm ∂g rn ∂g mn + m −  2  ∂x n ∂x ∂x r

 . 

z Γmn =  z  = g rz Γmnr . î mn

Christoffel Symbol of the second kind:

Derivation of the Riemann Curvature Tensor In general, a second order differentiation on a covariant vector is independent of the order in which it is carried out, i.e.:

∂ 2Vi ∂ 2Vi . = ∂x j ∂x k ∂x k ∂x j However, the presence of Christoffel symbols can have an effect on this statement. We investigate this by first finding the general second derivatives for both permutations of the differentiating parameters:

(V )

i ; j ;k

But

= Vi , jk − Γikr Vr , j − Γ jkr Vi ,r .

Vi ; j = Vi , j − ΓijsVs , ∴ (Vi: j ):k = Vi , jk − ⇔ (Vi: j ):k = Vi , jk −

∂Γijs

∂x ∂Γijs ∂x

k

k

[

]

[

Vs − ΓijsVs ,k − Γikr Vr , j − Γrjs Vs − Γ jkr Vi ,r − Γirs Vs

]

Vs − ΓijsVs ,k − Γikr Vr , j + Γikr Γrjs Vs − Γ jkr Vi , r + Γ jkr Γirs Vs

Now we interchange j and k (which is the other possible way of determining this second derivative):

(Vi:k ): j

= Vi ,kj

∂Γiks − j Vs − Γiks Vs , j − ΓijrVr ,k + Γijr Γrks Vs − Γkjr Vi ,r + Γkjr Γirs Vs . ∂x

We now find the difference between these two. On the RHS, the first, third, fourth, sixth, and seventh terms cancel out, thus giving the result:

Vi: jk − Vi:kj = −

 ∂Γiks ∂Γijs  ∂Γijs ∂Γiks r s r s r s r s V V V V + Γ Γ + − Γ Γ =  j − k + Γik Γrj − Γij Γrk Vs . s ik rj s s ij rk s k j ∂x ∂x ∂x  ∂x 

We define the Riemann (or Riemann-Christoffel) Curvature tensor by: s ∂Γiks ∂Γij R = − + Γikr Γrjs − Γijr Γrks . ∂x j ∂x k s The difference between the covariant derivatives can thus be written as Vi: jk − Vi:kj = Rijk Vs . The Riemann s ijk

tensor used in this equation is called the Riemann curvature tensor of the second kind. The curvature tensor of the first kind is defined as:

Rijkl = g ir R rjkl . Symmetry Properties: First skew symmetry Second skew symmetry

Courtney James Mewton

Rijkl = − R jikl Rijkl = − Rijlk

Page 1

GR, Tensor Analysis & Geometry

GENERAL RELATIVITY, TENSOR ANALYSIS AND GEOMETRY

Block symmetry

Rijkl = Rklij

Bianchi’s identity

Rijkl + Riklj + Riljk = 0

The Ricci Tensor The Ricci tensor of the first kind is simply a contraction of the Riemann tensor:

Rij = Rijkk . The last index can be raised to yield the Ricci tensor of the second kind:

Ri j = g ik Rki . If this tensor is finally contracted by letting I = j, we get the Ricci curvature scalar. If it is zero, the space is flat. From the first of the two equations above the Ricci tensor of the first kind can be calculated directly by:

Rij = Rijkk =

k ∂Γikk ∂Γij − + Γikr Γrjk − Γijr Γrkk . j k ∂x ∂x

Transformation Of A Geodesic From Parameter u To v, Where v = f(u) Given a particular geodesic in terms of a parameter u, in this section the geodesic equation will be transformed so that it is in terms of a new parameter v.

D  dx a  Start with du  du dx a du D  ∂x a  then du  ∂v Substitute

b c  d 2 xa a dx dx  = + Γbc =0. 2 du du  du dx a dv = , dv du ∂x b dv dx c ∂x a d 2 v dv  ∂x a dv  = + Γbca + =0 du  ∂u∂v du ∂v du du ∂v du 2

∂x a ⇔ ∂u∂v ∂x a ⇔ ∂u∂v

b c dv ∂x a d 2 v a ∂x dx dv + Γbc =− du ∂v du du ∂v du 2 b c dv  du du  ∂x a d 2 v  du du  a ∂x dx dv  du du    + Γbc  =−   du  dv dv  ∂v du du  dv dv  ∂v du 2  dv dv  2

∂x a ∂x b dx c ∂x a d 2 v  du  ⇔ 2 + Γbca =−   . ∂v dv ∂v du 2  dv  ∂v Since the LHS is now in terms of v, partial differentiation can be replaced by normal differentiation:

dx a dx b dx c dx a d 2 v ⇔ 2 + Γbca =− dv dv dv du 2 dv By letting

λ=−

d 2v du 2

2

 dv    .  du 

2

 dv    , we arrive at the final equation:  du  b c dx a dx a a dx dx λ + Γ = . bc dv dv dv dv 2

General Relativity The Metric Tensor for Special Relativity


Page 2



0 0 1 0 0 1 0 0 . −  = 0 0 − 1 0  0 0 0 − 1

η µν

Einstein's Law of Gravitation Simply stated, Einstein's law of gravitation is: Rµν = 0 . This condition holds when the local space is completely devoid of all forms of matter and energy.

Derivation of the Schwarzchild Solution In this section the line-element solution to the field equations for a quasi-static gravitational field produced by a spherical body will be derived. We start by setting up a general line element employing spherical coordinates:

(

)

c 2 dτ 2 = Ac 2 dt 2 − Bdr 2 − Wr 2 dθ 2 + sin 2 θdφ 2 . Before we continue, a few assumptions need to be made: The space is asymptotically flat. This means that A = B → 1 as r → ∞ . The gravitational field only affects time and radial distance, so W = 1. We can immediately define the metric tensor:

g 00 = Ac 2 , g11 = − B , g 22 = − r 2 and g 33 = − r 2 sin 2 θ . Since we are dealing with the empty space surrounding the body, the Ricci tensor needs to equal zero. With this in mind, the derivation begins. We first calculate the Christoffel symbols. Note that since a Christoffel symbol of the second kind is defined as z Γmn =  z  = g rz Γmnr î mn

we need only calculate them for values when r = z.

Γ100 = g 00 Γ100 = 12 A −1c −2 [g 00,1 + g10 , 0 − g10, 0 ] = 12 A −1 A′

1 Γ00 = g 11Γ001 = − 12 B −1 [g 10 ,0 + g 01, 0 − g 00,1 ] = 12 B −1 g 00,1 = 12 B −1 A′c 2

Γ111 = g 11Γ111 = − 12 B −1 [g11,1 + g 11,1 − g11,1 ] = − 12 B −1 g 11,1 = 12 B −1 B ′

1 Γ22 = g 11Γ221 = − 12 B −1 [g12 , 2 + g 21, 2 − g 22 ,1 ] = − 12 B −1 g 22,1 = − 12 B −1 ⋅ (2r ) = − rB −1

(

)

1 Γ33 = g 11Γ331 = − 12 B −1 [g13,3 + g 31,3 − g 33,1 ] = − 12 B −1 (− g 33,1 ) = − 12 B −1 2r sin 2 θ = − rB −1 sin 2 θ

Γ212 = g 22 Γ212 = − 12 r −2 [g 21, 2 + g 22,1 − g 21, 2 ] = − 12 r −2 ⋅ (− 2r ) = r −1

Γ332 = g 22 Γ332 = − 12 r −2 [g 23,3 + g 32,3 − g 33, 2 ] = − 12 r −2 ⋅ 2r 2 cosθ sin θ = − cosθ sin θ

(

)

Γ133 = g 33 Γ133 = − 12 r −2 sin −2 θ [g 33,1 + g13,3 − g 13,3 ] = − 12 r −2 sin −2 θ ⋅ − 2r sin 2 θ = r −1

(

)

Γ = g Γ233 = − 12 r sin θ [g 33, 2 + g 23,3 − g 23,3 ] = − 12 r sin θ ⋅ − 2r cosθ sin θ = cot θ 3 23

−2

33

−2

−2

We now solve the field equations:

[

(

) (

) (

−2

2

) (

1 1 R00 = Γ000 ,0 − Γ000 ,0 + Γ000 Γ000 − Γ000 Γ000 + Γ00 Γ100 − Γ00 Γ100 + Γ002 Γ200 − Γ002 Γ200 + Γ003 Γ300 − Γ003 Γ300

[ + [Γ + [Γ

(

)(

)

(

) ( − Γ Γ ) + (Γ Γ − Γ Γ ) + (Γ Γ

1 1 0 1 0 1 1 1 1 1 2 1 2 1 3 1 3 1 + Γ01 , 0 − Γ00 ,1 + Γ01 Γ00 − Γ00 Γ01 + Γ01 Γ10 − Γ00 Γ11 + Γ01 Γ20 − Γ00 Γ21 + Γ01 Γ30 − Γ00 Γ31

( + (Γ

) ( )+ (Γ

)( )+ (Γ Γ

2 02 , 0

1 1 − Γ002 , 2 + Γ020 Γ002 − Γ000 Γ022 + Γ02 Γ102 − Γ00 Γ122 + Γ022 Γ202

3 03, 0

− Γ003 ,3

0 03

Γ003 − Γ000 Γ033


1 Γ − Γ00 Γ133

1 3 03 10

Page 3

2 03

3 20

2 00

2 22

3 02

2 00

3 23

3 03

2 30

3 30

)]

)] )]

− Γ003 Γ322 − Γ003 Γ333


)]


)(

(

)

(

) (

R11 = Γ100 ,1 − Γ110 ,0 + Γ100 Γ010 − Γ110 Γ000 + Γ101 Γ110 − Γ111 Γ100 + Γ102 Γ210 − Γ112 Γ200 + Γ103 Γ310 − Γ113 Γ300  1 1 1 1 1 1 1 + Γ11,1 − Γ111 ,1 + Γ110 Γ01 − Γ110 Γ01 + Γ111 Γ111 − Γ111 Γ111 + Γ112 Γ21 − Γ112 Γ21 + Γ113 Γ31 − Γ113 Γ31

[

(

(

) ( Γ ) + (Γ

(

)

) (

)] Γ )]

) ( Γ )+ (Γ

)( )(

1 2 1 2 2 2 2 2 3 2 3 2 + Γ122 ,1 − Γ112 , 2 + Γ120 Γ012 − Γ110 022 12 Γ11 − Γ11 Γ12 + Γ12 Γ21 − Γ11 22 12 Γ31 − Γ11 32  + Γ133 ,1 − Γ113 ,3 + Γ130 Γ013 − Γ110 Γ033 + Γ131 Γ113 − Γ111 Γ133 + Γ132 Γ213 − Γ112 Γ233 + Γ133 Γ313 − Γ113 Γ333 

[

(

(

)

(

)

(

)

(

)]

)]

) (

1 1 R22 = Γ200 , 2 − Γ220 ,0 + Γ200 Γ020 − Γ220 Γ000 + Γ20 Γ120 − Γ22 Γ100 + Γ202 Γ220 − Γ222 Γ200 + Γ203 Γ320 − Γ223 Γ300

)]

)( )( )] −Γ Γ − Γ Γ ) + (Γ Γ − Γ Γ )+ (Γ Γ − Γ Γ ) + (Γ Γ − Γ Γ )] −Γ Γ − Γ Γ ) + (Γ Γ − Γ Γ )+ (Γ Γ − Γ Γ ) + (Γ Γ − Γ Γ )] R = [Γ − Γ + (Γ Γ − Γ Γ ) + (Γ Γ − Γ Γ )+ (Γ Γ − Γ Γ ) + (Γ Γ − Γ Γ )] + [Γ − Γ + (Γ Γ − Γ Γ ) + (Γ Γ − Γ Γ )+ (Γ Γ − Γ Γ ) + (Γ Γ − Γ Γ )] + [Γ − Γ + (Γ Γ − Γ Γ ) + (Γ Γ − Γ Γ )+ (Γ Γ − Γ Γ ) + (Γ Γ − Γ Γ )] [ + [Γ + [Γ

( + (Γ + (Γ

)

(

1 1 0 1 0 1 1 1 1 1 2 1 2 1 3 1 3 1 + Γ21 , 2 − Γ22 ,1 + Γ21 Γ02 − Γ22 Γ01 + Γ21 Γ12 − Γ22 Γ11 + Γ21 Γ22 − Γ22 Γ21 + Γ21 Γ32 − Γ22 Γ31 2 22 , 2

2 22 , 2

3 23, 2

3 22 , 3

0 30 , 3

33

0 22

2 02

0 22

2 02

1 2 22 12

0 23

3 02

0 22

3 03

1 3 23 12

0 33, 0

0 30

1 31, 3

1 33,1

0 31

2 32 , 3

2 33, 2

0 32

[

1 03

2 03

(

0 03

0 33

0 33

0 33

0 00

1 01

1 32

2 13

2 22

1 3 22 13

1 0 30 13

1 1 31 13

2 02

1 2 22 12

2 23

2 30

2 31

1 2 33 12

2 32

2 22

3 22

1 0 33 10

1 1 33 11

) (

2 22

1 23

2 23

2 22

2 22

3 23

0 23 2 33

2 33

3 22

3 23

2 33

1 21

3 30

3 1 31 33 3 32

3 22

3 32

0 20

2 22

) (

2 32

2 33

) (

3 22

2 32

3 33

0 33

3 33

3 33

3 33

0 30

1 31

2 32

1 + Γ333 ,3 − Γ333 ,3 + Γ330 Γ033 − Γ330 Γ033 + Γ331 Γ133 − Γ33 Γ133 + Γ332 Γ233 − Γ332 Γ233 + Γ333 Γ333 − Γ333 Γ333

)]

We are left with 1 0 1 1 1 1 2 1 3 R00 = −Γ00 ,1 + Γ01Γ00 − Γ00 Γ11 − Γ00 Γ12 − Γ00 Γ13 ,

R11 = Γ100 ,1 + Γ100 Γ010 − Γ111 Γ100 + Γ122 ,1 − Γ111 Γ122 + Γ122 Γ212 + Γ133 ,1 − Γ111 Γ133 + Γ133 Γ313 , 1 1 1 1 2 1 3 1 3 3 3 R22 = −Γ22 Γ100 − Γ22 ,1 − Γ22 Γ11 + Γ21Γ22 + Γ23, 2 − Γ22 Γ13 + Γ23Γ32 , 1 1 − Γ332 , 2 − Γ33 Γ122 + Γ323 Γ332 . R33 = −Γ331 Γ100 − Γ331 ,1 − Γ331 Γ111 + Γ313 Γ33

These equations must equal zero, thus after substitution, we have:

 B′A′ A′′ A′A′ A′  1 R00 =  − + −  , 2 4A r B  4B 2 A′ A′′ A′B′ B′A R11 = − , + − − 4A 2 4B Br A′r B′r 1 R22 = − + −1, 2 AB 2 B 2 B B′r 1  2  A′r R33 =  − + − 1 sin θ = R22 sin 2 θ . 2 2 2 AB B B   Equation 2 becomes:

A′2 A′′ A′B′ B′A B′A A′2 A′′ A′B′ 0=− . + − − ⇒ =− + − 4A 2 4B 4A 2 4B Br Br This can be substituted into equation 1 to give:

B′A A′ B′ A′  B′A A′  1 0 = − −  ⇒ =− ⇒ =− . r B Br r B A  Br This can also be written as:


Page 4


(1)

(2) (3) (4)


1 dB dA dB dA =− ⇔ =− ⇒ B = ⇒ AB = 1 , Bdr Adr B A A upon solving the simple differential equation. We use this solution to simplify equation 3:

0=

A′r A′r ′ + + A − 1 ⇔ 1 = A′r + A = A′r + Ar ′ = (Ar ) 2 2A

We now integrate:

∫ dr = ∫ dr (Ar )dr ⇒ r + k = Ar , d

where k is an integration constant. The equation can be rearranged to find A:

A =1+ Since

k . r

B = A −1 , −1

 k B = 1 +  . r  The value of k is the last thing to obtain. In the next section on the approximation of Newtonian gravitation, g 00 = 1 + h00 . It can immediately be seen that the h00 is equivalent to k. In the Newtonian approximation, it is required that the Newtonian gravitational potential

V = 12 c 2 h00 .

Using the Newtonian potential

V = − GM r , this gives a value of k = − 2GM rc . By substituting A and B back into the original lineelement equation at the start of this section, we have the Schwarzchild solution: 2

 2GM c 2 dτ 2 =  1 − rc 2 

−1

(

)

 2 2  2GM  dr 2 − r 2 dθ 2 + sin 2 θdφ 2 . c dt − 1 − 2  rc   

Utilising Geodesic Equation to Find GR Approximation of Newtonian Gravity A particle travels through spacetime along a geodesic, given by the equation: ν σ dx µ µ dx dx + Γ = 0. νσ dτ dτ dτ 2

τ is the time experienced relative to the particle. The equation simply states that relative to a free particle, it experiences no net acceleration (though other objects appear to accelerate if a gravitational field is present). We wish to determine the motion of the particle relative to coordinate time, denoted by t. The equation would give the path of the particle in accordance to what other observers would see if they thought they were in a gravitational field. With the above said, the following equation can be immediately written, transforming from proper time to coordinate time: ν σ  d 2t d 2xµ µ dx dx  + Γ = νσ  dτ 2 dt dt dt 2 

2 µ  dt   dx .  2   dτ   dt

First, we expand and consider its spatial components: 0 0 0 j k k  d 2t d 2 xi i dx dx i dx dx i dx dx + Γ jk + 2Γ0 k + Γ00 = dt dt dt dt dt dt  dτ 2 dt 2 0 j i dx dx = Γ0i k One of the Christoffel symbols has a coefficient of two since Γ j 0 dt dt

2 i  dt   dx .  2   dτ   dt dx 0 dx k . We simplify the dt dt

equation: j k  d 2t d 2 xi dx k 2 i i dx dx i  2 + Γ + Γ + Γ = c c 0k 00 jk  dτ 2 dt dt dt dt 2 


Page 5

2 i  dt   dx .  2   dτ   dt



We assume that the gravitational field is quasi-static, i.e. that it doesn't change with respect to time. Therefore, any derivatives of the metric tensor with respect to time can be left out. Now we evaluate the connection coefficients:

1 iρ g iρ  ∂g ρj ∂g ρk ∂g jk   Γ = g Γρjk = + − ρ  . These derivatives are quite small, and can be neglected.1 2 2  ∂x k ∂x j ∂x   ∂hρ 0 ∂h ρk ∂h  g iρ  ∂g ρ 0 ∂g ρk ∂g 0 k   k + 2Γ0i k = 2 − ρ  ≈ η iρ + h iρ  k + 0 − 0ρk  0 2  ∂x ∂x ∂x  ∂x ∂x   ∂x ∂hµν  ∂h0 ρ ∂h ρk ∂h  ∂h   ∂h ≈ η iρ + h iρ  k + 0 − 0ρk  ≈ −δ is  0ks − 0sk  on neglecting terms involving . ∂x ∂x  ∂x  ∂x 0  ∂x  ∂x g iκ  ∂g κ 0 ∂g 0κ ∂g 00  δ iz ∂h00 Γ00i = + − κ ≈ .  2  ∂x 0 2 ∂x z ∂x 0 ∂x  i jk

(

(

)

)

Now we need to evaluate the RHS of the equation. We start by first looking at the following line element:

1 1 dx µ dxν dx µ dxν  dτ  c dτ = g µν dx dx ⇔   = 2 (η µν + hµν ) = 2 (1 + hµν ) . dt dt dt dt c c  dt  2

2

2

µ

ν

Neglecting

terms involving the spatial components, which are small in comparison to the temporal components, we get:

dτ  dτ  12 = (1 + h00 ) ≈ (1 + 12 h00 ) .   = (1 + h00 ) ⇒ dt  dt  2 dh d τ dh00 = = c 000 , so the RHS of our major equation becomes: Now 2 dt dt dx dh 2 c 000 2 i   dh00 dx d t  dt  dx i dx i dx 1 ( ) 1 ≈ c 0 − 2 h00 .   ≈  dt  dτ 2  dτ 2   1 + 12 h00 dt dt dx 2

This is negligible since it involves temporal derivatives of the gravitational field. By plugging everything into our equation, we get: iz ∂h0 k  dx k ∂h00 d 2 xi 2 δ is  ∂h0 s δ − − + c =0.   2 k s 2 ∂x z dt ∂x  dt  ∂x

Through multiplying by mass m and rearranging the equation, we get:

∂h0 k  dx k δ iz ∂h00 d 2 xi is  ∂h0 s δ m 2 = − mc 2 + m − .   k 2 ∂x z dt ∂x s  dt  ∂x The term on the left is the force that the particle appears to experience. The first term on the right is some kind of potential of the field, since its temporal component is involved. The last term, which involves perpendicular velocities, is indicative of some sort of Coriolis force. We are not interested in the Coriolis effects, so we shall assume that we are in a non-rotating frame, so we get: iz ∂h00 d 2 xi 2 δ = − mc . 2 2 ∂x z dt 2 If we denote a potential by V = 12 c h00 , then the equation simply becomes F = − m∇V , or

m

d 2 xi ∂V = −δ iz z . We want the metric tensor to be flat when there is no gravitational field present. The 2 dt ∂x 1

The metric tensor

g µν ≈ η µν + h µν , where η µν is the familiar metric tensor of Special Relativity, and

h µν are small terms which include the action of any gravitational fields which may be present, and are small in µν comparison to the η in weak gravitational fields, as opposed to the awesome sucking power of a black hole! Courtney James Mewton

Page 6



equation for the potential leads us to the expression

g 00 = η 00 + h00 = 1 + 2V c 2 . To finally get the actual

expression for the potential in terms of mass, we need to use the line element from the Schwarzchild solution. The temporal component gives V = − GM r . With this expression, we can easily obtain an approximation of the gravitational force:

F = − m∇V = − GMm r 2 . Field Equations in the Presence of Matter: The Poisson Approximation Let us write the equation:

R µν − 12 g µν = κT µν , or, more compactly as:

G µν = κT µν , where G is the Einstein tensor. As a test for General Relativity, at velocities which are small in comparison to the speed of light, there must be an approximation to Poisson’s equation: ∇ V = 4πGρ . To achieve this requires the weak field approximation by leaving out negligible terms in the Ricci tensor.2 2

R µν

G µν = κT µν − 12 g µν R = κT µν

g µν R µν − 12 g µν g µν R = κT µν g µν R − 2 R = κT µν g µν ∴ R = −κT µν g µν We substitute this back into our original equation:

R µν = κT µν + 12 g µν R = κT µν − 12 g µν κT µν g µν We are approximating that the material energy tensor has a negligible value in for all µ, ν except when µ = ν = 0, so we get:

R 00 = κT 00 − 12 g 00κT 00 g 00 = 12 κT 00 . We now make the Ricci tensor covariant:

g 00 g 00 R 00 = 12 κT 00 g 00 g 00 ⇔ R00 = 12 κT00 Weak field approximation.

Rµν ≈

1 2

(g

αα , µν

− gνα , µα − g µα ,να + g µν ,αα ) ≈ 12 g µν ,αα

∴ R00 ≈ 12 g 00,αα ≈ 12 ∇ 2 g 00 = ∴

1 2 ∇V c2

1 2 ∇ V = 12 κT00 = 12 κρv0 v0 2 c

The particles are traveling at the travelling at the speed of light through time, so we get:

∇ 2V = 12 κρc 4 This must equal Poisson’s equation, stated earlier This gives a value

κ=

8πG . c4

2

Note that this particular calculation would be shorter if we took the Einstein tensor and the material energy tensor to be covariant as opposed to contravariant, but due to the actual form of the material energy tensor, I prefer it to be contravariant.


Page 7


Tensor Analysis and Geometry

Recommend Documents