Problem sets - General Relativity Solutions by Sergei Winitzki January 11, 2007
Contents 1 Coordin Coordinate atess and and 1-form 1-formss 1.1 Inv Invertibl ertiblee transf transformati ormations ons . . . . . . . . 1.2 Exampl Examples es of coordina coordinate te transf transformatio ormations ns 1.3 Bas Basis is in tang tangen entt spac spacee . . . . . . . . . . 1.4 Diffe Different rentials ials of function functionss as 1-form 1-formss . . . 1.5 Bas Basis is in in cotan cotangen gentt space space . . . . . . . . . 1.6 Linear Linearly ly indepen independen dentt 1-form 1-formss . . . . . . 1.7 Transfo ransformatio rmation n law law for for 1-forms 1-forms . . . . . 1.8 Exampl Examples es of transf transformati ormations ons . . . . . .
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2 2 2 3 3 3 4 4 4
2 Ten enso sors rs 2.1 Defi Definit nition ion of of tensor tensor prod product uct . . . 2.3 Exa Exampl mplee of ten tensor sor . . . . . . . . 2.5 Cont Contractio raction n of tensor tensor indice indicess . . 2.8 Exa Exampl mples es of spa spaces ces with with a metr metric ic
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7 7 7 7 8
3 The The 3.1 3.2 3.3 3.4 3.5
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9 9 10 10 10 11
Christoffe Christ offell symbo symboll Tran ransfo sforma rmatio tions ns 1 . . . . . . . Tran ransfo sforma rmatio tions ns 2 . . . . . . . Covarian Cov ariantt deriv derivativ atives es . . . . . . The Lei Leibni bnitz tz rul rulee . . . . . . . . Locally inert inertial ial refer reference ence frame
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4 Geodesics Geodesics and cur curv vatur ature e 4.11 Ge 4. Geode odesi sics cs . . . . . . . . . . . . . . . 4.1.1 4.1 .1 Fir First st der deriv ivati ation on . . . . . . . 4.1.2 4.1 .2 Sec Second ond der deriv ivatio ation n. . . . . . . 4.2 Comm Commutator utator of cov covarian ariantt deriva derivative tivess 4.3 Pa Paral rallel lel tran transpor sportt . . . . . . . . . . . 4.4 Rie Rieman mann n ten tensor sor . . . . . . . . . . . . 4.5 Loren Lorentz tz transf transformatio ormations ns . . . . . . .
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5 Gravita Gravitation tion theory theory app applied lied 5.11 Re 5. Reds dshi hift ft . . . . . . . . . . . . . . . . . 5.2 Energy Energy-mome -momentum ntum tensor 1 . . . . . . 5.3 Energy Energy-mome -momentum ntum tensor 2 . . . . . . 5.4 Weak gra gravit vity y . . . . . . . . . . . . . . 5.5 Equati Equations ons of motion from conser conserva vation tion
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11 11 11 12 12 13 13 14
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14 14 15 15 16 16
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17 17 17 17 20 22
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22 22 23 24 24
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6 The The gra gravit vitati ationa onall field 6.1 Deg Degree reess of fre freedo edom m . . . . . . . . . . . . . . . . 6.2 Spher Spherically ically symmetric symmetric space spacetime time . . . . . . . . 6.2.1 Straig Straightfo htforwar rward d soluti solution on . . . . . . . . . 6.2.2 Soluti Solution on using confor conformal mal transf transformati ormation on 6.3 Equ Equati ations ons of mot motion ion . . . . . . . . . . . . . . . 7 Weak gra gravita vitationa tionall fields 7.1 Gra Gravitatio vitational nal bending of light light 7.2 Ein Einste stein in tenso tensorr for weak weak field field 7.3 Gra Gravitatio vitational nal perturb perturbations ations I 7.4 Gra Gravitatio vitational nal perturb perturbations ations II
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8 Gravita Gravitationa tionall radiation radiation I 8.1 Gauge inv invarian ariantt variabl variables es . . . . . . . . . 8.2 Detec Detecting ting grav gravitation itational al wa waves ves . . . . . . . 8.2.1 Using distan distances ces betw between een partic particles les . 8.2.2 Using geodesi geodesicc deviat deviation ion equati equation on 8.3 Po Poiss isson on equ equatio ation n . . . . . . . . . . . . . . 8.4 Met Metric ric pert perturb urbati ations ons 1 . . . . . . . . . . . 8.5 Met Metric ric pert perturb urbati ations ons 2 . . . . . . . . . . .
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9 Gravita Gravitationa tionall radiat radiation ion II 9.1 Proje Projecti ction on of the matt matter er tenso tensorr . . . . . . . . . . 9.2 Mat Matter ter sou source rcess . . . . . . . . . . . . . . . . . . . 9.3 Energy Energy-mome -momentum ntum tensor tensor of of gravitational gravitational wa waves ves 9.4 Po Powe werr of emit emitted ted radi radiatio ation n. . . . . . . . . . . . .
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29 29 30 30 32
10 Derivation: Derivation: gravita gravitational tional waves waves in flat spacetime
33
A good textbook corresponding to the level of this course: Genera Generall Relativity: An Introduction Introduction for Physicists by M. P. Hobson, G. P. Efstathiou, and A. N. Lasenby (Cambridge University Press, 2006).
1 1.1
Coord Coo rdin inat ates es an and d 1-f 1-for orms ms Invertib Inv ertible le trans transform formation ationss
The inverse function theorem guarantees that the equations ξ α = ξ α (x) are solvable near a point x0 if det ∂ξ α (x)/∂x β = 0 at x0 . Under this condition, condition, the coordina coordinate te transformation transformation is invertible invertible at x0 . Not Note: e: we are are invert inverting ing not not just one function ξ = ξ (x), but we are determining x from a system of n equations, say ξ α (x) = C α , where C α are n given values.
1.2
Examp Exa mples les of coordi coordina nate te transf transform ormati ations ons
1. a) Since x = u 1 + v 2 + u3 /3, it is clear that x has range ( , + ) for any fixed value of v as u varies in the range ( , + ). Simila Similarly rly,, y has the range ( , + ). To verify that the coordinate system system (x, (x, y) covers the entire plane, it is sufficient to show that x has the full range at every fixed value of y . It is sufficient sufficient to consider y0 > 0 (else change v v). At fixed fixed y = y0 > 0, we have y0 = v + vu 2 + v3 /3 and thus the admissible values of u are from to + , 1 3 while the admissible values of v are from 0 to v = vmax such that y0 = vmax + 3 vmax . Then
−∞ ∞ →−
−∞ ∞
−∞ ∞
−∞
u= x= =
− ± ± ± y0
− 13 v3
(we have
y0
1 3 3v
v v 2
1+v +
−v− 3v
2 8 2 y0 + v + 3 9 3v
y0
− v − 13 v3 ≥ 0for 0 < v < v v
− y0
− y0
∞
max ),
− 13 v3
v v
− 13 v3 .
v v
We have now expressed x as a function of v, i.e. x = x(v). Wh When en v varies from 0 to vmax, x(v) varies from to 0. Since x(v) is nonsingular for v > 0, it follows that x has the full range. Therefore, the coordinates (x, ( x, y) cover the entire two-dimensional plane. b) The coordinate transformation is nonsingular if
±∞
det
∂ (x, y) ∂ ( = 0. 0. ∂ ((u, v) ∂
Compute: det
∂x ∂u ∂x ∂v
∂y ∂u ∂y ∂v
= det
1 + u2 + v 2 2uv
2uv 1 + u2 + v 2
= 1 + 2 u2 + v2 + u2
(1)
− v 2 2 > 0.
Therefore there are no singular points. 2. a) To determine the range, first consider φ = 0. Then x = r sinh θ, y = 0, z = r cosh θ. It is clear that z 2 x2 = r2 . Since r 0, the coordinates (x,y,z (x,y,z)) cover only the domain z > x . With arbitrar arbitrary y φ, it is clear that the coordinates 2 2 (x,y,z x,y,z)) cover the domain z > x + y .
≥
||
|| ||
2
−
b) Compute the determinant: det
sinh θ cos φ r cosh θ cos φ sinh θ sin φ r cosh θ sin φ cosh θ r sinh θ
−r sinh θ sin φ r sinh θ cos φ 0
= r2 sinh θ.
The coordinates are singular if r = 0 or θ = 0. c) The singularity at r = 0 is due to the fact that the set r = 0, θ , φ corresponds to a single point x = y = z = 0. This is similar to the singularity of the spherical coordinates at r = 0. Po Poin ints ts along along the cone z = x2 + y 2 are not covered because they correspond to θ , r 0. The singul singulari arity ty at θ = 0, r = 0 is due to the fact that the set r, θ = 0, φ corresponds to the point x = y = 0, z = r at fixed r = 0. This is similar to the polar coordinate singularity. 3. a) To determine determine the range, note that r sin θ 0 for the given range of θ of θ and r. However, this is immaterial since the factors cos φ and sin φ will make x, y cover the full range ( , + ). The coordinates coordinates (x,y,z (x,y,z)) are a slight modification of the standard spherical coordinates. These coordinates cover the whole space ( x,y,z ). x,y,z). b) Compute the determinant:
{
{
→∞ → ≥
}
det
− −
sin θ cos φ sin θ sin φ cos θ
}
||
−∞ ∞
−r cos θ cos φ r sin θ sin φ −r cos φ sin φ −r sin θ cos φ 0 −r sin θ
= r2 sin θ.
This is nonzero unless r = 0 or θ = 0. c) The singularities are completely analogous to those in the spherical coordinates.
1.3
Basi Ba siss in ta tang ngen entt spac space e
Suppose that the vectors eα = ∂x∂ α are linearly dependent, then there exist constants cα , not all zero, such that the vector cα eα equals zero. Act with this vector on the coordinate function x1 : cα eαx1 = cα
∂ 1 x = c1 . ∂x α
By assumption, cαeα = 0, therefore c1 = 0. It follows that every cα equals zero, contradicting the assumption.
1.4
Differe Diff eren ntia tials ls of funct function ionss as 1-for 1-forms ms d(x) = dx, d (xy xy)) = xdy + y dx, d 4x2 y + x3 z = 8xy + 3x 3x2 z dx + 4x 4x2 dy + x3 dz ,
Now let us compute dh by first finding d (ar (arcta ctan n (x
d arctan
x2
2x y2
± dy , ± y)) = 1 +dx(x ( x ± y )2
− −1
1
=
1+
−2 =
4x2 (x2 −y 2 −1)2 x2 + y 2 + 1
(x2
d x2 = 2xdx, d (x + y ) = dx + dy, xdx + ydy d 3 x2 + y 2 = 3 . x2 + y 2
2dx y2
ydy)) − ydy − x2 − − 1 (x2 − y 2 − 1)2 dx + 4xydy 4xydy
− y2 − 1)2 + 4x 4x2
4x (xdx
.
Adding these together and noting that
we get
2
1 + (x (x + y)
1 + (x (x
2
− y)
= x2
dh((x, y) = d arctan( dh arctan(x x + y ) + arctan(x arctan(x
4x2 , − y2 − 1 2 + 4x
2x y2
− y) + arctan x2 − − 1
= 0.
This means that h(x, y) is a constant. By using the tangent sum rule, we can easily show that h(x, y) = 0.
1.5
Basi Ba siss in cot cotan ange gen nt spac space e
Note that the relation
∂ dx , β ∂x α
3
= δαβ
is the definition of how the 1-form dxα acts on vectors ∂/∂xβ . Now, it is clear that any 1-form is decomposed as a linear combination of the 1-forms dx1 ,...,dxn . It rem remain ainss to sho show w that all the these se forms are linearl linearly y independe independent nt.. If this were α not so, there would exist a linear combination cαdx = 0 such that not all cα = 0. Act with this on a vector ∂/∂x1 and obtain ∂ ∂ 0 = 0, 1 = cα dxα , 1 = c1 . ∂x ∂x
Therefore c1 = 0. Similarly, we find that every other cα = 0, which contradicts the assumption.
1.6
Linear Lin early ly inde independ penden entt 1-for 1-forms ms
1. Two 1-form 1-formss d(ex cos y ), d(ex sin y ) are linearly independent for every x, y because d (ex cos y ) = ex cos ydx ex sin ydy, d (ex sin y ) = ex sin ydx + ex cos ydy,
−
and the following determinant is always nonzero, det 2. Two 1-forms 1-forms (1 + y )dx
ex cos y ex sin y
x
sin y x e cos y
−e
= e2x = 0.
xydy,, 8dx are linearly independent independent if the follo following wing determinant determinant is nonze nonzero, ro, − 2xydy 1 + y −2xy det = 16 16xy. xy.
8
0
This happens for xy = 0.
3. Three 1-form 1-formss dx + dy dy,, dx + dz dz,, dy + dz are always linearly independent. 4. Three 1-form 1-formss dx
1.7
dy,, dy − dz dz,, dz − dx are always linearly dependent (their sum is zero). − dy
Tran ransfo sforma rmatio tion n law law for 1-for 1-forms ms
The transf transformati ormation on law for 1-form 1-forms, s,
∂ x ∂ ˜ ˜α β dx ˜ = dx , ∂x β α
under a coordina coordinate te transf transformati ormation on xα x ˜α = x ˜α (xβ ), is merely a different interpretation of the definition of the 1-form dx ˜α (see Problem 1.4 1.4), ), where x ˜α (x) is considered a scalar function in the coordinates xβ .
→
1.8
Examp Exa mples les of tran transfo sform rmati ations ons
a) First compute dx and dy dy:: dx = 1 + u2 + v2 du + (2uv (2uv)) dv,
dy = (2 (2uv uv)) du + 1 + u2 + v2 dv.
Then it is easy to compute xdx + ydy, ydy,etc. etc. For instance, d
1 = x+y
− (dxx ++ydy)2 = −
(du + dv dv)) 1 + (u (u + v) (u + v)
2
b) The component transformation matrix is given in Eq. (1 (1).
1+
1 3
2
(u + v )
2
2.
1A Tangent plane If the tangent plane is at angle α with the horizontal, then the acceleration is g sin α (from elementary mechanics). Since 0 α < π2 , we need to maximize α or, equivalently, tan α, which equals
≤
∂z rh 1 2 −2 = 2 e− 2 r σ , ∂r σ
r
≡
x2 + y 2 .
The maximum of ∂z/∂r is at r0 = σ . For example, a point with maximum maximum acceleration acceleration is x0 = σ, y0 = 0, z0 = The tangent plane at a point (x ( x0 , y0 , z0 ) is given by the equation nx (x
− x0) + n
y
(y
− y0) + n 4
z
(z
− z0) = 0,
−he−
1 2
.
2
Tenso sor rs
2.1
Defin De finit itio ion n of tens tensor or prod produc uctt
a),b) A direct calculation using the property dxi ,∂/∂xk = δki gives:
ω1, v1 =
∂ dx + 2ydy, 2ydy, 3 ∂x
= 3, 3,
etc.
The results: T ((v1 , v1 ) = 0, T
T ((v1 , v2 ) = 6x. T
c) d) First, show that the sum of two linear functions functions is again a linear function: function: If If A A(x) and B (x) are linear functions, i.e. if A(x + λy) = A(x) + λA λA((y) and likewise for B , then A + B obviously has the same property. Now, since a tensor is defined as a multi-linear function, it is clear that tensors form a vector space.
2.3
Exam Ex ampl ple e of te tens nsor or
a) An obvious example of such T is the vector product, T T ((u, v) = u v, defined in three-dimensional space. To determine the rank of T T ,, we need to represent T as a multilinear number-valued function of some number of 1-forms and vectors, e.g. A(f 1 ,..., f r , v1 ,..., vs ). It is clear clear that that T T ((v1 , v2 ) itself is not such a function because it has vector values instead of scalar (number) values. So we need to add a 1-form to the list of arguments. We can define
×
A(f 1 , v1 , v2 ) = f 1 , T T ((v1 , v2 )
and then it’s clear that A is multilinear. Therefore T is a tensor of rank 1+2. α b) The calculation may go as follows. We need to determine the components T βγ such that α β γ [T T ((u, v)]α = T βγ u v .
So we rewrite the given definition of T T ((u, v) in the index notation, e.g. like this: α
[T T ((u, v)] = 2εα βγ uβ vγ
−u
α
nβ vβ .
α Now we would like to move uβ vγ out of the brackets and so determine T βγ . How Howeve ever, r, the expre expression ssion above above contains α β β γ α u v instea instead d of of u u v . Therefore we rename the index β to γ and also introduce a Kronecker symbol δβ , so as to rewrite identically uαnβ vβ = uβ vγ nγ δβα .
Therefore [T T ((u, v)]α = 2ε 2 εα βγ uβ vγ α T βγ = 2ε 2 εα βγ
2.5
−u
β γ
v nγ δβα = 2εα βγ
α γ β.
−n δ
α γ β
−n δ
uβ v γ ,
Contra Con tracti ction on of of tens tensor or indic indices es
a) Using the definition definition of a tensor as a mul multiline tilinear ar function, it is easy to show that linear combination combinationss of tensor tensorss are also multilinear functions. Tensor products and contractions are also multilinear. The arguments are much simpler than the proof of tensor transformation law for components. b) Contracting two lower indices, e.g. T ααβ ααβ , gives components of a quantity which is not a tensor because these components do not transform correctly under changes of basis. If T ααβ ααβ were a tensor it would transform as ∂x λ ˜ T ααγ T µµλ ααγ = µµλ . ∂ ˜ ∂ x ˜γ However, this does not agree with the contraction of the tensor T αβγ αβγ , which transforms as ∂x λ ∂x µ ∂x ν ˜αβγ T = T λµν αβγ λµν . ∂ ˜ ∂ x ˜α ∂ ∂ ˜ x ˜β ∂ ∂ ˜ x ˜γ The contraction over α = β yields
α
˜ααγ T ααγ =
α
∂x λ ∂x µ ∂x ν ∂x λ T = T µµλ λ λµν µν µµλ. ∂ ˜ ∂ x ˜α ∂ ∂ ˜ x ˜α ∂ ∂ ˜ x ˜γ ∂ x ∂ ˜ ˜γ
c) Calculation gives α T αγ = 2εααγ
because εα αγ = 0 and δαα = 3.
α γ α
−n δ 7
=
− 3n
γ
2.8
Exam Ex ampl ples es of spa space cess with with a metri metric c
a) We perform the calculation in components, g (u, v) = uα vα
α β γ λ µ αβγ ε λµ u n v n .
−ε
We would like to write g (u, v) = gαβ uα vβ , where gαβ are the components of the metric tensor. Using the known identity for the ε-symbol, εαβγ εα λµ = δβλ δγµ δβµ δγλ ,
−
we find g (u, v) = uα vα
βλ γµ
= uα v α Denote n2 nµ nµ The result is
≡
β
γ λ µ
− (δ δ − δ δ ) u n v −u v n n +u n v n . λ
λ
βµ γλ
µ
µ
λ
λ
µ
n
µ
α β
≡ g(n, n), and then we need to relabel indices such that v u g (u, v) = δ − n2 δ + n n v u .
Therefore
αβ
αβ
gαβ = δαβ
− n2 δ
α β
αβ
can be moved outside the parentheses.
α β
+ nα nβ .
To analyze the conditions under which det gαβ = 0, we can choose an orthonormal basis such that nα is parallel to the first basis vector. Then the components of the vector n in this basis are ( n , 0, 0) and the matrix gαβ has the following simple form: 1 0 0 2 0 1 n 0 gαβ = . 2 0 0 1 n
||
2
−
−
n2 . Therefore, the matrix gαβ is nondegenerate if n2 = 1.
−
Then it is clear that det gαβ = 1 b) Similar calculations give
gαβ =
3 + n2 0
0 3
.
Therefore the determinant of gαβ is always nonzero. c) Considerations are analogous to b), except that the size of the matrix gαβ is larger. d) The metric is (dx (dx))2 + (dy ( dy))2 + (dz (dz))2 , and we need to express dx dx,, dy dy,, dz through du and dv dv.. A calculation gives g = dx2 + dy2 + dz 2 = R2 sinh2 u + cosh2 u du2 + R2 cosh2 u dv2 .
2A Vector equations
a) The equation contains two given vectors Aα and Bα . The solu solutio tion n xα can be found as a linear combination of Aα , β γ Bα, and the cross product εαβγ A B with unknown coefficients. Using vector notation, we have γ ((A x = αA + β B + γ
× B) .
Substituting Substi tuting this expre expression ssion into the give given n equati equation, on, kx + x
× A = B,
and using the known identity (A
× B) × C = B (A · C) − A (B · C) ,
(2)
we find A (kα
− A · B) + B (kβ kβ + + A · A − 1) + (A × B) (kγ − β ) = 0.
On purpose, we write this equation in the form of linear combination of the three vectors A, B, and A B. Sin Since ce we we are considering the generic case, these three vectors are independent and so each of the coefficients above must be zero:
×
kα
− A · B = 0,
kβ + kβ +A A
· − 1 = 0,
kγ
− β = 0.
Solving this system of equations, we find (assuming k = 0 in the generic case)
α=
A B , k
·
β =
1
− A · A, k
γ =
b) We have in vector notation x
Multiply
×C:
× A = B,
(x
x C = k.
·
× A) × C = B × C. 8
1
− A · A. k2
Simplify using the identity (2 (2), A (x C)
· − x (A · C) = kA − x (A · C) = B × C.
Therefore x=
kA
− B × C. A·C
c), d) The equations have the form xαM βα = Aβ , where M βα is a matrix and Aβ is a known vector. vector. The solution solution is −1 −1 x = M A, where M is the inverse matrix (it exists in the generic case).
2B Tensor equations a) Since the vectors Aα and Bα are a basis in two-dimensional space (in the generic case they are linearly independent), then the symmetric tensor X αβ can be written as X αβ = f Aα Aβ + g Aα B β + Aβ Aα + hB α B β ,
where the coefficients f , g , h are unknown. unknown. It remains to determine determine these these coefficients. coefficients. Using AαB α = 0 and denoting Aα Aα A 2 , we get the system of equations
≡| |
X αβ Aα = B β X αβ Bα = 0 X αβ δαβ = 0 The result is X αβ =
|B |2 A |A|2 |B|2 − |A|2 1
α
(f + g ) A 2 = 1,
⇒ ⇒ ⇒
Aβ +
| | (g + h) |B |2 = 0, f |A|2 + h |B |2 = 0. 1
|B|2 − |A|2
B αB β
α
−A
Bβ
−B
α
Aβ .
b) An antis antisymmet ymmetric ric tensor X αβ in three dimensions can be always expressed as
X αβ = εαβγ uγ , where uγ is an unknown vector that we need to determine. We can now rewrite the conditions on X αβ in a vector form, X αβ Aα = B β X αβ Bα = 0 It follows that u is parallel to B and then the condition u is the only admissible solution.
×A=B u×B = 0 u
⇒ ⇒
αβ
× A = B leaves the only solution u = 0, and therefore X
=0
2C Degeneracy of the metric a) The metric can be written in the basis dx dx,, dy as the matrix
{
}
g=
y2 1 + x2
1 + x2 0
.
2
The determinant of this matrix is 1 + x2 which is always nonzero. b) In the basis where Aα is parallel to the first basis vector, the vector Aα has components (A, (A, 0, 0, 0,... ,...)) and therefore the metric gαβ has the form r2 0 0 ... 0 1 0 . 0 0 1 .. .. . .
−
−
The metric is degenerate if r = 0 (i.e. at the origin).
3 3.1
The Th e Ch Chri rist stoff offel el sy sym mbol Tran ransf sform ormati ations ons 1
We are considering a flat space where Euclidean coordinates exist. Suppose xα and x ˜α are tw twoo coordina coordinate te system systems, s, while ξ α is the standard Euclidean coordinate system. The Christoffel symbols are defined as
{ }
{ }
Γµαβ ˜µ Γ αβ
∂ 2 ξ ν = ∂x α ∂x β ∂ 2 ξ ν = ∂ ˜ ∂ x ˜α ∂ ∂ ˜ x ˜β 9
∂x µ , ∂ξ ν ∂ ˜ ∂ x ˜µ . ∂ξ ν
{ }
˜ and Γ can be found as follows. Assuming that the functions xα (˜ The relationship between Γ x) and also x ˜α (x) are known, we may express the partial derivative operators using the chain rule, ∂ ∂ ˜ ∂ x ˜α ∂ = , ∂x µ ∂x µ ∂ ∂ ˜ x ˜α ∂ ∂x α ∂ = . ∂ ˜ ∂ x ˜µ ∂ ˜ ∂ x ˜µ ∂x α Also we can express ∂ x ∂ ˜ ˜µ ∂ ˜ ∂ x ˜µ ∂x λ = . ∂ξ ν ∂x λ ∂ξ ν ˜ µ (when Γµ is known) as follows, Therefore we can calculate Γ αβ αβ ˜µ Γ αβ
˜µ ∂x λ ∂ x ∂ ˜ = ∂x λ ∂ξ ν
∂x γ ∂ ∂ ˜ ∂ x ˜α ∂x γ
∂x δ ∂ ∂ ˜ ∂ x ˜β ∂x δ
ξν
∂ x ∂ ˜ ˜µ ∂x λ ∂x γ ∂x δ ∂ 2 ξ ν ∂ 2 xδ ∂ξ ν = + ∂x λ ∂ξ ν ∂ ∂ ˜ x ˜α ∂ ˜ ∂ x ˜β ∂x γ ∂x δ ∂ x ∂ ˜ ˜β ∂ ∂ ˜ x ˜γ ∂x δ ∂ x ∂ ˜ ˜µ ∂x γ ∂x δ λ ∂ 2 xδ Γ + ∂x λ ∂ ∂ ˜ x ˜α ∂ ∂ ˜ x ˜β γδ ∂ ˜ ∂ x ˜β ∂ ∂ ˜ x ˜γ ∂ ˜ ∂ x ˜µ ∂x γ ∂x δ λ ∂ 2 xλ = Γ + ∂x λ ∂ ∂ ˜ x ˜α ∂ ∂ ˜ x ˜β γδ ∂ ∂ ˜ x ˜β ∂ ∂ ˜ x ˜γ =
∂ x ∂ ˜ ˜µ ∂x λ ∂ξ ν ∂x λ ∂ξ ν ∂x δ ∂ ˜ ∂ x ˜µ . ∂x λ
(3)
Note that the Euclidean coordinate system ξ ν is not needed to determine the transformation of Γ.
3.2
Tran ransf sform ormati ations ons 2
Consider a vector field uµ . Assume that
∇u ν
µ
obeys the correct transformation law for rank (1,1) tensors, µ
∇u ν
=
and substitute
µ ∂ ˜ x ˜α ˜ αu˜β ∂x ∂ , ∂ ˜ ∂ x ˜β ∂x ν
∇
∇ u ≡ ∂x∂ u ∇˜ u˜ ≡ ∂ ∂ ˜∂ x˜ u˜ µ
ν
µ
+ Γµαν uα,
µ
α ˜µ u +Γ αν ˜ .
ν
µ
ν
ν
˜ Note that We can now express Γ through Γ. ∂ x ∂ ˜ ˜α ∂x β = δγα β γ ∂x ∂ ∂ ˜ x ˜ because the matrices are ∂ ∂ ˜ x ˜α /∂x β and ∂x β /∂ /∂ ˜ x ˜γ are inverse to each other. The result is Γλαβ
3.3
˜ν ∂ ˜ρ ∂x λ ˜ν ∂ ˜ρ ∂ ˜ x ∂ ˜ x ∂ 2 xλ ∂ ∂ ˜ x ∂ ˜ x µ ∂ ˜ = Γνρ α β µ + ν ρ α β . ∂x ∂x ∂ ∂ ˜ x ˜ ∂ x ∂ ˜ ˜ ∂ ∂ ˜ x ˜ ∂x ∂x
Covaria Cov ariant nt deriv derivativ atives es
The rule is that every upper index gets a +Γ and every lower index gets a Γ. Each term with Γ replaces one index in the original tensor by one of the indices in Γ. Therefore we can write the answer as
−
β λβ αλ T αβ γδµ ;ν = ∂ ν T αβ γδµ + Γα λν T γδµ + Γ λν T γδµ λ αβ λδµ γν T
−Γ 3.4
λ αβ δν T γλµ
−Γ
λ αβ γδλ µν T
−Γ
The Th e Le Leib ibni nitz tz ru rule le
Perform an explicit calculation, β
Aα;γ B +
Aα B β;γ
Aα B β
This proves the required property.
;γ
=B
− −
β
∂ γ Aα
Γλαγ Aλ
β
+ Aα ∂ γ B +
Γβλγ B λ
Γλαγ Aα B β + Γβλγ Aα B λ .
= ∂ γ Aα B β
10
;
3.5
Locally Loca lly ine inertia rtiall refer referenc ence e fram frame e
In this problem (unlike problem 3.1 3.1)) the coordina coordinate te syste system m ξ α is not a flat Euclidean coordinate system, but it is just a coordinate system which which is like Euclidean Euclidean at one point. Now we want want to use the formula (3 (3), which will enable us to ˜ compute the Christoffel symbol Γ in the coordinate system ξ , given the Christoffel symbol Γ in the original coordinate system xα . To use that formula, we need to compute some derivatives. Denoting x ˜α ξ α , we find
{ }
{ }
∂ x ∂ ˜ ˜α = δβα + xµ ∂x β
{ }≡{ }
µ
− x(0)
α µ δβν Γα (0)µν = δβ + x
µ
− x(0)
Γα (0)µβ .
,
The inverse derivative, ∂x α/∂ /∂ ˜ x ˜β , can be found by inverting this matrix; the result can be found simply by assuming that ∂x α = δβα + xµ β ∂ ˜ ∂ x ˜
−
xµ(0) Aα µβ + O (x
where Aα µβ is an unknown matrix. So up to quadratic terms we find ∂x α = δβα β ∂ ˜ ∂ x ˜ Therefore
−
µ
x
µ
− x(0)
∂ 2 xλ = ∂ ˜ ∂ x ˜β ∂ ∂ ˜ x ˜γ
− x(0))2
Γα (0)µβ + O (x
λ
−Γ(0)
βγ
− x(0))2
.
−
+O x
Finally, we find
x(0) .
∂ x ˜µ ∂x γ ∂x δ λ ∂ 2 xλ ∂ ∂ ˜ x ˜µ ˜ µ = ∂ ˜ Γ Γ + αβ ∂x λ ∂ ∂ ˜ x ˜α ∂ ∂ ˜ x ˜β γδ ∂ ∂ ˜ x ˜β ∂ ∂ ˜ x ˜γ ∂x λ µ µ = Γαβ Γ(0)αβ + O x x(0) .
−
−
At x = x0 we have Γµαβ = Γ µ(0)αβ . It follows that the new Christoffel symbol is equal to zero at x = x0 .
˜ ... + ... Alternatively, one can use the transformation law for the Christoffel symbol in the inverse direction, Γ = Γ ...,, α α α α 2 i.e. one denotes x ξ , x˜ x . This has the advantage that only derivatives ∂ ξ/∂x∂x need to be computed, 2 and not the derivatives ∂ x/∂ξ∂ξ x/∂ξ∂ξ.. Since all derivativ derivatives es only need to be ev evaluate aluated d at x = x0 , the first-order derivatives α β α α ∂x/∂ξ at x = x0 can be found as the inverse matrix to ∂ξ /∂x = δβ , i.e. ∂x /∂ξ β = δβα . This considerably considerably simplifies simplifies the calculations.
{ }≡{ } { }≡{ }
4
Geode Ge odesi sics cs an and d cu curv rvat atur ure e
4.1 4. 1
Geod Ge odes esic icss
(a) Note that d/ds is the ordinary (not “covariant”) derivative in the direction of uα . The geodesi geodesicc equation equation can can be rewritten for the 1-form uα as duα uγ uα;γ = 0 = Γβαγ uβ uγ . ds An explicit formula for Γβαγ yields
−
Γβαγ uβ uγ = Γβ αγ uβ uγ = Note that gβα,γ gαγ,β is antisymmetric in (β (β The remai remaining ning term yield yieldss
−
↔ γ ).).
1 (gβα,γ + gβγ,α 2
−g
αγ,β ) u
β γ
u .
Therefore There fore these terms will cancel after a contraction contraction with uβ uγ .
1 gαβ,γ uβ uγ . 2 b) We give two derivations; the first one is direct and the second one uses the property (a). Γβαγ uβ uγ =
4.1.1 4.1 .1
First Fir st der deriv ivati ation on
Note that
d gαβ uαuβ = uγ gαβ uα uβ ,γ ds where we must use an ordinary derivative instead of the covariant derivative (according to the definition of d/ds d/ds). ). So we find 2gαβ uα uγ uβ,γ . uγ gαβ uα uβ ,γ = gαβ,γ uα uβ uγ + 2g
Now we need to simplify an expression containing uβ,γ . By assumption, the derivative of the vector field uα satisfies d α β γ γ α α β γ u + Γα βγ u u = u u,γ + Γ βγ u u = 0 ds 1 αλ = uγ uα (gλµ,γ + gλγ,µ ,γ + g 2 11
µγ,λ ) u
−g
γ µ
u .
Therefore
− 12 u 1 =− u 2
uα uγ uα ,γ =
αg λ
αλ
(gλµ,γ + gλγ,µ
(gλµ,γ + gλγ,µ
−g
−g
γ µ µγ,λ) u u
γ µ µγ,λ ) u u
=
− 12 g
λµ,γ u
λ µ γ
u u .
What remains is a straightforward computation: uγ gαβ uα uβ
4.1.2 4.1 .2
,γ
= gαβ,γ uα uβ uγ + 2g 2gαβ uαuγ uβ,γ = gαβ,γ uα uβ uγ = 0.
−g
λµ,γ u
λ µ γ
u u
Second Sec ond der deriv ivat ation ion
We write
d d αβ gαβ uα uβ = g uα uβ = uγ g αβ ,γ uα uβ + 2g 2g αβ uα,γ uβ uγ . ds ds γ Now we need to express uα,γ u duα /ds /ds.. To do that, we use the property derived in (a),
≡
duα 1 = gβγ,αuβ uγ , ds 2 and find
d αβ dg αβ dg αβ dgλγ λ γ αβ λ γ g uα uβ = uα uβ + g gλγ,αu uβ u = uα uβ + u u . ds ds ds ds
It remains to express the derivative of g αβ through the derivative of gαβ . We use the identity d αβ d α g gβγ = δ = 0, ds ds γ
thus
dg αβ gβγ = ds
Therefore
− dgds
βγ
dg αβ uαuβ = ds
g αβ
− dgds
λµ
and thus
4.2
⇒
dg αβ = ds
− dgds
λµ αλ βµ
g αλ g βµ uαuβ =
g
− dgds
λµ
g
.
uλ uµ ,
d αβ g uα uβ = 0. 0. ds
Commutat Comm utator or of cov covarian ariantt deriv derivativ atives es
First compute α α λ uα ;β = u,β + Γβλ u , α α λ uα ;βγ = u,β + Γβλ u
=
uα ,βγ
+
,γ
λ Γα βλ,γ u
µ µ λ + Γα µγ u,β + Γ βλ u
−
α λ Γµβγ uα ,µ + Γ µλ u
µ µ λ α α λ + Γα βλ u,γ + Γ µγ u,β + Γ µγ Γβλ u
−Γ
µ α βγ u;µ .
Since we want to compute the commutator uα uα (β ;βγ ;γβ , we can omit the terms that are symmetric in (β terms are the following: µ λ α uα Γα Γµβγ uα ,βγ , βλ u,γ + Γµγ u,β , ;µ .
−
The remaining terms are
µ α λ α λ uα ;βγ = Γ βλ,γ u + Γµγ Γβλ u + (symmetric in β
↔ γ ) ,
which yields uα ;βγ
α γβ
− u;
= uλ Γα λβ,γ
α λγ,β
−Γ
+ Γµλβ Γα γµ
12
µ α λγ Γβµ
−Γ
α = uλ Rλγβ .
↔ γ ).).
These The se
4.3
Paral Pa rallel lel tra trans nsport port
The parallel-transported vector can be represented by a 1-form Aα (s) such that dAα ds
−
Γβαg Aβ uγ
γ
= 0,
u
dxγ . ds
≡
However, the closed curve is assumed to cover only a very small neighborhood of one point x0 , so we can approximate Aα by a constant, Aα (x0 ), along the curve. Therefore δAα =
dAα ds = ds
Γβαγ (x)Aβ (x)dxγ
β (x0 )
≈A
Γβαγ (x)dxγ .
Now, in a locally inertial system at x0 we have Γβαγ (x0 ) = 0. Therefore we can Taylor expand Γ βαγ (x) near x0 as
− x0
Γβαγ,λ + O (x
λ
λ
Γβαγ (x) = xλ
λ
Therefore δAα
≈A
β (x0 )
Γβαγ,λ
x
− x0
γ
dx
≈
− x0 )2
.
Aβ (x0 )Γβαγ,λ (x0 )
xλ dxγ ,
where we have again approximated Γ βαγ,λ (x) by its value at x = x0 , and also used the identity
γ λ
d x x
Therefore we may rewrite δA α
4.4
≈
Aβ Γβαγ,λ
= 0=
γ
λ
x dx +
xλ dxγ .
1 x dx = Aβ Γβαγ,λ Γβαλ,γ 2 1 β = Aβ Rαλγ xλ dxγ . 2 λ
γ
−
dxγ = 0. Further,
xλ dxγ
Riem Ri eman ann n te tens nsor or
a) It is more convenient to consider the fully covariant tensor Rαβγδ . This tensor has the following symmetries, Rαβγδ = Rβαγδ = Rαβδγ , Rαβγδ + Rβγαδ + Rγαβδ = 0, 0, Rαβγδ = Rγδαβ .
−
−
(4) (5) (6)
However, it is known that the property (6 ( 6) follows from (4)-( )-(55), therefore it is sufficient to consider these two properties [note that (5 (5) does not follow from (4 (4), (6)]. Let us first con consid sider er the property property (4 (4). For fixed γ, δ , we have that Rαβγδ is an antis antisymmet ymmetric ric n n matrix (indices α, β ). ). This matrix has 12 n(n 1) indepen independent dent components. components. Like Likewise wise for fixed α, β . Therefore, the number of independent components of Rαβγδ is reduced to
×
−
1 N 1 = n(n 2
− 1)
2
.
Now we use the property (5 ( 5). Let us see whether the property (5 ( 5) is nontrivial at fixed δ . If If α α = γ , then the property (5 ( 5) becomes Rαβαδ + Rβααδ + Rααβδ = 0 (n (noo sum summa mati tion on)), which is already a consequence of (4 ( 4). Like Likewise wise for for β = γ or for α = β . There Therefore, fore, the property property (5 ( 5) is a new constraint only if all three indices α,β,γ are different (i.e. α = β , α = γ , β = γ ). ). Suppose that α,β,γ are different. There are
N 2 =
1 n(n 6
1)(n n − 2) − 1)(
choices of such α,β,γ . The Theref refore ore,, for each each δ = 1,...,n we obtain N 2 additi additional onal constraints. constraints. Final Finally ly,, let us che check ck that every such constraint is nontrivial for every δ (even if δ is equal to one of α,β,γ ). ). Suppose δ = α, then (5 (5) becomes Rαβγα + Rβγαα + Rγαβα = 0 (n (noo sum summa mati tion on)). This is a nontrivial constraint (equivalent to Rαβαγ = Rαγαβ ). There Therefore, fore, the number number of constraints constraints is nN 2 , and thus the total number of independent components of Rαβγδ is N = N 1
−
n2 n2 1 nN 2 = . 12
−
b) Weinberg, Chapter 6, 8
§
13
c) There is only one independent component of Rαβγδ in two dimensions. dimensions. For instance, we can choose R1212 as the independent parameter. Then we can express the Ricci tensor as Rαβ = g λµRλαµβ . Calculating component by component, we find R11 = g 22R1212 , R = g αβ Rαβ = Note that the matrix
−g12R1212, R22 = g11R1212 , 2g 11 g 22 − 2g 12g 21 R1212 = 2gR 2 gR1212 . R12 =
g 22 g 12
−g1112 g
−
is equal to the inverse matrix to g αβ (which is gαβ ), multiplied by the determinant det g αβ ; since det g αβ = 1/g 1 /g,, we have
Therefore Gαβ = Rαβ
4.5
− 12 g
g 22 g 12
−g1112 g
−
αβ R
= gg αβ .
= gg αβ R1212
− 12 g
αβ 2gR 1212
= 0.
Loren Lor entz tz tra transf nsform ormati ations ons
A Lorentz transformation is represented by a matrix Λ α β such that β Λα λ Λµ gαβ = gλµ .
Consider an infinitesimal Lorentz transformation, Λαλ = δ αλ + εH αλ . The number of parameters in Lorentz transformations is the same as the number of parameters in H αβ . The condi conditio tion n α for H β is (δαλ + εH αλ ) δβµ + εH βµ gαβ = gλµ .
Disregardin Disre gardingg terms of order ε2 , we find
0 = δ βµ H αλ gαβ + δλα H βµ gαβ = H µλ µλ + H λµ λµ . Therefore, H λµ n matrix, which has 12 n(n 1) indepe independen ndentt compone components. nts. For n = 4 we get 6 λµ is an antisymmetric n components. These can be interpreted as three spatial rotations and three Lorentz rotations (boosts).
×
5 5.1 5. 1
−
Gravi Gra vita tati tion on the theor ory y app appli lied ed Reds Re dshi hift ft
In the weak field limit, the Newtonian gravitational potential near a mass M is Φ=
GM , r
while the component g00 of the metric is
2Φ . c2 (We write the units explicitly.) Therefore the redshift factor z (r) at distance r from the center of the Earth is g00 = 1 +
z (r) =
1+
2GM c2 r
≈ 1 + GM . c2 r
To compare the redshift factors at the surface of the Earth, denote by RE the radius of the Earth. We know that the gravitation gravi tational al acceleration acceleration at the surfac surfacee is GM m gE = 2 9.81 2 . RE s
≈
Therefore, There fore, it is conv convenien enientt to expre express ss GM = gE R2E . For a vertical distance L between sender and receiver, we find z (RE ) 1 + gE RE c−2 = E z (RE + L) 1 + gE RE c−2 RR E +L Since in our problem L
≪R
E,
≈
gE RE 1+ c2
− 1
RE RE + L
we may approximate z (RE ) z (RE + L)
≈ 1 + gc2L ≈ 1 + 1.1.1 10L16m . E
14
= 1+
c2
gE RE L . (RE + L)
5.2
Energy Ene rgy-m -mome oment ntum um ten tensor sor 1
In the nonrelativistic nonrelativistic limit, we may disre disregard gard gravitation; gravitation; gαβ = ηαβ . The EMT of an ideal fluid is T αβ = pηαβ + ( p ( p + ρ) uα uβ ,
−
where uα is the 4-ve 4-velocity locity vector vector of the fluid motion. In the nonrel nonrelativi ativistic stic limit, uα velocity and v 1 in the units where c = 1. The conservation law is
| |≪
(1,, v), where v is the 3-vector of ≈ (1
0 = T αβ ,β = p,α + ( p ( p + ρ),β uα uβ + ( p ( p + ρ) uα uβ ,β + ( p ( p + ρ) uα,β uβ .
−
Let us simplify this expression by introducing the time derivative along the fluid flow, d dt
≡u
α
∂ α .
Then we find 0 = p,α + uα p˙ + ρ˙ + ( p ( p + ρ) uβ ,β + ( p ( p + ρ) u˙ α .
−
Contracting with uα and using uα u˙ α = 0, we find
ρ˙ + ( p ( p + ρ) uβ ,β = 0.
(7)
(8)
This is the relativistic continuity equation. Using this equation, we find from Eq. (7 (7) that 0 = p,α + uα p˙ + ( p ( p + ρ) u˙ α .
(9)
Now let us apply the nonrelativistic limit, uα have d ∂ ˙ X X X + X + ( (v ) X. dt ∂t The continuity equation (8) gives dρ + ( p ( p + ρ)div )divv = 0. 0. dt
In our notation, for any quantit quantity y X we
− ≈ (1 (1,, v), to Eqs. (8 (8) and (9 (9).
≡
≡
·∇
This is the ordinary, nonrelativistic continuity equation.1 Finally, Eq. (9 (9) gives ( p + ρ) v˙ = 0. ∇ p + v p˙ + ( p (Note that p,j =
j
−∇ p p.).) This is the Euler equation, dv 1 = dt p + ρ
5.3
−∇ − p
v
dp dt
.
Energy Ene rgy-m -mome oment ntum um ten tensor sor 2
Compute the covariant derivative, αβ
T
;α
;α
;β
= Φ Φ =Φ
;α
;β
;α Φ
−
1 αβ ;λ g Φ Φ;λ 2 ;α
+Φ Φ
;β
;α
= Φ ;α ;α Φ;β .
−g
;α αβ
Φ;λ Φ;λα
Here we used Φ;αβ = Φ;βα which follows from Γλαβ = Γλβα (note that Φ is a scalar; covariant derivatives do not commute when applied to vectors!) and also the property Φ;α ;β X α = Φ;αβ X α , which is due to gαβ ;µ = 0. Therefore, we get
Φ;α ;α Φ;β = 0;
;β = 0 then also Φ;α this entails Φ;α ;α = 0 (since Φ;β = 0 is a weaker condition than Φ ;;α ;α = 0, so it is α = 0, i.e. if Φ sufficient to write the latter). 1
Note that in the usual, nonrelativist nonrelativistic ic continuit continuity y equa equations tions as they are writt written en in most books, there is no p + ρ - just ρ. Th This is is is so because in most cases the matter is nonrelativistic, so p ≪ ρ and p + ρ ≈ ρ. This is, however, not true for relativistic matter, such as photons (electromagnetic radiation) for which p = 13 ρ.
15
5.4
Wea eak k gr gra avi vitty
A very short solution is to write R00 directly through Γ and note that only Γα 00,α comes in (if we disregard terms of α second order). order). Then compute compute Γ 00 explic explicitly itly through Φ. (Assu (Assume me that gαβ,0 = 0.) We may disregard terms terms of order ΓΓ because Γ is of order Φ, and also we may raise and lower indices using ηµν instead of gµν . (This is somewhat somewhat heuristic; heuristic; see below.) The calculation goes like this: R00 = Γα Γα ), 00,α 0α,0 + O( Γ Γ ), 1 αβ Γα (g0β, 0 + gβ0,0 g00,β ) = 00 = η 2
−
−
−η
αβ
Φ,β ,
therefore theref ore (using Φ,0 = 0) R00 =
−η
αβ
Φ,αβ =
∆Φ.. −Φ = −Φ 00 + Φ 11 + Φ 22 + Φ 33 = ∆Φ ,
,
,
,
Here is another, somewhat more comprehensive solution. In the weak field limit, we write gµν = ηµν + hµν . Then we only compute everything up to first order in hµν . Therefore, we may raise and lower indices using the Minkowski metric ηαβ inste instead ad of gαβ . Note: the Newtonian limit does not determine the components of gµν excep exceptt for g00 = 1 + 2Φ. The actual actual metric metric gµν is equal to ηµν plus a small first-order deviation, hµν , but this deviat deviation ion cannot be expre expressed ssed just through the Newt Newtonian onian 1 potential potent ial Φ 2 h00 . In principle, one needs to solve the full Einstein equations to find hµν ; in other words, one needs to determine determ ine other, “post-Newton “post-Newtonian ian potentials” potentials” and not just the Newt Newtonian onian potential Φ. How However ever,, when one only wants to compute effects of gravitation on motion of slow bodies, only g00 is necessary necessary.. So it is sufficient sufficient to compu compute te just the Newtonian potential Φ. But e.g. trajectories of light rays cannot be computed accurately in the Newtonian limit (because light does not move slowly). To compute trajectories of light rays, one needs all components of hµν , not just h00 . Let us do the computation through hµν in a more general way. First we compute the Christoffel symbol and the Ricci tensor: 1 Γλαν = η λγ (hαγ,ν + hνγ,α hαν,γ ) ; Γ O(Φ); 2 λ Rαβ = Rαλβ = Γλαβ,λ Γλαλ,β + ΓΓ ΓΓ Γλαβ,λ Γλαλ,β .
≡
− − ≈
−
⇒ ∼ −
(We may disregard the ΓΓ terms since they are second order in Φ.) Now we compute (again up to first order in Φ) 1 λγ η (hαγ,λβ + hλγ,αβ 2 1 = η λγ (hαγ,βλ + hβγ,αλ 2
Γλαλ,β = Rαβ
−h −h
αλ,γβ )
αβ,γλ
1 λγ η hλγ,αβ , 2
=
−h
λγ,αβ ) .
Let us now compute just the component R00 , recalling that hµν is time-independent (so hµν,0 = 0) and h00 = 2Φ: 1 λγ η (h0γ, 0λ + h0γ, 0λ 2 = η λγ Φ,λγ = ∆Φ ∆Φ..
R00 =
− h00 − h ,γλ
λγ, 00 )
=
− 12 η
λγ
h00,γλ
−
5.5
Equati Equ ations ons of motion motion from from conse conserv rvati ation on law law
We would like to rewrite the covariant conservation law T µν ;ν = 0 through ordinary derivatives. The given relations are useful; let’s derive them first. ∂ ∂x ν
√−g = − √1 2 −g 1√ = −gg Γµµν
∂ g ∂x ν
=
αβ gαβ,ν ; 2 1 = g µα (gµα,ν + gνα,µ 2
− 2√1−g −g
νµ,α )
gg αβ gαβ,ν
√1−g ∂x∂ √−g.
1 = g µα gµα,ν = 2
ν
Now rewrite the covariant derivative of T µν explicitly: µν T µν ;µ = T ,µ + Γµαµ T αν + ΓναµT µα =
Apply this to the given T µν : µν
0 = T
;µ
m0 = g
√− +
√m−0g
ds
√1−g √−gT
dxµ dxν (4) σ δ (x ds ds
α
ds
µν
β
σ
−x
+ Γναβ T αβ .
(s))
dx dx ν Γ (x)δ (4) (xσ ds ds αβ 16
,ν
,µ σ
−x
(s)) ))..
Since in the first term the dependence on xσ is only through δ(4) , we can rewrite dxµ ∂ δ (4) (xα ds ∂x µ
d (4) α δ (x ds
−
(s)) =
−
α
−x
(s)) =
α
−x
(s))
(this is easily understood if read from right to left) and then integrate by parts,
Finally, 0=
dxν d (4) σ ds δ (x ds ds
√ −g m0
µν
T
;µ
=
σ
−x
d2 xν (4) σ ds 2 δ (x ds
α β d2 xν ν dx dx ds + Γαβ δ(4) (xσ 2 ds ds ds
σ
−x
(s)) ))..
σ
−x
(s)) ))..
This is a function of xσ which should equal zero everywhere. Therefore, the integrand should vanish for every value of s, α β d2 xν ν dx dx + Γαβ = 0. ds2 ds ds
Remark: in general, equations of motion do not follow from conservation law, but they do follow if there is only one field. (e.g. one fluid, or one scalar field, or some number of point particles). The situation in ordinary mechanics is similar: e.g. the equation of motion for a particle follow from the conservation of energy only if the motion is in one dimension: mv 2 E = + V V ((x) = const, const, 2 dE 0= = (m ( mv˙ + V ′ (x)) v mv˙ = dt
⇒
−V ′(x).
However, equations of motion do not follow from conservation of energy if there is more than one degree of freedom. Similarly, equations of motion for say two scalar fields Φ, Φ , Ψ do not follow from the conservation of their combined T µν µν . These fields have two different equations of motion, and one cannot hope to derive them from a single conservation law.
6
The gr gra avi vita tati tion onal al fie field ld
6.1
Degr De gree eess of fr free eedo dom m
The electromagnetic field is described by a 4-vector potential Aµ (x). This would would give 4 degrees degrees of freedom. freedom. How However ever,, there is also a gauge symmetry, Aµ Aµ + φ,µ ,
→
where φ(x) is an arbitrary function of spacetime. Using this gauge symmetry, we may e.g. set the component A0 (x) = 0. Then only three functions of spacetime (A ( A1 , A2 , A3 ) are left. Hence the electromagn electromagnetic etic field has 3 degrees of freedom. freedom. There are additional gauge symmetries involving functions φ(x) that do not depend on time. Since these functions φ(x) are functions only of three arguments, they do not change the number of degrees of freedom.
6.2 6.2.1
Spheri Sph erical cally ly symm symmetr etric ic spa spacet cetime ime Straigh Stra ightfor tforwa ward rd solut solution ion
A direct computation computation listing all the possibl possiblee Chris Christoffel toffel symbols symbols and components components of the Ricci tensor is certa certainly inly straightstraightforward but very long. Here is a way to compute the curvature tensor without writing individual components. Since the metric has a diagonal form, let us denote gαβ = ηαβ Aα ,
g αβ = η αβ
where Aα
≡
1 Aα
(no summation!), summation!),
eN , eL , r2 , r2 sin2 θ
(10)
(11)
is a fixed array of four functions. For this calculation, calculation, we do not use the Einstein summation summation convention convention any more; every summation summat ion will be writte written n explicitly. explicitly. How Howeve ever, r, we make heavy use of the fact that ηαβ = 0 only if α = β , and that λλ ηλλ = η . At the end of the cal calcul culati ation on of the Ricci Ricci tensor Rαβ , we shall substitute the known functions Aα and use the resulting simplifications. We begin with the calculation of the Christoffel symbols,
Γλαβ =
µ
=
1 λµ 1 η [ηαµ Aµ,β + ηβµ Aµ,α 2 Aλ
1 λ Aλ,β 1 Aλ,α δα + δβλ 2 2 Aλ Aλ 17
− 12 η
λλ ηαβ
−η
αβ Aα,µ ]
Aα,λ . Aλ
(12)
Well, we are not going to finis finish h this calc calculati ulation on here, anyway anyway.. But this is roug roughly hly how it goes. Let us at least derive the useful formula below. The relationship relationship between the curv curvature ature tensors under a confor conformal mal transformation transformation is found as follo follows. ws. First we define the conformally transformed metric for convenience as follows,
g˜αβ = e2Ω gαβ . λ Then the Christoffel symbols receive a correction which is a tensor Bαβ ,
˜ λ = Γλ + B λ ; Γ αβ αβ αβ
λ Bαβ
≡δ
λ α Ω,β
+ δβλ Ω,α
−g
,λ αβ Ω .
The Riemann and the Ricci tensors are defined (in Landau-Lifshitz sign convention) by Rλαµβ = Γλαβ,µ
λ αµ,β
+ Γλµν Γναβ
λ λα,β
+ Γλλν Γναβ
−Γ −Γ
Rαβ = Rλαλβ = Γλαβ,λ
λ ν βν Γαµ ,
−Γ −Γ
λ ν βν Γαλ .
˜λ ˜λ The same relation holds for R αµβ and Rαβ through Γαβ (note that these relations do not involve the metric gαβ explicitly). Therefore λ ˜λ R Rλαµβ = Bαβ,µ αµβ λ ˜ αβ Rαβ = Bαβ,λ R
− −
λ αµ,β
λ ν λ ν + Bµν Γναβ + Γλµν Bαβ + Bµν Bαβ
λ λα,β
λ ν λ ν + Bλν Γναβ + Γλλν Bαβ + Bλν Bαβ
−B −B
λ ν βν Γαµ
−B −B
λ ν βν Γαλ
λ ν βν Bαµ
−Γ −Γ
λ ν βν Bαλ
λ ν βν Bαµ ;
−B −B
λ ν βν Bαλ .
We shall only compute the expression for the Ricci tensor Rαβ . As a preparation, we compute λ Bλα = δλλ Ω,α
≡ N N Ω Ω
,α ,
where N = δλλ = gαβ g αβ is the number of spacetime dimensions. We shall always raise and lower indices using the original metric gαβ . So we compute term by term, ˜ αβ R
−R
αβ
= δαλ Ω,β + δβλ Ω,α
,ν
λ βν
ν α
ν α
,β
ν λ
,α
βν
,β
,λ ,λ
λ λν
,αβ
,λ
,ν
,ν
,ν
ν αβ
ν αβ
λ λν
ν α
ν β
ν βλ
,λ
,ν
λ ν
,β
,ν
ν αβ
λ λα
,ν
,β
βν
,λ
βν
λ λα
,β
,α
,ν
αβ
ν αλ
ν α
ν λ
,λ
λ λβ
,β
ν αλ
βν
,λ
,β
,λ
,α
λ αβ
,α
αβ
λ λβ
,λ
,ν
αλ
λ λν
,ν
λ ,ν αλ βν Ω
,α
,λ
αβ
ν αβ
λ ν
,ν
,αβ
,ν
,λ
,ν
λ β
λ β
,λ
,ν
,ν
αν
,λ
αλ
αβ
αβ
ν αλ
,α
αβ
,αβ
αβ
αβ,λ
,β
,α
,α
,λ ,λ
αβ
αβ
ν β
,λ
,αβ
αβ Ω
,λ
− N N Ω Ω + N N Ω Ω Γ +Γ δ Ω +δ Ω −g Ω + N N Ω Ω δ Ω +δ Ω −g Ω − δ Ω +δ Ω −g Ω Γ − Γ (δ Ω + δ Ω − g Ω ) − δ Ω + δ Ω − g Ω (δ Ω + δ Ω − g Ω ) − g Ω − g Ω − N = 2Ω N Ω Ω + N N Ω Ω Γ +Γ Ω +Γ Ω −g Γ Ω + 2N 2N Ω Ω Ω − Ng Ω Ω − Γ Ω − Γ Ω + g Γ Ω − Γ Ω − Γ Ω + g Γ N )) Ω Ω + 2g 2g Ω Ω − (2 + N = −g Ω − g Γ Ω − N g Ω Ω + 2g 2g Ω Ω = − (N − 2) Ω (N − 2) Ω Ω − g (N − 2) Ω Ω + Ω + Γ Ω − Ω Γ + (N + g Γ Ω +g Γ Ω −g Ω .
−g
,ν
,ν
αβ
,α
,λ
αβ,λ
,λ
,λ
,β
αβ
,λ
,λ ,λ
,λ
λ λν
,ν
Now we note that some of the Γ terms can be absorbed into covariant derivatives, and also that the terms in the last bracket cancel, gβν ΓναλΩ,λ + gαν Γνβλ Ω,λ gαβ,λ Ω,λ = 0,
−
so the resulting formula can be written more concisely as ˜ αβ R
−R
αβ
= (N
− 2)[Ω
,α Ω,β
− Ω; ] − g αβ
αβ
The modified Ricci scalar is ˜ = g˜αβ R ˜ αβ = e−2Ωg αβ Rαβ + e−2Ωg αβ (N R
(N
− 2) Ω
,λ Ω
,λ
+ Ω;λ ;λ .
− 2)[Ω Ω − Ω; ] − g (N − 2) Ω = e−2ΩR + e−2Ω (N − 2)[Ω Ω − Ω; ; ] − N (N − 2) Ω Ω + Ω; ; = e−2Ω {R − (N − 2) (N − 1) Ω Ω − 2 (N − 1) Ω; ; } .
,α
,α
,α
α
,α
α
,α
,β
αβ
,λ
,λ
α
αβ
λ
,λ Ω
λ
α
,λ
+ Ω;λ ;λ
The Einstein tensor is modified as follows, ˜ αβ = R ˜ αβ G
− 12 g˜
− 12 g
− 2)[Ω
[R
,α
,α
,β
αβ
− Ω; ] − g (N − 2) Ω Ω + Ω; ; − 2 (N − 1) Ω; ; ] (N − 2) (N − 3) Ω Ω + (N (N − 2) Ω; ;
,α Ω,β
,α
− (N − 2) (N − 1) Ω Ω + (N (N − 2)[Ω Ω − Ω; ] + g αβ
= Gαβ
˜ = Rαβ + (N (N
αβ R
αβ
αβ
αβ
α
2
α
,λ
,α
,α
,λ
λ
α
λ
α
.
Note that there is no change in Gαβ in two dimensions (since the Einstein tensor is always equal to zero). 21
6.3
Equa Eq uati tion onss of mo motio tion n
The equation for the covariant component u1 (s) is du1 ds
Using the metric gαβ = diag f, 1/f, d dλ
−
r2,
− −
r2
− − 1 2
−1
f r˙
− 12 u
α β
2
u
∂ (gαβ ) = 0. ∂r
sin θ , where f df ˙2 t dr
−
d dr
≡ 1−
1 r˙ 2 f
˙ φ˙ , where ˙ r, rg /r /r,, and u = t, ˙ θ, where ˙ α
− 2rθ˙2 − 2rφ˙ 2 sin2 θ
d/dλ,, we find ≡ d/dλ
= 0.
(14)
To derive this equation from other equations given in the lecture, we transform in a clever way the expression d 0= dλ
K
d = f t˙2 dλ
Namely, we try to separate terms of the form
d dλ
−1 2
− f
r˙
2 ˙2
−r θ −r φ
For example,
− −
d d f t˙2 = 2t˙ f t˙ dλ dλ
We find d dλ
u1 u1
d t˙2 f, dλ
sin θ .
d g11 dλ
d dλ
(uαuα ) in the following way,
(no summation!). summation!).
etc.
K = dλd f t˙2 − f −1r˙2 − r2θ˙2 − r2 sin2 θφ˙ 2 d df d d d = 2 t˙ f t˙ − t˙2 2r˙ f −1 r˙ + r˙ 2 f −1 − 2θ˙ − dλ dλ dλ dλ dλ d d − 2φ˙ dλ r2 sin2 θφ˙ + φ˙ 2 dλ r2 sin2 θ .
0=
2
(uα ) out of the terms of the form
d d d u1 u1 = g11u1 u1 = 2u 2 u1 g11 u1 dλ dλ dλ
2 ˙2
d 2 r2 θ˙ + θ˙2 r dλ
Now we substitute the given equations (2)-(4), and also evaluate derivatives of the metric, e.g. df /dλ = f ′ r˙ , so 0=
−
= r˙
t˙2 f ′ r˙
−
− d dλ
d 2r˙ f −1 r˙ dλ −1
f ′ ˙ 2 φ˙ 2 sin θ cos θ + θ˙2 2rr˙ + φ˙ 2 2r sin2 θ r˙ + 2φ˙ 2 r2 θ˙ sin θ cos θ r˙ 2 r˙ 2θr f f ′ 2 r˙ + 2r 2rθ˙2 + 2r 2rφ˙ 2 sin2 θ . f 2
− − −
2f r˙
′ ˙2
f t
2
−
This is obviously equivalent to Eq. (14 (14). ). Note: the reason one of the equat equations ions follows follows from other equations equations is that the equation uα uα = const is a conse consequenc quencee β α of the four geodesic equations, u u ;β = 0, and the fact that gαβ ;µ = 0. There Therefore, fore, when we consider consider the four geodesic α equations and the equation uα u = const, any one of these five equations is a consequence of four others.
7 7.1
Wea eak k gra gravit vitat ation ional al fiel fields ds Gravit Gra vitati ationa onall bendin bending g of ligh lightt
In the lecture it was shown that the trajectory of a light ray in polar coordinates satisfies the equation d2 dφ2
1 r
+
1 3 rg = , r 2 r2
rg
≡ 2GM ≈ 3km 3km,, c2
where M is the mass of the Sun. Introduce an auxiliary variable v(φ) v ′′ + v =
≡ r−1 and solve the equation
3 rg v2 2
perturbatively, assuming that v is small, v(φ) = v0 (φ) + v1 (φ) + ... The unperturbed solution is
1 cos φ, R0 where R0 is the distance of closest approach to the Sun. Then v0 (φ) =
v1′′ + v1 =
3 rg 3 rg 2 cos φ = (1 + cos cos 2φ) . 2 R02 4 R02 22
The soluti solution on is found with undete undetermine rmined d coeffici coefficients ents,, v1 (φ) = A + B cos2 cos2φ, φ,
3 rg , 4 R02
A=
B=
− 14 Rr 2 . g
0
The total deflection angle is found as δ = φ1 φ2 π, where φ1,2 are fixed by the condition v(φ) = 0. We find a quadratic equation 2R0 R0 R02 cos2 φ cos φ 2 = 0, cos φ = + 2. 2. rg rg rg2
− −
−
−
±
Only the solution with the minus sign is meaningful (cos φ < 1). Since rg find rg cos φ + O(rg3 /R03 ). R0
Taylor series and ≪ R0, we may expand this in Taylor
≈−
Therefore, There fore, the angle φ is very close to π/ π/2, 2, φ1,2 =
±
This formula can be rewritten as
For the Sun we have R⊙ = 6, 96
π +ε , 2
ε
≈ Rr 0 , ⇒ g
δ = 2ε 2ε =
2rg . R0
2rg /R⊙ 2rg R⊙ δ= = . R0 /R⊙ R⊙ R0
× 105km and r
g
2rg /R⊙ = 8, 489
× 10−6 =
= 1, 751”
= 2, 954 km, therefore
8, 489
× 10−6 × 360◦/2π × 3600
[arc seconds]
(see R. Oloff “Geom “Geometrie etrie der Raumzeit,” Raumzeit,” 2nd German edition, edition, page 151).
7.2
Eins Ei nste tein in tens tensor or for for weak weak field field
For this problem Chapter 4 from the book Norbert Straumann “General Relativity and Relativistic Astrophysics” is useful. We have gµν = ηµν + hµν and Rµν = Γ λµν,λ Γλλµ,ν ,
−
where (... (...)),µ denotes a derivative ∂ µ (... ...). ). Here one can ask the students about the symmetry of this tensor. Furthermore 1 1 α Γαµν = η αβ [hµβ,ν + hνβ,µ hµν,β ] = hµ,ν + hα hµν,α , ν,µ 2 2
−
−
where as usual we use the convention convention that indic indices es are raise raised d or low lowered ered with η µν ; thus e.g. hαβ we have 1 λ Rµν = hµ,νλ + hλν,µλ hµν h,µν , 2 where
−
−
= η µν ∂ µ ∂ ν and h = hλλ = η λαhαλ . And for the Ricci scalar we obtain R = η µν Rµν = hλν,νλ
≡η
αλ
(15)
hλβ . Using Eq. (15 (15))
− h.
Thus in the linear approximation we have Gµν = Rµν
−
1 1 λ ηµν R = hµ,νλ + hλν,µλ 2 2
− h − h − µν
,µν
ηµν hλβ,βλ
+ ηµν h .
Let us introduce a new variable γ µν hµν 12 ηµν h. The traces traces of two tensors tensors h and γ are related by γ = µν 1 hµν γ µν η γ . Inserting the last expression for hµν in Gµν , we have µν 2 µν
≡
−
≡
Gµν
−
1 λ λ = γ µ,νλ + γ ν,µλ 2 1 ,λ ,λ = + γ νλ,µ γ 2 µλ,ν
or finally Gµν
µν µν
λβ µν γ ,βλ
µν µν
,λβ µν γ βλ
− γ − η − γ − η
1 µ,λ λ,µ = γ λ,ν + γ ν,λ 2
23
−
µ γ ν
−
β,λ δνµ γ λ,β
.
= ,
−h, thus
7.3
Gravit Gra vitati ationa onall pertur perturba bation tionss I
The metric is written as gµν = ηµν + δg µν , i.e. g00 = 1 + 2Φ, 2Φ, g0i = B,i + S i ,
gij =
−δ
ij
+ 2Ψδ 2Ψδij + 2E 2E ,ij ,ij + F i,j i,j + F j,i j,i + hij ,
(16)
where S i,i = F i ,i = hij,i = hij η ij = 0, hij = hji . We shall use the formula for Gµν derived in Problem 7.2 7.2.. All 3-dimensional indices are raised and lowered using δij , so we can write these indices in any position, as convenient: j
δgj0 = δg0j =
δg ij =
−δg0,
−δg
ij .
Also note that for any quantity X we have
˙ X ,j = X We need to write the components of γ νµ = δgνµ
− 12 δ
−X
,j ,j
µ¯ ν h,
¯h
.
≡ δg
µ µ,
using the 3+1 decomposition: ¯h = δg µ = η µν δgµν = 2 (Φ µ
− 3Ψ − ∆E ) , γ 00 = Φ + 3Ψ + ∆E, ∆E, γ 0 = B + S = −γ 0 = γ 0 , γ = − (Φ − Ψ − ∆E ) δ − 2E − F − F − h ,j
j
i j
ij
j
j
,ij ,ij
i,j i,j
j
j,i j,i
ij
= γ ij .
Now we compute γ λ0,λ = γ ˙ 00
j,j
B ;
−γ = −γ ˙ 0 − γ = −B˙ − S ˙ − (− (Φ − Ψ − ∆E ) δ − 2E − F − F − h ) ˙ + (Φ − Ψ + ∆E = −B˙ − S ∆ E ) + ∆F ∆F ; ¨ + 3Ψ ¨ + ∆ E ¨ − B˙ + −B˙ − S ˙ + (Φ − Ψ + ∆E = γ 0 + γ =Φ ∆E ) + ∆F ∆F 0 ¨ + 3Ψ ¨ + ∆ E ¨ − 2∆B˙ + ∆(Φ − Ψ + ∆E =Φ ∆ E ) .
γ λj,λ =
λ j,λ
j i,i
j
,j
β,λ γ λ,β
−
˙ + 3Ψ ˙ + ∆ E ˙ =Φ
− γ 0
,j
j
j
ij
j,λ λ
,
i,j i,j
j,i j,i
ij ,i
j
,j
,λ λ
,ij ,ij
,j
,j
j
j
,j
,j
Then we compute each component of Gµν separately: 0,λ 2G00 = 2γ λ, 0
−
β,λ λ,β
− γ 00 − δ00γ
¨ + 3Ψ ¨ + ∆E ¨ + 2∆B˙ Φ
= 4∆Ψ; 0,λ
2G0j = γ λ,j +
−
λ,0 γ j,λ
− γ 0 −
j
− −
¨ + 3Ψ ¨ + ∆ E ¨ =2 Φ
∂ 0 ∂ 0 (Φ + 3Ψ + ∆E ∆E ) + ∆(Φ + 3Ψ 3Ψ + ∆E ∆E )
∆ E ) − ∆ (Φ − Ψ + ∆E
β,λ δj0 γ λ,β
− − −
˙ + 3Ψ ˙ + ∆ E ˙ = Φ
¨,j + S ¨j + ∆ (B,j + S j ) = 4Ψ ˙ ,j + ∆S B ∆S j
2Gij
B˙
∗
B
,j
¨,j B
− S ¨ + j
˙ Φ
− Ψ˙ + ∆E ˙
,j
+ ∆F ˙j
− ∆F ˙ , ˙ + (Φ − Ψ + ∆E ˙ + (Φ − Ψ + ∆E = γ + γ − γ − δ γ = −B˙ − S ∆ E ) + ∆F ∆F + −B˙ − S ∆ E ) ¨ + 3Ψ ¨ + ∆ E ¨ − 2∆B˙ + ∆(Φ − Ψ + ∆E + ((Φ − Ψ − ∆E ) δ + 2E 2E + F + F + h ) − δ Φ ∆ E ) ¨ ¨ + F ¨ − S ˙ − S ˙ + h − 2δ 2Ψ ¨ + ∆ Φ − Ψ − B˙ + E ¨ . = 2 Φ − Ψ − B˙ + E + F i,λ λ,j
j,λ λ,i
i β,λ j λ,β
i j
ij
,ij
,j
,ij ,ij
i,j i,j
j
i,j i,j
j,i j,i
j
,j
j,i j,i
ij
i,j i,j
λ,0 = * - the origin of the minus sign here is γ j,λ
7.4
j
ij
j,i j,i
ij
ij
,i
,i
i
,i
+ ∆F ∆F i
,j
j,λ λ,
−γ 0.
Gravit Gra vitati ationa onall pertur perturba bation tionss II
Under an infinitesimal transformation xµ gαβ
µ
µ
→ x + ξ , the metric changes as →g −g ξ −g ξ =g −ξ −ξ αβ
γ αγ ,β
γ βγ ,α
αβ
β,α .
α,β
(17)
(This can be easily found from the standard formula for the change of coordinages, involving ∂ ∂ ˜ x ˜µ /∂x ν .) Now let let us write Eq. (17 (17)) in full, using the perturbation variables (16 ( 16), ), the covariant components ξµ , and the decomposition ξµ = ξ 0 , ξ⊥i + ζ ,i transformati ormation on of g of gαβ component by component using the 3+1 decomposition, and ,i . We can write the transf we use the fact that the background metric is diagonal,
g00
→ g00 − 2ξ00; ,
g0i
→ g0 − ξ 0 − ξ 0 ; i
,i
24
i,
gij
→g −ξ −ξ ij
i,j
j,i .
To simplify calculations, we adopt the convention of raising and lowering the spatial indices i,j,... by the Euclidean spatial metric δij rather than by ηij . This will get rid of some minus signs. We also denote ∂ 0 ∂ t by the overdot. Thus we have
≡
g00
→ g00 − 2ξ˙0;
→ g0 − ξ0 − ξ˙ ;
g0i
i
gij
i
,i
→g −ξ −ξ ij
j,i .
i,j
Substituting Substi tuting the perturb perturbation ation variables variables from Eq. (16 16), ), we get
→ Φ − ξ˙0, B + S → B + S − ξ 0 − ξ˙⊥ − ζ ˙ , + F + h → 2Ψ 2Ψδδ + 2E 2E + F + F Φ
,i
2Ψδδij + 2E 2Ψ 2E ,ij ,ij + F i,j i,j
i
j,i j,i
,i
ij
i
ij
,i
i
,ij ,ij
i,j i,j
(18) (19)
,i ,i
j,i j,i
+ hij
− ξ⊥ − ξ⊥ − 2ζ i,j
,,ij ij .
j,i
(20)
Now we need to separate these equations and derive the transformation laws for the individual perturbation variables. This is easy to do if we perform a Fourier transform of Eqs. ( 18 18)-( )-(20 20)) and pass to the Fourier space (where every variable is a function of a 3-vector k). A vector V i is decomposed into scalar and vector components as follows, (V )
V j = ik i kj V (S ) + V j
≡ V kk2 ,
(V )
l l
V (S )
;
V j
≡ V − ik j
j
V l kl = V j k2
− ik V ( j
S)
.
(21)
The idea is first, to project the given vector V i (k ) onto the direction of ki , and second, to subtract the projection from V i and to obtain the component of V i which is transversal to ki . The imaginary imaginary unit factors are added as coeffici coefficients ents at 21)) translates to real space as kj for convenience: with these factors, the decomposition ( 21 V j = ∂ j V (S ) + V j(
V )
.
The same procedure applied to a symmetric tensor T ij ij leads to a decomposition into scalar, vector, and tensor components. Let us go through this procedure in more detail. First, we subtract the trace and obtain the traceless part T (1) of the tensor T T ,, 1 (1) (1) T ij T ij δij T ll T ii = 0. 0. ij ll ; 3
≡
Note the coefficient
1 3
−
(1)
that depends on the num number ber of spatial dimensions dimensions (three). (three). Now we project T ij onto ki kj and (2)
obtain the scalar component component T (S ) proportional to ki kj and the tensor T ij orthogonal to ki kj : (1)
T ij
≡ −
(1)
1 ki kj + δij k 2 T (S ) + T ij(2); 3
(S )
T
≡−
3 ki kj T ij ; 2 k4
T ij(2)ki kj = 0.
Note that T ij(2) is again a tracetrace-free free tensor, T ii(2) = 0, due to the subtraction of 13 k 2 δij in the first term. Final Finally ly,, we project pro ject T ij(2) onto ki and kj separa separately tely,, to obtain a “vec “vector” tor” part T j( (T )
part T ij
(T )
(T )
such that T ij kj = 0 and T ii T ij(2) = iki T j(
V )
+ ik ikj T i(
V )
V )
such that T j(
V )
kj = 0, and a completely traceless (“tensor”)
= 0:
+ T ij( ); T
T j(
V )
≡ i kk2 T (2),
T ij(
T )
l
jl
≡ T (2) − i ij
In real space, the full decomposition is
≡
−
−
ki T j(
1 1 V V T T ij T ll δij ∆ T (S ) + ∂ i T j( ) + ∂ j T i( ) + T ij( ) . ij = ll δij + ∂ i ∂ j 3 3 3 1 1 1 (2) T (S ) ∂ i ∂ j T ij T ll T ij T ij T ll ∂ i ∂ j ij ll δij ; ij ll δij 2 2∆ 3 3 1 (V ) (2) (T ) (2) (V ) (V ) T j = ∂ i T ij , T ij = T ij ∂ i T j ∂ j T i . ∆
≡
−
−
−
−
V )
−
+ kj T i(
V )
.
1 δij ∆ T (S ) ; 3
It may be conv convenien enientt to gather the “trace” “trace” terms (the terms containing containing δij ) as one term, (V )
(tr) T ij δij + ∂ i ∂ j T (S ) + ∂ i T j ij = T
(V )
+ ∂ j T i
(T )
+ T ij ,
T (tr)
≡ 13 T − 13 ∆T ( ll ll
S)
.
Note tha Note thatt the pert perturb urbati ation on vari ariabl ables es Ψ, E , Fi , hij are obt obtain ained ed by thi thiss dec decompo omposit sition ion met method, hod, sta starti rting ng fro from m the symmetric perturbation tensor δg ij , with slight slight modifications: modifications: there are some cosmetic cosmetic factors of 2 and some minus signs. Applying the decomposition method to Eqs. (18 18)-( )-(20 20), ), we get the following transformation laws for the perturbation variables, Φ E
→ Φ − ξ˙0, → E − ζ,
→ B − ξ0 − ζ,˙ S → S − ξ˙⊥ , → Ψ, F → F − ξ⊥ , h → h
B Ψ
i
i
25
i
i
i
i
ij
ij .
Remarks:
1. It is clear clear that one can set set F i = 0, B = E = 0 with a coordina coordinate te transformation transformation.. Other components components will then show whether wheth er the geometry is really perturbed or it’s just a coordin coordinate ate transformatio transformation n of a flat space. In general, there will remain 6 independent components of perturbations (Φ, (Φ, Ψ, S i , hij ). 2. These considera considerations tions depend depend rather crucially on the silently made assumption that all the metric perturbations vanish, δg µν 0, at spatial infinity. These boundary conditions are implicitly used when defining the Fourier transforms necessary for the tensor/vector/scalar decompositions (a Fourier transform is undefined undefined without this boundary b oundary condition). Alternatively, one may do without Fourier transforms but then one still needs boundary conditions to solve the relevant Poisson equations for components. Without boundary conditions, there is no unique decomposition of the form
→
X i = A,i + Bi ,
A=
1 X i,i i,i , ∆
because the function A is defined up to solutions of ∆A ∆ A = 0. So the tensor/vector/s tensor/vector/scalar calar decomposi decomposition tion is actually without ut a fixed assumption assumption about the boundary conditions. conditions. The boundary conditions conditions δg µν 0 at spatial undefined witho infinity infinit y is a natura natural, l, phys physically ically motivated motivated set of boundary conditions. conditions. An explic explicit it coun counterex terexample ample where these boundary conditions are not satisfied: gµν = diag(A, diag(A, B, B, B ), whe where re A = 1, B = 1 ar aree consta constant nts. s. Th This is metri metricc is flat but one cannot see this by using the perturbation perturbation formalism! formalism! (The component component Ψ = 0 cannot be removed by a gauge transformation.) The reason is that this gµν is a “per “pertur turbati bation” on” of flat metric metric with Φ and Ψ tha thatt do not decay decay to zero at spatial infinity infinity. So a coordin coordinate ate transformation transformation with ξ µ decaying to zero cannot bring this metric to ηµν .
→
− − −
8
Gravi Gra vita tati tion onal al ra radi diat atio ion n I
8.1
Gauge Gaug e inv invaria ariant nt varia ariables bles
¨ and S i Using the equations derived in Problem 7.4 7.4,, it is very easy to verify that D = Φ Ψ B˙ + E under infinitesimal changes of coordinates (i.e. invariant under infinitesimal gauge transformations).
− −
8.2 8.2.1
− F ˙ are invariant i
Detecting Dete cting gra gravitati vitational onal wa wave vess Using Usin g distan distances ces betw between een partic particles les
(This solution follows Hobson-Efstathiou-Lasenby [2006], 18.4.) Consider a plane wave moving in the z direc direction, tion, (all other components components of hµν are zero)
§
hxx =
−h
yy
= A+ e−iω(t−z) ,
hxy = hyx = A× e−iω(t−z) .
(22)
To detect the presence of this gravitational wave, let us imagine a cloud of particles initially at rest at different positions. The 4-vectors describing the particles are uµ = (1 (1,, 0, 0, 0), so one can easily see that these particles move along geodesics: uν uµ ;ν = uν uµ ,ν + Γµνα uν uα = Γµ00 , 1 Γµνα = η λµ (hλν,α + hλα,ν hαν,λ ) , 2 1 µ Γ00 = η λµ (hλ0,0 + hλ0,0 h00,λ ) = 0. 2
− −
Therefore the coordinates xµ of each particle remain remain constant with time. How Howeve ever, r, the distan distance ce between between each pair of particles partic les is deter determined mined through the space spacelike like vector vector ∆xµ xµ(1) xµ(2) as
≡
∆L2
≈ (η
µν
−
+ hµν ) ∆xµ ∆xν
and will change with time because of the dependence on hµν . Sin Since ce the onl only y non nonzer zeroo componen components ts of hµν are the x, y components, it is clear that only changing lengths are between particles that have some separation in the x, y directions. Therefore it is sufficient to consider a ring of particles situated in the x y plane plane.. The physically physically measured distances distances between the particles in the ring will change with time, i.e. the ring will experience a deformation. To visualize the deformation, it is convenient to make a local coordinate transformation (local in the neighborhood of the ring) such that the metric becomes flat, gµν xµ xν = ηµν x ˜µ x ˜ν (up to second-order terms). The trick that performs this transf transformati ormation on is the follo following, wing, 1 1 x ˜µ = xµ + hµλ xλ xµ + hαλ xλ η αµ . 2 2 It is easy to check that gµν xµ xν (ηµν + hµν ) xµ xν = ηµν x ˜µ x ˜ν + O(h2 ).
−
≡
≡
Therefore, x Therefore, ˜µ can be understood as the (approximate) Cartesian coordinates where the length is given by the usual Pythago Pyt hagorea rean n formula. formula. No Now w if we compute compute the shap shapee of the ring in the these se coor coordin dinate ates, s, it wil willl be eas easy y to in inter terpre prett thi thiss shape in a straightforward way. 26
Consider Consid er a partic particle le with constant 3-coordinates 3-coordinates (x,y,z x,y,z). ). After the coordinate transformation, we have 1 (A+ x + A× y ) e−iω(t−z) , 2 1 y˜ = y + (A× x A+ y ) e−iω(t−z) , 2 z˜ = z.
x ˜ =x+
−
To visualize the deformation, it is convenient to consider first the case A+ = 0, A× = 0 and then the opposite case. The deformation deformation of the ring is squee squeezing zing in one direct direction ion and expansion in the orthogo orthogonal nal direction. direction. It follows that A+ describes a deformation in the two vertical directions, while A× describes a deformation in the directions at 45◦ . Note that the defor deformation mationss change the shape of the ring in the same way, way, except for the rotated orientation orientation.. This π can be verified by performing a rotation by 4 ,
→ √ 1 2
x ˜ y˜
1 1
−1
x ˜ y˜
1
,
and then it is straightforward to see that this will exchange A+ and A× . 8.2.2
Using Usin g geodesic geodesic devia deviation tion equa equation tion
PLEASE NOTE: The commonly found arguments that use the geodesic deviation equation are suspect because the geodesic deviation equation uses coordinates ξ µ rather than gauge-inv gauge-invarian ariantt quantities. quantities. A cloud of particles particles at rest in the gravitational field hµν described by Eq. (22 ( 22)) will stay indefinitely at rest in the coordinate system (ξ ( ξ µ = const) even though the distances between between particles will change with time. See arxiv:gr-qc/06050 arxiv:gr-qc/0605033 33 for nice explanations. explanations. The solution given above is simple and straightforward. The argument using the geodesic deviation (see Carroll, Chapter 6, p. 152-154) goes like this: The geodesic deviation equation can be simplified for a deviation vector S σ corresponding to nonrelativistic (almost stationary) particles moving with 4-velocity approximately equal to (1, (1 , 0, 0, 0),
d2 S σ = Rσ00λ S λ . 2 dt The Riemann tensor to first order in h can be expressed as ¨σ Rσ00λ = h λ (note that hµ0 = 0). Therefore, the geodesic deviation equation becomes ¨σ = h ¨ σ S λ , S λ x ¨ S = ω 2 (hxx S x + hxy S y ) = ω 2 (A+ S x + A× S y ) e−iω(t−z) , ¨y = ω 2 (hyx S x + hyy S y ) = ω 2 (A× S x A+ S y ) e−iω(t−z) , S
−
and there is no change in the z direction.
8.3
Pois Po isso son n eq equa uati tion on
The general solution of the Poisson equation, ∆φ = 4πρ, with boundary conditions φ
→ 0 at infinity, is easy to find using the Fourier transform:
−k2φ(k) = 4πρ πρ((k), d3 k φ(x) = − e (2π (2 π )3
πρ((k) ik·x 4πρ
k2
=
−
d3 k
ik·x 4π
(2π (2 π)
k2
3e
3
d ye
−ik·y
ρ(y) =
d3 y ρ(y)G(x
− y),
where G(x) is the Green’s function, G(x) =
−
d3 k
ik·x 4π
e (2π (2 π)3
k2
=
−
Here we used the known integral
∞
0
Therefore
1 π
∞
π
dk
0
0
sin z 1 dz = z 2 φ(x) =
dθ sin θ e
ik|x| cos θ
+∞
−
−∞
− ||
∞
0
dk sin kx = k
− |x1 | .
sin z 1 dz = π. z 2
d3 y ρ(y). x y
| − |
27
=
2 π x
(23)
One can denote this integral more concisely, φ = 4π
1 ρ, ∆
1 where the operator ∆ is just a shorthand notation for the integral in Eq. (23 (23). ). Note that the function ρ must fall off sufficiently rapidly as x or else the integral (23 (23)) will not converge. It is −2−ε sufficient that ρ(x) at large x (where ε > 0). x
|
8.4
|∼| |
| |→∞
||
Metr Me tric ic per pertu turb rbat atio ions ns 1
An arbitrary 3-vector X i (such as T 0i ) is decomposed into scalar and vector parts as follows, X i = a,i + bi ,
bi,i = 0.
To determine an explicit expression for a, let us compute the divergence of X i , X i,i i,i = a,ii = ∆a. Therefore a(x) =
−
1 4π
One can write more concisely
a=
8.5
d3 y X i,i i,i (y). x y
| − |
1 X i,i i,i . ∆
Metr Me tric ic per pertu turb rbat atio ions ns 2
The energy-momentum tensor T µν µν is decomposed as 0
T i = α,i + β i ,
α
≡
1 0 T , ∆ k,k
β i
0
≡ T
i
(T )i
T ki = µδik + λ,ik + σi,k + σk,i + T k T i(
β i,i 0, i,i = σi,i = 0, We need to verify that the equation
T )i
= 0, 0,
− 1 0 T ∆ k,k
;
,i
,
( ) T k,i = 0. 0. T i
1 ˙ − F ¨)=σ , − 16 (S 16πG πG i
i
(24)
i
which represents the vector part of the spatial Einstein equation (here σi is the vector part of the spatial T ij ij ), also follows from the conservation of T µν and from the other Einstein equations. µν To calculate the components λ, µ of the EMT, we compute T ii = ∆λ ∆ λ + 3µ, 3µ, i T k,i = ∆λ ∆ λ,k + µ,k + ∆σ ∆σk , i T k,ik = ∆∆λ ∆∆ λ + ∆µ. ∆µ.
Now we solve this system of equations and find
1 µ= T ii 2 1 i σj = T j,i ∆
− −
1 i 3 1 i 1 i T k,ik , λ = T k,ik T , ∆ 2∆ 2 i 1 1 i T k,ik , (T ) T ki = T ki µδki ∆ ∆ ,j
−
−
−λ −σ −σ ,ik
i,k
k,i .
Note that the operator ∆12 applied to a function f f ((x) is defined only if the function f has a sufficiently fast decay at −3−ε . It is sufficien sufficientt that f f ((x) with ε > 0 at large x . This is a faster decay decay than that required by by the x x 1 operator ∆ . The Einstein equations are
| |→∞
|
D,ij
−δ
ij
¨ ∆D + 2Ψ
|∼| |
2∆Ψ = 8πGT 8πGT 00 , ˙ ,i + 1 ∆S ˜i = 8πGT i0 = 8πG (α,i + β i ) , 2Ψ 2 1 ˜˙ 1 ˜˙ j,i + hij = 8πGT ji = 8πG µδik + λ,ik + σi,k + σk,i + T (T )i , S i,j i,j + S j,i k 2 2
−
where we have denoted for brevity
||
˜i S
≡ S − F ˙ , i
i
D
≡ Φ − Ψ + B − E,˙
28
which are gauge-invariant variables. In the 3+1 decomposition, the Einstein equations become ∆Ψ = 4πGT 4πGT 00 , ˙ = 4πGα, Ψ
(25) (26)
˜i = 16 ∆S 16πGβ πGβ i , D = 8πGλ, ¨ = 8πGµ, ∆D + 2Ψ ˜˙ i = 16 S 16πGσ πGσi ,
− −
hij
= 16 16πGT πGT j(
(27) (28) (29) (30)
T )i
.
(31)
The conservation law of the EMT in 3+1 decomposition looks like this, T 00,0 + T 0j,j = 0, This gives therefore
j T i,00 + T i,j = 0. 0.
(32)
T ˙00 = ∆α,
˙ i + ∆λ α˙ ,i + β ∆λ,i + µ,i + ∆σ ∆σi = 0, 0,
T ˙00 = ∆α, ∆ α,
α˙ + ∆λ ∆λ + µ = 0, 0,
˙ i + ∆σ β ∆σi = 0.
(33)
Then it is easy to see that Eqs. (26 26)), (29 29), ), and (30 (30)) are conse conseque quence ncess of Eqs. (25 25), ), (27 27), ), (28 28), ), and the conservation laws (33 (33). ). In particular, 1 ˜˙ i = ∂ t 1 16 S 16πGβ πGβ i = 16πG 16 πG∆ ∆σi = 16 16πGσ πGσi . ∆ ∆
−
9 9.1
−
Gravi Gra vita tati tion onal al ra radi diat atio ion n II Projec Pr ojecti tion on of the the mat matte ter r tenso tensor r
a) First note that P ab ab is a projector, P ab ab P bc bc = P ac ac , and its image has dimension 2, that is, the trace of P ab ab is 2, P ii ii = 3
−n n
i i
= 2.
Therefore, for any X ab ab we have (T )
X ii ii = P ia ia X ab ab P bi bi
− 12 P P ii ii
ab a b X ab ab
= P ab ab X ab ab
− 12 2P
ab a b X ab ab
= 0. 0.
b) We compute (T )
X ik,i ik,i =
P ia ia X ab ab P bk bk
−
1 P ik ik P ab ab X ab ab 2
=
P ia ia P bk bk
,i
−
1 P ik ik P ab ab 2
X ab ab + P ia ia P bk bk
,i
−
1 P ik ik P ab ab X ab,i ab,i . 2
(34)
Note that the projection kills any component proportional to Ri because P ia ia Ri = 0. At the same time, X ab,i ab,i is proportional to Ri because ) = Ri X ′ . X ab,i R ab,i = X (t R ,i Therefore the second term in Eq. (34 ( 34)) vanishes:
P ia ia P bk bk
−| |
−
−
1 P ik 0. ik P ab ab Ri = 0. 2
So only the first term remains, (T )
X ik,i P ia ik,i = ia P bk bk
−
29
1 P ik ik P ab ab 2
X ab ab . ,i
However, this term contains derivatives of P ab ab , which are also sometimes proportional to Ri . We compute P ik,a ik,a =
−n
i,a nk
P ia ia P bk bk
− 12 P
ik P ab ik ab
i k,a ;
− R1 (P
P ik,a ik,a =
−n n
ai nk ai
= P ia,i ia,i P bk bk ,i
ni,a =
+ P ak ak ni ) ,
− 12 P
ik,i P ab ik,i ab
Ri R
P ik,i ik,i =
+ P ia ia P bk,i bk,i
,a
a
=
−
1 R
bk bk
k
a
bk bk
ab ab
k
i
ik ik
,a
=
δia R
− RRR3 i
a
=
1 P ia ia , R
(note that P ak ak nk = 0)
k
ab,i ab,i
1 P ik ik (P ai ai nb + P bi bi na ) 2R 1 1 ak nb + P ak ak ak nb + P bk bk na 2 2
ia ia
ab ab
− RR2 |R|
− R2 n , − 12 P P
− R2 n P + R1 n P − R1 P (P n 1 = −2n P + n P − P n − P R =
Ri,a R
=
bi k bi
ab k ab
+ P ik ik nb ) +
3 1 P bk bk na + P ak ak nb . 2 2
than P ab This is higher-order in 1/ 1/ R ab , as required.
| |
9.2
Matt Ma tter er so sour urce cess
The question is to verify the following property, (T )
(T ) X ik Qik , ik =
where Qik = X ik ik
−
1 δik r2 T 00 d3 r. 3
It is easy to see that X ik transverse-tracele aceless ss part of δij is ik differs from Qik only by a term of the form A(t, R)δij . The transverse-tr zero, 1 P ia P ik 0. ia P bk bk ik P ab ab δab = 0. 2
−
Therefore the transverse-traceless parts of X ik ik and Qik are the same.
9.3
Energy-m Ener gy-momen omentum tum tensor tensor of grav gravitatio itational nal wav waves es
See Hobson-Efstathiou-Lasenby [2006], 17.11. We need to compute the secon second-orde d-orderr terms in the Einstein tensor. tensor. The idea is to separate the second-order second-order terms already alread y in the Ricci tensor. We will also try to simpl simplify ify things by using the fact that hµν is purely transverse-traceless; h0α = 0, hii = 0, hik,i = 0. It follows that η µν hµν = 0, hµν ,µ = 0.
§
Also, it is given that the EMT of matter vanishes, T µν µν = 0, which we will use below. First we decompose the metric, gµν = ηµν + hµν , g µν = η µν hµν ;
−
note that now indices are always raised and lowered using ηµν . We need to comput computee the Ricci tensor to second order. order. The Christoffel symbol up to second order is 1 λµ λ (2)λ η hλµ (hµα,β + hµβ,α hαβ,µ ) = Γ(1) αβ + Γαβ , 2 1 = η λµ (hαµ,β + hβµ,α hαβ,µ) , 2 1 λµ = h (hαµ,β + hβµ,α hαβ,µ ) . 2
Γλαβ = Γ(1) αβ
λ
Γ(2) αβ
λ
−
−
−
−
−
The Ricci tensor is Rαβ = Γλαβ,λ (1)
(1)λ
(2)
(2)λ
Rαβ = Γαβ,λ Rαβ = Γαβ,λ
λ λα,β
−Γ − Γ(1) − Γ(2)
+ Γλλν Γναβ
λ λα,β , λ λα,β
λ ν βν Γαλ
−Γ
(1)λ (1)ν
+ Γλν Γαβ
(1) (2) = Rαβ + Rαβ ,
− Γ(1) Γ(1) . λ
βν
ν
αλ
Let us now evaluate these expressions and simplify as much as possible, as early as possible: (1)λ
1 λµ 1 λµ η hλµ,α = η hλµ ,α = 0, 0, 2 2 1 λµ λ λ = Γ (1) Γ(1) η (hαµ,βλ + hβµ,αλ αβ,λ λα,β = 2
Γαλ = (1)
Rαβ
−
30
−h
αβ,µλ )
=
− 12 h
αβ ,
(1)
because of the transverse traceless property of hµν . Now, since Rαβ is found from the first-order Einstein equation (1)
Rαβ
− 12 η
αβ R
(1)
= 8πGT αβ αβ ,
and it is given that T αβ αβ = 0. Hence, we have hαβ = 0. Let us now evaluate derivatives of the second-order terms in the Christoffel symbols: (2)λ
Γαλ =
−Γ(2)
λ αλ,β
− 12 h
λµ
(hαµ,λ + hλµ,α
−h
αλ,µ )
=
− 12 h
λµ
hλµ,α ,
1 λµ 1 h hλµ,αβ + hλµ hλµ,β , 2 2 ,α 1 λµ = h (hαµ,βλ + hβµ,αλ hαβ,µλ ) 2 =
(2)λ
Γαβ,λ
−
−
(1)λ (1)ν
(in the last line we used hλµ ,λ = 0). Finally, we tackle the term Γ βν Γαλ . In this term, it helps to write 1 λ hα,β + hλβ,α h,λ αβ , 2 where again the indices are raised via η µν since we only need this term to first order. Then we can simplify this expression by grouping together terms where α, β appear in similar positions:
(1)λ
Γαβ =
(1)λ (1)ν
4Γβν Γαλ =
hλβ,ν
+
hλν,β
h,λ βν
−
hνα,λ + hνλ,α λ β,ν
,ν αλ ,ν αλ
βν,λ
λν ,α
(expand brackets) = hλβ,ν hνα,λ + hλβ,ν hνλ,α ,λ (move, rename λ, ν ) = h,ν βλ hαν + hβλ,ν
(symmetry (symme try of of h hµν ) =
+
,ν αλ
−h
λ ν ν,β α,λ
+ hλν,β hνλ,α
λν ,β αν,λ
+ hλν ,β hλν,α
−h h +h h h −h h +h h − h ) h − 2h h h − 2h h . λν ,α
,λ (gather terms) = 2h 2h,ν (hβλ,ν βλ hαν + (h
hλν,αhλν ,β
λ β,ν
,λ 2h,ν 2h βλ αν
λ β,ν
−
λ β,ν
,ν αλ
,ν λ ν,β αλ
λν ,β αλ,ν
+ hλν ,β (hαν,λ
,ν αλ
,λ βν
,λ βν
ν α,λ
−h h −h h −h h −h h −h h −h −h )+h h λ β,ν
λν ,β
αλ,ν
,ν λα
ν λ,α
,ν + h,λ βν hαλ
λν βν,λ h,α
,λ + h,ν βλ hαν
λν,α
(2)
Finally Final ly,, we put togeth together er the expre expression ssion for Rαβ : 1 1 1 ,λ Rαβ = hλµ ( hαµ,βλ hβµ,αλ + hαβ,µλ + hλµ,αβ ) + hλµ hλν,α hλν 2h,ν 2hλβ,ν h,ν ,α hλµ,β ,β + 2h βλ hαν αλ 2 2 4 1 1 1 ,ν λ ,λ = hλµ ( hαµ,βλ hβµ,αλ + hαβ,µλ + hλµ,αβ ) + hλµ hαλ hβ,ν h,ν (35) ,α hλµ,β + βλ hαν . 2 4 2 The Ricci scalar is 1 1 1 ,ν αλ 3 1 αλ,ν (2) ,α R(2) = η αβ Rαβ = hλµ hλµ + hλµ hαλ h,ν hαλ,ν hαν,λ = hλµ,α hλµ,α h hαν,λ , (36) ,α hλµ + 2 4 2 4 2 where we again used the transverse traceless property of hµν and also hαβ = 0. Note that the first-order first-order Ricci scalar scalar αβ (1) is zero, R = 0, since T αβ this reason reason we we may use η in Eq. (36 (36), ), otherwise we would have to write αβ = 0. For this (1) (2) (1) (η + h) R + R and pick up a second-order term hR . (1) Again, since R = 0, we may use ηµν rather than gµν to compute the Einstein tensor: (2)
− −
− −
−
−
(2) G(2) αβ = Rαβ
− 12 η
αβ R
(2)
−
−
−
.
We do not write the answer explicitly since it is a combination of Eqs. (35 35)) and (36 (36). ). (2) Now let us perform an averaging of the quantity Gαβ ov over er both space and time. In other words, we integrate integrate G(2) αβ over a 4-dimensional region such that hµν = 0 and hµν,α = 0 on the boundary of that region. Then ∂ µ (... ...)) = 0 and so we may integrate by parts, for example Aµ Bν,α = Aµ,α Bν ,
−
as long as Aµ Bν cont contains ains first powe powers rs of hαβ or hαβ,γ , so that boundary terms vanish. Then, for example, hλµ hλµ,αβ =
hλµ ,α hλµ,β ,
hλµ hαν,βλ =
hλµ ,λ hαν,β = 0,
h
− −
,ν αλ hβµ,ν
= − h
αλ hβµ
(37)
= 0,
by hαβ = 0 and by the transverse traceless property of hµν . Man Many y terms cancel in this way; for instance, instance, R(2) = 0. Finally, we get
1 1 ,ν λ Gαβ = hβµ,αλ + hαβ,µλ + hλµ,αβ ) + hλµ h h ,α hλµ,β + 4 2 αλ β,ν 1 λµ 1 λµ 1 λµ = h hλµ,αβ + h,α hλµ,β = h hλµ,β 2 4 4 ,α using Eq. (37 (37). ). Finally, we obtain the required equation, (2)
1 λµ h ( hαµ,βλ 2
−
−
(GW)
T µν µν =
1 − 8πG
− (2) Gαβ =
1 1 hλµ hij hij,β . ,α hλµ,β = 32πG 32 πG 32πG ,α 32πG 31
−
,λ h,ν βλ hαν
9.4
Powe Po wer r of em emitte itted d radia radiatio tion n
To derive the relations dΩ 1 = δ lm , 4π 3 dΩ 1 lm kr nl nm nk nr = (δ δ + δlk δmr + δ lk δmr ), 4π 15
nl nm
let us consider the generating function gΩ (ql )
≡
dΩ exp 4π
inl ql ,
−
which is a function of a vector argument ql . After computing gΩ(ql ) it will be easy to obtain integrals such as the above:
nl nm
dΩ ∂ ∂ =i i gΩ (qj ) 4π ∂q l qm
,
etc.
qj =0
The computation is easy if we introduce spherical coordinates with the z axis parallel to the vector ql , then nl ql = q cos θ, where q ql ql , and then we have
| |≡√
2π 1 gΩ (ql ) = dφ 4π 0 sin q = =1 q
π
dθ sin θ exp[ i q cos θ] =
−||
0
− 3!1 q q
|| ||
||
l l
+
1 (ql ql )2 5!
1 2isin q 2π 4π i q
−
− 7!1 (q q )3 + ...
|| −||
l l
We have used the Taylor expansion for convenience of evaluating derivatives at q = 0. These deriv derivative ativess can be found as follows,
||
∂g Ω ∂q l 2 ∂ gΩ ∂q k ∂q l ∂ 3 gΩ ∂q j ∂q k ∂q l ∂ 4 gΩ ∂q j ∂q k ∂q l ∂q m
−
||
2 4 1 1 ql + ql q 2 ... = + q 3! 5! 3 30 1 1 1 + q 2 δkl + ql qk , 3 30 15
=
−
=
−
|| −
||
2
ql ,
1 (qj δkl + qk δjl + ql δjk ) , 15 1 = (δjm δkl + δkm δjl + δlmδjk ) . 15 =
Now we compute compute the inte intensit nsity y of radiat radiation. ion. The flux of radiat radiation ion in the direc direction tion nk is integrate this flux through a sphere of radius R: dE = R2 dt
2
d Ω
(GW )
R2 T 0k n = 32πG 32 πG k
2
d Ω
hij ,0 hij,k
(GW ) T nk , 0k
and we need to
nk .
The perturbation hij is found from the Einstein equation. It was derived in the lecture that, in the leading order in 1 /R /R,, we have (T T ) Q ¨
ij (t
hij = 2G (T T )
Qij
− |R |) ,
|R | − 12 P
≡
P ai ai P bj bj
ab P ij ab ij
Qab .
The projection tensor P ij 9.1.. The tensor Qij is defined by ij is defined in Problem 9.1 Qij (t)
≡ 3
d x
xi xj
1 2 x δij T 00 00 (x, t) 3
− ||
and is by definition trace-free, Qii = 0. Thus we have 16πG (T T ) ¨ 16πG Rk Qij (t R ) , R R ... ... dE G = d2 Ω (T T ) Qij (T T ) Qij . dt 8π
hij,k =
−| |
It remains to compute the average over the sphere of (T T )
... (T T ) ... Qij Qij . 32
... Consider any symmetric, trace-free tensor Aij inste instead ad of Qij ; the transv transverse-t erse-tracel raceless ess part of Aij is defined by (T T )
Since Aii = 0, we have Aab P ab ab = (T T )
−A
ab na nb
Aij (T T ) Aij = =
≡
Aij
P ai ai P bj bj
−
1 P ab ab P ij ij Aab . 2
and so
P ai ai P bj bj
−
P ac ac P bd bd
= Aab Aab
−
−
1 1 P ab P ci P cd ab P ij ij ci P dj cd P ij ij Aab Acd 2 2 1 1 P ab P ac na nb nc nd Aab Acd ab P cd cd Aab Acd = ac P bd bd 2 2 1 2AacAbc na nb + Aab Acd na nb nc nd . 2
−
−
After integration over the sphere, according to formulas derived above, we have (again note that δab Aab = 0 and Aab = Aba ) 1 4π 1 4π and thus
1 1 Aac Abc δab = Aab Aab , 3 3 1 2 d2 Ω Aab Acd na nb nc nd = Aab Acd (δabδcd + δac δbd + δadδbc ) = Aab Aab, 15 15
1 4π
d2 Ω Aac Abc na nb =
2
d Ω
(T T )
Aij
(T T )
−
Aij = Aab Aab 1
... Finally, substituting Qij inste instead ad of of A Aij , we find dE G 1 = dt 2 4π
2
d Ω
(T T )
2 1 2 + 3 2 15
=
2 AabAab . 5
... (T T ) ... G ... ... Qij Qij = Qij Qij . 5
(38)
The angular brackets indicate that we must perform an averaging over spacetime domains. This means, for us, that we need to average over time (since Qij is a funct function ion only of time) time).. Ave Averaging raging is performed performed over timescales timescales larger than the typical timescale of change in the source. For instance, if the source is a rotating body, then averaging must be performed overr seve ove several ral periods of rotati rotation. on.
10
Deriv Der ivati ation: on: gra gravit vitat ation ional al wav waves in flat spac spaceti etime me
The metric is assumed to be of the form gµν = ηµν + hµν , where ηµν = diag(1, diag(1, 1, 1, 1) is the Minkowski metric for flat space and hµν is a small perturbation which is assumed to fall off to zero quickly quickly at infinit infinity y. We start with a 3+1 decomposition decomposition of the metric perturbation perturbation hµν and compute the Einstein tensor (see Problems 7.2 7.2,, 7.3 7.3,, 7.4 7.4)) in terms of the perturbation variables Φ, Ψ, etc. We also decompose the matter energy-momentum tensor T µν and obtain the Einstein equations separately for each component (8.5 (8.5). ). The result µν is that (a) the variables E , B , F i can be set to zero by choosing a coordinate system; (b) if there is no matter (vacuum) the scalar and vector components of the metric perturbation are equal to zero; (c) the tensor component hij satisfies the wave equation (31 (31). ). Solutions of the wave equation in four dimensions with retarded boundary condition can be written using the known Green’s function. For instance, if
− − −
f ((t, r) f
= A(t, r)
f ((t, R) = f
⇒
−
1 4π
d3 r
A(t
− |r − R| , r) . |r − R|
We will use this formula for f (T ) hij and A 16 16πG πG(T ) T ji . Now, we are interested in describing the radiation sent far away by a matter distribution, so we take the limit R r , and then we can approximately set
≡
≡
(T )
hij
| |≫| |
4G R
≈ −| |
d3 r (T ) T ji (t
− |r − R| , r).
Now we use a trick (See Hobson-Efstathiou-Lasenby, 17.9) to express the components T ij through T 00; it is much easier to compute with T 00 beca because use this is jus justt the energy energy density density of mat matter ter.. Con Consid sider er first the ten tensor sor T ji rather than its transverse-traceless part (T ) T ji . The trick is to write the integral (out of sheer luck)
§
3
i j
d r ∂ a ∂ b r r
ab
T
33
=2
d3 r T ij .
Then we integrate by parts and use the conservation laws (32 ( 32), ), i T j,i =
2
3
ij
d r T =
−T 00,
−T 00 r r =−
i T j,ij =
j,
3
ab
d r T
,ab
j, j
i j
≡ T ¨00;
= T 00,00
T ij ,ij =
−T ¨00.
¨0. d3 r ri rj T 0
Now, we need to obt Now, obtain ain the tra transv nsvers erse-t e-trac racele eless ss part of the tensor. tensor. In pri princi nciple ple,, we have have the formula formulass for this (se (seee Problem 8.5 8.5). ). But they are very very complicat complicated. ed. A sho shortc rtcut ut is to notice notice tha thatt the projectio projection n operator operator P ab does the job ab (Problems 9.1 and 9.2 9.2), ), at least in the leading order in 1/ R . (W (Wee are only interested interested in everything everything to leadin leadingg order in 1/ R since all smaller terms will not give any flux of radiated energy.) The result is
| |
| |
(T )
hik (R, t) = Qik (t)
2G (T T ) ¨ Qik (t R
− |R|),
| |
≡
3
d
r T 00 (r, t)
ri rk
−
1 2 r δik . 3
(39)
The tensor Qik is the quadrupole moment of energy distribution; it is a traceless and symmetric tensor. In principle, we could just use the integral
d3 r T 00 (r, t)ri rk ,
(40)
because the transv transverse-t erse-tracel raceless ess parts of (40 40)) and of Q of Qik are the same, but it is more convenient to use Qik . T ) ( Since we found the tensor perturbation hij , now we would like to compute the energy radiated in the gravitational waves. wa ves. For this we need the energ energy-mome y-momentum ntum tensor tensor of gravi gravitation tational al waves. waves. This is a rather nontrivial nontrivial object, ob ject, since in gen genera erall the gra gravit vitati ationa onall fiel field d does not ha have ve any ene energy rgy-mo -momen mentum tum tensor. tensor. In the cas casee of gra gravit vitatio ational nal waves waves in GW ) ( flat back background ground spacetime, spacetime, one can define some quan quantity tity T µν µν which looks like the energy-momentum tensor of gravitation gravi tational al waves waves (but actual actually ly is not even a genera generally lly covarian covariantt tensor). We will compute this quantity quantity below b elow.. This quantity is useful because it gives the correct value of the energy after one integrates over a large region of spacetime. The real justification for using this procedure is complicated and is beyond the scope of this introductory course of General Gener al Relativity Relativity. We will only show a heuris heuristic tic justification, justification, which is the following. following. Grav Gravitatio itation n is sensit sensitive ive to eve every ry kind of energy, because the energy-momentum tensor acts as a “source” for gravity (it is on the right-hand side of the Einstein Einste in equation). equation). So gravitation gravitation should be also sensitive sensitive to the energ energy y in gravi gravitationa tationall waves. waves. One expects that the GW ) ( energy-momentum tensor for gravitational waves, T µν µν (if we know how to compute it), will act as an additional source for gravity, gravity, like every every other energ energy-mome y-momentum ntum tensor tensor for other kinds of matter. We will guess the form formula ula for (GW ) T as follows. We can write the Einstein equation and expand it in powers of the perturbation h : µν µν µν (1)α
Gα β [ηµν + hµν ] = Gβ
(2)α
[h] + Gβ
[h] + ... = 8πGT βα .
(41)
Here G(1) is the first-order Einstein tensor, G(2) is the secon second-orde d-orderr etc. First we solve solve only to first-o first-order rder in h (this is (1) what we have been doing so far) and then we will get an approximate solution hµν : G(1) [h(1) ] = 8πGT 8 πGT βα . β α
(42)
This solution disregards the effect of gravitational waves and only takes into account the effect of matter T βα. We can can (2) try to get a more precise solution by using the second-order terms in Eq. (41 ( 41). ). Then we will get a correction h to the (1) (2) solution; soluti on; the soluti solution on g = η + h + h will be more precise. From Eq. (41 ( 41)) we find (1)α
Gβ
(2)α
[h(1) + h(2) ] + Gβ
[h(1) ] = 8πGT βα.
Now this is similar to Eq. (42 (42), ), but it looks as if there is an additional term in the energy-momentum tensor, which we may rewrite as (1)α
Gβ
(1)
[h
+h
(GW)
(2)
] = 8πG
T βα
T βα
+
1 (2) ≡ − 8πG G
α
β
(GW)
T βα
[h(1) ].
,
This motivates us to say that the EMT for gravitational waves is given by this formula. But of course this is not a real derivation because this does not show why the quantity (GW) T µν µν has anything to do with the energy carried by waves. (2)α The second-order terms Gβ are computed in Problem 9.3 9.3.. The result result is use used d to compute compute the pow power er radiate radiated d in (2)α
gravitational waves (Problem 9.4 9.4). ). Note that the calculation of Gβ uses averaging over spacetime in an essential way. Thus, the result is an averaged power radiated during a long time—much longer than the typical time scale of change in the sources—and averaged over large distances, much larger than the typical length scale of the sources. This kind of averaging is assumed in Eq. (38 ( 38). ). It remains unclear exactly how one performs averaging over space and time. One can use the formula (38 (38)) to compu compute te the grav gravitation itational al radiation emitted emitted by nonrelativistic nonrelativistic matter far aw away ay from 0 those places where the matter is cont contained ained.. The distribution distribution of the energy density density, T 0 (r, t), should should be giv given. en. The Then n one computes the tensor Qij according to Eq. (39 (39), ), by integrating integrating over over space where the matte matterr is cont contained. ained. Final Finally ly,, one ... computes the third derivative Qij , the trace, and averages over long time, as indicated in Eq. (38 ( 38). ). If we want to insert factorss of c, we replace G by Gc−9 . factor 34
WiSe06 T VI – General Relativity
Problem Sheet 1
http://www.theorie.physik.uni-muenchen.de/~serge/WS06-T6/
1
Coord Coordina inates tes and 1-forms 1-forms
1.1 Inverti Invertible ble transfo transformat rmation ionss Under what conditions is a coordinate transformation ξ α = ξ α(xα ) invertible in a neighborhood of some point xα ?
1.2 Examples Examples of coordinate coordinate transformation transformationss The following coordinate transformations are given, mapping the standard Euclidean coordinates (x, y ) or (x,y,z ) into new coordinates. 1. In a two-dimensio two-dimensional nal plane, (x, y ) → (u, v), where −∞ < u, u, v < +∞: 1 3 u , 3 1 y = v + vu 2 + v 3 . 3
x = u + uv 2 +
2. In a three-dimensio three-dimensional nal space, (x,y,z ) → (r,θ,φ), where −∞ < r < +∞, 0 ≤ θ < +∞, 0 ≤ φ < 2π: x = r sinh θ cos φ, y = r sinh θ sin φ, z = r cosh θ.
3. In a three-dimensio three-dimensional nal space, (x,y,z ) → (r,θ,φ), where 0 ≤ r < +∞, 0 ≤ θ
≤ π,
0 ≤ φ < 2π :
x = −r sin θ cos φ, y = −r sin θ sin φ, z = r cos θ.
The following questions must be answered in all three cases: (a) Find the subdomain subdomain covered covered by the new coordina coordinates. tes. Hint: Consider Consider e.g. the range of x at constant value of y . (b) Find the points where the new coordinates do not specify a one-to-one invertible transformation (sin- ( sin- gular points). points ). (c) If singular points exist, give a geometric interpretation.
1.3 Basis Basis in tangen tangentt space space Prove that the vectors eα =
∂ ∂x
α
are linearly independent.
1.4 Differentials Differentials of functions functions as 1-forms If f (xα ) is a function of coordinates xα , then one defines the 1-form df (called the differential of the function f ) as df ≡
α
∂f dxα . ∂x α
(1)
Compute d(x), d(x2 ), d(xy ), d(x + y ). Compute the 1-forms df , dg , dh, where the functions f , g , h are defined as follows, f (x,y,z ) =4x2 y + x3 z,
g (x, y ) =3
x2 + y 2 ,
h(x, y ) =arctan(x + y ) + arctan(x − y )
+ arctan
2x x2 − y 2 − 1
.
1.5 Basis Basis in cotang cotangent ent space space Show that the 1-forms dx1 , ...,dx ...,dxn comprise a basis in the space of 1-forms at any point M . Show that < dxα ,
∂ >= δ αβ . β ∂x
1.6 Linearly Linearly independen independentt 1-forms Check whether whether the following following sets of 1-forms are linearly linearly independent independent at each point point of the 2-dimension 2-dimensional al or the 3-dimensional space respectively. If not, determine the points where these sets are linearly dependent. 1. Two 1-forms d(ex cos y ), d(ex sin y ). 2. Two 1-forms (1 + y )dx − 2xydy , 8dx. 3. Three 1-forms 1-forms dx + dy , dx + dz , dy + dz . 4. Three 1-forms 1-forms dx − dy , dy − dz , dz − dx.
1.7 Transformation ransformation law for 1-forms Derive the transformation law for 1-forms, dx ˜α =
under a coordinate transformation xα
˜α →x
˜α β ∂ x dx , ∂x β
(2)
=x ˜α (xβ ).
1.8 Example Exampless of transfor transformati mations ons Consider the coordinate transformation (x, y ) 1-form,
→
(u, v) defined in Problem Problem 1.2(1). Transform ransform the follo following wing
ω=d
1 , x+y
into the coordinates (u, v ) in two ways: (a) By a direct substitution of the new coordinates. (b) By using the transformation law (2).
2
WiSe WiSe06 06 T VI – Gene Genera rall Re Rela lati tivi vitty
Supp Supple leme men ntal tal Prob Proble lem m Shee Sheett 1
http://www.theorie.physik.uni-muenchen.de/~ http://www.theorie.physik.uni -muenchen.de/~serge/WS06-T serge/WS06-T6/ 6/
2D surfaces embedded in 3D Euclidean space 1A Tangent plane Consider the surface given by z = −h exp − 2 1 (x2 + y 2 ) . If gravity acts in the negative z -direction, -direction, at what points will a ball rolling along this surface experience the greatest acceleration? Find the tangent plane at one of these points.
2 σ
1B Induced metric Find the metric for the surface given parametrically by
y
= a sin2 θ cos φ, = a sin2 θ sin φ,
z
= a cos θ sin θ,
x
where, as usual, θ defined there?
∈ [0, π) and φ ∈ [0, 2π). Is the metric well defined at θ = 0? Do you think the surface is well
1C Embedding waves 1. Sketch Sketch the surface given given by x
=
y
=
z
=
√ cos v , 2 − sin u √ sin v , 2 − sin u √ cos u , 2 − sin u
where u, v ∈ [0, 2π ). (Hint: Consider the intersection of the surface with the plane y = 0. What happens for general v ?) 2. Find the normal vector vector and the tangent plane to this surface at point ( u, v). 3. Determine Determine the induced induced metric on the surface. surface. Then consider consider the 2D vector V = (cos v, sin v), i.e. a
V = cos v
∂ ∂ + sin v , ∂v ∂u
defined within the surface. Is V a unit vector? What are the 3D Euclidean components of the vector V in the 3D space? space? Show Show that that the 3D com compone ponent ntss of the vecto vectorr V everywhere lie in the tangent plane to the surface. a
WiSe06 T VI – General Relativity
Problem Sheet 2
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2 2.1 2.1
Tensors Defin De finit ition ion of ten tenso sor r produ product ct
If ω1 and ω2 are 1-forms, their tensor product ω1 ⊗ ω2 is defined as a function on pairs of vectors: (ω1 ⊗ ω2 )(v )(v1 , v2 ) = ω1 , v1 ω2, v2 .
(1)
Let ω1 = dx + 2ydy 2ydy,, ω2 = −2dy be 1-forms on a 2-dimensional space and v1 = 3∂/∂x, ∂/∂x, v2 = −x (∂/∂x + ∂/∂y) ∂/∂y) be vector fields (also defined in this 2-dimensional space). Just for this problem, let us denote T ≡ ω1 ⊗ ω2 . (a) Compute T ( T (v1, v1 ). (b) Compute T ( T (v1 , v2 ). (c) Show that T ( T (a + λb, u) = T ( T (a, u) + λT ( λT (b, u), (2) where a, b, u are vectors and λ is a number). The same property holds for the second argument of T . T . Such functions T are called called bilinear. bilinear. (d) Show that all bilinear functions of pairs of 2-dimensional vectors belong to a vector space of such functions. Show that the tensor products dx ⊗ dx, dx, dx ⊗ dy, dy, dy ⊗ dx, dx, dy ⊗ dy form a basis in that space. (That space is called the space of tensors of rank 0 + 2.)
2.2 2.2
Gene Genera rall tens tensor orss
(a) A general tensor of rank r + s is defined as a multilinear function on sets of r 1-forms f j and s vectors vj (multilinear means linear in every argument). An example of a tensor of rank r + s is a tensor product of r vectors e1 , ..., er , and s 1-forms ω1 , ..., ωs , denoted by e1 ⊗ ... ⊗ er ⊗ ω1 ⊗ ... ⊗ ωs . This tensor is a function that acts on a set of r of r 1-forms f j and s vectors vj via the formula e1 ⊗ ... ⊗ er ⊗ ω1 ⊗ ... ⊗ ωs (f 1 ,...,f r , v1 ,..., vs ) = f 1 , e1 ... f r , er ω1 , v1 ... ωs , vs . (This is a generalization of Eq. (1) to tensors of rank r + s.) Show that this function is linear in every argument. Such functions are called r + s-linear functions. Show that all r + s-linear functions form a vector space. This vector space is called the space of tensors of rank r + s. (b) Let ej , j = 1,...,N , and ω j , j = 1,...,N are bases in the space of vectors and in the space of 1-forms respectively (both spaces have dimension N ). N ). Show that the set of tensors eα ⊗ ... ⊗ eα ⊗ ω β ⊗ ... ⊗ ω β 1
1
s
r
(3)
form a basis in the space of r + s-tensors (where αj and β j exhaust all possible combinations of indices). Note that this set contains nr+s basis tensors.
2.3 2.3
Exa Examp mple le
(a) Let T be a bilinear function of two vectors with vector values, i.e. T(v1 , v2 ) is a vector if v1 , v2 are vectors. Give a simple example of such T as a tensor and determine its rank. (b) A particular example of such a tensor T in 3-dimensiona 3-dimensionall Euclidean Euclidean space is the following following function, T(v1 , v2 ) = 2v 2 v1 × v2 − v1 (n · v2 ),
(4)
where n is a fixed vector. Show that the function T is bilinear in v1 , v2. Determine the components T αβγ of the tensor T in an orthogonal basis where the vector n has the components nα ≡ (n1 , n2 , n3 ). 1
2.4
Transf ransform ormati ation on law law
Derive Derive the transformati transformation on law for the components components T α
1
2.5
...α
β1 ...β
r
s
of a tensor of rank r + s.
Contra Contractio ctions ns of tenso tensor r indic indices es
(a) Show that the results of addition, multiplication by scalar, tensor multiplication, and index contraction of tensors are again tensors. Use the definition of tensor from Problem 2.2. (b) Show that a contraction of indices in the same position (e.g. lower indices with lower indices, T ααβ ααβ ) does not generally yield a tensor. (c) Consider the tensor T αβγ defined in Problem 2.3(b) and compute the contraction T ααβ . Is the result a tensor? If so, determine its rank.
2.6
Inv Invarianc ariance e of the the int interv erval al
Show that the spacetime interval ds2 ≡ gαβ dxα dxβ is invariant under coordinate transformations xα → x ˜α if gαβ are components of a tensor transforming according to the tensor transformation law gαβ → g˜αβ =
2.7
∂x µ ∂x ν gµν . ∂ ˜ ∂ x ˜α ∂ ˜ ∂ x ˜β
(5)
Corres Correspond pondenc ence e between between vect vectors ors and and 1-forms 1-forms
For a given metric gαβ , each vector v has a corresponding 1-form which we shall denote ω . This 1-form is defined by its action on an arbitrary vector x as follows, v
ω , x = v · x, v
(6)
where the scalar product v · x is defined through the metric gαβ . Show that the components of the 1-form ω in the basis dxα are related to the components of the vector v in the basis ∂/∂xα by ωα = gαβ vβ .
2.8 2.8
v
(7)
Exam Exampl ples es of spa space cess with with a metr metric ic
(a) Consider Consider the usual, Euclidean Euclidean 3-dimensiona 3-dimensionall space with the metric g (v1 , v2 ) = v1 · v2 − (n × v1 ) · (n × v2 ),
(8)
where v1 · v2 is the usual scalar product, a × b is the cross product, and n is a fixed vector with components nα . Compute the components of the tensor gαβ . For which vectors n is the metric g nondegenerate (i.e. det gαβ = 0)? (b) Answer the same questions for the 2-dimensional Euclidean space with the metric g (v1 , v2 ) = 3v 3 v1 · v2 + (n (n · v1 )(n )(n · v2 ).
(9)
Note that the cross product is undefined in the 2-dimensional space. (c)* Answer the same questions for the metric (9) now defined in an r-dimensional Euclidean space, r ≥ 3. (d)* Consider Consider a 2-dimensional 2-dimensional surface embedded embedded in the 3-dimensional 3-dimensional Euclidean Euclidean space, x = R cosh u cos v, y = R cosh u sin v,
(10) (11)
z = R sinh u.
(12)
Determine the 2-dimensional metric gαβ in the basis du,dv. du,dv.
2
WiSe06 WiSe06 T VI – General General Relativit Relativity y Supplemen Supplemental tal Problem Problem Sheet 2
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Calculations with tensor indices 1A Vector equations In the following equations, the vector xα is unknown unknown and all other quantities quantities are known. The symbol εαβγ denotes the completely completely antisymmetri antisymmetricc tensor. Determine Determine the unknown unknown vector xα from the given given data. In every every case, assume assume the “generic” “generic” choice choice of data. This means means that every every β given scalar, vector and tensor is nonzero (k, Aα , B , ...), there are no accidental cancellations or linear dependence between given vectors, matrices are nondegenerate, etc. (a) kxα + εαβγ xβ Aγ = B α (3-dimensional vectors). The assumption of the “generic” case is k= 0 and Aα and Bα linearly independent. (b) εαβγ xβ Aγ = Bα , xα C α = k (3-dimensional vectors). (c) xα Aα = k , xβ Bβ = l (2-dimensional vectors). (d) xα Aαβ = Bβ (3-dimensional vectors and a given tensor Aαβ ).
1B Tensor equations In the following equations, the tensor X αβ is unknown and all other quantities are known. The dimens dimension ionali ality ty of the (Euclid (Euclidean) ean) space space is indicat indicated. ed. Determi Determine ne X αβ under the assumption that all given quantities are generic. (a) X αβ = X βα , X αβ Aα = B β , X αα = 0, X αβ Bα = 0, where Aα Bα = 0 (2-dimensiona (2-dimensional). l). αβ βα αβ β αβ α (b) X = −X , X Aα = B , X Bα = 0, where A Bα = 0 (3-dimensional vectors).
1C Degeneracy of the metric (a) A two-dimensional space with coordinates (x, y ) has the metric given as a bilinear form g = y 2dx ⊗ dx + (x2 + 1)(dx ⊗ dy + dy ⊗ dx).
(1)
Is the metric nondegenerate at all points (x, y )? (b) The same question for the n-dimensional metric of the form
gαβ = δαβ − 1 + r
2
Aα Aβ ,
where Aα is a given vector and r2 ≡ δαβ xα xβ is the squared Euclidean distance.
(2)
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3 3.1
The The Chri Christ stoff offel el sym symbol
µ
Γαβ
Transf ransform ormati ations ons 1
In flat space with standard Euclidean coordinates ξ µ and arbitrary coordinates xµ = xµ (ξ α ), the Christoffel symbol can be found as µ Γαβ =
∂ 2 ξ ν ∂x µ . ∂x α ∂x β ∂ξ ν
µ Derive the transformation law for Γ αβ between arbitrary coordinate systems xµ and x ˜µ :
∂ 2 xσ ∂ x ˜α ∂x ρ ∂x σ ˜α ˜ α = Γµ ∂ x . Γ + βγ ρσ ˜β ∂ x ˜γ ˜β ∂ x ˜γ ∂x σ ∂x µ ∂ x ∂ x
3.2
(1)
Transf ransform ormati ations ons 2
Show that the Christoffel symbol must transform according to Eq. (1) not only in flat space but also in arbitrary space. Hint: consider the covariant derivative of a vector field, Aα;β =
∂A α µ Aµ , − Γαβ β ∂x
and demand that the components Aα;β transform as a tensor.
3.3
Covaria Covariant nt deriv derivatives atives
Derive the explicit form of the covariant derivative T αβγδµ ;ν
for a tensor field T αβγδµ .
3.4 3.4
The The Leib Leibni nitz tz rule rule
Prove the Leibnitz rule in the following specific case,
A
αB
3.5
β
;γ
= Aα;γ B β + Aα B β;γ .
Locally Locally iner inertia tiall refer referenc ence e frame frame
α α Suppose that the Christoffel symbol at a point x(0) in some coordinate system xα has the value Γ(0) and is µν α α symmetric, Γ(0)µν = Γ(0)νµ . Then a locally inertial system at point x0 can be constructed by defining the new coordinates 1 µ µ α ν α + Γ(0) ξ α (x) = xα − x(0) x − x(0) xν − x(0) µν . 2 The point x0 in the new coordinates is the origin ξ α = 0. Prove explicitly that the Christoffel symbol, when transformed to the new coordinates, is equal to zero at the point ξ α = 0.
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4 4.1 4.1
Geode Geodesi sics cs and and curv curvat atur ure e Geod Geodes esic icss
(a) Show that the geodesic equation can be written in the following form, duα 1 ∂g βγ β γ u u = 0. − 2 ∂x α ds
(1)
(b) Show that gαβ uαuβ is constant along a geodesic.
4.2
Comm Commutator utator of cov covarian ariantt deriv derivative ativess
Show that uα ;β ;γ − uα ;γ ;β = Rα δγ β uδ ,
(2)
where the Riemann tensor is defined by ∂ Γα δβ ∂ Γα δγ R δγβ = − + Γα σγ Γσ δβ − Γα σβ Γσ δγ . γ β ∂x ∂x α
4.3 4.3
(3)
Paral aralle lell trans transport port
Consider a vector Aα parallel-trans parallel-transported ported along a small closed curve xµ (s). Show that the change in Aα after the parallel transport can be approximately expressed as δAα ≡
γ Γβ αγ (x)Aβ dx
≈
1 δ R αβγ Aδ 2
xβ dxγ ,
(4)
where it is assumed that the area within the closed curve is very small. Hint:
α Use a locally inertial coordinate system where Γβγ = 0 at one point. Also, show that
α
β
x dx = −
xβ dxα .
(5)
4.4 4.4
Riem Rieman ann n tens tensor or
(a) Using the symmetry properties of the Riemann tensor Rαβγδ , compute the number of independent components of Rαβγδ in an n-dimensional space (n ≥ 2). (b) Prove the Bianchi identity: Rα βγ δ;σ + Rα βσγ ;δ + Rα βδσ ;γ = 0. (c) Compute the Einstein Einstein tensor Gα β in an arbitrary two-dimensional space. the independent components of Rαβγδ .
4.5 4.5
Hint:
First determine
Lore Lo ren ntz trans transfo form rmati ation onss
Determine the number of independent parameters in Lorentz transformations x˜µ = Λµα xα , given by matrices Λβ α , and interpret these parameters. Hint: It is easier to consider infinitesimal Lorentz α transformations transformations Λβ = δβ α + εH β α, where ε ≪ 1 and so ε2 can be disregarded.
2
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5 5.1 5.1
Gra Gravita vitati tion on the theor ory y appl applie ied d Reds Redshi hift ft
Calculate the gravitational redshift at the surface of the Earth for the vertical distance of 1m between the sender and the receiver. Same question for 1 km.
5.2
Energy Energy-m -mome oment ntum um tensor tensor 1
Rewrite the conservation law T αβ ;β = 0 explicitly in the nonrelativistic limit for an ideal fluid, and show that these equations equations coincide with the continui continuity ty equation and the Euler equation. equation.
5.3
Energy Energy-m -mome oment ntum um tensor tensor 2
The EMT for a massless scalar field is T α β = Φ ;α Φ;β
− 12 δ α β Φ;γ Φ;γ .
Show (using the conservation law) that the equation of motion the field is Φ ;α ;α = 0.
5.4 5.4
Weak eak gra gravit vity
Show that in the limit of weak static gravitational field (g00 = 1 + 2Φ(x,y,z ), and gµν is independent of t) the following relation holds, R00 ≈ ∆Φ + O(Φ2 ) , where ∆ is the ordinary Laplace operator, ∆
5.5
≡ ∂ xx xx + ∂ yy yy + ∂ zz zz .
Equati Equations ons of motion motion from from conser conserv vation ation law law
The EMT for a point particle of mass m0 moving along a worldline xσ (s) can be expressed as µν
T
=
1
√−g m0
ds
dxµ dxν (4) σ δ (x ds ds
− xσ (s)) .
Show that the conservation law T µν ;ν = 0 implies the geodesic equation for xσ (s). Hint:
First derive the relations relations
√1−g ∂x∂ µ (√−g) , ∂ √ 1 µν µ ρσ T µν ;ν = √ −g ∂x ν ( −gT ) + Γ ρσ T . Γν µν =
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6 6.1 6.1
The The gra gravita vitati tion onal al field field Degr De gree eess o off free freedo dom m
Using the scheme developed in the lecture, compute the number of degrees of freedom in the electromagnetic field, taking into account the presence of charges and currents.
6.2
Spheri Spherical cally ly symmetr symmetric ic spaceti spacetime me
α Compute the Ricci tensor Rβ and the curvature scalar R for a spherically spherically symmetric gravitational gravitational field. Assume that the metric has the form ν (t,r )
gµν
e 0 = 0 0
0
−eλ(t,r) 0 0
0 0 0 0 2 0 −r 2 0 −r sin2 θ
.
Write the corresponding Einstein equations in vacuum (T (T µν µν = 0).
6.3
Motion Motion in in Schw Schwarz arzsc schil hild d space spacetim time e
Derive the equation for the covariant component u1 of the 4-velocity of a particle in Schwarzschild spacetime (u (u1(λ) ≡ −f −1 (r )r˙ , f ( f (r ) = 1 − rg /r). /r). Verify that this equation follows from Eqs. (1)–(4) given in the lecture: f t˙2 − f −1 r˙ 2 − r2 θ˙2 − r 2 sin2 θφ˙ 2 d f t˙ dλ d −r 2 θ˙ + r2 sin θ cos θφ˙ 2 dλ d r 2 sin2 θφ˙ dλ
(1)
= 0,
(u0)
(2 )
= 0,
(u2)
(3 )
= 0,
(u3)
(4 )
= K (= uα uα ),
where the overdot (˙) denotes d/dλ and the spherical coordinates are {x0, x1 , x2 , x3 } ≡ {t,r,θ,φ }.
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7 7.1 7.1
Weak eak gra gravita vitati tion onal al field fieldss Gra Gravita vitati tiona onall bendin bending g of ligh lightt
Verify that the gravitational bending of light passing near the Sun is δ = 1.75′′
R⊙ R
where R is the distance at which the light passes from the center of the Sun and R⊙ is the radius of the Sun.
7.2 7.2
Einst Einstei ein n tenso tensor r for for wea weak k field field
Derive the following expression for the Einstein tensor due to a weak gravitational field, Gµ ν =
1 ¯ µ ,α ¯ α β ,β ,α + h ¯ µ α ,α ,ν + h ¯ α ν ,µ ,α + O (h2 ), −h ν ,α − δ µ ν h 2
(1)
¯ µ ν = hµν − 1 δ µ ν h. where h 2
7.3
Gravit Gravitati ationa onall perturba perturbatio tions ns I
Derive the expressions (shown in the lecture) for the Einstein tensor Gµν in terms of the scalar, vector, and tensor perturbations of the gravitational field. The background is the flat Minkowski spacetime, (0) gαβ = ηαβ , and the metric is g00 = 1 + 2Φ, g0i = B,i + S i ,
7.4
gij = −δij + 2Ψ δij + 2E ,ij ,ij + F i,j i,j + F j,i + hij .
(2)
Gravit Gravitati ationa onall perturba perturbatio tions ns II
Derive Derive the transformation laws laws for the scalar, vector, vector, and tensor perturbations of the gravitationa gravitationall field, under an infinitesimal change of the coordinates, x ˜µ = xµ + ξ µ (x).
(3)
i Note: It is convenient to decompose ξ µ as ξ µ = (ξ 0, ξ⊥ + ζ ,i ), where ξ 0 and ζ are scalar functions i and ξ⊥ ,i = 0.
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8 8.1
Gra Gravita vitati tion onal al radi radiat atio ion n I Gauge Gauge invaria invariant nt variables ariables
¨ and S i − F ˙ i, are gauge-invariant. Verify that the following combinations of metric perturbations, D = Φ − Ψ − B˙ + E
8.2
Detecting Detecting gravitati gravitational onal waves waves
Light noninteracting particles are situated in the x − y plane in free space. A plane gravitational wave propagating in the z direction passes through the ring. The metric is of the form gµν = ηµν + hµν , where hµn contains only the pure tensor component, 0 0 0 0 0 A+ A 0 hµν = exp[− exp[−iω (t − z )] . (1) 0 A −A+ 0 0 0 0 0
×
×
Describe Describe the deformation deformation of the shape of the ring due to the gravitation gravitational al wave. wave. Consider Consider cases A+ = 0, A = 0 and A+ = 0 , A = 0. ×
×
8.3 8.3
Pois Poisso son n equa equati tion on
Derive Derive the solution of the following differential differential equation, ∆φ(x) = 4πρ (x),
(2)
with boundary conditions φ → 0 at |x| → ∞.
8.4 8.4
Metr Metric ic pertu perturb rbat ation ionss 1
Determine an explicit expression for α through T 0 i,i, where ∂ i α represents the scalar part of T 0 i .
8.5 8.5
Metr Metric ic pertu perturb rbat ation ionss 2
Verify that the equation
1 ˙ ¨ − (S (3) i − F i ) = 8πGσ i 2 which which follows follows from vector vector part of the spatial spatial Einstein Einstein equation, also follows from other components components of the Einstein equation and from the conservation law (as derived in the lecture).
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9 9.1 9.1
Gra Gravita vitati tion onal al radi radiat atio ion n II Projec Projectio tion n of the the matt matter er tens tensor or
The projection operator P ij ij is defined by ni ni = 1,
P ij ij = δij − ni nj ,
Show that the projected tensor (T )
(T )
ni ≡
Ri . R
(1)
X ik ik (t, |R|) defined by
1 X ik ik = P ia ia X ab ab P bk bk − P ik ik P ab ab X ab ab , 2
X ik ik (t, R) ≡
d3 r ri rk T 00 00 (t − |R| , r),
(2)
has the following following properties, a) ( ) X ii ii = 0;
(3)
T
−1
b) ( ) X ik,i ik,i = O (X |R| T
that is,
9.2 9.2
(T ) X
ik ik
(4)
is transverse-traceless up to terms of order |R|−1 .
Matt Matter er sour source cess
Verify that
(T )
( ) X ik Qik , where ik = T
(T )
X ik ik (t − |R|) is the projected tensor defined in Problem 9.1 and (T )
1 P ik ik P ab ab Qab , 2
(5)
1 (ri rk − δik r2 ) T 0 0 d3 r. 3
(6)
Qik = P ia ia Qab P bk bk −
Qik ≡
9.3
),
Energy-m Energy-momen omentum tum tensor tensor of gra gravitati vitational onal wav waves es
Compute the second-order terms G(2)α β , i.e. terms quadratic in hµν , of the Einstein tensor Gα perturbaβ for small perturba( ) tions in flat space, gµν = ηµν + hµν , where only the transverse and traceless part hik is nonzero. Verify that the energy-momentum tensor of gravitational waves in vacuum ( T µν µν = 0 for matter) is T
(GW)
9.4
T α β ≡ −
1
G = (2)α
8πG
β
1 32πG
(T )
hi k ,α ( ) hi k ,β . T
(7)
Powe Power r of emitte emitted d radia radiatio tion n
Show that the rate of energy loss (energy lost per unit time) is ... ... dE G G ... ... d2 Ω( ) Qik ( ) Qik = − Qik Qik . =− 8π 5 dt
T
T
(8)
Here the integration goes over all directions ni in 2-sphere. In the calculation, derive and use the following relations, d2 Ω 1 = δlm , 4π 3 2Ω 1 lm kr d = (δ δ + δlk δ mr + δlk δmr ) . nl nm nk nr 4π 15
nl nm
(9) (10)
