general relativity problem set 1
DESCRIPTION
Solutions to problems in Wald's General Relativity typed up in latexTRANSCRIPT
Problem Set 1
Aaron HillmanPHYS 538
February 2, 2015
Problem 1.
(a) Due to symmetry we can consider one i and j and for f+i ◦ f+−1
j and the proof will be
complete. Consider f+1 ◦ f+−1
2 . This map is clearly
f+1 ◦ f+−1
2 : (x1, x3) 7→(√
1− (x21 + x23), x3
)Since the preimage of this map is the open unit disc, we do not have x21 + x23 = 1, thereforeall partials with respect to the first output are continuous. Clearly all derivatives of x3 arealso continuous. Thus the overlap, and all other overlap maps, are C∞.
Problem 2.
(a) We have that the commutator for two vector fields X and Y and a real valued functionf has the form
[X, Y ](f) = X[Y (f)]− Y [X(f)]
We can test the linearity property by considering the following equation and using thelinearity of vector fields
[X, Y ](af + bg) = X[Y (af + bg)]− Y [X(af + bg)]
= X[aY (f) + bY (f)]− Y [aX(f) + bX(f)]
= aX[Y (f)] + bX[Y (f)]− aY [X(f)]− bY [X(f)]
Thus the commutator is linear. We can now see that the commutator satisfies the Leibnizrule at some point p in a manifold M
[X, Y ](fg) = X[Y (fg)]− Y [X(fg)]
= X[f(p)Y (g) + g(p)Y (f)]− Y [f(p)X(g) + g(p)X(f)]
= f(p)X[Y (g)] + g(p)X[Y (f)]− f(p)Y [X(g)]− g(p)Y [X(f)]
1
Thus the Leibniz rule is satisfied.(b) Consider
[[X, Y ], Z] = [XY − Y X,Z] = (XY − Y X)Z − Z(XY − Y X)
= XY Z − Y XZ − ZXY + ZY X
We can then consider [[Z,X], Y ] which is
ZXY −XZY − Y ZX + Y XZ
Lastly we have [[Y, Z], X]
Y ZX − ZY X −XY Z +XZY
Summing this three commutators, we have that the terms cancel pairwise like so
(ZXY − ZXY ) + (XY Z −XY Z) + (Y XZ − Y XZ) + (ZY X − ZY X) = 0
Thus
[[X, Y ], Z] + [[Z,X], Y ] + [[Y, Z], X] = 0
(c) First consider the commutator
[[Yα, Yβ], Yδ] =
[∑γ
CγαβYγ, Yδ
]By linearity we can ”take out” the sum from the bracket and now we can apply the sumdefinition again to yield
[[Yα, Yβ], Yδ] =∑γ,µ
CγαβC
µγδYµ
If we then consider the Jacobi identity proved in part (b), we can write∑γ,µ
CγαβC
µγδ + Cγ
δαCµγβ + Cγ
βδCµγα = 0
Problem 3.
(a) In any coordinate basis {∂/∂xi} we have for two vector fields, V and W that
V [W ] =∑µ
V [W ]µ =∑α
∑ν
V ν ∂
∂xν
(Wα ∂
∂xα
)Thus if we desire the µth component of V [W ] with respect to the basis then we have
V [W ]µ =∑ν
V ν ∂Wµ
∂xν+ V νW µ ∂
∂xν
2
We have that W [V ] is merely the operation flipped, and the cross term with the product ofV and W cancel, thus the µth component of V [W ] −W [V ] and thus of the commutator ofthe two fields is
[V,W ]µ =∑ν
(V ν ∂
∂xνW µ −W ν ∂
∂xνV µ
)
(b) First contract the right side of the equation to get∑α,β
Cγαβ(Y α∗)µ(Y β∗)ν(Yσ)µ(Yρ)
ν = Cγσρ
We can now contract the left side of the equation to get
∂(Y γ∗)µ∂xν
(Yσ)µ(Yρ)ν − ∂(Y γ∗)ν
∂xµ(Yσ)µ(Yρ)
ν
Note the following: ∂ν((Yγ∗)µ(Yσ)µ) = ∂ν((Y
γ∗)µ)(Yσ)µ + ∂ν((Yσ)µ))(Y γ∗)µ = 0. This allowsus to write
(Yσ)µ∂(Yρ)
ν
∂xµ(Y γ∗)ν − (Yρ)
ν ∂(Yσ)µ
∂xν(Y γ∗)µ
This is expression is equivalent to [Yσ, Yρ]µ(Y γ∗)µ =
∑γ
Cγσρ(Y
γ∗)µ(Y γ)µ = Cγσρ. We have
those shown that both sides are equal under the same contraction, showing that both sidesare equal.
Problem 4.
Consider existing coordinates in Rn, (x1, ..., xn) and a chart ψ which maps from an open setO containing p to an open set U ⊂ Rn. Also, consider coordinates (y1, ..., yn). We can usethe coordinate transformation law to find the dual basis to the tangent space containing pfor the y coordinates. This law states
dyµ =∂yµ
∂xνdxν
If we are to have that Yµ = ∂∂yµ
. Then we must also have Y µ = dyµ, as the vector basis
directly induces a dual basis. We can then consider the νth component of dyµ with respectto the basis {dxν}. This is
(dyµ)ν = (Y µ)ν =∂yµ
∂xν
We can then consider the fact provided in the hint, which states that if
(Y µ)ν =∂yµ
∂xν
3
is to have a solution, then we must have
∂(Y µ)ν∂xσ
=∂(Y µ)σ∂xν
If we then invoke the equation proved in 4b, then we know we must have
∂(Y µ)ν∂xσ
− ∂(Y µ)σ∂xν
=∑α,β
Cµαβ(Y β∗)ν(Y
β∗)σ = 0
We can show the middle portion is equal to 0. Consider the equation from 3c (where thereis no relationship between variables used below and those in the above paragraph)
[Yα, Yβ] =∑γ
CγαβYγ
[Yα, Yβ](Y γ∗) =∑γ
CγαβYγ(Y
γ∗) = Cγαβ
[Yα, Yβ](Y γ∗)(Y α∗)µ(Y β∗)ν =∑α,β
Cγαβ(Y α∗)µ(Y β∗)ν
We know by construction in the problem that for all α, β we have [Yα, Yβ] = 0, therefore weknow ∑
α,β
Cγαβ(Y α∗)µ(Y β∗)ν = 0
Knowing this, we have satisfied the condition
∂(Y µ)ν∂xσ
=∂(Y µ)σ∂xν
And we know that there exist coordinates (y1, ..., yn) such that Yµ = ∂∂yµ
in a neighborhoodof any p ∈M .
Problem 5.
We begin with the tensor transformation law for a (0, 2) tensor. This takes the form
g′
µ′ν′ =∑µ,ν
∂xµ
∂x′µ′∂xν
∂x′ν′ gµν
For Euclidean 3-space we know that the components of the metric tensor can be representedby gµν = δµν where δµν is the Kronecker delta. Using this, and the equation for the metric interms of its components we have
g′= ds2 =
∑µ
(∂xµ
∂x′µ′
)2
(dx′µ′
)2
4
Applying this to a transformation for Cartesian to spherical coordinates, we can considerthe relation
x = r sin(θ) cos(φ)
y = r sin(θ) sin(φ)
z = r cos(θ)
Considering the partials, and the equation for the metric above we have
ds2 =
(sin2(θ)(cos2(φ) + sin2(φ) + cos2(θ)
)dr2 +
(r2 sin2(θ)(cos2(φ) + sin2(φ))
)dφ2+(
r2 cos2(θ)(cos2(φ) + sin2(φ)) + r2 sin( θ)
)dθ2
=dr2 + r2dθ2 + r2 sin2(θ)dφ2
Problem 6.
First we can consider the equivalent rotation relation in cylindrical coordinates. Thisyields
t = t′
r = r′
φ = φ′+ ωt
We know from 8a that in matrix form this metric (ignoring the z coordinate) is−1 0 00 1 00 0 r2
If we now transform to rotating coordinates and consider the tensor transformation law,noting that the non-rotating metric has no off diagonal components
gij =∂xµ
∂x′ i∂xµ
∂x′jgµµ
The only components sharing dependence are φ and t thus they will be the only aspectschanged by the metric. The new metric, in cylindrical is
ds2 = −(1− r2ω2)dt2 + dr2 + r2dφ2 + 2r2ωdtdφ
Since this cylindrical metric has already accounted for the rotation of the frame, we canconvert to cartesian using the tradition relations below
φ = tan−1(y
x
)r =
√x2 + y2
5
We have, considering the total differential
dφ =xdy − ydx
r2
dr =xdx+ ydy
rdt = dt
We can now plug this differentials into the rotating cylindrical metric
ds2 = −(1− r2ω2)dt2 +
(xdx+ ydy
r
)2
+(xdy − ydx)2
r2+ 2ω(xdy − ydx)dt
We can collect terms and reintroduce the unchanged z differential to yield
g = ds2 = −(1− r2ω2)dt2 + dx2 + dy2 + dz2 + 2ωxdydt− 2ωydxdt
We can find the inverse metric by inverting the matrix representing the metric. This matrixis
g =
−(1− r2ω2) −wy wx 0−wy 1 0 0wx 0 1 00 0 0 1
We can row-reduce the metric to take the form of the identity, and perform the same op-erations on the identity matrix to yield the inverse metric. These operations, in order, areR1 = −(R1 +ωyR2−ωxR3), R2 = R2 +ωyR1, R3 = R3−ωxR1. We have that the invertedmatrix is
g−1 =
−1 −wy wx 0−wy 1− w2y2 w2xy 0wx w2xy 1− w2x2 00 0 0 1
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