general relativity problem set 1

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Problem Set 1 Aaron Hillman PHYS 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 :(x 1 ,x 3 ) 7 q 1 - (x 2 1 + x 2 3 ),x 3 Since the preimage of this map is the open unit disc, we do not have x 2 1 + x 2 3 = 1, therefore all partials with respect to the first output are continuous. Clearly all derivatives of x 3 are also 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 function f 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 the linearity 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 Leibniz rule 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

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Solutions to problems in Wald's General Relativity typed up in latex

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Page 1: General Relativity Problem Set 1

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)]

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Page 2: General Relativity Problem Set 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

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Page 3: General Relativity Problem Set 1

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

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Page 4: General Relativity Problem Set 1

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

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Page 5: General Relativity Problem Set 1

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

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Page 6: General Relativity Problem Set 1

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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