General Relativity: HW#8 Solutions

Katrin Schenk

Problem 1.

Part a. Let the transformation from the coordinate system {x

} to the system {x

} be

x

= x

(x

). (1)

Define the matrices

Λ

≡ ∂x

∂x

(2)

and

Λ

≡ ∂x

∂x

, (3)

which satisfy Λ

Λ

= δ

. It follows that

∂

= Λ

∂

. (4)

From the definition of the connection coefficients we now obtain

∇

∂

= Γ

∂

= ∇

β∂_λ)

(Λ

∂

) = Λ

∇

(Λ

∂

)

= Λ

(Λ

∂

+ Λ

Γ

∂

)

= Λ

(Λ

Λ

+ Λ

Λ

Γ

)∂

, (5)

and thus

Γ

= Λ

Λ

Λ

Γ

+ Λ

Λ

Λ

, (6) which can be rewritten as

Γ

= ∂x

∂x

∂x

∂x

∂x

∂x

Γ

+ ∂x

∂x

∂

x

∂x

∂x

. (7)

Part b. As an example, we will calculate one of the connection coefficients, Γ

, by transforming this connection coefficient from Cartesian coordinates [x

= (x, y, z)] to spherical polar coordinates [x

= (r, θ, ϕ)] using the result of Part (a). In Cartesian coordinates, the connection vanishes, so only the second term in Eq. (6) contributes. Hence we obtain

Γ

= ∂x

∂x

∂

x

∂x

∂x

. (8)

Now using the transformation (x, y, z) = (r sin θ cos ϕ, r sin θ sin ϕ, r cos θ) we get

Γ

= ∂r

∂x

∂

x

∂ϕ∂ϕ

= ∂r

∂x

∂

x

∂ϕ∂ϕ + ∂r

∂y

∂

y

∂ϕ∂ϕ

= x

r

∂

x

∂ϕ∂ϕ + y r

∂

y

∂ϕ∂ϕ

= − (x

+ y

) r

= −r sin

θ. (9)

As a check, lets make sure this matches the Γ

calculated in the standard way from the metric in spherical coordinates

ds

= dr

+ r

dθ

+ r

sin

θdϕ

. (10)

From this metric we get

Γ

= 1

2 g

(g

+ g

− g

)

= − 1

2 g

g

= −r sin

θ. (11)

The other connection coefficients are

Γ

= −r

Γ

= − cos θ sin θ Γ

= 1

r Γ

= cot θ Γ

= 1

r , (12)

all others being zero except those related to the above coefficients by interchanging the two covariant indices. To calculate the divergence of a vector in this coordinate system we write

∇

v

= v

+ Γ

v

= v

+ v

+ v

+ Γ

v

+ Γ

v

+ Γ

v

= v

+ v

+ v

+ Γ

v

+ Γ

v

+ Γ

v

= v

+ 2

r v

+ v

+ cot θv

+ v

= 1

r

∂

∂r (r

v

) + 1 sin θ

∂

∂θ (sin θv

) + ∂v

∂ϕ . (13)

Question: Why is this answer different than the result for ∇ · v that you would find, say, on the back cover of Jackson?

Problem 2:

Part a. We want to prove that if g is a non-degenerate, symmetric, covariant, two index tensor on a vector space V , that one can always find a basis θ

of V

such that g = g

θ

⊗ θ

, where g

is diagonal with each element being ±1. [Note that the hypothesis of symmetry was omitted from the question, it should have been included]. From the fact that g is symmetric and non-degenerate we know that it is diagonalizable and has a set of real, non-zero eigenvalues {λ

} α = 1, . . . , n, and a corresponding set of linearly independent eigenvectors, {~v

}, such that

g(~ v

, ~ v

) = δ

λ

. (14)

By linearity we are free to choose the normalization of these eigenvectors. If we choose

~ v

≡ ~ v

q

|λ

| (15)

then we get

g(~ v

, ~ v

) = ±δ

. (16) From the orthonormal basis vectors, {~v

} we can construct the dual basis {θ

} in the usual way, from which the result follows.

Part b. Pick an arbitrary coordinate system {x

} in a neighborhood of P with x

( P) = 0.

Then, the basis vectors ~ e

can be written as

~ e

= e

(x

) ∂

∂x

.

Define a new coordinate system x

to be linearly related to the old coordinate system, via x

= e

(0)x

,

where the matrix appearing in this equation is e

evaluated at the point P. It now follows that

~

e

( P) = ∂

∂x

!

P

. (17)

Part c. (Solution by Takemi Okamoto.) Let g

be a metric on a manifold M and let T

(M ) be the tangent space at a point P. From part (a) we know that there exists a basis θ

of the dual space T

(M )

, such that g = g

θ

⊗ θ

with g

diagonal with elements

±1. Let the basis { ~ e

} be the basis of the tangent space T

(M ) that is dual to the {θ

}.

(i.e. θ

( ~ e

) = δ

) By part (b), given this basis we can always find a coordinate system {x

}

such that Eq. (17) holds. Thus, given a metric and a point P, we can find coordinates {x

}

such that g

( P) is diagonal with diagonal elements ±1.

Part d. (Solution by Jeremy Darling.) Lets start off in a coordinate system {x

} with x

( P) = 0. Consider now any other coordinate system {x

} related to the first by some smooth transformation

x

= x

(x

). (18)

We can perform a Taylor expansion and write the mapping (18) as

x

= a

+ b

x

+ C

x

x

+ O(x

), (19) for some constants a

, b

, and C

. We now use the transformation law

g

= g

∂x

∂x

∂x

∂x

(20)

to calculate g

. Using the comma notation for derivatives (i.e.

≡ x

), we obtain from Eq. (19) the components of the transformation matrix

x

= b

+ 2C

x

+ O(x

), (21) where we have used the fact that C

is symmetric on α and β. Inserting Eq. (21) into Eq.

(20) we find

g

= g

(b

+ 2C

x

+ O(x

))(b

+ 2C

x

+ O(x

))

= g

(b

b

+ 2b

C

x

+ 2b

C

x

+ O(x

)). (22) Upon taking a derivative and evaluating at the point P we get

g

( P) = g

( P)b

b

+ g

( P)(2b

C

+ 2b

C

). (23) Now without loss of generality we can take g

( P) = η

. Lets choose b

= δ

. This yields

g

( P) = M

+ 4η

C

+ 4η

C

, (24) where M

is a fixed set of numbers. Setting g

( P) = 0 gives us an equation that we can solve to determine C

uniquely. The solution is

C

= 1

8 η

[ −M

− M

+ M

] .

Problem 3 Part a. (Solution due to Paul Wiggins.) The equation of motion for particles in circular motion is

d

~ x

dt

= a = −v

e

r . (25)

From the definition of r = r sin θ cos φˆi + r sin θ sin φˆ j + r cos θˆ k we can calculate the components of ¨ r ≡ a to be (setting r = 1 and thus ˙r = 0)

a

= − ˙θ

− sin

θ ˙ φ

(26)

a

= θ ¨ − sin θ cos θ ˙φ

(27)

a

= φ + 2 cot θ ˙ ¨ θ ˙ φ. (28)

(Note that we have resolved a in terms of the coordinate spherical basis, NOT the orthonormal one you are probably more familiar with.) Using the expressions (26)–(28) for the acceleration components in the equation of motion (25), we obtain

0 = θ ¨ − sin θ cos θ ˙φ

0 = φ + 2 cot θ ˙ ¨ θ ˙ φ. (29)

Now, using the geodesic equation

¨

x

+ Γ

˙x

˙x

= 0 (30)

as a template, we can read off the connection coefficients:

Γ

= − sin θ cos θ

Γ

= Γ

= cot θ, (31)

all others being zero. This connection is NOT flat since its associated Riemann tensor is non-zero. (See MTW pages 340-341 for a straight forward, detailed calculation.)

Part b. Use the coordinates x

= (θ, ϕ) which are good coordinates away from the poles.

We define a connection ∇ by giving its coefficients on the coordinate system {x

}:

Γ

= 0,

for all A, B and C in {1, 2}. Its clear that this connection is flat. It is defined only in the region of the two-sphere that excludes the poles, and hence is not a globally defined connection. Note that this connection is not compatible with the usual metric on the 2-sphere, but it is compatible with the metric

ds

= dθ

+ dϕ

,

which again is only defined away from the poles.