Solutions "General relativity - the theoretical minimum" (part II)

Exercise list

Lecture 4

Exercise page 123

Exercise page 132

Exercise 4.1

Exercise page 143

Exercise page 145

Exercise page 148

Exercise 4.2

Exercise page 151

Exercise page 161

Exercise page 123

There are few intermediate proofs requested to arrive to the formula for the covariant derivative of a contravariant vector.

We start from the relationship between covariant and contravariant vectors:

V^m = g^{mp} V_p

The covariant derivative satisfy the product rule. In the best set of coordinates we have:

D_r \left(V_m U_n\right) = \partial_r \left(V_m U_n\right) = \partial_r \left(V_m \right) U_n + V_m \partial_r \left(U_n\right) = D_r V_m + D_r U_n

We can write:

D_r V^m = \left(D_r g^{mp}\right) V_p + g^{mp} \left(D_r V_p \right)

We can demonstrate that the covariant derivative inverse metric tensor is zero.

We consider the best coordinates (where the Christoffel symbols are zero) and we begin with the definition of the contravariant metric tensor, g^{np}, as the inverse of the covariant metric tensor, g_{mn}:

g_{mn}g^{np} = {\delta^p}_m

Now, we take the partial derivative \partial_r:

Since the product rule applies to differentiation, we have:

\partial_r (g_{mn}g^{np}) = (\partial_r g_{mn}) g^{np} + g_{mn} (\partial_r g^{np}) = \partial_r ({\delta^p}_m) = 0

In the best coordinate system the components of the covariant metric tensor g_{mn} are constants, which means their partial derivatives are zero:

\partial_r g_{mn} = 0

Therefore:

g_{mn} (\partial_r g^{np}) = 0

We can contract the last equation with the contravariant metric tensor g^{mk}:

\begin{aligned} & g^{mk} g_{mn} (\partial_r g^{np}) = g^{mk} (0) \\ & {\delta^k}_n (\partial_r g^{np}) = 0 \\ & \partial_r g^{kp} = 0 \end{aligned}

As the effect of the Kronecker delta in this contraction is to replace the index n with k.

As k and p can be any indices, all components of the partial derivative of the contravariant metric tensor are zero.

Then the first term is then zero, and using the covariant derivative of V_p:

\begin{aligned} D_r V^m & = g^{mp} \left(\partial_r V_p - {\Gamma^t}_{rp} V_t \right) \\ & = g^{mp}\partial_r V_p - g^{mp} {\Gamma^t}_{rp} V_t \end{aligned}

since:

\partial_r V^m = \partial_r\left(g^{mp} V_p \right) = g^{mp}\partial_r V_p + \partial g^{mp} V_p

We can substitute:

D_r V^m = \partial_r V^m - \partial_r g^{mp} V_p - g^{mp} {\Gamma^t}_{rp} V_t

Now we use:

\begin{aligned} & V_p = g_{pk} V^k \\ & V_t = g_{tk} V^k \end{aligned}

Substituting:

\begin{aligned} D_r V^m & = \partial_r V^m - \partial_r g^{mp} V_p - g^{mp} {\Gamma^t}_{rp} V_t \\ & = \partial_r V^m - \partial_r g^{mp} g_{pk} V^k - g^{mp} {\Gamma^t}_{rp} g_{tk} V^k \\ & = \partial_r V^m + \left[ - \partial_r g^{mp} g_{pk} - g^{mp} {\Gamma^t}_{rp} g_{tk}\right] V^k \\ \end{aligned}

Let’s work on rewriting this term:

- \partial_r g^{mp} g_{pk} - g^{mp} {\Gamma^t}_{rp} g_{tk}

We start with the identity of the metric and its inverse:

g^{mp}g_{pk} = {\delta^m}_k

We differentiate and since {\delta^m}_k is just a matrix of constants its derivative is zero:

\partial_r(g^{mp}g_{pk}) = 0

We apply the product rule to the left-hand side:

\begin{aligned} & (\partial_r g^{mp})g_{pk} + g^{mp}(\partial_r g_{pk}) = 0 \\ & (\partial_r g^{mp})g_{pk} = -g^{mp}(\partial_r g_{pk}) \end{aligned}

We can substitute this result back:

-(\partial_r g^{mp}) g_{pk} = - \left( -g^{mp}(\partial_r g_{pk}) \right) = g^{mp}(\partial_r g_{pk})

The expression inside the bracket is now:

\left[ g^{mp}(\partial_r g_{pk}) - g^{mp} {\Gamma^t}_{rp} g_{tk} \right] = g^{mp} \left[(\partial_r g_{pk}) - {\Gamma^t}_{rp} g_{tk} \right]

The next step is to replace the Christoffel symbol {\Gamma^t}_{rp} with its definition in terms of derivatives of the metric tensor. Its definition is:

{\Gamma^t}_{rp} = \frac{1}{2} g^{tq} \left( \partial_r g_{qp} + \partial_p g_{qr} - \partial_q g_{rp} \right)

We substitute this into your expression for the term {\Gamma^t}_{rp} g_{tk}:

{\Gamma^t}_{rp} g_{tk} = \left[ \frac{1}{2} g^{tq} \left( \partial_r g_{qp} + \partial_p g_{qr} - \partial_q g_{rp} \right) \right] g_{tk}

The term g^{tq} is contracted with g_{tk}. This product is the Kronecker delta:

g^{tq} g_{tk} = {\delta^q}_k

When the Kronecker delta {\delta^q}_k multiplies the rest of the terms, it replaces every index q with an index k:

{\Gamma^t}_{rp} g_{tk} = \frac{1}{2} {\delta^q}_k \left( \partial_r g_{qp} + \partial_p g_{qr} - \partial_q g_{rp} \right) = \frac{1}{2} \left( \partial_r g_{kp} + \partial_p g_{kr} - \partial_k g_{rp} \right)

We have now successfully eliminated the Christoffel symbol from inside the bracket. The expression is now purely in terms of metric derivatives:

g^{mp} \left[ \partial_r g_{pk} - \frac{1}{2} \left( \partial_r g_{kp} + \partial_p g_{kr} - \partial_k g_{rp} \right) \right]

We can finally do some algebraic steps:

\begin{aligned} & g^{mp} \left[ \partial_r g_{pk} - \frac{1}{2} \left( \partial_r g_{kp} + \partial_p g_{kr} - \partial_k g_{rp} \right) \right] \\ & = g^{mp} \left[ \partial_r g_{pk} - \frac{1}{2} \partial_r g_{kp} - \frac{1}{2} \partial_p g_{kr} + \frac{1}{2} \partial_k g_{rp} \right] \\ & = g^{mp} \left[ \partial_r g_{pk} - \frac{1}{2} \partial_r g_{pk} - \frac{1}{2} \partial_p g_{kr} + \frac{1}{2} \partial_k g_{rp} \right] \\ & = g^{mp} \left[\frac{1}{2} \partial_r g_{pk} - \frac{1}{2} \partial_p g_{kr} + \frac{1}{2} \partial_k g_{rp} \right] \\ & = \frac{1}{2} g^{mp} \left[ \partial_r g_{pk} + \partial_k g_{rp} - \partial_p g_{kr} \right] \\ & = {\Gamma^m}_{rk} \end{aligned}

This final expression is the definition of {\Gamma^m}_{rk}. So:

\begin{aligned} D_r V^m & = \partial_r V^m + \left[ - \partial_r g^{mp} g_{pk} - g^{mp} {\Gamma^t}_{rp} g_{tk}\right] V^k \\ & = \partial_r V^m + {\Gamma^m}_{rk} V^k \end{aligned}

And finally renaming the dummy index k \to t:

D_r V^m = \partial_r V^m + {\Gamma^m}_{rt} V^t

Exercise page 132

The square of the length of a vector is defined as:

\left| V^m \right|^2 = V^m \cdot V_m

Applying this definition to the tangent vector:

\left| t^m \right|^2 = t^m \cdot t_m = \frac{\mathrm dX^m}{\mathrm dS} \frac{\mathrm dX_m}{\mathrm dS} = \frac{1}{\mathrm dS^2} \mathrm dX^m \mathrm dX_m

We then use the formula for relating the contravariant component of a vector with its covariant components:

V_n = g_{mn} V^m

Therefore:

\left| t^m \right|^2 = \frac{1}{\mathrm dS^2} \mathrm dX^m \mathrm dX_m = \frac{1}{\mathrm dS^2} \mathrm dX^m g_{mn} \mathrm dX^n = \frac{1}{\mathrm dS^2} g_{mn} \mathrm dX^m \mathrm dX^n

Finally, \mathrm dS^2 is defined as:

\mathrm dS^2 = g_{mn} \mathrm dX^m \mathrm dX^n

which leads to the desired result:

\left| t^m \right|^2 = \frac{1}{\mathrm dS^2} g_{mn} \mathrm dX^m \mathrm dX^n = \frac{\mathrm dS^2}{\mathrm dS^2} = 1

Exercise 4.1

This exercise continue a previous exercise (here). I will repeat some of these calculation for the specific case of a 2-sphere (R=1).

1. metric tensor

A point P in space is expressed in Cartesian coordinates using spherical coordinates (\theta, \phi) with \theta as latitude:

\begin{aligned} x &= \cos(\theta) \cos(\phi) \\ y &= \cos(\theta) \sin(\phi) \\ z &= \sin(\theta) \end{aligned}

The Jacobian matrix J transforms an infinitesimal displacement in spherical coordinates, \mathrm dY = [\mathrm d\theta, \mathrm d\phi]^T, to Cartesian coordinates, \mathrm dX = [\mathrm dx, \mathrm dy, \mathrm dz]^T.

The components of the Jacobian are J^i_{\ j} = \dfrac{\partial X^i}{\partial Y^j}. Calculating the partial derivatives:

J = \begin{bmatrix} \dfrac{\partial x}{\partial \theta} & \dfrac{\partial x}{\partial \phi} \\[10px] \dfrac{\partial y}{\partial \theta} & \dfrac{\partial y}{\partial \phi} \\[10px] \dfrac{\partial z}{\partial \theta} & \dfrac{\partial z}{\partial \phi} \end{bmatrix} = \begin{bmatrix} -\sin(\theta) \cos(\phi) & -\cos(\theta) \sin(\phi) \\ -\sin(\theta) \sin(\phi) & \cos(\theta) \cos(\phi) \\ \cos(\theta) & 0 \end{bmatrix}

The infinitesimal displacements are given by the relation \mathrm dX = \mathbf{J} \, \mathrm dY, which expands to:

\begin{aligned} \mathrm dx &= -\sin(\theta) \cos(\phi) \, \mathrm d\theta - \cos(\theta) \sin(\phi) \,\mathrm d\phi \\ \mathrm dy &= -\sin(\theta) \sin(\phi) \, \mathrm d\theta + \cos(\theta) \cos(\phi) \, \mathrm d\phi \\ \mathrm dz &= \cos(\theta) \, \mathrm d\theta \end{aligned}

The square of the infinitesimal line element, \mathrm ds^2, is \mathrm dx^2 + \mathrm dy^2 + \mathrm dz^2. Performing the calculation:

\begin{aligned} \mathrm ds^2 &= \left( -\sin(\theta) \cos(\phi) \, \mathrm d\theta - \cos(\theta) \sin(\phi) \, \mathrm d\phi \right)^2 + \left( -\sin(\theta) \sin(\phi) \, \mathrm d\theta + \cos(\theta) \cos(\phi) \, \mathrm d\phi \right)^2 + \left( \cos(\theta) \, \mathrm d\theta \right)^2 \\ &= \mathrm d\theta^2 (\sin^2 (\theta)\cos^2 (\phi) + \sin^2 (\theta)\sin^2\phi + \cos^2 (\theta)) + \mathrm d\phi^2(\cos^2 (\theta)\sin^2\phi + \cos^2 (\theta)\cos^2 (\phi)) \\ &= \mathrm d\theta^2 (\sin^2 (\theta) + \cos^2 (\theta)) + \mathrm d\phi^2(\cos^2 (\theta)) \\ &= \mathrm \, d\theta^2 + \cos^2(\theta) \, \mathrm d\phi^2 \end{aligned}

The final expression for the line element is:

\mathrm ds^2 = \mathrm d\theta^2 + \cos^2(\theta) \mathrm \, d\phi^2

The metric tensor g_{mn} is defined by the line element \mathrm ds^2 according to the formula:

\mathrm ds^2 = g_{mn} \mathrm dY^m \mathrm dY^n

where the coordinates are Y^1 = \theta and Y^2 = \phi.

By comparing the formula to the given line element:

\mathrm ds^2 = \mathrm d\theta^2 + (\cos^2(\theta)) \,\mathrm d\phi^2

we can identify the components of the metric tensor by matching the coefficients.

The coefficient of \mathrm d\theta^2 is g_{11}:

g_{11} = 1

The coefficient of \mathrm d\phi^2 is g_{22}:

g_{22} = \cos^2(\theta)

The coefficient of the cross-term \mathrm d\theta \mathrm d\phi is g_{12} + g_{21}. Since this term is absent, the off-diagonal components are zero:

g_{12} = g_{21} = 0

Therefore, the metric tensor g_{mn} in matrix form is:

g_{mn} = \begin{bmatrix} 1 & 0 \\ 0 & \cos^2(\theta) \end{bmatrix}

The inverse metric tensor g^{mn} is defined as the matrix inverse of the metric tensor g_{mn}. It satisfies the relation:

g^{mk} g_{kn} = {\delta^m}_n

Since the metric tensor is a diagonal matrix, its inverse is found by taking the reciprocal of each element on the main diagonal:

g^{mn} = \begin{bmatrix} 1 & 0 \\ 0 & \dfrac{1}{\cos^2(\theta)} \end{bmatrix}

2. Christoffel symbols

The Christoffel symbols are given by the formula:

{\Gamma^t}_{mn} = \frac{1}{2} g^{rt} \left[ \partial_n g_{rm} + \partial_m g_{rn} - \partial_r g_{mn} \right]

Here, the indices t, m, n, r range over our coordinates, \{\theta, \phi\}.

We start with \Gamma^1_{22}, where t=1, m=2, n=2. The summation index is r:

{\Gamma^1}_{22} = \frac{1}{2} g^{r1} \left[ \partial_\phi g_{r2} + \partial_\phi g_{r2} - \partial_r g_{22} \right]

Since g^{r1} is non-zero only for r=1, the sum collapses to a single term:

\begin{aligned} {\Gamma^1}_{22} & = \frac{1}{2} g^{11} \left[ \partial_\phi g_{12} + \partial_\phi g_{12} - \partial_\theta g_{22} \right] \\ & = \frac{1}{2} g^{11} \left[ - \partial_\theta g_{22} \right] = \frac{1}{2} \left( 2 \sin(\theta) \cos(\theta) \right) \\ & = \sin(\theta) \cos(\theta) \end{aligned}

as g_{12} = \partial_\phi g_{12} = 0.

We compute now \Gamma^2_{12}, where t=2, m=1, n=2. The summation index is r:

{\Gamma^2}_{12} = \frac{1}{2} g^{r2} \left[ \partial_\phi g_{r1} + \partial_\theta g_{r2} - \partial_r g_{12} \right]

Since g^{r2} is non-zero only for r=2, the sum collapses to a single term:

\begin{aligned} {\Gamma^2}_{12} & = \frac{1}{2} g^{22} \left[ \partial_\phi g_{21} + \partial_\theta g_{22} - \partial_\phi g_{12} \right] \\ & = \frac{1}{2} g^{22} \left[ \partial_\theta g_{22} \right] \\ & = \frac{1}{2} \left( \frac{1}{\cos^2(\theta)} \right) \left( -2 \sin(\theta) \cos(\theta) \right) \\ & = -\frac{\sin(\theta)}{\cos(\theta)} = -\tan\theta \end{aligned}

Since g_{21} = g_{12} = \partial_\phi g_{21} = \partial_\phi g_{12} are zero. By symmetry of the lower indices in the formula, {\Gamma^2}_{21} = {\Gamma^2}_{12}.

We compute now \Gamma^1_{11}, where t=1, m=1, n=1. The summation index is r:

{\Gamma^1}_{11} = \frac{1}{2} g^{r1} \left[ \partial_\theta g_{r1} + \partial_\theta g_{r1} - \partial_r g_{11} \right]

Since g^{r1} is non-zero only for r=1, the sum reduces to:

\begin{aligned} {\Gamma^1}_{11} & = \frac{1}{2} g^{11} \left[ \partial_\theta g_{11} + \partial_\theta g_{11} - \partial_\theta g_{11} \right] \\ & = \frac{1}{2} g^{11} \left[ \partial_\theta g_{11} \right] \end{aligned}

As g_{11} = 1, its derivative with respect to \theta is zero, \partial_\theta g_{11} = 0:

{\Gamma^1}_{11} = 0

We compute now \Gamma^1_{12}, where t=1, m=1, n=2. The summation index is r:

{\Gamma^1}_{12} = \frac{1}{2} g^{r1} \left[ \partial_\phi g_{r1} + \partial_\theta g_{r2} - \partial_r g_{12} \right]

The sum is non-zero only for the term where r=1:

\begin{aligned} {\Gamma^1}_{12} & = \frac{1}{2} g^{11} \left[ \partial_\phi g_{11} + \partial_\theta g_{12} - \partial_\theta g_{12} \right] \\ & = \frac{1}{2} g^{11} \left[ \partial_\phi g_{11} \right] \end{aligned}

Since g_{11}=1, its derivative with respect to \phi is zero, \partial_\phi g_{11} = 0:

{\Gamma^1}_{12} = 0

By symmetry of the lower indices, {\Gamma^1}_{21} = {\Gamma^1}_{12} = 0.

We compute now \Gamma^2_{11}, where t=2, m=1, n=1. The summation index is r:

{\Gamma^2}_{11} = \frac{1}{2} g^{r2} \left[ \partial_\theta g_{r1} + \partial_\theta g_{r1} - \partial_r g_{11} \right]

This sum is non-zero only for the term where r=2:

\begin{aligned} {\Gamma^2}_{11} & = \frac{1}{2} g^{22} \left[ \partial_\theta g_{21} + \partial_\theta g_{21} - \partial_\phi g_{11} \right] \end{aligned}

Since g_{21} = 0 and g_{11} = 1, their respective derivatives \partial_\theta g_{21} and \partial_\phi g_{11} are zero:

{\Gamma^2}_{11} = 0

We compute now \Gamma^2_{22}, where t=2, m=2, n=2. The summation index is r:

{\Gamma^2}_{22} = \frac{1}{2} g^{r2} \left[ \partial_\phi g_{r2} + \partial_\phi g_{r2} - \partial_r g_{22} \right]

The sum collapses to the single term where r=2:

\begin{aligned} {\Gamma^2}_{22} & = \frac{1}{2} g^{22} \left[ \partial_\phi g_{22} + \partial_\phi g_{22} - \partial_\phi g_{22} \right] \\ & = \frac{1}{2} g^{22} \left[ \partial_\phi g_{22} \right] \end{aligned}

As g_{22} = \cos^2(\theta), it does not depend on \phi, so its derivative \partial_\phi g_{22} = 0:

{\Gamma^2}_{22} = 0

To summarize, the non-zero Christoffel symbols at a point (\theta, \phi) on the sphere are:

\begin{aligned} & {\Gamma^1}_{22} = \sin(\theta) \cos(\theta) \\ & {\Gamma^2}_{12} = {\Gamma^2}_{21} = -\tan\theta \end{aligned}

All other symbols are zero:

{\Gamma^1}_{11} = {\Gamma^1}_{12} = {\Gamma^1}_{21} = {\Gamma^2}_{11} = {\Gamma^2}_{22} = 0

3. Tangent vector to a meridian

A meridian is defined as circle of constant longitude passing through a given point on the surface and the poles, so \phi = \phi_0 constant.

A tangent vector is defined as:

t^m = \frac{\mathrm dY^m}{\mathrm ds}

where:

\mathrm ds^2 = g_{mn} \mathrm dY^m \mathrm dY^n = \mathrm d\theta^2 + (\cos^2(\theta)) \,\mathrm d\phi^2 = \mathrm d\theta^2

The coordinates are Y^1 = \theta and Y^2 = \phi since \phi = \phi_0 constant, \mathrm d\phi= 0, so:

\mathrm ds = \sqrt{ \mathrm ds^2}= \mathrm d\theta

For t=1 we then have:

t^1 = \frac{\mathrm dY^1}{\mathrm ds} = \frac{\mathrm d\theta}{\mathrm d\theta} = 1

For t=2 we then have:

t^2 = \frac{\mathrm dY^2}{\mathrm ds} = \frac{\mathrm d\phi}{\mathrm d\theta} = 0

4. Covariant derivative of the tangent vector

We define the vector covariant derivative as:

D_r V^m \equiv \partial_r V^m + {\Gamma^m}_{rt} V^t

Given the vector:

t = \begin{bmatrix} t^1 \\ t^2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}

The partial derivatives are all zero because the vector is constant, and as we have all the Christoffel symbols we can compute this tensor:

\begin{aligned} & D_1 t^1 = {\Gamma^{1}}_{11} t^1 + {\Gamma^{1}}_{12} t^2 = (0)(1) + (0)(0) = 0 \\ & D_2 t^1 = {\Gamma^{1}}_{21} t^1 + {\Gamma^{1}}_{22} t^2 = (0)(1) + (\sin(\theta)\cos(\theta))(0) = 0 \\ & D_1 t^2 = {\Gamma^{2}}_{11} t^1 + {\Gamma^{2}}_{12} t^2 = (0)(1) + (-\tan(\theta))(0) = 0 \\ & D_2 t^2 = {\Gamma^{2}}_{21} t^1 + {\Gamma^{2}}_{22} t^2 = (-\tan(\theta))(1) + (0)(0) = -\tan(\theta) \end{aligned}

So the covariant derivative tensor is:

D_r t^m = \begin{bmatrix} D_1 t^1 & D_2 t^1 \\ D_1 t^2 & D_2 t^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & - \tan(\theta) \end{bmatrix}

4. Covariant change of the tangent vector along a meridian

The covariant change of a vector is defined as:

DV^m \equiv D_r V^m \mathrm dY^r

For the tangent vector is:

Dt^m = D_r t^m \mathrm dY^r = D_1 t^m \mathrm dY^1 + D_2 t^m \mathrm dY^2 = D_1 t^m \mathrm d \theta + D_2 t^m \mathrm d \phi

Using the covariant derivative tensor and the fact that along a meridian \phi = \phi_0 constant so \mathrm d\phi = 0:

\begin{aligned} & Dt^1 = D_1 t^1 \mathrm d \theta + D_2 t^1 \mathrm d \phi = (0) \mathrm d \theta + (0) \mathrm d \phi = (0) \mathrm d \theta + (0)(0) = 0 \\ & Dt^2 = D_1 t^2 \mathrm d \theta + D_2 t^2 \mathrm d \phi = (0) \mathrm d \theta + (-\tan(\theta))\mathrm d \phi = (0) \mathrm d \theta + (-\tan(\theta))(0) = 0 \end{aligned}

So the covariant change of the tangent vector is zero.

Exercise page 143

We can verify it by direct substitution:

\begin{aligned} \cos^2(\theta) + \sin^2(\theta) & = \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^2 + \left(\frac{e^{i\theta} - e^{-i\theta}}{2i} \right)^2 \\ & = \frac{e^{2i\theta} + 2e^{i\theta} e^{-i\theta} + e^{-2i\theta}}{4} + \frac{e^{2i\theta} - 2e^{i\theta} e^{-i\theta} + e^{-2i\theta}}{4i^2} \\ & = \frac{e^{2i\theta} + 2 + e^{-2i\theta}}{4} + (-1) \frac{e^{2i\theta} - 2 + e^{-2i\theta}}{4} \\ & = \frac{1}{2} + \frac{1}{2} = 1 \end{aligned}

Exercise page 145

We can verify it by direct substitution:

\begin{aligned} \cosh^2(\omega) - \sinh^2(\omega) & = \left(\frac{e^{\omega} + e^{-\omega}}{2}\right)^2 - \left(\frac{e^{\omega} - e^{-\omega}}{2} \right)^2 \\ & = \frac{e^{2\omega} + 2e^{\omega} e^{-\omega} + e^{-2\omega}}{4} - \frac{e^{2\omega} - 2e^{\omega} e^{-\omega} + e^{-2\omega}}{4} \\ & = \frac{e^{2\omega} + 2 + e^{-2\omega}}{4} - \frac{e^{2\omega} - 2 + e^{-2\omega}}{4} \\ & = \frac{e^{2\omega} + 2 + e^{-2\omega} - e^{2\omega} + 2 - e^{-2\omega}}{4} \\ & = \frac{4}{4} \\ & = 1 \end{aligned}

Exercise page 148

Considering c = 1, the hyperbolic coordinates are given by:

\begin{aligned} & X = r\cosh(\omega) \\ & T = r\sinh(\omega) \end{aligned}

If r_1 = 1, r_2=2, r_3=3, by construction |MN| = r_2 - r_1 = 1 and |NP| = r_3 - r_2 = 1 = |MN|.

The hyperbole is defined as:

X^2 - T^2=r^2

We consider a fixed \omega=\omega_c, the events on the three hyperbolas are:

\begin{aligned} & R = [X_1, T_1]^T =[r_1\cosh(\omega_c),\ r_1\sinh(\omega_c)]^T \\ & S = [X_2, T_2]^T =[r_2\cosh(\omega_c),\ r_2\sinh(\omega_c)]^T \\ & T = [X_3, T_3]^T =[r_3\cosh(\omega_c),\ r_3\sinh(\omega_c)]^T \end{aligned}

The three events are collinear:

[X_1,T_1]^T = \frac{r_1}{r_2}\,[X_2,T_2]^T = \frac{r_1}{r_3}\,[X_3,T_3]^T

so O,R,S, T lie on the same straight line, with a common slope:

\frac{T}{X}=\tanh(\omega_c)

Along a ray \omega_c is constant so:

\begin{aligned} & \mathrm dX = \cosh(\omega_c) \, \mathrm dr \\ & \mathrm dT = \sinh(\omega_c) \, \mathrm dr \end{aligned}

So the square of the proper distance along a ray is:

\mathrm dS^2 = \mathrm dX^2 - \mathrm dT^2 = (\cosh^2(\omega_c) - \sinh^2(\omega_c))\,\mathrm dr^2 = \mathrm dr^2

The spacelike proper distance for a finite interval:

\sqrt{\Delta S^2}=|\,\Delta r\,|= |r_2 - r_1| = |RS| = 1

Similarly:

\sqrt{\Delta S^2}=|\,\Delta r\,|= |r_3 - r_1| = |RT| = |RS| + |ST| = 2

so that:

|ST| = 2 - |RS| = 1 = |RS| = |MN| = |NP|

Exercise 4.2

We continue the previous exercise here and use the results obtained.

The simultaneity condition with a boost speed v is:

\begin{aligned} & T^\prime_1 = T^\prime_2 \\ & \gamma( T_1 - v X_1) = \gamma(T_2 - v X_2) \\ & v = \frac{T_1 - T_2}{X_1 - X_2} \end{aligned}

For points on the same ray:

\frac{T}{X} = v = \tanh (\omega_c)

With this v:

\begin{aligned} \gamma & = \frac{1}{\sqrt{1 - v^2}} = \frac{1}{\sqrt{1 - \tanh^2 (\omega_c)}} \\ & = \frac{1}{\sqrt{1 - \frac{\sinh^2 (\omega_c)}{\cosh^2(\omega_c)}}} = \frac{1}{\sqrt{\frac{\cosh^2(\omega_c) - \sinh^2 (\omega_c)}{\cosh^2(\omega_c)}}} \\ & = \frac{1}{\sqrt{\frac{1}{\cosh^2(\omega_c)}}} = \cosh(\omega_c) \end{aligned}

and the Lorentz transform gives:

\begin{aligned} & T^\prime_i = \gamma(T_i - v X_i)= 0 \\ & X^\prime_i = \gamma(X_i - v T_i)= r_i \end{aligned}

Therefore the observers who see R and S as simultaneous move at speed v=\tanh\omega_c relative to the fixed frame, and in that frame the spatial separation is r_2-r_1.

Similar calculation can be performed for the point T, which will give the same speed and a separation r_3 - r_2.

Exercise page 151

Each observer traveling along a hyperbola of constant r experiences a constant proper acceleration. The proper acceleration is the acceleration measured in the observer’s own instantaneous rest frame, and its magnitude is a relativistic invariant.

To determine its value, we relate the derivatives with respect to proper time \tau to derivatives with respect to the parameter \omega. The differential interval of proper time is given by:

\mathrm d\tau^2 = \mathrm dT^2 - \mathrm dX^2

Substituting the differentials we find the relationship between \mathrm d\tau and \mathrm d\omega:

\mathrm d\tau^2 = (r\cosh(\omega)\,\mathrm d\omega)^2 - (r\sinh(\omega)\,\mathrm d\omega)^2 = r^2(\cosh^2(\omega) - \sinh^2(\omega))\,\mathrm d\omega^2 = r^2\,\mathrm d\omega^2

This provides the conversion factor:

\begin{aligned} & \mathrm d\tau = r\,\mathrm d\omega \\ & \frac{\mathrm d\omega}{\mathrm d\tau} = \frac{1}{r} \end{aligned}

The position four-vector is x = [T, X]^T:

x(\omega) = [r\sinh(\omega),\ r\cosh(\omega)]^T

The four-velocity u is the derivative of x with respect to proper time \tau:

\begin{aligned} u & = \frac{\mathrm dx}{\mathrm d\tau} = \frac{\mathrm dx}{\mathrm d\omega}\frac{\mathrm d\omega}{\mathrm d\tau} \\ & = \frac{1}{r}[r\cosh(\omega),\ r\sinh(\omega)]^T \\ & = [\cosh(\omega),\ \sinh(\omega)]^T \end{aligned}

The four-acceleration a is the derivative of the four-velocity with respect to \tau:

\begin{aligned} a & = \frac{\mathrm du}{\mathrm d\tau} = \frac{\mathrm du}{\mathrm d\omega}\frac{\mathrm d\omega}{\mathrm d\tau} \\ & = \frac{1}{r} [\sinh(\omega),\ \cosh(\omega)]^T \end{aligned}

The magnitude of the proper acceleration is:

\begin{aligned} |a|^2 & = \left(a^1\right)^2 - \left(a^0\right)^2 \\ & = \left(\frac{1}{r}\cosh(\omega)\right)^2 - \left(\frac{1}{r}\sinh(\omega)\right)^2 \\ & = \frac{1}{r^2}(\cosh^2(\omega) - \sinh^2(\omega)) = \frac{1}{r^2} \end{aligned}

The proper acceleration for an observer on a hyperbola of parameter r is constant throughout their motion, with a magnitude of \alpha = 1/r. Therefore, the observers located at R, S, and T have distinct proper accelerations of 1/r_1, 1/r_2, and 1/r_3, respectively.

Exercise page 161

Equation (42), since it is an approximation at the first order, it actually gives a non-zero curvature.

Approximated equation

Christoffel symbols

Starting from the equation:

\mathrm d\tau^2 = (1 + 2gy) \mathrm dt^2 - \mathrm dy^2

We begin with a two-dimensional spacetime described by the coordinates (t, y). The geometry of this spacetime is defined by the following metric tensor g_{\mu\nu}:

g_{\mu\nu} = \begin{bmatrix} 1 + 2gy & 0 \\ 0 & -1 \end{bmatrix}

From this matrix, we can identify the individual components of the metric tensor. With the convention that x^0 = t and x^1 = y, the components are:

\begin{aligned} g_{tt} &= 1 + 2gy \\ g_{yy} &= -1 \\ g_{ty} &= g_{yt} = 0 \end{aligned}

The Christoffel symbols {\Gamma^\sigma}_{\mu\nu}, are computed from the derivatives of the metric tensor. The general formula is:

{\Gamma^\sigma}_{\mu\nu} = \frac{1}{2} g^{\sigma\rho} \left(\partial_\mu g_{\rho\nu} + \partial_\nu g_{\rho\mu} - \partial_\rho g_{\mu\nu} \right)

To proceed, we need the inverse metric tensor, g^{\mu\nu}, and the partial derivatives of the metric components. The inverse metric is found by taking the inverse of the (diagonal) matrix g_{\mu\nu}:

g^{\mu\nu} = \begin{bmatrix} \frac{1}{1 + 2gy} & 0 \\ 0 & -1 \end{bmatrix}

The non-zero components of the inverse metric are g^{tt} = (1 + 2gy)^{-1} and g^{yy} = -1.

Next, we calculate the partial derivatives of the metric tensor components with respect to the coordinates t and y. The only component that is not constant is g_{tt}.

\begin{aligned} & \partial_y g_{tt} = 2g \\ & \partial_y g_{\mu\nu} = 0 \quad \forall \mu, \nu \neq \{\mu=\nu=0\} \\ & \partial_t g_{\mu\nu} = 0 \quad \forall \mu, \nu \end{aligned}

All other partial derivatives are zero.

We start from the y components:

{\Gamma^y}_{\mu\nu} = \frac{1}{2} g^{y\rho} \left(\partial_\mu g_{\rho\nu} + \partial_\nu g_{\rho\mu} - \partial_\rho g_{\mu\nu} \right)

Computation of {\Gamma^y}_{\mu \nu}

Since the metric is diagonal, the only non-zero term in the sum over \rho is for \rho = y. The component g^{yy} is equal to -1:

\begin{aligned} {\Gamma^y}_{\mu\nu} & = \frac{1}{2} g^{yy} \left(\partial_\mu g_{\nu y} + \partial_\nu g_{\mu y} - \partial_yg_{\mu\nu} \right) \\ & = -\frac{1}{2} \left(\partial_\mu g_{\nu y} + \partial_\nu g_{\mu y} - \partial_yg_{\mu\nu} \right) \end{aligned}

We now calculate each component of {\Gamma^y}_{\mu \nu} individually.

Component {\Gamma^y}_{tt}

For this component, we set \mu = t and \nu = t:

\begin{aligned} {\Gamma^y}_{tt} & = -\frac{1}{2} \left(\partial_t g_{t y} + \partial_t g_{t y} - \partial_yg_{tt} \right) \\ &= -\frac{1}{2} \left( 0 + 0 - 2g \right) \end{aligned}

This gives the result:

{\Gamma^y}_{tt} = g

Component {\Gamma^y}_{ty}

For this component, we set \mu = t and \nu = y:

{\Gamma^y}_{ty} = -\frac{1}{2} \left(\partial_t g_{yy} + \partial_y g_{ty} - \partial_yg_{ty} \right) = 0

This gives the result:

{\Gamma^y}_{ty} = 0

Due to the symmetry of the lower indices of the Christoffel symbols, {\Gamma^y}_{yt} is also zero.

Component {\Gamma^y}_{yy}

Finally, we set \mu = y and \nu = y:

{\Gamma^y}_{yy} = -\frac{1}{2} \left(\partial_y g_{yy} + \partial_y g_{yy} - \partial_yg_{yy} \right) = 0

The result for this component is:

{\Gamma^y}_{yy} = 0

We compute then the t components:

{\Gamma^t}_{\mu\nu} = \frac{1}{2} g^{t\rho} \left(\partial_\mu g_{\rho\nu} + \partial_\nu g_{\rho\mu} - \partial_\rho g_{\mu\nu} \right)

Computation of {\Gamma^t}_{\mu \nu}

Since the metric is diagonal, the only non-zero term in the sum over \rho is for \rho = t. The component g^{tt} is equal to (1 + 2gy)^{-1}:

\begin{aligned} {\Gamma^t}_{\mu\nu} & = \frac{1}{2} g^{tt} \left(\partial_\mu g_{\nu t} + \partial_\nu g_{\mu t} - \partial_t g_{\mu\nu} \right) \\ & = \frac{1}{2(1 + 2gy)} \left(\partial_\mu g_{\nu t} + \partial_\nu g_{\mu t} - \partial_t g_{\mu\nu} \right) \end{aligned}

We now calculate each component of {\Gamma^t}_{\mu \nu} individually.

Component {\Gamma^t}_{tt}

For this component, we set \mu = t and \nu = t:

{\Gamma^t}_{tt} = \frac{1}{2(1 + 2gy)} \left(\partial_t g_{tt} + \partial_t g_{tt} - \partial_t g_{tt} \right) = 0

This gives the result:

{\Gamma^t}_{tt} = 0

Component {\Gamma^t}_{ty}

For this component, we set \mu = t and \nu = y:

\begin{aligned} {\Gamma^t}_{ty} & = \frac{1}{2(1 + 2gy)} \left(\partial_y g_{tt} + \partial_t g_{ty} - \partial_t g_{ty} \right) \\ &= \frac{1}{2(1 + 2gy)} \left( 2g + 0 - 0 \right) = \frac{g}{1 + 2gy} \end{aligned}

This gives the result:

{\Gamma^t}_{ty} = \frac{g}{1 + 2gy}

Due to the symmetry of the lower indices of the Christoffel symbols, {\Gamma^t}_{yt} is also \frac{g}{1 + 2gy}.

Component {\Gamma^t}_{yy}

Finally, we set \mu = y and \nu = y:

{\Gamma^t}_{yy} = \frac{1}{2(1 + 2gy)} \left(\partial_y g_{ty} + \partial_y g_{yt} - \partial_t g_{yy} \right) = 0

The result for this component is:

{\Gamma^t}_{yy} = 0

The calculations show that for the given metric tensor, most of the {\Gamma^\sigma}_{\mu \nu} components are zero. The only non-zero components are:

\begin{aligned} & {\Gamma^y}_{tt} = g \\ & {\Gamma^t}_{ty} = {\Gamma^t}_{yt} = \frac{g}{1 + 2gy} \end{aligned}

Riemann curvature tensor

The Riemann curvature tensor, {R^t}_{srn}, is defined in terms of the Christoffel symbols and their derivatives:

{R^t}_{srn} = \partial_r{\Gamma^t}_{sn} - \partial_s {\Gamma^t}_{rn} + {\Gamma^p}_{sn}{\Gamma^t}_{pr} - {\Gamma^p}_{rn} {\Gamma^t}_{ps}

In a two-dimensional spacetime, the Riemann tensor has only one independent component. Therefore, it is sufficient to calculate a single component, such as {R^t}_{yty}, to test for curvature:

{R^t}_{y t y} = \partial_t {\Gamma^t}_{yy} - \partial_y {\Gamma^t}_{yt} + {\Gamma^p}_{yy} {\Gamma^t}_{p t} - {\Gamma^p}_{yt} {\Gamma^t}_{p y}

Expanding the summation over p = (t, y) and substituting the known Christoffel symbols:

\begin{aligned} {R^t}_{y t y} &= 0 - \partial_y\left(\frac{g}{1 + 2gy}\right) + \left({\Gamma^t}_{yy}{\Gamma^t}_{tt} + {\Gamma^y}_{yy}{\Gamma^t}_{yt}\right) - \left({\Gamma^t}_{yt}{\Gamma^t}_{ty} + {\Gamma^y}_{yt}{\Gamma^t}_{yy}\right) \\ &= - \left(-\frac{g}{(1+2gy)^2}(2g) \right) + (0+0) - \left( \left(\frac{g}{1+2gy}\right)\left(\frac{g}{1+2gy}\right) + 0 \right) \\ &= \frac{2g^2}{(1+2gy)^2} - \frac{g^2}{(1+2gy)^2} = \frac{g^2}{(1+2gy)^2} \end{aligned}

This gives the component:

{R^t}_{y t y} = \frac{g^2}{(1+2gy)^2}

Since the only independent component of the curvature tensor (in two dimensions) is not zero, the space is curved.

Exact equation

Christoffel symbols

Starting from the equation:

\mathrm d\tau^2 = (1 + gy)^2 \mathrm dt^2 - \mathrm dy^2

We begin with a two-dimensional spacetime described by the coordinates (t, y). The geometry of this spacetime is defined by the following metric tensor g_{\mu\nu}:

g_{\mu\nu} = \begin{bmatrix} (1 + gy)^2 & 0 \\ 0 & -1 \end{bmatrix}

From this matrix, we can identify the individual components of the metric tensor. With the convention that x^0 = t and x^1 = y, the components are:

\begin{aligned} & g_{tt} = (1 + gy)^2 \\ & g_{yy} = -1 \\ & g_{ty} = g_{yt} = 0 \end{aligned}

The Christoffel symbols {\Gamma^\sigma}_{\mu\nu}, are computed from the derivatives of the metric tensor. The general formula is:

{\Gamma^\sigma}_{\mu\nu} = \frac{1}{2} g^{\sigma\rho} \left(\partial_\mu g_{\rho\nu} + \partial_\nu g_{\rho\mu} - \partial_\rho g_{\mu\nu} \right)

To proceed, we need the inverse metric tensor, g^{\mu\nu}, and the partial derivatives of the metric components. The inverse metric is found by taking the inverse of the (diagonal) matrix g_{\mu\nu}:

g^{\mu\nu} = \begin{bmatrix} \frac{1}{(1 + gy)^2} & 0 \\ 0 & -1 \end{bmatrix}

The non-zero components of the inverse metric are g^{tt} = (1 + gy)^{-2} and g^{yy} = -1.

Next, we calculate the partial derivatives of the metric tensor components with respect to the coordinates t and y. The only component that is not constant is g_{tt}.

\begin{aligned} & \partial_y g_{tt} = 2g(1 + gy) \\ & \partial_y g_{\mu\nu} = 0 \quad \forall \mu, \nu \neq \{\mu=\nu=0\} \\ & \partial_t g_{\mu\nu} = 0 \quad \forall \mu, \nu \end{aligned}

All other partial derivatives are zero.

We start from the y components:

{\Gamma^y}_{\mu\nu} = \frac{1}{2} g^{y\rho} \left(\partial_\mu g_{\rho\nu} + \partial_\nu g_{\rho\mu} - \partial_\rho g_{\mu\nu} \right)

Computation of {\Gamma^y}_{\mu \nu}

Since the metric is diagonal, the only non-zero term in the sum over \rho is for \rho = y. The component g^{yy} is equal to -1:

\begin{aligned} {\Gamma^y}_{\mu\nu} & = \frac{1}{2} g^{yy} \left(\partial_\mu g_{\nu y} + \partial_\nu g_{\mu y} - \partial_yg_{\mu\nu} \right) \\ & = -\frac{1}{2} \left(\partial_\mu g_{\nu y} + \partial_\nu g_{\mu y} - \partial_yg_{\mu\nu} \right) \end{aligned}

We now calculate each component of {\Gamma^y}_{\mu \nu} individually.

Component {\Gamma^y}_{tt}

For this component, we set \mu = t and \nu = t:

\begin{aligned} {\Gamma^y}_{tt} & = -\frac{1}{2} \left(\partial_t g_{t y} + \partial_t g_{t y} - \partial_yg_{tt} \right) \\ &= -\frac{1}{2} \left( 0 + 0 - 2g(1 + gy) \right) \end{aligned}

This gives the result:

{\Gamma^y}_{tt} = g(1 + gy)

Component {\Gamma^y}_{ty}

For this component, we set \mu = t and \nu = y:

{\Gamma^y}_{ty} = -\frac{1}{2} \left(\partial_t g_{yy} + \partial_y g_{ty} - \partial_yg_{ty} \right) = 0

This gives the result:

{\Gamma^y}_{ty} = 0

Due to the symmetry of the lower indices of the Christoffel symbols, {\Gamma^y}_{yt} is also zero.

Component {\Gamma^y}_{yy}

Finally, we set \mu = y and \nu = y:

{\Gamma^y}_{yy} = -\frac{1}{2} \left(\partial_y g_{yy} + \partial_y g_{yy} - \partial_yg_{yy} \right) = 0

The result for this component is:

{\Gamma^y}_{yy} = 0

We compute then the t components:

{\Gamma^t}_{\mu\nu} = \frac{1}{2} g^{t\rho} \left(\partial_\mu g_{\rho\nu} + \partial_\nu g_{\rho\mu} - \partial_\rho g_{\mu\nu} \right)

Computation of {\Gamma^t}_{\mu \nu}

Since the metric is diagonal, the only non-zero term in the sum over \rho is for \rho = t. The component g^{tt} is equal to (1 + gy)^{-2}:

\begin{aligned} {\Gamma^t}_{\mu\nu} & = \frac{1}{2} g^{tt} \left(\partial_\mu g_{\nu t} + \partial_\nu g_{\mu t} - \partial_t g_{\mu\nu} \right) \\ & = \frac{1}{2(1 + gy)^2} \left(\partial_\mu g_{\nu t} + \partial_\nu g_{\mu t} - \partial_t g_{\mu\nu} \right) \end{aligned}

We now calculate each component of {\Gamma^t}_{\mu \nu} individually.

Component {\Gamma^t}_{tt}

For this component, we set \mu = t and \nu = t:

{\Gamma^t}_{tt} = \frac{1}{2(1 + gy)^2} \left(\partial_t g_{tt} + \partial_t g_{tt} - \partial_t g_{tt} \right) = 0

This gives the result:

{\Gamma^t}_{tt} = 0

Component {\Gamma^t}_{ty}

For this component, we set \mu = t and \nu = y:

\begin{aligned} {\Gamma^t}_{ty} & = \frac{1}{2(1 + gy)^2} \left(\partial_y g_{tt} + \partial_t g_{ty} - \partial_t g_{ty} \right) \\ &= \frac{1}{2(1 + gy)^2} \left(2g(1 + gy) + 0 - 0 \right) = \frac{g}{1 + gy} \end{aligned}

This gives the result:

{\Gamma^t}_{ty} = \frac{g}{1 + gy}

Due to the symmetry of the lower indices of the Christoffel symbols, {\Gamma^t}_{yt} is also \frac{g}{1 + gy}.

Component {\Gamma^t}_{yy}

Finally, we set \mu = y and \nu = y:

{\Gamma^t}_{yy} = \frac{1}{2(1 + gy)^2} \left(\partial_y g_{ty} + \partial_y g_{yt} - \partial_t g_{yy} \right) = 0

The result for this component is:

{\Gamma^t}_{yy} = 0

The calculations show that for the given metric tensor, most of the {\Gamma^y}_{\mu \nu} components are zero. The only non-zero components are:

\begin{aligned} & {\Gamma^y}_{tt} = g(1 + gy) \\ & {\Gamma^t}_{ty} = {\Gamma^t}_{yt} = \frac{g}{1 + gy} \end{aligned}

Riemann curvature tensor

The Riemann curvature tensor, {R^t}_{srn}, is defined in terms of the Christoffel symbols and their derivatives:

{R^t}_{srn} = \partial_r{\Gamma^t}_{sn} - \partial_s {\Gamma^t}_{rn} + {\Gamma^p}_{sn}{\Gamma^t}_{pr} - {\Gamma^p}_{rn} {\Gamma^t}_{ps}

In a two-dimensional spacetime, the Riemann tensor has only one independent component. Therefore, it is sufficient to calculate a single component, such as {R^t}_{yty}, to test for curvature:

{R^t}_{y t y} = \partial_t {\Gamma^t}_{yy} - \partial_y {\Gamma^t}_{yt} + {\Gamma^p}_{yy} {\Gamma^t}_{p t} - {\Gamma^p}_{yt} {\Gamma^t}_{p y}

Expanding the summation over p = (t, y) and substituting the known Christoffel symbols:

\begin{aligned} {R^t}_{y t y} &= 0 - \partial_y\left(\frac{g}{1 + gy}\right) + \left({\Gamma^t}_{yy}{\Gamma^t}_{tt} + {\Gamma^y}_{yy}{\Gamma^t}_{yt}\right) - \left({\Gamma^t}_{yt}{\Gamma^t}_{ty} + {\Gamma^y}_{yt}{\Gamma^t}_{yy}\right) \\ &= - \left(-\frac{g}{(1+gy)^2}(g) \right) + (0+0) - \left( \left(\frac{g}{1 + gy}\right)\left(\frac{g}{1 + gy}\right) + 0 \right) \\ &= \frac{g^2}{(1+gy)^2} - \frac{g^2}{(1+gy)^2} = 0 \end{aligned}

This gives the component:

{R^t}_{y t y} = 0

Since the only independent component of the curvature tensor (in two dimensions) is zero, the space is flat.

Coordinates transformation

Now that we found that the space is flat, we can find the transformation which gives the coordinates where the matrix has the simplified form \mathrm dT^2 - \mathrm dX^2.

The starting metric is:

\mathrm d\tau^2 = (1+gy)^2 \mathrm dt^2 - \mathrm dy^2 And our destination is the flat Minkowski metric in some new coordinates (T, X):

\mathrm d\tau^2 = \mathrm dT^2 - \mathrm dX^2

we can try to use the flat coordinates (T, X) using two new variables, which we can call r and \omega, in a way that resembles polar coordinates but with hyperbolic functions:

\begin{aligned} T &= r \sinh(\omega) \\ X &= r \cosh(\omega) \end{aligned}

First, let’s find what the Minkowski metric looks like in these (r, \omega) coordinates. We compute the differentials of T and X:

\begin{aligned} \mathrm dT &= \frac{\partial T}{\partial r} \mathrm dr + \frac{\partial T}{\partial \omega} \mathrm d\omega = \sinh(\omega) \mathrm dr + r \cosh(\omega) \mathrm d\omega \\ \mathrm dX &= \frac{\partial X}{\partial r} \mathrm dr + \frac{\partial X}{\partial \omega} \mathrm d\omega = \cosh(\omega) \mathrm dr + r \sinh(\omega) \mathrm d\omega \end{aligned}

Now we construct the line element \mathrm dT^2 - \mathrm dX^2:

\begin{aligned} \mathrm dT^2 - \mathrm dX^2 = & \left( \sinh(\omega) \mathrm dr + r \cosh(\omega) \mathrm d\omega \right)^2 - \left( \cosh(\omega) \mathrm dr + r \sinh(\omega) \mathrm d\omega \right)^2 \\ = & \left( \sinh^2(\omega) \mathrm dr^2 + r^2 \cosh^2(\omega) \mathrm d\omega^2 \right) - \left( \cosh^2(\omega) \mathrm dr^2 + r^2 \sinh^2(\omega) \mathrm d\omega^2 \right) \\ = & (\sinh^2(\omega) - \cosh^2(\omega))\mathrm dr^2 + (r^2 \cosh^2(\omega) - r^2 \sinh^2(\omega))\mathrm d\omega^2 \\ = & r^2 \mathrm d\omega^2 - \mathrm dr^2 \end{aligned}

Now we have the original metric and the target metric expressed in a new form. For the coordinate transformation to exist, these two forms must be equal:

(1+gy)^2 \mathrm dt^2 - \mathrm dy^2 = r^2 \mathrm d\omega^2 - \mathrm dr^2

The structure of this equation is similar on both sides and we can directly compare the corresponding terms:

\begin{aligned} & \mathrm dr^2 = \mathrm dy^2 \\ & r^2 \mathrm d\omega^2 = (1+gy)^2 \mathrm dt^2 \end{aligned}

From the first equation, we get \mathrm dr = \mathrm dy. Integrating this gives r as a function of y. To simplify the second equation, it is useful to define r in a way that relates it to the coefficient (1+gy). Let’s try the identification:

r = \frac{1}{g} (1+gy)

Differentiating this gives \mathrm dr = \mathrm dy, which satisfies our first equation. Now, we substitute this definition of r into the second equation:

\left( \frac{1}{g} (1+gy) \right)^2 \mathrm d\omega^2 = (1+gy)^2 \mathrm dt^2

We can cancel the (1+gy)^2 term from both sides, which leaves:

\frac{1}{g^2} \mathrm d\omega^2 = \mathrm dt^2

Taking the square root gives \frac{1}{g} \mathrm d\omega = \mathrm dt. Integrating this relation gives:

\omega = gt

We have now successfully found the intermediate variables r and \omega in terms of our original coordinates y and t:

\begin{aligned} & r(y) = \frac{1}{g} (1+gy) \\ & \omega(t) = gt \end{aligned}

The final step is to substitute these expressions back into our initial hyperbolic definition of T and X:

\begin{aligned} & T = r \sinh(\omega) = \frac{1}{g} (1+gy) \sinh(gt) \\ & X = r \cosh(\omega) = \frac{1}{g} (1+gy) \cosh(gt) \end{aligned}

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