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Ch. 6 Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Jacobian of 2D Polar Coordinates

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Solution strategy: Directly compute the determinant of the Jacobian matrix.

The Jacobian matrix is:

\[ J = \begin{pmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial \theta} \\[8pt] \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial \theta} \end{pmatrix} = \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix} \]

Computing the determinant:

\[ \det J = \cos\theta \cdot r\cos\theta - (-r\sin\theta) \cdot \sin\theta = r\cos^2\theta + r\sin^2\theta = r(\cos^2\theta + \sin^2\theta) \]
\[ \boxed{\det J = r} \]

Verification: This is consistent with the Jacobian being \(r\) in the polar coordinate area element \(dA = r\,dr\,d\theta\). Furthermore, the transformation is non-singular for \(r > 0\) and degenerates at \(r = 0\) (the origin), which is geometrically correct.

B-2. Inverse Metric in 3D Spherical Coordinates

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Solution strategy: Since the metric tensor is a diagonal matrix, the inverse is obtained by taking the reciprocal of each diagonal component.

The metric tensor is:

\[ g_{ij} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2\sin^2\theta \end{pmatrix} \]

Since the inverse of a diagonal matrix has reciprocals of each diagonal component:

\[ \boxed{g^{ij} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \dfrac{1}{r^2} & 0 \\ 0 & 0 & \dfrac{1}{r^2\sin^2\theta} \end{pmatrix}} \]

Verification: We confirm that \(g_{ij}\,g^{jk} = \delta_i^k\). For example, the \((2,2)\) component: \(r^2 \cdot \dfrac{1}{r^2} = 1\). The \((3,3)\) component: \(r^2\sin^2\theta \cdot \dfrac{1}{r^2\sin^2\theta} = 1\). All off-diagonal components are 0. This indeed gives the identity matrix. ✓

B-3. Matrix Representation of a General 2-Dimensional Metric

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Solution strategy: Expand \(ds^2 = g_{ij}\,du^i\,du^j\) and compare coefficients.

Expanding the line element:

\[ ds^2 = g_{11}\,du^2 + (g_{12} + g_{21})\,du\,dv + g_{22}\,dv^2 \]

Using the symmetry of the metric tensor \(g_{12} = g_{21}\), the coefficient of the cross term is \(2g_{12}\).

Comparing with the given line element:

  • \(g_{11} = 1 + u^2\)
  • \(2g_{12} = 2uv \implies g_{12} = uv\)
  • \(g_{22} = 1 + v^2\)
\[ \boxed{g_{ij} = \begin{pmatrix} 1 + u^2 & uv \\ uv & 1 + v^2 \end{pmatrix}} \]

Verification: The determinant \(\det g = (1+u^2)(1+v^2) - u^2v^2 = 1 + u^2 + v^2 > 0\), so the matrix is positive definite, which is consistent with a Riemannian metric. ✓

B-4. Metric Tensor on a Sphere at a Specific Point

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Solution strategy: Substitute \(\theta = \pi/3\) into the metric tensor components.

The metric tensor of the sphere is:

\[ g_{ij} = \begin{pmatrix} a^2 & 0 \\ 0 & a^2\sin^2\theta \end{pmatrix} \]

At \(\theta = \pi/3\), since \(\sin(\pi/3) = \dfrac{\sqrt{3}}{2}\), we have \(\sin^2(\pi/3) = \dfrac{3}{4}\).

\[ \boxed{g_{ij}\bigg|_{(\pi/3,\,0)} = \begin{pmatrix} a^2 & 0 \\ 0 & \dfrac{3}{4}\,a^2 \end{pmatrix}} \]

Verification: At \(\theta = \pi/2\) (the equator), \(g_{22} = a^2\) is maximized; at \(\theta = 0\) (the north pole), \(g_{22} = 0\) and the metric degenerates. Since \(\theta = \pi/3\) lies between the equator and the north pole, \(g_{22} = \frac{3}{4}a^2\) falls between \(0\) and \(a^2\), which is consistent. ✓

B-5. Jacobian Matrix of a Linear Coordinate Transformation

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Solution strategy: Compute each partial derivative directly.

From \(x = u + v\), \(y = u - v\):

\[ \frac{\partial x}{\partial u} = 1, \quad \frac{\partial x}{\partial v} = 1, \quad \frac{\partial y}{\partial u} = 1, \quad \frac{\partial y}{\partial v} = -1 \]
\[ \boxed{J = \frac{\partial x^i}{\partial u^j} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}} \]

Verification: \(\det J = 1 \cdot (-1) - 1 \cdot 1 = -2 \neq 0\), so it is nonsingular. This is consistent with the existence of the inverse transformation \(u = \frac{x+y}{2}\), \(v = \frac{x-y}{2}\). ✓

B-6. Application of the Metric Tensor Transformation Law

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Solution strategy: Compute \(g' = J^T g\,J = J^T J\) (since \(g = \delta_{ij}\)).

\[ g' = J^T J = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}^T \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \]

Computing each component:

  • \((1,1)\): \(1 \cdot 1 + 1 \cdot 1 = 2\)
  • \((1,2)\): \(1 \cdot 1 + 1 \cdot (-1) = 0\)
  • \((2,1)\): \(1 \cdot 1 + (-1) \cdot 1 = 0\)
  • \((2,2)\): \(1 \cdot 1 + (-1)(-1) = 2\)
\[ \boxed{g'_{kl} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}} \]

That is, the line element becomes \(ds^2 = 2\,du^2 + 2\,dv^2\).

Verification: Confirm by direct substitution. From \(dx = du + dv\), \(dy = du - dv\):

\[ ds^2 = dx^2 + dy^2 = (du+dv)^2 + (du-dv)^2 = 2\,du^2 + 2\,dv^2 \quad \checkmark \]

B-7. Cartesian Components of Coordinate Basis Vectors

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Solution Strategy: Find the Cartesian components of \(\boldsymbol{e}_\theta\) using partial derivatives, then substitute specific values.

From the definition of coordinate basis vectors:

\[ \boldsymbol{e}_\theta = \frac{\partial x}{\partial \theta}\,\boldsymbol{e}_x + \frac{\partial y}{\partial \theta}\,\boldsymbol{e}_y \]

From \(x = r\cos\theta\), \(y = r\sin\theta\):

\[ \frac{\partial x}{\partial \theta} = -r\sin\theta, \quad \frac{\partial y}{\partial \theta} = r\cos\theta \]

Substituting \(r = 3\), \(\theta = \pi/4\), and using \(\sin(\pi/4) = \cos(\pi/4) = \dfrac{\sqrt{2}}{2}\):

\[ (\boldsymbol{e}_\theta)^x = -3 \cdot \frac{\sqrt{2}}{2} = -\frac{3\sqrt{2}}{2} \]
\[ (\boldsymbol{e}_\theta)^y = 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \]
\[ \boxed{(\boldsymbol{e}_\theta)^x = -\frac{3\sqrt{2}}{2}, \quad (\boldsymbol{e}_\theta)^y = \frac{3\sqrt{2}}{2}} \]

Verification: The squared magnitude of \(\boldsymbol{e}_\theta\) is \(|\boldsymbol{e}_\theta|^2 = \left(\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{3\sqrt{2}}{2}\right)^2 = \frac{18}{4} + \frac{18}{4} = 9 = r^2\). This is consistent with the metric tensor component \(g_{\theta\theta} = r^2 = 9\). ✓

B-8. Determinant of the Jacobian Matrix of the Inverse Transformation

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Solution strategy: Derive the determinant relationship from the relation \(J\tilde{J} = I\).

Taking the determinant of both sides of \(J\tilde{J} = I\):

\[ \det J \cdot \det \tilde{J} = \det I = 1 \]

From Problem B-1. Jacobian of 2D Polar Coordinates, \(\det J = r\), so:

\[ \det \tilde{J} = \frac{1}{\det J} = \frac{1}{r} \]
\[ \boxed{\det \tilde{J} = \frac{1}{r}} \]

\(\det J\) and \(\det \tilde{J}\) are reciprocals of each other.

Verification: We confirm this by directly using the Jacobian matrix of the inverse transformation. From the main text:

\[ \tilde{J} = \begin{pmatrix} \cos\theta & \sin\theta \\[4pt] -\dfrac{\sin\theta}{r} & \dfrac{\cos\theta}{r} \end{pmatrix} \]
\[ \det \tilde{J} = \frac{\cos^2\theta}{r} + \frac{\sin^2\theta}{r} = \frac{1}{r} \quad \checkmark \]

Medium

M-1. Coordinate Transformation \((u, v)\) and the Metric Tensor

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Solution strategy: Find the Jacobian matrix of the inverse transformation, derive the metric tensor using the transformation law, and verify by direct substitution.

(a) Jacobian matrix \(\partial(x, y)/\partial(u, v)\)

We find the inverse of the coordinate transformation \(u = x + y\), \(v = x - y\). Solving the system of equations gives:

\[ x = \frac{u + v}{2}, \qquad y = \frac{u - v}{2} \]

The Jacobian matrix is:

\[ \frac{\partial(x, y)}{\partial(u, v)} = \begin{pmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[8pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{pmatrix} = \begin{pmatrix} \dfrac{1}{2} & \dfrac{1}{2} \\[8pt] \dfrac{1}{2} & -\dfrac{1}{2} \end{pmatrix} \]
\[ \boxed{J = \frac{\partial(x, y)}{\partial(u, v)} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}} \]

(b) Derivation of the metric tensor \(g'_{ij}\) using the transformation law

We use the transformation law \(g'_{kl} = \dfrac{\partial x^i}{\partial u^k}\dfrac{\partial x^j}{\partial u^l}\,g_{ij}\). Since the metric in the original coordinates is \(g_{ij} = \delta_{ij}\) (Cartesian coordinates), we simply compute \(g' = J^T J\).

\[ g' = J^T J = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}^T \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix} \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix} \]

Computing each component:

  • \((1,1)\): \(\dfrac{1}{2}\cdot\dfrac{1}{2} + \dfrac{1}{2}\cdot\dfrac{1}{2} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}\)

  • \((1,2)\): \(\dfrac{1}{2}\cdot\dfrac{1}{2} + \dfrac{1}{2}\cdot\left(-\dfrac{1}{2}\right) = \dfrac{1}{4} - \dfrac{1}{4} = 0\)

  • \((2,1)\): \(\dfrac{1}{2}\cdot\dfrac{1}{2} + \left(-\dfrac{1}{2}\right)\cdot\dfrac{1}{2} = \dfrac{1}{4} - \dfrac{1}{4} = 0\)

  • \((2,2)\): \(\dfrac{1}{2}\cdot\dfrac{1}{2} + \left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right) = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}\)

\[ \boxed{g'_{ij} = \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix}} \]

(c) Verification by direct substitution

From \(x = (u+v)/2\), \(y = (u-v)/2\), we have:

\[ dx = \frac{du + dv}{2}, \qquad dy = \frac{du - dv}{2} \]

Substituting into \(ds^2 = dx^2 + dy^2\):

\[ ds^2 = \left(\frac{du + dv}{2}\right)^2 + \left(\frac{du - dv}{2}\right)^2 \]
\[ = \frac{du^2 + 2\,du\,dv + dv^2}{4} + \frac{du^2 - 2\,du\,dv + dv^2}{4} \]
\[ = \frac{2\,du^2 + 2\,dv^2}{4} = \frac{1}{2}\,du^2 + \frac{1}{2}\,dv^2 \]
\[ \boxed{ds^2 = \frac{1}{2}\,du^2 + \frac{1}{2}\,dv^2} \]

This is in perfect agreement with the metric tensor \(g'_{ij} = \text{diag}(1/2,\, 1/2)\) obtained in (b). ✓

Consistency check: \(\det J = \dfrac{1}{2}\cdot\left(-\dfrac{1}{2}\right) - \dfrac{1}{2}\cdot\dfrac{1}{2} = -\dfrac{1}{2} \neq 0\), so the matrix is non-singular. The relation \(\det g' = 1/4 = (\det J)^2\) is also consistent. Since the metric is diagonal and constant, this coordinate system is orthogonal, and the space is flat (consistent with the discussion in Problem M-5. Metric and Flatness of a Cylindrical Surface). ✓

M-2. Derivation of the Line Element in Spherical Coordinates

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Solution strategy: Find the total differentials of \(x = r\sin\theta\cos\varphi\), \(y = r\sin\theta\sin\varphi\), \(z = r\cos\theta\), and substitute into \(ds^2 = dx^2 + dy^2 + dz^2\).

Computing the total differentials

Computing \(dx\):

\[ dx = \frac{\partial x}{\partial r}\,dr + \frac{\partial x}{\partial \theta}\,d\theta + \frac{\partial x}{\partial \varphi}\,d\varphi \]
\[ dx = \sin\theta\cos\varphi\,dr + r\cos\theta\cos\varphi\,d\theta - r\sin\theta\sin\varphi\,d\varphi \]

Computing \(dy\):

\[ dy = \sin\theta\sin\varphi\,dr + r\cos\theta\sin\varphi\,d\theta + r\sin\theta\cos\varphi\,d\varphi \]

Computing \(dz\):

\[ dz = \cos\theta\,dr - r\sin\theta\,d\theta \]

Expanding \(dx^2\)

\[ dx^2 = \sin^2\theta\cos^2\varphi\,dr^2 + r^2\cos^2\theta\cos^2\varphi\,d\theta^2 + r^2\sin^2\theta\sin^2\varphi\,d\varphi^2 $$ $$ + 2r\sin\theta\cos\theta\cos^2\varphi\,dr\,d\theta - 2r\sin^2\theta\sin\varphi\cos\varphi\,dr\,d\varphi - 2r^2\sin\theta\cos\theta\sin\varphi\cos\varphi\,d\theta\,d\varphi \]

Expanding \(dy^2\)

\[ dy^2 = \sin^2\theta\sin^2\varphi\,dr^2 + r^2\cos^2\theta\sin^2\varphi\,d\theta^2 + r^2\sin^2\theta\cos^2\varphi\,d\varphi^2 $$ $$ + 2r\sin\theta\cos\theta\sin^2\varphi\,dr\,d\theta + 2r\sin^2\theta\sin\varphi\cos\varphi\,dr\,d\varphi + 2r^2\sin\theta\cos\theta\sin\varphi\cos\varphi\,d\theta\,d\varphi \]

Expanding \(dz^2\)

\[ dz^2 = \cos^2\theta\,dr^2 + r^2\sin^2\theta\,d\theta^2 - 2r\sin\theta\cos\theta\,dr\,d\theta \]

Summing the terms

Coefficient of \(dr^2\):

\[ \sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta = \sin^2\theta(\cos^2\varphi + \sin^2\varphi) + \cos^2\theta = \sin^2\theta + \cos^2\theta = 1 \]

Coefficient of \(d\theta^2\):

\[ r^2\cos^2\theta\cos^2\varphi + r^2\cos^2\theta\sin^2\varphi + r^2\sin^2\theta = r^2\cos^2\theta(\cos^2\varphi + \sin^2\varphi) + r^2\sin^2\theta = r^2 \]

Coefficient of \(d\varphi^2\):

\[ r^2\sin^2\theta\sin^2\varphi + r^2\sin^2\theta\cos^2\varphi = r^2\sin^2\theta(\sin^2\varphi + \cos^2\varphi) = r^2\sin^2\theta \]

Coefficient of \(dr\,d\theta\):

\[ 2r\sin\theta\cos\theta\cos^2\varphi + 2r\sin\theta\cos\theta\sin^2\varphi - 2r\sin\theta\cos\theta $$ $$ = 2r\sin\theta\cos\theta(\cos^2\varphi + \sin^2\varphi) - 2r\sin\theta\cos\theta = 2r\sin\theta\cos\theta - 2r\sin\theta\cos\theta = 0 \]

Coefficient of \(dr\,d\varphi\):

\[ -2r\sin^2\theta\sin\varphi\cos\varphi + 2r\sin^2\theta\sin\varphi\cos\varphi = 0 \]

Coefficient of \(d\theta\,d\varphi\):

\[ -2r^2\sin\theta\cos\theta\sin\varphi\cos\varphi + 2r^2\sin\theta\cos\theta\sin\varphi\cos\varphi = 0 \]

Final result

All cross terms vanish:

\[ \boxed{ds^2 = dr^2 + r^2\,d\theta^2 + r^2\sin^2\theta\,d\varphi^2} \]

Verification:

  • Restricting to \(\theta = \pi/2\) (the \(xy\) plane) gives \(ds^2 = dr^2 + r^2\,d\varphi^2\), which matches the line element in 2D polar coordinates. ✓
  • Dimensional analysis: each term has dimensions of [length]\(^2\). ✓

M-3. Metric Tensor in Parabolic Coordinates

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(a) Jacobian Matrix

From \(x = \sigma\tau\), \(y = \frac{1}{2}(\tau^2 - \sigma^2)\):

\[ J = \begin{pmatrix} \dfrac{\partial x}{\partial \sigma} & \dfrac{\partial x}{\partial \tau} \\[8pt] \dfrac{\partial y}{\partial \sigma} & \dfrac{\partial y}{\partial \tau} \end{pmatrix} = \begin{pmatrix} \tau & \sigma \\ -\sigma & \tau \end{pmatrix} \]
\[ \boxed{J = \begin{pmatrix} \tau & \sigma \\ -\sigma & \tau \end{pmatrix}} \]

(b) Derivation of the Metric Tensor

Computing \(g' = J^T J\):

\[ g' = \begin{pmatrix} \tau & -\sigma \\ \sigma & \tau \end{pmatrix} \begin{pmatrix} \tau & \sigma \\ -\sigma & \tau \end{pmatrix} \]

Each component:

  • \((1,1)\): \(\tau^2 + \sigma^2\)
  • \((1,2)\): \(\tau\sigma - \sigma\tau = 0\)
  • \((2,1)\): \(\sigma\tau - \tau\sigma = 0\)
  • \((2,2)\): \(\sigma^2 + \tau^2\)
\[ \boxed{g'_{kl} = \begin{pmatrix} \sigma^2 + \tau^2 & 0 \\ 0 & \sigma^2 + \tau^2 \end{pmatrix}} \]

Therefore the line element is:

\[ \boxed{ds^2 = (\sigma^2 + \tau^2)(d\sigma^2 + d\tau^2)} \]

Verification: The metric takes the conformally flat form \(ds^2 = \Omega^2(d\sigma^2 + d\tau^2)\). This means that parabolic coordinates form an orthogonal coordinate system (\(g_{12} = 0\)), which is consistent with the properties of conformal mappings.

Furthermore, we can verify that \(\det J = \tau^2 + \sigma^2\) and \(\det g' = (\sigma^2 + \tau^2)^2 = (\det J)^2\) holds. ✓

M-4. Proof of Symmetry of the Metric Tensor

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Solution Strategy: Swap indices on the right-hand side of the transformation law and use \(g_{ij} = g_{ji}\).

The transformation law for the metric tensor is:

\[ g'_{kl} = \frac{\partial x^i}{\partial u^k}\frac{\partial x^j}{\partial u^l}\,g_{ij} \]

Writing out \(g'_{lk}\):

\[ g'_{lk} = \frac{\partial x^i}{\partial u^l}\frac{\partial x^j}{\partial u^k}\,g_{ij} \]

Relabeling the dummy indices \(i\) and \(j\) on the right-hand side (simply renaming the summation indices does not change the value):

\[ g'_{lk} = \frac{\partial x^j}{\partial u^l}\frac{\partial x^i}{\partial u^k}\,g_{ji} \]

Using \(g_{ji} = g_{ij}\) (the symmetry assumption in the original coordinate system):

\[ g'_{lk} = \frac{\partial x^i}{\partial u^k}\frac{\partial x^j}{\partial u^l}\,g_{ij} = g'_{kl} \]
\[ \boxed{g'_{kl} = g'_{lk}} \]

Therefore, if the metric tensor is symmetric in the original coordinate system, symmetry is preserved under any coordinate transformation. \(\blacksquare\)

Verification: We confirm using the concrete example in Problem B-6. Application of the Metric Tensor Transformation Law. \(g'_{kl} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\) is clearly symmetric. In the parabolic coordinates of Problem M-3. Metric Tensor in Parabolic Coordinates, \(g'_{12} = g'_{21} = 0\), which is also symmetric. ✓

M-5. Metric and Flatness of a Cylindrical Surface

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(a) Matrix Representation of the Metric Tensor

From the line element \(ds^2 = a^2\,d\varphi^2 + dz^2\):

\[ \boxed{g_{ij} = \begin{pmatrix} a^2 & 0 \\ 0 & 1 \end{pmatrix}} \]

(b) Relationship Between Constant Metric Tensor and Flatness

Conclusion: The statement "metric tensor components are constant ⇒ the space is flat" is correct.

The reasoning is given below.

The cylinder surface is flat:

A cylinder surface can be made by rolling up a sheet of paper. The rolling operation does not change any distance relationships on the paper (no stretching or compression), so the intrinsic geometry of the cylinder surface is the same as that of a flat plane. Indeed, performing the coordinate transformation \(s = a\varphi\):

\[ ds^2 = ds_{\text{arc}}^2 + dz^2 \]

which is precisely the line element of a plane in Cartesian coordinates.

Meaning of constant metric tensor components:

When all components of the metric tensor are constant, the metric can be transformed to \(\delta_{ij}\) (the identity matrix) by a coordinate transformation (a linear transformation with a constant matrix). Specifically, if \(g_{ij}\) is a positive-definite symmetric constant matrix, then there exists an appropriate constant matrix \(A\) such that \(A^T g A = I\). The coordinates after this transformation are Cartesian coordinates themselves, and the space is flat.

Contrast with the sphere:

The metric tensor of a sphere \(g_{ij} = \begin{pmatrix} a^2 & 0 \\ 0 & a^2\sin^2\theta \end{pmatrix}\) depends on \(\theta\). However, as discussed in this chapter, the mere fact that the metric tensor depends on coordinates does not imply that the space is curved (even in polar coordinates on a flat plane, \(g_{22} = r^2\) depends on coordinates). Whether a space is truly curved is determined by whether the metric tensor can be made constant (or equal to \(\delta_{ij}\)) through a coordinate transformation. In the case of the sphere, no coordinate transformation can make the metric tensor constant, which reflects the fact that the sphere is intrinsically curved (strictly speaking, this is determined by the Riemann curvature tensor).

\[ \boxed{\text{All metric tensor components are constant} \implies \text{the space is flat}} \]

However, the converse does not hold (even in a flat space, the metric tensor depends on coordinates if curvilinear coordinates are used).

M-6. Geometry of "Circles" on a Sphere

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Strategy: Use the metric of the sphere \(ds^2 = a^2(d\theta^2 + \sin^2\theta\,d\varphi^2)\) to calculate the circumference and distance.

(a) Calculating the circumference \(C\)

On the line \(\theta = \theta_0\) = constant, \(d\theta = 0\), so the line element is

\[ ds = a\sin\theta_0\,d\varphi \]

Integrating \(\varphi\) from \(0\) to \(2\pi\),

\[ C = \int_0^{2\pi} a\sin\theta_0\,d\varphi = 2\pi a\sin\theta_0 \]
\[ \boxed{C = 2\pi a\sin\theta_0} \]

(b) Distance \(r\) from the north pole along the sphere

The distance along the sphere from the north pole (\(\theta = 0\)) to the line \(\theta = \theta_0\) is measured along a great circle with \(\varphi\) = constant. Since \(d\varphi = 0\),

\[ ds = a\,d\theta \]

Integrating \(\theta\) from \(0\) to \(\theta_0\),

\[ r = \int_0^{\theta_0} a\,d\theta = a\theta_0 \]
\[ \boxed{r = a\theta_0} \]

(c) Calculating \(C/(2\pi r)\)

\[ \frac{C}{2\pi r} = \frac{2\pi a\sin\theta_0}{2\pi \cdot a\theta_0} = \frac{\sin\theta_0}{\theta_0} \]
\[ \boxed{\frac{C}{2\pi r} = \frac{\sin\theta_0}{\theta_0}} \]

Verifying that this becomes less than 1 as \(\theta_0\) increases:

For \(\theta_0 > 0\), the inequality \(\sin\theta_0 < \theta_0\) holds (\(\sin x < x\) for \(x > 0\)). Therefore,

\[ \frac{\sin\theta_0}{\theta_0} < 1 \quad (\theta_0 > 0) \]

In the limit \(\theta_0 \to 0\), \(\sin\theta_0/\theta_0 \to 1\), so for sufficiently small circles the flat-space ratio \(C = 2\pi r\) is recovered. As \(\theta_0\) increases, the ratio decreases from 1; at \(\theta_0 = \pi/2\) (the equator), \(C/(2\pi r) = \sin(\pi/2)/(\pi/2) = 2/\pi \approx 0.637\).

Physical interpretation: In flat space, the ratio of circumference to radius is always \(C = 2\pi r\) (\(C/(2\pi r) = 1\)). The fact that this ratio is less than 1 on the sphere is a direct manifestation of the sphere's positive curvature. In a space with positive curvature, the circumference is "deficient" relative to the distance from the center—it is shorter than in the flat-space case. This is an intrinsic geometric property of the sphere, a quantity that can be calculated directly from the metric tensor.

Consistency check: At \(\theta_0 = \pi\) (the "circle" enclosing the south pole), \(C = 2\pi a\sin\pi = 0\) and \(r = a\pi\). Since the south pole is a single point, the circumference should indeed be zero. ✓

Taylor expanding for \(\theta_0 \to 0\) gives \(\sin\theta_0 \approx \theta_0 - \theta_0^3/6\), so \(C/(2\pi r) \approx 1 - \theta_0^2/6\). It is known that for a sphere with curvature \(K = 1/a^2\), the relation \(C/(2\pi r) \approx 1 - Kr^2/6\) holds. Substituting \(r = a\theta_0\) gives \(1 - (1/a^2)(a\theta_0)^2/6 = 1 - \theta_0^2/6\), which agrees. ✓

M-7. Derivation of \(g'_{33}\) using the metric tensor transformation law

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Solution Strategy: Starting from the relationship between spherical coordinates \((r, \theta, \varphi)\) and Cartesian coordinates \((x, y, z)\), derive \(g'_{33} = r^2\sin^2\theta\) using the transformation law.

Calculation

Coordinate correspondence: let \(u^1 = r\), \(u^2 = \theta\), \(u^3 = \varphi\), and \(x^1 = x\), \(x^2 = y\), \(x^3 = z\).

The transformation law gives:

\[ g'_{33} = \frac{\partial x^i}{\partial u^3}\frac{\partial x^j}{\partial u^3}\,g_{ij} = \sum_{i=1}^{3}\sum_{j=1}^{3}\frac{\partial x^i}{\partial \varphi}\frac{\partial x^j}{\partial \varphi}\,\delta_{ij} = \sum_{i=1}^{3}\left(\frac{\partial x^i}{\partial \varphi}\right)^2 \]

(Since \(g_{ij} = \delta_{ij}\), only the \(i = j\) terms survive.)

From the coordinate transformation \(x = r\sin\theta\cos\varphi\), \(y = r\sin\theta\sin\varphi\), \(z = r\cos\theta\), we compute each partial derivative:

\[ \frac{\partial x}{\partial \varphi} = -r\sin\theta\sin\varphi \]
\[ \frac{\partial y}{\partial \varphi} = r\sin\theta\cos\varphi \]
\[ \frac{\partial z}{\partial \varphi} = 0 \]

Substituting into the transformation law:

\[ g'_{33} = (-r\sin\theta\sin\varphi)^2 + (r\sin\theta\cos\varphi)^2 + 0^2 \]
\[ = r^2\sin^2\theta\sin^2\varphi + r^2\sin^2\theta\cos^2\varphi \]
\[ = r^2\sin^2\theta(\sin^2\varphi + \cos^2\varphi) \]
\[ = r^2\sin^2\theta \cdot 1 \]
\[ \boxed{g'_{33} = r^2\sin^2\theta} \]

This agrees with the known value of the \((3,3)\) component of the metric tensor in spherical coordinates.

Verification: This matches the coefficient of \(d\varphi^2\) in the spherical coordinate line element \(ds^2 = dr^2 + r^2\,d\theta^2 + r^2\sin^2\theta\,d\varphi^2\) derived by direct substitution in Problem M-2. Derivation of the Line Element in Spherical Coordinates. ✓

At \(\theta = \pi/2\) (the equatorial plane), \(g'_{33} = r^2\), which reduces to \(g_{\varphi\varphi} = r^2\) for 2-dimensional polar coordinates. ✓

At \(\theta = 0\) (on the \(z\)-axis), \(g'_{33} = 0\), which is consistent with the fact that the \(\varphi\) coordinate degenerates (changes in \(\varphi\) do not contribute to distance on the \(z\)-axis). ✓

Advanced

A-1. Metric Tensor in General Curvilinear Coordinates

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(a) Derivation of the Metric Tensor

For \(x = f(u^1, u^2)\), \(y = g(u^1, u^2)\), we transform the Cartesian metric \(ds^2 = dx^2 + dy^2\).

The total differentials are:

\[ dx = \frac{\partial f}{\partial u^i}\,du^i, \quad dy = \frac{\partial g}{\partial u^i}\,du^i \]

From \(ds^2 = dx^2 + dy^2 = g_{ij}\,du^i\,du^j\):

\[ g_{ij} = \frac{\partial f}{\partial u^i}\frac{\partial f}{\partial u^j} + \frac{\partial g}{\partial u^i}\frac{\partial g}{\partial u^j} \]

Using the shorthand notation \(f_i \equiv \dfrac{\partial f}{\partial u^i}\), \(g_i \equiv \dfrac{\partial g}{\partial u^i}\):

\[ \boxed{g_{ij} = f_i f_j + g_i g_j} \]

Writing out the components explicitly:

\[ g_{11} = f_1^2 + g_1^2, \quad g_{12} = g_{21} = f_1 f_2 + g_1 g_2, \quad g_{22} = f_2^2 + g_2^2 \]

(b) Orthogonality Condition

The condition for the coordinate basis vectors to be orthogonal, \(g_{12} = 0\), is:

\[ \boxed{f_1 f_2 + g_1 g_2 = 0 \quad \text{i.e.} \quad \frac{\partial f}{\partial u^1}\frac{\partial f}{\partial u^2} + \frac{\partial g}{\partial u^1}\frac{\partial g}{\partial u^2} = 0} \]

This means that the inner product of the two coordinate basis vectors \(\boldsymbol{e}_1 = (f_1, g_1)\) and \(\boldsymbol{e}_2 = (f_2, g_2)\) is zero.

(c) Verification with Specific Examples

Polar coordinates \((u^1, u^2) = (r, \theta)\):

From \(f = r\cos\theta\), \(g = r\sin\theta\):

\[ f_1 = \cos\theta, \quad f_2 = -r\sin\theta, \quad g_1 = \sin\theta, \quad g_2 = r\cos\theta \]

Checking the orthogonality condition:

\[ f_1 f_2 + g_1 g_2 = \cos\theta \cdot (-r\sin\theta) + \sin\theta \cdot r\cos\theta = -r\sin\theta\cos\theta + r\sin\theta\cos\theta = 0 \quad \checkmark \]

Polar coordinates form an orthogonal coordinate system.

Parabolic coordinates \((u^1, u^2) = (\sigma, \tau)\):

From \(f = \sigma\tau\), \(g = \frac{1}{2}(\tau^2 - \sigma^2)\):

\[ f_1 = \tau, \quad f_2 = \sigma, \quad g_1 = -\sigma, \quad g_2 = \tau \]

Checking the orthogonality condition:

\[ f_1 f_2 + g_1 g_2 = \tau \cdot \sigma + (-\sigma) \cdot \tau = \sigma\tau - \sigma\tau = 0 \quad \checkmark \]

Parabolic coordinates also form an orthogonal coordinate system.

Verification: This is consistent with the fact that the metric tensor for parabolic coordinates obtained in Problem M-3. Metric Tensor in Parabolic Coordinates is a diagonal matrix (\(g_{12} = 0\)). The polar coordinate metric \(ds^2 = dr^2 + r^2\,d\theta^2\) is also diagonal, which is consistent. ✓

A-2. Rindler Coordinates

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(a) Components of the Jacobian Matrix

From \(ct = \xi\sinh\eta\), \(x = \xi\cosh\eta\):

\[ \frac{\partial(ct)}{\partial \eta} = \xi\cosh\eta, \quad \frac{\partial(ct)}{\partial \xi} = \sinh\eta \]
\[ \frac{\partial x}{\partial \eta} = \xi\sinh\eta, \quad \frac{\partial x}{\partial \xi} = \cosh\eta \]
\[ \boxed{J = \begin{pmatrix} \xi\cosh\eta & \sinh\eta \\ \xi\sinh\eta & \cosh\eta \end{pmatrix}} \]

(Rows correspond to \((ct, x)\), columns correspond to \((\eta, \xi)\))

(b) Derivation of the Line Element in Rindler Coordinates

The Minkowski metric is \(g_{\mu\nu} = \text{diag}(-1, +1, +1, +1)\). Since the \(y\) and \(z\) directions are not transformed, we consider only the \((ct, x)\) part.

Applying the transformation rule \(g'_{kl} = \dfrac{\partial x^\mu}{\partial u^k}\dfrac{\partial x^\nu}{\partial u^l}\,g_{\mu\nu}\):

Calculation of \(g'_{\eta\eta}\):

\[ g'_{\eta\eta} = \left(\frac{\partial(ct)}{\partial \eta}\right)^2 g_{00} + \left(\frac{\partial x}{\partial \eta}\right)^2 g_{11} \]
\[ = (\xi\cosh\eta)^2 \cdot (-1) + (\xi\sinh\eta)^2 \cdot (+1) \]
\[ = -\xi^2\cosh^2\eta + \xi^2\sinh^2\eta = -\xi^2(\cosh^2\eta - \sinh^2\eta) = -\xi^2 \]

Here we used the hyperbolic identity \(\cosh^2\eta - \sinh^2\eta = 1\).

Calculation of \(g'_{\xi\xi}\):

\[ g'_{\xi\xi} = (\sinh\eta)^2 \cdot (-1) + (\cosh\eta)^2 \cdot (+1) \]
\[ = -\sinh^2\eta + \cosh^2\eta = \cosh^2\eta - \sinh^2\eta = 1 \]

Calculation of \(g'_{\eta\xi}\):

\[ g'_{\eta\xi} = \xi\cosh\eta \cdot \sinh\eta \cdot (-1) + \xi\sinh\eta \cdot \cosh\eta \cdot (+1) \]
\[ = -\xi\cosh\eta\sinh\eta + \xi\sinh\eta\cosh\eta = 0 \]

Since the \(y\) and \(z\) directions are not transformed, \(g'_{yy} = 1\), \(g'_{zz} = 1\).

Therefore:

\[ \boxed{ds^2 = -\xi^2\,d\eta^2 + d\xi^2 + dy^2 + dz^2} \]

Verification: \(\det J = \xi\cosh^2\eta - \xi\sinh^2\eta = \xi\), so the transformation is non-singular for \(\xi > 0\). Also, the worldline with \(\xi = \text{const}\) satisfies \(x^2 - c^2t^2 = \xi^2\), which is a hyperbola corresponding to the worldline of uniformly accelerated motion. ✓

(c) Discussion of Physical Meaning

Rindler coordinates describe the coordinate system of an observer undergoing uniform acceleration with magnitude \(a = c^2/\xi\). By the equivalence principle, this coordinate system is locally equivalent to a stationary coordinate system in a uniform gravitational field.

Physical meaning of \(g_{\eta\eta} = -\xi^2\) depending on \(\xi\):

The relationship between coordinate time \(\eta\) and proper time \(\tau\) for a stationary observer (\(d\xi = dy = dz = 0\)) is:

\[ ds^2 = -c^2\,d\tau^2 = -\xi^2\,d\eta^2 \quad \implies \quad d\tau = \frac{\xi}{c}\,d\eta \]

(In units where \(c = 1\): \(d\tau = \xi\,d\eta\))

This means that even when the same coordinate time \(d\eta\) elapses, proper time advances at different rates at locations with different values of \(\xi\). Specifically:

  • At locations with large \(\xi\) ("upward" in the acceleration direction; in the gravitational analogy, where the gravitational potential is higher), proper time advances faster
  • At locations with small \(\xi\) ("downward"; where the gravitational potential is lower), proper time advances slower
  • As \(\xi \to 0\), \(g_{\eta\eta} \to 0\), and the advance of proper time stops (the Rindler horizon)

This is precisely the manifestation of gravitational time dilation discussed in Ch. 5. According to the equivalence principle, in a uniform gravitational field, "time passes faster at higher locations." The \(\xi\) dependence of \(g_{\eta\eta} = -\xi^2\) in the Rindler metric precisely expresses this effect in the language of the metric tensor.

Relation to gravitational redshift:

When two observers at different \(\xi\) positions exchange signals with the same coordinate time interval \(\Delta\eta\), their respective proper time intervals are \(\Delta\tau_1 = \xi_1\,\Delta\eta\) and \(\Delta\tau_2 = \xi_2\,\Delta\eta\). When \(\xi_1 < \xi_2\), we have \(\Delta\tau_1 < \Delta\tau_2\), meaning that light emitted from "below" appears redshifted to the observer "above." This is gravitational redshift, a direct consequence derived from the equivalence principle.

\[ \boxed{\text{The $\xi$ dependence of $g_{\eta\eta}$ reflects gravitational time dilation in a uniformly accelerating frame ($\approx$ uniform gravitational field)}} \]

Supplementary note: \(\xi = 0\) is called the Rindler horizon, corresponding to the boundary that is causally inaccessible from the perspective of the uniformly accelerating observer. This shares similarities with the event horizon of a black hole and serves as an important model for understanding the physics of horizons in general relativity.