Repeat the argument that led to Eq. (5.1) under more realistic assumptions: suppose a fraction ε of the kinetic energy of the mass at the bottom can be converted into a photon and sent back up, the remaining energy staying at ground level in a useful form. Devise a perpetual motion engine if Eq. (5.1) is violated.

Trivial.

Explain why a uniform external gravitational field would raise no tides on Earth.

Trivial.

1. Show that the coordinate transformation \( (x, y) \to (\psi , \eta) \) with \( \psi = x \) and \( \eta = 1 \) violates Eq. (5.6).

   Trivial.

2. Are the following coordinate transformations good ones? Compute the Jacobian and list any points at which the transformations fail.

   1. \( \psi = (x^2 + y^2)^{1/2}, \eta = \arctan (y/x) \);

      No. It would be fine with the restriction \( x > 0 \).

   2. \( \psi = \ln x, \eta = y \);

      Same as (i).

   3. \( \psi = \arctan (y/x), \eta = (x^2 + y^2)^{−1/2} \).

      Same as (i).

A curve is defined by \( \{ x = f (\lambda), y = g(\lambda), 0 \leq λ \leq 1 \} \). Show that the tangent vector \( (\mathrm{d} x/\mathrm{d} λ, \mathrm{d} y/\mathrm{d} λ) \) does actually lie tangent to the curve.

Trivial.

Sketch the following curves. Which have the same paths? Find also their tangent vectors where the parameter equals zero.

(a) \( x = \sin \lambda, y = \cos \lambda \); (b) \( x = \cos (2 \pi t^2), y = \sin (2 \pi t^2 + \pi) \); (c) \( x = s, y = s + 4 \); (d) \( x = s^2, y = −(s − 2)(s + 2) \); (e) \( x = \mu, y = 1 \).

Trivial. (a) and (b) have the same path.

Justify the pictures in Fig. 5.5.

Trivial.

Calculate all elements of the transformation matrices \( {\Lambda^{\alpha'}}_\beta \) and \( {\Lambda^\mu}_{\nu'} \) for the transformation from Cartesian \( (x, y) \) – the unprimed indices – to polar \( (r, \theta) \) – the primed indices.

The transformation matrix of \( {\Lambda^{\alpha'}}_\beta \) is \( \begin{pmatrix} \frac{x}{r} & \frac{y}{r} \\ -\frac{y}{r^2} & \frac{x}{r^2} \end{pmatrix} \). The transformation matrix of \( {\Lambda^\mu}_{\nu'} \) is \( \begin{pmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{pmatrix} \).

1. (Uses the result of Exer. 7.) Let \( f = x^2 + y^2 + 2 x y \), and in Cartesian coordinates \( \vec{V} \to (x^2 + 3 y, y^2 + 3 x) \), \( \vec{W} \to (1, 1) \). Compute \( f \) as a function of \( r \) and \( \theta \), and find the components of \( \vec{V} \) and \( \vec{W} \) on the polar basis, expressing them as functions of \( r \) and \( \theta \).

   It is easy to verify that \( f = r^2 (1 + 2 \cos \theta \sin \theta) \).

   Further, \( \vec{V} \to_{(r, \theta)} (r^2 (\cos^3 \theta + \sin^3 \theta) + 6 r \cos \theta \sin \theta, r (\cos \theta \sin^2 \theta - \cos^2 \theta \sin \theta) + 3 \cos^2 \theta - 3 \sin^2 \theta) \), and \( \vec{W} \to_{(r, \theta)} (\cos \theta + \sin \theta, \frac{\cos \theta - \sin \theta}{r}) \).

2. Find the components of \( \tilde{\mathrm{d}} f \) in Cartesian coordinates and obtain them in polars (i) by direct calculation in polars, and (ii) by transforming components from Cartesian.

   \( \tilde{\mathrm{d}} f \to_{(x, y)} (2x + 2y, 2x + 2y) \) and \( \tilde{\mathrm{d}} f \to_{(r, \theta)} (2r (\cos \theta + \sin \theta)^2, 2r^2 (\cos^2 \theta - \sin^2 \theta)) \) through both (i) and (ii). One should keep in mind that the components of \( \tilde{\mathrm{d}} f \) should be treated as a row vector and left-multiplied to the matrix form of \( {\Lambda^\mu}_{\nu'} \) in (ii).

3. (i) Use the metric tensor in polar coordinates to find the polar components of the one-forms \( \tilde{V} \) and \( \tilde{W} \) associated with \( \vec{V} \) and \( \vec{W} \). (ii) Obtain the polar components of \( \tilde{V} \) and \( \tilde{W} \) by transformation of their Cartesian components.

   \( \tilde{V} \to_{(r, \theta)} (r^2 (\cos^3 \theta + \sin^3 \theta) + 6 r \cos \theta \sin \theta, r^3 (\cos \theta \sin^2 \theta - \cos^2 \theta \sin \theta) + 3 r^2 \cos^2 \theta - 3 r^2 \sin^2 \theta ) \), and \( \tilde{W} \to_{(r, \theta)} (\cos \theta + \sin \theta, -r \sin \theta + r \cos \theta) \) through both methods.

Draw a diagram similar to Fig. 5.6 to explain Eq. (5.38).

Do it on a piece of paper instead of using LaTex!

Prove that \( \nabla \vec{V} \), defined in Eq. (5.52), is a \( \begin{pmatrix} 1 \\1 \end{pmatrix} \) tensor.

The exercise itself is unclear. \( \nabla \vec{V} \) is defined as a \( \begin{pmatrix} 1 \\1 \end{pmatrix} \) tensor which returns vector \( \frac{\partial \vec{V}}{\partial x^\beta} \) when operating on \( \vec{e}_\beta \). Eq. (5.52) simply gives its components consistent with the definition. Eq. (5.52) should not be understood as the definition of \( \nabla \vec{V} \).

The intention of the author is likely to let the student show that \( {V^\alpha}_{; \beta} \) satisfies the tensor transformation rule, i.e., that \( {V^{\mu'}}_{; \nu'} = {\Lambda^{\mu'}}_\alpha {\Lambda^{\beta}}_{\nu'} {V^\alpha}_{; \beta} \). This can be proved as follows.

By definition, \( {V^{\mu'}}_{; \nu'} \) is the \( \mu' \)-th component of \( \frac{\partial \nabla \vec{V}}{\partial x^{\nu'}} = \frac{\partial \nabla \vec{V}}{\partial x^\beta} \frac{\partial \beta}{\partial \nu'} = \frac{\partial \nabla \vec{V}}{\partial x^\beta} {\Lambda^\beta}_{\nu'} \). Since \( \frac{\partial \nabla \vec{V}}{\partial x^{\nu'}} \) is itself a vector, it follows the tensor transformation rule and

\[ \begin{eqnarray} && {V^{\mu'}}_{; \nu'} \\ &=& (\frac{\partial \nabla \vec{V}}{\partial x^{\nu'}})^{\mu'} \\ &=& (\frac{\partial \nabla \vec{V}}{\partial x^{\nu'}})^\alpha {\Lambda^{\mu'}}_\alpha \\ &=& (\frac{\partial \nabla \vec{V}}{\partial x^\beta} {\Lambda^\beta}_{\nu'})^\alpha {\Lambda^{\mu'}}_\alpha \\ &=& (\frac{\partial \nabla \vec{V}}{\partial x^\beta})^\alpha {\Lambda^\beta}_{\nu'} {\Lambda^{\mu'}}_\alpha \\ &=& {V^\alpha}_{; \beta} {\Lambda^\beta}_{\nu'} {\Lambda^{\mu'}}_\alpha. \end{eqnarray} \]

(Uses the result of Exers. 7 and 8.) For the vector field \( \vec{V} \) whose Cartesian components are \( (x^2 + 3 y, y^2 + 3 x) \), compute: (a) \( {V^\alpha}_{, \beta} \) in Cartesian; (b) the transformation \( {\Lambda^{\mu'}}_\alpha {\Lambda^\beta}_{\nu'} {V^\alpha}_{, \beta} \) to polars; (c) the components \( {V^{\mu'}}_{; \nu'} \) directly in polars using the Christoffel symbols, Eq. (5.45), in Eq. (5.50); (d) the divergence \( {V^\alpha}_{, \alpha} \) using your results in (a); (e) the divergence \( {V^{\mu'}}_{; \mu'} \) using your results in either (b) or (c); (f) the divergence \( {V^{\mu'}}_{; \mu'} \) using Eq. (5.56) directly.

(a) \( \begin{pmatrix} 2x & 3 \\ 3 & 2y \end{pmatrix} \). (b) \( \begin{pmatrix} 2r (\cos^3 \theta + \sin^3 \theta) + 6 \cos \theta \sin \theta & 2 r^2 (\cos \theta \sin^2 \theta - \cos^2 \theta \sin \theta) + 3 r \cos^2 \theta - 3 r \sin^2 \theta \\ 2 \cos \theta \sin^2 \theta - 2 \cos^2 \theta \sin \theta + \frac{3 \cos^2 \theta - 3 \sin^2 \theta}{r} & 2 r (\cos^2 \theta \sin \theta + \cos \theta \sin^2 \theta) - 6 \cos \theta \sin \theta \end{pmatrix} \). (c) Same as (b). (d) \( 2x + 2y \). (e) Same as (d).

For the one-form field \( \tilde{p} \) whose Cartesian components are \( (x^2 + 3 y, y^2 + 3 x) \), compute: (a) \( p_{\alpha, \beta} \) in Cartesian; (b) the transformation \( {\Lambda^\alpha}_{\mu'} {\Lambda^\beta}_{\nu'} p_{\alpha, \beta} \) to polars; (c) the components \( p_{\mu'; \nu'} \) directly in polars using the Christoffel symbols, Eq. (5.45), in Eq. (5.63).

(a) \( \begin{pmatrix} 2x & 3 \\ 3 & 2y \end{pmatrix} \). (b) \( \begin{pmatrix} 2r (\cos^3 \theta + \sin^3 \theta) + 6 \cos \theta \sin \theta & 2 r^2 (\cos \theta \sin^2 \theta - \cos^2 \theta \sin \theta) + 3 r \cos^2 \theta - 3 r \sin^2 \theta \\ 2 r^2 (\cos \theta \sin^2 \theta - \cos^2 \theta \sin \theta) + 3 r \cos^2 \theta - 3 r \sin^2 \theta & 2 r^3 (\cos^2 \theta \sin \theta + \cos \theta \sin^2 \theta) - 6 r^2 \cos \theta \sin \theta \end{pmatrix} \). (c) Same as (b).

For those who have done both Exers. 11 and 12, show in polars that \( g_{\mu' \alpha'} {V^{\alpha'}}_{; \nu'} = p_{\mu'; \nu'} \).

Trivial.

For the tensor whose polar components are \( (A^{rr} = r^2, A^{r \theta} = r \sin \theta, A^{\theta r} = r \cos \theta, A^{\theta \theta} = \tan \theta) \), compute Eq. (5.65) in polars for all possible indices.

\( \nabla_r A^{r r} = 2r \), \( \nabla_\theta A^{r r} = -r^2 (\cos \theta + \sin \theta) \), \( \nabla_r A^{r \theta} = 2 \sin \theta \), \( \nabla_\theta A^{r \theta} = r (1 + \cos \theta - \tan \theta) \), \( \nabla_r A^{\theta r} = 2 \cos \theta \), \( \nabla_\theta A^{\theta r} = r (1 -\sin \theta - \tan \theta) \), \( \nabla_r A^{\theta \theta} = \frac{2 \tan \theta}{r} \), \( \nabla_\theta A^{\theta \theta} = \cos \theta + \sin \theta + \frac{1}{\cos^2 \theta} \).

For the vector whose polar components are \( (V^r = 1, V^\theta = \theta) \), compute in polars all components of the second covariant derivative \( {V^\alpha}_{; \mu ; \nu} \). (Hint: to find the second derivative, treat the first derivative \( {V^\alpha}_{; \mu} \) as any \( \begin{pmatrix} 1 \\1 \end{pmatrix} \) tensor: Eq. (5.66).)

First we have \( {V^r}_{; r} = {V^r}_{; \theta} = {V^\theta}_{; r} = 0 \) and \( {V^\theta}_{; \theta} = \frac{1}{r} \). Then \( {V^r}_{;r ;r} = {V^r}_{;r ;\theta} = {V^r}_{;\theta ;r} = 0 \), \( {V^r}_{;\theta ;\theta} = -1 \), \( {V^\theta}_{;r ;r} = 0 \), \( {V^\theta}_{;r ;\theta} = {V^\theta}_{;\theta ;r} = - \frac{1}{r^2} \), and \( {V^\theta}_{;\theta ;\theta} = 0 \).

Fill in all the missing steps leading from Eq. (5.74) to Eq. (5.75).

There are really no missing steps from from Eq. (5.74) to Eq. (5.75), except for the most obvious ones.

Discover how each expression \( {V^\beta}_{, \alpha} \) and \( V^\mu {\Gamma^{\beta}}_{\mu \alpha} \) separately transforms under a change of coordinates (for \( {\Gamma^{\beta}}_{\mu \alpha} \), begin with Eq. (5.44)). Show that neither is the standard tensor law, but that their sum does obey the standard law.

We have

\[ \begin{eqnarray} && {V^{\nu'}}_{, \mu'} \\ &=& {({\Lambda^{\nu'}}_\beta V^\beta)}_{, \mu'} \\ &=& {\Lambda^{\nu'}}_\beta {V^\beta}_{, \mu'} + {({\Lambda^{\nu'}}_\beta)}_{, \mu'} V^\beta \\ &=& {V^\beta}_{, \alpha} {\Lambda^{\nu'}}_\beta {\Lambda^\alpha}_{, \mu'} + {({\Lambda^{\nu'}}_\beta)}_{, \mu'} V^\beta, \end{eqnarray} \]

and

\[ \begin{eqnarray} && V^{\sigma'} {\Gamma^{\nu'}}_{\sigma' \mu'} \\ &=& V^{\sigma'} {(\frac{\partial \vec{e}_{\sigma'}}{\partial x^{\mu'}})}^{\nu'} \\ &=& V^{\sigma'} {(\frac{\partial (\vec{e}_{\omega} {\Lambda^\omega}_{\sigma'})}{\partial x^{\mu'}})}^{\beta} {\Lambda^{\nu'}}_\beta \\ &=& V^{\sigma'} {(\frac{\partial \vec{e}_{\omega}}{\partial x^{\mu'}} {\Lambda^\omega}_{\sigma'} + \vec{e}_{\omega} \frac{\partial {\Lambda^\omega}_{\sigma'}}{\partial x^{\mu'}})}^{\beta} {\Lambda^{\nu'}}_\beta \\ &=& V^{\sigma'} {(\frac{\partial \vec{e}_{\omega}}{\partial x^{\alpha}} {\Lambda^\alpha}_{\mu'} {\Lambda^\omega}_{\sigma'} + \vec{e}_{\omega} \frac{\partial {\Lambda^\omega}_{\sigma'}}{\partial x^{\mu'}})}^{\beta} {\Lambda^{\nu'}}_\beta \\ &=& V^{\sigma'} {\Gamma^\beta}_{\omega \alpha} {\Lambda^\alpha}_{\mu'} {\Lambda^\omega}_{\sigma'} {\Lambda^{\nu'}}_\beta + V^{\sigma'} {\delta^\beta}_{\omega} ({\Lambda^\omega}_{\sigma'})_{,\mu'} {\Lambda^{\nu'}}_\beta \\ &=& V^{\omega} {\Gamma^\beta}_{\omega \alpha} {\Lambda^{\nu'}}_\beta {\Lambda^\alpha}_{\mu'} + ({\Lambda^\beta}_{\sigma'})_{,\mu'} {\Lambda^{\nu'}}_\beta V^{\sigma'}. \end{eqnarray} \]

Clearly neither \( {V^\beta}_{, \alpha} \) or \( V^\mu {\Gamma^{\beta}}_{\mu \alpha} \) transforms according to the standard tensor law due to the second term in the above equations. Their sum, however, does transform according to the standard tensor law, as shown below.

\[ \begin{eqnarray} && {V^{\nu'}}_{, \mu'} + V^{\sigma'} {\Gamma^{\nu'}}_{\sigma' \mu'} \\ &=& {V^\beta}_{, \alpha} {\Lambda^{\nu'}}_\beta {\Lambda^\alpha}_{, \mu'} + {({\Lambda^{\nu'}}_\beta)}_{, \mu'} V^\beta + V^{\omega} {\Gamma^\beta}_{\omega \alpha} {\Lambda^{\nu'}}_\beta {\Lambda^\alpha}_{\mu'} + ({\Lambda^\beta}_{\sigma'})_{,\mu'} {\Lambda^{\nu'}}_\beta V^{\sigma'} \\ &=& ({V^\beta}_{, \alpha} + V^{\omega} {\Gamma^\beta}_{\omega \alpha}) {\Lambda^{\nu'}}_\beta {\Lambda^\alpha}_{, \mu'} + {({\Lambda^{\nu'}}_\beta)}_{, \mu'} {\Lambda^\beta}_{\sigma'} V^{\sigma'} + ({\Lambda^\beta}_{\sigma'})_{,\mu'} {\Lambda^{\nu'}}_\beta V^{\sigma'} \\ &=& ({V^\beta}_{, \alpha} + V^{\omega} {\Gamma^\beta}_{\omega \alpha}) {\Lambda^{\nu'}}_\beta {\Lambda^\alpha}_{, \mu'} + ({({\Lambda^{\nu'}}_\beta)}_{, \mu'} {\Lambda^\beta}_{\sigma'} + {\Lambda^{\nu'}}_\beta ({\Lambda^\beta}_{\sigma'})_{,\mu'}) V^{\sigma'} \\ &=& ({V^\beta}_{, \alpha} + V^{\omega} {\Gamma^\beta}_{\omega \alpha}) {\Lambda^{\nu'}}_\beta {\Lambda^\alpha}_{, \mu'} + {({\Lambda^{\nu'}}_\beta {\Lambda^\beta}_{\sigma'})}_{, \mu'} V^{\sigma'} \\ &=& ({V^\beta}_{, \alpha} + V^{\omega} {\Gamma^\beta}_{\omega \alpha}) {\Lambda^{\nu'}}_\beta {\Lambda^\alpha}_{, \mu'} + {({\delta^{\nu'}}_{\sigma'})}_{, \mu'} V^{\sigma'} \\ &=& ({V^\beta}_{, \alpha} + V^{\omega} {\Gamma^\beta}_{\omega \alpha}) {\Lambda^{\nu'}}_\beta {\Lambda^\alpha}_{, \mu'} \end{eqnarray} \]

The last equality is because \( {\delta^{\nu'}}_{\sigma'} \) is a constant.

Verify Eq. (5.78).

Trivial.

Verify that the calculation from Eq. (5.81) to Eq. (5.84), when repeated for \( \tilde{\mathrm{d}}r \) and \( \tilde{\mathrm{d}} \theta \), shows them to be a coordinate basis.

For \( \tilde{\mathrm{d}}r \), the unhatted version is the same as the hatted version. For \( \tilde{\mathrm{d}} \theta \), one has \( \tilde{\omega}^\theta = \tilde{\mathrm{d}} \theta = - \frac{\sin \theta}{r} \tilde{\mathrm{d}} x + \frac{\cos \theta}{r} \tilde{\mathrm{d}} y \). It is easy to verify that \( \frac{\partial}{\partial y} (\frac{y}{x^2 + y^2}) + \frac{\partial}{\partial x} (\frac{x}{x^2 + y^2}) = 0 \).

For a noncoordinate basis \( \{ \vec{e}_\mu \} \), define \( \nabla_{\vec{e}_\mu} \vec{e}_{\nu} - \nabla_{\vec{e}_\nu} \vec{e}_{\mu} := {c^\alpha}_{\mu \nu} \vec{e}_\alpha \) and use this in place of Eq. (5.74) to generalize Eq. (5.75).

Now Eq. (5.74) must be replaced by \( {\Gamma^\mu}_{\alpha \beta} - {\Gamma^\mu}_{\beta \alpha} = {c^\mu}_{\beta \alpha} \). Further we have \( g_{\alpha \beta, \mu} + g_{\alpha \mu, \beta} - g_{\beta \mu, \alpha} = {c^\nu}_{\mu \alpha} g_{\nu \beta} + {c^\nu}_{\beta \alpha} g_{\nu \mu} + {c^\nu}_{\beta \mu} g_{\alpha \nu} + 2 g_{\alpha \nu} {\Gamma^\nu}_{\beta \mu} \) and finally \( \frac{1}{2} g^{\alpha \gamma} (g_{\alpha \beta, \mu} + g_{\alpha \mu, \beta} - g_{\beta \mu, \alpha} - {c^\nu}_{\mu \alpha} g_{\nu \beta} - {c^\nu}_{\beta \alpha} g_{\nu \mu} - {c^\nu}_{\beta \mu} g_{\alpha \nu}) = {\Gamma^\gamma}_{\beta \mu} \).

Consider the \( x − t \) plane of an inertial observer in SR. A certain uniformly accelerated observer wishes to set up an orthonormal coordinate system. By Exer. 21, § 2.9, his world line is

\[ t(\lambda) = a \sinh \lambda, x(\lambda) = a \cosh \lambda, \]

where \( a \) is a constant and \( a \lambda \) is his proper time (clock time on his wrist watch).

1. Show that the spacelike line described by Eq. (5.96) with \( a \) as the variable parameter and \( \lambda \) fixed is orthogonal to his world line where they intersect. Changing \( \lambda \) in Eq. (5.96) then generates a family of such lines.

   The world line's tangent vector \( \vec{U} \) has components \( U^t = \frac{\partial t}{\partial \lambda} \frac{\partial \lambda}{\partial \tau} = \cosh \lambda \) and \( U^x = \sinh \lambda \).

   With \( a \) as the variable parameter, the spacelike vector \( \vec{V} \) has components \( V^t = \sinh \lambda \) and \( V^x = \cosh \lambda \).

   It is easy to verify that \( \vec{V} \cdot \vec{U} = 0 \).

2. Show that Eq. (5.96) defines a transformation from coordinates \( (t, x) \) to coordinates \( (\lambda, a) \), which form an orthogonal coordinate system. Draw these coordinates and show that they cover only one half of the original \( t − x \) plane. Show that the coordinates are bad on the lines \( |x| = |t| \), so they really cover two disjoint quadrants.

   We have \( a = \pm \sqrt{x^2 - t^2} \) and \( \lambda = \tanh^{-1} \frac{t}{x} \) and both values are well defined when \( |x| > |t| \). But \( \lambda \) is undefined on \( |x| = |t| \).

3. Find the metric tensor and all the Christoffel symbols in this coordinate system. This observer will do a perfectly good job, provided that he always uses Christoffel symbols appropriately and sticks to events in his quadrant. In this sense, SR admits accelerated observers. The right-hand quadrant in these coordinates is sometimes called Rindler space, and the boundary lines \( x = \pm t \) bear some resemblance to the black-hole horizons we will study later.

   First, \( \vec{e}_\lambda = {\Lambda^t}_\lambda \vec{e}_t + {\Lambda^x}_\lambda \vec{e}_x = a \cosh \lambda \vec{e}_t + a \sinh \lambda \vec{e}_x \), and \( \vec{e}_a = {\Lambda^t}_a \vec{e}_t + {\Lambda^x}_a \vec{e}_x = \sinh \lambda \vec{e}_t + \cosh \lambda \vec{e}_x \).

   Consequently, \( \frac{\partial \vec{e}_\lambda}{\partial \lambda} = a \sinh \lambda \vec{e}_t + a \cosh \lambda \vec{e}_x = a \vec{e}_a \), \( \frac{\partial \vec{e}_\lambda}{\partial a} = \cosh \lambda \vec{e}_t + \sinh \lambda \vec{e}_x = \frac{\vec{e}_\lambda}{a} \), \( \frac{\partial \vec{e}_a}{\partial \lambda} = \cosh \lambda \vec{e}_t + \sinh \lambda \vec{e}_x = \frac{\vec{e}_\lambda}{a} \), and \( \frac{\partial \vec{e}_a}{\partial a} = 0 \).

   Finally, \( {\Gamma^\lambda}_{\lambda \lambda} = 0 \), \( {\Gamma^a}_{\lambda \lambda} = a \), \( {\Gamma^\lambda}_{\lambda a} = \frac{1}{a} \), \( {\Gamma^a}_{\lambda a} = 0 \), \( {\Gamma^\lambda}_{a \lambda} = \frac{1}{a} \), \( {\Gamma^a}_{a \lambda} = 0 \), and \( {\Gamma^\lambda}_{a a} = {\Gamma^a}_{a a} = 0 \).

Show that if \( U^\alpha \nabla_\alpha V^\beta = W^\beta \), then \( U^\alpha \nabla_\alpha V_\beta = W_\beta \).

If \( U^\alpha \nabla_\alpha V^\beta = W^\beta \), we have

\[ \begin{eqnarray} && U^\alpha \nabla_\alpha V_\beta \\ &=& U^\alpha \nabla_\alpha (g_{\beta \mu} V^\mu) \\ &=& U^\alpha \nabla_\alpha g_{\beta \mu} V^\mu + U^\alpha g_{\beta \mu} \nabla_\alpha V^\mu \\ &=& g_{\beta \mu} U^\alpha \nabla_\alpha V^\mu \\ &=& g_{\beta \mu} W^\mu \\ &=& W_\beta. \end{eqnarray} \]

The second equality is beause covariant differentiation obeys the same sort of product rule as Eq. (5.60). THe third equality is because \( \boldsymbol{\mathrm{g}} \) is constant.