Wald ch 4 problem 3a, derive equation (4.4.24)

[etotheipi]

Homework Statement

To linear order in the velocity of the test body, and neglecting stresses, show that the linearised Einstein equation yields$$\mathbf{a} = -\mathbf{E} - 4\mathbf{v} \times \mathbf{B}$$

Relevant Equations

N/A

I'm just confused, no idea really what to do. Since the time derivatives of the $\bar{\gamma}_{\mu \nu}$ are assumed to be zero, and the space-space components are also assumed negligible, we have
$$\begin{align*}
\nabla^2 \bar{\gamma}_{00} &= 16 \pi J_0 = -16\pi T_{a0} t^a \\
\nabla^2 \bar{\gamma}_{0i} &= 16 \pi J_i = -16\pi T_{ai} t^a \\
\end{align*}$$where here $\nabla$ denotes the purely spatial Laplacian operator. Once we have solved for $\bar{\gamma}_{\mu \nu}$ it should be possible to extract the equations of motion from the simplified geodesic equation$$\frac{d^2 x^{\mu}}{dt^2} = - \Gamma^{\mu}_{00} = - \frac{1}{2} \eta^{\mu d}(2\partial_0 \gamma_{0d} - \partial_d \gamma_{00})$$where the symmetry of $\gamma_{ab} = g_{ab} - \eta_{ab}$ was used. My issue is to solve for the $\bar{\gamma}_{\mu \nu}$ in the first place. I know we can write down$$T_{ab} = 2t_{(a} J_{b)} - \rho t_a t_b = -2t_{(a} T_{|c|b)} t^c - \rho t_a t_b$$but I don't know how to solve that for $T_{ab}$, which I need to do in order to substitute back into the first two equations...

How should I proceed? Thanks

 

[TSny]

Once we have solved for $\bar{\gamma}_{\mu \nu}$...

Just below equation (4.4.23), Wald defines $$A_a \equiv -\frac{1}{4}\overline{\gamma}_{ab}t^b$$ $A_a$ is the gravitational analog of the electromagnetic 4-vector potential. You can use this relation to solve for $\overline{\gamma}_{\mu 0}$ in terms of $A_\mu$. The spatial components $\overline{\gamma}_{ij}$ are all zero. You can show this by first showing $\nabla^2 \overline{\gamma}_{jk} = 0$ and then using the argument given by Wald at the very bottom of page 76.

Then you can express the geodesic equation in terms of the $A_\mu$. In analogy to EM, define the spatial vectors $\vec{E}$ and $\vec{B}$ in terms of the $A_\mu$. Finally, see if you can write the geodesic equation in terms of $\vec{E}$ and $\vec{B}$.

... it should be possible to extract the equations of motion from the simplified geodesic equation$$\frac{d^2 x^{\mu}}{dt^2} = - \Gamma^{\mu}_{00}$$

For this exercise, you need to keep terms of first-order in the test particle's spatial velocity $\frac{dx^k}{dt}$. So, there are additional terms that need to be included in the geodesic equation.

 

[etotheipi]

Thanks! I think I understand the problem better now. Let me write up what I've gotten so far. So, Wald defined:$$A_{\mu} := -\frac{1}{4} \bar{\gamma}_{\mu \nu} t^{\nu}$$however the vector $t^a = (\partial/\partial x^0)^a$ clearly only has one non-zero component, namely $t^0 = 1$, so that equation can just be inverted to obtain$$\bar{\gamma}_{\mu 0} = -4A_{\mu}$$Now let's consider the geodesic equation (by the way, I noticed that I wrote "$t$" instead of "$\tau$" in the geodesic equation in my first post - I'll revert back to "$\tau$" to avoid confusion between the coordinate $t = x^0$ and the affine parameter $\tau$). Keeping in mind that we need to retain terms to first order in the spatial velocity, but still approximating $dx^0 / d\tau = 1$, i.e. $dx^a/d\tau = (1, dx^1/d\tau, dx^2/d\tau, dx^3/d\tau)$, we obtain

$$
\begin{align*}

\frac{d^2 x^{\mu}}{d\tau^2} &= -\Gamma^{\mu}_{00} \frac{dx^0}{d\tau} \frac{dx^0}{d\tau} - \Gamma^{\mu}_{0 \rho} \frac{dx^0}{d\tau} \frac{dx^{\rho}}{d\tau} - \Gamma^{\mu}_{\rho 0} \frac{dx^{\rho}}{d\tau} \frac{dx^0}{d\tau} \\ \\

&= -\Gamma^{\mu}_{00} - \Gamma^{\mu}_{0 \rho} \frac{dx^{\rho}}{d\tau} - \Gamma^{\mu}_{\rho 0} \frac{dx^{\rho}}{d\tau} \\ \\

&= - \Gamma^{\mu}_{00} - 2\Gamma^{\mu}_{0 \rho} \frac{dx^{\rho}}{d\tau}

\end{align*}
$$Now we need to work out the Christoffel symbols! The definition in Wald is given in terms of the $\gamma_{\mu \nu}$, so we first need to express these in terms of our $\bar{\gamma}_{\mu \nu}$. Luckily this is given in equation (4.4.16),$$\gamma_{\mu \nu} = \bar{\gamma}_{\mu \nu} - \frac{1}{2} \eta_{\mu \nu} \bar{\gamma}$$The trace $\bar{\gamma} = \bar{\gamma}_{00}$ since $\bar{\gamma}_{11} = \bar{\gamma}_{22} = \bar{\gamma}_{33} = 0$. Using the $(-+++)$ metric signature for $\eta_{ab}$, which means $\eta_{00} = -1$, we have$$\begin{align*}
\gamma_{00} &= \bar{\gamma}_{00} + \frac{1}{2} \bar{\gamma}_{00} = \frac{3}{2} \bar{\gamma}_{00} = -6A_0 \\ \\

\gamma_{0i} &= \bar{\gamma}_{0i} - \frac{1}{2} \eta_{0i} \bar{\gamma} = \bar{\gamma}_{0i} = -4A_i , \quad i \in \{1,2,3 \}
\end{align*}
$$And we can write for $\Gamma^{\mu}_{00}$
$$
\begin{align*}
\Gamma^{\mu}_{00} &= \frac{1}{2} \eta^{\mu \nu} (2 \partial_0 \gamma_{0 \nu} - \partial_{\nu} \gamma_{00} ) \\

&= \frac{1}{2} \eta^{\mu \mu} (2\partial_0 \gamma_{0 \mu} + 6 \partial_{\mu} A_{0})
\end{align*}
$$and then for $\Gamma^{\mu}_{0 \rho}$$$
\begin{align*}
\Gamma^{\mu}_{0\rho} &= \frac{1}{2} \eta^{\mu \nu} (\partial_0 \gamma_{\rho \nu} + \partial_{\rho} \gamma_{0 \nu} - \partial_{\nu} \gamma_{0 \rho}) \\

&= \frac{1}{2} \eta^{\mu \mu} (\partial_0 \gamma_{\rho \mu} + \partial_{\rho} \gamma_{0 \mu} - \partial_{\mu} \gamma_{0 \rho})
\end{align*}
$$I'm going to go to sleep now but does that look okay so far? I guess the last part is just considering the different cases for $\mu$, and expressing the Christoffel symbols in terms of $A_{\mu}$.

 

[TSny]

The trace $\bar{\gamma} = \bar{\gamma}_{00}$

Check the sign here. The trace is ${\overline \gamma^a}_a$.

 

[etotheipi]

Oh yeah, thanks! I hate signs $$\bar{\gamma} = {\bar{\gamma}^{\mu}}_{\mu} = {\bar{\gamma}^{0}}_0 = - \bar{\gamma}_{00}$$I think I figured out which are the relevant Christoffel symbols,$$
\begin{align*}
\Gamma^{i}_{00} &= \frac{1}{2} \eta^{i\nu}(\partial_0 \gamma_{0 \nu} + \partial_0 \gamma_{0\nu} - \partial_{\nu} \gamma_{00}) \\

&= \frac{1}{2}(2\partial_0 \gamma_{0i} - \partial_i \gamma_{00}) \\

&= \partial_i A_0
\end{align*}$$and$$
\begin{align*}
{\Gamma^i}_{0j} &= \frac{1}{2} \eta^{i\nu}(\partial_0 \gamma_{j \nu} + \partial_j \gamma_{0 \nu} - \partial_{\nu} \gamma_{0j}) \\

&= \frac{1}{2}(\partial_j \gamma_{0i} - \partial_i \gamma_{0j}) \\

&= 2(\partial_i A_j - \partial_j A_i) := 2F_{ij}
\end{align*}
$$I get for the geodesic equation$$\ddot{x}^i = -\partial_i A_0 - 4F_{ij} \dot{x}^j$$which looks a bit promising, but not quite there yet. I know from equation (4.2.22) that, in the observer's frame where their own 4-velocity is $v^a = (1,0,0,0)$, their measured magnetic field is$$B_i = -\frac{1}{2}{\epsilon_{i0}}^{\rho \sigma} F_{\rho \sigma}$$Also, because the signature is $(-+++)$ I can be afford to be bit cavalier with the spatial indices. Usually for the cross product I would write$$(\mathbf{v} \times \mathbf{B})_i = \epsilon_{ijk} v_j B_k$$but that seems a bit problematic here, because it's using a rank 3 levi civita symbol instead of a rank 4 levi civita symbol. Is there an easy way to fix that up?

Also, I'm not totally sure what to do with the $\partial_i A_0$ bit. We know that $E_i = F_{i0} = \partial_i A_0 - \partial_0 A_i$, so that first term will only be what we want if $\partial_0 A_i = 0$

 

[TSny]

\begin{align*}
{\Gamma^i}_{0j} &= \frac{1}{2} \eta^{i\nu}(\partial_0 \gamma_{j \nu} + \partial_j \gamma_{0 \nu} - \partial_{\nu} \gamma_{0j}) \\

&= \frac{1}{2}(\partial_j \gamma_{0i} - \partial_i \gamma_{0j}) \\

&= 2(\partial_i A_j - \partial_j A_i) := 2F_{ij}
\end{align*}

In electromagnetism $\vec B$ is related to $\vec A$ as $\vec B = \vec \nabla \times \vec A$.
So, $B^i = B_i = \varepsilon_{ijk}\partial_j A_k$ (where $i$ = 1, 2, or 3 and we sum over $j$, $k = 1, 2, 3)$.

For this problem, define $\vec B$ by this relation. From this, show that $\varepsilon_{ijk}B_k = \partial_i A_j - \partial_j A_i$.
Use this to express $\Gamma^i_{0j}$ in terms of the components of $\vec B$.
(This avoids working with the rank 4 Levi-Civita symbol.)

Also, I'm not totally sure what to do with the $\partial_i A_0$ bit. We know that $E_i = F_{i0} = \partial_i A_0 - \partial_0 A_i$, so that first term will only be what we want if $\partial_0 A_i = 0$

Remember that the time derivative of any of the $\gamma_{\mu \nu}$ is assumed to be negligible. So, $\partial_0 A_i$ can be neglected. I'm not sure if this is what you're uncertain about.

 

[etotheipi]

Ah, that's so smooth!$$\epsilon_{ijk} B_k = \epsilon_{ijk} \epsilon_{klm} \partial_l A_m = (\delta_{il} \delta_{jm} - \delta_{jl} \delta_{im})\partial_l A_m = \partial_i A_j - \partial_j A_i$$in which case $\Gamma^{i}_{0j} = 2\epsilon_{ijk} B_k$ and$$\ddot{x}^i = -\partial_i A_0 - 4\epsilon_{ijk} \dot{x}_j B_k = -E_i - 4\epsilon_{ijk} \dot{x}_j B_k \implies \ddot{\mathbf{x}} = -\mathbf{E} - 4\dot{\mathbf{x}} \times \mathbf{B}$$where I used that $\partial_i A_0 = \partial_i A_0 + \partial_0 A_i$ (I forgot that we were neglecting the time derivatives ).

Thanks a bunch!
(maybe I'll try the other parts of the question sometime, but I'll probably leave them for a little while because they look a bit intimidating )

 

[TSny]

Looks good!

 

[etotheipi]

Hey @TSny, I decided to have a go at the rest just for completeness' sake. Part (b)'s being a little bit stubborn, so I started with (c) and just assumed the results of (b). I wondered if you had some tips for going about solving (b)?

Starting with part (c). Since $S^a$ is parallel-transported along his worldline, which is a geodesic of tangent vector $u^a$, we have$$\begin{align*}
\frac{dS^{\nu}}{dt} + u^{\mu} \Gamma^{\nu}_{\mu \lambda} S^{\lambda} = 0 \quad &\implies \quad \frac{dS^{\nu}}{dt} = - \Gamma^{\nu}_{0 \lambda} S^{\lambda} \\

&\implies \quad \frac{dS^{i}}{dt} = - \Gamma^{i}_{0 \lambda} S^{\lambda}, \quad i \in \{1,2,3 \}

\end{align*}$$Now consider the quantity $\mathbf{B} \times \mathbf{S}$;$$\begin{align*}
(\mathbf{B} \times \mathbf{S})^i = {\epsilon^{i}}_{jk} B^j S^k &= (\delta_{kl} \delta_{im} - \delta_{il} \delta_{km})S^k \partial_l A_m \\
&= -S^k(\partial_i A_k - \partial_k A_i) = -\frac{1}{2} \Gamma^i_{0k} S^k
\end{align*}$$Hence $(\boldsymbol{\Omega} \times \mathbf{S})^i = -\Gamma^i_{0k} S^k$ which completes the proof of $d\mathbf{S}/dt = \boldsymbol{\Omega} \times \mathbf{S}$.

The previous bit, part (b), is just fiddly. Since $\partial^a \partial_a A_b = -4\pi j_b$, and since the temporal derivatives are assumed to vanish, the equation to solve is$$\nabla^2 A_{\mu} = -4\pi j_{\mu}$$Furthermore, for this solution,$$T_{ab} = 2t_{(a} j_{b)} - \rho t_a t_b \implies T_{00} = 2t_0 j_0 - \rho t_0 t_0 = -2j_0 - \rho$$and thus $j_0 = -T_{0 \nu} t^{\nu} = -T_{00} = 2j_0 + \rho \implies j_0 = - \rho$.

Let's first try to solve the equation for the spatial part $\mathbf{A}(\mathbf{x})$. I tried orienting a spherical coordinate system with the $\theta = 0$ axis parallel to $\boldsymbol{\omega}$, so that$$\mathbf{A}(\mathbf{x}) = \int_{\partial \zeta} \frac{\sigma \mathbf{v}'}{|\mathbf{x} - \mathbf{x}'|} da'$$where $$\sigma \mathbf{v} = \sigma \omega \hat{\mathbf{z}} \times \mathbf{x}' = \sigma \omega R (-\sin{\theta} \sin{\varphi}, \sin{\theta} \cos{\varphi}, 0)$$whilst $da' = R^2 \sin{\theta} d\theta d\varphi$ and $|\mathbf{x} - \mathbf{x}'| = \sqrt{(\mathbf{x} - \mathbf{x}') \cdot (\mathbf{x} - \mathbf{x}')} = \sqrt{R^2 + r^2 - 2rR \cos{\xi}}$, where $\xi$ is the angle between $\mathbf{x}$ and $\mathbf{x}'$.

I'm having some trouble figuring out how to express this angle $\xi$ in terms of the spherical coordinates... how's best to proceed?

 

[TSny]

Starting with part (c). Since $S^a$ is parallel-transported along his worldline, which is a geodesic of tangent vector $u^a$, we have$$\begin{align*}
\frac{dS^{\nu}}{dt} + u^{\mu} \Gamma^{\nu}_{\mu \lambda} S^{\lambda} = 0 \quad &\implies \quad \frac{dS^{\nu}}{dt} = - \Gamma^{\nu}_{0 \lambda} S^{\lambda} \\

&\implies \quad \frac{dS^{i}}{dt} = - \Gamma^{i}_{0 \lambda} S^{\lambda}, \quad i \in \{1,2,3 \}

\end{align*}$$Now consider the quantity $\mathbf{B} \times \mathbf{S}$;$$\begin{align*}
(\mathbf{B} \times \mathbf{S})^i = {\epsilon^{i}}_{jk} B^j S^k &= (\delta_{kl} \delta_{im} - \delta_{il} \delta_{km})S^k \partial_l A_m \\
&= -S^k(\partial_i A_k - \partial_k A_i) = -\frac{1}{2} \Gamma^i_{0k} S^k
\end{align*}$$Hence $(\boldsymbol{\Omega} \times \mathbf{S})^i = -\Gamma^i_{0k} S^k$ which completes the proof of $d\mathbf{S}/dt = \boldsymbol{\Omega} \times \mathbf{S}$.

That all looks good to me.

The previous bit, part (b), is just fiddly. Since $\partial^a \partial_a A_b = -4\pi j_b$, and since the temporal derivatives are assumed to vanish, the equation to solve is$$\nabla^2 A_{\mu} = -4\pi j_{\mu}$$Furthermore, for this solution,$$T_{ab} = 2t_{(a} j_{b)} - \rho t_a t_b \implies T_{00} = 2t_0 j_0 - \rho t_0 t_0 = -2j_0 - \rho$$and thus $j_0 = -T_{0 \nu} t^{\nu} = -T_{00} = 2j_0 + \rho \implies j_0 = - \rho$.

OK

Let's first try to solve the equation for the spatial part $\mathbf{A}(\mathbf{x})$. I tried orienting a spherical coordinate system with the $\theta = 0$ axis parallel to $\boldsymbol{\omega}$, so that$$\mathbf{A}(\mathbf{x}) = \int_{\partial \zeta} \frac{\sigma \mathbf{v}'}{|\mathbf{x} - \mathbf{x}'|} da'$$where $$\sigma \mathbf{v} = \sigma \omega \hat{\mathbf{z}} \times \mathbf{x}' = \sigma \omega R (-\sin{\theta} \sin{\varphi}, \sin{\theta} \cos{\varphi}, 0)$$

This integral occurs in the standard EM problem of finding the B field of a rotating, uniformly charged spherical shell. The trick is to choose the $\theta = 0$ axis along the vector $\mathbf{x}$ instead of along $\boldsymbol{\omega}$. This makes the components of $\boldsymbol{\omega}$ and $\mathbf{v}$ more complicated, but overall it makes the integration easier.

 

[etotheipi]

I think I got it... I'll redefine $\xi$ as the angle between $\boldsymbol{\omega}$ and $\mathbf{x}$, and orient the Cartesian coordinate system such that $\mathbf{x} \parallel \mathbf{e}_3$ and $\boldsymbol{\omega} = (0, \omega \sin \xi, \omega \cos \xi)$. In that case
$$\mathbf{v}' = \boldsymbol{\omega} \times \mathbf{x}' = R\omega
\begin{pmatrix}
\sin{\xi} \cos{\theta} - \sin{\theta} \sin{\varphi} \cos{\xi} \\ \cos{\xi} \sin{\theta} \cos{\varphi} \\-\sin{\xi} \sin{\theta} \cos{\varphi}
\end{pmatrix}
$$The integral is just$$\mathbf{A}(\mathbf{x}) = \int_{\partial \zeta} \frac{ \sigma \mathbf{v}' }{|\mathbf{x} - \mathbf{x}'|} da' = \sigma R^2 \int_{\varphi = 0}^{2\pi} \int_{\theta = 0}^{\pi} \frac{ \mathbf{v}' \sin{\theta}}{|\mathbf{x} - \mathbf{x}'|} d\theta d\varphi$$where $|\mathbf{x} - \mathbf{x}'| = \sqrt{R^2 + r^2 - 2Rr\cos{\theta}}$. This turns out to be separable in $\theta$ and $\varphi$, and since $\varphi \in [0,2\pi]$ the only non-zero term is going to be, defining $C: = 2\pi \sigma R^3 \omega \sin{\xi}$,

$$
\begin{align*}
A_1(\mathbf{x}) &= 2\pi \sigma R^3 \omega \sin{\xi} \int_{0}^{\pi} \frac{\sin{\theta} \cos{\theta}}{\sqrt{R^2 + r^2 - 2Rr\cos{\theta}}} d\theta \\ \\

&= C \int_{-1}^{1} \frac{\lambda}{\sqrt{R^2 + r^2 - 2Rr\lambda}} d\lambda \\ \\

&= \frac{C}{4R^2 r^2} \int_{\mu(\lambda = -1)}^{\mu( \lambda = 1)} \left( \sqrt{\mu} - \frac{(R^2 + r^2)}{\sqrt{\mu}} \right) d\mu \\ \\

&= \frac{C}{4R^2 r^2} \left[ \sqrt{R^2 + r^2 - 2Rr \lambda} \left( \frac{2}{3} (R^2 + r^2 - 2Rr\lambda) - 2(R^2 + r^2) \right) \right]_{-1}^{1} \\ \\

&= - \frac{C}{3R^2 r^2} \left[ (R^2 + r^2 -Rr\lambda) \sqrt{R^2 + r^2 -2Rr\lambda} \right]_{-1}^{1}\end{align*}$$Since $R>r$ we have $\sqrt{R^2 + r^2 - 2Rr} = |R - r| = R- r$, and hence

$$A_1(\mathbf{x}) = - \frac{C}{3R^2 r^2} \left( 2R^2 r - 2r(R^2 + r^2) \right) = \frac{2rC}{3R^2}$$That is, the vector potential is $$\mathbf{A}(\mathbf{x}) = \frac{4\pi \sigma \omega rR \sin{\xi}}{3} \boldsymbol{e}_1$$and the magnetic field is $$\mathbf{B}(\mathbf{x}) = \nabla \times \mathbf{A}(\mathbf{x}) = \frac{4\pi \sigma \omega R \sin{\xi}}{3} \nabla \times (r\mathbf{e}_1)$$where $\nabla \times (r \mathbf{e}_1)= \frac{1}{r}(0, z, -y)$. I'll check a bit later on if that agrees with the form in the book, but right now I'm shattered and need to go to sleep, so I'm just pretty happy to have an answer

 

[TSny]

$$
\begin{align*}
A_1(\mathbf{x}) &= 2\pi \sigma R^3 \omega \sin{\xi} \int_{0}^{\pi} \frac{\sin{\theta} \cos{\theta}}{\sqrt{R^2 + r^2 - 2Rr\cos{\theta}}} d\theta \\ \\

&= C \int_{-1}^{1} \frac{\lambda}{\sqrt{R^2 + r^2 - 2Rr\lambda}} d\lambda \\ \\

&= \frac{C}{4R^2 r^2} \int_{\mu(\lambda = -1)}^{\mu( \lambda = 1)} \left( \sqrt{\mu} - \frac{(R^2 + r^2)}{\sqrt{\mu}} \right) d\mu \\ \\

&= \frac{C}{4R^2 r^2} \left[ \sqrt{R^2 + r^2 - 2Rr \lambda} \left( \frac{2}{3} (R^2 + r^2 - 2Rr\lambda) - 2(R^2 + r^2) \right) \right]_{-1}^{1} \\ \\

&= - \frac{C}{3R^2 r^2} \left[ (R^2 + r^2 -Rr\lambda) \sqrt{R^2 + r^2 -2Rr\lambda} \right]_{-1}^{1}\end{align*}$$

In the last line, the factor $(R^2 + r^2 -Rr\lambda)$ should be $(R^2 + r^2 + Rr\lambda)$. This is probably just a typographical error because it looks like your final result for $A_1(\mathbf{x})$ is correct. Everything else in the post looks good to me.

 

[etotheipi]

Thanks for going through it! Yeah, in my notes I've got the plus sign so I figure it's a mistype

I'm not quite sure why this last part's not working, I guess there's something I'm misunderstanding. In these coordinates, with $\mathbf{x} \parallel \mathbf{e}_3$, we have
$$\mathbf{B}(\mathbf{x}) = \nabla \times \mathbf{A}(\mathbf{x}) = \frac{4\pi \sigma \omega R \sin{\xi}}{3} \nabla \times (r\mathbf{e}_1) = \frac{4\pi \sigma \omega R \sin{\xi}}{3r} \begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$$But the curl is evaluated at $\mathbf{x} = r \mathbf{e}_3$, where $z = r$ and $y=0$. So it would just seem that, also given $\sigma = M/4\pi R^2$, that we'd have$$\mathbf{B}(\mathbf{x}) =\frac{M \omega \sin{\xi}}{3R} \mathbf{e}_2 $$which isn't a multiple of $\boldsymbol{\omega} = (0, \omega \sin{\xi}, \omega \cos{\xi})$ ... can you see where I messed up?

 

[TSny]

I'm not quite sure why this last part's not working, I guess there's something I'm misunderstanding. In these coordinates, with $\mathbf{x} \parallel \mathbf{e}_3$, we have
$$\mathbf{B}(\mathbf{x}) = \nabla \times \mathbf{A}(\mathbf{x}) = \frac{4\pi \sigma \omega R \sin{\xi}}{3} \nabla \times (r\mathbf{e}_1) = \frac{4\pi \sigma \omega R \sin{\xi}}{3r} \begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$$But the curl is evaluated at $\mathbf{x} = r \mathbf{e}_3$, where $z = r$ and $y=0$. So it would just seem that, also given $\sigma = M/4\pi R^2$, that we'd have$$\mathbf{B}(\mathbf{x}) =\frac{M \omega \sin{\xi}}{3R} \mathbf{e}_2 $$which isn't a multiple of $\boldsymbol{\omega} = (0, \omega \sin{\xi}, \omega \cos{\xi})$ ... can you see where I messed up?

The angle $\xi$ depends on $\mathbf{x}$. So, when taking the curl of $\mathbf{A}(\mathbf{x})$, $\xi$ cannot be treated as a constant.

Likewise, I don't think you can treat $\mathbf{e}_1$ as a fixed vector when taking the curl of $r \mathbf{e}_1$.

However, note that $\omega r \sin{\xi} \, \mathbf{e}_1 = \boldsymbol{\omega} \times \mathbf{x}$. So, $\mathbf{A}(\mathbf{x})$ can be written in terms of the coordinate-independent expression $\boldsymbol{\omega} \times \mathbf{x}$. It's not hard to find the curl of this expression.

 

[etotheipi]

Ah, okay, I see now where I went wrong! I could do it using your suggestion; keeping in mind $\boldsymbol{\omega}$ is constant,$$
\begin{align*}

3R \mathbf{B} = \nabla \times (\boldsymbol{\omega} \times \mathbf{x}) &= \epsilon_{ijk} \mathbf{e}_i \partial_j (\boldsymbol{\omega} \times \mathbf{x})_k \\

&= \epsilon_{ijk} \mathbf{e}_i \partial_j \epsilon_{klm} \omega_l x_m \\

&= \epsilon_{kij} \epsilon_{klm} \mathbf{e}_i \partial_j \omega_l x_m \\

&= (\delta_{il} \delta_{jm} - \delta_{jl} \delta_{im}) \mathbf{e}_i \partial_j \omega_l x_m \\

&= (\omega_i \partial_m x_m - \omega_j \partial_j x_i)\mathbf{e}_i \\

&= [(\nabla \cdot \mathbf{x})\omega_i - \omega_j \partial_j x_i ] \mathbf{e}_i\\
&= (3\omega_i - \omega_i) \mathbf{e}_i = 2\omega_i \mathbf{e}_i = 2\boldsymbol{\omega}\end{align*}
$$which then gives the desired result. Thanks for all your help, I think I learned more than I bargained for with this question!

 

[TSny]

That looks good. I was able to help you only because I had already seen the solution for the B field of a rotating sphere with uniform surface charge density.