Stress-energy tensor for a rotating sphere

[Haorong Wu]

Homework Statement

From Bernard Schutz's A first course in general relativity, exercise 19, in chapter 8.6:
Suppose a spherical body of uniform density $\rho$ and radius $R$ rotates ridigly about the $x^3$ axis with constant angular velocity $\Omega$. Write down the components $T^{0\nu}$ in a Lorentz frame at rest with respect to the center of mass of the body, assuming $\rho$, $\Omega$, and $R$ are independent of time. For each component, work to the lowest nonvanishing order in $\Omega R$.

Relevant Equations

For dust, $T= \rho U \otimes U$.

For a perfect fluid, $T=(\rho + p) U \otimes U+p g^{-1}$

The answer with no details is given by

$T^{00}=\rho, T^{01}=-\rho \Omega x^2, T^{02}=\rho \Omega x^1, T^{03}=0.$ The components $T^{ij}$ are not fully determined by the given information, but they must be of order $\rho v^i v^j$, i.e. of order $\rho \Omega^2 R^2$.

First, I considered a spherical shell because I thought the velocities at different radius $r$ will be different and hence the four-momentum will be different, as well.

Then, I writed down the linear momenta by $$\epsilon^{ijk} r_i p_j = L_k$$ with $\vec r = R \cos \Omega t \hat x +R \sin \Omega t \hat y$, $\vec L = L_3 \hat z=\frac {4 \pi R^3 \rho \Omega ^2 d R} 3 \hat z$. So $$P_x=\frac {4 \pi R^3 \rho \Omega ^2 d R \cos \Omega t} {3} \rm{~and}$$ $$P_y=- \frac {4 \pi R^3 \rho \Omega ^2 d R \sin \Omega t} {3}.$$

So the velocity is $$V_x=\frac R 3 \Omega^2 \cos \Omega t, V_y=- \frac R 3 \Omega^2 \sin \Omega t,$$ and the four-velocity will be given by $U=(\gamma, \gamma v_x, \gamma v_y, 0)$ with $\gamma=1/\sqrt {1-v_x^2-v_y^2}$.

I am not sure what to do next. I tried to use the equations for dust that $T^{00}=\rho U^0 U^0$, that would yield $\rho \gamma ^2$. From the given answer, I see that I should really work in the MCRF with $\gamma=1$. But in such a frame, $U^i=0$. Then I would not derive the correct form of $T^{0i}$ in the answer.

Looking for some hints. Thanks.

 

[etotheipi]

Since the body undergoes rigid rotation about $Ox^3$, you may write $\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$ where $\boldsymbol{\omega} = \Omega \mathbf{e}_3$. In suffix notation, $v_i = \epsilon_{ijk} (\Omega\mathbf{e}_3)_j x_k = \Omega \epsilon_{ijk} \delta_{3j} x_k = \Omega \epsilon_{i3k} x_k$. You correctly write the stress energy tensor for dust, which is $T^{\mu \nu} = \rho u^{\mu} u^{\nu}$. Note that $v = |\boldsymbol{\omega} \times \mathbf{r}| = \Omega r \sin{\theta}$. Considering the components in the local Lorentz frame, where $u = \gamma(1, \mathbf{v})$, then\begin{align*}
T^{00} = \rho \gamma^2 = \frac{\rho}{1-v^2} = \rho \sum_{k=0}^{\infty} v^{2k} &= \rho \sum_{k=0}^{\infty} (\Omega r \sin{\theta})^{2k} \\

&= \rho + \mathcal{O}(\Omega^2 r^2)
\end{align*}\begin{align*}
T^{i0} = \rho \gamma^2 v^i = \frac{\rho \Omega \epsilon_{i3k} x_k}{1-v^2} &= \rho \Omega \epsilon_{i3k} x_k \sum_{k=0}^{\infty} (\Omega r \sin{\theta})^{2k} \\

&=\rho \Omega \epsilon_{i3k} x_k + \mathcal{O}(\Omega^3 r^3)
\end{align*}You can evaluate each of the $T^{i0}$,\begin{align*}
T^{10} &= \rho \Omega \epsilon_{13k} x_k + \mathcal{O}(\Omega^3 r^3) = -\rho \Omega x_2 + \mathcal{O}(\Omega^3 r^3)\\
T^{20} &= \rho \Omega \epsilon_{23k} x_k + \mathcal{O}(\Omega^3 r^3) = \rho \Omega x_1 + \mathcal{O}(\Omega^3 r^3) \\
T^{30} &= \rho \Omega \epsilon_{33k} x_k + \mathcal{O}(\Omega^3 r^3) = 0+ \mathcal{O}(\Omega^3 r^3)
\end{align*}