Moment of Inertia in the Slowly-Rotating Approximation

The differential equations for slow rigid rotation are solved by o2scl::tov_solve if o2scl::tov_solve::ang_vel is set to true.

In the case of slow rigid rotation with angular velocity $ \Omega $, the moment of inertia is

\[ I = \frac{8 \pi}{3} \int_0^R dr~r^4\left(\varepsilon+P\right) \left(\frac{\bar{\omega}}{\Omega}\right) e^{\Lambda-\Phi} = \frac{8 \pi}{3} \int_0^R dr~r^4\left(\varepsilon+P\right) \left(\frac{\bar{\omega}}{\Omega}\right) \left(1-\frac{2 G m}{r}\right)^{-1/2} e^{-\Phi} \]

where $ \omega(r) $ is the rotation rate of the inertial frame, $ \Omega $ is the angular velocity in the fluid frame, and $ \bar{\omega}(r) \equiv \Omega - \omega(r) $ is the angular velocity of a fluid element at infinity. The function $ \bar{\omega}(r) $ is the solution of

\[ \frac{d}{dr} \left( r^4 j \frac{d \bar{\omega}}{dr}\right) + 4 r^3 \frac{d j}{dr} \bar{\omega} = 0 \]

where the function $ j(r) $ is defined by

\[ j = e^{-\Lambda-\Phi} = \left( 1-\frac{2 G m}{r} \right)^{1/2} e^{-\Phi} \, . \]

Note that $ j(r=R) = 1 $. The boundary conditions for $ \bar{\omega} $ are $ d \bar{\omega}/dr = 0 $ at $ r=0 $ and

\[ \bar{\omega}(R) = \Omega - \left(\frac{R}{3}\right) \left(\frac{d \bar{\omega}}{dr}\right)_{r=R} \, . \]

One can use the TOV equation to rewrite the moment of inertia as

\[ I= \left(\frac{d \bar{\omega}}{dr}\right)_{r=R} \frac{R^4}{6 G \Omega} \, . \]

The star's angular momentum is just $ J = I \Omega $.

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