Gravitational Collapse of Star

A star can be thought of as a spherically symmetric body that rotates about an axis passing through its center. The spherical symmetry of the star implies that all three of its principal moments of inertia are equal to one another, and that any axis that passes through the center is a principal axis of rotation. (See Section 1.7.2.)

Suppose that the star in question is rotating about the $z$-axis. Its moment of inertia is

$$I= \sum_i m_i\,(x_i^{\,2}+y_{i}^{\,2}),$$ (1.219)

where the sum is over all of the mass elements that make up the star. If we model the star as a body of uniform mass density $\rho$ then the previous equation becomes

$$I = \rho\int (x^2+y^2)\,dV,$$ (1.220)

where $dV$ is a volume element, and the integral is over the volume of the star. In spherical polar coordinates, $x^2+y^2 = r^2\,\sin^2\theta$ and $dV = r^2\,\sin\theta\,dr\,d\theta\,d\phi$. (See Section A.23.) Hence,

$$I = \rho\int_0^a \int_0^\pi\int_0^{2\pi} r^4\,\sin^3\theta\,dr\,d\theta\,d\phi,$$ (1.221)

where $a$ is the radius of the star. It follows that

$$I =2\pi\,\rho\int_0^a r^4\,dr\int_0^\pi \sin^3\theta\,d\theta = 2\pi\,\rho\,\frac{a^5}{5}\,\frac{4}{3}= \frac{8\pi\,\rho\,a^5}{15}.$$ (1.222)

However,

$$M = \frac{4\pi\,\rho\,a^3}{3},$$ (1.223)

where $M$ is the mass of the star. The previous two equations yield

$$I = \frac{2}{5}\,M\,a^2.$$ (1.224)

The angular momentum of the star is

$$L = I\,\omega,$$ (1.225)

where $\omega$ is its angular velocity. If the star is isolated, such that it is not subject to an external torque, then its angular equation of motion, (1.180), reduces to

$$\frac{dL}{dt} = 0.$$ (1.226)

In other words, the angular momentum of the star is a conserved quantity.

Consider what would happen if a star such as the Sun collapsed gravitationally until it became a neutron star. The radius of the Sun-like star is about $10^6$ km. On the other hand, the radius of a neutron star is only about $10$ km. Moreover, the Sun rotates at about 1 revolution per month, which corresponds to an angular velocity of $2.5\times 10^{-6}\,{\rm rad\,s^{-1}}$. Thus, we are considering a process in which the star's initial radius and angular velocity are $a_1=10^6$ km and $\omega_1=2.5\times 10^{-6}\,{\rm rad\,s^{-1}}$, respectively, and its final radius is $a_2=10$ km. The question is what is the star's final angular velocity, $\omega_2$. Well, it is clear from Equations (1.224) and (1.225) that if angular momentum is to be conserved during the collapse then we require

$$a_1^{\,2}\,\omega_1 = a_2^{\,2}\,\omega_2,$$ (1.227)

or

$$\omega_2 = \left(\frac{a_1}{a_2}\right)^2\omega_1 =\left(\frac{10^6}{10}\right)^2 2.5\times 10^{-6} =2.5\times 10^4\,{\rm rad\,s^{-1}},$$ (1.228)

which corresponds to about 4000 revolutions per second. Thus, we deduce that neutron stars are likely to rotate thousands of times a second (as is indeed the case) as a consequence of the conservation of angular momentum during their formation. Finally, the star's rotational kinetic energy is $K =(1/2)\,I\,\omega^2\propto a^2\,\omega^2$, so the ratio of the final to the initial kinetic energy of the star is

$$\frac{K_2}{K_1} = \left(\frac{a_2}{a_1}\right)^2\left(\frac{\omega_2}{\omega_1}\right)^2= \left(\frac{a_1}{a_2}\right)^2 = 10^{10}.$$ (1.229)