IPhO 1990 · Theory problem 3

The rotating neutron star

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A 'millisecond pulsar' is a source of radiation in the universe that emits very short pulses with a period of one to several milliseconds. This radiation is in the radio range of wavelengths; and a suitable radio receiver can be used to detect the separate pulses and thereby to measure the period with great accuracy.

These radio pulses originate from the surface of a particular sort of star, the so-called neutron star. These stars are very compact: they have a mass of the same order of magnitude as that of the sun, but their radius is only a few tens of kilometers. They spin very quickly. Because of the fast rotation, a neutron star is slightly flattened (oblate). Assume the axial cross-section of the surface to be an ellipse with almost equal axes. Let \(r_p\) be the polar and \(r_e\) the equatorial radii; and let us define the flattening factor by:

\[ \epsilon=\frac{(r_e-r_p)}{r_p} \]

Consider a neutron star with

a mass of \(2.0\cdot10^{30}\,\mathrm{kg}\),
an average radius of \(1.0\cdot10^4\,\mathrm{m}\),
and a rotation period of \(2.0\cdot10^{-2}\,\mathrm{s}\).

a - Calculate the flattening factor, given that the gravitational constant is \(6.67\cdot10^{-11}\,\mathrm{N.m^2.kg^{-2}}\).

In the long run (over many years) the rotation of the star slows down, due to energy loss, and this leads to a decrease in the flattening. The star has however a solid crust that floats on a liquid interior. The solid crust resists a continuous adjustment to equilibrium shape. Instead, starquakes occur with sudden changes in the shape of the crust towards equilibrium. During and after such a star-quake the angular velocity is observed to change according to figure 1.

Figure 1

A sudden change in the shape of the crust of a neutron star results in a sudden change of the angular velocity.

b - Calculate the average radius of the liquid interior, using the data of Fig. 1. Make the approximation that the densities of the crust and the interior are the same. (Ignore the change in shape of the interior).

Official English text of the problem.

Answers

  • a. In the spherical gravitational-field approximation, \[ \boxed{\epsilon\simeq3.7\times10^{-4}} \] (dimensionless).
  • b. The average radius of the liquid interior is approximately \[ \boxed{r_{\mathrm{liquid}}\simeq9.7\times10^3\,\mathrm{m}=9.7\,\mathrm{km}}. \]

Solution

a. Flattening factor

Let \(M\), \(R\), and \(T\) denote the mass, average radius, and rotation period. The angular velocity is

\[ \omega=\frac{2\pi}{T} =\frac{2\pi}{2.0\times10^{-2}} \simeq314.16\,\mathrm{s^{-1}}. \]

For the equilibrium shape, model the star as a rotating fluid and approximate its gravitational potential by the spherical expression \(-GM/r\). In the frame rotating with the star, the effective potential per unit mass is

\[ \Phi_{\mathrm{eff}}=-\frac{GM}{r}-\frac12\omega^2s^2, \]

where \(s\) is the distance from the rotation axis.

The equilibrium surface is an equipotential: otherwise matter would flow along it. At the pole \(s=0\), while at the equator \(s=r_e\). Hence

\[ -\frac{GM}{r_p} =-\frac{GM}{r_e}-\frac12\omega^2r_e^2, \]

or

\[ GM\left(\frac1{r_p}-\frac1{r_e}\right) =\frac12\omega^2r_e^2. \]

Using the definition of \(\epsilon\),

\[ \frac1{r_p}-\frac1{r_e} =\frac{r_e-r_p}{r_pr_e} =\frac{\epsilon}{r_e}, \]

so that

\[ \epsilon=\frac{\omega^2r_e^3}{2GM}. \]

Since the flattening is small, we may replace \(r_e\) by the average radius \(R\) in this expression:

\[ \epsilon\simeq\frac{\omega^2R^3}{2GM} =\frac{2\pi^2R^3}{GMT^2}. \]

Numerically,

\[ \epsilon\simeq \frac{2\pi^2(1.0\times10^4)^3} {(6.67\times10^{-11})(2.0\times10^{30})(2.0\times10^{-2})^2} =3.699\times10^{-4}. \]

Thus

\[ \boxed{\epsilon\simeq3.7\times10^{-4}}. \]

This corresponds to \(r_e-r_p\simeq\epsilon R\simeq3.7\,\mathrm{m}\), consistent with nearly equal radii.

b. Radius of the liquid interior

Before the quake, assume that the crust and liquid interior rotate together at angular velocity \(\omega_0\). During the sudden deformation, the crust’s moment of inertia decreases. There is insufficient time for appreciable angular momentum transfer to the liquid, so the crust spins up to \(\omega_1\), while the liquid initially remains at \(\omega_0\).

Over the following days, internal friction transfers angular momentum from the crust to the liquid until both rotate at \(\omega_2\). External angular momentum loss is negligible over this interval compared with the secular evolution over years.

Let \(I_c\) be the crust’s moment of inertia after the quake and \(I_l\) that of the liquid. The liquid’s moment of inertia is unchanged, as stipulated. Conservation of angular momentum during the relaxation gives

\[ I_c\omega_1+I_l\omega_0=(I_c+I_l)\omega_2. \]

Rearranging,

\[ I_c(\omega_1-\omega_2)=I_l(\omega_2-\omega_0). \]

Consequently,

\[ \boxed{\frac{I_l}{I_c+I_l} =\frac{\omega_1-\omega_2}{\omega_1-\omega_0}}. \]

The required conserved quantity is angular momentum; rotational kinetic energy is dissipated during the relaxation.

To relate the moments of inertia to radii, neglect the small oblateness and use the common density \(\rho\) specified in the question. A spherical shell of radius \(r\) and thickness \(dr\) has

\[ dm=4\pi\rho r^2\,dr, \qquad dI=\frac23r^2\,dm. \]

Thus a uniform sphere of radius \(a\) has

\[ I(a)=\frac{8\pi\rho}{3}\int_0^a r^4\,dr =\frac{8\pi\rho}{15}a^5. \]

Writing the liquid radius as \(r_l\), we obtain

\[ \frac{I_l}{I_c+I_l} =\frac{r_l^5}{R^5}. \]

Therefore,

\[ \boxed{ r_l=R\left( \frac{\omega_1-\omega_2}{\omega_1-\omega_0} \right)^{1/5} }. \]

Approximate readings from the graph are

\[ \omega_0\simeq314.1593\,\mathrm{s^{-1}},\qquad \omega_1\simeq314.1640\,\mathrm{s^{-1}},\qquad \omega_2\simeq314.1599\,\mathrm{s^{-1}}. \]

These give

\[ \left(\frac{r_l}{R}\right)^5 \simeq\frac{0.0041}{0.0047} \simeq0.872, \]

and hence

\[ r_l\simeq(1.0\times10^4)(0.872)^{1/5} \simeq9.73\times10^3\,\mathrm{m}. \]

To sensible precision,

\[ \boxed{r_l\simeq9.7\,\mathrm{km}}. \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.