A neutron star forms when a stellar core collapses to an extremely small radius. If external torques are negligible
during the collapse, angular momentum is approximately conserved. This makes the final object spin much faster.
\[
L_i=L_f.
\]
1. Angular momentum conservation
For a rotating object,
\[
L=I\omega.
\]
Therefore, conservation of angular momentum gives
\[
I_i\omega_i=I_f\omega_f.
\]
Here, \(I\) is moment of inertia and \(\omega\) is angular speed.
2. Moment of inertia scaling
For a roughly spherical rotating body, the moment of inertia can be written as
\[
I=kMR^2.
\]
The constant \(k\) depends on the internal mass distribution. If the mass and structure factor are treated as
unchanged, then \(I\propto R^2\).
More generally, we can keep a correction factor:
\[
q=\frac{k_fM_f}{k_iM_i}.
\]
Then
\[
I_f=q\,I_i\left(\frac{R_f}{R_i}\right)^2
\]
in the sense needed for the period formula below.
3. Relating angular speed to period
Angular speed and period are related by
\[
\omega=\frac{2\pi}{T}.
\]
Substitute \(I=kMR^2\) and \(\omega=2\pi/T\) into angular momentum conservation:
\[
k_iM_iR_i^2\left(\frac{2\pi}{T_i}\right)
=
k_fM_fR_f^2\left(\frac{2\pi}{T_f}\right).
\]
Cancel \(2\pi\):
\[
\frac{k_iM_iR_i^2}{T_i}
=
\frac{k_fM_fR_f^2}{T_f}.
\]
Solve for the final period:
\[
T_f
=
T_i
\left(\frac{k_fM_f}{k_iM_i}\right)
\left(\frac{R_f}{R_i}\right)^2.
\]
Using \(q=(k_fM_f)/(k_iM_i)\):
\[
\boxed{
T_f=T_iq\left(\frac{R_f}{R_i}\right)^2
}.
\]
In the simplest textbook model, \(q=1\):
\[
\boxed{
T_f=T_i\left(\frac{R_f}{R_i}\right)^2
}.
\]
4. Spin-up factor
The spin-up factor compares the initial and final periods:
\[
\text{spin-up factor}=\frac{T_i}{T_f}.
\]
For \(q=1\),
\[
\frac{T_i}{T_f}
=
\left(\frac{R_i}{R_f}\right)^2.
\]
This square dependence is why collapse causes a dramatic increase in rotation speed. If the radius shrinks by a
factor of \(1000\), the period shrinks by a factor of \(1000^2=10^6\).
5. Angular speeds
The initial angular speed is
\[
\omega_i=\frac{2\pi}{T_i}.
\]
The final angular speed is
\[
\omega_f=\frac{2\pi}{T_f}.
\]
Since \(T_f\) can become very small, \(\omega_f\) can become very large.
6. Inverse forms
The same equation can be rearranged to solve other unknowns.
Final radius:
\[
R_f=R_i\sqrt{\frac{T_f}{T_iq}}.
\]
Initial radius:
\[
R_i=R_f\sqrt{\frac{T_iq}{T_f}}.
\]
Initial period:
\[
T_i=\frac{T_f}{q(R_f/R_i)^2}.
\]
Mass/structure factor:
\[
q=\frac{T_f}{T_i(R_f/R_i)^2}.
\]
7. Worked example: Sun-like radius to neutron star radius
Suppose a star has an initial radius
\[
R_i=696340\ \mathrm{km}
\]
and collapses to a neutron star radius
\[
R_f=12\ \mathrm{km}.
\]
Let the initial rotation period be
\[
T_i=25\ \mathrm{days}.
\]
Using the idealized formula with \(q=1\):
\[
T_f=T_i\left(\frac{R_f}{R_i}\right)^2.
\]
Convert \(25\) days to seconds:
\[
T_i=25(86400)=2.16\times10^6\ \mathrm{s}.
\]
Substitute:
\[
T_f=
(2.16\times10^6)
\left(\frac{12}{696340}\right)^2.
\]
\[
T_f\approx6.42\times10^{-4}\ \mathrm{s}.
\]
In milliseconds:
\[
T_f\approx0.642\ \mathrm{ms}.
\]
This is an idealized estimate. Real neutron star periods depend on mass loss, internal structure, magnetic torques,
and the detailed dynamics of collapse.
8. Interpretation
The contraction is so dramatic because moment of inertia depends on \(R^2\). A neutron star has a tiny radius
compared with its progenitor, so conservation of angular momentum predicts a huge increase in spin rate.
Key idea: shrinking radius reduces moment of inertia; if angular momentum is conserved, angular speed increases
and the rotation period becomes much shorter.
