Pulsar rotation linear velocity

Question

I recently came upon millisecond pulsars. I knew of pulsars, I had never guessed though that they can spin so fast.

I believe I had read about a pulsar rotating at about 10-15Hz.

But I just read about J1748-2446ad, which rotates at the astonishing 716Hz.

Brief research and I find its radius is about 16km, making its rotation circumference:

$c = 2πr ≈ 100.5km$

Given that the pulsar rotates at 716Hz, its linear velocity at the surface should be:

$v = 716c/s ≈ 71900km/s$

Assuming these calculations are nearly correct, could it be that any point on the rotation circumference moves at about 24% of the speed of light? Can this be possible?

Thanks for any answers that will satisfy my curiosity.

Answer 1

I'm not sure where you get the 17 km radius from. It's possible, although 10 km is a more usually assumed radius for a neutron star. It might be a little bigger because of the rotation rate.

Is it possible? Well, yes it is because the rotation rate is unambiguously 716 Hz and the radius of a neutron star cannot be much smaller than about 10 km before becoming unstable and collapsing into a black hole.

Millisecond pulsars spin very fast because they accrete material from a companion. The accreted material lands on the surface with angular momentum and this adds to the angular momentum of the neutron star, spinning it up. The angular rotation rate is just the total angular momentum divided by the moment of inertia, and because the latter is proportional to the square of the radius, and neutron stars are small, this leads to a very fast rotation rate.

You probably shouldn't be all that surprised. For comparison, if you drop something from a long way off onto the surface of the neutron star, it will be travelling at speeds of about $0.5c$ when it impacts the surface.