Particle motion due to frame dragging near a pulsar?

Question

PSR J1748−2446ad is the fastest known pulsar rotating at 716 times per second. This neutron star is estimated to have a mass of less than two solar masses and a radius of less than 16km. The linear velocity at the equator is approximately 24% the speed of light or over 70,000 km/second. As such, this object would exhibit an extreme degree of frame dragging.

How would the trajectory of an incoming electrically neutral particle (say a neutron) initially approaching on the equatorial plane and perpendicular to the rotational axis be modified due to the frame dragging effect? Would the frame dragging effect be so great that the particle would spiral down revolving around the neutron star multiple times before impacting the surface, or would its impact point be shifted off a pathway perpendicular to the rotational axis by a far smaller amount?

Please assume a neutron star mass of two solar masses, a radius of 16km, an initial approach velocity of the particle toward the neutron of zero m/s (such that its relative motion would be initiated due to gravitational attraction), and an initial distance between the neutron and the surface of the neutron star of any arbitrary amount you deem reasonable (perhaps 100km or 1000km?) to demonstrate the magnitude of the frame dragging effect. Also, given these initial conditions, would the frame dragging affect the neutron path taking it off the equatorial plane or would it be constrained to that plane? It would be great if an equation of the actual path could be provided, or at least an equation giving the vector force on the neutron as a function of time, such that this (presumably) differential equation could subsequently be solved either exactly, or by numerical methods, by myself or others to be subsequently posted.

I understand frame dragging qualitatively but not quantitatively, and the math is beyond me. If a frame dragging expert could address this specific extreme though real example, I think it could help the community gain a better understanding of this phenomenon. Thanks in advance for responses and insights!

Answer 1

To answer this question one would need to be somewhat of an expert in numerical general relativity (GR) or know some literature there. Why? Because such an extremely fast rotating star has to be computed numerically. Analytic or semi-analytic approaches to compute the metric of such an object do not exist (to my knowledge or at least very complicated/rare). For someone (like me) without a running numerical GR code for neutron stars (like e.g. LORENE) one is left with two simpler options: use the Kerr-metric or assume slow rotation and use the somewhat simpler Hartle-Thorne metric of slow rotating neutron Stars -- ($O(J^1)$) approximation to the Kerr metric. I would argue that the Kerr metric is a decent approximation for the metric outside a fast rotating pulsar (at the very least at some distance from the star). I think it will suffice to get an idea about the magnitude of the frame dragging effect.

To get trajectories of particles moving in the curved spaces described by a metric one has to solve the corresponding geodesic equations which at least numerically is not difficult if one has access to a numerical solver for ordinary differential equations. The geodesic equations in the Kerr metric are quite lengthy (due to the involved Christoffel symbols) but they can be compute quite easily using the computer algebra system of ones choice. Looking up some estimates for the angular momentum of a pulsar like PSR J1748−2446ad I found $J\approx1.0\,G M_\odot^2/c=8.8\times10^{41}\mathrm{kg}\,\mathrm{m}^2\,\mathrm{s}^{-1}$ here. Plugging this into the Kerr metric ($M=2\,M_\odot$, $J=1\,G M_\odot^2/c$): a particle falling from rest at $(r=50\,\mathrm{km},\theta=\pi/2,\phi=0)$ $\Leftrightarrow(x=50\,\mathrm{km},y=0,z=0)$ in the equatorial plane gets deflected only by $y\approx 240\,\mathrm{m}$ once it reaches $r=16\,\mathrm{km}$. Increasing the angular momentum to the theoretical maximum for the Kerr metric in this situation ($M=2\,M_\odot$, $J=4\,G M_\odot^2/c$) yields a deflection of $y\approx 980\,\mathrm{m}$ at $r=16\,\mathrm{km}$.

Frame dragging does not deflect the particle out of the equatorial plane if we do not give an initial velocity in $z$-direction. This can be shown quite easily from the geodesic equation for $\theta''(\lambda)$ at the initial point $\lambda=0$ with $\theta(0)=\pi/2$ it simply reads $$\theta ''(0)+\frac{2 \theta '(0) r'(0)}{r(0)}=0,$$ thus for $\theta'(0)=0$ we get no acceleration in $z$-direction. Outside the equatorial plane the situation is different and we can expect acceleration $\theta ''(\lambda)\neq0$.

So even in those rather extreme scenarios the deflection is not extreme and in both scenarios the particle would also hit the event horizon of the black hole. That being said the deflection for objects with such an extreme angular momentum is visible and most certainly enough to have a drastic effect on particle orbits. For more details one could look into literature, pictures or animations for particle orbits around stellar mass rotating black holes: in terms of angular momentum those could be quite similar to the extreme pulsar we considered here with a spin parameter of $a=J/M=1/2$.

For less extreme scenarios frame dragging is a very subtle effect.