A 'low' gravity pulsar? (Math nerd thread)

Factual Questions
1

http://en.wikipedia.org/wiki/PSR_J1748-2446ad

According to Wikipedia, this pulsar is travelling at 24% the speed of light at the equator. Given a neutron at surface on the equator, what forces are applied to it (downwards from gravity vs upwards from rotation)? Assume a radius of 16km.

If the speed was increased, when would the neutron leave the pulsar? Would these neutrons revert back to (carbon?) atoms if they were no longer under this extreme force?

Also, if I had an extra 50 IQ points and were calculating this myself, would I need to take into account time? For example, if it appears from earth to move at speed A, but time travels slower for it because of it’s speed and the gravitational pull, then it may appear (from the perception of the neutron) to travel at a different speed. Therefore, does it feel a different force upwards?

2

Well… if we’re going to be nerds about this issue, then I should start by pointing out that there is no upwards force in the scenario you describe. Each particle will be trying to move in a straight line per Newton’s laws, and remains put because forces act upon it to continually change the velocity. If those forces stopped, the particle would move away from the star, but on a tangent to the surface, not upward.

If you could separate the neutrons, they’d definitely revert back to ordinary matter. Wikipedia has some good articles on degenerate matter. Neutron-degenerate matter would revert to electron-degenerate matter, which would ultimately turn back into ordinary matter. Individual neutrons would turn into protons (hydrogen), but I’m sure that breaking up a neutron star would get you elements in all kinds of varieties.

I’ll leave the actual math to someone else, but any answer would be a guess at best… we don’t know how much neutrons in a neutron star try to stick together (or even, possibly, repel each other).

3

I doubt this very much. Neutrons are neutrons, and I know of no way they would “remember” that some of them were formerly protons and electrons that got merged. Furthermore neutrons by themselves don’t bond into stable “nuclei” by the strong nuclear force. At least two of them don’t and I don’t imagine more do.

So once a blob of nuclear matter left the neutron star, the “nuclei” would fission into individual neutrons and unless captured by other real nuclei, the individual neutrons would decay into protons and electrons (and anti electron-neutrinos). These might bond into a hydrogen atom, but I don’t think you’d get anything beyond that.

4

I think dracoi is correct that you’d get larger elements, because some of the neutrons would be decaying while the matter was still very dense, much denser than the Sun’s core.

No idea what you mean by neutrons having to “remember” what they were, or what that’s referring to.

5

I suspect there might be a film of ordinary matter on the surface of a neutron star, but don’t know

6

Rotating with a linear speed of 1/4 the speed of light, over a 15 km radius, the necessary acceleration is 6 x 10^14 m/sec^2. The net force on any particle that keeps it in that orbit is the acceleration times the particle’s mass.
Now that’s the net force. There’s a huge force from gravity pulling the particle in, and a smaller force from all neutrons underneath pushing the particle back out.

For relatavistic effects, the linear speed alone isn’t enough to make a big difference, something like 3% or so. However, the concentrated mass of the neutron star might have a bigger effect, which I can’t calculate offhand.

7

Neutrons spontaneously decay into protons with a half-life of about 10 minutes. You describe that yourself, so I think we’re in agreement on this. My post wasn’t about the neutrons remembering what they were - just describing the natural decay. A lone proton can be legitimately described as hydrogen (just an ionized version thereof).

I speculated that a whole clump of neutrons in degenerate matter might decay so that they form other elements. If you have four neutrons bound together and two of them decay into protons, you now have a stable helium ion and the other two neutrons would be tightly bound and stable. But, again, this is not a matter of “remembering” anything - just the way I’d expect a bunch of tightly compressed neutrons would operate. But I might be wrong on that point.

8

Fun problem. Smells like someone’s homework set. Still, I see four questions here(and I’ll do my best until someone like Chronos or Stranger on a Train shows up):

[ol]
[li]What are the forces on a neutron at the equator?[/li][li]How fast would the neutron star have to spin to eject a neutron at the equator from spin velocity alone?[/li][li]Will neutrons transmute when away from the parent neutron star? [Already answered by OldGuy and ZenBeam][/li][li]How is time affected, by either the rotational velocity of the neutron star or the gravitational field at the star’s surface?[/li][/ol]
For #1, I think Quercus and dracoi got there already, but here’s how I’d lay it out. First, I’m ignoring any neutron-neutron interactions. There are papers out there laying out hypothetical structures of neutron-degenerate matter and how they’d behave. As dracoi pointed out, I don’t know if we have a good idea of how neutrons in a neutron star would behave. So ignore friction. That leaves the gravitational force between the neutron and the star (Fg), the normal force—the force of the star pushing back on the neutron (Fn), and the centripetal force necessary to keep the neutron in a rotating path (Fc) and not flying off at the tangent that dracoi mentioned.

Classically, F(g) =the product of the two masses, divided by the square of the distance between their centers, all multiplied by Newton’s Gravitational Constant. For the OP’s star with given radius 16km, an ex rectum neutron star mass of 1.5 solar masses (2.98 X 10^30 kg), and a neutron mass of 1.67 X 10^-27 kg, I get a gravitational force of 1.3 X 10^-15 newtons. Plugging that into F=ma, yields an acceleration on the neutron of 7.78 X 10^11 m/s^2, or 7.94 X 10^10 gravities. Classical centripetal acceleration equals the square of the rotational velocity, divided by the radius. The rotational velocity is equal to 2 * radius *pi * rotational frequency. So, 2 * 16 km * pi * 716 cycles/sec = 7.20 X 10^7 m/s, or 24% of the speed of light, which agrees with wiki. From that, I got an acceleration of 3.24 X 10^11 m/s^2. Subtracting this centrifugal acceleration from gravity’s acceleration yields a net acceleration of 4.54 X 10^11m/s^2 towards the center of the star. (Yes, I know there’s no centrifugal acceleration, I’m weak. Obligatory xkcd reference here.)

Classically, the neutron therefore “feels” like it weighs 58.4% of what it normally would. For #2, I set f(c) = f(g), and came up with a velocity of 1.12 X 10^8 m/s, or 37.2% of the speed of light, required to cause a neutron to leave the star.

As Quercus notes, at a velocity of 24% of light, the Lorentz factor used in special relativity calculations is only 1.03. (I used this site for a calculator.) That doesn’t strike me as that big of a change for the OP’s purposes, compared to the significant figures I used. Of course, you’d have to take it into account when analyzing spectra from the star. 37.2 % of light speed has a Lorentz factor of 1.077 and I’ll leave it to someone much more skilled to use special relativity to determine how much that changes the classical answer to the OP’s problem.

General relativity looks exceedingly unpleasant to play with. My WAG from wiki, is that you’d want to look at the Kerr Metric to determine the effects of the star’s gravity on local spacetime. (I don’t think neutron stars are charged—how could they be, when they solely consist of neutral particles?—so the Kerr Metric I think would be where you’d start.) Treating the neutron star as a point source, and messing around with the equations for finding the ergosphere, I get a Schwartzchild radius of 4.43 km and an outer ergosphere boundary of 6.65 km, both comfortably within the surface of the neutron star, as you’d expect. Further messing around with the simplest case of frame dragging, I got an “omega” of 1.24 X 10^-2, which I’m guessing is radians per second. I wonder if you’d notice that degree of frame dragging? This paper goes into more detail about time difference between rotating clocks, and later applies it to time dilation from orbiting local astronomical features, such as orbits about Jupiter, the Sun and the Earth.

This equation yields time dilation due to a gravitational field. I think their equation comes from the Schwartzschild Metric for Einstein’s field equations for general relativity, where the field is non-rotating. Still, using that equation, I get a time dilation from gravity alone of 0.850 T(outside observer), which is much larger than the dilation from special relativity we’ve already gone through. I’d think that having a rotating field would exacerbate the dilation, but I couldn’t begin to tell you how much, or how that would affect the forces on our neutron at the surface. You might also find this guy’s blog interesting, where he goes into some detail about the calculations for frame-dragging for rotating masses of various sizes.

As I said, fun problem…

9

Awesome information Gray Ghost! I assure you it’s not homework! :slight_smile:

10

I’ll see if I can rustle up some exact answers next time I’m up at the office with my books. Meanwhile:

There is normal matter (or at least, more normal) on the surface of a neutron star. You can still get distinct atomic nuclei (albeit extremely neutron-rich nuclei) as far as a kilometer deep or so. And even in the inner portions, you’ll still have about 1 proton (with a corresponding electron floating around somewhere) per 10 neutrons (in fact, most of the degeneracy pressure comes from the electrons, not from the neutrons). So it would be possible in principle for a neutron star to have a charge, but in practice, it’s never going to be a significant charge compared to its mass. If you managed to shred a neutron star violently enough that you were bringing up the neutronium from the insides, I don’t know what it would decay to, but my guess would be mostly iron. For anything violent enough to do that, you wouldn’t be getting individual neutrons coming up, but huge splatters of material.

The Kerr metric is certainly a better approximation here than the Schwarzschild, but it still wouldn’t be quite right, since unlike a black hole, a neutron star is made out of stuff, and therefore has messy things like quadrupole moment. That said, though, determining exactly what the quadrupole moment is would require knowledge of the neutron star equation of state and other messy physics, so it’s probably best just to ignore that.

Sure, if you want to look at it that way, but then, there’s no downward force, either. In general relativity, gravity and centrifugal force are both equally real, and this is definitely a case where GR should be used.

11

OK, if the star were right at the point where material on the surface would start flying off, that surface material would effectively be in orbit, so we can use orbital equations. As I stated before, I’ll use the Kerr metric as an approximation, and the angular speed of a prograde circular orbit in the Kerr metric is given by

Omega = sqrt(M)/(r^(3/2) + a*sqrt(M))

a is the angular momentum per mass, so if the star is a sphere of uniform density (another approximation), a = 2/5 * r^2*Omega . Plugging this back into the first equation and solving for Omega, we find

Omega = (-r^3/2 + sqrt(r^3 + 8/5 r^2M)) / (2/5 r^2sqrt(M))

The mass and radius of this object really ought to be known well, since it’s in an eclipsing binary system (and man, you can learn everything about an eclipsing binary), but I don’t feel like doing a literature search to find them, and the Wikipedia page is a stub. So I’ll assume r = 16 km, and M = 2 km. Plugging those in gives us Omega = 12.6 kHz, or 2000 rotations per second (compared to the star’s observed rotation at 716 rotations per second).

Note that I made a number of assumptions here, most notably that the star is spherical, that its density is uniform, and that the spacetime around it is Kerr. These are only approximations, and in fact are probably mutually inconsistent. The result, however, is I think good enough for our purposes.