# Increasing mass, decreasing force?

1. Sep 14, 2004

### Rothiemurchus

As a neutron star collapses further and neutron degeneracy pressure causes
neutrons to increase in mass (by virtue of high speeds),
if that mass increase weakens the force of gravity it could stop
a singularity from forming.Is there any evidence that particles with large mass
increases due to high speed experience a weaker gravitational field?

2. Sep 14, 2004

### pervect

Staff Emeritus
As the neutron star collapses, both the density and pressure will increase. Both of these will increase the gravitational field, because both the density and the pressure contribute to the stress-energy tensor in a manner that will increase the Einstein curvature.

It's best not to talk about "increases in mass" of neutrons. To say that the mass increases as written is incorrect - saying that the "realtivsitc mass increases" would not be wrong, but the usage is debatable (we have an off-and-on debate about that far too often already, let's not start another one).

3. Sep 14, 2004

### humanino

Where's Pete

4. Sep 15, 2004

### Rothiemurchus

The reason I asked is because in the Bohr model of hydrogen a 1s electron orbits faster than a 2s electron and so then 1s electron would have more relativistic mass.
If this is associated with a weaker force - in this case a weaker electric force than expected between the proton and electron - then it might account for the lamb shift. I was just wondering if speed could weaken gravity.Those singularities have to be stopped from forming - they don't make sense to me.

5. Sep 15, 2004

### zefram_c

The Bohr model has long been discarded - try something a little more modern?

In any event, if you look hard enough, you can find the so-called "relativistic correction" calculation using perturbation theory. The first-order correction is larger than the Lamb shift by about a factor of the fine structure constant. The second and higher order corrections are insignificant.

You can also get an 'exact' solution from the Dirac equation for the electron which fully accounts for spin-orbit and relativistic structure. It doesn't predict a Lamb shift.
I think I'll stick with the account provided by QED for now
Higher speed(momentum) increases gravity, as has been mentioned.

6. Sep 17, 2004

### pmb_phy

Something sounds fishy to me here. I don't think this is true. As the star collapses the gravitational potential energy of the matter decreases. The volume decreases as well. So while the pressure increases and the energy density increases the volume decreases (less matter to integrate over) and the gravitational potential energy decreases. Thus the gravitational field at a set distance would seem to remain unchanged.

Pete

7. Sep 17, 2004

### pervect

Staff Emeritus
I'm not sure how you are defining "gravitational potential energy", but I think I agree with your conclusion if not the steps you take to reach it.

If an observer at infinity in flat space-time looks at the gravitational field of a collapsing star, he won't see any change in the "gravitational force", and the total mass of the star won't change, either. Well, I have to qualify that a bit - there's some possibiity that gravitational waves could carry away some energy, also the emission of light / electromagnetic radiation from the collapsing star could also carry away some energy. Other than losses due to radiation, the distant observer won't see any change in the "gravitational force" or the total mass as the star collapses.

From the point of view of preventing collapse, though (which was what the original question is about), the energy per unit volume is going up, and the pressure is going up as well, and both of these promote further collapse.

The fact that pressure causes gravity is one of the big differences between Einstein's theory and Newton's. This is what makes collapse to a singularity unstoppable in Einstein's theory. The pressure needed to resist collapse simply increases the field, overcoming the pressure, no matter how large the pressure gets. Not that by "field" here I'm not talking about the field as measured by a distant observer, which we both agree remains constant, but the Einstein curvature tensor - or the "thing that makes the ball of coffe grounds shrink" from Baez's GR tutorial.
http://math.ucr.edu/home/baez/gr/outline2.html

8. Sep 17, 2004

### Rothiemurchus

The only thing then that can stop a singularity forming is if the force
gets smaller or stays constant.This happens for the colour force as quarks get closer together, so gravity might have something in common with it.

Liked Baez's tutorial - it gives a clear picture of the Ricci curvature tensor.

Last edited: Sep 17, 2004
9. Sep 23, 2004

### pmb_phy

Look up Birchoff's theorem. It implies that any body which is spherically symmeteric anc collapses in a spherical manner then the metric will remain the same outside the matter. And such a body does not radiate gravitational waves.

See Relativity; Special, General and Cosmological, Wolfgang Rindler, Oxford, (2002) page 230

Pete

10. Sep 23, 2004

### pervect

Staff Emeritus
Interesting, but I had the impression that people are or will be looking (when LIGO comes online) for the gravitational waves emitted by collapsing stars. I believe the culprit here (as far as getting around Birkhoff's theorem) is the rotation of the star, though I'm not positive.

http://www.mpa-garching.mpg.de/HIGHLIGHT/2001/highlight0111_e.html [Broken]

Last edited by a moderator: May 1, 2017
11. Sep 23, 2004

### pmb_phy

I don't know. But if a collapsing star collapses symmetrically and is symmetric to begin with then the gravitational radiation would be monopole radiation and in GR there is no monopole radiation.

I have more time to quote Rindler now
Okay. Off I go to the butcher's block. :surprised See you if I make it out in one piece! :rofl:

Pete

Last edited by a moderator: May 1, 2017
12. Sep 24, 2004

### pervect

Staff Emeritus
But is a rotating star "spherically symmetrical" in the sense of Birkhoff's theorem?

Wald says wrt Birkhoff's theorem
But of course we have the Kerr solution not equal to the Schwarzschild solution. The rotation spoils the spherical symmetry.

13. Sep 24, 2004

Staff Emeritus
Schwarzschild doesn't have $$R_{ab}=0$$, it has the Einstein tensor = 0. That differs from Ricci by $$\frac{1}{2} g_{ab}R$$.

14. Sep 25, 2004

### pervect

Staff Emeritus
If Rab = 0, the trace Raa is zero, so Gab = Rab-.5*R*gab = 0.

BTW, the original quote is on pg 125 of Wald, "General Relativity".

15. Sep 25, 2004

### Nereid

Staff Emeritus
In the real world, massive stars with iron cores which undergo collapse do so somewhat asymmetrically ... indeed, IIRC it wasn't until 2D (or was it 3D?) models were done that the core could collapse at all! In 1D (and maybe 2D?), the collapse stalls, no BH forms, etc. Apparently it's the tiny instabilities - deviations from perfect spherical symmetry? - which allow the collapse to proceed, somewhat like why water from an upturned bucket doesn't fall as a perfectly flat sheet?

The net result is that a core collapse SN *always* generates gravitational waves, because it does not (and cannot) collapse in a perfectly symmetrical fashion.

16. Sep 26, 2004

### pmb_phy

I doubt that there'd be a measurable effect. You're talking about an extremely small deviation from the Schwarzschild spacetime so any changes in the gravitational field would be extremly hard to even measure.

Pete