# Gravitational Mass

1. Jan 26, 2007

### Fallen Seraph

Does gravitational mass increase as inertial mass does in relativistic situations? Or does GR make this idea irrelevant?

2. Jan 26, 2007

### nakurusil

Inertial mass does not "increase in relativistic situations". Neither does gravitational mass.

3. Jan 26, 2007

### disregardthat

May I ask what that is increasing then?

4. Jan 26, 2007

### nakurusil

5. Jan 26, 2007

### quantum123

Relativistic mass.

6. Jan 27, 2007

### disregardthat

Has relativistic mass any effect of an object?

Or is it just to explain why it is harder to accelerate an object at high velocities?

7. Jan 27, 2007

### Hootenanny

Staff Emeritus
This has been discussed to tedium in the SR forums; an FAQ exists which discusses it further.

8. Jan 27, 2007

### pmb_phy

Yes. It does. For the calculation which shows this please see

http://www.geocities.com/physics_world/gr/force_falling_particle

See also "Measuring the active gravitational mass on a moving object," by D.W. Olson and R.C. Guarino, Am. J. Phys., 53(7), July 1985. The abstract reads
In the first case above is the situation of particle which is falling in the field of a stationary planet while in the later situation the object is falling in a gravitational field of a moving planet.

Pete

9. Jan 27, 2007

### Chris Hillman

Just wanted to point out that this is one example where it would be quite confusing to speak of "relativistic mass" (bad terminology on many levels) instead of "relativistic kinetic energy" (good terminology).

A WP article I wrote on the Aichelburg-Sexl impulsive pp wave, an exact axisymmetric vacuum solution which has the character of a gravitational wave, but whcih arises from a (not uniqely defined) limiting process called an "ultrarelativistic boost of the Schwarzschild vacuum" (see http://en.wikipedia.org/w/index.php?title=Aichelburg-Sexl_ultraboost&oldid=45333920 )may suggest to the original poster how he might wish to rephrase his question.

(As usual, note that I linked to the last version I edited; more recent versions might be somewhat better or might be much worse!)

The AS pp wave has some interesting features: its curvature is concentrated in a single wavefront, and within that wavefront it falls of much less rapidly than does the "Coulomb tidal force field" which is characteristic of slowly moving and approximately spherically symmetric massive objects.

Note well: considering an ultra-relativistic is not appropriate for studying the gravitational field of a massive object which is moving at a nonrelativistic velocity with respect to the observer of interest, such as a planet moving at a nonrelativistic velocity with respect the observer of interest. Rather, the idea is that you can treat a massive but very rapidly moving Schwarzschild object either in the obvious way--- write down the parametrized geodesic of your ultrarelativistic observer, and compute the tidal tensor in his frame--- or you can take a stationary observer in the AS model, who thinks he is in flat spacetime until the wave whizzes by him at the speed of light. IOW, the AS model is only an approximation, but a very suggestive one.

Last edited: Jan 27, 2007
10. Jan 27, 2007

### nakurusil

Yes, of course. Unfortunately certain people are still stuck on "relativistic mass", the notion that "refuses" to go away. It is really unfortunate that people still attribute physical meaning to the mathematical entity with no ulterior physical meaning $$\gamma*m_0$$ that appears the the relativistic momentum $$\gamma*m_0*v$$ and in the relativistic energy $$\gamma*m_0*c^2$$. It would be nice if all textbooks standartised on the use of proper mass $$m_0$$ ONLY and put an end to the accursed "relativistic mass".

Last edited: Jan 27, 2007
11. Jan 27, 2007

### quantum123

Can an ultra-relativistic mass turn into a black hole?

12. Jan 27, 2007

### nakurusil

If I stick some paper wings to an elephant, will it learn how to fly? Or will you need to teach it?

13. Jan 27, 2007

### quantum123

That question belongs more to biology, not physics.

14. Jan 27, 2007

### Staff: Mentor

15. Jan 27, 2007

### MeJennifer

Interestingly on the referenced webpage is the following statement:

What momentum is ignored in the Schwarzschild solution?

16. Jan 27, 2007

### pervect

Staff Emeritus
A key point (IMO THE key point) in this derivation is the defintion of "gravitational force" in

http://www.geocities.com/physics_world/gr/grav_force.htm

It seems to me that Pete basically admits in this webpage that his defintion is not covariant - his statement that makes me say this is

G being the "gravitational force".

I do not believe that it has been demonstrated that this definition of "gravitational force" on this webpage has been proposed or adopted by any author other than Pete. Note specifically that it is different from the defintion offered by Olson and Guaranio (also cited).

The issue of whether or not the "gravitational force" is covariant may seem abstract, but it is very important. A non-covariant defintion, among other effects, means that the defintion of "force" depends on the specific choice of coordinates by the observer!

[add]I should add that there is a reasonably well-accepted defintion of "gravitational force" for stationary observers or even observers moving at low velocities. (See for instance the gravitoelectric field in GEM, i.e. http://arxiv.org/abs/gr-qc/9912027.pdf)

There are some significant defintional problems for even defining "gravitational force" for rapidly moving observers.

Last edited: Jan 27, 2007
17. Jan 27, 2007

### nakurusil

Objects do not have any increased tendency to form black holes due to their extra energy of motion. In a frame of reference stationary with respect to the object, it has only rest mass energy and will not form a black hole unless its rest mass is sufficient. If it is not a black hole in one reference frame, then it cannot be a black hole in any other reference frame.

18. Jan 27, 2007

### tehno

And It may also seem ridicilous.Especially having in mind covariant spirit of General relativity.

19. Jan 27, 2007

### pervect

Staff Emeritus
If you are suggesting that non-covariant definitions of force are "ridicilous", I would say that you are on the right general track, though it seems a bit harsh put that way.

And excuse me, but I can't help but observe - isn't it a bit ridicilous to misspell ridiculous ? :-)

Anyway, I hope we can keep this thread focused on the issues.

Last edited: Jan 27, 2007
20. Jan 27, 2007

### tehno

If certain authors suggesting a non-covariant definition of forces,I think they are on the right track to return to pre Maxwell time era where cog wheels ruled.
May covarince long live 'coz covarince rulz ("rulz" misspelld intentionaly )

21. Jan 27, 2007

### Chris Hillman

My contribution to physics?

As of a few minutes ago, Baez's website now sports a nifty new search tool, courtesy of Hack 88 from Google Hacks by Dornfest et al. That's really what I wanted to say... er, what were we talking about? Oh, yes, the Schwarzschild vacuum solution. Well, in deriving this we assume that the source of the field is an isolated massive object at rest in our coordinates, so we should expect that in any reasonable sense it will have zero momentum. And indeed, there are definitions of relativistic momentum (valid for asymptotically flat vacuum solutions) which apply and give the expected result.

The point is this: a rapidly moving object certainly doesn't turn into a black hole (of course not! ), but its gravitational field will appear to a suitable observer, relative to whom our object is moving very rapidly, very much like an axisymmetric pp wave with curvature concentrated in one wavefront, in fact near the center of that wavefront. This point of view has been widely exploited, e.g. by 't Hooft, in studying collisions of black holes.

22. Jan 27, 2007

### MeJennifer

Right, I simply do not see why momentum is in anyway relevant as the above mentioned quote suggests. Angular momentum, yes, the dynamics of space-time, I suppose that means the initial value problem, then yes, but not momentum.

On the contrary, the idea of momentum being of any influence seems more to affirm the misunderstanding that a rapidly moving object can become a black hole than debunk it.

23. Jan 27, 2007

### pervect

Staff Emeritus
I'm going to take the opportunity to "sharpen up" my objections to Pete's derivation. Pete and I have gone over this before, I think, though possibly not with exactly the current set of derivations. I'm not sure if Pete still has "his fingers in his ears", at one time he was mainly dealing with my objections by ignoring them.

Suppose we have some observer following a specific path through space-time. This observer has a 4-acceleration, which is a fully covariant 4-vector.

This 4-vector,scaled by the invariant mass m, is in fact what Pete computes as $$\frac{d \,P_u}{d \tau}$$ in his derivation.

It is also quite reasonable to take the spatial part of this 4-vector, where the spatial hypersurface is taken as that surface orthogonal to the proper time of the observer following a specific path. And it's reasonable to multiply this accleration by a mass and call it a force. So far, so good.

The spatial part of 4-accelerations can in fact be measured, by devices known as accelerometers.

This gives us an operational way to attempt to define the "gravitational field" of a moving observer, and to highlight the resulting problems. Suppose we have someone, moving "in a straight line with constant velocity" (more on this later), who carries along an accelerometer. This accelerometer will have some particular reading, a 3-vector, which we can interpret as the "gravitational field" for that observer.

Note that this defintion says that the "gravitational field" for someone standing on the Earth would be the reading of an accelerometer for a static observer who is stationary with respect to the Earth. Thus far there is no problem - the notion of "stationary with repsect to the Earth" can be well defined.

The problem is in the quoted phrase - what does "moving in a straight line with a constant velocity" actually mean?

It's certainly not a geodesic path. That path would have a zero 4-acceleration.

It's not the path that a light beam follows either.

In fact, there isn't, as far as I know, any sensible coordinate indepedent notion of what "travelling along a straight line at a constant velocity" actually means.

The statement makes sense in a flat geometry, such as a the Minkowskian geometry of SR. But there isn't any clear meaning for what "moving in a straight line at a constant velocity" means in the non-flat spacetimes of GR.

So when Pete computes the "gravitational force", he is implicitly using some coordinate based defintion of what a "straight line" is. The problem is that this notion has no geometrical significance - it's dependent totally on the choice of coordinates, a rather bad state of affairs.

Last edited: Jan 27, 2007