# Raindrops and Gravity

1. Sep 24, 2007

### oldman

My understanding is that the phenomenon of aberration does not affect the gravitational interaction between relatively moving bodies, whether such interaction be described from a Newtonian or General Relativistic perspective.

Aberration is an observer-dependent phenomenon. But so is relativistic mass change. Consider observers on two relatively moving bodies. who each perceive the inertial mass of the other body to increase with velocity.

Do such relatively moving observers also perceive a change in gravitational mass and if so, would they also perceive an increase in gravitational interaction? Or do special and general relativity conspire to leave this interaction unaffected, like aberration, by relative motion?

Last edited: Sep 24, 2007
2. Sep 24, 2007

### pervect

Staff Emeritus
There are a bunch of different concepts of mass defined in GR- it is not a single concept. As far as I know, though, none of the various sorts of masses commonly used in GR (ADM mass, Bondi mass, and Komar mass) are based on "relativistic mass", which is one reason I think relativistic mass is a dead-end concept.

See for instance http://en.wikipedia.org/w/index.php?title=Mass_in_general_relativity&oldid=159600370 for some discussion of how mass is treated in GR.

For instance, SR gives any point particle an energy momentum 4-vector. The ADM mass concept, which is one of many mass concepts in GR, is one of the more flexible concepts. It basically applies to any isolated system surrounded by a sufficiently large vacuum region (more formally, it applies to asymptotically flat space-times).

The ADM mass essentially gives you something that acts like the energy-momentum 4-vector of SR, but it's defined for a gravitating system "at infinity" where space-time is asymptotically flat. (Specifically, it's a space-like infinity, in case you care). So you can think of it as a meaningful way to talk about the total momentum and total energy of a gravitating system, as long as that system is isolated.

The ADM mass is not defined in terms of gravity as a force. You can basically describe how bodies move by the geodesic equation. At low velocities, this geodesic equation looks like a Newtonian force law, at higher velocities it starts to deviate in a way that cannot in genral be simply explained as forces.

On the other hand, it is clear that if you have an ideal gas consisting of randomly moving particles, said gas being contained in a pressure vessel, you can model the gravity from the assembly as a Newtonian force, even though you can't necessarily model the gravitational field of any individual particle by Newtonian forces. (We are assuming the particles in the gas are moving at relativistic velocities.)

The mass of the assembly taken as an entire unit can be seen to be affected by the motion of the particles of the gas - when you heat up the gas, the system gains energy, and this shows up as an increased mass.

See the Wikipedia article above for more. Note that the Komar (and ADM and SR) masses are all defined for this system, when taken as a whole, an they all agree. Also note that if you divide the system up into parts (the "gas alone" and the "pressure vessel alone") that you get different answers depending on which definition of mass you are using.

The Komar mass in this case is more closely related to the gravitational field, which GR predicts will be stronger just inside the interior of the pressure vessel than Newtonian theory does. (The system is nice because you can still think of gravity as a force for the entire system, even though you can't think of gravity as a force for any of the individual particles).

GR also predicts the same value for the gravitational field outside the pressure vessel, so you can think of the pressure in the interior causing "extra" gravity, and the tension in the pressure vessel causing "reduced" gravity, if you use the Komar approach.

If you use the SR approach, you'll won't see this redistribution of mass - so you'll be left scratching your head as to how to explain the increased gravitational field in the interior.

Last edited: Sep 24, 2007
3. Sep 24, 2007

### oldman

Thanks for your detailed reply to my rather naive question, Pervect, and for the Wikipedia reference.

It seems that the answer to my question " would they also perceive an increase in gravitational interaction?" is YES, judging by your remark:

and the Wikipedia comment in its Q & A section :

"Doesn't this imply that a moving particle has "more gravity" than a stationary particle? :This remark is probably true in essence..." .

I accept also that my referring to "relativistic mass" is too crude. The sophisticated distinctions folk draw between various GR definitions of mass make this clear. Although simple-minded questions of the sort I asked have no practical applications, say in designing accelerators, they may have a place in cosmology, where gravity rules and rapid motions do occur.

4. Sep 24, 2007

### pervect

Staff Emeritus
"One of the most astonishing features of the history of physics is the confusion which surrounds the definition of the key term in dynamics, mass". (Max Jammer, quoting G. Burniston Brown in an American Journal of Physics article.)

Unfortunately this tends to produce long answers to seemingly simple questions. So I guess the main point is that "what is mass" is actually a rather deep question, much moreso than one might think.

5. Sep 28, 2007

### oldman

I am still puzzling about the gravity of a moving object. Does its gravity increase with its velocity, or not? I interpreted the helpful replies of Pervect to my OP here to imply that YES, it does. But when I took this view in considering the anomalous acceleration of the Pioneer probe in Garth’s thread “Critique of Mainstream Cosmology”, over at the cosmology forum (post # 30), Garth told me that I was “trying to solve a GR problem (gravitational acceleration) by cobbling together Newtonian gravity with SR (which) does not work........”.

I think he is probably correct. But I can feel the sands of my understanding of relativity shifting beneath my feet:

In designing say a linac SR tells us that we must take into account the fact that the inertial masses of particles increase with their velocity. This is “perceived reality” for physicist/engineers who build accelerators. SR also tells us that if there existed observers moving with the linac exit beam, they would perceive a different reality, which would include smaller particle masses.

I believe that the central truth conveyed by SR is that there is no unique reality for all observers. What we call “reality” is in fact a masquerade of observer-dependent perspectives on events in space and time. So much for SR.

Now GR. There is no experimental justification or theoretical necessity for physicists to
distinguish between inertial and gravitational mass. We don’t understand why this is so, and therefore raise our ignorance to the status of a principle – the “weak” equivalence principle, which is deeply incorporated in GR. It seems to me that on the basis of this principle it must then be accepted that the gravitational masses (however one defines them, Pervect) of accelerated particles must increase along with their inertial masses. Of course this view is of little practical interest to physicist/engineers, but it has caused me confusion.

As I understand it, GR tells us that spacetime is distorted by mass/energy in a way that is described by the Riemann curvature tensor. Now here come the shifting sands: I had thought that while the components of this tensor are observer-dependent, the distortion that the tensor describes is not. But of course I was wrong. For if the inertial masses of particles in an accelerator are observer -dependent, their gravitational masses must also be observer dependent, together with the minute distortions they produce in spacetime. I guess this is because the components of the stress-energy tensor are also observer-dependent.

Which means that the distortions in spacetime produced by massive objects like the sun are similarly observer-dependent, and that rapid relative motions will cause them to be differently perceived.

Is this nonsense, or “is” there “really” no observer-independent “reality” in nature?

6. Sep 28, 2007

### Ich

There's a perfect observer-independent reality in nature. It is described in four dimensions, and every physical entity (objects, fields etc.) are represented by four-dimensional geometric objects.
For example, there is an observer-independent vector the length of which we call "mass". Unfortunately, you chose to call its time component "mass", which is common in popular physics. Most physicists prefer to call it "energy". This unfortunate choice hides from you the observer-independent nature of mass and lets you think that there is no independent reality.
GR uses a continuum description, where the stress-energy tensor is kind of a highly sophisticated mass density. It is observer-independent, and tightly connected with the observer-independent spacetime curvature.
But it has 10 independent components, so it cannot simply be replaced by a single number, that one would call mass density. You always lose important information. Even in the simple case of point particles, you need to know not only the mass, but also the velocity of each particle to calculate anything.
Further, if you know the stress-energy tensor, it is not generally possible to integrate it to give "the mass" of an extended object. There are situations where this reduction to a single number "mass" is useful in one way or the other. But in some situations, "the mass" is ambiguous and of limited use. GR doesn't need the whole concept, it works locally with locally defined quantities.
As you can see, this formulation of the equivalence principle is not easily adapted to GR. Inertial mass is ill-defined in exactly the same way as is gravitational mass. It is always constrained to small test particles of different composition. This is sufficient to develop a theory where the equality is not "raised to a principle" in ignorance, but where it is explained by the equivalence of accelerated frames without gravitation and static frames with gravitation. This latter formulation has a precise meaning in all circumstances.

7. Sep 28, 2007

### pervect

Staff Emeritus
Right. And you can think of the Riemann curvature tensor as being the tidal force. This gives a good clue as to a "better" way to think about gravity. Don't think about the force generated by a mass, think about the tidal force generated by a mass, instead.

Then we can rigorously answer the question about the tidal forces produced by a rapidly moving mass, without having to oversimplify anything.

One clarification has to be made. The equivalence between the Riemann and the tidal force works directly and easily only when the tidal force is measured in a non-rotating, non-accelerating frame.

That's correct.

Oh-oh. No, you were right the first time. The components of the Riemann tensor are observer dependent, but the Riemann itself can be regarded as a geometric object that is observer independent.

For a hopefully familiar example, think of the energy-momentum 4-vector of a point particle. The energy of an object is observer dependent. The momentum of an object is observer dependent. What's observer independent is the norm of the vector.

Observe independence basically means that the Riemann transforms in a standard way. It's a little like an object in an OOP programming environment - if you specify the components of the Riemann in one frame, you have the necessary information for the object to represent itself in any frame, because the transformation rules follow "standard" conventions.

8. Sep 29, 2007

### oldman

Thanks, Ich and Pervect, for straightening out my deviant thinking. It's reassuring to find that one of my original views wasn't quite wrong after all. Oversimplified, yes, but not too far from the truth. I can now go back to thinking of spacetime being like a strained elastic solid, with the analogue of the strain tensor being the Riemann curvature tensor and that of the the stress tensor being the stress-energy tensor.

I had suspected that the only invariant, observer-independent property of the Riemann could be the Ricci scalar, which I imagine (wrongly?) to be some kind of average curvature of spacetime taken over all directions. But is there a better invariant and possibly measurable characterisation of the Riemann (a "geometric object" in the intangible construct that is spacetime, as you explained) which one can fix one's mind on?

9. Oct 1, 2007

### pervect

Staff Emeritus
10. Oct 1, 2007

### Jorrie

Gravitational Acceleration and Velocity

Apart from the solid advice that was given so far, you may also want to look at a derivation by Pervect in a previous (closed) thread on radial and angular accelerations. It gives a very clear picture of how velocities influence the gravitational accelerations in Schwarzschild coordinates.

I have also sent you a PM with an alternative way to look at these accelerations, giving the same results from a different perspective.

Last edited: Oct 1, 2007