# Is it possible to decouple inertial and gravitational mass?

1. Aug 3, 2014

### martix

Or stated otherwise:
Is it possible, due to other effects than mass for a non-spinning object with inertial mass n to exert a gravitational force characteristic of another non-spinning object with inertial mass different than n - theoretically or practically?
I hope that covers all the caveats that might arise.

Also, my background in physics is rather basic, so apologies if it's nonsensical :)

2. Aug 3, 2014

### PAllen

Per a long series of ever more precise exeriments, no. Per Newtonian gravity or General relativity, no. One caveat: in general relativity, this is clearly true only if matter satisfies the dominant energy condition, which is equivalent to "locally consistent with special relativity". If you allow exotic matter that violates this condition, then a small body of it, in free fall, need not travel on timelike a geodesic, which means it 'falls differently' than other matter. Falling differently implies a different relation between gravitational and inertial mass.

While no one has proposed a theory of how to make a 'body' of exotic matter, it's existence in limited forms is required by QFT. All the same, there remains no experimental evidence of any matter that violates the equivalence principle (i.e. that falls differently than other matter).

3. Aug 4, 2014

### pervect

Staff Emeritus
Would you explain more? I don't see why exotic matter wouldn't follow a geodesic the same as any other sort of matter. In the Newtonian limit, for instance, it'd have a negative inertial and gravitational mass, so the force would change sign, but the acceleration wouldn't.

4. Aug 4, 2014

### PAllen

The theorems that derive, from the EFE, that the trajectory a small body approaches a (timelike) geodesic need to assume the body obeys the dominant energy condition (or something equivalent). Further, there are other papers in this research family that demonstrate the converse - that if the body violates the DEC, then it need not follow a timelike geodesic in free fall.

Here are a couple of examples (there are others):

http://arxiv.org/abs/1106.2336
http://philsci-archive.pitt.edu/5072/1/GeodesicLaw.pdf

[edit: To see that the issue is still relevant for more modern work on motion of small bodies, see note 8 on p.19 of:

http://arxiv.org/abs/0806.3293

This is an example of having to assume something equivalent to DEC.]

Last edited: Aug 4, 2014
5. Aug 4, 2014

### DrStupid

That's quite simple: An object composed of two concentric spherical shells exerts only the gravitational force of the inner shell to objects within the gap between the shells.

Not as simple: An object moving with relativistic velocity will exert an higher gravitational force compared to an object at rest with the same inertial mass.

6. Aug 4, 2014

### PAllen

Neither of these are violations of gravitational mass = inertial mass because they are both independent of composition. In the first case, the outer shell is irrelevant and only gravitational=inertial mass of inner shell matter.

In the second case, imagine a container of relativistically moving particles (i.e. very high temperature). The inertial and gravitational mass of this container are identical. The KE contributes both to inertial mass and gravitational mass.

If you want to consider only isolated particles, what you are really dealing with is velocity dependence with rules that are not so simple. For example, if two bodies are comoving at near c relative to some observer, their KE in that frame contributes not at all to their mutual attraction. If they are moving opposite each other, it does contribute, but that is a case of both inertial mass and gravitational mass of the system of two particles being increased by the KE.

7. Aug 4, 2014

### pervect

Staff Emeritus
I would say that the whole gravitational mass = inertial mass idea applies to Newtonian theory and/or to various weak field limits of GR (including some of PPN theory) but the topic doesn't really make sense in the context of GR itself.

GR doesn't and can't have separate concepts of "gravitational" and "inertial" mass - for good reason, the strong principle of equivalence would make them the same.

However, properly definiing mass in GR turns out to be rather tricky, the theory as it stands doesn't have a single defintion of the term, there are multiple definitions of the term, for instance ADM, Bondi, and Komar masses (among others).

Additionally, mass is not the source of gravity in GR - the stress-energy tensor has that role.

8. Aug 4, 2014

### PAllen

At a crude level, you could say energy contributes equally to both gravitational mass and inertia. Thus, rapidly moving particles do NOT violate such equivalence. I take the OP as asking if you can have an analog of varying charge to mass ratio as you can with EM, with different ions following different trajectories in the same field. The answer is emphatically no per experiment so far, and theoretically no per GR with the possible exception of 'bodies of exotic matter', for which no one has a recipe.

9. Aug 4, 2014

### martix

True. A good analogy.

10. Aug 5, 2014

### DrStupid

In the setup above the mass of the outer shell is irrelevant for gravity but not for inertia.

It's not simple but not unsolvable:

https://home.comcast.net/~peter.m.brown/ref/mass_articles/Olson_Guarino_1985.pdf

Under the conditions assumed in this paper the gravitational mass is twice as high as the classical inertial mass (Newton's "quantity of matter") if the speed gets close to c.

11. Aug 5, 2014

### PAllen

So, trying to understand your point, if the shells are connected by rigid rods, than for interaction between a body between the shells and the connected shells, the inertial of the connected shells includes both their masses, while the gravitational mass includes only the inner shell. I can see how you can look at it this way. But:

this is really a question of configuration of gravitational mass. It is analogous to noting that the electric field inside a charged conducting sphere is zero. This does not mean the charge is zero. I would still say gravitational mass (charge) = inertia mass for the assembly.

And so is the inertia (defined relativistically). All high speed particles interact the same with other matter and energy, gravitationally, the same way. [FYI: I am well aware of that paper and have read it a couple of times over the years.]

Anyway, the OP confirmed that my way of understanding the question was what they had in mind. For that, the only exception I know of is the slight possibility of a body of matter that violates the DEC.

[Edit: While that paper is a pioneering work, it does not deal with other interesting cases, like:

1) two dust beams with particles going the same direction at the same speed (in some reference frame) have no additional mutual attraction compared Newtonian expectations.

2) dust beam with particle moving near c will attract a test body stationary in that frame by a factor near 2 compared to Newtonian (this is analogous to the case they focus on).

3) two dust beams moving in opposite directions, near c, in some frame, will attract each other nearly 4 times Newtonian expectation. ]

Last edited: Aug 5, 2014
12. Aug 6, 2014

### DrStupid

That's another question. The OP asked for an object with inertial mass n that exerts a gravitational force characteristic of another object with inertial mass different than n. That's what my example refers to.

How does this definition looks like?

I do not see how this is possible for force or acceleration (I do not know what you mean with "attraction"). Both meet the Newtonian expectation in the rest frame of the beams but in contrast to classical mechanics they are frame dependent in SR. Therefore there should be a speed-dependent difference between the classical and the relativistic result.

13. Aug 6, 2014

### PAllen

Energy contributes to inertia as well as being a source of gravity. Both go up by gamma for a moving particle. Thus a box of elastically bouncing balls, all with speed v in the COM frame, will have both inertial and gravitational mass increased by gamma of the balls (assuming the walls are negligible).
I was imprecise here. The results are more complex than I remembered. Here is an analysis covering dust and light:

http://arxiv.org/pdf/gr-qc/9811052v1.pdf

Last edited: Aug 6, 2014
14. Aug 6, 2014

### DrStupid

That's the inertial mass as used by Newton in his definition of momentum. The gamma factor results from the replacement of Galilean transformation by Lorentz transformation in SR. According to the paper above there is an additional factor for the gravitational mass of particles on a hyperbolic path in the far field of another point mass. Therefore inertial mass and gravitational mass are not equal under these conditions.

15. Aug 6, 2014

### PAllen

Agreed, the beam results could lead to such conclusions. I think the following is true from the results of this paper's analysis:

Consider two parallel dust beams. Imagine a sequence of such experiments where the total energy density of the beams is held constant, while the speed increases (which would require decreasing the rest density of the beams in the sequence). As the speed approaches c, the beam's mutual deflection approaches zero. [This has a rather simple heuristic explanation: analyzed in a frame where the beams are very slow, the mutual deflection stays the same. Transforming back to a frame where the beams are near c, the deflection rate is reduce by gamma.]

In the anti-parallel case, the beam's mutual deflection increases by a factor of 4 as the limit of c is approached (holding total energy density constant).

In the case of mutual deflection between a very slow dust beam and a beam approaching c, with energy density for both constant in the sequence of tests, the deflection doubles as the fast beam is allowed to approach c.

Last edited: Aug 6, 2014
16. Aug 6, 2014

### pervect

Staff Emeritus
I don't see how you can have the mutual deflection remaining the same if you lower the rest density.

In the rest frame, the lower density will mean lower attraction in the rest frame. In the ultrarelativistic limit, the density will approach zero, which means no attraction.

However, this doesn't affect your conclusion that the deflection approaches zero as you approach a situation where the beam consists totally of energy with zero rest mass - it supports that conclusion.

Going on to more general remarks about the thread, which are not a direct reply to PAllen's post.

The conclusion that I would like to guide people towards is that naive notions of replacing "mass" with "energy" or "mass density" with "energy density" do not give exactly correct results in General Relativity. If one wants to get truly correct results, one needs to dig deeper into the theory, the idea of working it out in Newtonian terms by replacing "mass" with "energy" specifically isn't going to work.

The paper I usually cite on this is Measuring the active gravitational mass of a moving object .

If the idea that we replaced mass by energy were correct, we'd expect that the so-called M_res would be $\gamma M$ , but there is an additional velocity dependent factor that approaches 2 at high velocities.

Formally, I'd say that gravity in GR does not depend on "mass", or "energy", but the stress energy tensor, which includes terms due to momentum and pressure as well as energy.

Note that the term "active gravitational mass" as introduced by the author isn't something that's widely taught, it's just a concept that this author created to make his point. (The author himself mentions this in section IV where he discusses some of the more fundamental definitions of mass that are actually used in GR). I think it's a convenient concept, because the measurements can all be carried out in the flat spacetimes "before" and "after" the flyby, so one doesn't have to get involved with the complex issues of curved spacetime, furthermore the idea of what is being measured is widely understood. It's also a concept that is "good enough" to illustrate the failure of the "energy is mass" idea to give correct experimental results.

To recap, replacing "mass" with "energy" into the Newtonian formula for gravitation just won't work. It is a practice that should be avoided if one expects accurate results.

One can also show that the gravitational field due to an ultrarelativistic flyby is not spherically symmetric, but the arguments are more complex, the cleanest arguments involve considering not gravity, but tidal gravity. Using this approach, one can analyze the tidal gravity of an ultrarelativistic flyby, and find that the tidal field has an impulsive nature. If there is interest in this point, I'll attempt to dig up the paper. However, one can probably find simlar information by researching the point below for oneself.

The GR solution for the ultra-relativistic flyby is known as the Aichelburg–Sexl ultraboost for those who want to do more research.

17. Aug 6, 2014

### PAllen

Oops, yes I was adding a quick afterthought (not from the paper) providing intuitive justification. Too quick. The approach to zero attraction for the scenario will be of order 1/γ3 rather than 1/γ.
I am well familiar with this paper, but it covers only one configuration. It makes no mention of the parallel versus anti-parallel behavior described in the paper I referenced. However, I see no inconsistency. In the case of deflection between a slow beam and a ultra-relativistic beam, the paper I reference finds a factor 2 times the deflection considering the energy density alone.
The rest of what you say, I agree with.

18. Aug 7, 2014

### PAllen

Let me make one other summary point, irrespective of complexities discussed:

There is nothing you can put in a box A versus box B such that their COM will fall or orbit the sun differently. Particles, energy, gas, plasma, whatever.

The only caveats: boxes too big ( so they feel tidal forces significantly), boxes too massive (enough to perturb the sun), boxes sufficiently charged that EM radiation reaction is detectable (to my knowledge, this has never been measured for a macroscopic body), boxes contain hypothetical exotic matter (phenomenon of exotic matter falling differently has not been observed, nor is there any proposal for how to do so). (I don't need to mention gravitational waves, because they wouldn't be relevant for a body small enough not to perturb the sun).

19. Aug 7, 2014

### pervect

Staff Emeritus
I have a question about the exotic matter caveat. From my quick reading of the papers you cited earlier, there doesn't appear to be any definite prediction of how the exotic matter would fall. Rather, the claim seems to be that it "might not" fall the same as other matter. In other words, it appears to be possible (as far as I can tell) that some form of hypothetical exotic matter could exist that would fall normally, this possibility isn't ruled out mathematically. It's just that it doesn't "have to" fall normally from the stated assumptions. Would you agree with this interpretation?

20. Aug 7, 2014

### PAllen

Yes, I would agree, but it seems almost expected to me that generic matter violating DEC might have a spacelike trajectory; and conversely, that matter satisfying DEC cannot. What DEC formalizes is that locally neither matter nor energy can have flows faster than c. Matter violating DEC does have at least an energy flow faster than c. An example is Hawking radiation treated classically - you have mass flowing out of a BH; this requires a spacelike mass flow. This is why Hawking radiation violates all the energy conditions. The counterexamples to the geodesic motion derivations amount to formally establishing that ftl locally encoded in the stress energy tensor can produce ftl motion integrated up to the path of a small body. This is something you want to see a proof for, but it also seems expected rather than surprising to me.

On the other hand, I personally expect that at some point it will be established that there are constraints on exotic matter such that ftl motion of macroscopic bodies, warp drives, and traversible wormholes will be impossible. Yet I must admit that significant work by fair number of researchers has failed to establish this.

21. Aug 12, 2014

### rjbeery

Does NASA's "impossible space engine" qualify as a caveat as well? If true, it apparently violates conservation of momentum!

22. Aug 12, 2014

### pervect

Staff Emeritus
My not-very-good understanding of the Casimir effect is that the vacuum between two conducting plates violates the DEC when one includes quantum effects, because the rest energy density between the plates is lower than that of a classical vacuum, i.e.. negative.

However, I wouldn't expect that a pair of closely spaced conducting plates would fall any differently than a single conducting plate.

My suspicion is that violating the DEC is necessary but not sufficient, to cause objects to violate from geodesic trajectories.

23. Aug 14, 2014

### timmdeeg

I'm not sure about that, since Jaffe showed that the Casimir force can be calculated without reference to the quantum vacuum.

http://arxiv.org/PS_cache/hep-th/pdf/0503/0503158v1.pdf

However even anticipating a lower energy density between the plates relative to the outside would this necessarily mean a negative energy density? This conclusion seems to imply that the energy density outside is zero. But isn't this just per definitionem?