# Does mass really increase with speed

1. Dec 9, 2011

### lowemack

If mass really increases with speed then if 2 rockets leave earth, travelling parallel and close together, at close to the speed of light, then from an Earth point of view, as their mass increases, so should their gravity and they should be drawn together. But from the spaceship point of view their gravitational attraction should be minimal.

2. Dec 9, 2011

### D H

Staff Emeritus
Relativistic mass is an outdated concept. This is yet another example of where the concept of relativistic mass confuses rather than clarifies.

3. Dec 9, 2011

### nitsuj

That's true, from the spaceship's perspective gravity (and the mass of the ships) doesn't change.

From the point of view of Earth how is it determind the mass of the ships has increased?

I think it's only kinetic energy that increases, not the actual mass.

4. Dec 9, 2011

### QuArK21343

Relativistic mass is just the historical name for the quantity:

$$\frac{m}{\sqrt{1-v^2/c^2}}$$

but it is the so called rest-mass $m$ that really has a physical significance. Rest mass is the magnitude of the four-momentum vector:

$$m^2 c^2=\frac{E^2}{c^2}-p^2$$

It is an invariant quantity, meaning it is the same in every inertial frame of reference (this is a different from conservation of four-momentum: conservation of four-momentum refers to the fact that four-momentum, as a vector, is conserved in any interaction, as viewed in a particular reference frame; here, we are saying that mass is an invariant quantity, i.e. it is the same when computed in all possible r.f.). I think there is even a frequently-asked-question post that addresses this topic, so check it out.

I am not sure it is legal to talk about gravitational forces in special relativity, so I believe it is not right to say that since the masses increase (they don't!), their gravitational interaction increases too.

Last edited: Dec 9, 2011
5. Dec 9, 2011

### Staff: Mentor

It is not right even in the context of GR, which does deal with gravity. The "amount of gravity produced by an object" is frame-invariant; it doesn't matter what your state of motion is relative to the object.

6. Dec 9, 2011

### DrStupid

In GR the stress–energy tensor is the source of gravitation and it is not frame-invariant.

7. Dec 10, 2011

### dtyarbrough

ELROCH:
Thanks for clearing that up. So its okay for electrons to orbit near the speed of light around a nucleus moving at the speed of light. Einstein would not approve.

8. Dec 10, 2011

### torquil

What would Einstein not approve of? He was of course aware of the relativistic law of addition of velocities, so he would know that this does not lead to any violation of the speed of light limit. Or are you referring to something else?

9. Dec 10, 2011

### atyy

So could one instead say that this is a case where the relativistic mass (which is just another name for energy) clarifies, indicating that non-tidal gravity can be transformed away, consistent with the Principle of Equivalence?

The main problem seems not the coordinate variance of the description, rather the question seems to assume some sort of superposition, which may not hold in GR because it is a nonlinear theory.

Here is an interesting comment by Yau in this spirit (bcrowell has I think made a similar point many times, not sure if it's the same, but he's who I learnt it from, if I'm not misinterpreting him): The total energy in general relativity cannot be obtained by integrating any local density along a hypersurface. The reason is that the density would depend on the first order differentiation of the metric gij . But there is a coordinate system where such quantities are zero at that point."

Last edited: Dec 10, 2011
10. Dec 10, 2011

### Naty1

Hi lowemack..

Instead of saying 'mass' increases'' let's instead say 'energy'...that is, their kinetic energy appears to increase as observed from earth....but that is illusory since neither spaceship observes such an increase. They see earth's KE as increasing not their own.

So who is right...well, they all are because they have different reference frames. So that alone won't tell us what's going on. Such comparisons can be confusing.

Peter's comment is the key:

That results from the mathematics of relativity...it is NOT obvious.

Another way to remember this:
Any situation where you ask about a rapidly moving massive body's gravitational effect and a 'stationary' observer can be transformed to an equivalent question about the interaction between a rapidly moving observer a 'stationary' massive body. So all observations relating to a rapidly moving massive body can be answered as if the body is stationary...as if all measures are local. Local measures trump distant measures.

I think this would also be accurate: (any experts??)

If the spaceships have accelerated at the same rate together for the same time as they travel, they remain stationary relative to each other, their gravitational attraction remains as if they were stationary....But if one temporarily rockets ahead faster, for example, and remains ahead, now there has been relative motion between them, kinetic energy is observed for this period, and so they would have somewhat different gravitational attractions...and their paths would now vary for future travel.....gravitational curvature IS affected. And their recorded elapsed times would also differ were they to later come together and compare elapsed times.

Last edited: Dec 10, 2011
11. Dec 10, 2011

### chingel

But how does an earth based observer explain it? To him, the spaceships have increased mass, so shouldn't they have increased gravitational attraction, and if not, why not?

12. Dec 10, 2011

### atyy

Would it help to think of this in terms of some exact solution in GR? How about the LCDM model, where we say other "galaxies (local groups)" are "moving" away from us with increasing "acceleration"? Is there any way to make sense of all those terms as well as assign them "energy" and "gravitational attraction"?

13. Dec 10, 2011

### danR

You have to read the opening answer(s): the spaceships do not have increased mass, that notion is obsolete. They have increased total energy. Perhaps you are thinking E=mc2, and c is constant, and E is increasing (yes), therefore m must increasing.

The 'equals' sign does not indicate a 'same thing as' relationship, but an equivalence.

If the real masses did increase, it would entail an irresolvable paradox:

I have installed a nanogram-sensitive strain-balance between the two spaceships. I observe the readout through a powerful telescope and I see as they go faster, the readout increases. But the pilots radio back and say the readout does not change.

This is not the same sort of length/time 'paradox' that can be resolved on carefully reasoned relativistic grounds, it is just an impossibility. If the pilots don't measure a mass increase, there is none. It is constant.

Last edited: Dec 10, 2011
14. Dec 10, 2011

### Per Oni

Suppose we charge the rockets with 2 opposite charges, so that in addition we also have an attractive Coulomb force.

In the past I have read arguments concluding that the force between the rockets will become smaller due to the 2 magnetic fields as observed from earth. The readout should decrease according to these arguments.

15. Dec 10, 2011

### danR

Well, I'd have to look at the arguments a bit more fleshed-out than that. But do you mean the the pilots and the stationary observer see a different readout? I don't think so.

Is there greater attraction between the two? Consider two simplified spaceships: an electron and a positron traveling parallel down a linear accelerator at relativistic velocities. Do their charges change? Charge, like spin, is a fixed property. As far as I know, it doesn't change with velocity, which latter between the two is zero anyway: they have the same reference frame.

16. Dec 10, 2011

### Staff: Mentor

If you mean "not frame-invariant" in the sense that it's not a scalar, true. But it is certainly "frame-invariant" in the sense that it's a tensor and transforms accordingly when you change coordinates, so contracting it with other tensors yields frame-invariant scalars. Any actual observable that tells you about the "amount of gravity produced" by an object will be such a scalar, so it will be frame-invariant, as I said.

No. The "amount of gravity produced" by the object is not a function of its energy, it's a function of its stress-energy tensor, of which energy is only one component. In a frame in which the object is moving, there will be other non-zero components of the SET as well as the energy, and their effects will offset the apparent "effect" of the increase in energy, so the final result will be the same as it is for a frame in which the object is at rest.

This is true; so far I've only talked about a single object. Since GR is nonlinear, two solutions do not add up to another solution, so I can't just take the solution for each body in isolation and add them to get a solution for both bodies. That's why we don't have a closed form solution for, e.g., binary pulsars, but have to calculate their expected orbital changes due to the radiation of gravitational waves numerically.

However, we don't have to get into that to resolve the question of whether an object being in motion changes the gravity it produces. It doesn't.

This is key: this is almost always the quickest method of figuring out how much gravity an object produces. Transform to its rest frame, in which the stress-energy tensor will usually have its simplest form. Actually, in the case given in the OP, it's even simpler, since each object is isolated so the individual solution for the gravity produced by the object, in the vacuum region outside the object, is just the Schwarzschild solution with the object's mass M. Transforming that solution to a frame in which M is moving does not change any of its physical predictions, it just makes it look more complicated while still giving the same answers.

17. Dec 10, 2011

### atyy

Not for non-tidal gravity.

18. Dec 11, 2011

### Staff: Mentor

I may not have made myself clear. By "final result" I mean the calculated prediction for a scalar, i.e., an observable number. When you talk about being able to set up a local inertial frame in which "non-tidal gravity" vanishes, what makes that possible is the fact that a particular observable number, the acceleration felt by an observer moving on a geodesic worldline, is zero. Setting up a local inertial frame is simply setting up a frame in which the formulas that give you this observable number look as simple as possible.

The observable number is invariant. How it is interpreted may vary, but the number itself does not change. Nor does it change depending on whether we do the calculation in a frame where the source of gravity is at rest, or a frame in which the source is moving. Similar remarks apply to any other observable number.

19. Dec 11, 2011

### atyy

No, you made yourself clear. I don't disagree with what you wrote, but the point of the post you replied to was that the ability to ask the question does not go away if one uses instead the energy or even the stress-energy tensor instead of the relativistic mass.

So certainly we can say that gravity (in an appropriate sense) can be made to disappear with an appropriate choice of coordinates. We can also say that gravity (in another appropriate sense) is the same regardless of choice of coordinates. Even with requiring scalars to define gravity, one can still say that it is observer dependent since the metric needs at least two vectors to make a scalar, and one of the vectors could be the observer's tangent vector.

Anyway, no physics disagreement. More seriously, the parts of the question that don't make sense to me - and I'd like to know whether it is possible to make sense of them - are:

Does GR have a notion of the "gravity" between distant objects?

Does GR have any solutions where massive objects can move parallel to each other?

20. Dec 11, 2011

### Staff: Mentor

And the answer is still the same: if one is inferring from either the relativistic mass or the stress-energy tensor that an object's behavior as a source of gravity depends on its state of motion relative to you, one is inferring incorrectly.

Yes, no dispute here, although the first sense of "gravity" is not encouraged in GR for precisely this reason: that it is not a direct observable, because it can be made to vanish by a change of coordinates. (More precisely, it can be made to vanish *at a particular event* by a change of coordinates: you can't make it vanish over an extended patch of spacetime.)

But then you're talking about *different* scalars. If you compute the "effect of gravity" on observers with different tangent vectors, obviously you will get different answers (you're contracting the same "source" with different vectors). But that's not because the source of gravity changed; it's because the observer changed.

To put this another way, when we talk about an object as a "source" of gravity, we're talking about what kind of spacetime curvature it produces, and spacetime curvature is independent of the state of motion of observers that probe it. When we talk about the kind of "gravity" that can be made to vanish at an event by a change of coordinates, we're talking about a property of the observer's worldline: that its 4-acceleration (which is a direct observable, the reading of an accelerometer) is zero, i.e., it's moving on a geodesic. When we talk about the kind of "gravity" that is produced by an object's stress-energy tensor, we're talking about the spacetime as a whole, the geometric structure that determines which particular worldlines are geodesics. The two are related, but they are not the same.

I would say yes, but the question probably needs to be more specifically defined. I can think of at least two ways in which it does:

(1) There are numerical solutions for many-body systems (e.g., binary pulsars) which show the bodies orbiting each other, similar to the known Newtonian analytic solutions for the two-body problem. The difference is that GR includes the emission of gravitational waves, so the two bodies' orbits about each other are not constant; they slowly spiral inwards towards each other.

(2) For an extended system with nonzero stress-energy such as a perfect fluid, the stress-energy gravitates; e.g., an expanding FRW solution with zero cosmological constant has the expansion constantly slowing down, while a contracting FRW solution with zero cosmological constant has the contraction constantly speeding up. The expanding case corresponds to all the massive objects in the universe pulling on each other and slowing the expansion down.

Wouldn't this be equivalent to a solution where two massive objects are at rest relative to each other? Or do you mean moving on parallel worldlines but in opposite directions?

I believe the Chazy-Curzon vacuum is a solution (an unphysical one) for a pair of masses held at fixed distance apart by a massless "strut" (it's unphysical because the strut has no mass but it still can exert force on the two masses). I don't know if there are any physically realistic solutions of this sort; it's hard to see how there could be since two masses without any other mass present should fall towards each other.