- #1

- 13

- 0

- Thread starter lowemack
- Start date

- #1

- 13

- 0

- #2

- 15,393

- 685

- #3

- 1,352

- 90

That's true, from the spaceship's perspective gravity (and the mass of the ships) doesn't change.But from the spaceship point of view their gravitational attraction should be minimal.

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

- 47

- 0

Relativistic mass is just the historical name for the quantity:

[tex] \frac{m}{\sqrt{1-v^2/c^2}} [/tex]

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

[tex]m^2 c^2=\frac{E^2}{c^2}-p^2[/tex]

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.

[tex] \frac{m}{\sqrt{1-v^2/c^2}} [/tex]

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

[tex]m^2 c^2=\frac{E^2}{c^2}-p^2[/tex]

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:

- #5

- 31,641

- 10,385

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

- #6

- 1,989

- 386

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

- #7

- 6

- 0

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

- 649

- 2

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?

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.

- #9

atyy

Science Advisor

- 14,065

- 2,366

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

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:

- #10

- 5,601

- 40

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:

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.

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.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.

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:

- #11

- 300

- 22

- #12

atyy

Science Advisor

- 14,065

- 2,366

- #13

- 351

- 4

You have to read the opening answer(s): the spaceships do

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:

- #14

- 261

- 1

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

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.

I’m stuck. An irresolvable paradox?

- #15

- 351

- 4

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.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.

I’m stuck. An irresolvable paradox?

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

- 31,641

- 10,385

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

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.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?

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.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.

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.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

- #17

atyy

Science Advisor

- 14,065

- 2,366

Not for non-tidal gravity.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.

- #18

- 31,641

- 10,385

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.Not for non-tidal gravity.

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

atyy

Science Advisor

- 14,065

- 2,366

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.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.

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

- 31,641

- 10,385

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.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.

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.)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.

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.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.

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

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

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.

- #21

zonde

Gold Member

- 2,941

- 213

This is wishful thinking. Try to produce any reference that gives that transformation.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.

- #22

- 9,897

- 1,078

You'll find quite a lot on calculating the trajectories of a rapidly moving observer near a large mass. Which is the easy way to think about the problem, and most any basic textbook will cover it.

So Naty1's advice is quite sound, and he's pointing out that if you "shift gears" to calculating the problem from the POV of the massive body , the problem is quite a bit easier and within the grasp of a serious beginning student. A total layman might not be able to even calculate this much - it seems to be the case that they get weird ideas as a result of being unable to calculate things (which provides in the end a necessary cross-check on the concepts involve). Unforutnatley, unless/until such time as the layperson actually become at least a beginning students who can start to calculate things it's rather difficult to have a truly serious discussion :-(.

The basic tensor transformation laws do give you the tools you'd need to convert to ANY sort of coordinate system you want, in theory. Whether or not you get closed form solutions is another story.

As far as picking the "right" coordinate system goes, ones that have some sort of physical significance, it might not be obvious at all how to do th is at first. But if you' look, you'll also find the discussion of how to construct the Fermi-normal coordinates, and/or various extensions therof, and some discusssion here and there about the physical interpretation of said coordinates, why they are a "good choice", and some remarks about their physical interpretation.

See for instance MTW's "Gravitation", and online you'll find http://arxiv.org/PS_cache/arxiv/pdf/0901/0901.4465v2.pdf, which has some interpretational remarks as well.

This is a rather advanced topic, though, and unfortunately tends not to be particularly helpful to the beginning student. I tend to agree with Misner that the easisst solution is just to let go of the whole idea of "observers" for the beginning student, they turn out to require more advanced math to handle than is required to actually compute useful solutions. After one gets a handle on how to calculate really basic things without dragging in the idea of "observers", the concept can be re-introduced at a more advanced level.

I will remark, though, that to the second order, what an "observer" sees is describable in terms of tidal forces, something that most textooks introduce at a starting level. So, if one is not particularly argumentative, one can use some of this information to good effect for what an "observer" sees. If one is argumentative (this seems to happen a lot), well, one has to wade through the necessary math to achieve this understanding.

So Naty1's advice is quite sound, and he's pointing out that if you "shift gears" to calculating the problem from the POV of the massive body , the problem is quite a bit easier and within the grasp of a serious beginning student. A total layman might not be able to even calculate this much - it seems to be the case that they get weird ideas as a result of being unable to calculate things (which provides in the end a necessary cross-check on the concepts involve). Unforutnatley, unless/until such time as the layperson actually become at least a beginning students who can start to calculate things it's rather difficult to have a truly serious discussion :-(.

The basic tensor transformation laws do give you the tools you'd need to convert to ANY sort of coordinate system you want, in theory. Whether or not you get closed form solutions is another story.

As far as picking the "right" coordinate system goes, ones that have some sort of physical significance, it might not be obvious at all how to do th is at first. But if you' look, you'll also find the discussion of how to construct the Fermi-normal coordinates, and/or various extensions therof, and some discusssion here and there about the physical interpretation of said coordinates, why they are a "good choice", and some remarks about their physical interpretation.

See for instance MTW's "Gravitation", and online you'll find http://arxiv.org/PS_cache/arxiv/pdf/0901/0901.4465v2.pdf, which has some interpretational remarks as well.

This is a rather advanced topic, though, and unfortunately tends not to be particularly helpful to the beginning student. I tend to agree with Misner that the easisst solution is just to let go of the whole idea of "observers" for the beginning student, they turn out to require more advanced math to handle than is required to actually compute useful solutions. After one gets a handle on how to calculate really basic things without dragging in the idea of "observers", the concept can be re-introduced at a more advanced level.

I will remark, though, that to the second order, what an "observer" sees is describable in terms of tidal forces, something that most textooks introduce at a starting level. So, if one is not particularly argumentative, one can use some of this information to good effect for what an "observer" sees. If one is argumentative (this seems to happen a lot), well, one has to wade through the necessary math to achieve this understanding.

Last edited:

- #23

- 1,989

- 386

Please specify "amount of gravity produced".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.

- #24

- 261

- 1

Charge is an invariant although the distribution of the field is modified when a charge is travelling.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.

This thread reminded me of a chapter in: “Special relativity A.P.French”

On page 244 there is an example of 2 charges travelling side by side in a rest frame, not unlike the OP.

The conclusion is that we need to take into account a force transformation with the result that indeed different forces are experienced, depending which frame we are in. For the case of charges this is explained away by introducing magnetic fields.

We should however be able to apply these same force transformations for the case stated in the OP, with the result that we have 2 different readouts on our balance.

But in the end I don’t know and I’m still stuck.

- #25

- 31,641

- 10,385

Any direct observable that tells you about gravity. For example, suppose I am "hovering" at a constant height above a gravitating body, and I want to know how much I have to accelerate--how hard I have to fire my rocket engines--to do so. The answer is simple: it's the 4-acceleration of my worldline, which is a scalar invariant. The formula I use to calculate that number will look different in different coordinate systems: for example, it will look different in coordinates in which the source of gravity is moving, than it will in coordinates in which the source of gravity is at rest. But the final answer will be the same either way.Please specify "amount of gravity produced".

The above case may be too simple, though, because I don't need to know the stress-energy tensor inside the source to calculate it; I only need to know the source's total mass. But even in this case, it's the source's *rest* mass, *not* its relativistic mass, that comes into play.

Let's take a more complicated case: the expansion of the universe. Here there is a non-zero stress-energy tensor throughout the spacetime. Suppose I want to know how the expansion of the universe is changing with time; this depends on how much gravity the stress-energy in the universe is producing. We'll take the simpler case with zero cosmological constant (which does not seem to be true in the actual universe, but it's simpler to understand). In this case, the expansion of the universe is decelerating, and the rate at which it decelerates depends on the amount of stress-energy present. We can directly observe the deceleration by observing the Hubble constant and seeing that it decreases with time; equivalently, we can watch the observed redshift of a particular galaxy, which indicates how fast it is moving away from us (provided we take a galaxy that is far enough away that it is not part of our local galaxy cluster, so its motion relative to us is solely due to the expansion of the universe), and see that the redshift decreases with time.

So we have a direct observable that tells us "how much gravity" the universe is producing. And we will arrive at the same prediction for this observable regardless of what system of coordinates we do the calculation in; normally we would do it in the standard FRW coordinates, but the Earth is not at rest in those coordinates, so we could, if we wanted to, transform to the Earth's rest frame and re-do the calculation, and we would get the same answer.