Does mass really increase with speed


by lowemack
Tags: increase, mass, speed
lowemack
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#1
Dec9-11, 10:22 AM
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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.
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D H
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#2
Dec9-11, 11:45 AM
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Relativistic mass is an outdated concept. This is yet another example of where the concept of relativistic mass confuses rather than clarifies.
nitsuj
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#3
Dec9-11, 11:52 AM
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Quote Quote by lowemack View Post
But from the spaceship point of view their gravitational attraction should be minimal.
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.

QuArK21343
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#4
Dec9-11, 02:26 PM
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Does mass really increase with speed


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.
PeterDonis
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#5
Dec9-11, 02:36 PM
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Quote Quote by QuArK21343 View Post
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.
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.
DrStupid
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#6
Dec9-11, 03:05 PM
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Quote Quote by PeterDonis View Post
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.
In GR the stress–energy tensor is the source of gravitation and it is not frame-invariant.
dtyarbrough
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#7
Dec10-11, 08:10 AM
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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.
torquil
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#8
Dec10-11, 08:31 AM
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Quote Quote by dtyarbrough View Post
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.
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?
atyy
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#9
Dec10-11, 10:23 AM
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Quote Quote by D H View Post
Relativistic mass is an outdated concept. This is yet another example of where the concept of relativistic mass confuses rather than clarifies.
Quote Quote by DrStupid View Post
In GR the stress–energy tensor is the source of gravitation and it is not frame-invariant.
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."
Naty1
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#10
Dec10-11, 11:10 AM
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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:

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.
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.
chingel
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#11
Dec10-11, 11:24 AM
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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?
atyy
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#12
Dec10-11, 12:09 PM
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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"?
danR
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#13
Dec10-11, 01:42 PM
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Quote Quote by chingel View Post
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?
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.
Per Oni
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#14
Dec10-11, 03:21 PM
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Quote Quote by danR View Post
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.
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?
danR
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#15
Dec10-11, 05:36 PM
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Quote Quote by Per Oni View Post
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?
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.
PeterDonis
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#16
Dec10-11, 06:59 PM
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Quote Quote by DrStupid View Post
In GR the stress–energy tensor is the source of gravitation and it is not frame-invariant.
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.

Quote Quote by atyy View Post
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?
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.

Quote Quote by atyy View Post
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.
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.

Quote Quote by Naty1 View Post
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
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.
atyy
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#17
Dec10-11, 07:13 PM
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Quote Quote by PeterDonis View Post
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.
Not for non-tidal gravity.
PeterDonis
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#18
Dec11-11, 12:11 AM
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Quote Quote by atyy View Post
Not for non-tidal gravity.
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.


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