Questions about gravity at relativistic velocities

[nearc]

The ‘speed’ of gravity, like everything else, is also limited to the speed of light. So if two bodies [with mass] are traveling in the same direction with one trailing directly behind the other by some distance and both bodies are traveling at .999 C how does the gravity work between them? For example: If body A is in front and body B is behind. Then A will not feel the gravitational effects for some time. In fact if A is 1 light year ahead of B then it will take 999 years before A will feel the gravitational effects from B, however B will ‘run into’ A gravitational effects in about half a year?

 

[starthaus]

The ‘speed’ of gravity, like everything else, is also limited to the speed of light. So if two bodies [with mass] are traveling in the same direction with one trailing directly behind the other by some distance and both bodies are traveling at .999 C how does the gravity work between them? For example: If body A is in front and body B is behind. Then A will not feel the gravitational effects for some time. In fact if A is 1 light year ahead of B then it will take 999 years before A will feel the gravitational effects from B, however B will ‘run into’ A gravitational effects in about half a year?

Let v=0.999 c, L=1ly is the distance between bodies, then:

ct=L+vt i.e. B moves away from A by vt while the gravitational effects need to cover L+vt

So, t=L/(c-v) which is 1/0.001=1000 years

In the other case:

ct'+vt'=L

so

t'=L/(c+v) and this is about 0.5 years.

Now, exactly as in the MMX experiment we need to remember that L needs to be "length contracted", when viewed from the frame that measures A and B moving at v i.e. it needs to be multiplied by 1/\gamma=\sqrt {1-(v/c)^2}=1/22.4

So, the above times become:

1000/22.4=44.8 yr

Another way of solving this is to use the full-fledged Lorentz transforms. In the comoving frame, the gravitational effect cover the distance \Delta x=1ly in the time \Delta t=1yr
Then, in a frame wrt the two bodies move at v away from the observer the time separation is:

\Delta t'= \gamma (\Delta t+v\Delta x/c^2)=\gamma*2=44.8 yrs

If A and B move towards the observer, the formula changes:\Delta t'= \gamma (\Delta t-v\Delta x/c^2)=\gamma*/1000=0.0224 yrs

So, we can get the correct result through two different methods. As you can see, the correct results are quite different from your initial intuition.

 

[Mentz114]

So if two bodies [with mass] are traveling in the same direction with one trailing directly behind the other by some distance and both bodies are traveling at .999 C how does the gravity work between them?

So the two bodies have zero relative velocity ? They won't notice anything different in the gravitational interaction between them.

You say they are traveling at .999 c or whatever but not from which frame ( which does not enter the picture in any case ) since the bodies are stationary in each others frames.

 

[bcrowell]

So the two bodies have zero relative velocity ? They won't notice anything different in the gravitational interaction between them.

You say they are traveling at .999 c or whatever but not from which frame ( which does not enter the picture in any case ) since the bodies are stationary in each others frames.

I think the OP meant that in some frame K, bodies A and B are observed to be moving with the same velocity in the same direction, and the description of the time delays is meant to be in frame K.

For example: If body A is in front and body B is behind. Then A will not feel the gravitational effects for some time. In fact if A is 1 light year ahead of B then it will take 999 years before A will feel the gravitational effects from B, however B will ‘run into’ A gravitational effects in about half a year?

Yes, this is a correct description in frame K.

But Lut is also correct -- in the frame K' in which A and B are at rest, nothing unusual happens.

 

[Matterwave]

Can't we just see this from the point of time dilation? In the frame in which A and B don't move, the propagation is obviously 1 year. Therefore, in the frame in which A and B move, wouldn't the propagation time be t=1yr/sqrt(1-v^2/c^2)=22.36yrs? And this would then be the propagation from A to B and from B to A.

What is wrong with my analysis?

 

[bcrowell]

Can't we just see this from the point of time dilation? In the frame in which A and B don't move, the propagation is obviously 1 year. Therefore, in the frame in which A and B move, wouldn't the propagation time be t=1yr/sqrt(1-v^2/c^2)=22.36yrs? And this would then be the propagation from A to B and from B to A.

What is wrong with my analysis?

If you start from the Lorentz transformation, you can then show that \gamma gives the time dilation between frames K and K' for events that occurred *at the same place* in one of those frames. In the example we're talking about here, the emission and receipt of the gravitational "signal" doesn't occur at the same place in either frame.

 

[starthaus]

Can't we just see this from the point of time dilation? In the frame in which A and B don't move, the propagation is obviously 1 year. Therefore, in the frame in which A and B move, wouldn't the propagation time be t=1yr/sqrt(1-v^2/c^2)=22.36yrs? And this would then be the propagation from A to B and from B to A.

What is wrong with my analysis?

You need to factor in the spatial separation as well, you can't use straight time dilation because the start point of the signal (A) is separated from the endpoint (B) by 1 lyr in the comoving frame:

\Delta t'= \gamma (\Delta t+v\Delta x/c^2)=\gamma*2=44.8 yrs

See post #2 for the complete reasoning.

 

[Mentz114]

I don't know about any of this . Everyone seems to be answering the question as if the two bodies are receding/approaching each other at 0.999c. But they have parallel worldlines in all frames. I'm baffled .

What am I missing ?

 

[starthaus]

I don't know about any of this . Everyone seems to be answering the question as if the two bodies are receding/approaching each other at 0.999c. But they have parallel worldlines in all frames. I'm baffled .

What am I missing ?

That the gravitational effect propagates in a time interval that is frame dependent. In the comoving frame this means 1ly, in the frame that measures A and B moving away at v, this means a larger number.

 

[bcrowell]

I don't know about any of this . Everyone seems to be answering the question as if the two bodies are receding/approaching each other at 0.999c. But they have parallel worldlines in all frames. I'm baffled .

What am I missing ?

The long time delay in frame K is simply a time dilation.

One thing that may be causing confusion is that since the two bodies are at rest with respect to one another, there is actually no information being propagated by gravitational waves. If we want the OP's scenario to make a little more sense, we should actually let A and B each be, say, binary neutron stars, so that each one emits gravitational waves that can be detected by the other. Unless we have some mechanism like this for radiation, the field is exactly constant everywhere in frame K'.

 

[Matterwave]

Wait so now I'm confused. To the 2 bodies which are moving wrt me, they obviously see only a 1ly lag. To me...I see a 44.8 year lag? Or a 1000 year lag?

 

[starthaus]

Wait so now I'm confused. To the 2 bodies which are moving wrt me, they obviously see only a 1ly lag. To me...I see a 44.8 year lag? Or a 1000 year lag?

44.8, the calculations are all there.

 

[Dale]

I agree with bcrowell, in this situation there are no gravitational waves that propagate. Without something propagating it is confusing to talk about lag.

 

[starthaus]

I agree with bcrowell, in this situation there are no gravitational waves that propagate. Without something propagating it is confusing to talk about lag.

Replace the gravitational field with an em field. The problem is about calculating propagation time as a frame-dependent quantity.