I Spacelike hyperbolae -- accelerated reference frames

[smodak]

Uniformly accelerated reference frames in special relativity are represented in the space-like region of a lightcone as hyperbolae. These hyperbolae represent worldline of accelerated observers. However, there could not be be causal relationship between two events on the spacelike section unless information travels faster than light. Shouldn't that mean accelerated ref frames represent faster than light travel? What am I missing?

 

[PeterDonis]

there could not be be causal relationship between two events on the spacelike section unless information travels faster than light.

This is not correct. The hyperbolas representing the worldlines of accelerated observers are timelike, not spacelike. Your terminology, describing the "right wedge" of Minkowski spacetime as "the spacelike region", is misleading you; not all pairs of events in this region are spacelike separated. They are all spacelike separated from the origin (the apex of the light cone), but that's not the same thing.

 

[smodak]

This is not correct. The hyperbolas representing the worldlines of accelerated observers are timelike, not spacelike. Your terminology, describing the "right wedge" of Minkowski spacetime as "the spacelike region", is misleading you; not all pairs of events in this region are spacelike separated. They are all spacelike separated from the origin (the apex of the light cone), but that's not the same thing.

I think I get it. Thank you!

 

[pervect]

Uniformly accelerated reference frames in special relativity are represented in the space-like region of a lightcone as hyperbolae. These hyperbolae represent worldline of accelerated observers. However, there could not be be causal relationship between two events on the spacelike section unless information travels faster than light. Shouldn't that mean accelerated ref frames represent faster than light travel? What am I missing?

Consider the reference frame of an unaccelerated observer in flat space-time. You can draw the worldlines of observers - they're straight lines rather than hyperbola - but you can make the same remark about space-like separated events not having a causal relationship.

But I don't understand how you are interpreting these facts (which apply to an unaccelerated observer as well as an accelerated one) as faster than light travel. Everything you've said applies equally well to an inertial frame.

 

[smodak]

Consider the reference frame of an unaccelerated observer in flat space-time. You can draw the worldlines of observers - they're straight lines rather than hyperbola - but you can make the same remark about space-like separated events not having a causal relationship.

But I don't understand how you are interpreting these facts (which apply to an unaccelerated observer as well as an accelerated one) as faster than light travel. Everything you've said applies equally well to an inertial frame.

You are right. PeterDonis corrected my confusion. For some strange reason, I was making a connection that does not exist. Thanks for the help!

 

[Orodruin]

Note that the hyperbolae of the form $t^2-x^2=k^2$ (not $-k^2$) are space-like, but they do not describe world lines of accelerated observers and have time like separation from the origin.

 

[smodak]

Note that the hyperbolae of the form $t^2-x^2=k^2$ (not $-k^2$) are space-like, but they do not describe world lines of accelerated observers and have time like separation from the origin.

Not sure I understand that. I thought the Rindler Hyperbolea of the form $t^2-x^2=k^2$ represent uniformly accelerated reference frames.

 

[Orodruin]

Not sure I understand that. I thought the Rindler Hyperbolea of the form $t^2-x^2=k^2$ represent uniformly accelerated reference frames.

No, those would be the hyperbolae $x^2-t^2=k^2$.

 

[PeterDonis]

I thought the Rindler Hyperbolea of the form $t^2-x^2=k^2$ represent uniformly accelerated reference frames.

They do if $k^2$ is negative, in the form you've written it; the more usual form is $x^2 - t^2 = k^2$, where $k^2$ is positive (and is equal to $1 / a^2$, where $a$ is the proper acceleration of the worldline), because it is usually assumed that we are working with real numbers only.

Hyperbolas of the form $t^2 - x^2 = k^2$ where $k^2$ is positive are in the future or past light cones of the origin; they represent sets of events that are at the same proper time from the origin.

 

[m4r35n357]

What am I missing?

The hyperbolae in those diagrams are each shifted so that they share a common focus at (I think) $$x = -1 / a $$ the even horizon. If you un-shift each one to pass throught the origin, it is easy to see that they are always timelike.

 

[PeterDonis]

The hyperbolae in those diagrams are each shifted so that they share a common focus at (I think)
$$
x = -1 / a
$$

the even horizon.

The focus of the hyperbolas is at the spacetime origin, $x = t = 0$. Also, the horizon in this case (which is the lines $x = \pm t$, the asymptotes of the hyperbolas) is not an event horizon; it's the Rindler horizon.

 

[Orodruin]

The focus of the hyperbolas is at the spacetime origin, $x = t = 0$.

No they are not. You are thinking of the centre of the hyperbolas. The foci of the hyperbolae tend to $x\to \pm \infty$ as the proper acceleration goes to zero.

 

[PeterDonis]

You are thinking of the centre of the hyperbolas.

Oops, yes, you're right.

The foci of the hyperbolae tend to $x\to \pm \infty$ as the proper acceleration goes to zero.

This means the hyperbolas don't all have the same focus, even though they all have the same center, correct?

 

[m4r35n357]

No they are not. You are thinking of the centre of the hyperbolas. The foci of the hyperbolae tend to $x\to \pm \infty$ as the proper acceleration goes to zero.

Sorry folks, I messed that up! How about they are all shifted by$$+1/a$$ so that they share a common centre at the origin? If that is wrong then I give up . . . ;)

 

[Orodruin]

Oops, yes, you're right.

Despite my history of early morning posts when going to work - I do try some times

Edit: Also, I cheated. I discussed hyperbolic trajectories in a Kepler potential recently so I rehearsed the terminology ...

This means the hyperbolas don't all have the same focus, even though they all have the same center, correct?

Correct. The foci are further away from the centre than the point where a hyperbola crosses the x-axis.

Sorry folks, I messed that up! How about they are all shifted by$$+1/a$$ so that they share a common centre at the origin? If that is wrong then I give up . . . ;)

Well, it depends on what you mean by "shifted". They share the same centre so in that respect they are not shifted at all ... However, you need to shift them from there in order for them to go through the origin. But yes, in that respect they are shifted relative to where they would be if they all went through the origin.

 

[m4r35n357]

But yes, in that respect they are shifted relative to where they would be if they all went through the origin.

Yes, I only meant in the sense that a constant velocity line is typically drawn through the origin when illustrating its slope wrt light. I thought that was really what the OP was asking when he mentioned that accelerated world lines occupied the "space-like" zone.

 

[smodak]

Thanks everyone for replying. Now a different question. If you are staying on a specific hyperbola and asymptotically moving up towards x= t (or x = ct), you are not moving through space just through time, right? Where is the acceleration coming from? Is it only a temporal acceleration with no spatial acceleration? If, so, how then these coordinates represent accelerating reference frames in general?

 

[PeterDonis]

If you are staying on a specific hyperbola and asymptotically moving up towards x= t (or x = ct), you are not moving through space just through time, right?

There is no such thing in an absolute sense; whether or not you are "moving through space" or not depends on the coordinates you choose. In inertial coordinates, you are "moving through space" if your worldline is one of these hyperbolas; in Rindler coordinates, you're not.

 

[pervect]

Thanks everyone for replying. Now a different question. If you are staying on a specific hyperbola and asymptotically moving up towards x= t (or x = ct), you are not moving through space just through time, right?

Sorry, I'm not really following the question :(.

Any observer, accelerating or not, has some worldline that represents their history through space and time.

If we create some inertial frame of reference S, then in that inertial frame S a non-accelerating observer moves along a straight line. This can be regarded as a tautology rather than any law of physics - we can regard it as defining what an "inertial frame of reference" means in operational terms.

Where is the acceleration coming from? Is it only a temporal acceleration with no spatial acceleration? If, so, how then these coordinates represent accelerating reference frames in general?

I can't quite figure out what you're doing or asking here, my closest guess is that you're confused about what the difference between an accelerating frame of reference and an inertial frame of reference is. If we ignore gravity for the time being, imagining that we are out far enough from any massive body so gravity is not important, then we can regard the motion of a force-free body as sort of test. If the force-free, isolated body moves in a straight line, we have created an inertial frame of reference. If the force-free, isolated body does not move in a straight line, then we have not created an inertial frame of reference, we've created a non-inertial frame of reference.

Note that we (or at least I) regard a frame of reference as something we create, a human invention, not something that is "handed to us" or has any reality on its own. I suspect you may be coming from some other philosophical viewpoint, but I don't quite understand what that viewpoint may be, or how to reconcile your viewpoint (whatever it is) with mine so that we can meaningfully talk about physical facts.

[add]
I talked about what happens if we ignore gravity. If gravity is part of the question, things just get more complicated. Strictly speaking, inertial frames of reference simply don't exist if you have gravity. Non-strictly speaking, thigns aren't quite so bad, but I won't digress into that, I think we have enough to sort out already.

 

[smodak]

I think both of you have answered my question. Uniformly accelerated particles undergo hyperbolic motion in minkowski space in inertial coordinates. When you use Rindler coordinates, they are actually at rest. Thank you!

 

[vanhees71]

Some math about hyperbolic motion can also be found in my PF FAQ article here:

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[smodak]

Great Stuff. Thanks.