Relativistic simultaneity and symmetry problem

[Chris Miller]

With both their clocks synced at 00:00 (12 AM), spaceships A and B leave Earth in oposite directions, each at 17308257.5 m/sec (.57735027c), So the Lorentz factor of their relative velocity 259627884.49 m/sec (.866025404c) is 2.

They travel in opposite directions at this velocity for 2 hours by their own clocks. At which point each (somehow) stops and reverses course at this same velocity. Looking through a powerful telescope and allowing for c, A should see that it's 1 AM on B's clock, and B that it's 1 AM on A's.

So for A's next hour, from her frame of ref, she's traveling in the same direction as B. And for B's next hour, she's traveling in the same direction as A. I.e., for the next hour each sees the other traveling away from Earth while they are traveling towards it. So their relative velocity is now zero and each should see the other's clock to go from 1 AM to 2 AM while theirs goes from 2 AM to 3 AM.

Then, in A's next hour, when they are traveling towards each other at a relative velocity of .866025404c (for Lorentz factor of 2), her clock will run to 4 AM, at which time she will be back at earth. And vice versa. However, wouldn't each, during their final hour see (through that super powerful telescope) the other's clock run from 2 AM to 2:30 AM?

I'm having trouble getting them both back at the same time, despite their completely symmetrical trips.

 

[Orodruin]

Looking through a powerful telescope and allowing for c, A should see that it's 1 AM on B's clock, and B that it's 1 AM on A's.

No. You are not taking the travel time of light into account.

When an observer changes velocity, the notion of simultaneity in its rest frame changes. While 2am for A was simultaneous with 1am for B in A's original rest frame - it will no longer be the case in the new rest frame of A.

 

[Chris Miller]

No. You are not taking the travel time of light into account.

With "and allowing for c" I thought did. But I know the whole observing of the other's clock thing is problematic. You know what I mean. Maybe just, what each would expect the other's clock to read or something.

When an observer changes velocity, the notion of simultaneity in its rest frame changes. While 2am for A was simultaneous with 1am for B in A's original rest frame - it will no longer be the case in the new rest frame of A.

So after each stops at 2 hours, what time is it in the other's frame?

 

[Vitro]

With both their clocks synced at 00:00 (12 AM), spaceships A and B leave Earth in oposite directions, each at 17308257.5 m/sec (.57735027c), So the Lorentz factor of their relative velocity 259627884.49 m/sec (.866025404c) is 2.

They travel in opposite directions at this velocity for 2 hours by their own clocks. At which point each (somehow) stops and reverses course at this same velocity.

This is where your analysis starts going wrong. Forget everything about "seeing", it just complicates things.

Immediately before A's turnaround, based on A's simultaneity convention, B's clock shows 1am and B is still traveling away form Earth. Immediately after A's turnaround, based on A's new simultaneity convention, B's clock reads 3am and B is well on his way toward Earth already. By the time A and B get to Earth, at the same time, both their clocks will read 4am.

 

[Janus]

With both their clocks synced at 00:00 (12 AM), spaceships A and B leave Earth in oposite directions, each at 17308257.5 m/sec (.57735027c), So the Lorentz factor of their relative velocity 259627884.49 m/sec (.866025404c) is 2.

They travel in opposite directions at this velocity for 2 hours by their own clocks. At which point each (somehow) stops and reverses course at this same velocity. Looking through a powerful telescope and allowing for c, A should see that it's 1 AM on B's clock, and B that it's 1 AM on A's.

So for A's next hour, from her frame of ref, she's traveling in the same direction as B. And for B's next hour, she's traveling in the same direction as A. I.e., for the next hour each sees the other traveling away from Earth while they are traveling towards it. So their relative velocity is now zero and each should see the other's clock to go from 1 AM to 2 AM while theirs goes from 2 AM to 3 AM.

Then, in A's next hour, when they are traveling towards each other at a relative velocity of .866025404c (for Lorentz factor of 2), her clock will run to 4 AM, at which time she will be back at earth. And vice versa. However, wouldn't each, during their final hour see (through that super powerful telescope) the other's clock run from 2 AM to 2:30 AM?

I'm having trouble getting them both back at the same time, despite their completely symmetrical trips.

Here are some space-time diagrams to help.
First we show the frame in which both ships are traveling at .577c

The yellow lines are information traveling at c. Note that for each ship, the light that arrives from the other ship at turn around left that ship when its clock read 0.5 hrs. Thus at turn around, ship A will see ship B's clock reading 0.5 hrs and vice versa. Then 1 1/2 hours later when their clocks read 3.5 hrs, they will see the other ship turn around while seeing that ship's clock as reading 2 hrs.

Here's the diagram drawn from the frame moving at 0.577c relative to the last diagram.

If we assume that the Green and light blue lines are the two legs of ship A's trip, this is drawn from the rest frame of A's out bound trip. Note again that upon reaching turn around A sees B's clock as reading 0.5 hrs. At this moment, Ship B's clock actually reads 1 hr according to A.

Now we switch to the return leg of A.

At the start of the leg, A still sees B's clock as reading 0.5 hr, however, B's clock now actually reads 3 hrs at this moment according to A (relativity of simultaneity). For 1.5 hours A sees B's clock tick at the same rate as his own until his own clock reads 3.5 hrs, and he sees B's clock advance to 2 hrs and also sees B turn around. For the next 0.5 hours, he will see B's clock run fast (Doppler shift) until it reads 4 hrs when they meet, the same as his.

Note that the second image above is also for the rest frame of B's return trip and the third one works for B's outbound trip, so you can use the same diagrams to analyze what B sees during the trip also by reversing the order in which we use the diagrams.

 

[Chris Miller]

This is where your analysis starts going wrong. Forget everything about "seeing", it just complicates things.

Immediately before A's turnaround, based on A's simultaneity convention, B's clock shows 1am and B is still traveling away form Earth. Immediately after A's turnaround, based on A's new simultaneity convention, B's clock reads 3am and B is well on his way toward Earth already. By the time A and B get to Earth, at the same time, both their clocks will read 4am.

This is very helpful and interesting. So immediately after each turns around, the other appears (as in is expected) to jump instantaneously through time and space?

 

[Chris Miller]

Thanks Janus. Using c to determine what each "sees" at various junctures sort of confuses the exercise for me. I'm more concerned with what SR dictates each would expect is the situation (time/location/direction) in the other's FOR at any given point... which I think I do now, though it blows my mind a little.

 

[robphy]

Here's an extension of @Janus 's spacetime diagrams.
The ticks of each observer are drawn.
The dotted lines show lines of simultaneity [which are Minkowski-orthogonal to the worldlines]... these are parallel to the spacelike diagonals of that observer's light-clock diamond.

Follow those dotted spacelike lines to know that observer will assign for a time coordinate, although at a "kink" in the worldline, there is an abrupt change in coordinates [not experienced by an inertial observer].
Follow the cyan lightlike lines [along the light cones] to determine what might be visually "seen" by an observer.

 

[Vitro]

This is very helpful and interesting. So immediately after each turns around, the other appears (as in is expected) to jump instantaneously through time and space?

With strong emphasis on "appears" you could say that. But that's a misleading way to think about it because it implies some action or influence on B while in fact A's turnaround has no influence on B whatsoever.

What's changing is A's perspective on B; the intersection point of A's line of simultaneity (what A calls "now" or "at the same time") with B's worldline sweeps during A's turnaround from a crossing point where B's clock reads 1am to a different crossing point where it reads 3am. Yet from B's perspective there was no jump at all in his clock reading, it ticked normally for two hours between 1am and 3am.

 

[Chris Miller]

With strong emphasis on "appears" you could say that. But that's a misleading way to think about it because it implies some action or influence on B while in fact A's turnaround has no influence on B whatsoever.

What's changing is A's perspective on B; the intersection point of A's line of simultaneity (what A calls "now" or "at the same time") with B's worldline sweeps during A's turnaround from a crossing point where B's clock reads 1am to a different crossing point where it reads 3am. Yet from B's perspective there was no jump at all in his clock reading, it ticked normally for two hours between 1am and 3am.

Thanks again. Yes, I understand that the "jump" is only from the other's perspective. Still gives me twisty head. Guess it's not possible for two ships traveling towards Earth at the same (relativistic) velocity to appear simultaneously equidistant from Earth in each's FOR, only from earth's.

 

[Janus]

Thanks again. Yes, I understand that the "jump" is only from the other's perspective. Still gives me twisty head. Guess it's not possible for two ships traveling towards Earth at the same (relativistic) velocity to appear simultaneously equidistant from Earth in each's FOR, only from earth's.

One thing to keep in mind is that the apparent instantaneous nature of the "jump" shown in the Space-time diagrams are a result of assuming an instantaneous change in velocity on the part of Ship's A and B at turn around. To do this, A and B would have to undergo an infinite acceleration. And anytime you introduce an infinity, you can get unusual results.
If we instead assume a more realistic finite acceleration for some non-zero time, that "jump" becomes a more gradual transition. Thus when Ship A is undergoing its turn around acceleration, according to it, Ship B's clock runs fast for that period which is why it starts behind A's clock and ends up ahead of A's clock. This means that Ship B passes through its turn around rather than "skipping over it" as it seems to do in the space-time diagrams.

How fast B's clock runs according to A at any given moment will depend on three factors: The magnitude of the acceleration, the distance separating A and B, and the velocity difference between A and B. The acceleration can be a constant, but the other two will not be, as the relative speed and distance changes throughout A's acceleration maneuver.

 

[Bartolomeo]

They travel in opposite directions at this velocity for 2 hours by their own clocks. At which point each (somehow) stops and reverses course at this same velocity. Looking through a powerful telescope and allowing for c, A should see that it's 1 AM on B's clock, and B that it's 1 AM on A's.

So for A's next hour, from her frame of ref, she's traveling in the same direction as B. And for B's next hour, she's traveling in the same direction as A. I.e., for the next hour each sees the other traveling away from Earth while they are traveling towards it. So their relative velocity is now zero and each should see the other's clock to go from 1 AM to 2 AM while theirs goes from 2 AM to 3 AM.

As far as I know:
1. Observer in Special Relativity do not travel anywhere himself. Observer in Special Relativity is always at rest and observes how all other stuff travels in his rest frame.
2. Observer in Special Relativity do not admit that he was traveling for two hours by his own clock. He was at rest for two hours and observable object was traveling for two hours.
3. Observer in Special Relativity do not “looks through powerful telescope and allows c” to take readings of distant clock. At least I have never heard about that from official sources.
4. Observer in Special Relativity compares readings of moving clock with any Einstein synchronized clock of his reference frame, a.k.a. rest frame of observer.

Let’s observer A is at rest. After two hours observer A compares readings of clock B with clock A2, which is Einstein - synchronized with clock A.

Observer B is at rest also. After two hours he compares readings of clock A with clock B2 which is Einstein – synchronized with clock B.

https://en.wikipedia.org/wiki/Observer_(special_relativity)
“each observer makes observations in their immediate vicinity, where delays are negligible, cooperating with the rest of the team to set up synchronized clocks across the entire region of observation, and all team members sending their various results back to a data collector for synthesis”

http://www.pstcc.edu/departments/natural_behavioral_sciences/Web%20Physics/Chapter039.htm
Now, let's see how the measurement goes in frame S, in which the clock moves at velocity v. Here the time interval Δt is measured by two observers A and B at different positions.

 

[Bartolomeo]

At the start of the leg, A still sees B's clock as reading 0.5 hr, however, B's clock now actually reads 3 hrs at this moment according to A (relativity of simultaneity). For 1.5 hours A sees B's clock tick at the same rate as his own until his own clock reads 3.5 hrs, and he sees B's clock advance to 2 hrs and also sees B turn around. For the next 0.5 hours, he will see B's clock run fast (Doppler shift) until it reads 4 hrs when they meet, the same as his.

It is very interesting note. To my knowledge, special relativity claims, that if two observers (clocks) A and B are in uniform relative motion,clock A measures that clock B dilates, and clock B measures that clock A dilates.

https://en.wikipedia.org/wiki/Time_dilation
"In the special theory of relativity, a moving clock is found to be ticking slowly with respect to the observer's clock. If Sam and Abigail are on different trains in near-lightspeed relative motion, Sam measures (by all methods of measurement) clocks on Abigail's train to be running slowly and similarly, Abigail measures clocks on Sam's train to be running slowly."

I have never heard from official sources, that relativistic observer can measure the same rate of relatively moving clock or even greater one.
As far as I know relativistic Doppler shift appears in different colors: blueshift if source approaches and redshift when source recedes. Maybe I misinterpret you, but you mean that relativistic observer measures that clock tick faster when it approaches (due blueshift) and ticks slower when it recedes (due to redshift)? That confuses me a little bit. Could you please to clarify?

 

[Janus]

It is very interesting note. To my knowledge, special relativity claims, that if two observers (clocks) A and B are in uniform relative motion,clock A measures that clock B dilates, and clock B measures that clock A dilates.

https://en.wikipedia.org/wiki/Time_dilation
"In the special theory of relativity, a moving clock is found to be ticking slowly with respect to the observer's clock. If Sam and Abigail are on different trains in near-lightspeed relative motion, Sam measures (by all methods of measurement) clocks on Abigail's train to be running slowly and similarly, Abigail measures clocks on Sam's train to be running slowly."

I have never heard from official sources, that relativistic observer can measure the same rate of relatively moving clock or even greater one.
As far as I know relativistic Doppler shift appears in different colors: blueshift if source approaches and redshift when source recedes. Maybe I misinterpret you, but you mean that relativistic observer measures that clock tick faster when it approaches (due blueshift) and ticks slower when it recedes (due to redshift)? That confuses me a little bit. Could you please to clarify?

Yes, the Doppler shift does effect the tick rate at which you would see a clock receding or approaching. Relativistic Doppler shift is due to the combination of time dilation and the changing propagation delay due to changing distance between source and observer.
Let's say you have a clock that is traveling towards you at 0.5 c and at the moment it is light hour from you it reads 12:00. Thus 1 hr later, the light from that moment reaches you and you see it reading 12:00 . At that moment (by your clock) the other clock will 1/2 light hr from you and will read ~12:52(time dilation), the light from the clock at this moment will arrive at you 1/2 hr later. Thus you will see the clock read 12:00, and then 30 min later you will see it read ~12:52. You will see the other clock tick off ~52 min while only 30 have ticked off on your own, or in other words, you will see it tick at a rate ~1.733 times faster than your own clock.
The Relativistic Doppler shift gives a shift of 1.732 for 0.5c You will see the clock tick fast by the same factor as you see a shift in the light's frequency ( the only reason my answers don't match exactly is that I rounded to the nearest minute in my example.)

 

[Bartolomeo]

Yes, the Doppler shift does effect the tick rate at which you would see a clock receding or approaching. Relativistic Doppler shift is due to the combination of time dilation and the changing propagation delay due to changing distance between source and observer.
Let's say you have a clock that is traveling towards you at 0.5 c and at the moment it is light hour from you it reads 12:00. Thus 1 hr later, the light from that moment reaches you and you see it reading 12:00 . At that moment (by your clock) the other clock will 1/2 light hr from you and will read ~12:52(time dilation), the light from the clock at this moment will arrive at you 1/2 hr later. Thus you will see the clock read 12:00, and then 30 min later you will see it read ~12:52. You will see the other clock tick off ~52 min while only 30 have ticked off on your own, or in other words, you will see it tick at a rate ~1.733 times faster than your own clock.
The Relativistic Doppler shift gives a shift of 1.732 for 0.5c You will see the clock tick fast by the same factor as you see a shift in the light's frequency ( the only reason my answers don't match exactly is that I rounded to the nearest minute in my example.)

Thank you very much for your clarification. Sure, you mean "see".

 

[Battlemage!]

Yes, the Doppler shift does effect the tick rate at which you would see a clock receding or approaching. Relativistic Doppler shift is due to the combination of time dilation and the changing propagation delay due to changing distance between source and observer.
Let's say you have a clock that is traveling towards you at 0.5 c and at the moment it is light hour from you it reads 12:00. Thus 1 hr later, the light from that moment reaches you and you see it reading 12:00 . At that moment (by your clock) the other clock will 1/2 light hr from you and will read ~12:52(time dilation), the light from the clock at this moment will arrive at you 1/2 hr later. Thus you will see the clock read 12:00, and then 30 min later you will see it read ~12:52. You will see the other clock tick off ~52 min while only 30 have ticked off on your own, or in other words, you will see it tick at a rate ~1.733 times faster than your own clock.
The Relativistic Doppler shift gives a shift of 1.732 for 0.5c You will see the clock tick fast by the same factor as you see a shift in the light's frequency ( the only reason my answers don't match exactly is that I rounded to the nearest minute in my example.)

I think we had a discussion of the meaning of "see" on this topic before.

But on a slightly off topic tangent, to be honest I'm kind of confused as to the utility of time dilation versus this effect (what you actually see their clock read due to the Doppler shift). Any observer is going to be making measurements in his or her own reference frame, so why does what someone's clock "actually" reads "when" your clock ticks a certain time matter more than what you see their clock read? What I mean is, if 10 years have passed for me while they left Earth and then came back, which is of more practical value? The fact that since they were moving the whole time their clock ran slower than mine (moving clocks run slow), or that when I watched their clock at one point it was moving slower than mine (on the way to the destination) and at another point it was moving faster (on the way back from the destination), due to the Doppler effect, with the net result being that they age less than me? Both are true, if I understand it correctly, but one seems more practical to me (assuming they have powerful enough telescopes to see the clocks).

Or at least one is something I can actually measure, and the other is something I can only calculate.

 

[robphy]

Time dilation is associated with an observer's sense of "now". It is used in assigning time and space coordinates to distant events.

Merely receiving a light flash is not enough to assign coordinates. However, being able to see the image of a pre-calibrated is helpful.

To make predictions of what will happen in some part of our future often require initial conditions prepared or specified on events that are not causally accessible to us "now".

 

[nitsuj]

I think we had a discussion of the meaning of "see" on this topic before.

But on a slightly off topic tangent, to be honest I'm kind of confused as to the utility of time dilation versus this effect (what you actually see their clock read due to the Doppler shift). Any observer is going to be making measurements in his or her own reference frame, so why does what someone's clock "actually" reads "when" your clock ticks a certain time matter more than what you see their clock read? What I mean is, if 10 years have passed for me while they left Earth and then came back, which is of more practical value? The fact that since they were moving the whole time their clock ran slower than mine (moving clocks run slow), or that when I watched their clock at one point it was moving slower than mine (on the way to the destination) and at another point it was moving faster (on the way back from the destination), due to the Doppler effect, with the net result being that they age less than me? Both are true, if I understand it correctly, but one seems more practical to me (assuming they have powerful enough telescopes to see the clocks).

Or at least one is something I can actually measure, and the other is something I can only calculate.

It was always going c , what ever this concept of time and length measurements born of some idealized "now" moment...causality rules.

 

[Dale]

So for A's next hour, from her frame of ref, she's traveling in the same direction as B. And for B's next hour, she's traveling in the same direction as A. I.e., for the next hour each sees the other traveling away from Earth while they are traveling towards it. So their relative velocity is now zero

It may be too little too late, but I would point out that A's frame and B's frame are non inertial. So the usual reasoning and formulas go out the window. You would need to carefully define what you mean by their frame.

An example is here (note that figure 9 does wind up with something like what you are saying)

https://arxiv.org/abs/gr-qc/0104077

What most people have been describing is not A's frame but a series of momentarily comoving inertial frames (MCIF). The "jumps" described are not allowed in a single frame. They indicate a transformation from one MCIF to another MCIF. There is nothing wrong with using a series of MCIFs but it is not the same as the non inertial observers frame

 

[m4r35n357]

Or at least one is something I can actually measure, and the other is something I can only calculate.

Or to put in another way, you can do physics with the former, but "only" mathematics with the latter.

 

[Chris Miller]

One thing to keep in mind is that the apparent instantaneous nature of the "jump" shown in the Space-time diagrams are a result of assuming an instantaneous change in velocity on the part of Ship's A and B at turn around. To do this, A and B would have to undergo an infinite acceleration. And anytime you introduce an infinity, you can get unusual results.
If we instead assume a more realistic finite acceleration for some non-zero time, that "jump" becomes a more gradual transition. Thus when Ship A is undergoing its turn around acceleration, according to it, Ship B's clock runs fast for that period which is why it starts behind A's clock and ends up ahead of A's clock. This means that Ship B passes through its turn around rather than "skipping over it" as it seems to do in the space-time diagrams.

How fast B's clock runs according to A at any given moment will depend on three factors: The magnitude of the acceleration, the distance separating A and B, and the velocity difference between A and B. The acceleration can be a constant, but the other two will not be, as the relative speed and distance changes throughout A's acceleration maneuver.

Thanks for this further helpful clarification, Janus. I see that acceleration approaches infinity as t approaches 0. Would you agree that no matter how "gradual" a transition the other's jump, it's still a jump, or series of jumps? As I see it, A and B share infinite "nows," but A's infinity will be larger (and of course different) from her perspective/FOR, and B's from hers.

 

[Janus]

Thanks for this further helpful clarification, Janus. I see that acceleration approaches infinity as t approaches 0. Would you agree that no matter how "gradual" a transition the other's jump, it's still a jump, or series of jumps? As I see it, A and B share infinite "nows," but A's infinity will be larger (and of course different) from her perspective/FOR, and B's from hers.

Do you consider your own clock's time as advancing in a "a series of Jump's"? If not, whey would you consider the progression of the other ship's clock as advancing in a series of jumps just because it run's fast compared to your own while you are under acceleration?

 

[nitsuj]

Thanks for this further helpful clarification, Janus. I see that acceleration approaches infinity as t approaches 0. Would you agree that no matter how "gradual" a transition the other's jump, it's still a jump, or series of jumps? As I see it, A and B share infinite "nows," but A's infinity will be larger (and of course different) from her perspective/FOR, and B's from hers.

Time (geometry) is too "noble" to be considered as "discrete", only the mere things observed "within" it are up for being thought of as discrete.

Note that Lorentz gamma formula doesn't yield results in "steps".

Perhaps look at the fact that the fundamental forces "mediate" in quanta, completely different from "time / length or the applied "index" of a gamma factor.

I figure accelerating a ship would involve one or more of these fundamental forces, bound by being mediated in quanta of energy.

It may be equally fair to ask if you feel an angle is made up of discrete "steps", consider pi.

Rolex gets it ;)

 

[Chris Miller]

"1) There is no conclusive evidence that time is quantized, but 2) certain theoretical studies suggest that in order to unify general relativity (gravitation) with the theories of quantum physics that describe fundamental particles and forces, it may be necessary to quantize space and perhaps time as well." - William G. Tifft

It would seem to me that if their are no instantaneous jumps (as in the original hypothetical t=0 example) that given sufficient acceleration and distance, A could expect B's velocity relative to her ship to easily exceed c? And vice versa.

 

[Chris Miller]

It may be equally fair to ask if you feel an angle is made up of discrete "steps", consider pi.

Yes, but an infinite number of them (each merrily skipping over an infinite number of irrationals [like pi, sqrt(2), etc.] not present as points on on the angle's rational arc).

 

[Dale]

It would seem to me that if their are no instantaneous jumps (as in the original hypothetical t=0 example) that given sufficient acceleration and distance, A could expect B's velocity relative to her ship to easily exceed c? And vice versa.

I am not sure where you are getting this. Calculating the velocity for the case of constant proper acceleration is a typical homework problem in SR courses, and the expected result is always v<c. The result is in the relativistic rocket page which may include the derivation too. If A and B are expecting that then they are making a mistake.

 

[Chris Miller]

I am not sure where you are getting this. Calculating the velocity for the case of constant proper acceleration is a typical homework problem in SR courses, and the expected result is always v<c. The result is in the relativistic rocket page which may include the derivation too. If A and B are expecting that then they are making a mistake.

Clearly I'm confused, because say A, instead of reversing her velocity and direction instantaneously at her 2 hour mark, she does so over the course of 1 second? Would not B still be expected to have traveled from halfway out (to her 2 hour mark) to halfway back (to where she is) over the course of this second?

 

[Dale]

Clearly I'm confused, because say A, instead of reversing her velocity and direction instantaneously at her 2 hour mark, she does so over the course of 1 second? Would not B still be expected to have traveled from halfway out (to her 2 hour mark) to halfway back (to where she is) over the course of this second?

Ah, I misunderstood what you were saying. I thought that you were talking about the path of each traveler from an inertial frame, such as any of the momentarily comoving inertial frames discussed by others.

But here it sounds like you are talking about a true non inertial frame. You can certainly have v>c in non inertial frames. There is nothing wrong with that.

 

[Chris Miller]

Would it be wrong to say that in non inertial frames, v may approach infinity?

 

[robphy]

But here it sounds like you are talking about a true non inertial frame. You can certainly have v>c in non inertial frames. There is nothing wrong with that.

Would it be wrong to say that in non inertial frames, v may approach infinity?

What is the meaning of "v" in this context?
Regardless of the nature of the frame of reference, an observer's 4-velocity is restricted to pointing into the interior of the future light-cone [and thus its spatial component is always smaller than that of a light-signal].

 

[Chris Miller]

What is the meaning of "v" in this context?
Regardless of the nature of the frame of reference, an observer's 4-velocity is restricted to pointing into the interior of the future light-cone [and thus its spatial component is always smaller than that of a light-signal].

I think v would be the expected relative velocity of a pair of non inertial frames (as in ships A and B's reversing course in this example).

 

[robphy]

I think v would be the expected relative velocity of a pair of non inertial frames (as in ships A and B's reversing course in this example).

That relative-velocity has a magnitude strictly less than c.

 

[Chris Miller]

That relative-velocity has a magnitude strictly less than c.

You can certainly have v>c in non inertial frames. There is nothing wrong with that.

?

 

[PeroK]

?

Just think about the observed speed of a star far enough from the Earth, which in the Earth's frame orbits every 24 hours much further than a light day.

Or, if you simply spin round in a second or two, in your frame the Sun will have traveled about 500 million km.

 

[robphy]

?

Just think about the observed speed of a star far enough from the Earth, which in the Earth's frame orbits every 24 hours much further than a light day.

Or, if you simply spin round in a second or two, in your frame the Sun will have traveled about 500 million km.

In this case, that's not the usual relative-velocity
but instead some quantity Frankensteined from two coordinate systems [which is fine as long at it's properly defined... and not misinterpreted].