# Yet another SR question

1. Jun 18, 2006

### Al68

Let's say we have a ship following a buoy toward earth. The ship and buoy are at rest relative to each other, a distance of 10 light years apart, in the ship's frame. The distance between the ship and buoy is 5 light years in earth's frame. They are approaching earth at a relative velocity of 0.866c. When the buoy passes earth, an observer on earth starts his clock at t = 0, and an observer on the buoy starts his clock at t' = 0. An observer on the ship also starts his clock at t' = 0, since he is at rest with the buoy, and they planned this ahead of time. At t = 5.77 years in earth's frame, the ship reaches earth and turns around. At t' = 11.55 years in the frame of the ship and buoy, the ship reaches earth and turns around and leaves earth at 0.866c. There is a second buoy approaching earth from the opposite direction of the ship, now at rest with the ship, at 0.866c relative to earth. This second buoy is now 10 light years behind the ship, in the ship's and the second buoy's frame. This distance between the ship and the second buoy is now 5 light years in earth's frame. At t = 11.55 years in earth's frame, the second buoy passes earth. At t' = 23.1 years, the second buoy passes earth in the ship's frame.

The total time between the event of the first buoy passing earth and the second buoy passing earth (in the opposite direction) is 11.55 years in earth's frame, and 23.1 years in the ship's frame. At t = t' = 0, the ship was 10 light years from earth in the ship's frame and the ship was 5 light years from earth in earth's frame. At t = 11.55 years, the ship was 5 light years from earth again in earth's frame. At t' = 23.1 years, the ship was again 10 light years from earth in the ship's frame.

Would these results be correct?

Thanks,
Alan

Last edited by a moderator: Jun 19, 2006
2. Jun 19, 2006

### actionintegral

You are confusing yourself by using actual numbers. The answer to most
simultaneity questions can be understood by looking at t' = t-vx.

When you reverse the sign of the velocity, look at how you reverse the
change in proper time as a function of space! And the difference increases
as x increases!

3. Jun 19, 2006

### JesseM

Presumably you mean the ship-observer starts his clock simultaneously with the buoy's clock in the ship-buoy frame? When dealing with issues of simultaneity, it's important to be explicit about what frame you're referring to.
The ship has not moved inertially since its clock was started, so I don't see how you can say "in the ship's frame" here. If you just mean its current inertial rest frame (ie the frame of the second buoy), then you need to specify what event is going to be defined as the origin of this coordinate system before you can say at what time-coordinate the second buoy passes earth--in the simplest version of the Lorentz transform you're assuming all the frames you look at have the same origin, which in this case would mean the origin of the second buoy's coordinate system would have to be the event of the first buoy passing earth, since that was the origin of the first two coordinate systems. In this case, I don't think the event of the second buoy passing earth will have the coordinate 23.1 years, although I haven't checked. Also, if this is understood to be a third coordinate system you should introduce a third notation for its time coordinate, like t''.
Yes.
What is this "ship's frame"? If you're assuming something like the NI coordinate system we discussed on the last thread, I've said many times that it's inappropriate to refer to this as "the ship's frame". It's just one choice of non-inertial coordinate systems among many. And in any case, if you're using non-inertial coordinate systems, I don't know if I'd call this an "SR question" at all (if you state the laws of SR in tensor form it can work in arbitrary non-inertial coordinate systems, but in the regular non-tensor form you're familiar with it can only be used to talk about what happens in inertial coordinate systems).
If by "ship's frame" you just mean the inertial rest frame of the ship and the first buoy at that point, yes. If not, you have to specify what you mean by "the ship's frame".
Yes.
See above.

4. Jun 19, 2006

### RandallB

NO
It is not that “You are confusing yourself by using actual numbers”
And need to surrender to the use of formulas like ( t' = t-vx )
That will work but it’s not the double checking of understanding I suspect you’re going for.

DejaVu I’m sure we’ve been here before; If you want to track this kind of thing with actual numbers you must follow all of them carefully and correctly. You’re only using two Frames in your measurements. Just t & t’ you also need t’’ for the second buoy reference frame. The ship doesn’t get a frame – it is the one changing frames. You need to detail WHERE and WHEN in t’’ and d’’ the ship starts at, turns at, and ends at.

The more detail you give yourself for every element, (earth and buoy1 and buoy 2) in all three frames, as well as the ship, so you can total up the elapse time the ship experiences, the clearer the simultaneity point will start to dawn on you.

5. Jun 19, 2006

### Al68

Jesse, you are right again, I should not have used the term “ship’s frame”. I have rephrased everything to remove this term, and just use an observer on the ship, and the ship’s clock.

Let's say we have a ship following a buoy toward earth. The ship and buoy are at rest relative to each other, a distance of 10 light years apart, as measured by an observer on the ship. The distance between the ship and buoy is 5 light years in earth's frame. They are approaching earth at a relative velocity of 0.866c. When the buoy passes earth, an observer on earth starts his clock at t = 0, and an observer on the buoy starts his clock at t' = 0. The ship-observer starts his clock simultaneously with the buoy's clock in the ship-buoy frame at t’ = 0. At t = 5.77 years in earth's frame, the ship reaches earth and turns around. At t' = 11.55 on the ship’s clock, the ship reaches earth and turns around and leaves earth at 0.866c. There is a second buoy approaching earth from the opposite direction of the ship, now at rest with the ship, at 0.866c relative to earth. This second buoy is now 10 light years behind the ship, as measured by an observer on the ship. This distance between the ship and the second buoy is now 5 light years in earth's frame. At t = 11.55 years in earth's frame, the second buoy passes earth. At t' = 23.1 years, the second buoy passes earth as observed by the ship’s observer and the ship’s clock.

The total time between the event of the first buoy passing earth and the second buoy passing earth (in the opposite direction) is 11.55 years in earth's frame, and 23.1 years on the ship's clock. At t = t' = 0, the ship was 10 light years from earth as measured by an observer on the ship, and the ship was 5 light years from earth in earth's frame. At t = 11.55 years, the ship was 5 light years from earth again in earth's frame. At t' = 23.1 years, the ship was again 10 light years from earth as measured by an observer on the ship.

I’m using t’ here to indicate the reading on the ship’s clock, not to represent time in a particular frame.

Would these results be correct?

Thanks,
Alan

6. Jun 20, 2006

### JesseM

Looks right to me! I didn't do a detailed check with the Lorentz transform but based on how distances and times are related in the different frames, that should be correct. My only minor quibble is that when you say stuff like "At t' = 23.1 years, the second buoy passes earth as observed by the ship’s observer and the ship’s clock", since the ship has not stuck to a single inertial frame it might be better to say something like "When the ship's clock reads 23.1 years, this is simultaneous with the event of the second buoy passing earth as observed in the frame where the ship is at rest after the turnaround", but I assume this is what you meant anyway.

7. Jun 20, 2006

### Al68

Jesse,

It looks like, between these defined events, we have the inertial observer aging less than the non-inertial observer. Is it safe to say that this is because I specified that the events were a certain distance apart as measured by the ship-observer?

Of course in my example, if we were to specify events that were a certain distance apart in earth's frame (by introducing a space station at rest with earth that the ship passes), the time elapsed between these events would be less for the ship's observer than for an earth observer.

Also, in my example, all three defined events (the first buoy passing earth, the ship turnaround, and the second buoy passing earth) were local to earth. In the Twins Paradox, all three events (ship leaves earth, ship turnaround, and ship return to earth) are local to the ship.

It looks like how we define the events is more important than which observer is inertial.

Thanks,
Alan

8. Jun 20, 2006

### George Jones

Staff Emeritus
Yes.

Suppose (in SR) events A and B are experienced by an inertial observer. If events A and B are experienced also by an non-inertial observer, then the time elapsed between A and B will be greater on the inertial observer's watch. If there is no way to compare directly watches, i.e., if both watches don't experience the same pair of events, then there is no way to say which watch really ages more.

This is not the situation in the example you give here, but it is the situation in the twin paradox.

In particular, both twins experience both the ship leaving the Earth and the ship returning to the Earth.

9. Jun 20, 2006

### JesseM

In what frame? Remember, the inertial frame where the ship's clock reads 0 "at the same time" that the first buoy passes earth is a different one then the frame where the ship's clock reads 23.1 years "at the same time" that the second buoy passes earth. I'm not sure whether it would be possible to find an inertial frame where the ship ages more between the times of the two buoys passing earth in that frame--I only know it's guaranteed to age less when the two events are the ship leaving earth and reuiniting with earth, in which case all frames must agree on the exact amount of time each one aged between the events, and it will always be the non-inertial observer who aged less.
The ship leaving earth and the ship reuiniting with earth are local to both the ship and the earth (in the sense that they lie on both the ship and the earth's worldline). And those are the two events for which you want to know how much each twin aged between them--again, since they depart and reunite from a common location, there will be an objective answer to this question, whereas in your example or any other where you're not picking two events that lie at points where their worldlines cross, the answer depends on your choice of coordinate systems, it isn't really a physical question at all.

10. Jun 21, 2006

### Al68

I purposely made my example different from the Twins Paradox. And I purposely made the events local to earth and not the ship. I did this in order to have a non-inertial observer age more than an inertial observer between two defined events. I only wanted to point out that just because the inertial twin ages more in the Twins Paradox between the specified events, that doesn't mean that an inertial observer will always age more than a non-inertial observer between any two specified events.
Jesse,

As you have pointed out to me, this comparison is not between two different frames, just between two different observers. I would also point out that in the Twins Paradox, if we looked at either inertial frame of the ship, the elapsed time between the ship leaving earth and the ship returning to earth in either ship frame would be more than the elapsed time between these two events in earth's frame. It is only when we compare the observers' clocks that the ship's twin ages less than the earth twin. This is why I just wanted to compare observers' elapsed time between two events in my example, instead of the proper time of each inertial frame. I do appreciate your help in trying to word everything properly. I never realized how important it was until lately.

How about this example: One twin on earth and one twin goes on a ship in a highly elliptical orbit around earth. Both twins' paths are geodisics. After many orbits, the twins reunite. Does SR predict differential aging in this situation?

Thanks,
Alan

11. Jun 21, 2006

### JesseM

But the two events you're talking about don't occur on the ship's worldline, so there is no single answer to how much time elapses between them according to the ship's clock--that depends on your definition of simultaneity, which itself depends on what coordinate system you use. If you use something like the non-inertial system I called "NI" on the last thread, where the ship's definition of simultaneity always matches that of his instantaneous inertial rest frame, then the time between the events of the two buoys passing earth would indeed be 23.1 years. But this is just a statement about that particular coordinate system, not about objective physical reality. If the ship uses some other coordinate system, the answer to which events on his worldline are simultaneous with these events near earth can be different, therefore the amount of time elapsed on his clock between the events is different.
It's not the wording itself that's important though, it's what the words suggest about your conceptual understanding. To me it seems like your words often suggest confusion between statements which depend on an arbitrary choice of coordinate system and statements which actually reflect objective physical realities.
Well, orbits are only geodesics in curved spacetime, so you must use GR rather than SR here, unless you just want to know about ships travelling in ellipses in flat spacetime, but then these elliptical paths wouldn't be geodesics. Either way, the aging would probably be different, although I think it'd depend on the details of the situation (perhaps it would be possible to come up with a combination of orbits such that when they reunite they will be the same age, I'm not sure...this wouldn't be true in general though).

12. Jun 21, 2006

### RandallB

Of Course the inertial observer will always age more than a non-inertial observer. The conclusion you’ve drawn above only means you’ve made in error in working the data, to enable an invalid comparison.

Since simultaneity cannot be defined across distances in any one frame (Unless you have found the elusive preferred frame) the only frame that is useful in making the age comparison is the one where simultaneity is not affected by distance. That means the one the one where your two events are displaced by a distance of zero, so that only time separates them. As measured in that frame the non-inertial observer will always age less than the inertial one.
That is the point of being sure the twins come back together – to enable a valid measurement more easily. You’re not working enough detail correctly to see the point.