Some questions about relativistic effects.

[RussDill]

No. Acceleration is equivalent to a UNIFORM gravitational field. Special relativity can handle a uniform gravitational field exactly the same as it can handle acceleration. You're just constantly changing reference frames. However, since every gravitational field of any interest is NOT uniform, you cannot treat gravity globally as a simple acceleration, and special relativity falls apart. Two objects sitting on opposite sides of a planet cannot be treated as if they are accelerating away from each other when they clearly aren't. It is this nonuniformity of gravity which requires general relativity. Absent that, you can do everything you want with special relativity.

The two reference frames are experiencing different strengths of gravitational fields, yes?

 

[RussDill]

I was thinking something like this. The original ship was 1 light second across. If the blink occurs where its right in my line of sight I'll see it 2 seconds after the flash but it will have missed the mirror which has moved forward by 1 light second by the time the light reaches it.

The flash would have to be pointed forward at 45 degrees for it to hit the mirror if it were traveling at c. Am I thinking about that right?

Don't forget that photons have momentum, for conservation of momentum to occur, the photons leave the strobe at what seems to be a 45 degree angle.

 

[Ziggurat]

The two reference frames are experiencing different strengths of gravitational fields, yes?

Which two frames are you refering to? The two that I gave in my example, with two objects sitting on opposite sides of a planet? In that case the strength of the gravitational field is identical, only the direction changes. But in any case, the distinction between special relativity and general relativity is still that the former cannot handle non-uniform gravitational fields. That is its ONLY limit compared to general relativity.

 

[RussDill]

Which two frames are you refering to? The two that I gave in my example, with two objects sitting on opposite sides of a planet? In that case the strength of the gravitational field is identical, only the direction changes. But in any case, the distinction between special relativity and general relativity is still that the former cannot handle non-uniform gravitational fields. That is its ONLY limit compared to general relativity.

I'm refering to the stationary twin, vs the twin accelerating away. Certainly, there are really 6 frames of reference in the twin paradox.

a) stationary twin
b) twin accelerating away (Actualy, a contium of reference frames)
c) twin moving away at a constant velocity
d) twin turning around (again, a continum of reference frames)
e) twin moving back at a constant velocity
f) twin slowing to a stop

b, d, and f, would seem to require integration, and I think general relativity, set me straight here.

 

[Stimpson J. Cat]

RussDill,

I'm refering to the stationary twin, vs the twin accelerating away. Certainly, there are really 6 frames of reference in the twin paradox.

a) stationary twin
b) twin accelerating away (Actualy, a contium of reference frames)
c) twin moving away at a constant velocity
d) twin turning around (again, a continum of reference frames)
e) twin moving back at a constant velocity
f) twin slowing to a stop

b, d, and f, would seem to require integration, and I think general relativity, set me straight here.

They do require integration, but special relativity is sufficient. Anyway, when it comes to the Twin Paradox, the acceleration is kind of a red herring. After all, you can always take the limit in which the period of acceleration is so short compared to the total time of the trip that it makes no difference.

The appearance of the paradox comes from the misconception that the situation is symmetric for the two twins. Of course it is not. But the breaking of the symmetry is not in the fact that one observer is accelerating and the other isn't. After all, from the other frame of reference the opposite is true. The break in symmetry comes from the fact that the two evens which mark the beginning and ending of the experiment are in the same inertial frame that one of the twins is in for the duration of the experiment, while the other twin goes from that frame of reference to a different one, and back again.

If you think about it, the question "how much time passes between event A and event B for the two twins?" is only well defined for some specific reference frame. In this case, the frame which the two events are defined in, is the frame of the twin on Earth. You could do the experiment from any frame, and even try to make the experiment symmetric between the two twins by observing the whole thing from a 3rd reference frame. In that case, the relative ages of the two twins at the end of the experiment would depend on their velocities relative to that frame. For example, if we send both twins away from the Earth in opposite directions, and have them come back after going the same distance at the same speed, then they will be the same age when they return, even though from each of their points of view the other twin was moving relative to them. In this case the experiment is symmetric, so they both age the same amount and there is no paradox. The paradox only appears to be there when the situation is not symmetric, but superficially appears to be.


Dr. Stupid

 

[Ziggurat]

I'm refering to the stationary twin, vs the twin accelerating away. Certainly, there are really 6 frames of reference in the twin paradox.

a) stationary twin
b) twin accelerating away (Actualy, a contium of reference frames)
c) twin moving away at a constant velocity
d) twin turning around (again, a continum of reference frames)
e) twin moving back at a constant velocity
f) twin slowing to a stop

b, d, and f, would seem to require integration, and I think general relativity, set me straight here.

b,d, and f all require integration to do correctly, and the math can get considerably more complicated than the simple Lorenz transformations people usually deal with (although they're still the basis for the calculus involved. But none of those cases requires general relativity.

I think I might see what's tripping you up. First, take it as a complete given that special relativity can handle acceleration, any kind of acceleration, including everything you need to solve the twin problem. And it can handle acceleration because nothing about acceleration breaks special relativity (if it did, the theory would not have gotten much attention). But gravity does break special relativity, in a way that acceleration does not.

The acceleration = gravity is in some senses secondary: what is perhaps more fundamental is FALLING in a uniform gravitational field is exactly equivalent to being in an inertial reference frame. If you let your coordinate system fall, it is an inertial coordinate system. But of course, you can't make such a reference frame globally with special relativity once the gravitational field is non-uniform, because different parts of your reference frame then need to "fall" at different rates and in different directions, so it "falls" apart (ha ha!... sorry). In the twin problem, we've got various inertial reference frames, and accelerations which constantly change WHICH inertial reference frame any one twin is in at any one moment. But each of the infinite possible inertial reference frames we can choose is still itself constant, and they all maintain their relative velocities at all times, so special relativity has no problems with any of this. In short: acceleration is just changing inertial reference frames, which special relativity has no problem with, but non-uniform gravity means you can no longer even describe an inertial reference frame with special relativity on anything more than a local level.

So falling reference frames are inertial reference frames. That means if you're standing still on the surface of the earth, you're constantly accelerating compared to a reference frame that is falling. Hence gravity = acceleration, but only to the extent that locally the gravitational field is pretty damned close to uniform. And that's actually always the case locally, if you look at a small enough patch of space, except in the case of singularities, just as the surface of a sphere looks like a plane if you zoom up close enough, but what's true for a plane (parallel lines never cross) is no longer true for your curved surface on a global scale. The relationship of general relativity to special relativity is much the same: aside from singularities, special relativity still describes the tangent space for general relativity.

Hope that clarifies things a bit.

 

[Atlas]

Don't forget that photons have momentum, for conservation of momentum to occur, the photons leave the strobe at what seems to be a 45 degree angle.

Now there's a crazy idea. I wouldn't have thought of that. I still am wondering though... because photons are massless would they really have an X directional component if they were being aimed in the Y coordinate.

I'm asking. I learn something new each time these ideas are presented here.

Just one more comment on the diagram I presented earlier. I mentioned that if the craft was moving at speed c along the x coordinate that I thought the strobe should be aimed forward at 45 degrees to lead the target. But at speed c the hypotenuse of the triangle is longer than the the distance along the x coordinate. With the craft flying at speed c there is no angle at which a flash could be projected such that it would be received at the opposite wing tip.

Is that true or is there some relativistic weirdness that makes me wrong once more?

(edit: I know at any speed the hypotenuse of a right triangle is longer than either leg but what I meant was that because of that longer distance the light traveling at top speed c will miss the target which is traveling a shorter distance at the same speed c. )

 

[RussDill]

Hope that clarifies things a bit.

Doesn't time dialation occur in an accelerating reference frame? (vs a non accelerating reference frame)? How does special relativity address this dialation?

 

[RussDill]

Just one more comment on the diagram I presented earlier. I mentioned that if the craft was moving at speed c along the x coordinate that I thought the strobe should be aimed forward at 45 degrees to lead the target. But at speed c the hypotenuse of the triangle is longer than the the distance along the x coordinate. With the craft flying at speed c there is no angle at which a flash could be projected such that it would be received at the opposite wing tip.
I

right, if some hypothetical spaceship was traveling at c, then you could only hit targets behind you. Of course, spaceships can't travel at c, so you don't have this problem. (something similar does occur in a blackhole though).

 

[Ziggurat]

Doesn't time dialation occur in an accelerating reference frame? (vs a non accelerating reference frame)? How does special relativity address this dialation?

ds^2 = dx^2 + dy^2 + dz^2 - dt^2

That's your metric. If there's no acceleration, all you need are start and end coordinates, and you can calculate time along a path with very simple algebra. If the path is curved (acceleration), then you need to do some integration, and that can get messy very fast, but it's still just special relativity, regardless of how twisty you make that trajectory. The difference between calculating it for an accelerating trajectory versus a non-accelerating trajectory is simply the difference between calculating the length of a curved line and a straight line. In this sense, there's nothing special about time dilation for an accelerating object compared to a non-accelerating object.

The reason "acceleration" comes into the twin problem is that one of the twins has a curved trajectory, while one of them doesn't (the form of the curve doesn't matter). In Euclidean geometry, the shortest distance between two points is a straight line. In special relativity, the longest, rather than shortest, interval between two events is a straight line (play around with the metric a little bit, draw out line segment paths and calculate distances and you'll see what I mean). The earth-bound twin experiences no acceleration, and so traces out a straight line in space-time, while the traveling twin traces out a curved (or segmented, however you want to deal with it) trajectory. So people like to say that the time difference is because one of the tins accelerates, and in a sense that's correct. But it's not because acceleration has to be treated fundamentally any differently when calculating time dilation, any more than the length of curves in Euclidean geometry is any different than the length of straight lines. The calculus is still the same.

 

[RussDill]

Again, from what I understand, time dialation occurs to a reference frame in a gravitational well (not free falling). On earth, it is the same time dialation that would occur to a reference frame accelerating at 9.8m/s^2. Is this not correct?

 

[Ziggurat]

Again, from what I understand, time dialation occurs to a reference frame in a gravitational well (not free falling). On earth, it is the same time dialation that would occur to a reference frame accelerating at 9.8m/s^2. Is this not correct?

Ah, now I see what you're getting at.

The difficulty though is you always need to ask, time dilation compared to what? With the earth, you could ask, for example, time dilation compared to a distant spot far from the earth. But the answer depends on the entire gravitational well geometry, not just the local gravitational field. If you hollowed out a spot at the center of the earth, where gravity was zero, the time dilation there would still be "stronger" than the time dilation at the surface of the earth. To get the right answer, there's a sort of integration process you need to carry out along the path - which also means that if your path length is essentially zero, there's no relevant time dilation regardless of your acceleration. And you can't pick a path length between two inertial reference frames in special relativity, since all inertial reference frames are infinite. It never makes sense to talk about the time dilation due to a given acceleration, because it doesn't exist. There's no special time dilation needed to account for acceleration in special relativity.

The reason gravitational time dilation pops up has to do with making a reference frame that doesn't move with respect to your gravitational source, and hence is not really an inertial reference frame. It gets messy, and I don't think it's really worth getting into now.

 

[69dodge]

Originally posted by Stimpson J. Cat
The appearance of the paradox comes from the misconception that the situation is symmetric for the two twins. Of course it is not. But the breaking of the symmetry is not in the fact that one observer is accelerating and the other isn't. After all, from the other frame of reference the opposite is true.

Well. The other frame of reference is not inertial, so it doesn't count.

That's what people mean when they say that one observer is accelerating: he is accelerating relative to an inertial reference frame. Obviously, he is not accelerating relative to the reference frame in which he is motionless.

The break in symmetry comes from the fact that the two evens which mark the beginning and ending of the experiment are in the same inertial frame that one of the twins is in for the duration of the experiment, while the other twin goes from that frame of reference to a different one, and back again.

Events aren't in one reference frame or another; they're just events. The difference between the twins is that one doesn't accelerate relative to an inertial reference frame and the other one does.

If you think about it, the question "how much time passes between event A and event B for the two twins?" is only well defined for some specific reference frame. In this case, the frame which the two events are defined in, is the frame of the twin on Earth.

Huh? The travelling twin is just as involved in the events of his departure and arrival as the Earthbound twin is. More involved, arguably. Events aren't defined in any particular reference frame. They can be referred to any reference frame.

The time between two events does depend on which reference frame that time is measured in. Between the departure and arrival events, more time passes in the inertial reference frame in which the Earthbound twin is motionless than in the noninertial reference frame in which the travelling twin is motionless.

You could do the experiment from any frame, and even try to make the experiment symmetric between the two twins by observing the whole thing from a 3rd reference frame. In that case, the relative ages of the two twins at the end of the experiment would depend on their velocities relative to that frame.

Observing the whole thing from a different reference frame won't change the relative ages of the twins when they reunite. Every observer will agree that the twin who left Earth is younger on his return than the twin who stayed on Earth, regardless of the observer's own motion.

For example, if we send both twins away from the Earth in opposite directions, and have them come back after going the same distance at the same speed, then they will be the same age when they return, even though from each of their points of view the other twin was moving relative to them. In this case the experiment is symmetric, so they both age the same amount and there is no paradox.

This experiment is symmetric, but it's a different experiment, not merely the same experiment viewed from a different reference frame. Before, one twin's motion was inertial; now, neither's is.

 

[epepke]

The size of the ship isn't really important here, at least, not for what I'm trying to find out. I'm wondering about how time dilation works. I don't understand when things appear to speed up and down.

In that case, you're overcomplicating the problem.

Instead of picking D as the time for the light to get to the mirror, pick D' as the time for the light to get to the mirror and back again, which would be 2 seconds. Then it's easy. D>2 as seen by an observer in all cases other than rest (provided that the measurements take into account the change in distance that the spaceship travels with respect to the observer.)

The reason that it's overcomplicating is that in the case where the path of the light is even partially along the axis of travel of the spaceship, the time for light to get to the mirror will be different for another observer than the time it takes to get back.

As for how time dilation works, it's extremely simple. Have the light going at right angles to the direction of travel of the ship. At rest WRT an observer, the light will appear to bounce side to side in the spaceship. When the spaceship is moving, the light will go in a zig-zag pattern. A zig-zag pattern is longer. Because the speed of light is constant, if the light is going in a longer (zig-zag) pattern, it will take longer to get there. Therefore, that clock made with the light will appear to go slower.

All clocks have to go slower, too, because if not, you could determine something's absolute motion by comparing clocks, which would violate relativity.

 

[epepke]

Doesn't time dialation occur in an accelerating reference frame? (vs a non accelerating reference frame)? How does special relativity address this dialation?

Basic time dilation occurs in a moving reference frame. Another aspect of time dilation occurs in an accelerating reference frame.

Special relativity can be used to handle lots of accelerating cases, but historically, this wasn't done much until general relativity. And the so-called "twin paradox" can be resolved with SR ignoring the acceleration entirely.

As for your other question, the acceleration-equivalent time effects of gravitation work out to be exactly what you would get if you just dropped a clock (no air resistance, of course). So as long as you know what speed the clock is going when it passes you, you can work it out without using GR in lots of cases.

 

[Stimpson J. Cat]

69Dodge,

I think you are misunderstanding what I am saying.

Consider this alternative formulation of the "paradox":

A spaceship is coming from a very distant star towards Earth at 0.8c. An observer on Earth figures that the ship will pass very close to Earth, so he sets up a marker in space 4 light minutes away in the direction the ship is coming from, and 4 light minutes away in the direction the ship will moving away from the Earth in. Note that there will be no acceleration in this experiment.

He then looks at a clock in the alien ship, using a telescope, and the alien does likewise.

When he sees the ship pass marker A, he looks at the clock in the ship, and the clock in his office. Let's say his clock says 12:04PM, and the ship clock says 2:00PM. He then subtracts 4 minutes from the time on his clock, because he knows it took 4 minutes for the light from marker A to reach him. So his correct starting time is 12:00.

When the ship passes marker B, he looks again. This time his clock will clearly say 12:14PM (12:10, when he subtracts the 4 minute light-travel time), because at 0.8c it will take 10 minutes to travel 8 light minutes. Because of time dilation the spaceship clock will say 2:06PM, because (1-sqrt(0.8)^2)=0.6.

But what will the alien see? From his POV he is stationary. An inertial frame containing both markers, and the Earth, is moving past him at 0.8c. Because of length contraction, he will see the distance from marker A to marker B as being not 8 light minutes, but instead 4.8 light minutes. At 0.8c he will therefore think the trip takes 6 minutes. And of course he will agree with observer on Earth as to what times his clock says when he passes the markers (2:00PM and 2:06PM).

But he can not agree with observer on Earth that the amount of time which passes on Earth is 10 minutes. On the contrary, because of time dilation, he must say that only 3.6 minutes pass on Earth. Thus we have an apparent paradox, similar to that of the twin paradox, but with no acceleration to save it (as I said, though, acceleration is not the issue in the twin paradox either).

The resolution comes from the fact that, while both observers agree on what the alien's clock said at the beginning and ending of the experiment, they do not agree on what the Earth observer's clock said. Remember that the way the Earth observer knows what the alien's clock said, is by looking at the alien's clock when he passes the marker. The alien must do the same.

So the alien crosses marker A and is looking at the Earth. When the marker passes him, he knows that the Earth is 2.4 light minutes away (length contraction). Of course, at that time he sees the Earth further away, but he can correct for the time it takes the light to reach him. This correction tells him that the light emitted by the Earth when the marker passed him will reach him in 2.4 minutes, just 0.6 minutes before the Earth reaches him. So at that time, the Earth is only 0.48 light minutes away.

So now we can figure out what his clock will say. When that light from Earth reaches the alien, in the frame of Earth, the alien will be 0.8 light minutes away. So that means that the light from Earth currently reaching the alien will have departed only 0.8 minutes ago. Furthermore, in the frame of the Earth, the alien will have travelled 3.2 light minutes since the beginning of the experiment, which means that 4 minutes have already passed. Thus the clock on Earth will say 12:04, which means that the alien will calculate that the Earth clock said it was 12:03.2 when he passed marker A.

Likewise when marker B passes the alien, the Earth will be 2.4 light minutes away. Again, it will take 2.4 minutes for that light to reach the alien. By that time the earth will have moved away another 1.92 light minutes, giving a total distance of 4.32 light minutes (in the alien's frame). In the Earth frame, the alien will therefore be 7.2 light minutes from the Earth. This means that the light then reaching the alien left the Earth 7.2 minutes ago. And since, in the Earth frame, the alien passed marker B at 12:10, and has travelled 3.2 light minutes since then, the current time will be 12:14. Subtracting 7.2 minutes we get 12:06.8.

Thus from the alien's POV, the clock on Earth said 12:03.2 when the experiment started, and 12:06.8 when it finished. Total time: 3.6 minutes, just as time dilation predicts.

The point of all this is that the apparent contradiction arises from the problem that the two observers cannot agree on which events occurred simultaneously. They both agree that the start of the experiment occurred simultaneously with the alien's clock saying 2:00, and that the end of the experiment occurred simultaneously with the alien's clock saying 2:06. But the alien says that the beginning occurred simultaneously with the Earth clock saying 12:03.2, and the Earth observer says that the beginning occurred simultaneously with the Earth clock saying 12:00. Likewise they disagree on when the experiment ended in the Earth frame.

Again, there is no acceleration here. In the twin paradox, acceleration is not the issue either. In that frame, they can agree on what the clocks both said at the beginning and ending of the experiment, because they are both in the same frames at those times. Acceleration is only a factor in that one or both of them must accelerate for motion to occur. But the amount of time which passes during acceleration can be negligible in both frames. In fact, you could have the entire Earth-star frame do the accelerating instead of the ship, and it wouldn't make any difference. The twin is younger not because he did the accelerating, but because the beginning and end points of the trip are the Earth and star, and that distance is always longest in the frame of the Earth and star. Do the math, and you'll see. You get exactly the same answer either way.


Dr. Stupid

 

[69dodge]

Originally posted by Stimpson J. Cat
[...]

The point of all this is that the apparent contradiction arises from the problem that the two observers cannot agree on which events occurred simultaneously. They both agree that the start of the experiment occurred simultaneously with the alien's clock saying 2:00, and that the end of the experiment occurred simultaneously with the alien's clock saying 2:06. But the alien says that the beginning occurred simultaneously with the Earth clock saying 12:03.2, and the Earth observer says that the beginning occurred simultaneously with the Earth clock saying 12:00. Likewise they disagree on when the experiment ended in the Earth frame.

Again, there is no acceleration here.

Right. Relativity of simultaneity at a distance.

In the twin paradox, acceleration is not the issue either. In that frame, they can agree on what the clocks both said at the beginning and ending of the experiment, because they are both in the same frames at those times. Acceleration is only a factor in that one or both of them must accelerate for motion to occur. But the amount of time which passes during acceleration can be negligible in both frames. In fact, you could have the entire Earth-star frame do the accelerating instead of the ship, and it wouldn't make any difference. The twin is younger not because he did the accelerating, but because the beginning and end points of the trip are the Earth and star, and that distance is always longest in the frame of the Earth and star. Do the math, and you'll see. You get exactly the same answer either way.

Star? What star? One twin stays on Earth; the other leaves and returns. The trip starts and ends on Earth. There's no distance to mess up simultaneity. Both at the beginning of the trip and at the end, the two twins are in the same place so they can look directly at each other's clocks. Yet their clocks still disagree.

From the point of view of the travelling twin, a great deal of Earth time passes while he's turning around (i.e., accelerating)---more than enough time to make up for all the rest of his trip, during which, again from his point of view, Earth's clocks go slower than his.

 

[Stimpson J. Cat]

69dodge,

Star? What star? One twin stays on Earth; the other leaves and returns. The trip starts and ends on Earth.

Sorry, I should have been more specific. I was referring to the standard formulation of the twin paradox, where the twin travels to some nearby star and back again. Substitute "turning point" for star.

There's no distance to mess up simultaneity. Both at the beginning of the trip and at the end, the two twins are in the same place so they can look directly at each other's clocks. Yet their clocks still disagree.

There is a distance. The distance to whatever point the twin travels to, before he turns around and comes back. In fact, since you can synchronize clocks on the Earth and whatever point in space he turns around at, you can break the experiment into two parts. The first half of the trip takes less time for the twin, as does the second half. When the twin stops (relative to Earth) at the turning point, he can check the clock there which he knows is synchronized (in the frame of the Earth) to the clock on Earth. Sure enough, that clock will say more time has passed than has passed for the twin. The same thing happens on the return half of the trip.

From the point of view of the travelling twin, a great deal of Earth time passes while he's turning around (i.e., accelerating)---more than enough time to make up for all the rest of his trip, during which, again from his point of view, Earth's clocks go slower than his.

This is incorrect. The accelerating and decelerating time can be made as small as we want, so that it accounts for a negligible portion of the total trip time. For example, we could easily construct the experiment so that only 1 second of time passes (in the Earth frame) during all the combined acceleration and deceleration periods. Clearly the time difference is accumulated over the entire trip, during uniform motion, and not during that brief acceleration period.

Again, it doesn't even matter who does the accelerating. Imagine this experiment:

A huge ruler is built in space, which is 4 light minutes long. One twin is on one end of the ruler. The other in a spaceship next to his twin. He takes off for the other end at 0.8c, stops there and turns around to come back, also at 0.8c. Assume that all acceleration times are negligibly small.

From twin A's POV (on the ruler), the trip takes 10 minutes (8 light minutes / 0.8c). From twin B's POV, the total distance traveled by the ruler during the first part is 2.4 light minutes, and then 2.4 more during the second part, giving a total of 4.8 light minutes, for a total trip time of 6 minutes.

But you can switch things around. Leave twin B stationary, and move the ruler instead. You will get exactly the same result. Twin B will still only experience 6 minutes, while twin A experiences 10, even though twin A did the accelerating.

The reason is simple. The distance traveled is always 8 light minutes from the POV of twin A, and always 4.8 light minutes from the POV of twin B, regardless of who is actually doing the accelerating.

The acceleration is only important in the respect that somebody has to accelerate for the experiment to happen. It is not the case that the time discrepancy accumulates during the acceleration period. Of course, if the acceleration period is not negligible, some time difference will accumulate there too, but it does not all accumulate there.


Dr. Stupid

 

[69dodge]

Originally posted by Stimpson J. Cat
But you can switch things around. Leave twin B stationary, and move the ruler instead. You will get exactly the same result. Twin B will still only experience 6 minutes, while twin A experiences 10, even though twin A did the accelerating.

This cannot be right. An observer's proper time depends on whether he's dragging a ruler behind him? What if neither A nor B is attached to a ruler? That's the original question anyway. The only difference between the twins is that one accelerates and one doesn't. It obviously doesn't matter which one we call A and which one we call B.

 

[Stimpson J. Cat]

69Dodge,

This cannot be right. An observer's proper time depends on whether he's dragging a ruler behind him? What if neither A nor B is attached to a ruler? That's the original question anyway. The only difference between the twins is that one accelerates and one doesn't. It obviously doesn't matter which one we call A and which one we call B.

That is not the only difference, nor is it the relevant difference. The relevant difference is what events mark the beginning and end of the trip, and how they are defined.

Note that in the twin paradox, what you really have are two separate trips. One to the destination, and one back. For both of these trips, the events which mark the beginning and ending of the trip are defined to be when the twin leaves Earth, arrives at some destination, leaves that destination, and arrives back at Earth. Both the Earth and the destination locations are in the same inertial frame (we are assuming that they are not moving relative to each other). In the Earth-destination inertial frame, the distance between the Earth and the destination is the largest it will be in any inertial frame. That is why the twin who spends the duration of the trip in the Earth frame experiences a longer passage of time than the other one. It really has nothing to do with who does the accelerating.

You asked what if neither one is attached to the ruler. I am not sure what you mean. In the original thought experiment there is no "ruler", but the Earth-bound twin is still "attached" to the Earth-destination frame, just as twin A is "attached" to the frame of the ruler in my variant. That is why he experiences more time passing.

Do you mean what happens if both twins are moving relative to the Earth-destination frame? If that is the case, then the results are going to be different. The obvious example being the triplets variant of the experiment. In this case, Triplet A stays on the Earth, and triplets B and C each head off in opposite directions at the same speed, and for the same distance, and come back likewise.

In such a case, both B and C will be younger than A, but they will both be the same age as each other in spite of the fact that they were moving relative to each other. In this case the experiment really is symmetric with respect to B and C, so they must experience the same passage of time. Again the relevant issue is that the events which determine the beginning and endings of the trips are defined in the frame of the Earth, so the distance travelled is longest in that frame.


Dr. Stupid