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1. May 28, 2015

### Stephanus

Dear PF Forum,

Supposed T is a star 100 ly from earth.
If B travels to T from earth
A. Is the symmetry broken?
B. If B watches A's clock at Earth and A watches B's clock at T, do they see the other clock late by 100 years, no matter how fast B travels?

Now B travels back to A, and now B brings a companion C, to travel with him, sitting beside B
C. Is the symmetry broken for A and B?
D. If the symmetry is broken for A and B, what about for B and C?
1. Does B sees C age faster along the journey?
2. Does B sees C gesture/act in fast motion?
3. Does C sees B gesture/act in slow motion?

Thanks for the answer, and pardon me for this twins paradox again.

2. May 28, 2015

### Staff: Mentor

Which symmetry?
You can use the equations to calculate it. Or have a look at one of the many examples in other threads and other websites.
Again, which symmetry?
B and C do not move relative to each other, what do you expect? Do you see your neighbor age faster or slower than you?

3. May 28, 2015

### Staff: Mentor

When you say "see" do you mean literally what visual signal they would actually see, or do you mean what would they calculate in their reference frame.

4. May 28, 2015

### Stephanus

If B travels and A stays, but B doesn't come back. The clock symmetry is maintained, right?
But if B changing his inertial frame of reference, the symmetry is broken. Is that so?

5. May 28, 2015

### Stephanus

They ACTUALLY see. Using a very magnificent telescope.
A can see B's clock 100 ly away. And B uses telescope can see A's clock 100 ly away.

6. May 28, 2015

### Staff: Mentor

There is no symmetry at all, that is the point of the twin "paradox".

Then you have to take light travel times into account which makes the analysis a bit more complicated. Did you draw a Minkowski diagram?

7. May 28, 2015

### Staff: Mentor

8. May 28, 2015

### Stephanus

No, I wasn't drawing Minkowski Diagram. I haven't studied it, yet. It's just that they say that Twin Paradox symmetry is not broken when one of the participant travels. But the symmetry is broken is the other participant, changes its inertial frame of reference. That's why I want to get a full grasp of it first, before I ask about Universe Prefered Frame of Reference (again).

9. May 28, 2015

### ash64449

Symmetry is based on that A and B don't play the same role. One keeps changing frames while other does not.

Considering the case of original twin paradox,Here B is changing frames.

Overall, first A and B are in the same frame.But as B travels,B changes frame.

Now for the return trip,B changes the frame again(different from that of the frame which B occupies when B leaves A).

So here B keeps changing frames while A doesn't

10. May 28, 2015

### Staff: Mentor

How is C's clock set to start out? Is it synchronized with clocks on T, or with B's clock? I'll assume C's clock is synchronized with T's clocks to start out. And I'll assume that B and C leave T for Earth when T's clocks read 1000 years, and they travel at 87% of the speed of light, just as B did when traveling from Earth to T.

Then, when B and C start out, they emit a light signal to Earth showing both of their clock readings. Those will be 1000 years (for C) and 942.25 years (for B). When A receives this signal, his clock will read 1100 years.

When B and C arrive on Earth and meet A, B's clock will read 1000 years, C's clock will read 1057.75 years, and A's clock will read 1115.5 years.

11. May 28, 2015

### Stephanus

But "see" here, means that B has stopped at some point. At this example B stops at T. And A and B watches each other clock at the same frame of reference using a very powerful telescope. I think no Doppler effect there.

12. May 28, 2015

### Stephanus

Thanks Ash64449 for the answer. I mean A and B watch each other clock BEFORE B returns.

So B has stopped and (I think) they are at the same frame of reference.
1. Are they seeing the other clock late by 100 years?
2. Is there a doppler effect after B stops?

Thanks

13. May 28, 2015

### Staff: Mentor

They are at rest relative to each other. They are both "in" any frame of reference you choose; they are just not at rest relative to other frames besides the Earth-T frame.

No; the clock offsets they actually observe are different than that. See my post #10.

No.

14. May 28, 2015

### Staff: Mentor

You can still apply the Doppler analysis. It may even be the best/simplest way of analyzing these "what do you see throiugh a telescope" thought experiments because you're tracking the light from the event that is being seen to the telescope and eyes of the person who is doing the seeing when the light reaches them.

While they're moving apart, A sees fewer flashes of light from B and vice versa (if you you don't want to count flashes of light, you can count the number of times that the sweep second hand of the clock passes a tick mark on the face of the clock, or you can count the number of times that the rightmost digit on a digital clock display changes - they'll all yield the same result as you have to wait for the light to travel from the "tick" event to your telescope before you can see it).

When B reaches T and stops, the Doppler effect will go away as you say, so A sees B's clock going back to its old undilated tick rate and vice versa after they've both come to rest relative to one another. It's a good exercise to draw a spacetime diagram showing both the light flashes that A sees as he looks a clock that is at rest relative to him at T (ticks at the same rate as his own, but he sees it consistently 100 years later because of the light travel time) and the clock that B is carrying along his trip.

Last edited: May 28, 2015
15. May 28, 2015

### Stephanus

Thanks PeterDonis for your invaluable help.

Yes, yes. I supposed I forgot to mention some parameters. I tought it goes without saying about twin paradox. Clock synchronization at the begining and travel near the speed of light.

But, why A's clock runs faster than C's clock. From C point of view, it's A who travels, right?
Is it the rocket acceleration? But, they say acceleration has small effect on time dilation. Only changing inertial frame of refence does.

"Arrive on Earth" suggests that B is changing it's frame of reference, right?
"Meet A" suggests that either (B and C) or (A) are moving, but then again motion is relative, right?

16. May 28, 2015

### Stephanus

Thanks Nugatory, I'll do it.

17. May 28, 2015

### Stephanus

If B travels from Earth to T, why B is changing frame? Motion is relatvie, right?
In B point of view, isn't A who changing frame?

18. May 28, 2015

### Staff: Mentor

Whether or not that's true depends on which frame you use, the frame in which A, Earth, and T are always at rest, or the frame in which B and C are at rest while they are moving (but not before they start or after they arrive). Changing frames means changing simultaneity conventions; in the frame in which A, Earth, and T are at rest, B and C start out from T "at the same time" as A's clock reads 1000 years. But in the frame in which B and C are at rest while they are moving, B and C start out from T "at the same time" as A's clock reads 1086.6 years, so in this frame, A's clock only advances by 28.87 years during the journey, half of the 57.75 years that B's and C's clocks advance. This is just standard relativity of simultaneity.

No, it says that B changes his state of motion. Changing state of motion is not the same as "changing frame of reference". Any frame of reference can be used to describe B's entire motion, before and after he arrives on Earth.

No, it says that A's, B's, and C's worldlines all coincide. There is no assumption about who is "moving" and who is not.

19. May 28, 2015

### Staff: Mentor

No, B changes his state of motion relative to A, Earth, and T.

No, B again changes his state of motion relative to A, Earth, and T.

Neither one is "changing frame". They are moving relative to each other. There is no absolute sense in which either one is "moving" or "at rest".

As a general point: thinking of changing relative motion as "changing frames" is just going to cause confusion. It's important to keep inertial frames conceptually distinct from the states of relative motion of various objects. You can describe the entire scenario using just one frame; there is no need to "change frames" when someone changes their state of motion relative to someone else.

20. May 28, 2015

### ash64449

Yes. Both are in same reference now.