# B The Twins "Paradox" Redux?

1. Nov 16, 2016

### Chris Miller

Twins leave earth in opposite directions, accelerate at the same rate until their relative v = ~c, hold this v for 20 years (or however long), then decelerate at the same rate to v=0, and accelerate at the same rate back towards earth until their relative v again =~c, hold for same amount of time, then decelerate at the same rate until they reunite where they started. For the entire journey, wouldn't each have seen the other as aging much more slowly? Are they the same age now? How much time has passed on earth?

2. Nov 16, 2016

### Ibix

They'd be the same age. A triplet at home would be older. Look up Orodruin's Insight on the geometric resolution of the twin paradox.

Edit: one of the learning points of the twin paradox is that time dilation isn't anywhere near the whole story.

3. Nov 16, 2016

### Chris Miller

Thanks, that's what I thought. So even though each twin "saw" the other as aging more slowly for the entire trip, they are the same age when they reunite. It's like their triplet's/earth's frame or reference retroactively reverses their in-trip observations.

4. Nov 16, 2016

### Orodruin

Staff Emeritus
To boil it down: The time dilation formula is based on the assumption of the simultaneity in a particular frame. This notion of simultaneity will differ from frame to frame and will therefore change as an observer accelerates.

Note that time dilation does not describe what an observer actually sees, but what it can deduce happened. In order to find out how it would actually look you would need to take the finite travel time of light into account.

5. Nov 16, 2016

### Ibix

No. Their observations are always consistent with the notion that the travellers will be the same age when they return. The Earth frame does nothing. It's just a choice of coordinates.

6. Nov 16, 2016

### Orodruin

Staff Emeritus
Oh, and the easiest way to find this text is by following the "My Insights" link in my signature.

7. Nov 16, 2016

### rede96

If it were possible for each twin to observe the other for the entire trip then they couldn't see each other as ageing slower for the entire duration. That would be like Twin A watching Twin B's clock running slower for the whole trip and then suddenly at the end of the journey the clock jumps forward to the same time as Twin A

I'm not too sure what twin A would observe if he could watch Twin B's clock for the entire journey but what ever he sees it must symmetrical (e.g. if he sees twin B's clock running slower then at some point he must see it running equally faster.) as the two clocks have to read the same time again at the end of their journeys.

8. Nov 16, 2016

### Bandersnatch

9. Nov 16, 2016

### Chris Miller

Thanks, Orodruin. I get that acceleration impacts t dilation, which is why I kept it symmetrical. I also get that neither can really observe in "real" time the other's clocks (biological or mechanical). But, for the purposes of this thought experiment, can I not assume some sort of instantaneous (say entanglement based) method of observation? In any case, what both deduced was occurring, via SR, clearly was not, or was retroactively undone?

10. Nov 16, 2016

### PeroK

If you are serious about learning SR, you need to knuckle down and study the relativity of simultaneity and the so-called leading clocks lag rule, which lies at the heart of these problems.

Coming up with a pseudo explanation like this is not the real understanding.

11. Nov 16, 2016

### Orodruin

Staff Emeritus
This is exactly what you will find if you compute what the observers will actually see, as mentioned in my previous post. This is not what time dilation is about. Time dilation is about what time coordinates are assigned to different events in a particular frame and its comparison to the proper time.

This is based on the notion of simultaneity in the particular frame and changes when you change reference frame by accelerating.

12. Nov 16, 2016

### Chris Miller

And yet this (each seeing the other's clocks running slower for the entire trip) is exactly what SR predicts they would each observe, which is why it's called a paradox, I guess.

13. Nov 16, 2016

### PeroK

If you are serious about learning SR, you need to knuckle down and study the relativity of simultaneity and the so-called leading clocks lag rule, which lies at the heart of these problems.

Coming up with a pseudo explanation like this is not the real understanding.

14. Nov 16, 2016

### Orodruin

Staff Emeritus
No

Even if you could it would depend on which frame you want it to be instantaneous in. If it is instantaneous in one frame it will not be in another (it will be an FTL communication that could a priori go back in time). This is at the very core of relativity and is called the relativity of simultaneity: what is "at the same time" in one frame is not "at the same time" in another.

Failing to understand the relativity of simultaneity is likely responsible for more than 90 % of the misconceptions beginners have of relativity. Understanding it is crucial if you want to have any chance of understanding SR at a deeper level than popular science.

15. Nov 16, 2016

### PeroK

It's what you think SR predicts, because you only know about time dilation. SR is more than just time dilation.

16. Nov 16, 2016

### jbriggs444

Acceleration does not affect time dilation. Acceleration results in a change of synchronization convention.

17. Nov 16, 2016

### Chris Miller

so acceleration is not the same as a gravity (feels the same though), been wondering

18. Nov 16, 2016

### Chris Miller

I'm not sure I have the grey matter or mathematical chops to understand SR well. It'd be nice not to totally embarrass myself writing SF, though. And, so far, this site's been helpful in that regard.

PS

Aren't entanglement's changes simultaneous across any distance? And so FTL?

19. Nov 16, 2016

### jbriggs444

No. You continue to misunderstand.

Gravity is the same as an acceleration. That's the equivalence principle. But gravitational acceleration is NOT associated with time dilation. Gravitational potential is. The distinction is that potential is a combination of an acceleration and a distance.

Accelerations in special relativity can be interpreted as an incremental change of reference frame and an associated incremental change of clock synchronization. The amount of clock change scales with distance. A mere acceleration is not enough. You need an acceleration and a remote clock some distance away.

20. Nov 16, 2016

### Mister T

It's not a real paradox, it's an apparent paradox. This is why you often see it referred to as the Twin Trip or some such, as it really doesn't deserve the name Twin Paradox. A less complex apparent paradox is the symmetry of time dilation, which really ought to be resolved before tackling the twin paradox as glossing over it means you'll forever find the twin paradox baffling.

But to address the issue you raise, imagine each twin with a flashing strobe light. Let's call these strobe lights clocks, as they send out the flashes at regular intervals, say once per minute. When each is at rest relative to the other, the rate at which the flashes are seen by one twin is equal to the rate at which they are sent by the other. But if there is relative motion between the twins, that will not be the case. When moving apart each will see the other's flashes as slow, but when they approach each other, each will see the other's as fast. Now this speeding up and slowing down of clock rates is due only partially to time dilation, the other part being due to light travel time. So each will see the other's clock as running fast or slow depending on whether they are approaching or receding, but if you subtract off the part of the effect due to light travel time, then each will observe that the other's clock is running slow, regardless of whether they are approaching or receding.

So what you see is different from what you observe.

Anyway, if you go through the details you will find that after the twins reunite to compare notes, the number of flashes that each sent will equal the number of flashes that the other received. But they may not agree on the time that elapsed between the flashes, and when the total time between flashes is added up, they may get different sums.

Now, if the journeys are symmetrical in the way you propose, the sums will match. But a triplet that stayed home will get a larger sum.

There's a hokey old cartoon that can you find by searching YouTube for "Paul Hewitt Twin Trip". The physics is solid.

21. Nov 16, 2016

### Staff: Mentor

No, this is not at all what SR predicts. Not only is it not what they visually observe, there is also no reference frame where that is what they calculate to be true.

What they would observe is the other twins clock being redshifted during the first part, normal in the middle, then blueshifted during the last part. They end with the same elapsed time.

In the earth's frame they are always equal. They end with the same elapsed time.

In an inertial frame where one twin is initially at rest then the other will start slow, but the first will end slow. They end with the same elapsed time.

In a non inertial frame the math is complicated. They end with the same elapsed time.

No matter what frame you pick, if you actually do the math... They end with the same time.

No, it is nothing like that at all.

Last edited: Nov 16, 2016
22. Nov 17, 2016

### Chris Miller

That's why I quoted it in my question.

This, and your entire explanation was extremely helpful/interesting (even the cartoon!). Although I still don't understand how, if each (after discounting light travel time) sees the other's clock running slow for the entire journey (as SR would predict) how their (discounted) sums match at the end.

23. Nov 17, 2016

### jbriggs444

Read up on the twin's paradox. The explanation that I gravitate to is that the accounting from the travelling twin's point of view skips part of the aging process of the stay-at-home twin. At turnaround his notion of what age his twin is "right now" changes. He matches up the first bit of his twin's aging to his outbound trip. He matches up the last bit of his twin's aging to his return trip and never accounts for the part in between.

24. Nov 17, 2016

### robphy

Here's a diagram based on my Relativity on Rotated Graph Paper Insight.
(You can use https://www.geogebra.org/m/HYD7hB9v to draw parts of the diagram below.)

In the frame of the stay-at-home twin (on which the gridlines of the rotated graph paper are based),
we draw the ticks for each "traveling" twin (each with speed (3/5)c).
Then, I show the light signals sent by the "positive-x" twin... Think of it as the "negative-x" twin watching an 8 hour tv show broadcast by the "positive-x" twin.
Due to their relative motions, the viewer sees that show with irregular timing.
(You can imagine that each has the same 8-hr movie started at separation...and that one twin is comparing the received broadcast with her own local playback of the movie.)

The ticks of the stay-at-home twin are not shaded in... but you can shade them in to realize that
at reunion, 10 ticks elapse for the stay-at-home twin while 8 ticks elapsed for each of these symmetrically-traveling traveling-twins.

You could also analyze what each (traveling or stay-at-home) twin sees from the other two twins.
You could also try asymetrical twins, as well as trips with asymmetrical there and back legs.

(One could try to interpret the situation using light-travel time corrections and change of lines of simultaneity, etc... but I feel that this presentation is direct and to the point.)

Last edited: Nov 17, 2016
25. Nov 17, 2016

### Staff: Mentor

The quick answer is that special relativity does not predict that each one SEES the other's clock running slow . We've already discussed what the twins SEE at some length, and there's also a good explanation of what they do SEE in in the "Doppler analysis" section of the twin paradox FAQ.

If you haven't already read that FAQ in its entirety, do so now. It covers the asymmetric case in which one twin stays on earth while the other flies out and back, but until you clearly understand that case you won't be ready to take on the symmetric case.