# I have no wish to reignite the twins debate but..

1. Jan 4, 2016

### fxfocus

Hi, I'm a pensioner with little formal education, still struggling with SR and with no knowledge at all of GR.

When people ask why the twins age differently in a situation which appears symmetrical depending on which frame of reference you use; I see answers along the lines of...

Only the travelling twin experiences a change of reference or acceleration.

Which I accept, but...

What I haven't seen is any reference to the fact that from either perspective, only the travelling twin experiences length contraction, which shortens the length and thus duration of the journey?

Have I got that wrong?

2. Jan 4, 2016

### Staff: Mentor

No, this is not correct. From the traveling twin's perspective, the stay-at-home twin is moving, and the distance through which he is moving is length contracted.

I strongly suggest reading the Usenet Physics FAQ article on the twin paradox:

3. Jan 4, 2016

### phinds

Length contraction is irrelevant to the twin "paradox". All that matters is that you realize that they have taken different world lines and one of those world lines has more clock ticks than the other.

4. Jan 4, 2016

### Ibix

Nobody experiences length contraction or time dilation. It is, genuinely, something that always happens to other people. So, from the stay-at-home's frame, the rocket is contracted and time-dilated but neither the planets nor the space between them are contracted. Meanwhile, from the traveller's frame, the rocket is not contracted but the planets and the space between them are.

It turns out that the elapsed time on a path through spacetime is equal to a quantity called the "interval", which is closely analogous to the length of the path. The reason the twins are different ages is that they followed paths of different "lengths". At least one path must include a bend (acceleration) if the two are to meet again, which is where the "acceleration causes the difference" story comes from. It works in this example, but the path length explanation is more general.

5. Jan 4, 2016

### fxfocus

It's really the length of the path I'm trying to get at. It seems to me that the travelling twin takes the shorter path due (at least in part) to length contraction. Let's say we place a pebble at the turn around point. This pebble would always be at rest with respect to the stay at home twin but whether the travelling twin moves to the pebble or the pebble moves to the travelling twin only the travelling twin would perceive the journey to be shorter.

Fully expecting and happy to be told I'm wrong but just can't yet see why?

6. Jan 4, 2016

### Ibix

That is the traveller's explanation for why her clocks show less elapsed time than a Newtonian prediction. That isn't the paradox, however. The paradox is that, in the traveller's frame(s), the Earth is moving so its clocks tick slower, so surely the Earthbound twin should be younger when they meet up.

One resolution to this is to note that when the traveller changes direction she changes her definition of "now, on Earth" in a way that accounts for the error. The other is to use the interval argument I made in my last post.

7. Jan 4, 2016

### pervect

Staff Emeritus
Yes and no.

You're wrong if you think that the situation is symmetrical. Without using GR, we need two inertial frames of reference to describe the travelling twin, not one. Two and one are not "symmetrical".

You're mostly right when you point out that length contraction and time dilation combine to give a consistent picture where they don't make a consistent picture alone. You're mostly right instead of "spot on" because there's a third effect that you haven't taken into account yet, the relativity of simultaneity.

The notion of proper distance and proper time would be helpful in your understanding. The proper distance between two points is equal to the distance between two points in a frame of reference where they are both not moving. The distance between the two points in either of the two moving frames for the travelling twin is length contracted when compared to the proper distance. "Proper time", which can be throught of as "wristwatch time" is time that is measured along a worldline by a single clock - it's one of the most basic notions of time, as it does not require any notion of a way to "synchoronize" clocks. It does require you to specify the path of the particular clock that measures it.

In terms of proper time, the point is that the sum of the proper times of the traveller heading out and heading back in (this time is the sum of two proper times in two different inertial frames) is lower than the proper time of the traveller who stays put.

Good luck - I'm afraid that you'll find that re-opening this discusion yields the usual results - there isn't that much advantage to opening a new discussion when there's zillions of old ones, the same thing tends to keep happening.

8. Jan 4, 2016

### fxfocus

Apologies to all. I should have made it clearer my objective was not to resurrect a discussion of the twin paradox but to simply verify whether or not the stay at home twin would measure a longer distance to the turnaround point than would the travelling twin when viewed from either perspective. By which I mean whichever frame of reference you consider to be moving. I'm assuming if we marked the turn around point with a pebble then this pebble would always be at rest with respect to the stay at home twin.

The reason for asking is that I was under the impression for a long time that the twins scenario was symmetrical (excluding turn around) but now I'm thinking the pebble (or turn around point) breaks the symmetry because it can be considered attached to the stay at home twin.

Hope this helps, if not I promise not to labour the point further ;o)

Edit: I should have said that for this purpose I'm not taking into account any acceleration associated with the turn around.

Last edited: Jan 4, 2016
9. Jan 4, 2016

### Staff: Mentor

If you assume this, then yes, the distance from the stay-at-home twin to the turnaround point will be larger in the stay-at-home twin's frame than it will be in the traveling twin's frame. But that asymmetry is because you made an asymmetric assumption: you assumed that the turnaround point is at rest relative to the stay-at-home twin.

Suppose, instead, that we assume that the traveling twin himself marks the "turnaround point" by firing his rockets, and that the way we will measure the distance from the turnaround point to the stay-at-home twin is to use a pebble that is moving at the same velocity, relative to the stay-at-home twin, as the traveling twin on his outbound leg, and which is just passing the stay-at-home twin at the instant, in the traveling twin's frame on his outbound leg, that the traveling twin reaches the turnaround point. The distance between this pebble and the traveling twin, in the traveling twin's frame, will be larger than the distance between them in the stay-at-home twin's frame (which he can measure by simply measuring the time from when the traveling twin leaves to when the pebble passes him, and multiplying by the known speed of the two relative to him); so in this frame, it is the stay-at-home twin whose distance to the turnaround point is "length contracted".

The key point to take away from all this is that length contraction, time dilation, and relativity of simultaneity are all frame-dependent, so you have to be careful trying to reason about them, because you can make them appear to change by just changing which frame you pick. But changing which frame you pick can't change any of the actual physics. The actual physics is contained in invariants--things that don't change when you change frames. For example, the fact that, when the traveling twin returns, he has aged less than the stay-at-home twin, is an invariant. But things like the "distance to the turnaround point" are not.

10. Jan 4, 2016

### fxfocus

Thanks for that. I think it was when I first realised the turnaround point was at rest relative to the stay at home twin that the penny dropped for me. I understand you might construct scenarios where that isn't the case but I had not considered those. Many scenarios have the travelling twin visiting a distant planet before returning and perhaps that's the example I should have used.

So If I read your answer right, then just the length contraction alone would result in the travelling twin being younger even if we discounted the acceleration effects?

I think the most important thing I have learnt from this discussion is to spend some time carefully phrasing my questions in future. ;o)

11. Jan 4, 2016

### Staff: Mentor

No. Length contraction is best understood as an effect, not a cause. (The same is true for time dilation and relativity of simultaneity.) The cause of the traveling twin being younger when he returns is that he has followed a different path through spacetime, one which has a shorter elapsed time than that of the stay-at-home twin. In other words, it's the geometry of spacetime. Length contraction, time dilation, and relativity of simultaneity are just particular manifestations of the geometry of spacetime; in themselves they don't cause anything.

12. Jan 5, 2016

### fxfocus

Thanks for that. Seems I just keep asking the wrong questions so I'll stop now.

I am left with the impression that excluding the acceleration effects the travelling twin will still be younger in the scenario where the pebble is fixed.

Thanks to all who contributed to this thread.

13. Jan 5, 2016

### fxfocus

Marvellous isn't it..!

Within minutes of my last post I came across a post on another forum which I think elucidates my thinking much better than my clumsy attempts here.

I'm not sure if I'm allowed to place links to other forums so I'll take the liberty of quoting the post...

Edit: Which I think pretty much ties in with my original post )

Last edited: Jan 5, 2016
14. Jan 5, 2016

### Mister T

In addition to the pebble you've got at the turn-around point, imagine a stone tethered to the rocket and trailing behind it.

Earth frame: Pebble-Earth distance is fixed and equal to $L$.

Rocket frame: Stone-rocket distance is fixed and equal to $L$.

People in the Earth frame will observe the stone-rocket distance contracted. People in the rocket frame will observe the pebble-Earth distance contracted.

15. Jan 5, 2016

### fxfocus

I don't see the rocket at rest with respect to the stone. From either perspective the rocket must move between earth and the stone, otherwise (I think) you are comparing apples with oranges.

Edit: Woops..! I misread your post and confused the stone with the pebble. Your absolutely right of course.

Last edited: Jan 5, 2016