Confused about Velocity Based Time Dilation

[Salbris]

I've been twisting and turning my brain to figure this out but I need some explanation:

Following the twin paradox, I've learned that aging does occur, but I thought that time dilation was relative to the observer and the 'observee'? So if aging does occur then doesn't that mean that there is some absolute rest frame? However, I've only ever seen people discount any possibility of some absolute frame.

- Thanks in advance.

 

[A.T.]

Following the twin paradox, I've learned that aging does occur, but I thought that time dilation was relative to the observer and the 'observee'? So if aging does occur then doesn't that mean that there is some absolute rest frame?

And which frame would that be, since all inertial frames come to the same result about the age difference? Note that one of the twins is not inertial, but you can construct his view from two inertial frames (leave & return):
http://www.adamtoons.de/physics/twins.swf

However, I've only ever seen people discount any possibility of some absolute frame.

You cannot discount any possibility of it, but it is not needed to explain the age difference.

 

[Salbris]

Sorry, I'm stilling trying to understand this. How does the Twin B (the one that left the Earth) eventually measure aging in himself?

 

[A.T.]

How does the Twin B (the one that left the Earth) eventually measure aging in himself?

With a clock traveling with him.

Or did you mean how he computes the age difference? He can pick any inertial frame (unfortunately not his rest frame) and apply the SR formulas to the world lines there.

 

[Salbris]

With a clock traveling with him.

Or did you mean how he computes the age difference? He can pick any inertial frame (unfortunately not his rest frame) and apply the SR formulas to the world lines there.

And he will compute that he ages slower than Twin A?


Also, what about the case of two clocks in two different cases with a third clock at the start point:

1. Both clocks move at a velocity away from the third clock and each other at the same speed.

2. One clock stays with the third clock, while another goes at double the speed as before away from the other two.


Won't the clocks measure aging on each other as they move away from each other? How could they ever know that in the second case that the clock that moved would age relative the third but the other doesn't?

 

[A.T.]

And he will compute that he ages slower than Twin A?

He will compute that he will have aged less after the entire trip. Not necessarily that he ages slower at every time point during the entire trip.

Also, what about the case of two clocks in two different cases with a third clock at the start point.

In your examples the clocks never come back to compare passed times? The situation is symmetrical then. Everyone sees the other age slower.

 

[granpa]

its all about relativity of simultaneity. that's what confuses all beginners

 

[Salbris]

In your examples the clocks never come back to compare passed times? The situation is symmetrical then. Everyone sees the other age slower.

But that's not what I'm asking... Although you explained something else which is important, so the observer of both clocks notes the other as aging slower, but in actuality doesn't the clock that stayed with the third clock in the 2nd example actual age faster than the one that moved away?

 

[sylas]

But that's not what I'm asking... Although you explained something else which is important, so the observer of both clocks notes the other as aging slower, but in actuality doesn't the clock that stayed with the third clock in the 2nd example actual age faster than the one that moved away?

No. There is no "actuality" which let's you make an absolute determination of which observer is "actually" correct. They are both correct, and they are measuring for two different inertial frames. Neither frame is better than the other.

Cheers -- sylas

 

[Salbris]

No. There is no "actuality" which let's you make an absolute determination of which observer is "actually" correct. They are both correct, and they are measuring for two different inertial frames. Neither frame is better than the other.

Cheers -- sylas

But when the other returns, they get confirmation that, yes indeed he was aging slower, no?

 

[sylas]

But when the other returns, they get confirmation that, yes indeed he was aging slower, no?

Not really. You see, in order to return, the other has to be in two different inertial frames. So you can't compare which one is running more slowly at a particular time... there isn't a common notion of the particular instants that are simultaneous at the two clocks through the entire trip.

You can certainly talk about the elapsed time of any clock.

Here's my favourite example again. Suppose there is a star 6 light years distant. Clock A remains inertial, 6 light years away from that star.

Clock B moves to the star at 60% light speed, and then turns around and comes back. They are able to turn around very quickly, so that they are pretty much moving away from home at one instant, and then turned around to come back in the next instant.

The whole trip of clock B takes 20 years, as measured by clock A. And, from the point of view of clock A, the moving clock is running 80% slow the whole time, and returns showing 16 years.

Now... what does clock A experience? From their perspective, clock B is the one that is moving, and there is a star moving towards them, at 60% light speed.

But the distance is less. The star is 4.8 light years away in this frame, and it takes 8 years to reach clock B. When the star reaches clock B, clock B can see clock A still receding in the distance. The light they see is coming from 3 light years away, and so left clock A 3 years previously... that is, 5 years after clock A started moving away. And clock A is running 80% slow, from clock B's perspective, so clock A is showing 4 years as seen in the light coming to clock B.

Clock B can infer, of course, that clock A will be "now" another 1.8 light years on its way, 4.8 light years altogether; and is presumable "now" showing 6.4 years.

But then, something odd happens. The star suddenly reverses and flies away again at 60% light speed. The distant clock A is suddenly approaching again, at 60% light speed. Worse... this all takes place at once! The light with which clock B sees clock A is now coming from 12 light years distant; and hence (since the clock A is approaching) it is inferred to be "now" 6.4 light years distant; though what clock B actually sees is the older light. And, the clock A is still indicating the 4 years with the light used to see it.

To come 12 light years at 60% light speed takes 20 years... although because clock A is running slow, there will be another 16 years elapsed, on top of the 4 seen in the light from that clock.

When clock A arrives and clock B, 8 years later, it shows 20 years elapsed, as expected.

In this case, there's something very odd that takes place half way through the trip. What happens is that clock B turns around... moves into a new inertial frame. In that frame, everything changes. In an instant, the same light that was coming from 3 light years away is suddenly coming from 12 light years away... and that is correct! The size of clock A in the sky will also suddenly reduce, because it is suddenly so much further away in this new perspective.

There's more discussion, with spacetime diagrams, in [post=2199430]msg #50[/post] of "Twin Paradox- a quick(ish) question"; and in other posts of the thread.

Cheers -- sylas