Time Dilation in Twin Paradox: Exploring the Puzzling Reality

[system downs]

You are reffering to time dilation in an inertial reference frame, but the rocket twin is undergoing acceleration and thus is not in an inertial reference frame.

 

[Al68]

I never said your method was incorrect! If you look at post #225 I just agreed with DaleSpam that the method of taking the Earth's time at the turnaround and doubling it only works if you're using the Earth's rest frame. If you are, you do get the right answer. But this doesn't really address the question which is the basis for the twin paradox, which is "why can't you consider things from the traveling twin's perspective instead of the Earth-twin's perspective"? And the answer to that is that you can analyze the problem from the perspective a frame which is different from the Earth's frame, but if you pick an inertial frame then the traveling twin will change speed in this frame and for part of the journey his clock will be running slower than the Earth twin's, whereas if you try to use a coordinate system where the traveling twin is at rest throughout the journey, this is a non-inertial coordinate system so you can't assume the usual time dilation formula still works.

I must have missed something somewhere, I don't see where we disagree. My only point was that the twins paradox could be analyzed by just analyzing two one way trips. Same problem. Same result.

The question "why can't you consider things from the traveling twin's perspective instead of the Earth-twin's perspective?" has the same answer either way.

Al

 

[neopolitan]

I don't know if you are thinking the right thing and just writing it wrong, but no, what you wrote is not correct. First, proper times are frame invariant, so it doesn't make much sense to say "the proper time in a single inertial frame". Second, simultaneity is frame variant, so you need to specify which frame. What you have described here only works in the rest frame of one of the worldlines, in any other rest frame you will have to consider the distance traveled by both worldlines. Third, I don't know if this is general for arbitrary paths or if it only applies for straight worldlines. Finally, it sounds like you are considering scenarios where the twins do not start together and reunite at some other time. If so, then there are 4 events of interest, the two events of each worldline starting and the two events of each worldline ending. Those two starting events must be simultaneous with each other in the rest frame of one worldline as must the two ending events.

What I wrote was this (with emphasis added):

Now, for the purposes of explaining to the student, I would have thought we could say that we know that the square of the proper time in a single inertial frame which is taken to be at rest is equal to the sum of the squares of the proper time and coordinate distance in another inertial frame (or set of inertial frames) - between any two simultaneous events (simultaneous as defined by Einstein).

Initially, I thought about making the red "a" the definite article "the", to keep it to the twin paradox scenario, but then I thought that in reality it doesn't have to be.

As I read my paragraph again, I see I left ambiguity. I did indeed mean four simultaneous events, in two pairs. I should have written "two pairs of simultaneous events". My error. Also, to remove less obvious ambiguity, I should point out that the coordinate distance in the other inertial frame is according to the first mentioned "taken to be at rest" frame, and the simultaneous events are simultaneous in the "taken to be at rest frame". I took these latter two to be obvious, but I shouldn't do that.

So, rephrasing:

Given an inertial frame as our reference, a frame which is taken to be at rest (thus having a coordinate distance of zero), and another inertial frame, or set of contiguous inertial frames, if we consider two pairs of simultaneous events (each pair of events has one event to each frame) and if we use values according to an observer in our reference frame then: we know that the square of the proper time in former inertial frame which is taken to be at rest is equal to the sum of the squares of the proper time and coordinate distance in the other inertial frame (or set of inertial frames).

Then this would work for the twin paradox scenario, with two pairs of events which are actually just two events (departure and return) or pairs of events which are not collocated in spacetime, or a mix (like Al68 has sort of been discussing when breaking the twin paradox into two legs).

Is this less ambiguous and, being less ambiguous, correct?

If correct, could we be less exacting in order to convey the concept to the student, and then slowly build up the understanding of the conditions under which the equation is correct? (Sort of how one would explain a rainbow in steps, without at first talking about suspension of droplets of water of the correct size to refract rays of light, or the relative placement of those droplets, or how the image of the rainbow is formed in the eye and is not something external to our perceptions of it.)

cheers,

neopolitan

 

[Dale]

I'm too tired to untangle your rephrasing.

Given two (straight) worldlines in some inertial coordinate system the spacetime interval of those lines can be written (using my convention above and units where c=1):

St² = Tt² - Xt²
Sh² = Th² - Xh²

If Tt = Th (e.g. the coordinate time between two pairs of simultaneous events) then

St² - Sh² = Xh² - Xt²
or
St² + Xt² = Sh² + Xh²

That is all. Just a little algebra on the spacetime interval formula in the special case where the coordinate times are equal for two straight worldlines.

So what? What is the value? How is it in any way preferable to the more general spacetime interval formula?

 

[neopolitan]

I'm too tired to untangle your rephrasing.

Given two (straight) worldlines in some inertial coordinate system the spacetime interval of those lines can be written (using my convention above and units where c=1):

St² = Tt² - Xt²
Sh² = Th² - Xh²

If Tt = Th (e.g. the coordinate time between two pairs of simultaneous events) then

St² - Sh² = Xh² - Xt²
or
St² + Xt² = Sh² + Xh²

That is all. Just a little algebra on the spacetime interval formula in the special case where the coordinate times are equal for two straight worldlines.

So what? What is the value? How is it in any way preferable to the more general spacetime interval formula?

Basically all I am saying is in the first equation you raised.

St² = Tt² - Xt² (where Xh is zero and Tt = Th, which is all my convoluted phrasing said)

And, yes, it is in the third and fourth equations as well. The point is, I guess, that we are talking about spacetime intervals which are the same, so long as you use a consistent inertial frame. Which gets us back to the original point, if you don't use a consistent inertial frame it doesn't work.

This seems to be the one factor common to all approaches for explaining the twin paradox. If you use a consistent inertial frame it works. If you don't, you need to do something extra (which can be to consider simultaneity, or to note that there is a missing amount of time equal in magnitude to the disagreed distance travelled, which can be calculated as shown in an earlier post, or you can show that the line is bent in a spacetime vector diagram).

I don't think we are getting any further than that, and I don't really think there is any further to go than that.

cheers,

neopolitan

 

[Dale]

This seems to be the one factor common to all approaches for explaining the twin paradox. If you use a consistent inertial frame it works. If you don't, you need to do something extra ...
I don't think we are getting any further than that, and I don't really think there is any further to go than that.

I agree 100% with that. And really, that is the whole point of the relativity postulate: if you use any inertial frame everything works the same. (it even rhymes )