What Causes Time Dilation in the Twin Paradox?

[Dale]

Hi JesseM,

OK, here is one last attempt.

My objection to simultaneity-based resolutions is not a technical one, it is a pedagogical one. I understand the simultaneity explanations and their validity. However, it seems that you do not accept the validity or completeness of the spacetime geometric explanation. If the spacetime geometric explanation is valid and complete then, IMO, it is preferable pedagogically for several reasons:

1) it avoids explicit simultaneity issues which is the most difficult concept for students to learn
2) it allows an opportunity to teach students how to identify and avoid non-inertial coordinate systems
3) it is generally applicable including non-inertial coordinate systems, curved spacetimes, and arbitrary worldlines
4) it reinforces Minkowski geometry and the modern way to think about relativity

OK, having explained my motivation, here is how I would present the spacetime geometric approach to the twin paradox to students.

First, I would draw the spacetime diagram in the stay-at-home twin's rest frame and calculate the spacetime intervals for both twin's paths. I would spend some time talking about Minkowski geometry and how (in contrast to Euclidean geometry) the longest timelike interval between two events is a striaght line.

Second, I would talk about the traveler's point of view and draw the "mirror" spacetime diagram where the traveling twin has a vertical worldline. I would calculate the spacetime intervals and obtain the twin paradox. I would point out that the reference frame that I drew was a non inertial reference frame and mention that the usual laws, including the formula for the spacetime interval, only apply for inertial frames. At that point I would expect a rather lengthy discussion about non-inertial frames including physical features and geometrical features.

Third, I would Lorentz transform the original spacetime diagram into the inertial frame where the traveler was at rest during the first leg, and the inertial frame where the traveler was at rest during the second leg. I would show that the conclusion is the same in each inertial frame. I would then emphasize the point that the special theory of relativity says that the laws of physics are the same in all inertial frames, but not non-inertial frames. I would further mention that there is no inertial frame where the traveler is at rest the whole time.

If a student wanted to do the "two inertial frames and add them up" approach I would ask them to work the problem. If they correctly transform from the "first leg" frame to the "last leg" frame then they will get the correct answer, no paradox. If they do not correctly do the transform then I would point out how they are accidentally using the non-inertial frame described above.

 

[neopolitan]

Hi JesseM,

OK, here is one last attempt.

My objection to simultaneity-based resolutions is not a technical one, it is a pedagogical one. I understand the simultaneity explanations and their validity. However, it seems that you do not accept the validity or completeness of the spacetime geometric explanation. If the spacetime geometric explanation is valid and complete then, IMO, it is preferable pedagogically for several reasons:

1) it avoids explicit simultaneity issues which is the most difficult concept for students to learn
2) it allows an opportunity to teach students how to identify and avoid non-inertial coordinate systems
3) it is generally applicable including non-inertial coordinate systems, curved spacetimes, and arbitrary worldlines
4) it reinforces Minkowski geometry and the modern way to think about relativity

OK, having explained my motivation, here is how I would present the spacetime geometric approach to the twin paradox to students.

First, I would draw the spacetime diagram in the stay-at-home twin's rest frame and calculate the spacetime intervals for both twin's paths. I would spend some time talking about Minkowski geometry and how (in contrast to Euclidean geometry) the longest timelike interval between two events is a striaght line.

Second, I would talk about the traveler's point of view and draw the "mirror" spacetime diagram where the traveling twin has a vertical worldline. I would calculate the spacetime intervals and obtain the twin paradox. I would point out that the reference frame that I drew was a non inertial reference frame and mention that the usual laws, including the formula for the spacetime interval, only apply for inertial frames. At that point I would expect a rather lengthy discussion about non-inertial frames including physical features and geometrical features.

Third, I would Lorentz transform the original spacetime diagram into the inertial frame where the traveler was at rest during the first leg, and the inertial frame where the traveler was at rest during the second leg. I would show that the conclusion is the same in each inertial frame. I would then emphasize the point that the special theory of relativity says that the laws of physics are the same in all inertial frames, but not non-inertial frames. I would further mention that there is no inertial frame where the traveler is at rest the whole time.

If a student wanted to do the "two inertial frames and add them up" approach I would ask them to work the problem. If they correctly transform from the "first leg" frame to the "last leg" frame then they will get the correct answer, no paradox. If they do not correctly do the transform then I would point out how they are accidentally using the non-inertial frame described above.

Very good post, Dale.

The only thing I would add is that it might be worth discussing the fact that:

(total time elapsed for stayathome)² = (total time elapsed for traveller)² plus (total distance traveled by traveller, according to stayathome)²

or, alternatively ...

minus (total time elapsed for stayathome)² plus (total distance traveled by traveller, according to stayathome)² = minus (total time elapsed for traveller)²

There are two reasons for doing this, one, because it is true and, two, because some numerically minded people like myself will work this out anyway, and they should be taught the correct interpretation of this.

That may be S² = -t² + x² ... etc, I am not totally sure what the "correct interpretation" is.

cheers,

neopolitan

PS I just want to clarify that:
"total distance traveled by traveller, according to stayathome" = velocity of separation x time elapsed according to stayathome

In Fredrik's original diagram that is: v = 0.8c, t = 40 years and distance traveled = 32 lightyears.

I think that if you do it the wrong way, then you end up with an incorrect equation (which would work only if the stayathome twin ages 14.4 years). Due to treating a non-inertial frame as an inertial frame. I need to take some time to jot down figures, but not right now.
PPS Jotted down the figures, worked it out and, lo and behold, you get the missing years. That is (in Fredrik's example):

minus (total time elapsed for traveller)² plus (total distance traveled by stayathome, according to traveller)²
= - (24²) + (0.8 x 24)² = -655.63 = - (25.6²)

Add the 14.4 years, then you have 40 years.

How can that be interpreted? or is it a huge coincidence?

 

[phyti]

Referring to the drawing, A moves vertically while B moves along t1 with a spatial
axis of simultaneous events s1. As A reverses for the return leg, his s-axis
rotates to the position s2. In the real world this rotation would occur for a
finite time interval, and B would perceive A's rate of activity speed up. For the
purpose of resolving the twin aging issue, it is instantaneous, and thus referred
to as a 'jump', misleading but still accounting for A's time between the two
positions of B's s-axis. The reason for this approach being, the aging issue can be
explained within the framework of SR in terms of time dilation. Even though
acceleration is required to change time lines, it can be arbitrarily small in
comparison to the total distance, and not significantly alter the results.

As an alternate approach, B could synchronize his clock with A while moving past A,
a third person C could synchronize his clock with B at the reversal point, while
moving on the return leg, i.e., electronic communication of times and uniform
constant motion for all involved.

The purpose of the solution of the problem is to demonstrate the relationship of
time rate with path taken. If the path between two events is longer, the speed must
be greater, therefore the time is shorter because of time dilation (a function of v/c).

 

[matheinste]

Hello phyti.

I agree with all that you say and do not have any problem with the resolution of the twin paradox in its various forms. The problem that some people will have to your third person addition to the scenario is that at the event where B and C meet or cross, what B and c will observe will be different, as they are in different inertial frames. C will of course observe what B would have observed immediately after the turnaround event. This is in no way a criticism of what you have said but in the light of objections earlier in this thread i think that this will be an objection from those who do not like the idea of the consequences of a very fast turnaround ( a rapid ageing of A ), albeit not discontinuous, in the first place.

Matheinste

 

[boysherpa]

General Question re Twin Paradox

I read through this thread - interesting. Nice discussions.

Please note that I am not confused by this discussion, nor am I attempting to disprove it. I am fully aware of its implications, as in how the GPS satellite system corrects for relativistic effects to provide clock data. I simply am trying to understand what is meant by time in the context of the discussion. Specifically, is a separate time "dimension" (for lack of a better term) necessary to the discussion, or is time merely a convenience, and is that what creates the seeming "paradox".

Let's use Einstein's ruler example. Suppose we wish to measure space, but the ruler, being in space, is coupled to the thing being measured. Can its measurement be taken as able to reveal a change in that thing? What is the objective, incorruptible measurement device, and what is the thing actually being measured?

Well, to me, time is what clocks measure. So, I don't know what it means to say time flows differently in the two reference frames. Which observer determines this, or is it only inferred by the twins when they meet up? The latter seems to be the case.

I think the clock with the traveling twin has no choice but to tic differently, not because it measures a different time, but because its operation is dilated. The human body might be considered a clock, with chemical and telomeric tics, and thus travel-twin experience dilated "age". But does this mean there is a time which is slowed? I don't think this is proven on spacetime diagrams, i think it is weakly inferred but not substantiated.

But then, I could be full of beans.

 

[Fredrik]

I simply am trying to understand what is meant by time in the context of the discussion. Specifically, is a separate time "dimension" (for lack of a better term) necessary to the discussion, or is time merely a convenience, and is that what creates the seeming "paradox".

You could draw spacetime diagrams for things that happen in non-relativistic mechanics too. If we do, the time "dimension" is just a convenience. Everyone agrees which "slices" of this spacetime should be thought of as space. This is however not the case in SR.

Suppose we wish to measure space, but the ruler, being in space...

Your question is already ill-defined. In SR, what "slice" of spacetime you would consider space depends on your velocity. What's "space" to you is a mixture of space and time to me.

Let's be a bit more precise: If we both assign spatial and temporal coordinates to events using a grid of rulers and synchronized clocks (your grid being at rest relative to you, and mine being at rest relative to me), and your grid assigns the same time coordinate to events X and Y, my grid would assign different time coordinates to them, unless we're moving at the same velocity.

Well, to me, time is what clocks measure. So, I don't know what it means to say time flows differently in the two reference frames. Which observer determines this, or is it only inferred by the twins when they meet up? The latter seems to be the case.

What clocks measure is called "proper time". That's a coordinate-independent concept, but it's only well-defined along some curves in spacetime. No one says that time "flows" differently in two frames, or that time "flows" at all. What we're saying is that if X and Y are two specific events, with time coordinates t_X and t_Y in one inertial frame, and time coordinates t'_X and t'_Y in another, t_X-t_Y is usually not equal to t'_X-t'_Y. This is where SR differs from pre-relativistic physics, where we would always have t_X-t_Y=t'_X-t'_Y.

I think the clock with the traveling twin has no choice but to tic differently, not because it measures a different time, but because its operation is dilated. The human body might be considered a clock, with chemical and telomeric tics, and thus travel-twin experience dilated "age". But does this mean there is a time which is slowed? I don't think this is proven on spacetime diagrams, i think it is weakly inferred but not substantiated.

The traveling twin will be younger when they meet again because what a clock measures is the integral of \sqrt{dt^2-dx^2} along the curve that represents its motion. (dx is =0 everywhere along A's world line, but not along B's).

 

[Dale]

The only thing I would add is that it might be worth discussing the fact that:

(total time elapsed for stayathome)² = (total time elapsed for traveller)² plus (total distance traveled by traveller, according to stayathome)²

Hi neopolitan,

Sorry about the delay, finally had some time to consider this, and it seems to me that your formula can be derived from the spacetime intervals in the stay-at-home frame. I will use S for spacetime interval (or proper time), T for coordinate time, and X for coordinate distance. A following h is for the stay-at-home twin and a following t is for the traveling twin. And units where c=1.

Sh² = Th² - Xh²
St² = Tt² - Xt²

Since Xh = 0 and Th = Tt we can subtract the two equations and get

Sh² - St² = Xt²
or
Sh² = St² + Xt²

Which is what you have written above.

 

[neopolitan]

Hi neopolitan,

Sorry about the delay, finally had some time to consider this, and it seems to me that your formula can be derived from the spacetime intervals in the stay-at-home frame. I will use S for spacetime interval (or proper time), T for coordinate time, and X for coordinate distance. A following h is for the stay-at-home twin and a following t is for the traveling twin. And units where c=1.

Sh² = Th² - Xh²
St² = Tt² - Xt²

Since Xh = 0 and Th = Tt we can subtract the two equations and get

Sh² - St² = Xt²
or
Sh² = St² + Xt²

Which is what you have written above.

I certainly get "Xh = 0", but I am a bit lost with "Th = Tt" since I am trying to think about a time interval about which "A" and "B" will agree, and I can't come up with one. According to wikipedia ...

In special relativity, the coordinate time (relative to an inertial observer) at an event is the proper time measured by a clock that is at the same location as the event, that is stationary relative to the observer and that has been synchronised to the observer's clock using the Einstein synchronisation convention.

http://en.wikipedia.org/wiki/Coordinate_time"

Wrt to spacetime intervals, which are the basis of the S² equation, wikipedia's entry doesn't really lend itself to pasting here, but http://en.wikipedia.org/wiki/Spacetime_interval#Space-time_intervals" is the link.

I am not saying you are wrong, I follow all the logic and agree with it, but I can't quite get my head about one of your antecedants.

Then we get to the difficult part of interpreting the result of subtracting one equation from another to end up with the result. Is the method you present actually useful, pedagogically?

I was sort of hoping to have some useful interpretation of the numbers.

My shot from the hip, a while back, was a sort of economic approach. I'll rehash it for new readers.

You could think of it as if "A" and "B" have a certain amount of spacetime currency. This currency is enough to "pay" for the travel from one event in spacetime to another event. Say further that if both "A" and "B" stay at rest relative to each other, then these events share the same spatial location, that is they only differ in the time axis.

"A" decides to do just that and exchanges spacetime currency for the maximum amount of time currency, no space currency. "B", on the other hand, decides to do some traveling and exchanges spacetime currency for a mix of time currency and space currency.

The total amounts, however, add up to the same - using the equation we are discussing (using Dale's version):

Sh^² = St^² + Xt^²

Note that it doesn't matter a jot what sort of accelerations or manoeuvres "B" carries out - so long as we accept that Xt here is a simplification and "B" is not restricted to a single axis.

cheers,

neopolitan

PS I need to clarify something - the idea that "A" gets all time currency should be understood to be exchanging to what is, in "A"'s frame, purely time currency. Another observer may see that "A" was never really stationary at all, and had some space currency (ie an inertial velocity). This would be the same as the third observer being away that "A" has a little nest egg which is never counted by "A" in everyday financial considerations. The third observer will note that "B" would also have the same little nest egg, and the calculations would end up like this:

Sh^² + (little nest egg to a third observer)^² = St^² + Xt^² + (little nest egg according to a third observer)^²

Sorry about being pedantic, but it helps to cover yourself sometimes

 

[Dale]

I certainly get "Xh = 0", but I am a bit lost with "Th = Tt" since I am trying to think about a time interval about which "A" and "B" will agree, and I can't come up with one.

They don't agree on any coordinate time interval (T), they only agree on proper times (S).

Th and Tt are the time between the "departure" and "reunion" events in the stay-at-home twin's frame. None of the stuff in your formula or in my derivation of it has any reference to any measurement made in any frame other than the stay-at-home twin's frame.

 

[neopolitan]

They don't agree on any coordinate time interval (T), they only agree on proper times (S).

Th and Tt are the time between the "departure" and "reunion" events in the stay-at-home twin's frame. None of the stuff in your formula or in my derivation of it has any reference to any measurement made in any frame other than the stay-at-home twin's frame.

Ok.

Given that all measurements are made in the stayathome twin's frame, where is the pedagogical advantage in ignoring that, under the circumstances, Sh = Tt = Th and so then substituting we can arrive at:

St² = Sh² - Xt²

and so

Sh² = St² + Xt²

cheers,

neopolitan

 

[Dale]

Ok.

Given that all measurements are made in the stayathome twin's frame, where is the pedagogical advantage in ignoring that, under the circumstances, Sh = Tt = Th and so then substituting we can arrive at:

St² = Sh² - Xt²

and so

Sh² = St² + Xt²

Well, if you do that then the first equation reduces to the trivial 0=0. But in general I don't think there is a pedagogical advantage. I would just stick with the standard expression for the spacetime interval.

 

[neopolitan]

Well, if you do that then the first equation reduces to the trivial 0=0. But in general I don't think there is a pedagogical advantage. I would just stick with the standard expression for the spacetime interval.

Which first equation? Do you mean ... Sh² = Th² - Xh²

In this situation, what advantage is there in discussing Th and Tt at all? You have from experimentation (in the mind experiment at least) the values of Sh, St and Xh.

Additionally, you have the concept that Th, Tt and Sh are all equal from the perspective of the person doing the maths. Why go through the process of discussing Th and Tt, when you don't need to?

The only advantage that I can possibly come up with is forcing the use of a pseudo-Riemannian manifold and dissuading the student from using a more intuitive, bog standard vector concept.

The more intuitive, bog standard vector concept bring with it something that most people here will not be happy with, namely a geometry which requires hyperspherical curvature. The use of the pseudo-Reimannian manifold effectively flattens the hypersphere. And I know, not using the pseudo-Reimannian manifold could equally be said to create the hypersphere. The thing is that both methods work, so I am not sure that we can dismiss either.

(Think about the equations for working out the figures on a surface which is a plane that advances through spacetime with t, then think about doing the equations for working out the figures on the surface of a sphere which expands with t. Alternatively, work out what a vector joining two events will look like in both geometries and compare that to a pair of non-parallel vectors which together join those two events.)

I am sure there are arguments for why the standard equation is used, but I would prefer more proof than 1. it works, if applied correctly (since the other method works too, if applied correctly) and 2. it is what we use (tradition does not always give us the best way to do things).

Of course to use another method, one would have to present a good reason to do so. I have a reason, it was presented at the end of a very long thread on simultaneity and I reproduce it here:

While we have covered a lot of ground in this thread, and brought in a lot of different issues, some of which I have possibly not been as careful with as I could have been, I have tried to be very consistent about how I talk about dimensions. I didn't talk about going from 2 dimensions to 3 dimensions, or from 3 to four. I have tried to always talk about it in terms of 2 dimensions to 2+1 dimensions, or 3 to 3+1.

I have done this on purpose. The reason for it is that while we can nominate an x, y and z axis at random, or select axes which are most convenient for us, we can't do that with time.

You have done the same, at least effectively. You remove a dimension to make it easier to grasp what is being modeled, but you only ever take away a spacelike dimension, never the timelike dimension.

You can't take an inertial perspective (an inertial frame) and choose your four axes at random. There are three dimensions in which you can select axes however you like and one which is inviolate. Say you and I are at rest relative to each other. There is also a television in our frame, at rest relative to both of us and not lying on the line defined by our two positions. I could choose me-TV as my x axis, with myself as the origin. You could chose you-TV as your x-axis with the television as the origin. We could then assign internally consistent orthogonal y and z axes that are not common to each other. Your x, y and z axes would be a blend of my x, y and z axes. What we would be extremely unlikely to do is chose axes such that your x, y and z axes correspond to a blend of my x, y, x and t axes. If we did, then everything would have to be moving in order to stay still in this strange coordinate system. Can you see that is a problem?

So, what I am saying is that time is special, you have to treat it specially.

Now if time could be represented by just another othogonal plane, you could look at it from another perspective and end up with the problem of having blended spacelike and timelike axes.

If the timelike dimension has more of a circular (really hyperspherical) nature then, no matter what perspective you took, the timelike dimension would be unaffected. Yes, your altered perspective would affect the spacelike dimensions, making my x-axis a blend of your x,y and z axes. But our timelike dimension would be unaffected.

Now this might be something completely new, but I sincerely doubt it. I am probably just using clumsy almost physics-like terminology to express something that is already accepted. In any event, this is the physical aspect of what I am discussing. It leads to "an interesting coordinate system" but I think that coordinate system does make sense, even if it may be difficult to grasp.

For the context, see https://www.physicsforums.com/showpost.php?p=1681656&postcount=279" post in the original thread.

cheers,

neopolitan

 

[Dale]

In this situation, what advantage is there in discussing Th and Tt at all?

This whole economic model is your idea and I have already told you that I don't think there is any advantage to it. I was just trying to help you out by showing you how to derive it.

 

[neopolitan]

This whole economic model is your idea and I have already told you that I don't think there is any advantage to it. I was just trying to help you out by showing you how to derive it.

Ok, ignore the economic model.

Even so, why go through this process:

Student says "I see three quantities, time elapsed for A, time elapsed for B and distance traveled by B according to A. Or that could be four, if I say distance traveled by A according to A, but that is zero. Using the three non-zero values I can make an equation which makes sense to me. Can you tell me what it means?"

Teacher says "Ah yes. Here is another way you can make that equation. First you make two other equations, but you have to add two more variables, both of which the same value as one of the variables which you have already used. Then you can subtract one of these equations from the other, to eliminate those two variables I just added, and then you end up with the same equation as the one you just came up with. Now do you understand?"

Student looks at the teacher in a new light. Emboldened by the teacher's masterful complication of what seemed quite straightforward, could the student then ask:

"Can you please explain something else? I am aware that B's journey consists of two inertial journeys, not one single one. Even so, I was interested to see what happens if I run the same sort of equation as we have been discussing, using B's perspective. When I do that, I get an odd figure, 25.6 years (assuming that B travels for 24 shipboard years, at 0.8c, 12 years in each direction - according to B though, this means that A has a speed of 0.8c for 24 years). This is odd because this is precisely the difference between the proper time experienced by A (40 years) and the dilated time that B would calculate that A experiences in 24 years of traveling at 0.8c (14.4 years). What does it mean?"

Will the teacher give the same response as I got to my earlier post when I posed that very same question?

(It may be difficult to answer and take some time to formulate a response, I am happy with that. I am sure the teacher is allowed to say "I don't know, but it is important to me that people understand SR properly, so I will get back to you when I have figured it out myself." Science is strengthened, not weakened, by its proponents admitting when they are not certain about their subject material. )

cheers,

neopolitan

 

[Dale]

Using the three non-zero values I can make an equation which makes sense to me. Can you tell me what it means?"
...
Emboldened by the teacher's masterful complication of what seemed quite straightforward

I'm sorry but I don't know how I can be more clear about this. It is your equation developed by you. It is a product of your mind written and invented by you. As you say, it "makes sense" to you. So it is up to you to explain the meaning of your "straightforward" equation.

A derivation doesn't explain or give meaning (nor does it complicate anything), it only establishes an equation's validity and any underlying assumptions. You are acting as though you have never seen a derivation before, and I know that is not the case.

 

[neopolitan]

I'm sorry but I don't know how I can be more clear about this. It is your equation developed by you. It is a product of your mind written and invented by you. As you say, it "makes sense" to you. So it is up to you to explain the meaning of your "straightforward" equation.

A derivation doesn't explain or give meaning (nor does it complicate anything), it only establishes an equation's validity and any underlying assumptions. You are acting as though you have never seen a derivation before, and I know that is not the case.

This site is screwing with me again. It logs me out while I am posting a reply and deletes the content. Very annoying.

Dale,

I am clearly not making myself clear. It is the equation that needs some interpreting, not the derivation. The difference between a physics equation and a mathematics equation is that the physics one has a real application. The numbers mean something.

Let's continue to use the scenario Fredrik gave right at the beginning, twins A and B, B travels off at (relative) 0.8c for 12 years shipboard time then turns around and travels back at 0.8c, A stays at home. We can use subscript _h for values pertaining to A, and subscript _t for values pertaining to B.

The equation I had the student wanting to understand was, in my words:

(total time elapsed for stayathome)² = (total time elapsed for traveller)² plus (total distance traveled by traveller, according to stayathome)²

You presented an equation, in your words:

S_h² = T_h² - X_h²

and stated that T_h = T_t and, effectively, that T_h = T_t = S_t

so that, with a rearrangement

S_t² = S_h² + X_h²

where S_t is B's proper time (time elapsed on B's clock), S_h is A's proper time (time elapsed on A's clock) and X_h is the coordinate distance (distance traveled by B, according to A).

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).

Is that true (even if clumsily worded)?

I would think it would be useful to explain to the student why you can do this from A's perspective, and it works, but you can't from B's perspective, because it doesn't quite work out the same way.

If you try to do the same thing from B's perspective you end up with this equation (using time elapsed on clocks - proper time; coordinate time and the names A and B):

(total time elapsed on B's clock)² minus (total distance traveled by A, according to B)²= (coordinate time for A, according to B)²

which I think is:

S_t² - X_h² = T_h²

or in figures

24² - 19.2² = 14.4²

I imagine that this results from T_h = T_t = S_h, from B's perspective. (It certainly works at first glance, following the pretty much the same logic as the earlier derivation. The problem I come up with though is that T_h does not equal S_h if S_h is time elapsed on A's clock, since in the scenario that is 40, not 14.4. My understanding would be that this is a result of treating a non-inertial frame as an inertial frame.)

Interestingly enough I made a huge blunder in an earlier post and no one challenged me on it, so I get the pleasure of admitting I was wrong.

in https://www.physicsforums.com/showpost.php?p=1901524&postcount=202" I was fiddling with figures and came up with something which I both misrepresented and also typed incorrectly.

minus (total time elapsed for traveller)² plus (total distance traveled by stayathome, according to traveller)²
= - (24²) + (0.8 x 24)² = -655.63 = - (25.6²)

Add the 14.4 years, then you have 40 years.

I can only say in my defense that it was typed at about 2am.

I have since worked out where the figures come from, they are:

minus (coordinate distance, according to A)² plus (coordinate distance, according to B)² = minus (missing years)²

Now that I have the right label on the figures, it makes a lot more sense. If there is spatial disagreement (about distances travelled), this can be balanced by an equal magnitude temporal disagreement.

cheers,

neopolitan

 

[Dale]

I am clearly not making myself clear. It is the equation that needs some interpreting, not the derivation. The difference between a physics equation and a mathematics equation is that the physics one has a real application. The numbers mean something.

I understand all that. Perhaps I am too used to dealing with my 2-year-old where the concept of ownership is clear and has preeminence in all interactions. This is your equation so it is your job to do the explaining and interpreting.

Please, repeat after my 2-year-old, "mine, mine, mine"

I would think it would be useful to explain to the student why you can do this from A's perspective, and it works, but you can't from B's perspective, because it doesn't quite work out the same way.

B's frame is non-inertial. I wouldn't expect it to work out the same. You should worry about if it works in other inertial frames, not non-inertial frames.

 

[neopolitan]

I understand all that. Perhaps I am too used to dealing with my 2-year-old where the concept of ownership is clear and has preeminence in all interactions. This is your equation so it is your job to do the explaining and interpreting.

I did, at least in my last post. Then I asked for opinion with "Is that true?". It's not your job, but many like to critique here, so I thought I might get some response.

I didn't specifically ask for comment on the idea of two equal magnitude disagreements cancelling out, but I thought I might get some there anyway.

I guess it is possible that no-one feels qualified to comment if it is not part of the standard curriculum. (Note that "not being part of the standard curriculum" does not necessarily mean "wrong". It may imply that, but there are plenty of right things, and valid perspectives, which are not taught because lecturers don't have the luxury of determing the best way to teach the subject for each student, or even the best way for students in general, or the time to cover all the myriad ways that equations can be derived and interpreted. I suspect that much is taught in the way that is most convenient to the lecturer and the teaching environment. It's the poor tutors who usually have to put up with awkward questions like mine ).

cheers,

neopolitan

 

[Al68]

Wouldn't it greatly simplify things to just calculate a one-way trip, ship leaves Earth and comes to rest with Earth a specified distance away, then multiply the results by 2? Not only simpler, but some aspects of the problem would then have to be explained instead of avoided.

Al

 

[neopolitan]

Wouldn't it greatly simplify things to just calculate a one-way trip, ship leaves Earth and comes to rest with Earth a specified distance away, then multiply the results by 2? Not only simpler, but some aspects of the problem would then have to be explained instead of avoided.

Al

I have tried that elsewhere and was a rigorous as I could be, trying to eliminate the acceleration entirely and only consider the two one-way trips there and back again. I was able to show that if B travels from a spot which is at rest in A's frame (where A is) to another spot which is at rest in A's frame, then back to A, then both will agree with the timings on the clocks.

What they won't agree about is distances traveled and when events occured.

Naturally, people didn't like that.

cheers,

neopolitan

 

[Al68]

I have tried that elsewhere and was a rigorous as I could be, trying to eliminate the acceleration entirely and only consider the two one-way trips there and back again. I was able to show that if B travels from a spot which is at rest in A's frame (where A is) to another spot which is at rest in A's frame, then back to A, then both will agree with the timings on the clocks.

What they won't agree about is distances traveled and when events occured.

Naturally, people didn't like that.

cheers,

neopolitan

Everyone would agree on the distance traveled in each frame, and when the events occur. For example, if the ship travels 10 ly at 0.8c, everyone will agree that the distance is 10 ly in Earth's frame, 6 ly in the ship's (inertial) frame, 12.5 yrs on Earth's clock, 7.5 yrs on ship's clock. The ship and Earth end up in the same frame with the ship's twin younger. Of course people don't like it because the standard resolution doesn't fit.

And with any realistic acceleration, the ship's clock reading 7.5 yrs is simultaneous with Earth's clock reading 12.5 yrs. This is exactly what you have pointed out before, neopolitan, in other posts, referring to the midpoint of the turnaround. If we just cut the trip in half, we could avoid obfuscating the issue with simultaneity issues associated with the turnaround.

I don't know why anyone would object to analyzing the problem as a one way trip. Then to get results for a 2 way trip, just double the results.

It really could be that simple, if done correctly.

Al

 

[Dale]

I did, at least in my last post. Then I asked for opinion with "Is that true?"

sorry I missed that. I will go back and look, but not today.

 

[Dale]

Wouldn't it greatly simplify things to just calculate a one-way trip, ship leaves Earth and comes to rest with Earth a specified distance away, then multiply the results by 2?

The only frame where that works is the stay-at-home twin's frame. In all other inertial frames the journey is asymmetric so you can't just multiply by two.

 

[Al68]

The only frame where that works is the stay-at-home twin's frame. In all other inertial frames the journey is asymmetric so you can't just multiply by two.

Well, if the distance in Earth's frame is 10 ly, v=0.8c, then a one-way trip will take 12.5 yrs Earth's clock, 7.5 yrs ship's clock. A two-way trip will be 25 yrs Earth clock, 15 yrs ship's clock. Looks like multiplying by two works just fine to me.

Al

 

[JesseM]

Well, if the distance in Earth's frame is 10 ly, v=0.8c, then a one-way trip will take 12.5 yrs Earth's clock, 7.5 yrs ship's clock. A two-way trip will be 25 yrs Earth clock, 15 yrs ship's clock. Looks like multiplying by two works just fine to me.

That's why DaleSpam said "The only frame where that works is the stay-at-home twin's frame", because it does work in the Earth's frame, just not in other frames. And the whole point of the twin paradox is to ask "what if we look at things from the traveling twin's perspective rather than the Earth-twin's perspective?"

 

[neopolitan]

Well, if the distance in Earth's frame is 10 ly, v=0.8c, then a one-way trip will take 12.5 yrs Earth's clock, 7.5 yrs ship's clock. A two-way trip will be 25 yrs Earth clock, 15 yrs ship's clock. Looks like multiplying by two works just fine to me.

Al

A confusion comes in when you try to talk about the distance from the non-travelling twin to the turnaround point and are not rigorous about which frame you are using to discuss this distance.

In the frame of the traveling twin, the maximum separation obtained is 0.8x7.5=6 light years.

If you are able to think about the traveling twin actively trying to travel to a spot 10 light years away (ie start at rest with non-travelling twin, obtain a velocity of 0.8c, travel for 12.5 years) then you will end up with other figures.

We normally use the figures from the non-travelling twin's perspective, with the assumption that despite the fact that the traveling twin does all the work, that that twin is somehow passive, or at least purely reactive.

I think it is that combination which leads to confusion.

You could try to think of it like this: I am the traveling twin, I accelerate to 0.8c, I travel 10 lightyears (12.5 years traveling in my rest frame), I turn around and I travel back 10 lightyears. Then I can see what my non-travelling twin thinks.

The non-travelling twin will think that I traveled for 20.83333 years and must have traveled a total of 16.6666 lightyears. Which is right, in his frame. If I had stopped, relative to my non-travelling twin at maximum separation, I will find that a message from him will take 16.6666 years to get to me.

Now, go back to thinking about it the normal way. To travel 10 lightyears in my non-travelling twin's rest frame, I need not travel 10 lightyears in my chosen cruising inertial frame ("cruising rest frame" just doesn't sound right). I need to travel for 12.5*0.6 years, or 7.5 years, covering a distance, in my frame, of 6 light years.

Yep, I guess it is confusing :)

The positive thing is that when we do get interstellar transport at significant percentages of c, we won't have to travel for as long (shipboard time or distance) to get to our destinations. The question will then arise as to how you measure the mileage on such a spaceship, which frame?

cheers,

neopolitan

 

[granpa]

perhaps using proper velocity would make it less confusing

 

[neopolitan]

perhaps using proper velocity would make it less confusing

Proper velocity only applies if there is a third observer, doesn't it? I made an error. I was thinking of relativistic composite velocity (since simple addition doesn't work). edit by neopolitan

If that is us, and we consider either of the twins to be at rest (and A is the de facto privileged frame since it is the only contiguously inertial frame), then the proper velocity is the velocity according to that twin. And that is the same magnitude as, different direction to, the velocity of the other twin.

Or are you thinking of some sort of post facto velocity calculation? B comes to rest at the rest point, relative to A, notes the separation, divides the separation by the proper time elapsed (time on the clock) and comes up with a higher velocity?

I don't like that.

That would mean superluminal speeds. And that is not good.

(Using Al's figures, traveling 10ly at 0.8v in A's rest frame, B reaches that in 7.5 years giving a post facto speed of 1.333c. Really, really not good.)

cheers,

neopolitan

PS I looked up proper velocity, aka celerity, and found that yes, it is coordinate distance divided by proper time of traveller and can, therefore, be superluminal. Its usefulness seems to be set a few paygrades above an undergraduate when first introduced to the twins' paradox though.

Personally I am not that keen on calling a velocity which is a mix of coordinate time in one frame and time elapsed in another frame "proper". "Improper" velocity sounds more sensible to me. But, if I ever need to use it, I will call it celerity.

 

[Al68]

That's why DaleSpam said "The only frame where that works is the stay-at-home twin's frame", because it does work in the Earth's frame, just not in other frames. And the whole point of the twin paradox is to ask "what if we look at things from the traveling twin's perspective rather than the Earth-twin's perspective?"

It works in every frame. All frames would agree on the result.

Al

 

[Al68]

A confusion comes in when you try to talk about the distance from the non-travelling twin to the turnaround point and are not rigorous about which frame you are using to discuss this distance.

Hi neopolitan,

Sure, but that's just more reason to recognize the simple fact that the result is identical if we just consider half the trip at a time. It's the exact same problem with the same answer.

Al

 

[JesseM]

It works in every frame. All frames would agree on the result.

Not if you use the coordinates of a different frame! You said:

Well, if the distance in Earth's frame is 10 ly, v=0.8c, then a one-way trip will take 12.5 yrs Earth's clock, 7.5 yrs ship's clock. A two-way trip will be 25 yrs Earth clock, 15 yrs ship's clock. Looks like multiplying by two works just fine to me.

OK, now let's do the same thing, but using the coordinates of the frame where the ship is at rest during the outbound trip and the Earth and the destination are moving at 0.8c. In this frame, the distance from the Earth to the destination is only 10*sqrt(1 - 0.8^2) = 6 light years, so the ship's clock reads 6/0.8 = 7.5 years when it reaches the destination in this frame, and the Earth's clock reads 7.5*sqrt(1 - 0.8^2) = 4.5 years at the same moment in this frame (i.e. according to this frame's definition of simultaneity). If you double these numbers you still get 15 years for the ship's clock, but only 9 years for the Earth's clock, which is not the right answer to what the Earth's clock reads when the ship returns to Earth.

 

[Al68]

Not if you use the coordinates of a different frame! You said:

OK, now let's do the same thing, but using the coordinates of the frame where the ship is at rest during the outbound trip and the Earth and the destination are moving at 0.8c. In this frame, the distance from the Earth to the destination is only 10*sqrt(1 - 0.8^2) = 6 light years, so the ship's clock reads 6/0.8 = 7.5 years when it reaches the destination in this frame, and the Earth's clock reads 7.5*sqrt(1 - 0.8^2) = 4.5 years at the same moment in this frame (i.e. according to this frame's definition of simultaneity). If you double these numbers you still get 15 years for the ship's clock, but only 9 years for the Earth's clock, which is not the right answer to what the Earth's clock reads when the ship returns to Earth.

You only get the right answer considering a one way trip if you do it correctly, not if it's done incorrectly. Which is why I think it would be very useful.

If we did use the 4.5 yrs, we would have to add 8 yrs to that when the ship came to rest to get 12.5 yrs. The ship would see the Earth clock "jump ahead" by 8 yrs when the ship comes to rest with earth. Only after the ship comes to rest with Earth is a one way trip half of a two way trip.

Al

 

[JesseM]

You only get the right answer considering a one way trip if you do it correctly, not if it's done incorrectly. Which is why I think it would be very useful.

But you said it would work even if you didn't use the Earth's frame--so, explain how you could do it "correctly" in the frame where the ship is at rest during the outbound trip and the Earth is moving at 0.8c. Do you deny that according to this frame's definition of simultaneity, when the ship's clock reads 7.5 years the Earth's clock reads 4.5 years? If you want to instead use the definition of simultaneity in the Earth frame, then you're admitting that your method only works if you make use of the Earth frame rather than any other frame.

 

[Dale]

If we did use the 4.5 yrs, we would have to add 8 yrs to that when the ship came to rest to get 12.5 yrs. The ship would see the Earth clock "jump ahead" by 8 yrs when the ship comes to rest with earth. Only after the ship comes to rest with Earth is a one way trip half of a two way trip.

That's not an inertial frame. Obviously you can make it work there, but also obviously you won't be using the standard intertial frame rules.

 

[JesseM]

You didn't add this until after I had already responded:

If we did use the 4.5 yrs, we would have to add 8 yrs to that when the ship came to rest to get 12.5 yrs. The ship would see the Earth clock "jump ahead" by 8 yrs when the ship comes to rest with earth. Only after the ship comes to rest with Earth is a one way trip half of a two way trip.

Based on what? You certainly aren't using a single inertial frame here. If we allow arbitrary non-inertial coordinate systems, there's no reason that the coordinate system's definition of simultaneity at the moment the ship comes to rest relative to the Earth would have to match up with the definition of simultaneity in the inertial frame where the ship is at rest at that moment.

 

[Dale]

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).

Is that true (even if clumsily worded)?

I think that you are thinking the right thing and just writing it wrong (clumsily), 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.

 

[Al68]

But you said it would work even if you didn't use the Earth's frame--so, explain how you could do it "correctly" in the frame where the ship is at rest during the outbound trip and the Earth is moving at 0.8c. Do you deny that according to this frame's definition of simultaneity, when the ship's clock reads 7.5 years the Earth's clock reads 4.5 years?

No. I would just point out that after the ship is at rest with earth, everyone will agree that Earth's clock reads 12.5 yrs. The difference in simultaneity between the two frames is 8 yrs. 4.5 + 8 = 12.5.

If you want to instead use the definition of simultaneity in the Earth frame, then you're admitting that your method only works if you make use of the Earth frame rather than any other frame.

My method in this example is the same method used in the standard resolution. The twins end up in the same frame with the ship's twin 5 yrs younger than Earth's twin.

Al

 

[Al68]

That's not an inertial frame. Obviously you can make it work there, but also obviously you won't be using the standard intertial frame rules.

I've never heard of a twins paradox variation where the ship stays in a single inertial frame. I'm using the same rules as the standard resolution.

Al

 

[JesseM]

No. I would just point out that after the ship is at rest with earth, everyone will agree that Earth's clock reads 12.5 yrs.

Only if you choose to use the inertial frame where the Earth is at rest, and the traveler is instantaneously at rest at the midpoint of the acceleration. But this was exactly my point, that your method only works if you use the coordinates of a single inertial frame, which you seemed to be disagreeing with when you said "It works in every frame. All frames would agree on the result." If you still maintain it works in every frame, tell me how it works in the inertial frame where the traveler is at rest during the outbound journey before accelerating (which is not the same as a non-inertial frame where the traveler is at rest at every moment, including during the acceleration).

 

[Al68]

You didn't add this until after I had already responded:

Based on what? You certainly aren't using a single inertial frame here. If we allow arbitrary non-inertial coordinate systems, there's no reason that the coordinate system's definition of simultaneity at the moment the ship comes to rest relative to the Earth would have to match up with the definition of simultaneity in the inertial frame where the ship is at rest at that moment.

Well if the ship is at rest with earth, they're in the same frame by definition. t=12.5 is simultaneous with the ship's clock reading 7.5 yrs after the ship comes to rest at the end of the trip.

Al

 

[JesseM]

Well if the ship is at rest with earth, they're in the same frame by definition. t=12.5 is simultaneous with the ship's clock reading 7.5 yrs after the ship comes to rest at the end of the trip.

You're not answering my question about other frames though--it's not like in relativity you're forced to only use frames that objects are at rest in, you can use any inertial frame you want! Are you agreeing that your method can only work in the frame where the Earth is at rest, that it can't be generalized to an inertial frame where the Earth is in motion?

 

[Al68]

Only if you choose to use the inertial frame where the Earth is at rest, and the traveler is instantaneously at rest at the midpoint of the acceleration. But this was exactly my point, that your method only works if you use the coordinates of a single inertial frame, which you seemed to be disagreeing with when you said "It works in every frame. All frames would agree on the result." If you still maintain it works in every frame, tell me how it works in the inertial frame where the traveler is at rest during the outbound journey before accelerating (which is not the same as a non-inertial frame where the traveler is at rest at every moment, including during the acceleration).

It works the same in that frame as in the standard resolutions. The ship's twin realizes that if the Earth sends a signal at t=4.5, and if his ship broke so he could never accelerate, when he receives this signal and subtracts the time it had to take for the signal to reach him at c, he would conclude that the signal was sent at t'= 7.5 yrs in his frame. And he would realize that if he had decelerated to come to rest with earth, then Earth's clock would read 12.5 yrs simultaneously with the ship's clock reading 7.5 yrs.

Al

 

[JesseM]

It works the same in that frame as in the standard resolutions. The ship's twin realizes that if the Earth sends a signal at t=4.5, and if his ship broke so he could never accelerate, when he receives this signal and subtracts the time it had to take for the signal to reach him at c, he would conclude that the signal was sent at t'= 7.5 yrs in his frame.

Yes, of course. But this 7.5 years doesn't enter into the calculations for the total time elapsed on Earth when the traveler returns, if we use your method.

And he would realize that if he had decelerated to come to rest with earth, then Earth's clock would read 12.5 yrs simultaneously with the ship's clock reading 7.5 yrs.

It would read 7.5 years simultaneously with the Earth's clock reading 12.5 years in the frame of the Earth. So what you're telling me has nothing to do with using your method in the other frame where the ship was at rest during the outbound journey--all you're saying is that regardless of what the traveler's current rest frame is, he can calculate that in the frame of the Earth the turnaround happens when the Earth clock reads 12.5 years, and then your method is to double that. Of course any observer, no matter what his physical state of motion, can calculate how things will work in the Earth's rest frame, but when I (and DaleSpam) said that your method only works in the Earth's rest frame, I was talking about calculations, the physical motion of the observer doing the calculations is irrelevant.

 

[Al68]

Yes, of course. But this 7.5 years doesn't enter into the calculations for the total time elapsed on Earth when the traveler returns, if we use your method.

It would read 7.5 years simultaneously with the Earth's clock reading 12.5 years in the frame of the Earth. So what you're telling me has nothing to do with using your method in the other frame where the ship was at rest during the outbound journey--all you're saying is that regardless of what the traveler's current rest frame is, he can calculate that in the frame of the Earth the turnaround happens when the Earth clock reads 12.5 years, and then your method is to double that. Of course any observer, no matter what his physical state of motion, can calculate how things will work in the Earth's rest frame, but when I (and DaleSpam) said that your method only works in the Earth's rest frame, I was talking about calculations, the physical motion of the observer doing the calculations is irrelevant.

Well, I must not understand what you're saying. "My" method here is the same method used in the standard resolutions. How would you explain a one way trip where the ship ended up at rest with Earth with the ship's twin younger than the Earth twin?

Al

 

[JesseM]

Well, I must not understand what you're saying. "My" method here is the same method used in the standard resolutions. How would you explain a one way trip where the ship ended up at rest with Earth with the ship's twin younger than the Earth twin?

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.

 

[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