Is it time to "retire" time dilation and length contraction?

[Smattering]

"Differential aging".

O.k., then let's call it differential aging. Still, the concept will appear magical to most people who are not used to it.

And the answer is: because the length of the worldline is the "age".

Hm ... then I really need to look up how the length of a world line is defined, because I would have expected that the younger twin is the one with the longer world line and the older twin is the one with the shorter world line.

In other words, the term "different rates" implies that there is some absolute standard according to which "rates" are measured without regard to the lengths of worldlines, and there isn't.

I see what you mean, but I am not sure whether I can agree here. After all, there is no observer who sees them aging at the same average rate when averaging over the entire journey, is there? So although there is no absolute standard to which rates are measured, there is indeed an intersubjective agreement that the twins are aging at a different average rate between departure and reunification.

 

[Orodruin]

Hm ... then I really need to look up how the length of a world line is defined, because I would have expected that the younger twin is the one with the longer world line and the older twin is the one with the shorter world line.

The concept of "length" in space time does not follow the Pythagorean theorem you are used to but instead the Pythagorean theorem comes with a minus sign on the space component. This changes the behaviour of the geometry.

 

[PeterDonis]

let's call it differential aging. Still, the concept will appear magical to most people who are not used to it.

Sure, that's unavoidable. The only response is to show them the experimental evidence that says it's true, and then give them the theoretical model (SR and Minkowski spacetime) that correctly predicts the experimental results.

I really need to look up how the length of a world line is defined

It's defined as the integral of the line element $ds^2$ along the worldline. This isn't just true in spacetime; it's true in any Riemannian or pseudo-Riemannian manifold.

I would have expected that the younger twin is the one with the longer world line and the older twin is the one with the shorter world line.

I assume that's because you are looking at the worldlines as they are drawn on a spacetime diagram, and seeing that, in the Euclidean geometry of the diagram, the younger twin's worldline is longer. But the actual geometry of spacetime is not Euclidean; it's Minkowskian. So the spacetime diagram, which can only be drawn in a medium with Euclidean geometry, cannot possibly represent all worldlines in Minkowskian spacetime with their actual spacetime lengths. (This should be obvious if you consider the worldline of a light ray: it's a 45-degree line on a spacetime diagram, with some nonzero Euclidean length, but its actual Minkowskian length is zero.) That's why you need to compute the length of the worldline using the formula I gave above; you can't just eyeball it from the diagram.

there is no observer who sees them aging at the same average rate when averaging over the entire journey, is there?

No, of course not. But none of these observers can claim that their observations of the rates are the "right" ones, the ones we should use as an absolute standard. There is no absolute standard for observing "rates of aging". The only absolute is the actual differential aging that is observed when the twins meet up again, and that is not a rate, it's just two different ages.

although there is no absolute standard to which rates are measured, there is indeed an intersubjective agreement that the twins are aging at a different average rate between departure and reunification.

The "average rate" is not a direct observable; it's something that's calculated. The direct observable is the different ages of the twins when they meet up again. The "average rate" is really just a different way of representing this same observable.

Furthermore, "average rate" is not an absolute way of representing that observable, unlike the observable itself, the difference in ages. Why? Consider: what do you divide the different ages by in order to get the average rate? There is no answer; there is no absolute number that represents the "reference" amount of time elapsed between the two events (the twins separating and meeting up again). So there's no absolute number you can divide the different ages by to get different average rates. The different ages themselves are the only absolute numbers in the scenario.

 

[Vitro]

I assume that's because you are looking at the worldlines as they are drawn on a spacetime diagram, and seeing that, in the Euclidean geometry of the diagram, the younger twin's worldline is longer. But the actual geometry of spacetime is not Euclidean; it's Minkowskian. So the spacetime diagram, which can only be drawn in a medium with Euclidean geometry, cannot possibly represent all worldlines in Minkowskian spacetime with their actual spacetime lengths. (This should be obvious if you consider the worldline of a light ray: it's a 45-degree line on a spacetime diagram, with some nonzero Euclidean length, but its actual Minkowskian length is zero.) That's why you need to compute the length of the worldline using the formula I gave above; you can't just eyeball it from the diagram.

+1 on this, I was about to say the same thing. This is another common misunderstanding, confusing spacetime interval with the length of the line segment on a spacetime diagram.

 

[PWiz]

Hm ... then I really need to look up how the length of a world line is defined, because I would have expected that the younger twin is the one with the longer world line and the older twin is the one with the shorter world line.

It is the way you think - the accelerated twin has a "longer" worldline than the unaccelerated twin. Generalizing Newton's first law, we can say that the unaccelerated twin moves on a geodesic by taking the "shortest" route between two events (the event where the twins separate and the event where they meet again). But the "length" of this route is defined in a way different from your intuition: $ds^2 = dx^i dx^j - dt^2$ ($dx^i dx^j$ is the ordinary spatial distance between the two events). The minus sign in the equation is what is responsible for throwing you off; however, it is easy enough to see that if $ds^2$ is to be minimum (generalization of Newton's first law), the spatial distance must be minimum ( as expected) but $dt^2$ must be maximum. Any acceleration is going to result in $ds^2$ becoming greater than its minimum value, and $dt^2$ becoming less than its maximum value. This is why we say that the accelerated twin ages less. (Side note - the negative sign in the metric equation makes spacetime a pseudo Riemannian manifold, and it is the primary reason your intuition [which thinks in terms of Riemannian manifolds] fails.)

 

[PWiz]

Of course you are right, but I don't think Smattering is wrong either.

It seems to me that Smattering misinterpreted Peter - he didn't say "the longer the worldline, the older the twin".

However, I have to agree that Peter's original statement is a little obsure. Perhaps it would be better to say that the length of the twin's worldline is directly related to the "age" of the twin.

 

[m4r35n357]

It seems to me that Smattering misinterpreted Peter - he didn't say "the longer the worldline, the older the twin".

However, I have to agree that Peter's original statement is a little obsure. Perhaps it would be better to say that the length of the twin's worldline is directly related to the "age" of the twin.

Unfortunately you are replying to a deleted reply (To Oroduin's post #32), sorry about that. If anyone is wondering, I pointed out that Smattering's choice of words (quoted therein) indicated that he had understood correctly, but used phrases like "longer world line" where he should have just said "longer line on the diagram".

 

[Smattering]

If anyone is wondering, I pointed out that Smattering's choice of words (quoted therein) indicated that he had understood correctly, but used phrases like "longer world line" where he should have just said "longer line on the diagram".

Yes, exactly. Sorry for that. What I really meant was indeed the length of the respective lines on the diagram. When the length of a world line is definded as the proper time, then of course the age difference is (per definition) equivalent to length difference of the world lines.

However, I am still unsure why this should ease understanding. It seems to me a bit as in the following hypothetical dialog:

Learner: I have calculated the proper times for the two twins, and it appears to me that they will have aged differently when they reunite.
Mentor: That's correct.
Learner: But I do not understand why the twins will have aged differently when they reunite.
Mentor: You will understand that once you think about it from a geometrical point of view.
Learner: What geometry are you referring to?
Mentor: The geometry of spacetime of course. In this geometry, you just have to calculate the length of the twins' world lines.
Learner: So how is the length of the twins' world lines defined?
Mentor: It's defined as the proper time of the respective twin.

 

[m4r35n357]

Yes, exactly. Sorry for that. What I really meant was indeed the length of the respective lines on the diagram. When the length of a world line is definded as the proper time, then of course the age difference is (per definition) equivalent to length difference of the world lines.

However, I am still unsure why this should ease understanding. It seems to me a bit as in the following hypothetical dialog:

Learner: I have calculated the proper times for the two twins, and it appears to me that they will have aged differently when they reunite.
Mentor: That's correct.
Learner: But I do not understand why the twins will have aged differently when they reunite.
Mentor: You will understand that once you think about it from a geometrical point of view.
Learner: What geometry are you referring to?
Mentor: The geometry of spacetime of course. In this geometry, you just have to calculate the length of the twins' world lines.
Learner: So how is the length of the twins' world lines defined?
Mentor: It's defined as the proper time of the respective twin.

Yes, that's right, there is just one thing to calculate; dt^2 - dx^2 (give or take a summation or integral, of course). So what is the problem? BTW here is my version of the dialogue:

Learner: I have calculated the proper times for the two twins, and it appears to me that they will have aged differently when they reunite.
Mentor: That's correct. 10/10!
Learner: But I do not understand why the twins will have aged differently when they reunite.
Mentor: But you just told me you calculated it.

OK I am being flippant, but I am reminded of Einstein trying to explain to Lorentz that "yes you can use your ether theory to get the right answers, but my way is simpler and better defined so why bother?". Lorentz and Poincare would reply "but our way is more intuitive" and so on back & forth. Then two years later Minkowski comes along to Einstein and says "your physics approach is all very well but look at my way, it's so much simpler and better defined than yours" . . . . etc. etc. etc.

If you prefer to see things in terms of these "internal variables", time dilation and length contraction, that is your choice, you can still get the right answers.

 

[Smattering]

Learner: I have calculated the proper times for the two twins, and it appears to me that they will have aged differently when they reunite.
Mentor: That's correct. 10/10!
Learner: But I do not understand why the twins will have aged differently when they reunite.
Mentor: But you just told me you calculated it.

O.k. I guess that is the relativistic version of "shut up and calculate" then.

Probably, I was not getting your whole point. From your point of view being able to calculate something correctly seem to be the best we can expect anyway. So instead of bothering with foredoomed attempts of motivation, just teach learners the best defined approach to calculate something.

 

[m4r35n357]

O.k. I guess that is the relativistic version of "shut up and calculate" then.

It's a personal choice, as my last sentence was intended to say (and this whole thread was meant to discuss just that). I found the Minkowski approach allowed me to "see the wood for the trees" more easily, and move on to looking at more interesting problems. Horses for courses, I suppose.

You seem to be more aligned with Mr. Einstein ;) Here is a passage taken from my favourite source:

[Kevin Brown's words of introduction]
"Einstein was not immediately very appreciative of his former instructor's contribution, describing it as "superfluous learnedness", and joking that "since the mathematicians have tackled the relativity theory, I myself no longer understand it any more". He seems to have been at least partly serious when he later said "The people in Gottingen [where both Minkowski and Hilbert resided] sometimes strike me not as if they wanted to help one formulate something clearly, but as if they wanted only to show us physicists how much brighter they are than we". Of course, Einstein's appreciation subsequently increased when he found it necessary to use Minkowski's conceptual framework in order to develop general relativity. Still, even in his autobiographical notes, Einstein gives no indication that he thought Minkowski’s approach represented a profound transformation of special relativity."

[Einstein's words]
"Minkowski's important contribution to the theory lies in the following: Before Minkowski's investigation it was necessary to carry out a Lorentz transformation on a law in order to test its invariance under Lorentz transformations; but he succeeded in introducing a formalism so that the mathematical form of the law itself guarantees its invariance under Lorentz transformations."

 

[Herman Trivilino]

If you prefer to see things in terms of these "internal variables", time dilation and length contraction, that is your choice, you can still get the right answers.

The student must face the conflict between his intuition and the result of his calculation. There is no way around that with any approach.

In the twenty years between 1990 and 2010 we saw a highly significant drop in the appearance of relativistic mass in introductory physics textbooks. One of the things that contributed to that change was the fact that it was being used to explain why massive objects can't be accelerated to light speed. The authors were getting that physics wrong. So in future editions they were motivated by that to make changes. Yes, it's wonderful that that went along with what many of us considered to be an improved pedagogy.

I don't see the same thing happening with time dilation and length contraction. There isn't time in an introductory physics course for non-majors to go into the geometrical approach. The only hope of imparting a change in worldview is to have students confront their misconceptions in the allotted time.

For physics majors, yes, you can find a way to arrange the introductory course so that you do have time to go into the geometrical approach. Many instructors will not buy into spending that much time because they won't see the benefit weighed again the cost. I don't see any way to remove length contraction and time dilation from the lexicon of physics if it can't be removed from the introductory textbooks.

I don't see the situation with relativistic mass being in the same category. I would instead put relativistic mass in the same category as work done by friction, which is another topic you see disappearing from introductory physics textbooks. Maybe someday authors in the US will begin to realize that the pound is not defined as a unit of force.

 

[SlowThinker]

I calculate the "length" of the world line, t^2 - x^2 (and x = 0.9t). I certainly wouldn't go to the extra effort of calculating the Lorentz Transform, or bother to look up the formulas for time dilation or length contraction (I haven't memorized them either).

That is my point, in a nutshell.

I'm still trying to understand how this is intuitive.
My example: Alpha Centauri is 4.2 ly away. I fly there at 0.9c and want to find my age when I get there.
If I wanted to compute t^2-x^2 naively, I'd use t=4.2/0.9 and x=4.2, so we have
$$wrong=\sqrt{4.2^2 (1/0.9^2-1)}=2.03$$
Why would I use t=4.2 and x=4.2*0.9 (except to get the correct result, of course)?
$$correct=\sqrt{4.2^2 (1-0.9^2)}=1.83$$
Can you (try to) make me "see it"?

 

[Vitro]

O.k. I guess that is the relativistic version of "shut up and calculate" then.

Probably, I was not getting your whole point. From your point of view being able to calculate something correctly seem to be the best we can expect anyway. So instead of bothering with foredoomed attempts of motivation, just teach learners the best defined approach to calculate something.

Not just calculate something in the abstract but make a testable quantitative prediction that can be checked against experimental data.

The WHY only matters if it can lead to a testable difference. If you agree with the result of the calculation then the WHY becomes irrelevant, the answer is because our model says so and experiments say that our model is good.

 

[Fredrik]

Learner: I have calculated the proper times for the two twins, and it appears to me that they will have aged differently when they reunite.
Mentor: That's correct.
Learner: But I do not understand why the twins will have aged differently when they reunite.

That clocks measure proper time...should be viewed as part of the definition of SR. So to say that you don't understand why the twins have different ages, after proving that SR predicts that they do, is to say that you don't understand why the world behaves as described by SR instead of as described by pre-relativistic classical mechanics. No one really understands that. Only a better theory can explain why a theory agrees with experiments...and if we had a theory that explains why SR is a good theory, you'd probably be asking about that theory.

 

[m4r35n357]

I'm still trying to understand how this is intuitive.
My example: Alpha Centauri is 4.2 ly away. I fly there at 0.9c and want to find my age when I get there.
If I wanted to compute t^2-x^2 naively, I'd use t=4.2/0.9 and x=4.2, so we have
$$wrong=\sqrt{4.2^2 (1/0.9^2-1)}=2.03$$
Why would I use t=4.2 and x=4.2*0.9 (except to get the correct result, of course)?
$$correct=\sqrt{4.2^2 (1-0.9^2)}=1.83$$
Can you (try to) make me "see it"?

Don't ask me, I agree with your "wrong" answer! Your "correct" answer seems to treat light years as a unit of time . . .
BTW, remember I'm a student not a teacher ;)

 

[SlowThinker]

Don't ask me, I agree with your "wrong" answer! Your "correct" answer seems to treat light years as a unit of time . . .
BTW, remember I'm a student not a teacher ;)

Hmm maybe the "wrong" is actually correct...

Edit: Maybe it is simpler after all...

 

[bcrowell]

I've been deemphasizing length contraction and time dilation for about the last 5-10 years in my own teaching. My current approach is presented here http://www.lightandmatter.com/lm/ in ch. 22-24. This is for people in algebra-based physics (biology majors, etc.). Matrices are out of the question for these folks. I use graphs to discuss the Lorentz contraction. I've also used a similar approach in a gen ed class: http://www.lightandmatter.com/poets/ . I don't think this approach is any harder or more abstract than the one where you concentrate on length contraction and time dilation. In that type of approach, you run into all kinds of conceptual difficulties, such as the belief that all of relativity reduces to length contraction and time dilation.

Another presentation worth looking at, at a much higher level of math and abstraction, is Bertel Laurent, Introduction to spacetime: a first course on relativityhttp://[URL="http://www.lightandmatter.com/cgi-bin/meki?physics/relativity_special"]www.lightandmatter.com/cgi-bin/meki?physics/relativity_special[/URL] . His approach is aggressively coordinate-independent.

 

[m4r35n357]

I don't see the same thing happening with time dilation and length contraction. There isn't time in an introductory physics course for non-majors to go into the geometrical approach. The only hope of imparting a change in worldview is to have students confront their misconceptions in the allotted time.

I'm not necessarily advocating the full geometric approach, or an entire course, just this elementary algebra on the Lorentz Tranform:
$$
x' = \gamma (x - vt)
$$
$$
t' = \gamma (t - vx)
$$
Square the top and bottom equations and subtract the new top from the new bottom (pardon the names, I'm getting tired)
$$
t'^2 - x'^2 = \frac{(t^2 - x^2) - v^2(t^2 - x^2)}{1 - v^2} = t^2 - x^2 = \tau^2
$$
[EDIT] You may want to insert one or two steps ;)

 

[PWiz]

Can you (try to) make me "see it"?

Of course. You don't move in your rest frame, so we get $d\tau ^2 =- ds^2$. You are traveling at a velocity of 0.9 (geometrized units) relative to someone (an inertial observer I hope!). You can calculate the spacetime interval in this frame $ds^2 = dx^2 - dt^2$.
So $d\tau = \sqrt{(1-(\frac{dx}{dt})^2)} dt$
Just integrate both sides (after substituting $\frac{dx}{dt} = 0.9$ ) and substitute the limits 0 and 4.2/0.9 on the right, and bam, there's your answer.

It's just a question of rearranging the differentials. I don't understand why it appears so complicated to a student new to SR.

 

[m4r35n357]

Hmm maybe the "wrong" is actually correct...

Edit: Maybe it is simpler after all...

If it makes you feel any better, you had me worried for a while ;)

 

[PeterDonis]

It is the way you think - the accelerated twin has a "longer" worldline than the unaccelerated twin.

No, you have it backwards. The "accelerated" twin (the "traveling" twin would be a better term--the one that goes away and comes back) has a shorter worldline--he has aged less when they meet up again.

Generalizing Newton's first law, we can say that the unaccelerated twin moves on a geodesic by taking the "shortest" route between two events

No. A timelike geodesic in spacetime is the longest curve between two given events, not the shortest.

The rest of your post just compounds the same error. Please

It seems to me that Smattering misinterpreted Peter - he didn't say "the longer the worldline, the older the twin".

Yes, I did.

I have to agree that Peter's original statement is a little obsure. Perhaps it would be better to say that the length of the twin's worldline is directly related to the "age" of the twin.

No, my statement was that the length of the twin's worldline is the age of the twin, and that is the exact truth. Your understanding of how "length" works for timelike worldlines in spacetime is incorrect. See above.

 

[PeterDonis]

What I really meant was indeed the length of the respective lines on the diagram.

But that "length" has no physical meaning; the diagram is in a Euclidean medium, and spacetime is not Euclidean, so Euclidean lengths on the diagram do not represent spacetime lengths. They can't, as I explained in a previous post. Trying to use those Euclidean lengths to develop an interpretation of spacetime lengths is not going to work out well.

 

[Smattering]

But that "length" has no physical meaning; the diagram is in a Euclidean medium, and spacetime is not Euclidean, so Euclidean lengths on the diagram do not represent spacetime lengths. They can't, as I explained in a previous post. Trying to use those Euclidean lengths to develop an interpretation of spacetime lengths is not going to work out well.

I agree that the lengths of the lines in the diagram do not have a physical meaning quantitatively.

 

[PWiz]

The rest of your post just compounds the same error. Please

Um, I guess there's been some miscommunication here. By "length" of the worldline, I mean the length you make out in a spacetime diagram.

 

[Orodruin]

Um, I guess there's been some miscommunication here. By "length" of the worldline, I mean the length you make out in a spacetime diagram.

What do you even mean by this? It is a very imprecise statement. As Peter has been saying, the "length" of a world line has little to do with its length in the Euclidean medium in which you are drawing a space time diagram.

 

[PeterDonis]

I agree that the lengths of the lines in the diagram do not have a physical meaning quantitatively.

In what sense do you think they do have a physical meaning?

(Hint: the standard SR answer to this question is "none".)

 

[PeterDonis]

By "length" of the worldline, I mean the length you make out in a spacetime diagram.

And, as I've been telling Smattering, this length has no physical meaning. The physical length of the worldline is the one you calculate from the metric. The Euclidean length as you draw it on a spacetime diagram means nothing.

 

[m4r35n357]

What do you even mean by this? It is a very imprecise statement. As Peter has been saying, the "length" of a world line has little to do with its length in the Euclidean medium in which you are drawing a space time diagram.

I'm going to have a go at clearing this stuff up (wish me luck).

On a space time diagram, lines of equal proper time lie between (say) the origin and points on a U-shaped hyperbola. The lengths of the lines on the diagram are clearly not related to the constant proper time they actually represent. I believe this is what PeterDonis and others are enforcing, which is right, no question.

On the other hand, when we draw out the twin paradox, we draw lines between the origin and a various points on a horizontal line above it. In this specific case, the longer the line on the diagram, the shorter the proper time. This is of course not true in general.

The disagreements and misunderstanding here are all related to this IMO.

 

[Smattering]

In what sense do you think they do have a physical meaning?

(Hint: the standard SR answer to this question is "none".)

Are there cases such that the longer world line does not correspond to the shorter line in the diagram?