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

[m4r35n357]

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

Does post #59 help?

 

[Smattering]

Does post #59 help?

Yes, it seems to imply that there are counterexamples in the general case.

 

[PeterDonis]

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

Yes.

 

[Smattering]

Yes.

O.k., I certainly believe you when saying that there are such cases. Is there a simple one that can be described in a few lines such that I can reproduce it?

 

[PeterDonis]

Is there a simple one that can be described in a few lines such that I can reproduce it?

It was already implicit in post #59; you appeared to recognize that in post #60. But here goes: consider two lines, whose endpoints have the following (x, t) coordinates on a spacetime diagram:

Line 1: $(0, 0)$ to $(\sqrt{3}, 2)$.

Line 2: $(0, 0)$ to $(0, 0.9)$.

Line 2 is obviously shorter in the Euclidean sense; but if you compute the interval $dt^2 - dx^2$ for both lines above, you will see that it is also shorter in the spacetime sense.

 

[Herman Trivilino]

I'm not necessarily advocating the full geometric approach, or an entire course, just this elementary algebra on the Lorentz Tranform:

In the brief treatments I mentioned time dilation and length contraction are introduced before the Lorentz transformation. That's what you see in the textbooks, but I suspect many instructors of the non-majors classes do not even get to the Lorentz transformation equations.

 

[Orodruin]

In the brief treatments I mentioned time dilation and length contraction are introduced before the Lorentz transformation. That's what you see in the textbooks, but I suspect many instructors of the non-majors classes do not even get to the Lorentz transformation equations.

This is really a problem when it comes to "real" relativity classes. I spend the first three lectures of my SR class just clearing up all of the misconceptions the students have picked up from popular science and from their earlier introductory classes and doing things properly from a geometrical perspective and with constant analogies to the corresponding effects in Euclidean geometry (with the SR particulars underlined as being due to the indefinite metric). Looking back at the course literature from the introductory modern physics course I had as an undergrad, it even stated that atmospheric muons traveled ca 700 m in their rest frame before reaching the Earth ... At least I get to finally talk about electromagnetism and its formulation in terms of the field tensor tomorrow morning ...

 

[Smattering]

It was already implicit in post #59; you appeared to recognize that in post #60. But here goes: consider two lines, whose endpoints have the following (x, t) coordinates on a spacetime diagram:

Line 1: $(0, 0)$ to $(\sqrt{3}, 2)$.

Line 2: $(0, 0)$ to $(0, 0.9)$.

Line 2 is obviously shorter in the Euclidean sense; but if you compute the interval $dt^2 - dx^2$ for both lines above, you will see that it is also shorter in the spacetime sense.

Yes, obviously the length of the lines in the diagram does not have any physical meaning, as long as the lines do not reunite.

 

[Orodruin]

Yes, obviously the length of the lines in the diagram does not have any physical meaning, as long as the lines do not reunite.

No, this is wrong. The lines do have a physical meaning as the proper time elapsed from the start of the line to the end of the line. This is independent of reuniting with a different line (or even the existence of a different line).

 

[Herman Trivilino]

This is really a problem when it comes to "real" relativity classes.

Yes, but keep in mind that I was talking about the very brief treatment given to non-majors near the end of their two-semester sequence. And this is in the US where students have, usually at most, one year of pre-college physics.

Looking back at the course literature from the introductory modern physics course I had as an undergrad, it even stated that atmospheric muons traveled ca 700 m in their rest frame before reaching the Earth ... At least I get to finally talk about electromagnetism and its formulation in terms of the field tensor tomorrow morning ...

My modern physics professor treated only the energy and momentum aspects of special relativity, but other professors who taught that same course went into it in much more depth. As a junior in 1975 I took a course in SR. We used Taylor and Wheeler "Spacetime Physics" and N. David Mermin "Space and Time in Special Relativity". Mermin used a "5 rules" approach, one of which was time dilation, one was length contraction, and one was simultaneity. Years later, maybe around 2005 or so, he presented at a colloquium at the local university, a talk on a way to teach SR. The title included "Light Rectangles", a topic I'd earlier read about in a journal article he authored.

In the question-and-answer session afterwards I asked him something, I don't remember what, but his answer prompted me to follow up with a query about his "5 rules". He said that that is absolutely the wrong way to teach it. I indicated that it was the book I'd learned from and he just gave me a nod!

 

[SlowThinker]

It seems, until someone invents a hyperbolic, rather than Euclidean, paper, people will be confused...

Maybe it's possible to use this trick:
First demonstrate Euclidean distance using a circle.
Draw a line A-B at an angle, and ask "How do I measure its length?" Then draw a circle centered at A going through B, and on that circle mark points H to the right of A and V above A. Then both A-H and A-V are the length.

Then switch to Minkowski distance using a hyperbola.
Draw the asymptotes through A, and then a hyperbola through B, marking either X or T, depending on the angle of A-B. Then you have the way to show Minkowski distance. And you get "timelike", "spacelike" and "lightlike" for free.

Now you can draw the Twin paradox on a blackboard and it's clear that the traveling twin travels shorter Minkowski distance.

 

[PeterDonis]

obviously the length of the lines in the diagram does not have any physical meaning, as long as the lines do not reunite.

The Euclidean lengths don't have meaning even if the lines do reunite. The third line in the triangle given in my last post ends up being spacelike, and we want to restrict attention to timelike lines, so instead consider the triangle described by the three points A, B, C given by: (x, t) = $(0, 0)$, $(0, 0.25)$, $(\sqrt{3}, 2)$. The side lengths are AB = 0.25, AC = 1, BC = 0.25. So the spacetime length of BC is exactly the same as that of AB, yet the Euclidean length of BC is much longer than that of AB--in fact it's nearly as long as that of AC. So we have a total spacetime length of 0.5, AB + BC, represented by Euclidean lines that, taken together, are almost the same length as a Euclidean line, AC, that represents twice the total spacetime length.

 

[Smattering]

No, this is wrong. The lines do have a physical meaning as the proper time elapsed from the start of the line to the end of the line. This is independent of reuniting with a different line (or even the existence of a different line).

We were talking about the length of the diagram lines within the Euclidean medium here. Not about the world lines.

 

[Orodruin]

We were talking about the length of the diagram lines within the Euclidean medium here. Not about the world lines.

In that case it is still wrong. The Euclidean length never has physical meaning even when the curves reunite.

 

[PWiz]

I just looked back at post #35, and I see that there are (major) problems with it.

1) I've used "length" to describe the length of a worldline in a spacetime diagram in some parts of the post, while in some other parts I've used it to describe the "distance on the manifold" (which is represented by a family of hyperbolas in a spacetime diagram) without explicitly mentioning what I mean. Unfortunately, this is giving the impression that I believe that the Euclidean "length" of a worldline as calculated in a spacetime diagram has physical significance. This is not the case - I'm well aware that the spacetime diagram "length" is useless. Taking the 1+1 dimensional case for the spacetime diagram, the "length" of a worldline on the diagram will be given by $s = \int \sqrt{1+ \left ( \frac{dx}{dt} \right ) ^2 }dt$. This has no relation with the spacetime interval (in flat 1+1 spacetime): $s = \int \sqrt{1- \left ( \frac{dx}{dt} \right ) ^2 }dt$ (I've used the ---+ convention because its much more convenient to use to calculate the line element on the manifold for timelike intervals, as $ds^2 < 0$ in the +++- convention for such intervals, and we have to unnecessarily add a -ve sign in the integral to calculate $|s|$ in that case.) When I said

But the "length" of this route is defined in a way different from your intuition: ds2=dxidxj−dt2ds^2 = dx^i dx^j - dt^2 (dxidxjdx^i dx^j is the ordinary spatial distance between the two events).

I meant the line element on the manifold here, and I didn't mention what I was talking about in the original post - the Euclidean spacetime diagram length or the line element on the manifold. (A fatal flaw)

In fact, the spacetime diagram length and the actual spacetime interval can never be related even in non-flat spacetime, because the Euclidean metric tensor in the spacetime diagram (used for measuring the Euclidean length of a curve in the diagram) will always be positive-definite while the (pseudo) metric tensor of spacetime will always have a Lorentzian signature.
If we solve the Euler-Lagrange equation for the spacetime diagram "length", we get the function $x=At + B$ which acts as the extremizer, where $A$ and $B$ are constants. This function is a minimizer here, so $v =$ constant is going to result in the "shortest" worldline on a spacetime diagram. This is what I meant when I said

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).

If we solve the Euler-Lagrange equation for the spacetime interval case, we again get $x=At + B$ as the extremizer, but this time it acts as a maximizer (because of the Lorentzian signature of the spacetime pseudo metric tensor). A geodesic in flat spacetime is just constant velocity motion, so the spacetime interval for a geodesic must be maximum (and yes, I know this is true for non-flat spacetime geodesics too [we can derive the geodesic equation from the concept of maximized spacetime intervals]). This brings me to problem number 2.

2)

however, it is easy enough to see that if ds2ds^2 is to be minimum (generalization of Newton's first law), the spatial distance must be minimum ( as expected) but dt2dt^2 must be maximum. Any acceleration is going to result in ds2ds^2 becoming greater than its minimum value, and dt2dt^2 becoming less than its maximum value.

This is just wrong. I don't know what was going through my mind when I typed this, but I've already given the math above to show that what I typed here is wrong. Generalization of Newton's first law will mean that the spacetime interval will be maximum. Since $d\tau ^2 =ds^2$ (again, using the ---+ convention for convenience), proper time must be maximized too. In this sense, the length of the worldline (the line element on the manifold) is the age of the object. Agreed.

@PeterDonis @Orodruin It must have been an eyesore to see such a basic error. Have I made satisfactory reparations in this follow-up post?

P.S. I'm switching from soda to coffee now. Seriously.

 

[PeterDonis]

If we solve the Euler-Lagrange equation for the spacetime diagram "length", we get the function $x=At + B$ which acts as the extremizer, where AA and BB are constants. This function is a minimizer here, so v=v = constant is going to result in the "shortest" worldline on a spacetime diagram.

This is still wrong, because you're assuming that minimizing the Euclidean length is equivalent to maximizing the spacetime length. That is not true in general; it happens to be true for the particular case that you picked, but other cases have been given in this thread where it is not.

When we say the Euclidean length has no physical significance, we really mean "no". You can't use it for anything. You shouldn't even mention it at all. It just causes confusion if you do.

Have I made satisfactory reparations in this follow-up post?

Not entirely. See above. What you say about the spacetime length is fine, though

I'm switching from soda to coffee now.

IIRC, Mountain Dew has as much caffeine as coffee, so if you're really partial to soda, you still have an option.

 

[MarcusAgrippa]

Hmm so when I fly to Alpha Centauri at 0.9c, I only age 1.83 years.
How do you explain it without length contraction?
How can you even use Lorentz transformation without referring to length contraction?

Use proper time. This requires neither a Lorentz transformation, nor any calculation involving length contraction or time dilation.

Proper time is the time read by an on-board clock - the number of ticks of the clock, if you like. We all have an on-board clock, a ticker that one day will stop, short, never to go again. Our on-board clock is not an ideal clock - occasionally it skips a beat, or races when we see a pretty girl or hunk of a man, but the number of times it ticks determines our age. This ticker, over a long time, will tick on average at the same rate as an ideal clock - well, roughly so.

Proper time along two different world lines that begin at the same event A and end at the same event B display different proper times elapsed from A to B. The persons whose world lines they represent thus age by a different amounts. The longest proper time lapse is along the straight world line joining A to B. This is easy to show by a simple geometric argument requiring no calculation. The time lapse along any other curved (timelike) world line joining A to B is shorter, and so the person whose world line this is ages less between events A and B. This offers a simple and complete resolution of the twin paradox, without any computations, by simply understanding the physical significance of proper time.

 

[Buzz Bloom]

All they tell you is some unobservable weirdness associated with two moving objects

Hi m4r35n357:

I am a bit puzzled by your use of the word "unobservable" as applied to time dilation in SR. I am thinking of a particle of a type with a known measured half-life "living" much longer than its known (non-relativistic) half-life when traveling (relative to an observer) at relativistic speeds.

Regards,
Buzz

 

[bligh]

Great reading above!
As one of the "newbies" more intermediate in understanding, I would say. Where is there something to read that sums up consensus in early 21st century?
My incomplete knowledge is this.
1 absolute time and space are real but unobtainable. Therefore I suggest a locally accurate Universal Reference Point, as a point midway between 3 local galactic centers. It is held in place automatically at equal distances from the 3 gal. centers. The URP would be theoretical but its attachment to abs space and time is that locally it would be a proxy for "absolute" frame of reference. It would determine what is actually happening in terms of distance and velocity.
My understanding is that Time dilation is accepted as shown by GPS. My understanding is that a cesium clock on the GPS would tick at a slower rate physically relative to a URP. My understanding is that length contraction has never been demonstrated without good argument. As far as mass getting larger with increased velocity that is an idea related to inertia, and inertia is only acceleration relative to the underlying field that produces mass in the first place. And humans age based upon their own personal history which is based upon the Ref Frame they biologically evolved in. Proof is in physiological changes observed in our own "space" travelers on the International Space Station.
I am not a crackpot. If I misunderstand I would like a reference or two to set me straight. I is obvious to me that my observations may or may not be incorrect, judging from the preceding comments.
Bligh

 

[Alan McIntire]

I didn’t see any mention of Doppler shift here, so I’ll give my 2 cents worth based on a book I read by PJE Peebles.

Assume Alpha Centari is exactly 4 light years away, and one twin is
traveling
there at 4/5 speed of light. (Using a 3,4,5 triangle I avoid
irrational numbers in my computations.

Traveling at 4/5 the speed of light, from the point of view of the
stay at home twin, the trip will take 10 years, 5 years there, 5 years
back.

Time for the traveller T’ = T( sqrt( 1- (v^2/c^2))) = 3/5 T
Likewise, the distance for the traveler, D’ = 3/5 D

The traveler on the spaceship sees himself traveling a distance of
4*3/5 = 2 2/5 light years in a time of 3 years, and likewise the 2 2/5
light years back
in a time of 3 years, so the traveler will see the trip as lasting 6
years.

Say the twins have super telescopes and can see each other throughout
the trip.
As long as they are traveling apart, the twins will see each other as
aging at 1/3 speed. As long as they are traveling towards each other,
the twins will see each other as aging at triple speed.

The difference is, the traveling twin will see the stay at home twin
as aging at 1/3 speed for the 3 years to Alpha Centauri, for a total
of 1 year,
and at triple speed for the 3 year trip back to Earth = 3*3=9.
The traveling twin will see the stay at home age 1 year during the
trip out, and 9 years during the trip back, for a total of 10 years.

The stay at home twin will see the travel age at 1/3 speed for 9
years, the 5 years it takes the traveler to get to Alpha Centauri,
plus the 4 years it takes the light to get back to earth. Since the
total trip will take 10 years, the stay at home twin will see the
traveler age at triple speed during the one year he observes the
traveler coming back to earth. The Earth observer sees the traveler
age at 1/3 speed for 9 years, for a total of 3 years, and at triple
speed for 1 year, for another 3 years, giving 6 years for the round
trip.

Both observers see each other aging at the same slow rate while moving
apart, they see each other aging at the same fast rate while moving
together. The difference lies in one observer deliberately changes
the relative motion of his rocket from moving away from Earth to
moving towards earth, and the other observer remaining passive, and
not seeing the change until the light from the
traveler reaches earth. If the Earth could be accelerated like a
rocket ship, and the earthbound observer decided to change his frame
so the rocket appeared to be moving towards him at 4/5 lightspeed
rather that away at 4/5 lightspeed, while the rocket remained in
motion past Alpha Centauri, then it would have been the Earth twin who
appeared to age less.
Of course you could have some intermeditate situation where BOTH
observers decide to change their relative motion before they see the
other observer change his motion.

We get our prejudices on absolute time from living on earth, where communication is within 1/10,000 of a second. We cannot see stars or planets "now". we see them
light years away for stars, light minutes or hours away for planets.

 

[PWiz]

That is not true in general;

But I'm not talking about the general case; I'm only considering timelike trajectories in flat spacetime. Then minimizing the Euclidean distance (on the spacetime diagram) automatically maximizes the spacetime interval.

IIRC, Mountain Dew has as much caffeine as coffee, so if you're really partial to soda, you still have an option.

A tempting alternative indeed.

 

[PeterDonis]

I'm not talking about the general case; I'm only considering timelike trajectories in flat spacetime. Then minimizing the Euclidean distance (on the spacetime diagram) automatically maximizes the spacetime interval.

If you're only talking about one single timelike geodesic, yes, this is true. But the discussion in this thread is considering a wider scope than that; it is considering "triangles" in spacetime composed of three timelike geodesic sides. For that case, the rule breaks down. That being so, I don't think talking about the rule in Euclidean terms has any value, since questions along these lines always come up when considering "triangle" type scenarios like the twin paradox.

 

[SlowThinker]

As long as they are traveling apart, the twins will see each other as
aging at 1/3 speed. As long as they are traveling towards each other,
the twins will see each other as aging at triple speed.

Where do these numbers come from? The easiest way to find them seems to be working backward from the final result obtained in some other way.

Also, this thread isn't really about the Twin Paradox and the various ways to arrive at the result. We're debating if there is a scenario where the use of time dilation is useful. If you wanted to show a simple way to arrive at $\sqrt{(4/0.8)^2-4^2}=\sqrt{(4/4*5)^2-4^2}=3$, I'm afraid you failed.

 

[PAllen]

Where do these numbers come from? The easiest way to find them seems to be working backward from the final result obtained in some other way.

Also, this thread isn't really about the Twin Paradox and the various ways to arrive at the result. We're debating if there is a scenario where the use of time dilation is useful. If you wanted to show a simple way to arrive at $\sqrt{(4/0.8)^2-4^2}=\sqrt{(4/4*5)^2-4^2}=3$, I'm afraid you failed.

The standard formula for collinear relativistic Doppler factor is:

√((1+β)/(1-β)) for approach
and
√((1-β)/(1+β)) for separation

 

[polydigm]

First a question. Is everyone here setting c=1 in their calculations? I was taught ds²=c²dt²-dx², with ct as time axis in Minkowski space.

On the discussion earlier about rates of ageing, I agree with the notion that these are not different in different frames. The twin leaving at age 20 and travelling, to return at age 30 to find his sibling at age 50, will not have been able to achieve 30 years of effort while he is away. He will not have aged more slowly than his sibling. He will have traveled for 10 years in his own frame and it will have seemed like 10 years when he returns. So, it is definitely not about rates of ageing.

Medical science has effectively slowed rates of ageing over the last century or so. People live longer lives. Relativity will not enable you to live a longer life, so different rates of ageing is not a useful concept.

 

[Dale]

And what term other than "time dilation" would you propose to describe the fact that one person has aged at a different rate than the other or some physical process has progressed at a different rate?

I have thought a bit about this and (4 pages later) I am inclined to agree.

One reason to avoid the term "relativistic mass" is that we already have the term "total energy" which describes the same thing. As you point out, that is not true here. There is not another name referring to the same quantity as time dilation.

Certainly you could prepare your curriculum without overemphasizing it, but it seems to be at least a non-reduntant term.

 

[PeterDonis]

There is not another name referring to the same quantity as time dilation.

But "time dilation" is used to name two different concepts: the frame-dependent concept of "rates of time flow" being different from one frame to another, and the frame-independent concept of two different timelike worldlines between the same two events having different lengths. So to avoid ambiguity, one of those concepts needs to have a different name. That's why I proposed "differential aging" for the second one.

 

[Dale]

So to avoid ambiguity, one of those concepts needs to have a different name. That's why I proposed "differential aging" for the second one.

Yes, I think that is the better approach.

 

[Herman Trivilino]

But "time dilation" is used to name two different concepts: the frame-dependent concept of "rates of time flow" being different from one frame to another, and the frame-independent concept of two different timelike worldlines between the same two events having different lengths. So to avoid ambiguity, one of those concepts needs to have a different name. That's why I proposed "differential aging" for the second one.

Is it really that simple, though? For the traveling twin a proper time $\Delta \tau$ elapses during his entire journey. The stay-at-home twin calculates that the time $\gamma \Delta \tau$ will elapse on his clock. Is this a dilated time, differential aging, or both?

 

[Dale]

Is it really that simple, though? For the traveling twin a proper time $\Delta \tau$ elapses during his entire journey. The stay-at-home twin calculates that the time $\gamma \Delta \tau$ will elapse on his clock. Is this a dilated time, differential aging, or both?

That is invariant so it is differential aging.