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

[SlowThinker]

See the bit where it says "more"? It's bedtime in the UK so I'll let you get on with it for now . . .

Nope, no such button. It seems Apple and Google don't quite cooperate.

Edit: Same on Android.

 

[Mister T]

Now, if I have understood the Lorentz transform correctly, the "moving frame" length needs to be measured at different times,

If it were measured at the same time it would be proper length.

and the "moving frame" time at different positions.

If it were measured at the same position it would be proper time.

These conclusions follow immediately from invariance of the interval.

This is perhaps the root of my feeling that they are both unobservable, and rather contrived.

They're definitely observable. And although they may seem contrived they are very real. There are plenty of people with you, though, in thinking that they are not pedagogically advisable. This thread has illuminated alternative ways of presenting SR without them, and has definitely spawned a lot of informative discussion.

 

[Mister T]

I don't know that it's "cheating", but it might be confusing, since, as I said in a previous post, the term "time dilation" has two possible meanings. One is the invariant thing you describe. The other is something that is not invariant; it's frame-dependent (the fact that a moving clock "appears to run slow", which depends on your choice of frame).

It seems to me, then, that differential aging is just a difference between two proper times, something that will therefore always be invariant.

In SR the definition of dilated time is $\gamma \Delta \tau$ where $\Delta \tau$ is a proper time and $\gamma=(1-\beta^2)^{-\frac{1}{2}}$ where $\beta$ is the relative speed of the observer. That's a frame-dependent quantity, except in the example I gave above which is perhaps not a strictly valid example of time dilation because it involves a change in $\beta$. If not a "cheat" then maybe a slight of hand?

Therefore would it be better to say that during the first half of the traveling twin's journey a proper time $\Delta \tau$ elapses on his ship and the stay-at-home twin measures this to be the dilated time $\gamma \Delta \tau$? Likewise for the return trip.

 

[PeterDonis]

It seems to me, then, that differential aging is just a difference between two proper times, something that will therefore always be invariant.

Yes.

would it be better to say that during the first half of the traveling twin's journey a proper time $\Delta \tau$ elapses on his ship and the stay-at-home twin measures this to be the dilated time ##\gamma \Delta \tau##? Likewise for the return trip.

"Measures" is perhaps not the best word here, because this is not a direct measurement the stay-at-home twin makes. He can calculate, after the fact, that, relative to his inertial frame, the traveling twin turned around when his clock read $\gamma \Delta \tau$. But he can't observe that directly. The only direct measurements he can make are observations of light signals coming from the traveling twin, showing images of the traveling twin's clock; he can see what the traveling twin's clock reads in those images, and measure the Doppler shift of the signals. That's it. Everything else is calculated, and using the word "measurement" for something that's calculated seems like a bad idea to me.

 

[Mister T]

Everything else is calculated, and using the word "measurement" for something that's calculated seems like a bad idea to me.

Well, usually the word "observe" is used.

Say the traveling twin agrees to send a signal back home when he arrives. The stay-at-home twin gets the signal, subtracts off the travel time of the signal, and arrives at a result. That is the result of a calculation. But isn't it the case that most measurements are the result of calculations?

 

[PeterDonis]

usually the word "observe" is used.

Yes, and it has the same problems. Unfortunately, there isn't really a good word to describe this; I often try to say "calculate", but that's cumbersome. Sometimes I've tried "judge", but that doesn't seem to help much.

Say the traveling twin agrees to send a signal back home when he arrives. The stay-at-home twin gets the signal, subtracts off the travel time of the signal, and arrives at a result. That is the result of a calculation.

Yes, that's the sort of calculation I had in mind.

But isn't it the case that most measurements are the result of calculations?

Not in the same way. For example, suppose the stay-at-home twin measures the Doppler shift of the signal from the traveling twin. That measurement is the "result of a calculation", in the sense that he can't directly detect the shift; he can only detect the frequency. He calculates the shift as the difference between the frequency he detects and the (presumed known) frequency of emission.

Now contrast this with the stay-at-home twin's calculation of the time of emission of the signal the traveling twin sends when he turns around. He directly measures the time of arrival of the signal. He subtracts off the travel time--but how does he know the travel time? He can't measure it directly, and it's not a previously known constant like the emission frequency of the light signal. He has to calculate it. How does he calculate it? Well, he knows the traveling twin's speed--or at least he knows what speed the traveling twin said he was going to use, and how long the traveling twin intended to travel, by his own clock, before he turned around. Or, he can watch the traveling twin continuously, detecting light signals from him all during his outward trip (assuming the traveling twin is emitting such signals), and use the Doppler measurements from those signals to verify the traveling twin's speed away from him. When he then receives the turnaround signal (which he will detect by the sudden shift in Doppler from redshift to blueshift), he can then go back and put together all those measurements and calculate the travel time of the turnaround signal based on the distance the traveling twin was when he emitted it. But all that is a lot more complicated, and involves more variables, than the simple calculation of the difference between received frequency and known emitted frequency.

To an extent this is a judgment call, of course; but I think it's clear that there's a big difference between the "calculation" of the Doppler shift and the calculation of the traveling twin's turnaround time, big enough to justify using the term "measurement" for the first but not the second.

 

[m4r35n357]

Nope, no such button. It seems Apple and Google don't quite cooperate.
View attachment 91990
View attachment 91991
Edit: Same on Android.

OK, Here's the text (I intend to revisit this in the near future to tighten it up a bit):

The twin "paradox" (in quotes because it is NOT a real paradox!) is a valuable learning tool for Special Relativity: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_intro.html

This is a series of visualizations of the journey from the point of view of the traveling twin, who flies 20 light years away from his home station then returns. The journey consists of four parts, joined together. The first quarter is an acceleration away from the station. During the second and third quarters the ship accelerates towards the station (so that at the half way point the ship is stationary 20 light years away). In the fourth quarter the ship accelerates away from the station in order to come to rest there.

The total coordinate travel time (shown as a red dot in the top left HUD clock) is 43.711/58.918 years for acceleration at earth/moon gravity levels, whilst proper time (green dot) is 12.101/38.694 years. The yellow dot represents the time the traveller would see on the station clock face through a very powerful telescope! The octahedral stations (spaced one light year apart and one light year to the left of the flight path) are all synchronized to coordinate time and rotate once over the course of the whole journey. There is a 2x2 light year wall one light year beyond the far end of the journey, and large rectangular frames every 5 light years. Where a floor is shown it is 1 light year per stripe, and there are small 1 ly milestone spheres along the way, with a larger one every five light years.

The flights are rendered without relativistic effects and then for two values of acceleration (currently Earth gravity and moon gravity). The distortion artifacts are due to aberration of light, the Doppler effect and the headlight effect. The Earth gravity videos exhibit some nice penrose-terrel "rotation" effects. Magenta markers in a circle show where you really are in the scene (Doppler shift = gamma for this circle), and grey markers show the circle where the doppler shift is 1 (gamma is in theory directly observable on this circle).

These videos were made with POV-Ray (http://www.povray.org/) and avconv/ffmpeg (https://www.ffmpeg.org/) video encoding software.

 

[Mister T]

Yes, and it has the same problems. Unfortunately, there isn't really a good word to describe this; I often try to say "calculate", but that's cumbersome. Sometimes I've tried "judge", but that doesn't seem to help much.

Perhaps "determine"?

The issue is that we're talking about an amount of time that passes between two events that are separated in the observer's space. (Otherwise, of course, we'd be talking about proper time.) So in that sense it's a quantity that can't be measured "directly" by that observer only because he can't be in two places at the same time. But it's still a very observable, measurable, determinable, and real amount of time. What we're looking for here is a word that's least likely to induce misconceptions. Perhaps there is no such optimal word and the best thing is to have ready a vocabulary of alternatives to be used interchangeably, exposing the listener to the variety.

Now contrast this with the stay-at-home twin's calculation of the time of emission of the signal the traveling twin sends when he turns around. He directly measures the time of arrival of the signal. He subtracts off the travel time--but how does he know the travel time? He can't measure it directly, and it's not a previously known constant like the emission frequency of the light signal. He has to calculate it. How does he calculate it? Well, he knows the traveling twin's speed--or at least he knows what speed the traveling twin said he was going to use, and how long the traveling twin intended to travel, by his own clock, before he turned around.

That's if the distance had not already been measured by some other means beforehand. If, for example, the trip is to the moon, he already knows the distance and can calculate the delay time of a light signal in advance. Then, the only thing needed is one subtraction.

He could even get around that by setting a clock to that much time before zero. When the signal is received it automatically stops the clock and the time can be read directly.

 

[m4r35n357]

BTW sources for the ray-traced videos are available on GitHub

 

[valentin mano]

We have two synchronized clocks and one of them is acceleratated near the speed of light.The GR approach is neaded here,because SR is symmetric by the Lorenz Transforms.

 

[Dale]

We have two synchronized clocks and one of them is acceleratated near the speed of light.The GR approach is neaded here,because SR is symmetric by the Lorenz Transforms.

Back in Einstein's day and the early years of GR, that was a common way to look at it. In modern terms GR is usually only considered to be involved when spacetime is significantly curved (I.E. when tidal gravity is important). Simply using tensors and non inertial coordinates in flat spacetime is not considered GR except insofar as flat spacetime is a trivial solution of the Einstein field equations.

 

[valentin mano]

I was just trying to say,that the twin paradox has nothig to do with the Special Relativity.The principle of equivalence(no tidal forces)is implied here.

 

[PeterDonis]

I was just trying to say,that the twin paradox has nothig to do with the Special Relativity.

And that is incorrect. The standard twin paradox is set in flat spacetime, and SR is entirely sufficient to analyze it.

The principle of equivalence(no tidal forces)is implied here.

First, the principle of equivalence does not say "no tidal forces", period. It says "no tidal forces are detectable in a small enough region of spacetime".

Second, "no tidal forces" means "flat spacetime", which again means SR is entirely sufficient.

Third, why is the lack of tidal forces important in analyzing the twin paradox?

 

[Chestermiller]

I really like what you said in your original post. As a relative newcomer to relativity, I also struggled with the many texts that I tried to learn from, because the material was presented so badly (even though I was an experienced professional in engineering). Eventually, I was able to put the pieces of the puzzle together on my own, but not without considerable effort. So I totally agree that there has to be a better way of teaching this subject matter than the way it has traditionally been done.

I like the curriculum you laid out in your original post. In my judgement, the entry point to understanding relativity in depth starts with the Lorentz Transformation. So getting to the Lorentz Transformation as quickly as possible is a desirable goal. From there, one can then quickly deduce the basic 4D geometry of spacetime, which then leads to everything else.

In my judgement, before discussing the Lorentz Transformation, it is important for students to begin to develop awareness and comfort with some of the new phenomena they are going to encounter in SR. It doesn't have to be much; just enough to pique their interest. Any ideas on how to accomplish this?

As far as deriving the Lorentz Transformation is concerned, it leaves me cold to think that you might have in mind using Einstein's two postulates. These two postulates are, in my judgement, not the essence of SR. They are the effects of the unique 4D geometry of spacetime, and, although they are historically significant, are not the causes of anything. Did you have in mind deriving the Lorentz Transformation using something other than the 2 postulates?

Chet

 

[valentin mano]

Flat Spacetime does not mean "inertial frame of reference",which is the initial frame of Special Relativity.

 

[PeterDonis]

Flat Spacetime does not mean "inertial frame of reference",which is the initial frame of Special Relativity.

SR was originally formulated in terms of inertial frames, yes, but as DaleSpam pointed out in post #116, in the modern view SR is defined by flat spacetime, including the use of non-inertial frames in flat spacetime.

 

[valentin mano]

[Moderator's note: edited to fix quote tags.]

Third, why is the lack of tidal forces important in analyzing the twin paradox?

The tidal forces are not important,just the acceleration,that one of the clocks expiriences.It accelerates to near the speed of light,than returns
back to the one,that has not been moved.In SR there are two inertial frames,passing by each other and each frame sees its clock ticking
faster than the other.The same way EPR paradox is not solved in SR.

 

[Chestermiller]

Did you have something to say here, or did you like what Peter said so much that you decided to re-emphasize it?

But seriously, though, Chapter 6 of MTW is entitled Accelerated Observers. The first section in this chapter is entitled "Accelerated Observers can be Analyzed Using Special Relativity"

I think that says it all.

Chet

 

[valentin mano]

Sorry,I thought I was talking to a layman.English is not my native.

 

[PeterDonis]

The same way EPR paradox is not solved in SR.

I don't understand what any of this has to do with the EPR paradox. But this whole subthread is getting way off topic.

 

[PeterDonis]

The subject has been discussed thoroughly. Thread closed.