[SOLVED] variable speed of light

[vidovnan]

OK people, I'm sure you'll find this one easy. For myself, I have been working on it for a while now and am not getting far enough.

My work colleague has posed me this problem:

how can you fit a 5m(eter) car into a 3m garage?

Now, I know this has something to do with the variable speed of light. I know that the speed of light varies in a gravitational field. And I know that the physical dimensions of bodies vary when these are plunged into a gravitational field.

However, that's about all I know so far.

And my work colleague is soooo smug.

Then I found you...

Please help me challenge his smugness.

[chroot]

Light actually does not vary in speed. All observers will always measure light as moving through their local frame at c.

What your friend's question involves is actually a phenomenon called length contraction. You see, the length of an object (like the car) is not a fixed quantity. Different observers moving at different relative speeds to the car will actually measure its length differently.

Let's say you're standing still beside the 3 m garage. A physicist would say you are using the garage as your 'rest frame,' and you are at rest in it. The garage's length, 3 m, is called its 'proper length.' The car's proper length is 5 m -- this means that a person standing still beside it would measure it as being 5 m in length.

If you propel the car to a high enough velocity, it is possible that you, standing beside your garage, will measure the car's length as being shorter than its proper length of 5 m.

The equation governing this effect (called length contraction) is this:

l = l0 / γ

where l0 is the proper length of the car, and l is the measured length of the car.

γ (lowercase Greek gamma) is a unitless number that is always greater than or equal to one. When you are at rest with respect to the object you're measuring, γ = 1, and you will measure its length as its proper length.

γ can be found by plugging in v into the following definition:

γ = 1 / sqrt(1 - v2 / c2)

To find the velocity the car must have relative to the garage to cause an observer standing beside the garage to measure its length to be 3 m, you must solve the equation:

3 = 5 / γ

3 = 5 * sqrt(1 - v2 / c2)

v = (4/5) c, or four-fifths the speed of light.

If you drive your 5 m (proper) car at four-fifths the speed of light with respect to your 3 m (proper) garage, it will fit inside it -- before promptly crashing through the back wall!

- Warren

[jcsd]

Nothing to do with VSL theories (which are a far from conventional) or gravitational fields even, it's special relativty and length contraction, which is given by the following equation:

L = L0√(1 - v2/c2)

Solving for L = 3 and L0 = 5, we find that v = 0.8c. So in other words: to fit a 5m car into a 3m garage the car must be travelling at 4/5 the speed of light in a vacuum (relative to the garage).

edited to add: looks like chroot beat me to it!

[vidovnan]

Hey, Thanx people! Now that's what I call service.

Remind me to visit this board more often.

[zoobyshoe]

Warren,

If you get the car going fast enough to measure its length as 3m instead of 5m,and don't cra**** into the garage, how long is the car after you slow it back down to a halt in your rest frame?

-zoob

[jcsd]

5m of course, length contraction doesn't deform the object. Interestingly, due to the failure of simultaneity at distance, though an observber in the rest frame of the garage will see both the front and back end of the car in the garage at the same time, an observer in the rest frame of the car will not.

[zoobyshoe]

Originally posted by jcsd
5m of course, length contraction doesn't deform the object. Interestingly, due to the failure of simultaneity at distance, though an observber in the rest frame of the garage will see both the front and back end of the car in the garage at the same time, an observer in the rest frame of the car will not.
JCSD,

This being the case (which makes perfect sence to me) why is it that people claim a clock that is measured to be running slow when moving past an observer, will actually be found to have lost time when slowed down and returned to the observer's rest frame (which doesn't make sence to me)?

[Ivan Seeking]

Originally posted by zoobyshoe
JCSD,

This being the case (which makes perfect sence to me) why is it that people claim a clock that is measured to be running slow when moving past an observer, will actually be found to have lost time when slowed down and returned to the observer's rest frame (which doesn't make sence to me)?

Hey Zooby, I thought I would jump in since I'm here.
I suspect that you don't really believe the statements above. From the frame of the garage, the car really is shorter. Next, if observers in both frames of reference must observe light to travel with speed C, and since we know the length contraction is given by

L = L0√(1 - v2/c2)

and we know that for any observer, C = L0/t0 = L/t,

it can easily be shown that

t = t0√(1 - v2/c2)

Therefore just as length contracts, observers in the frame of the garage will see clocks running slowly in the car. Again, this is not just an illusion; this is real. Now, it gets interesting since for an observer in the car, the garage is in motion and the car is at rest. So, an observer in the car sees clocks in the length contracted garage running slowly. Whose clocks are right? Both. When someone accelerates, in this case when the car comes to a stop and comes back, we have chosen a preferred observer - the garage - and we find that the clocks in the car have lost time. It we speed up to catch the car, assuming we started out in motion with the car and synchronized our clocks, we would find that the clocks in the frame of the garage have lost time. Again, the one who changes their state of motion is the one whose clocks have lost time. Note that in reality, the presence of gravity complicates this situation.

[zoobyshoe]

If the clock in the car comes back into the rest frame of the garage having lost time, why hasn't the car lost length?

[Ivan Seeking]

Originally posted by zoobyshoe
If the clock in the car comes back into the rest frame of the garage having lost time, why hasn't the car lost length?

When the car is in motion, it is shorter in the frame of the garage. Likewise, from the frame of the garage, the car's clocks are running more slowly. When the car stops, that is, when the frame of the car coincides with the frame of the garage, the two lengths L and L0 agree. Likewise, it we compare the ticks of the clocks, we find that again they agree – they occur at the same rate. However, and this was a key test of relativity, we find that while the frames of the car and garage did not coincide, ie, while the car is in motion as viewed from the garage, the clocks in the car really were running more slowly...just as observed and predicted.

This was finally verified I think in the early sixties using two atomic clocks; one on the ground, and one in a jet. After flying one of the clocks around for a while, and after accounting for the effects of gravity, the clock on the plane had indeed lost time as predicted to within the accepted margins of error. This has since be replicated in many other ways. Also, we see the lifespan of subatomic particles increase according to Relativity and their relative speed – since their clocks run more slowly, we see them live longer. This is seen in particle accelerators as well as in nature.

[zoobyshoe]

Originally posted by Ivan Seeking
Again, the one who changes their state of motion is the one whose clocks have lost time.
Both parties see the other as in motion according to their own frame. Both parties see the other's clock as running slow.
When both parites come to be in the same frame again neither can say if it was he or the other who decelerated. Both must find the the other's clock has lost some time, in which case they will both find both clocks to agree both as to the time and the rate of timekeeping. The illusion of slow clocks only exists while the relative motion is occuring. When the relative motion stops the illusion stops.

[Ivan Seeking]

Originally posted by zoobyshoe
Both parties see the other as in motion according to their own frame. Both parties see the other's clock as running slow.
When both parites come to be in the same frame again neither can say if it was he or the other who decelerated. Both must find the the other's clock has lost some time, in which case they will both find both clocks to agree both as to the time and the rate of timekeeping. The illusion of slow clocks only exists while the relative motion is occuring. When the relative motion stops the illusion stops.

We can tell who accelerates [decelerates] - the one who experiences a force. If the relative motion of observer B changes wrt observer A, and if observer A experiences no forces, then A knows that B has changed his state of motion. Likewise, B feels a force and is also aware of whose frame of reference has changed. This indicates who is at rest -the preferred observer. This is no illusion; it is a 98 year old, well tested theory. The clocks do not agree when we compare the results in the same frame of reference; and the two clocks vary by the amount predicted by Special Relativity [or General Relativity if required].

Edit: A key concept here is that until someone experiences a force, indicating a change in their state of motion, there is no preferred observer. We can define either [or any] frame of reference to be at rest as long as the state of motion remains constant.

[zoobyshoe]

So, it seems that acceleration is the critical point.

If we accelerate the car away from the garage, what does each observe about the other's clock?

[Ivan Seeking]

Originally posted by zoobyshoe
So, it seems that acceleration is the critical point.

If we accelerate the car away from the garage, what does each observe about the other's clock?

The clocks in the car would be seen to run slowly as before; and due to the acceleration.

From General Relativity: Clocks run more slowly in gravity fields.

[Ambitwistor]

No, acceleration isn't the critical point. You can construct variants of the twin paradox in which nobody accelerates, but the twins come back with different ages. (e.g., going around a closed universe, or a gravitational slingshot back home). All that is required for different elapsed times is that, by whatever means (acceleration or not), the two twins take spacetime paths of different lengths.

[zoobyshoe]

Originally posted by Ambitwistor All that is required for different elapsed times is that, by whatever means (acceleration or not), the two twins take spacetime paths of different lengths.
In which case the one who takes the longer path ages less?

This is in General Relativity, right? I'm not finding a discussion that goes further than to say: "As a consequence of its motion the clock goes more slowly than when at rest." in SR.

[Ivan Seeking]

Originally posted by Ambitwistor
or a gravitational slingshot back home

How do we do this without any acceleration?

[Ivan Seeking]

Originally posted by zoobyshoe
[B]In which case the one who takes the longer path ages less?

More

This is in General Relativity, right? I'm not finding a discussion that goes further than to say: "As a consequence of its motion the clock goes more slowly than when at rest." in SR.

SR is only valid in the inertial frame - no forces.
We need GR to calculate the effects of acceleration [gravity].

[chroot]

Originally posted by zoobyshoe
In which case the one who takes the longer path ages less?

A path in spacetime is called an interval.

The length of an interval is called its proper time. If you have a clock follow some path, the clock will measure that much time having been elapsed when moved along that path.
This is in General Relativity, right? I'm not finding a discussion that goes further than to say: "As a consequence of its motion the clock goes more slowly than when at rest." in SR.
This is not really true -- a distant observer will measure a clock as running slowly when it is moving at high relative velocity to the observer. According to the clock, however, everything is just fine.

Imagine Picard is flying along in the Enterprise at 0.9c with respect to the Earth. An observer on the Earth will measure Picard's clock as running slow compared to an identical Earth-bound clock. Picard, however, will see everything on the bridge of the Enterprise as running completely normally, but will measure the Earth-bound clock as running slowly.

If you think about it, it has to be that way... if it weren't, then some cosmic ray particle moving at 0.9c with respect to you somewhere in the depths of space would somehow affect YOUR clock!

- Warren

[Ambitwistor]

Originally posted by zoobyshoe
In which case the one who takes the longer path ages less?

The longer path ages more, if you measure "length" using the spacetime interval.


This is in General Relativity, right?


Special relativity too.

Consider the case where the Earth twin stays at home for 10 years according to his own clock. The other twin travels at 80% of light speed, travelling 4 lightyears in 5 years according to the Earth twin, then returns the same way (after an instantaneous deceleration and acceleration back home).

The Earth twin's worldline is a line from (t,x) = (0,0) to (10,0). The travelling twin's worldline consists of two line segments, one from (0,0) to (5,4), the other from (5,4) to (10,0).

The proper time measured by the Earth twin is the spacetime length of his worldline,


\tau = \sqrt{{\Delta t}^2-{\Delta x}^2} = \sqrt{(10-0)^2-(0-0)^2} = 10


The proper time measured by the travelling twin is the spacetime length of his worldline,


\begin{equation*}
\begin{split}
\tau &= \tau_1+\tau_2 = \sqrt{{\Delta t_1}^2-{\Delta x_1}^2}+
\sqrt{{\Delta t_2}^2-{\Delta x_2}^2} \\
&= \sqrt{(5-0)^2-(4-0)^2}+\sqrt{(10-5)^2-(0-4)^2}\\
&= 3+3 = 6
\end{split}
\end{equation*}


The travelling twin ages 6 years to the Earth twin's 10 years.

[Ivan Seeking]

In order to declare one frame valid and the other not, i.e. if we are to determine whose clocks have lost time, someone has to accelerate. Until that happens, boths frames of reference are valid.

[chroot]

Originally posted by Ivan Seeking
In order to declare one frame valid and the other not, i.e. if we are to determine whose clocks have lost time, someone has to accelerate.
No. As Ambitwistor just demonstrated with some killer graphics [;)], all you need to do is calculate the proper times along the paths of both twins, and compare.

Both frames are perfectly "valid," as are all frames. There's no such thing as an invalid frame.

- Warren

[Ivan Seeking]

Zooby, do you feel spoiled? [:D]

[Ambitwistor]

Originally posted by Ivan Seeking
How do we do this without any acceleration?

I was speaking of proper acceleration.

[Ivan Seeking]

Originally posted by chroot
No. As Ambitwistor just demonstrated with some killer graphics [;)], all you need to do is calculate the proper times along the paths of both twins, and compare.

We still have no preferred observers in the inertial frame.


Both frames are perfectly "valid," as are all frames. There's no such thing as an invalid frame.

- Warren [/B]

If one accelerates, that frame is no longer valid under SR, and we can then make a distinction between the two systems.

[Ambitwistor]

Originally posted by Ivan Seeking
We still have no preferred observers in the inertial frame.

I'm not sure what point you're trying to make.


If one accelerates, that frame is no longer valid under SR, and we can then make a distinction between the two systems.

However, acceleration is not necessary to distinguish between the two systems. As I said, the only thing that determines whether two twins age asymmetrically is whether the spacetime lengths of their worldlines are different. The symmetry can be broken by acceleration, or by other means.

We can resolve this using pure SR, by the way: my calculation was performed entirely in an SR inertial frame (that of the Earth twin).

[chroot]

Originally posted by Ivan Seeking
We still have no preferred observers in the inertial frame.
I have no idea what this means.
If one accelerates, that frame is no longer valid under SR, and we can then make a distinction between the two systems.
This is also incorrect. Special relativity is all that's necessary to understand the twin paradox; Ambitwistor explained it nicely. The only thing you need general relativity to explain is gravitation.

- Warren

[Ivan Seeking]

Originally posted by chroot
I have no idea what this means.

There is no absolute reference frame. There is no absolute state of rest.

This is also incorrect. Special relativity is all that's necessary to understand the twin paradox; Ambitwistor explained it nicely. The only thing you need general relativity to explain is gravitation.

- Warren [/B]

How do we determine which twin is younger? One of them has to accelerate in order to leave earth; unless he was born in a state of relative motion as compared to his twin.

[Ambitwistor]

Originally posted by Ivan Seeking
There is no absolute reference frame. There is no absolute state of rest.

So?


How do we determine which twin is younger?

The younger twin is the one whose worldline is shorter. Which twin that is depends on the physical situation. I gave an example.

[Ivan Seeking]

One of them has to accelerate in order to leave earth; unless he was born in a relative state of motion as compared to his twin.

[zoobyshoe]

Well, Ambitwistor's is the best explanation of the twin paradox I have run into. It's the first time I even understood the importance of the "Spacetime Interval". (And, as Chroot said, the graphics were killer.)

This gives me a much better sence of what people are saying is the case. I am fairly certain I don't grasp it yet, but the "spacetime interval" must surely have been the missing link I needed to start putting this together in my mind.

The problem for me has always been that if the two people in relative motion measure each others clocks as slow it strikes me as proof positive both clocks are fine and would agree on the total time elapsed if compared later in the same frame. The difference in length of the spacetime interval finally introduces the asymetry that accounts for the differences in the time elapsed in the two different frames.

[chroot]

Ivan,

In case you haven't seen this:

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

- Warren

[Ambitwistor]

Originally posted by Ivan Seeking One of them has to accelerate in order to leave earth; unless he was born in a state of relative motion as compared to his twin.

Yes. So?

[chroot]

Originally posted by zoobyshoe
It's the first time I even understood the importance of the "Spacetime Interval".
The interval is a very important quantity in relativistic physics because it is invariant. No matter what coordinate system you use, or which observers you consider to be at rest, the interval they will measure for some path \Gamma is always the same. The interval is independent of observers and is a fixed quantity for any particular path through spacetime.

- Warren

[Ambitwistor]

Originally posted by zoobyshoe
The problem for me has always been that if the two people in relative motion measure each others clocks as slow it strikes me as proof positive both clocks are fine and would agree on the total time elapsed if compared later in the same frame. The difference in length of the spacetime interval finally introduces the asymetry that accounts for the differences in the time elapsed in the two different frames.

It's good to think in terms of geometry. Special relativity is just Euclidean geometry in disguise (with a slightly modified Pythagorean theorem).

Lorentz transformations are analogous to rotations. (Because of the minus sign in the Pythagorean theorem for distance, Lorentz transformations trace out hyperbolas in spacetime instead of circles in space.) So the statement that "there are no preferred inertial frames in SR" is really a disguised version of the statement, "there are no preferred directions in Euclidean geometry".

You can see that in this version of the twin paradox, the two worldlines make a triangle: the worldline of the Earth twin is one side, and the worldline of the travelling twin is the other two. People say, "why isn't the travelling twin's frame the same as the Earth twin's?" In Euclidean geometry, the analog of switching inertial frames is rotating. But you can see that just by a rotation, you can't turn the two bent line segments of the travelling twin into one line segment like the Earth twin's worldline: they are not geometrically equivalent to each other. Rotations don't affect lengths, so the two sides of the triangle will never be equal in length to the other side of the triangle, no matter what you do to it --- unless you start deforming it, but then that doesn't describe the same situation anymore.

[Ivan Seeking]

I have no idea why anyone has objected to anything I have said here. I will read the link and pick this up later.

[Ambitwistor]

Originally posted by zoobyshoe
Well, Ambitwistor's is the best explanation of the twin paradox I have run into. It's the first time I even understood the importance of the "Spacetime Interval".

Here's a secret: I almost never perform Lorentz transformations. I just work with 4-vectors, and extract physical information from them by finding their spacetime lengths or projections onto other 4-vectors.

[Ambitwistor]

Originally posted by Ivan Seeking
I have no idea why anyone has objected to anything I have said here.

It hasn't been clear to me what your points have been, but it has seemed at times that you have been implying that acceleration is necessary for the twins to experience different elapsed proper times, and/or general relativity is required to resolve the twin paradox in the presence of acceleration, neither of which is true.

[zoobyshoe]

Originally posted by chroot If you think about it, it has to be that way... if it weren't, then some cosmic ray particle moving at 0.9c with respect to you somewhere in the depths of space would somehow affect YOUR clock!

This part, I don't get. I thought cosmic rays were photons, and as such, could never be observed going less than C.

In other words, I have been under the impression that even if I am traveling at 0.9c all photons whose speed I measure going in any direction relative to mine will be clocked going at C. Is this not the case?

[chroot]

Originally posted by zoobyshoe
This part, I don't get. I thought cosmic rays were photons
No, sorry for the confusion -- some cosmic rays are photons -- but many are massive particles like protons and electrons. The photons, of course, will always be observed travelling at c, while the massive particles will always be < c.

- Warren

[zoobyshoe]

Originally posted by chroot some cosmic rays are photons -- but many are massive particles like protons and electrons.
Interesting. How did they all get the same name, being such different things?

[Ambitwistor]

"Cosmic ray" is just a catch-all term for "particles that come from outer space".

[zoobyshoe]

In Six Easy Pieces he mentions cosmic rays as the highest energy photons we are aware of. Is there a term that can be used to differentiate these from the "riff raff" cosmic rays?

[Ambitwistor]

Originally posted by zoobyshoe
In Six Easy Pieces he mentions cosmic rays as the highest energy photons we are aware of. Is there a term that can be used to differentiate these from the "riff raff" cosmic rays?

Ultrasuperduper high energy cosmic rays.

Actually, we've detected cosmic rays with much higher energies than the highest-energy cosmic-ray photons we've detected. The most energetic photons we've seen are from gamma-ray bursts, maybe on the order of 10 TeV (possibly 100's of TeV's; I'm not sure what the state of the art is). But we've seen charged cosmic rays up beyond 100 million TeV.

[Ivan Seeking]

Originally posted by Ambitwistor
It hasn't been clear to me what your points have been, but it has seemed at times that you have been implying that acceleration is necessary for the twins to experience different elapsed proper times, and/or general relativity is required to resolve the twin paradox in the presence of acceleration, neither of which is true.

Well, somehow we got off track I think...perhaps you are expecting Ivan the Terrible? [:D]

First, I was explaining the mechanics of the paradox; that's all.

Next, I addressed the issue of preferred observers; and I still think correctly so. Perhaps this language is out of favor, but specifically I meant that no absolute state of rest or motion exists. This is a still significant concept of SR; no?

Finally, I keep addressing the issue that if we wish to describe one frame of reference as preferred, this in response to Zooby's question about whose clocks run slowly - meaning to prefer one system over the other - a frame of rest must be defined. I meant this all within the context of SR. In all cases we must still define a frame of rest by which we determine who will age less quickly; true? With our twins, we know who is in motion - the one who leaves earth...and this requires acceleration. The page linked makes this assumption immediately

[Ambitwistor]

Originally posted by Ivan Seeking
Next, I addressed the issue of preferred observers; and I still think correctly so. Perhaps this language is out of favor, but specifically I meant that no absolute state of rest or motion exists. This is a still significant concept of SR; no?

Yes, it's why people find the twin paradox puzzling: they think that since there are no preferred frames, each observer should observe the same thing.

(Actually, there are senses in which states of motion are absolute in SR: you can absolutely determine the magnitude of your proper acceleration.)


Finally, I keep addressing the issue that if we wish to describe one frame of reference as preferred, this in response to Zooby's question about whose clocks run slowly - meaning to prefer one system over the other - a frame of rest must be defined.

Well, you don't have to work in the rest frame of some observer --- if you wanted, you could choose some bizarre curvilinear coordinate system --- but it's easiest to do so when such a global inertial frame exists (as in Minkowski spacetime).


In all cases we must still define a frame of rest by which we determine who will age less quickly; true? With our twins, we know who is in motion - the one who leaves earth...and this requires acceleration. The page linked makes this assumption immediately

I wouldn't say that we know that the Earth twin is "at rest", if that's what you're implying by defining a rest frame, but it is true that we know that acceleration breaks the symmetry between the two twins in this variant of the twin paradox. (There are other variants in which this is not the case.)

[S = k log w]

Originally posted by Ivan Seeking
When the car is in motion, it is shorter in the frame of the garage. Likewise, from the frame of the garage, the car's clocks are running more slowly. When the car stops, that is, when the frame of the car coincides with the frame of the garage, the two lengths L and L0 agree. Likewise, it we compare the ticks of the clocks, we find that again they agree – they occur at the same rate. However, and this was a key test of relativity, we find that while the frames of the car and garage did not coincide, ie, while the car is in motion as viewed from the garage, the clocks in the car really were running more slowly...just as observed and predicted.

This was finally verified I think in the early sixties using two atomic clocks; one on the ground, and one in a jet. After flying one of the clocks around for a while, and after accounting for the effects of gravity, the clock on the plane had indeed lost time as predicted to within the accepted margins of error. This has since be replicated in many other ways. Also, we see the lifespan of subatomic particles increase according to Relativity and their relative speed – since their clocks run more slowly, we see them live longer. This is seen in particle accelerators as well as in nature.

To the extent that isotopes have a 1/2t, does that in relativity,
an accelerating particle runs more slowly than than at rest, factor in when calculating the 1/2t of an isotope?

[jcsd]

Yes, a particle will appear to have a longer half-life when it is travelliong at relativistic speeds in some rest frame, indeed this is a famous example of a relativtic effect.

[Ivan Seeking]

Originally posted by Ambitwistor
I wouldn't say that we know that the Earth twin is "at rest", if that's what you're implying by defining a rest frame, but it is true that we know that acceleration breaks the symmetry between the two twins in this variant of the twin paradox. (There are other variants in which this is not the case.)

This was a poor choice of words on my part.

There are other variants in which acceleration does not break the symmetry between the two twins? I may not know what you mean. Could you give an example or two?

[Ambitwistor]

There are other variants in which acceleration does not break the symmetry between the two twins? I may not know what you mean. Could you give an example or two?

I already gave two examples in my first post to this thread. (Note that, as I clarified later, I am speaking of proper acceleration.)

[Ivan Seeking]

No, acceleration isn't the critical point. You can construct variants of the twin paradox in which nobody accelerates, but the twins come back with different ages.

How exactly does this apply to the twins paradox? I thought a key assumption is that they start and end in the same frame of reference. By this it would seem that the twin set into motion must experience a force in order to start, and to end his trip.

[Ambitwistor]

How exactly does this apply to the twins paradox? I thought a key assumption is that they start and end in the same frame of reference.

Not really; they just need to start and end at the same place. If you stick in an arbitrarily large acceleration, it changes the elapsed proper time by an arbitrarily small amount. In my example, if the travelling twin goes out and comes back, it still takes him only 6 subjective years to do that, regardless of whether he stops at the end or keeps going.

[S = k log w]

Originally posted by Ambitwistor
Not really; they just need to start and end at the same place. If you stick in an arbitrarily large acceleration, it changes the elapsed proper time by an arbitrarily small amount. In my example, if the travelling twin goes out and comes back, it still takes him only 6 subjective years to do that, regardless of whether he stops at the end or keeps going.

This poses an interesting question. IF, as you had said, there is both a theoretical model in which one of the twins does not accelerate, but non-the-less 'comes back' at a different time (or age, as the case may be), and also if it is so that an acceleration need not be 'large' (you have not defined a range for 'large'), but
an arbitrary 'small' (whatever 'small' may mean), the would, if string theory were true, the 'vibration' of the string be (by each string an element in a large dynamic consisting of the sum of all of the strings), would all matter exist in two or more times?

[kawikdx225]

If neither twin accelerates or decelerates then how do you determine which twin is younger since they can both argue the other is moving? Sorry if you already covered this.

Kaw

[Ambitwistor]

If neither twin accelerates or decelerates then how do you determine which twin is younger since they can both argue the other is moving? Sorry if you already covered this.

That's what much of this thread has been about. The short version: the younger twin is the one who travels along a shorter path in spacetime. For the long version, read the thread.

[russ_watters]

Originally posted by kawikdx225
If neither twin accelerates or decelerates then how do you determine which twin is younger since they can both argue the other is moving? Sorry if you already covered this.

Kaw That's what much of this thread has been about. The short version: the younger twin is the one who travels along a shorter path in spacetime. For the long version, read the thread. Actually, if neither twin accelerates, then they are both sitting next to each other not moving with respect to each other for their entire lives. They stay the same age relative to each other.

The whole point of them being "twins" is that at the starting point in their lives they are sitting next to each other in the same frame. To get an age difference, one MUST accelerate.

I think the problem got so complicated, the initial intent of a "twins paradox" may have been lost.

[Ambitwistor]

The whole point of them being "twins" is that at the starting point in their lives they are sitting next to each other in the same frame. To get an age difference, one MUST accelerate.

That's how you get them to follow different spacetime paths, if you assume they were originally following the same spacetime path. (There are versions of the twin paradox in which you don't assume that.) But the main point concerning acceleration in this thread was that some people say that acceleration is what causes the travelling twin to be younger, and that's not really accurate. The elapsed proper time experienced by an observer is not tied up in what happens at the beginning, end, or middle of the trip; it has to do with the geometry of the entire worldline.

[Ivan Seeking]

Originally posted by Ambitwistor
Not really; they just need to start and end at the same place. If you stick in an arbitrarily large acceleration, it changes the elapsed proper time by an arbitrarily small amount. In my example, if the travelling twin goes out and comes back, it still takes him only 6 subjective years to do that, regardless of whether he stops at the end or keeps going.

If we never break symmetry, then how does address the paradox? For example, suppose someone happens by planet earth while travelling near the SOL. I look through my telescope at this person and see someone who appears to be twenty years old. Assume the traveller follows the proper path - only proper acceleration - and then passes by earth in another eighty years or so and I look again. Through my ancient eyes I see a barely aged, 26 year old in the ship. He has also looked at me on first pass, and again on the return path. From his point of view, I was the one in motion. I am the younger. It seems that we have the original paradox. In other words, I don't see how we resolve the twins paradox without breaking symmetry.

[Ambitwistor]

If we never break symmetry, then how does address the paradox?

If there's no symmetry breaking, then there's no age difference. All I'm trying to say here is that acceleration is not the key to the twin paradox; you can have twin paradoxes in which acceleration breaks the symmetry, and paradoxes in which something else breaks the symmetry. All that matters is something break the symmetry.

(I was also pointing out that whether you break the symmetry by accelerating/decelerating at turnaround, you don't have to involve any inital or final acceleration or deceleration either; the end result depends on the entire history of the worldline, not just the endpoints.)

[Ivan Seeking]

Originally posted by Ambitwistor
If there's no symmetry breaking, then there's no age difference....the end result depends on the entire history of the worldline, not just the endpoints.)

OK, this is the crux of the paradox to me. Under SR, we have no absolute frame of reference; and the relative motion between our two observers dates back ultimately to the big bang. So, it seems that we can never find a common or preferred frame no matter how far back we look. So it would seem that the answer is not that there is no age difference, the answer is that unless we break symmetry, there is no unique answer to the question: Who ages less quickly? To me, this also implies that the age of the universe depends on the observer. I have never been clear on this point.

[Ambitwistor]

OK, this is the crux of the paradox to me. Under SR, we have no absolute frame of reference; and the relative motion between our two observers dates back ultimately to the big bang.

Okay. (Although if you're talking about Big Bangs, you're not doing SR anymore.)

So, it seems that we can never find a common or preferred frame no matter how far back we look.

There doesn't have to be a common or preferred frame to compare ages. However, the two twins' worldlines have to be able to intersect in two places in order to compare clocks.


So it would seem that the answer is not that there is no age difference, the answer is that unless we break symmetry, there is no unique answer to the question: Who ages less quickly?

If the situation is truly symmetric, you can uniquely answer the question: they both age at the same rate.

To me, this also implies that the age of the universe depends on the observer.

You're right, the age of the universe does depend on the observer. There is a preferred class of observers in cosmology that everyone uses when talking about the age of the universe: the ones who see the universe as maximally isotropic. The Earth is close to, but not quite, one of those preferred observers.

[chroot]

Originally posted by Ambitwistor
the ones who see the universe as maximally isotropic. The Earth is close to, but not quite, one of those preferred observers.
Just to add to what Ambi said: these preferred frames are just those frames in which the cosmic microwave background radiation is uniform in all directions. If you have some relative motion with respect to the CMBR, you see it as blueshifted along the direction you're moving, and redshifted in the opposite direction. A "comoving" observer is one who is just sort of going along with the flow of the expansion of the universe.

An analogy are two people floating in a river. One guy just floats and lets the water carry him -- he's a comoving observer, and sees the water moving the same speed all around him: zero.

Another guy swims hard in some arbitrary direction. He sees the water moving faster in some directions than in others.

There's nothing that violates relativity theory: this CMBR rest frame is not an absolute rest frame or anything like that. There's nothing special about it relativistically. There is, however, something special about it cosmologically.

Cosmologists measure the age of the universe as the proper time experienced by such comoving observers.

- Warren

[Ivan Seeking]

Originally posted by Ambitwistor
If the situation is truly symmetric, you can uniquely answer the question: they both age at the same rate.

Truly symmetric? I'm not sure what exceptions you refer too here.

So you are saying that in my example, when we use our telescopes during each of our two passes near each other, we see each other aging at the same rate? That is, on his second pass by earth, I don't see a 26 year old in my scope, I see a 100 year old person in the ship?

[Ambitwistor]

So you are saying that in my example, when we use our telescopes during each of our two passes near each other, we see each other aging at the same rate? That is, on his second pass by earth, I don't see a 26 year old in my scope, I see a 100 year old person in the ship?

I can't say for sure, since I asked and you didn't clarify the details of how the other twin is returning. But I would guess that the situation is not symmetric and nor will be their ages. If the travelling twin returns by accelerating/decelerating, or by gravitational slingshot, or by circumnavigating a closed universe, then his worldline is very different from the Earth twin's, so there is no symmetry (regardless of whether he starts or ends at rest with respect to the Earth twin).

[Ivan Seeking]

Originally posted by Ambitwistor
I can't say for sure, since I asked and you didn't clarify the details of how the other twin is returning. But I would guess that the situation is not symmetric and nor will be their ages. If the travelling twin returns by accelerating/decelerating, or by gravitational slingshot, or by circumnavigating a closed universe, then his worldline is very different from the Earth twin's, so there is no symmetry (regardless of whether he starts or ends at rest with respect to the Earth twin).

I mean the situation is which our world lines have never crossed since the big Band, and where our traveler follows a course having only proper accelerations - allowing a second pass by earth and a second comparison of our clocks, by telescope..

[Ambitwistor]

Originally posted by Ivan Seeking I mean the situation is which our world lines have never crossed since the big Band, and where our traveler follows a course having only proper accelerations

What does "having only proper accelerations" mean? That the traveller has a nonzero proper acceleration at all times? Does he accelerate and decelerate? Or what?

[Ivan Seeking]

Originally posted by Ambitwistor
What does "having only proper accelerations" mean? That the traveller has a nonzero proper acceleration at all times? Does he accelerate and decelerate? Or what?

I mean that we never break symmetry.

[Ambitwistor]

I mean that we never break symmetry.

Can you describe a physical scenario in which the travelling twin is able to pass the Earth twice, without breaking symmetry?

[Ivan Seeking]

Originally posted by Ambitwistor
Can you describe a physical scenario in which the travelling twin is able to pass the Earth twice, without breaking symmetry?


From some of your earlier comments, I thought that you knew of such a trick. If this is not possible, then all is well.

[Ambitwistor]

From some of your earlier comments, I thought that you knew of such a trick. If this is not possible, then all is well.

I can invent such a scenario (but not in pure special relativity); I just didn't know what scenario you were considering. For instance, both twins could circumnavigate a closed universe, with opposite velocities relative to a cosmological observer. In that case, their aging would be symmetric.

[Ivan Seeking]

Originally posted by Ambitwistor
I can invent such a scenario (but not in pure special relativity); I just didn't know what scenario you were considering. For instance, both twins could circumnavigate a closed universe, with opposite velocities relative to a cosmological observer. In that case, their aging would be symmetric.

So in this case, if on some near pass one twin accelerates and joins the other, we find that the twins are [almost exactly] the same age?

[Ivan Seeking]

Originally posted by chroot
Cosmologists measure the age of the universe as the proper time experienced by such comoving observers.

- Warren

Is there a maximum age for the universe as viewed by some class of observers; where all other observers will measure a lesser age?

[Ambitwistor]

So in this case, if on some near pass one twin accelerates and joins the other, we find that the twins are [almost exactly] the same age?

Yes, assuming that they were the same age during one of the passes, they will remain the same age on successive passes. Accelerating into the same frame doesn't change this (much).

[Ambitwistor]

Is there a maximum age for the universe as viewed by some class of observers; where all other observers will measure a lesser age?

Yes. The preferred comoving cosmological observers experience maximal proper time.

[S = k log w]

Originally posted by russ_watters
Actually, if neither twin accelerates, then they are both sitting next to each other not moving with respect to each other for their entire lives. They stay the same age relative to each other.

The whole point of them being "twins" is that at the starting point in their lives they are sitting next to each other in the same frame. To get an age difference, one MUST accelerate.

I think the problem got so complicated, the initial intent of a "twins paradox" may have been lost.

If neither twin accelerates, than can their either be no twin, or perhaps can both twins exist at the same time in the same place?

[Ivan Seeking]

Originally posted by Ambitwistor
Yes, assuming that they were the same age during one of the passes, they will remain the same age on successive passes. Accelerating into the same frame doesn't change this (much).

This is true since by circumnavigating a closed universe, they qualify as preferred comoving cosmological observers?

Maybe a better question, what about this path integral is unique?

[Ambitwistor]

This is true since by circumnavigating a closed universe, they qualify as preferred comoving cosmological observers?

In the scenario I was describing, both twins have opposite velocities with respect to a cosmological observer; neither one is cosmological. If one is cosmological and the other is not, then the situation is not symmetric and the cosmological observer will age more.

Maybe a better question, what about this path integral is unique?

What do you mean?

[Ivan Seeking]

Originally posted by Ambitwistor
What do you mean?

Is the only unique characteristic of this path that the observers can make two passes without breaking symmetry?

I was thinking that by circumnavigating the universe, they still move with the expansion of space [sorry, I don't know the correct language here other than to say in a manner similar to the defintion of a preferred cosmological observer]. While cicumnavigating the universe, the universe expands; so it seems that in order to maintain symmetry, these two observers must move with this expansion. This would seem to make their situtation unique.

[Ambitwistor]

Originally posted by Ivan Seeking
Is the only unique characteristic of this path that the observers can make two passes without breaking symmetry?

Well, that was my purpose in constructing it. I can't say whether this is "the only unique characteristic".

I was thinking that by circumnavigating the universe, they still move with the expansion of space [sorry, I don't know the correct language here other than to say in a manner similar to the defintion of a preferred cosmological observer].

Well, we don't have to consider an expanding universe or even a curved one, but anyway...


While cicumnavigating the universe, the universe expands; so it seems that in order to maintain symmetry, these two observers must move with this expansion. This would seem to make their situtation unique.

Unique, compared to what?

[Ivan Seeking]

First, Ambitwistor, I want to thank you for all of your great answers here.

I have been reviewing the discussion to get my bearings. In your example, when our two travelers pass each other, do they still measure [see] each other's clocks running slowly - assuming that we compensate for the communication delays and doppler shift.

[Ambitwistor]

I have been reviewing the discussion to get my bearings. In your example, when our two travelers pass each other, do they still measure [see] each other's clocks running slowly - assuming that we compensate for the communication delays and doppler shift.

If you're talking about the two moving twins in the closed universe example, then yes, they're each see each other's clocks running slowly by the usual gamma factor.

[Ivan Seeking]

Originally posted by Ambitwistor
If you're talking about the two moving twins in the closed universe example, then yes, they're each see each other's clocks running slowly by the usual gamma factor.

How does the frame of a preferred cosmological observer differ from that of an absolute rest frame? I find myself struggling to avoid this implication. It would seem that we can judge all frames of reference by the PCO frame, and the notion that we have no preferred observers then fails.

[chroot]

Ivan,

It happens to be a special reference frame because it was given to us by the universe, in a sense. However, it's not special from the perspective of relativity theory.

For example, the universe also gave us Bob. We are welcome to consider the reference frame in which Bob is at rest 'special,' in the sense that the universe gave us that frame. We can choose to judge all other frames of reference with respect to Bob.

There are, however, an infinite number of these 'special' reference frames, including the PCO, Bob, Frank, and Tommy too. That makes them seem, well, not so special after all.

- Warren

[Ivan Seeking]

Originally posted by chroot
Ivan,

It happens to be a special reference frame because it was given to us by the universe, in a sense. However, it's not special from the perspective of relativity theory.

For example, the universe also gave us Bob. We are welcome to consider the reference frame in which Bob is at rest 'special,' in the sense that the universe gave us that frame. We can choose to judge all other frames of reference with respect to Bob.

There are, however, an infinite number of these 'special' reference frames, including the PCO, Bob, Frank, and Tommy too. That makes them seem, well, not so special after all.

- Warren

So then the symmetry between our two navigators is not absolute, and who is the younger depends on ones frame of reference?

[chroot]

Originally posted by Ivan Seeking
So then the symmetry between our two navigators is not absolute, and who is the younger depends on ones frame of reference?
No, the proper time along a worldline is invariant, and will be measured the same by all observers.

- Warren

[Ivan Seeking]

Originally posted by chroot
No, the proper time along a worldline is invariant, and will be measured the same by all observers.

- Warren

Yes, after a few quick calculations I see that I can't create the paradox here that I thought. The real paradox seems to be the apparent conflict between this statement and what is observed on moving clocks. How do we reconcile this? How can we observe clocks running slowly when in fact they may or may not be?

[Ambitwistor]

Originally posted by Ivan Seeking
How does the frame of a preferred cosmological observer differ from that of an absolute rest frame?

Yes, that's a common question. You have to carefully examine what relativity actually says. It says that the laws of physics do not prefer any reference frame. But solutions of those laws certainly can.

Suppose you had a universe with a spherical mass in it. Certainly there is a preferred reference frame: the rest frame of the mass. An observer, through physical measurements of the spacetime, can determine whether he's in that frame (even if he can't see the mass). The same goes for a cosmological frame: if the universe is isotropic in your frame, you're a cosmological observer. If it's not, you aren't. Anybody can tell whether they're in that special frame.

(If you want to nitpick, even special relativity has preferred frames: they are the inertial frames; any observer can determine absolutely whether they are in an inertial frame or not, by seeing whether they are experiencing any proper acceleration. It's just that Minkowski spacetime has a lot more symmetry than Schwarzschild spacetime, or Friedmann spacetime, or whatever, so there are lots of preferred frames.)

[Ambitwistor]

How can we observe clocks running slowly when in fact they may or may not be?

What do you mean? The circumnavigating ("Magellanic") twin paradox I raised isn't any different in this respect than two observers in motion in ordinary Minkowski spacetime: each observer symmetrically sees the other's clock as running at a slower rate than their own.

[Ivan Seeking]

Originally posted by Ambitwistor
What do you mean? The circumnavigating ("Magellanic") twin paradox I raised isn't any different in this respect than two observers in motion in ordinary Minkowski spacetime: each observer symmetrically sees the other's clock as running at a slower rate than their own.

True. I thought I had a way to escape this particular paradox. You guys blocked my exit.

[Ivan Seeking]

In the Magellanic twin paradox, if each twin continually monitors the ticks of the other's clock throughout the trip, say by using LASER pulses, then we should be able to compare the number of ticks measured by each on any pass; and at the end of the trip. That the number of ticks measured by each would be the modified by the factor gamma, and since this agrees with the predicted value, I have always considered these to be real measurements.

You seem to be telling me that each twin can see the other's clocks running slowly relative to their own, by telescope, and by counting LASAR pulses, but each sees the other aging [by telescope] not only symmetrically, but at a rate that does not agree with the measured clocks?

[Hurkyl]

Consider this:

Every year, the eastbound twin sends a laser pulse to the west.
Every year, the westbound twin sends a laser pulse to the east.

When they meet again, a lot of the laser pulses are still in transit; neither twin will have counted a number of laser pulses equal to the number of years he (and his twin) has aged.

[Ivan Seeking]

Originally posted by Hurkyl
Consider this:

Every year, the eastbound twin sends a laser pulse to the west.
Every year, the westbound twin sends a laser pulse to the east.

When they meet again, a lot of the laser pulses are still in transit; neither twin will have counted a number of laser pulses equal to the number of years he (and his twin) has aged.

Eastbound sends his pulses East.
Westbound sends his west.

[Hurkyl]

Ok.

First, the twins go through a period of not seeing any pulses; they have to wait for the first pulse to go around the universe. Then, due to the doppler effect, each twin will receive more than one pulse per year, and the number of received laser pulses will match the number of years they've aged.


I think I've figured out the other part of your problem. The way Einstein synchronized measurement in Special Relativity interacts oddly with a closed, flat universe. Basically, your measurement of the time coordinate depends on how many times around the universe you are looking. E.G.

The westbound twin measures 10 years on his clock before they meet again.

However, if the westbound twin is always looking west in order to watch the eastbound twin, then the westbound twin might measure, according to the westbound twin's clock, that the eastbound twin has been travelling 15 years.

So, in this situation, the westbound twin will measure that the eastbound twin's clock is running 33% too slow, but that's exactly compensated by the fact the westbound twin measures the eastbound twin travelling for 50% more time.

[Ivan Seeking]

Originally posted by Hurkyl
Ok.

First, the twins go through a period of not seeing any pulses; they have to wait for the first pulse to go around the universe. Then, due to the doppler effect, each twin will receive more than one pulse per year, and the number of received laser pulses will match the number of years they've aged.

We already clarified earlier that we have compensated for communication delays and doppler shift. The final number of pulses counted over one full circulation is all that matters anyway. It really doesn't matter when we count them. The twins could just stop [relative to the PCO] when they meet and wait for the pulses.

[Ivan Seeking]

Originally posted by Hurkyl
I think I've figured out the other part of your problem. The way Einstein synchronized measurement in Special Relativity interacts oddly with a closed, flat universe. Basically, your measurement of the time coordinate depends on how many times around the universe you are looking. E.G.

The westbound twin measures 10 years on his clock before they meet again.

However, if the westbound twin is always looking west in order to watch the eastbound twin, then the westbound twin might measure, according to the westbound twin's clock, that the eastbound twin has been travelling 15 years.

I've been staring at this but I don't get it. The westbound twin and the eastbound twin both left the origin at t=0 for everyone. When they meet again, he measures 10 years on his clock. Clearly they have both been traveling for the same period of time; according to the westbound twin.

[Hurkyl]

Clearly they have both been traveling for the same period of time; according to the westbound twin.

You'd think! I've had a discussion with someone else on this very scenario and it took me a while to realize that this is not the case.

Have you drawn space-time diagrams before, and thus have seen how lines of simultaneity relate to one's motion and measuring system? This will make more sense if you have...


Let's actually use a model of our universe; grab a sheet of paper and roll it into a cylinder. Let's draw the simplest reference frame; pick a point near the middle of the cylinder to be the origin, draw a line down the axis of the cylinder to be the time axis, and draw a circle around the cylinder to be the x-axis.

Now, let's draw the worldline of one of the twins. Have him start at the origin of the special coordinate system. Draw a line at an angle of, say, 30 degrees to the left of the time axis. (so he's travelling north-north-west... where north is the time direction)

Now, let's draw a line of simultaneity on the cylinder. This will make a 60 degree angle to the time axis, so it runs in a west-north-westery direction.

Because this is a line of simultaneity for the westbound twin, all points on this line are measured to be occuring at the same time. The key feature to notice is that as the line does not meet itself to form a circle. The time at which the westbound twin measures an event depends on which direction he is looking, and how many times around the universe he considers the event to be happening!


In particular, if the westbound twin is looking west at his other twin, he'll measure that the eastbound twin has a big head-start. (Of course, he can't make this measurement until the first laser pulse reaches him)

[Ivan Seeking]

OK...bare with me here I'm struggling with this, are you saying that in order to integrate over all space, part of our path becomes purely temporal?

[Hurkyl]

I think I'm trying to say something about global reference frames.

If I understand correctly, the "right" answer to your questions is:

"There is no coordinate system which contains both the westbound and the eastbound twins' paths in their entirety."


Consider this: in a real special relativistic reference frame, the westbound twin should never meet the eastbound twin again; the eastbound twin keeps going east, east, and more east.

However, we've all played enough Asteroids to know how to patch this up; draw two vertical lines which are supposed to represent the same place in the universe and say that anyone that goes past the line on the right wraps around to the line on the left.


The problem is that Asteroids has mislead us! In the simplistic universe I described, those lines are generally not supposed to be vertical lines; they should be slanted lines... so when you "wrap around", you change both your spatial position and your temporal position.

(equivalently, you can draw the lines vertical, but when you wrap around, you are also shifted... so, for example, the event 100 light-years to the east could be the very same as the event 10 years in the future)


The right answer, I guess, is that you're supposed to use two different coordinate systems, one for, say, the first two-thirds of the trip and one for the last two-thirds of the trip, and get the right coordinate transformations in the region where they overlap.


And I'm saying all of this because I don't know what you mean in your last post. [:)]

[Ivan Seeking]

Sorry, I have to be careful with the language here. Judging by your answer I would say that you understood my question well enough. [:)]

The right answer, I guess, is that you're supposed to use two different coordinate systems, one for, say, the first two-thirds of the trip and one for the last two-thirds of the trip, and get the right coordinate transformations in the region where they overlap.


Does this solution require that path followed by each twin is infinite in length? This may sound obvious but I just want to be sure.

[Ivan Seeking]

After a fantastic review of the principles involved here, and thanks to all this has been absolutely great, I am still left without any way to resolve the contradiction between the elapsed time measured on a clocks in motion, and the time truly elapsed, the proper time, as measure by all observers. If we go back to the original question, way back on page one, we argue that we really can fit a 5 meter car in a 3 meter garage. This depends on length contraction being real while the car is in motion. Based on this line of reasoning, and since we tend to consider that measured as real, and since we can show that time dilation really does happen for clocks in motion, I have always considered that it is real. That is, observers in the inertial frame measure the true elapsed time for all clocks in relative motion [in the inertial frame]. I thought I knew why this is not a problem [really I just landed on another paradox], and I thought that this agreed with GR, but Chroot, Ambitwistor, and Einstein cut me off at the pass.

Now, either I have thought about this so long I can’t see straight any more, or it seems that for me this is the state of the art of the twins paradox: In the example given by Ambitwistor, on page 2, the twin in motion measures the earth based clock running slowly over the entire trip. We can make his trip as long as we want. The correction to the clock on earth as observed by the twin in motion never seems to happen except during the turn around and the final deceleration. This deceleration is independent of the length of the twin’s trip; that is, the effects of acceleration and deceleration seem to be independent of the correction needed for the earth based clock as viewed by the traveling twin. Is this problem resolved in some way that I have completely missed, or have I stated the problem incorrectly, or do we have a real discontinuity between the value for the earth based clock as measured by the twin in motion – while in flight - and the proper earth time elapsed on the earth based clock?.

[Hurkyl]

The discontinuity is the instantaneous acceleration. The net effect of the acceleration is that the spacebound twin instantaneously changes his reference frame.

If you compute the space and time coordinates of earth before and after this change, you'll find that there's a large difference in the time coordinate. (the difference is proportional to the distance to earth)

If, instead of instantaneously turning around, the spacebound twin accelerated gradually, he'd observe the clock on Earth is running really, really fast.

[Ambitwistor]

In the example given by Ambitwistor, on page 2, the twin in motion measures the earth based clock running slowly over the entire trip. We can make his trip as long as we want. The correction to the clock on earth as observed by the twin in motion never seems to happen except during the turn around and the final deceleration.

If you travel towards someone or away from someone, you will observe that person's clock as running more slowly than your own; that's ordinary SR. What happens at turnaround, though, is that you switch inertial frames, and so you switch surfaces of simultaneity: you can't forget the relativity of simultaneity. All of a sudden, what the distant clock reads "now" jumps, because you have suddenly redefined what "now" means by switching frames. The Earth-based clock doesn't do this. This is what the "time gap objection" in the FAQ is about.

[Ivan Seeking]

There is still one element of the time gap explanation that I have never been able to reconcile. Now, if I read the FAQ explanation correctly, the traveling twin sees the earth based twin age nearly the entire 14 years [in this case the real elapsed earth time] during the return leg of the trip. This throws me a bit.

Let’s make this simpler and assume that our traveling twin travels out from earth for a designated period of time, and he then stops and waits until he sees light pulses from earth. We plan all of this so that he should see the first pulse just after he reaches a complete stop. Then he accelerates to near C and heads home. Now, after compensating for the Doppler shift, according to the traveling twin the earth based clocks are running slowly. Right??? The linked page seems to say otherwise. Now, let’s further assume that we can decelerate the twin’s ship very quickly; say in just a few hundred meters. Doesn’t he see the earth clocks catch up to real earth time [ie. so that both twins count the same number of flashes over the entire return trip] entirely during the deceleration?

If true, this is the part that loses me:

Until the twin decelerates, where are the missing light pulses?

In the earth frame, the pulses left long ago and are clearly not contained in the space between us and the twin.

[Hurkyl]

From the spacebound twin's point of view (measurements of Earth are assumed to be corrected for the propagation of light, including the Doppler effect):

On the outward journey, he will observe earth's clocks running slowly.

As he decelerates to come to a "stop" (I presume you mean that he comes to rest in Earth's reference frame), he observes Earth's clocks running very fast.

When he comes to a stop, he will observe that Earth's clocks are now ahead of his own.

He receives the first laser pulse, so he begins his acceleration. During this acceleration, he will see Earth's clocks run very fast.

During the return journey, Earth's clocks are very nearly at a standstill, however, due to the doppler effect, he receives laser pulses at a rapid rate.

Finally, he decelerates when he gets back to Earth, which has virtually no effect on any observations about Earth or laser pulses.

[Ivan Seeking]

So I did read this correctly. It seems that the form of the relativistic Doppler shift equation never sank in back in college; and I never looked it up. This along with a fundamental misunderstanding of the space-time interval seems to have led me astray.

So, we can sum this up as follows:

1). First and foremost: measurements on clocks and rods in the inertial frame that are made by observers in the inertial frame are real. There are no illusions.

2). With the exception of gedunken experiments that require infinite path lengths or that are in some other way impractical, we observe no discontinuities in the behavior of clock and rods that require any tricks of math to resolve.

3). Finally, the apparent strangeness of the statement that each observer sees the other’s clock running slowly is really resolved by the Relativistic Doppler shift equation, and of course the relativity of simultaneity.

One interesting aside here is that objects in motion are viewed as rotated. A cube in motion appears rotated to reveal one side otherwise parallel to the observer’s line of sight. Spheres in motion do not appear contracted as ellipsoids; they still look like spheres!

For me there is only one question remaining. From what I understand of SR and GR, when we dig deep enough we find that the constancy of C for all observers lies at the base of both theories. Can we say why C is constant for all observers? Is there a more fundamental statement in this regard?

Thanks again Hurkyl, Chroot, and Ambitwistor. I don't think I have ever had the chance to so fully explore some of the finer point of SR. [a)]

[Ambitwistor]

From what I understand of SR and GR, when we dig deep enough we find that the constancy of C for all observers lies at the base of both theories. Can we say why C is constant for all observers? Is there a more fundamental statement in this regard?

Not really. There are other formulations of relativity in which the "speed of light is constant" axiom is derived, not postulated ... but you can always play that sort of game: pick a few theorems of a theory, take them as axioms, and derive the old "axioms" as theorems from the new axioms (previously theorems). It's not really anything deep that goes beyond what we already know. But it depends on your sense of aesthetics. For instance, if you really don't like instantaneous action at a distance, then you might take causality axioms as fundamental, and then that would "explain" the constancy of the speed of light. (In non-relativistic theories, effects can propagate at arbitrarily high speeds; if you set a speed limit, then you can assume some additional symmetry principles that force the speed limit to be universal for all observers.)