Why is the speed of light constant for all observers?

[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 traveling 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 traveling 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 traveling 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 traveling 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 occurring 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.