One-way speed of light and clock sync

[nikolakis]

Ok, i know this makes me look like a crackpot, but suppose for an instant there is an aether or absolute space or whatever!
Let me be on a rod that travels tranversely with respect to aether and let me be exactly on the middle point of this rod. I have a GPS device with me and send it upwards for some distance. It transmits two light signals back to the ends of the rod and I synchronize my clocks at both ends to read the same time accordingly. Light travels same distances at same speed so i am certain that my clocks are synchronized!
Now, very slowly I turn my rod so it travels longitudinally with respect to aether. Shouldn't my clocks remain synchronized and reading the same time? If so then I measure different light speeds for the inbound and outbound legs of a two-way trip of a beam of light!
Reichenbach and Grunbaum have stated that ε=1/2 factor Einstein uses to synchronize two
spatially remote clocks is an arbitrary plantation.
Is it true? Why the one-way speed of light will never be measured?

 

[mfb]

Shouldn't my clocks remain synchronized and reading the same time?

They perform a different motion relative to the aether. It depends on the way you want to use this aether. If you want to look at your own reference frame and assume physics is the same there: Sure, but where is the point in the aether then?

The other way of your one-way-experiment is the GPS signal. With an aether, were you assume that light has the same speed in all directions, you can measure the one-way-speed. This is not interesting, however, as you have assumed that it is constant anyway.

 

[nikolakis]

Thank you!
In the context of the old Michelson-Morley experiment the speed of light in aether is always constant. It is the velocity of the observer which should account for the different measurements of the speed of light we should measure and we did not! (as far as the two- way speed of light is concerned)
To clarify things here i don't care for aether! Call it absolute space if you like...
Furthermore, this experiment never refuted the existence of an ether. It accounted for an isotropic speed of light, by introducing the γ-contraction-factor for longitudinal lengths and time. See Arthur Eddington.
Anyway, since in my thought experiment i move tranversely wrt to absolute space it does not mind what the speed of light is! To synchronize two different clocks it suffices to know that the relative speed of light is the same for both clocks, which by symmetry holds true in my case.
And, when i rotate the rod i do it very slowly at almost 0 speed, which does not alter the flow of time of the clocks.

 

[mfb]

In the context of the old Michelson-Morley experiment the speed of light in aether is always constant.

Right. But if you have to assume that to perform your experiment at all, there is no point in measuring this speed in different directions. No theory predicts a deviation.

 

[nikolakis]

I am not assuming anything...
I was just pondering on how an ε=1/2 can lead to a theory of such inner beauty...

 

[nikolakis]

On re-reading my statement, i realized i have made a mistake.
I said that the rod is moving transversely with respect to aether. This is wrong. I should have said that the rod has a velocity directly perpendicular to its longitudinal axis of direction with respect to the static aether.
The rest of the argument holds valid.
Let's assume that I am an early 20th century physicist who believes that the speed of light in luminiferous motionless aether is constant.
I have also verified by experiment that rods contract a γ-Lorentz factor √(1-v²) when they move with velocity v longitudinal to their axis and clocks delay the same amount independent of velocity direction.
Also, forget about Earth's rotational velocity, its speed around the sun, the speed of the galaxy, and expansion of the universe. All these will amount to a resultant velocity at an angle θ say to the longitudinal axis of my rod. By slowly turning the axis of my rod, I can ensure that at some point this resultant velocity will be directly perpendicular to the longitudinal axis of my rod. How can I be sure of this?
By establishing my position exactly in the middle of the rod, and sending two wrist-watches ( both at same speed) - lol - at both ends of the rod and recording the times they read on their arrival. I shall postulate by a simple symmetry argument that if their respective reading times on arrival are identical, then the resultant relative velocity of my rod with respect to its longitudinal axis is directly perpendicular to it!
This postulate is in accordance even with SR (were not true!) that is on a static line in the horizontal direction in Minkowskian-spacetime, two identical clocks placed in its middle will arrive on the same time at its ends, were they to move with identical speeds.
I will synchronize, the clocks at both ends of the rod with the same identical reading of time of my moving watches, if i ever ever arrive at such a preposterous measurement and i will remain clinged to the idea that this is the only possible true synchronization!
Now, i very slowly rotate my rod. Length-contraction? Irrelevant! Who cares? By slow rotation, I have ensured that the rate of flow of time has not changed in any of the clocks.
But at what cost? Now, I am having an √(1-v²) length rod with all the clocks synchronized to it at the same time in the moving frame of the rod with respect to the static aether!
The two-way speed of light is always the same, but the one-way speed... oh no!

 

[pervect]

The way I look at it is this:

There can be only one clock synchronization for which Newton's laws work exactly.

Consider two objects of equal mass moving at the same velocity in opposite directions and coming to a stop. Assume that this happens with a fair, isotropic clock synchronization scheme as a starting point for the analysis.

Now, let's look at some clock synchronization other than this "fair" one, to see what happens - to find the effect of an "unfair" or "non-isotropic" clock synchronization.

In that case, the objects still come to a stop - the results of the experiment do not depend on how we syncrhonize our clocks. However, the measured velocities are no longer numerically equal, due to what one might call "clock synchronziation error" - the result of using something other than the isotropic clock synchronization.

So there is a unique clock synchronization in which equal masses moving at the same velocity in opposite directions stop.

This is theoretically sufficient to define what we want, but AFAIK it's hard to test experimentally.

What Einstein's postulate says is that this fair, isotropic, clock synch (which is what we want) can be achieved using the principle that the speed of light is constant in any frame rather than the tedious method of colliding equal masses and making sure they come to a complete stop.

 

[ghwellsjr]

Ahh--the slow transport of clocks--the myth that it is better than Einstein's synchronization convention. In fact, it is no different. Let me show you why.

First off, let me say that when you want to adopt a state of absolute aether, it is no different than selecting an Inertial Reference Frame (IRF) so you don't have to feel apologetic. But the good news is that you can use all of Einstein's Special Relativity to analyze what would happen in a state of absolute aether.

You have stated many equivalent things in this thread, including that you can synchronize two remote clocks with a signal that originates at a location equidistant from both clocks or that you can use this technique to confirm that two such clocks are in sync, including two clocks (or watches) that were slowly transported either from the end points of a rotating rod or from the midpoint of the rod to its extremities.

These are all different expressions of the same theme, slow transport of clocks. So in order to not have to analyze each one of them separately, I'm going to analyze just one scenario where we have three co-located clocks and then we slowly move two of them in opposite directions by the same distance and then we will use a signal sent from the stationary clock to confirm that they continue to remain in sync.

I'm going to use a distance measure of feet and I'm going to define the speed of light to be one foot per nanosecond. We're going to move the two clocks at a speed of 0.1c for 100 nanoseconds according to their own Proper Time which will move them about 10.05 feet. Just before they get there, we're going to send signals from the stationary clock and show that they arrive at the same time according to the times on the two slowly moved clocks. Here's a spacetime diagram to illustrate this in an IRF in which the clocks started out at rest:

The dots on each worldline represent Proper Time one-nanosecond ticks of each clock. They all started out at the origin of the frame set to zero so we see them 100 nanoseconds later where the last dot appears for each of them. We have to send the signal shown in green coming from the stationary clock at about 90.5 nanoseconds in order for them to arrive exactly when the two slowly moved clocks come to rest with readings of 100 nanoseconds on them.

Now we look at the same thing in an IRF moving at -0.6c with respect to the first one:

Here we see that even though the two slowly moved clocks receive the signals sent from the stationary clock at the same Proper Times for both of them, they are not in sync.

The point of this exercise is to illustrate that it is impossible to determine by light signals or by slowly moving clocks any sense of absolute synchronization. They will both agree but still not indicate any synchronization with respect to a state of absolute aether.

 

[nikolakis]

Ok, ghwellsjr, thank you!

I have been convinced! And by joining the two end-points of the red and black lines in the second diagram, we drop back to the rest frame of the moving rod from which we emitted the slow clocks!

Another question is: Is the Reichenbach assertion that the choice of ε=1/2 in clock synchronization is arbitrary, correct or is it a fallacy?

 

[ghwellsjr]

Ok, ghwellsjr, thank you!

You're welcome.

I have been convinced! And by joining the two end-points of the red and black lines in the second diagram, we drop back to the rest frame of the moving rod from which we emitted the slow clocks!

Another question is: Is the Reichenbach assertion that the choice of ε=1/2 in clock synchronization is arbitrary, correct or is it a fallacy?

I'm not familiar with Reichenbach or his work but it sounds very much like Einstein pointing out that we can apply any synchronization we want and so he divides the roundtrip time for a light signal in half to define the time on the remote clock.

 

[nikolakis]

First of all, thanks to everyone into this forum.

The reason that I've made such a foolish mistake, to think that we can find a universally transverse moving frame in aether to synchronize our clocks within, and question the principles of SR, is because i was thinking of a rotating rod and its associated flat metric space-time.

I am familiar with the concept of treating the helix described by any of its points as a geodesic so-to-speak, doing flat space-time tensor diff. geometry to trace how a light beam will bend to trace its course, and in the enlightment that an helix is not a true geodesic in flat spacetime, starting to think what kind of curved spacetime could create such an helix to take away all the fields creating centrifugal forces in it... hence GR.

I think it was the rotating rod first, this and the writings of Ernst Mach that probed Einstein to GR. Anyway...

Common knowledge has it in SR, that if we ever find ourselves in an accelerating frame ( when a rod rotates ) and suddenly the acceleration stops, then we re-synchronize our clocks to account for the discrepancies in the one-way speed of light.

And yet, the upsetting fact remains, I was thinking overnight and could not sleep, What if there is an "aether"? What if, by accident, we hit upon a frame where a rod's velocity is perpendicular to its axis as we move into "aether" and synchronize our clocks there, we will never be sure that we have found one, as ghwellsjr very wisely demonstrated to me, but what if? And, if we rotate our rod very slowly, nothing changes in our clock-readings. Only length-contraction occurs.

SR is a self-coherent theory, it contains no contradictions, we can't use its axioms to prove that it is wrong, but this does not prove or disprove anything...

My foolish mistake manifests that we can't say: Ok, let's suppose ε=1/2, this leads to Lorentz-transformations, the Minkowskian pseudo-Euclidean spacetime, the metrix dτ^2=ds^2-dt^2 --- let's see if we can be led to a reduction in absurdum using its axioms... no!

I haven't done the math yet, but another space-time with a different metric should be invented for cases other than ε=1/2. This would lead to another theory, taking none however from the validity of SR, it would be just another theory...

I have nothing more to say upon this matter, if anyone could shed more light into this problem, you are most welcome.

 

[Dale]

My foolish mistake manifests that we can't say: Ok, let's suppose ε=1/2, this leads to Lorentz-transformations, the Minkowskian pseudo-Euclidean spacetime, the metrix dτ^2=ds^2-dt^2 --- let's see if we can be led to a reduction in absurdum using its axioms... no!

I haven't done the math yet, but another space-time with a different metric should be invented for cases other than ε=1/2. This would lead to another theory, taking none however from the validity of SR, it would be just another theory...

It wouldn't even be another theory, just a new coordinate system (a non-inertial frame) in the existing theory. It could be handled the same way any non-inertial coordinate system is handled..

 

[nikolakis]

It wouldn't even be another theory, just a new coordinate system (a non-inertial frame) in the existing theory. It could be handled the same way any non-inertial coordinate system is handled..

Obviously, you are thinking how the rod contracts when it rotates, and you are still applying the axioms of SR with ε=1/2 when it does. Still, that's not a contradiction!

With another ε some other contraction could be valid, as well. I do not know...

 

[Dale]

Obviously, you are thinking how the rod contracts when it rotates, and you are still applying the axioms of SR with ε=1/2 when it does.

No, I am most defnitely not thinking that.

With another ε some other contraction could be valid, as well. I do not know...

Yes, the transformations between different non-inertial frames defined by ε≠1/2 would not be the same as the transforms between different inertial frames defined by ε=1/2. That would mean that the expressions for length contraction, time dilation, and relativity of simultaneity would be different also.

Nevertheless, they are just a different set of coordinates, like any other non-inertial set of coordinates, and don't lead to any new experimental predictions.

 

[nikolakis]

Nevertheless, they are just a different set of coordinates, like any other non-inertial set of coordinates, and don't lead to any new experimental predictions.

I am not certain as to the validity that they should be a non-inertial set of coordinates, i do not even now what kind of manifold we should be using to lead us to the same experimental predictions, namely same two-way light speed, same length-contraction and same time-dilation, take aside the one-way speed of light and simultaneity which must be different.

 

[Dale]

I am not certain as to the validity that they should be a non-inertial set of coordinates

Obviously they are non inertial. The one way speed of light is c in inertial frames, by the second postulate.

i do not even now what kind of manifold we should be using to lead us to the same experimental predictions,

You would use a flat manifold.

namely same two-way light speed, same length-contraction and same time-dilation, take aside the one-way speed of light and simultaneity which must be different.

Length contraction and time dilation are frame variant coordinate artifacts, they are not experimental predictions. Experimental predictions are all invariants.

 

[nikolakis]

Length contraction and time dilation are frame variant coordinate artifacts, they are not experimental predictions. Experimental predictions are all invariants.

Sorry, I was referring only in the sense of the Michelson-Morley experiment and the other experiment with the longer arm, i always forget its name, that established time-dilation.

I am trying to forget everything about SR, to see what happens with other ε's.

But, i don't have to argue about anything, about SR or its postulates.

 

[Dale]

I am trying to forget everything about SR, to see what happens with other ε's.

It isn't that difficult. Just write down the transform between an inertial frame and a non-inertial frame, and use that to derive the metric in the non-inertial frame. Once you have the metric, then everything else follows the normal rules.

 

[nikolakis]

It isn't that difficult. Just write down the transform between an inertial frame and a non-inertial frame, and use that to derive the metric in the non-inertial frame. Once you have the metric, then everything else follows the normal rules.

But, how can i do this since i haven't defined any flat space-time yet?

So far, all experiments have convinced me of the following 3 facts:

1. The invariance of the 2-way speed of light.

2. The existence of a Lorentz factor γ, for lengths and times in moving frames.

3. SR is an ε-dependent theory ( with ε=1/2 ) that makes accurate predictions.

If some-one has proven that ε=1/2 is the only possible value, because there can be no-other flat spacetime with properties 1 and 2, other than this with ε=1/2 ( the space-time of SR ), I would be very glad to accept it.

 

[DrGreg]

It isn't that difficult. Just write down the transform between an inertial frame and a non-inertial frame, and use that to derive the metric in the non-inertial frame. Once you have the metric, then everything else follows the normal rules.

But, how can i do this since i haven't defined any flat space-time yet.

But you don't need any of that to work out the transform. Just look at how ε is defined and you should be able to relate a coordinate system (t,x) in which ε=½ to another coordinate system (T,X) with arbitrary ε and get
\begin{align}<br /> cT &amp;= ct + (2\epsilon - 1) x \\<br /> X &amp;= x<br /> \end{align}You can then work out what c^2 dt^2 - dx^2 is in terms of dT and dX.

 

[Dale]

But, how can i do this since i haven't defined any flat space-time yet?

So define one. All it takes is a statement like "in the absence of gravity" or "neglecting gravity" or "far from any significant gravitational source".

I don't see what your remaining concern is. Take the transform posted by DrGreg, calculate the metric, and there you have a complete description of the physics.

 

[Austin0]

Hi I have a question.
Could you explain what you meant when you said the constancy of c wrt inertial frames was natural and not simply a convention?

 

[Dale]

Hi I have a question.
Could you explain what you meant when you said the constancy of c wrt inertial frames was natural and not simply a convention?

Hi Austin0, I am not sure who you are talking to, but I think that you must be referring to some other thread. Nobody has said anything like that in this one.

 

[nikolakis]

Ok, I'll go for DrGreg's transformations to find the null rays in the deformed frame (X,T):

\begin{align}<br /> T &amp;= t + (2\epsilon - 1) x \\<br /> X &amp;= x<br /> \end{align}

This can be done both ways: either by setting x=t and working out: X=something*T, or by setting the new interval ds=0 in (X,T) coordinates and thus arriving at: dX=something*dT, both expressions being equivalent.

I have one last stupid question before putting this thread to rest, and I must ask it:

What do I take the new speed of light to be? If i remember well from my calculations, there is an 4ε(ε-1) coefficient in the dX^2 factor in the new ds. Do I measure the new speed of light as: dX/dT, or do I take it: √(4ε(ε-1))dX/dT ?

I am almost certain that this must be: √(4ε(ε-1))dX/dT, but am I right?

Again, many thanks to DrGreg he was most helpful.

 

[nikolakis]

I think any question without LATEX is being frowned upon. I'll ask properly this time.

I will go for DrGreg's tranforms with c=1 and evaluate the new speed of light in the deformed frame.

\begin{align}<br /> T &amp;= t + (2\epsilon - 1) x \\<br /> X &amp;= x<br /> \end{align}

On setting x=t one gets: X = \frac{1}{2ε}T or upon evaluating the new ds=\sqrt{4ε(1-ε)dX^{2}-dT^{2}+2(2ε-1)dXdT} = 0 \Rightarrow dX = \frac{1}{2ε}dT, both expressions being equivalent.

How do I measure the new speed of light?

Is it \frac{dX}{dT} = \frac{1}{2ε}, or is it: \frac{\sqrt{4ε(1-ε)}dX}{dT} = \sqrt{\frac{1-ε}{ε}} ?

 

[nikolakis]

Better still, if I make use of the transform:

X=x
T=2t,

What do I take the new speed of light to be?

Since I would be measuring ds on the T-axis in the new metric space defined by the transformation, wouldn't that imply that the speed of light is the same and not 1/2?
When T=1, X=1/2. But according to the new ds on the T-axis: ds=dT/2, therefore c=1 !

 

[Dale]

Better still, if I make use of the transform:

X=x
T=2t,

What do I take the new speed of light to be?

Since I would be measuring ds on the T-axis in the new metric space defined by the transformation, wouldn't that imply that the speed of light is the same and not 1/2?
When T=1, X=1/2. But according to the new ds on the T-axis: ds=dT/2, therefore c=1 !

x=ct gives a speed dx/dt=c. Substituting in the transformation gives X=T c/2 which gives a speed dX/dT=c/2 \ne c

 

[Ookke]

Would it work to sync clocks with this procedure?
1. Start with two clocks at the same place, synchronized.
2. Accelerate both clocks to opposite directions with acceleration of same magnitude and duration (duration is measured in both clocks own frame).
3. Let clocks go the same amount of time, in their own frames.
4. Stop the clocks with brakings that exactly reverse the initial accelerations. The clocks become at rest.

After this, there should be no difference in clock readings, at least none that can be explained by acceleration.

 

[ghwellsjr]

Would it work to sync clocks with this procedure?
1. Start with two clocks at the same place, synchronized.
2. Accelerate both clocks to opposite directions with acceleration of same magnitude and duration (duration is measured in both clocks own frame).
3. Let clocks go the same amount of time, in their own frames.
4. Stop the clocks with brakings that exactly reverse the initial accelerations. The clocks become at rest.

After this, there should be no difference in clock readings, at least none that can be explained by acceleration.

Those two clocks will remain in sync during their trips according to their initial (and final) rest frame, but they will not be in sync with all the other clocks at different locations that remained at rest in that frame.

That is what I depicted in the first diagram from post #8:

Note that the corresponding black and red dots are always at the same coordinate times but they are at different coordinate times than the corresponding blue dots.

Furthermore, if you repeated the process with another pair of clocks that come to rest at different locations, they will not be in sync with the first pair.

So the answer to your question is no, that it not a way to synchronize a bunch of clocks. Einstein's method still prevails.

 

[nikolakis]

x=ct gives a speed dx/dt=c. Substituting in the transformation gives X=T c/2 which gives a speed dX/dT=c/2 \ne c

I know this! I have already said that c=1/2 for the dX/dT in the new metric space ! (Please use c=1 for the rest of this discussion.)

What I am saying is since the new metric is ds = \sqrt{dX^{2}-dT^{2}/4}, by measuring a difference dT=1 on the T-axis, this same ds gives ds=dT/2. Since I, now belong in the ds = \sqrt{dX^{2}-dT^{2}/4} metric space I shall be measuring ds on the T-axis, not dT!
Therefore v= \frac{dX}{dT/2} = 2dX/dT = 1.

 

[Dale]

I know this! I have already said that c=1/2 for the dX/dT in the new metric space ! (Please use c=1 for the rest of this discussion.)

What I am saying is since the new metric is ds = \sqrt{dX^{2}-dT^{2}/4}, by measuring a difference dT=1 on the T-axis, this same ds gives ds=dT/2. Since I, now belong in the ds = \sqrt{dX^{2}-dT^{2}/4} metric space I shall be measuring ds on the T-axis, not dT!
Therefore v= \frac{dX}{dT/2} = 2dX/dT = 1.

No, v=dX/dT, by definition.

 

[Ookke]

Those two clocks will remain in sync during their trips according to their initial (and final) rest frame, but they will not be in sync with all the other clocks at different locations that remained at rest in that frame.

...

So the answer to your question is no, that it not a way to synchronize a bunch of clocks. Einstein's method still prevails.

Ok, thanks. Even though this is not a method to synchronize multiple clocks, it might be of some use that the two clocks are sync in their final rest frame.

It bothers me in Einstein's clock synchronization that the speed of light is assumed to be constant in all directions. If we synchronize clocks based on this assumption, we cannot use the same clocks to measure one-way speed of light. Sounds a case of circular reasoning.

Syncronization by moving doesn't at least directly involve this assumption. The clock owners could agree beforehand to send a test signal at specific clock reading. Both owners would record the time when the signal from other clock arrives, then they would compare the results by changing information. I don't think it's likely to get different results for the two directions, I'm just curious if this kind of experiment would be valid in principle.

 

[nikolakis]

Ookke, there is a way to measure the one-way speed in light and synchronize two distant clocks instantaneously, in theory!

In cylindrical spacetime any light signal will return back to its origin, and any two extreme clocks on a line identify at the same point. Since their distance is now 0, synchronize them!
Then measure the one-way speed of light.
Of course, you must ensure that your space-time is a cylinder everywhere.

It's a fictitious example, but in theory at least, it can be measured.

i am very much worried at what DaleSpam said, that v=dX/dT "by definition".

 

[ghwellsjr]

Those two clocks will remain in sync during their trips according to their initial (and final) rest frame, but they will not be in sync with all the other clocks at different locations that remained at rest in that frame.

...

So the answer to your question is no, that it not a way to synchronize a bunch of clocks. Einstein's method still prevails.

Ok, thanks. Even though this is not a method to synchronize multiple clocks, it might be of some use that the two clocks are sync in their final rest frame.

Well, if you're only going to use two clocks, then you don't have to be make sure they follow the same acceleration profile. You can even have just one of them accelerate and when the two clocks are as far apart as you want, you can declare them to be in sync.

It bothers me in Einstein's clock synchronization that the speed of light is assumed to be constant in all directions. If we synchronize clocks based on this assumption, we cannot use the same clocks to measure one-way speed of light. Sounds a case of circular reasoning.

If you do what I just suggested (accelerate the two clocks differently), then you will find that when you measure the time it takes for light to traverse between them, it will be different in the two directions. Will that make you happy?

Syncronization by moving doesn't at least directly involve this assumption. The clock owners could agree beforehand to send a test signal at specific clock reading. Both owners would record the time when the signal from other clock arrives, then they would compare the results by changing information. I don't think it's likely to get different results for the two directions, I'm just curious if this kind of experiment would be valid in principle.

No, it would not be valid. When you move two clocks identically and then you declare them to be in sync, you are doing the same thing that Einstein says when you declare them to be in sync if two light signals from their midpoint arrive when the clocks read the same. If you move two clocks differently and then you declare them to be in sync you are following your own scheme and light signals sent from their midpoint will arrive when the clocks read different times.

Maybe you should go back and read post #8.

 

[Dale]

i am very much worried at what DaleSpam said, that v=dX/dT "by definition".

Why? It is a pretty standard definition, taught in the first week of most freshman physics courses.

 

[nikolakis]

Well, starting with the familiar ℝ^{3} manifold, let's examine what a natural representation of a metric space equipped with ds = \sqrt{dx^{2}+4dy^{2}} would be.

Let

X=x,
Y=y,
Z=\sqrt{3} y

ds = \sqrt{dX^{2}+dY^{2}+dZ^{2}} is inherited from ℝ^{3}, hence ds = \sqrt{dx^{2}+4dy^{2}} . It is the projection of Z=\sqrt{3} Y plane onto Z=0.

Any creature living on the Z=0 plane equipped with the aforementioned metric should measure the same lengths as one living on the Z=\sqrt{3} Y plane. For example, a distance of dy=1, dx=0 on the Z=0 plane, corresponds to ds=2 on the Z=\sqrt{3} Y plane.

But, since the tangent of any angle is nothing but the ratio of lengths of the sides of orthogonal triangles, the velocity expressed by this angle must remain unchanged.
Because the lengths don't change!

Therefore, v = 2dy/dx, not v=dy/dx !

 

[Dale]

Any creature living on the Z=0 plane equipped with the aforementioned metric should measure the same lengths

Sure. All actual experimental measurements are frame invariant. But according to the creatures using the aforementioned metric that is because the rulers are length-contracted in the Z direction and thus do not correctly measure distance in that direction.

But, since the tangent of any angle is nothing but the ratio of lengths of the sides of orthogonal triangles, the velocity expressed by this angle must remain unchanged.
Because the lengths don't change!

Therefore, v = 2dy/dx, not v=dy/dx !

This simply doesn't follow. This isn't a matter of reasoning from the metric, it is simply a definition. v=dx/dt. Velocity is a coordinate-dependent quantity. The very fact that your suggested adjustments are giving you coordinate-independence should be a big warning sign that the quantity you are looking at is not velocity.

 

[nikolakis]

This simply doesn't follow. This isn't a matter of reasoning from the metric, it is simply a definition. v=dx/dt. Velocity is a coordinate-dependent quantity. The very fact that your suggested adjustments are giving you coordinate-independence should be a big warning sign that the quantity you are looking at is not velocity.

There must be a big lapse in my understanding.

In my previous example, let us use:

X=x,
T=t,
Z=\sqrt{3} x

thus giving ds = \sqrt{dt^{2}+4dx^{2}}

You are saying that all ds's are frame invariant, aren't you?

How far, how long, a displacement in dx should be on Z=0, if not ds=2dx?
My contracted ruler in the X-direction (not Z-direction as you have mentioned) measures ds, not dx !
And by division, (by measuring with the same uncontracted ruler in the T-direction, an interval ds=dt) :

v=2dx/dt

What all these dx's and dy's and dt's have to do with the measurements of the interval on my ruler? Does the ruler know maths to calculate the interval?

 

[nikolakis]

The very fact that your suggested adjustments are giving you coordinate-independence should be a big warning sign that the quantity you are looking at is not velocity.

And there is certainly not co-ordinate independence in general, save for the metric in this particular example which I've chosen, deliberately. Look at my #25 post.

Isn't it here, a more expert in this field to answer this question? Thank you.

 

[Dale]

Isn't it here, a more expert in this field to answer this question?

Apparently not, otherwise they would have posted. However, I don't think that much expertise is required to tell you that the definition of v is dx/dt.

 

[Dale]

Let

X=x,
Y=y,
Z=\sqrt{3} y

let us use:

X=x,
T=t,
Z=\sqrt{3} x

I hadn't noticed this before, but did you mean to have Z=\sqrt{3} y in the first and Z=\sqrt{3} x in the second instead of Z=\sqrt{3} z in both?

If you intended to do that it is wrong. You wind up with 3 coordinates spanning a 2 dimensional space. That is not valid.

 

[pervect]

Why? It is a pretty standard definition, taught in the first week of most freshman physics courses.

v = dx/dt works fine in inertial frames. It gives bizarre results in non-inertial frames.

For instance, if you're heading for a star at a high relativistic velocity and de-accelerate, due to the Lorentz contraction factor changing, you can easily fins scenarios where dx/dt increase. But it makes little physical sense to say that you are moving away from the star.

So it's best to define the velocity as a vector in the tangent space. IMO.

 

[nikolakis]

I hadn't noticed this before, but did you mean to have Z=\sqrt{3} y in the first and Z=\sqrt{3} x in the second instead of Z=\sqrt{3} z in both?

If you intended to do that it is wrong. You wind up with 3 coordinates spanning a 2 dimensional space. That is not valid.

Har, har. Are you joking? How many coordinates span a 2-Sphere?

Both examples are correct. I've made no mistake. It's a valid representation of my metric.

 

[nikolakis]

v = dx/dt works fine in inertial frames. It gives bizarre results in non-inertial frames.

For instance, if you're heading for a star at a high relativistic velocity and de-accelerate, due to the Lorentz contraction factor changing, you can easily fins scenarios where dx/dt increase. But it makes little physical sense to say that you are moving away from the star.

So it's best to define the velocity as a vector in the tangent space. IMO.

Thank you, pervect!. This is exactly what I expected to hear. Another instance is the metric ds = \sqrt{dr^2 + r^2d\theta^2}, which describes what you've just said...
Many thanks!

 

[Dale]

Both examples are correct. I've made no mistake. It's a valid representation of my metric.

It's an invalid coordinate system.

To expand on why it is invalid, recall that a coordinate chart maps an open subset of a manifold to an open subset of Rn, where n is the dimension of the manifold. The two transforms you proposed map to a non-open subset of R3.

 

[Dale]

v = dx/dt works fine in inertial frames. It gives bizarre results in non-inertial frames.

Agreed.

So it's best to define the velocity as a vector in the tangent space. IMO.

Unless there is another related concept, I think you are referring to the definition of the four-velocity, which isn't the same as the velocity.

 

[pervect]

Agreed.

Unless there is another related concept, I think you are referring to the definition of the four-velocity, which isn't the same as the velocity.

The 3-velocity isn't a tensor (just part of one), but it's still defined / definable in the tangent space. You just omit the time component to get the 3-velocity in a locally Minkowskii normalized tangent space, just as you would in flat space-time.

It doesn't make sense in generalized coordinates to think of 3-velocity as distance / time because of the problem I mentioned earlier.

Finding the velocity in the tangent space removes the hiccups.

It also make sense to talk about the magnitude of the relative velocity as the angle between worldlines. The magnitude of the relative velocity is a scalar, which is the magnitude of the 3-vector.

You can express this geometrically in terms of the dot product of the four vectors, which is another quantification of "the angle between worldlines".

If you have two 4-vectors p and q, the dot product of p and q determines the magnitude of the relative velocity \beta = v/c by the relation

\gamma = \frac{1}{\sqrt{1-\beta^2}} = - p \cdot q

The easiest way to demonstrate this is to take a Minkowskii space, where p = (1,0,0,0) and q = gamma*(1, \vec{v})

(I believe I saw this formula first on PF, I found it handy. I don't know of a textbook that explicitly goes into the geometrical formulation of the magnitude of a three-velocity via the dot product, but I found it a useful relationship, both practically and conceptually).

Note that I've used geometric units.

Of course p and q have to be in the same tangent space - i.e. at the same point in space-time.

This doesn't matter in a flat space-time, because you can parallel transport either one of the vectors to the other unambiguously. In curved space-time, parallel transport is of course path dependent.

In a static space-time, one can find the relative velocity relative to the static space-time by replacing q with a renormalized Killing vector (i.e. q is a unit magnitude vector that points in the same direction as the Killing vector).

 

[Dale]

It also make sense to talk about the magnitude of the relative velocity as the angle between worldlines. The magnitude of the relative velocity is a scalar, which is the magnitude of the 3-vector.

You can express this geometrically in terms of the dot product of the four vectors, which is another quantification of "the angle between worldlines".

If you have two 4-vectors p and q, the dot product of p and q determines the magnitude of the relative velocity \beta = v/c by the relation

\gamma = \frac{1}{\sqrt{1-\beta^2}} = - p \cdot q

The easiest way to demonstrate this is to take a Minkowskii space, where p = (1,0,0,0) and q = gamma*(1, \vec{v})

I see what you are doing here, and it makes sense. But the dot product above is frame invariant, which to my understanding is not what the OP wants and is not what is usually meant by "velocity". Using this formulation the speed of light is c by definition, no matter what coordinates or synchronization convention you use. At that point, I would just go ahead and use four-vectors.

I do agree that, in general, four-vectors and other geometric approaches are far superior to coordinate-based approaches and I think that the OP would be well served to look into them and not worry so much about alternative synchronization conventions etc.

 

[nikolakis]

If OP stands for original poster, I think pervect makes perfect sense to me.

In his example and for a uniform accelerating or de-accelerating frame, I consider the metric space (in geometric units) given by ds = \sqrt{dr^{2}-r^{2}dτ^{2}}, τ is proper time as measured by the accelerating observer and r is the "proper" distance (so to speak, measured in new coordinate r) from the gravitating object, producing the uniform acceleration.

The constancy of the speed of light in the tangent space, automatically implies: c = \frac{dr}{rdτ}=1. In my #38 post, replace r=2, dτ=dt, dr=dx. DaleSpam has insisted that c = \frac{dr}{dτ}

As for the second part, I am not certain that Killing vectors are applicable to a shear with dilation transformation applied with the metric:

On setting x=t one gets: X = \frac{1}{2ε}T or upon evaluating the new ds=\sqrt{4ε(1-ε)dX^{2}-dT^{2}+2(2ε-1)dXdT} = 0 \Rightarrow dX = \frac{1}{2ε}dT, both expressions being equivalent.

How do I measure the new speed of light?

Is it \frac{dX}{dT} = \frac{1}{2ε}, or is it: \frac{\sqrt{4ε(1-ε)}dX}{dT} = \sqrt{\frac{1-ε}{ε}} ?

Can 4-velocity vectors be used here as well? My question remains.

I promise, i will look into 4-vectors and other geometric approaches, but for now, I need a quick answer beforehand.

Thank you.

 

[Dale]

DaleSpam has insisted that c = \frac{dr}{dτ}

No, DaleSpam has insisted and continues to insist that v = dr/dt. c is a defined constant. There may be an alternative standard definition for velocity, but I am not aware of it.

What pervect is discussing is (to the best of both his knowledge and mine) not the standard definition of velocity. As he explicitly mentioned:

(I believe I saw this formula first on PF, I found it handy. I don't know of a textbook that explicitly goes into the geometrical formulation of the magnitude of a three-velocity via the dot product, but I found it a useful relationship, both practically and conceptually).

For four-vectors, the magnitude of the four-velocity of a light pulse is always 0, regardless of the coordinates.