One-way speed of light and clock sync

[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.

 

[pervect]

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

Ah - well, you're right in that I assumed standard, isotropic clock synchronizations in my analysis and wasn't considering non-standard ones.

I would have to agree that with non-Einsteinian clock synchronizations, one marks out some straight course of known distance, synchronizes the two clocks at start and end of the course, and divide the distance by the change in clock readings (which represents the time interval according to the synchronization convention used) to get the velocity.

This will be the average velocity, one needs the additional step of taking the limit as the distance goes to zero in general - taking the limit gets rid of the effects of acceleration or curvature that I was concerned about.

 

[nikolakis]

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:

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

I am not happy with your definition that v = dr/dτ. I can't escape the idea that speed should be measured here, on my ds = rdτ worldline where I belong, not there where r=1. Is it a misconception of my part that velocity is a vector, whereas speed is a scalar? Has it anything to do, with our measurements in a 2-dimensional flat space-time ( cylindrical, conical, tilted planar ) embedded in 3-dimensions?

As for vectors, I think there exist manifolds where the dot product is not defined. ( Finslerian manifolds )

I believe that everything should follow from the metric, and from the metric only, since our clocks and rulers are measuring ds's and not arbitrary co-ordinate plantations.

Maybe, my question belongs to another thread... Anyway, thanks for the input.