Speed of light in NON-inertial frames

[Curl]

I haven't studied this very much, but how do EM waves behave in noninertial frames? Do photons have an acceleration in order to maintain constant speed c in the noninertial frame?

What happens to the Lorrent'z force in non-inertia frames? A charge moving in a magnetic field is equal to a magnetic field moving across a charge from an inertial observer; the velocity vector is relative. But what if a magnetic field source is accelerating across a charge?

Anyone got a good read on this, or can type up an explanation?

 

[Vanadium 50]

How do you define a "speed of anything" in a non-inertial frame?

Let's set aside light for a moment. You have an object at speed v. You are traveling at v'(t) - i.e. you are in a non-inertial frame. What's the (single) speed of this object in your frame?

 

[AJ Bentley]

It's a mistake to think of the non-inertial frame as 'affecting' anything. It's merely a point of view.

Picture yourself on a children's roundabout - you are an observer in a non-inertial frame.

How do photons behave? - Exactly the same as they did before - the fact that you happen to be spinning round as you look at them makes no difference to them.
Of course, what YOU see takes a bit of working out - but that's YOUR problem - not the photons.

 

[Dickfore]

The phase velocity of EM waves in a gravitational field changes.

 

[Curl]

in a small time interval, do you see photons moving away from you at C?

 

[JesseM]

in a small time interval, do you see photons moving away from you at C?

No, dx/dt for a photon can be different than c if x and t represent the coordinates of a non-inertial frame. A non-inertial frame is basically any arbitrary way of assigning position and time coordinates to different points in spacetime that doesn't happen to be an inertial frame--for example, even for a single non-inertial object, there are an infinite different number of ways of constructing a non-inertial "rest frame" for it.

 

[bcrowell]

I haven't studied this very much, but how do EM waves behave in noninertial frames?

The speed of light is always found to be c by any local measurement made by any observer, regardless of whether the observer is accelerating or not.

Do photons have an acceleration in order to maintain constant speed c in the noninertial frame?

No. Relativistic velocities don't add linearly. If you combine a velocity of c with some other velocity v, the relativistic equation for combination of velocity always gives c, not c+v.

 

[Fredrik]

How do you define a "speed of anything" in a non-inertial frame?

I think an even more important question is, how do you define an accelerating coordinate system? I mean, a coordinate system is just a ("nice") function from a region of spacetime into $$\mathbb R^4$$, but when we talk about "inertial" or "accelerating" coordinate systems, there's a specific synchronization procedure that we're supposed to use to associate a coordinate system with the motion of an object. And the standard procedure ensures that the speed of a ray of light through the origin in the accelerating coordinate system is =c.

However, if we try to measure that speed with an accelerating measuring device that was designed for local measurements (measurements in a region of spacetime that's so small that the effects of acceleration can be ignored), the result will be different from c. This doesn't mean that the speed of light is different from c in the accelerating coordinate system. It just means that the device isn't working properly when it's accelerating.

 

[Passionflower]

This doesn't mean that the speed of light is different from c in the accelerating coordinate system. It just means that the device isn't working properly when it's accelerating.

So see if I get this right concerning your opinion on this matter:

If I have a Born rigid spaceship undergoing constant acceleration and I measure the roundtrip time of light from the floor to the ceiling and back with a clock and an emitter on the floor and a mirror on the ceiling and find a different time from the case where the spaceship is inertial this is due, in your opinion, because the measuring setup is not working properly? And, out of curiosity, and again, in your opinion, is this setup also not working properly if we measure the roundtrip time from the top to the base of a tower standing on a massive planet?

As I wrote in another thread the measured speed of light depends on two factors here:

1. If the spaceship travels inertially.
2. The height of the spaceship.

That means for instance that if the spaceship is undergoing constant acceleration then in the limit, which is simply a special case, when the height of the ship approaches zero the measured result will be identical to the case where the spaceship travels inertially.

 

[starthaus]

in a small time interval, do you see photons moving away from you at C?

For a short interval, in a very small space vicinity, speed of light is "c" in rotating frames.
The science advisor "Demystifier" (H.Nikolic) has an excellent paper on this subject.

 

[Curl]

In Einstein's thought experiment, shooting a laser horizontally in an elevator that is accelerating upwards will cause the light to "bend" as seen by the observer in the elevator.

What I'm asking is, if I was in a train that is accelerating and I shoot a paintball gun head on, it would appear as I am "catching up" to the paintball since it stops accelerating after leaving the gun (ignore gravitational force) but *I* am accelerating, thus the speed of the paintball seems to decrease with respect to my eyes. The only way the speed will stay constant with respect to my face is if the paintball had an acceleration matching mine (and the train's which I'm standing on).

What if you do this same experiment with a laser instead of a paintball gun? If the speed of light is constant, then would it appear as if the photons have an acceleration matching mine?

P.S. I know I'm accelerating because my coffee will spill over if I hold it vertically, and I was once "at rest" but now I feel the acceleration.

 

[bcrowell]

What if you do this same experiment with a laser instead of a paintball gun? If the speed of light is constant, then would it appear as if the photons have an acceleration matching mine?

No, for the reason given in #7. The photons have velocity c according to any local measurement. In your frame they will get red-shifted as they fly, but they will always have velocity c.

 

[Fredrik]

If I have a Born rigid spaceship undergoing constant acceleration and I measure the roundtrip time of light from the floor to the ceiling and back with a clock and an emitter on the floor and a mirror on the ceiling and find a different time from the case where the spaceship is inertial this is due, in your opinion, because the measuring setup is not working properly?

If we want to measure the roundtrip time, then there's nothing wrong with the setup. If we're trying to measure the speed of light at the floor, then there is.

And, out of curiosity, and again, in your opinion, is this setup also not working properly if we measure the roundtrip time from the top to the base of a tower standing on a massive planet?

Same thing there. If we want to find the local speed of light at a specific height, this setup is less accurate the taller the tower is. If the setup "works" depends on how accurate we need the measurement to be.

As I wrote in another thread the measured speed of light depends on two factors here:

1. If the spaceship travels inertially.
2. The height of the spaceship.

That means for instance that if the spaceship is undergoing constant acceleration then in the limit, which is simply a special case, when the height of the ship approaches zero the measured result will be identical to the case where the spaceship travels inertially.

Yes, I think you have pretty accurately summarized my views on this. I admit that it's possible that I might have some detail wrong, because I haven't proved all of this rigorously to myself, but the calculations I've worked through (Hurkyl's calculation, my addition to it, and a calculation of the one-way travel time) all seem to support this view.

 

[AJ Bentley]

If the speed of light is constant, then would it appear as if the photons have an acceleration matching mine?

You have to be careful by what you mean by speed. especially when you yourself are accelerating.
In an accelerating frame, the instantaneous speed of light is constant. That means that if you measure it very,very quickly, over a short distance, you will get the universal constant c.
But if you take your time over it and measure it over a significant distance, your own motion messes with the values of distance and time you get. If you then apply a naive calculation of v= distance/time with your wobbly measurements you will get a silly answer.

It's exactly the same situation as if someone were to jog your arm and fiddle with your watch while you were measuring. In the lift, you are being joggled very smoothly and in a predicable way, but it's still a joggle.

 

[Passionflower]

You have to be careful by what you mean by speed. especially when you yourself are accelerating.
In an accelerating frame, the instantaneous speed of light is constant. That means that if you measure it very,very quickly, over a short distance, you will get the universal constant c.

Speed is something that one can measure, it is distance over time. One can measure it in the limit or over a larger region and time. One can measure it when one is traveling inertially or accelerating. Why do you think the definition of speed or velocity should depend on whether someone is traveling inertially or not?

But if you take your time over it and measure it over a significant distance, your own motion messes with the values of distance and time you get.

One of the first things to know about relativity is that motion is relative, in general relativity the physical equations take the same form in all coordinate systems and what an observer measures is real and significant to him! You cannot say, well yes you measure it but it is not real because something is messed up, that is not science but sheer nonsense!

If you then apply a naive calculation of v= distance/time with your wobbly measurements you will get a silly answer.

Are you saying one cannot measure anything when you are accelerating except in the limit? Do I, standing on earth, when I measure something get silly answers because I am accelerating?

Whatever an observer measures, independent whether he travels inertially or is accelerating, is of real scientific significance and is not silly or wobbly.

 

[JesseM]

No, for the reason given in #7. The photons have velocity c according to any local measurement. In your frame they will get red-shifted as they fly, but they will always have velocity c.

They only have a velocity of c in a locally inertial frame. If you use a coordinate system where your accelerating ship is at rest (like Rindler coordinates, where time is measured by accelerating clocks and distance is measured by rulers undergoing Born rigid acceleration) then light won't necessarily move at c.

 

[JesseM]

In an accelerating frame, the instantaneous speed of light is constant.

No it isn't. For example, in Rindler coordinates the instantaneous coordinate velocity of a light beam approaches zero in the limit as it approaches the Rindler horizon. Again, only if you are using a locally inertial frame is the instantaneous velocity guaranteed to be c.

 

[DrGreg]

No, for the reason given in #7. The photons have velocity c according to any local measurement. In your frame they will get red-shifted as they fly, but they will always have velocity c.

They only have a velocity of c in a locally inertial frame. If you use a coordinate system where your accelerating ship is at rest (like Rindler coordinates, where time is measured by accelerating clocks and distance is measured by rulers undergoing Born rigid acceleration) then light won't necessarily move at c.

In an accelerating frame, the instantaneous speed of light is constant.

No it isn't. For example, in Rindler coordinates the instantaneous coordinate velocity of a light beam approaches zero in the limit as it approaches the Rindler horizon. Again, only if you are using a locally inertial frame is the instantaneous velocity guaranteed to be c.

It depends what you mean by "speed". If you mean "coordinate speed" i.e. rate of change of coordinate distance with respect to coordinate time, then JesseM is correct.

But I suspect bcrowell and AJ Bentley didn't mean that. If you use a "local ruler" and a "local clock" then even accelerating observers measure an instantaneous speed of c (but not an average speed over a non-zero distance).

 

[JesseM]

But I suspect bcrowell and AJ Bentley didn't mean that. If you use a "local ruler" and a "local clock" then even accelerating observers measure an instantaneous speed of c (but not an average speed over a non-zero distance).

But in that case you aren't really measuring the speed in an "accelerating frame". Instead, you are getting an accelerating observer to measure the speed with an inertial ruler and clock that happen to be instantaneously at rest relative to himself (as opposed to using a local ruler and pair of clocks undergoing Born rigid acceleration, for example).

 

[Passionflower]

But in that case you aren't really measuring the speed in an "accelerating frame". Instead, you are getting an accelerating observer to measure the speed with an inertial ruler and clock that happen to be instantaneously at rest relative to himself (as opposed to using a local ruler and pair of clocks undergoing Born rigid acceleration, for example).

Exactly right!

 

[DrGreg]

But in that case you aren't really measuring the speed in an "accelerating frame". Instead, you are getting an accelerating observer to measure the speed with an inertial ruler and clock that happen to be instantaneously at rest relative to himself (as opposed to using a local ruler and pair of clocks undergoing Born rigid acceleration, for example).

Even using a local ruler and pair of clocks undergoing Born rigid acceleration, where the clocks measure proper time rather than Rindler coordinate time, would still measure a speed of c in the limit as the separation tends to zero.

 

[JesseM]

Even using a local ruler and pair of clocks undergoing Born rigid acceleration, where the clocks measure proper time rather than Rindler coordinate time, would still measure a speed of c in the limit as the separation tends to zero.

Ah, I hadn't realized that the Rindler time coordinate is different from the proper time of each clock at constant Rindler distance--p. 26 of this book mentions that proper time $$d\tau$$ is proportional to dt (difference in Rindler coordinate time) times the distance (in Rindler coordinates) from the horizon. So what would happen if we defined a different non-inertial coordinate system where lines of constant position coordinate looked just the same as for Rindler coordinates (a family of clocks accelerating in a Born rigid way), but the time coordinate for an event at any fixed position coordinate would just be equal to the proper time of a clock at that coordinate? (one of the family of accelerating clocks, where we can assume that all the clocks were set to zero at the moment they were all instantaneously at rest in some inertial frame) In this post kev illustrated some "lines of equal proper time for the Rindler observers", assuming his calculations were correct they look pretty similar to the lines of constant Rindler coordinate time shown in the second diagram here. Does what you say above imply that in such a not-quite-Rindler coordinate system, light beams would always have an instantaneous coordinate velocity of c? If so that doesn't really make sense to me, since it seems like it should still take an infinite coordinate time for anything to reach the Rindler horizon in this coordinate system, just like in Rindler coordinates...

edit: maybe the issue is that distance in Rindler coordinates does not really correspond to what would "naturally" be measured by a ruler undergoing Born rigid acceleration? Of course such a ruler has different parts moving at different velocities so you can't talk about it having a single inertial rest frame, but it would seem "natural" that in the limit as you pick markings closer and closer together on the accelerating ruler (so that the difference in velocities between the markings approaches zero), the coordinate distance between the markings should approach the distance measured between them by an inertial ruler instantaneously at rest relative to one of the markings. Perhaps if we defined a not-quite-Rindler coordinate system with a definition of distance that satisfied this requirement, the distance to the Rindler horizon would actually turn out to be infinite, so light could travel at a coordinate speed of c everywhere and yet never reach the Rindler horizon in finite coordinate time...

 

[Fredrik]

But in that case you aren't really measuring the speed in an "accelerating frame". Instead, you are getting an accelerating observer to measure the speed with an inertial ruler and clock that happen to be instantaneously at rest relative to himself (as opposed to using a local ruler and pair of clocks undergoing Born rigid acceleration, for example).

The accelerating coordinate system is defined so that a null geodesic through the origin has speed 1 at the origin, i.e. the same speed as in the comoving inertial frame. (The simultaneity lines of the accelerating coordinate system are chosen to ensure that this is the case).

It has been suggested a few times in this thread and the other that we can measure the speed of light in a Born rigid accelerating rocket's "accelerating frame" by measuring the roundtrip time τ of light that goes tail-head-tail, and calculating the speed as τ/2L, where L is the length of the rocket. But this is not the speed of light in the accelerating coordinate system. If the light is emitted at time coordinate t=0, and returns at t=T, the average speed in the coordinate system is T/2L, and the result of the non-local measurement is τ/2L, where τ is the proper time of the world line of the rear of the rocket, from the emission event to the detection event. So our non-local measurement only returns the average speed of light if T=τ, and this is certainly not true in general.

 

[Passionflower]

It has been suggested a few times in this thread and the other that we can measure the speed of light in a Born rigid accelerating rocket's "accelerating frame" by measuring the roundtrip time τ of light that goes tail-head-tail, and calculating the speed as τ/2L, where L is the length of the rocket. But this is not the speed of light in the accelerating coordinate system. If the light is emitted at time coordinate t=0, and returns at t=T, the average speed in the coordinate system is T/2L, and the result of the non-local measurement is τ/2L, where τ is the proper time of the world line of the rear of the rocket, from the emission event to the detection event. So our non-local measurement only returns the average speed of light if T=τ, and this is certainly not true in general.

Indeed, the measured speed of light from the back is not the same as measured from the front of a Born rigid spaceship that undergoes a constant proper acceleration. So what? There is nothing special about that. And furthermore the measurement is done by two different observers and in two different directions. So are you saying that if you cannot encapsulate that in one coordinate system it is not longer valid?

The world according to the observer at the front is just as real as the one at the back, they can do any measurement they like. They can setup a coordinate system where they are at rest and the rest of the world is moving and spinning around them, and that is pretty much how you and I see reality as well.

 

[Fredrik]

I'm just saying that the speed of light at the origin of the accelerating frame is 1 by definition of accelerating frame, if we try to measure that speed the way you (and others) have suggested, we will (almost) always get a result that's different from 1. So that way of measuring it clearly doesn't work.

 

[Passionflower]

I'm just saying that the speed of light at the origin of the accelerating frame is 1 by definition of accelerating frame, if we try to measure that speed the way you (and others) have suggested, we will (almost) always get a result that's different from 1. So that way of measuring it clearly doesn't work.

What exactly 'does not work' if I am at the top of a tower with an emitter and a clock and I send out a light pulse that is reflected at the foot of that tower and take the elapsed time / (2 * the height of the tower)?

 

[Fredrik]

What exactly 'does not work' if I am at the top of a tower with an emitter and a clock and I send out a light pulse that is reflected at the foot of that tower and take the elapsed time / (2 * the height of the tower)?

It depends on what you're trying to do. If I flap my arms, it "doesn't work" if I'm trying to fly, but it works just fine if I'm trying to look silly. And can we keep this a discussion about SR, please. Gravity is just an additional complication.

 

[Passionflower]

And can we keep this a discussion about SR, please. Gravity is just an additional complication.

Actually conform the equivalence principle if we ignore tidal effects the above mentioned situation is identical to that of an accelerating spaceship or the proverbial elevator.

 

[Fredrik]

The concept of "comoving inertial frame" also gets more complicated because there's more than one kind, and I'm not 100% sure that there are no extra complications regarding how to associate a coordinate system with an object's motion. I don't see a reason to introduce anything that might possibly confuse the issue.

You ignored the essential part of my post, where I explained that it's not possible to fail unless you try. I can't tell you what (if anything) is wrong with what you're doing unless you tell me what you're trying to do.

 

[Passionflower]

I can't tell you what (if anything) is wrong with what you're doing unless you tell me what you're trying to do.

All I am trying to do here is highlight that there is more to relativity than the 'mantras': "in the limit the speed is c" and "in the limit spacetime is flat" resp. for accelerating observers or observers in curved spacetimes. I think it is wrong to encourage people to stop thinking beyond this level.

Beyond that is where the interesting stuff is:

• What does the world look like for accelerating observers in flat spacetime?
• What does the world look like for observers in curved spacetime?

Just considering the limit is literally rather shortsighted.

 

[Fredrik]

And all I'm doing is to say that you can't just perform this measurement and say "look, the speed of light in the accelerating frame is 0.8". This would be to ignore the definition of one of the terms in the sentence you just used, and that's never a great idea.

 

[Passionflower]

And all I'm doing is to say that you can't just perform this measurement and say "look, the speed of light in the accelerating frame is 0.8". This would be to ignore the definition of one of the terms in the sentence you just used, and that's never a great idea.

First of all in this topic I never even mentioned the word frame. I never said anything about the speed 'in a frame', I am talking about the measured speed of light for a particular observer.

An observer can measure the time it takes for something to go from A to B and back, if we know the distance between A and B we can calculate the speed. If this something is light it is called the speed of light if it is a rock it is the speed of a rock. Really I do not see any problem with that.

But it seems we just have to agree to disagree.

 

[Dickfore]

I think people make an obvious oversight when it comes to the term "speed of light". In Landau, Classical Theory of Fields, they formulate Einstein's second postulate as:

"There is a finite limit speed c with which interaction can be transmitted."

According to the First Postulate (Principle of Relativity), this limit must have the same value in every inertial reference frame.

Then, they go on to prove that the space-time interval:

$$ds^{2} \equiv c^{2} \, dt^{2} - dx^{2} - dy^{2} - dz^{2}$$

is an invariant and derive everything else from there.

In particular, they build electromagnetism "from the ground up". It turns out from Maxwell's equations then, that electromagnetic fields can exist independent from any charges and currents, but have to be time-dependent. These fields propagate as waves and their speed of propagation in vacuum (free space) is equal to the same c as above. That is why this is called speed of light in vacuum.

However, as we all know, the speed of propagation of electromagnetic waves need not always be c, as in some materials, where it is also frequency dependent (dispersion).

For the case of non-inertial reference frames (or the case where a gravitational field is present), they simply define the space-time interval as a general quadratic form:

$$ds^{2} = g_{\mu \nu} \, dx^{\mu} \, dx^{\nu}$$

with $g_{\mu \nu} = g_{\nu \mu}$ - the metric tensor, containing all the information about the space-time and the particular coordinate system. Because at any point we may diagonalize this quadratic matrix, it must resemble the Minkowski metric tensor $g^{(0)}_{\mu \nu} = \mathrm{diag}(1, -1, -1, -1)$. In particular, for real space-time, we must have:

$$g < 0$$

The metric tensor can be used to deduce real time intervals, distances and synchronize clocks.

For example, if we are at a particular point in space ($x^{i} = \mathrm{const.} \Rightarrow dx^{i} = 0$) and consider two infinitesimally closed events, than, by analogy with SR we define the proper time interval as $ds = c \, d\tau$

$$ds^{2} = g_{0 0} \, (dx^{0})^{2} = (c \, d\tau)^{2}$$

$$g_{00} \ge 0 \Rightarrow d\tau = \frac{\sqrt{g_{0 0}}}{c} \, dx^{0}$$

The case $g_{0 0} < 0$ does not necessarily mean that that space-time is impossible, but simply that the particular coordinate system we are using is unsuitable.

Now comes the important point: Distances are defined through the same "radar procedure" using something that moves along null geodesics. Namely, let us consider two points A with space coordinates $x^{i}$ and B, which is infinitely close and with space coordinates $x^{i} + dx^{i}$. We shine a light ray from B at $x^{0} + (dx^{0})_{2}$, it propagates to A, reflects at x^{0} and arrives back at B at $x^{0} + (dx^{0})_{2}$. Since the wave is traveling along a null geodesic, we may find $(dx^{0})_{1/2}$ by equating $ds = 0$ and solving the quadratic equation:

$$g_{0 0} (dx^{0})^{2} + 2 \, g_{0 i} \, dx^{i} \, dx^{0} + g_{i j} \, dx^{i} \, dx^{j} = 0$$

where a summation from 1 to 3 over a repeated Latin superscript and subscript is implied. The solution of this equation is:

$$(dx^{0})_{1/2} = \frac{-g_{0 i} \, dx^{i} \mp \sqrt{(g_{0 i} \, g_{0 j} - g_{0 0} \, g_{i j}) \, dx^{i} \, dx^{j}}}{g_{0 0}}$$

According to what has been said above for proper time intervals, the round trip time, according to B is:

$$d\tau = \frac{\sqrt{g_{0 0}}}{c} \, \left((dx^{0})_{2} - (dx^{0})_{1}\right)$$

and this, by definition, corresponds to a distance:

$$dl = \frac{c \, d\tau}{2}$$

which gives the following:

$$dl^{2} = \gamma_{i j} \, dx^{i} \, dx^{j}, \gamma_{i j} = \frac{g_{0 i} \, g_{0 j}}{g_{0 0}} - g_{i j}$$

for the spatial distance and $\gamma_{i j} = \gamma_{j i}$ is the spatial metric tensor. For coordinate systems attainable by physical bodies, the quadratic form $dl^{2} \ge 0$ must be positive definite.

The moment of reflection of the signal at point A, according to B, corresponds to the time coordinate $x^{0} + \Delta x^{0}$, where:

$$\Delta x^{0} = \frac{(dx^{0})_{1} + (dx^{0})_{2}}{2} = g_{; i} \, dx^{i}, \; g_{; i} = -\frac{g_{0 i}}{g_{0 0}}$$

which is the synchronization offset. In this way, we can synchronize clocks along any open line in space, bu not, in general, over closed loops, since:

$$-\oint{\frac{g_{0 i}}{g_{0 0}} \, dx^{i}} \neq 0$$

in general. As a conclusion, signals that travel along null geodesics have speed c by definition.

But, if you write the equations of electromagnetism in curved spacetime, you will see that they predict propagation of electromagnetic waves at different speeds than c.

 

[Passionflower]

Now comes the important point: Distances are defined through the same "radar procedure" using something that moves along null geodesics. Namely, let us consider two points A with space coordinates $x^{i}$ and B, which is infinitely close and with space coordinates $x^{i} + dx^{i}$. We shine a light ray from B at $x^{0} + (dx^{0})_{2}$, it propagates to A, reflects at x^{0} and arrives back at B at $x^{0} + (dx^{0})_{2}$. Since the wave is traveling along a null geodesic, we may find $(dx^{0})_{1/2}$ by equating $ds = 0$ and solving the quadratic equation:

$$g_{0 0} (dx^{0})^{2} + 2 \, g_{0 i} \, dx^{i} \, dx^{0} + g_{i j} \, dx^{i} \, dx^{j} = 0$$

where a summation from 1 to 3 over a repeated Latin superscript and subscript is implied. The solution of this equation is:

$$(dx^{0})_{1/2} = \frac{-g_{0 i} \, dx^{i} \mp \sqrt{(g_{0 i} \, g_{0 j} - g_{0 0} \, g_{i j}) \, dx^{i} \, dx^{j}}}{g_{0 0}}$$

According to what has been said above for proper time intervals, the round trip time, according to B is:

$$d\tau = \frac{\sqrt{g_{0 0}}}{c} \, \left((dx^{0})_{2} - (dx^{0})_{1}\right)$$

and this, by definition, corresponds to a distance:

$$dl = \frac{c \, d\tau}{2}$$

which gives the following:

$$dl^{2} = \gamma_{i j} \, dx^{i} \, dx^{j}, \gamma_{i j} = \frac{g_{0 i} \, g_{0 j}}{g_{0 0}} - g_{i j}$$

for the spatial distance and $\gamma_{i j} = \gamma_{j i}$ is the spatial metric tensor. For coordinate systems attainable by physical bodies, the quadratic form $dl^{2} \ge 0$ must be positive definite.

The moment of reflection of the signal at point A, according to B, corresponds to the time coordinate $x^{0} + \Delta x^{0}$, where:

$$\Delta x^{0} = \frac{(dx^{0})_{1} + (dx^{0})_{2}}{2} = g_{; i} \, dx^{i}, \; g_{; i} = -\frac{g_{0 i}}{g_{0 0}}$$

which is the synchronization offset. In this way, we can synchronize clocks along any open line in space, bu not, in general, over closed loops, since:

$$-\oint{\frac{g_{0 i}}{g_{0 0}} \, dx^{i}} \neq 0$$

in general. As a conclusion, signals that travel along null geodesics have speed c by definition.

There are various way to define distance in the GTR. Radar distance is just one of them, ruler and optical distance are alternatives and usually they do not give the same values. Fermi coordinates also have some kind of distance.

According to you a 100 meter long Born rigid spaceship undergoing a proper acceleration is no longer 100 meter in fact its size is no longer unique.

 

[jcsd]

I don't see the problem, surely sometimes it is useful to use definitions of 'speed' that for light isn't always equal to c. Just need to be aware of exactly which definition you're using!