Length and distance in SR

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1. Jan 10, 2014

JVNY

I would appreciate views on the following understanding of length and distance in SR, restricted here to inertial movement. This is the beginning of a follow up to the thread: https://www.physicsforums.com/showthread.php?t=724025

1. The spatial length of an object is the spatial separation (distance) between its ends measured at the same time.

2. One can measure length (distance) using a number of methods, such as a specified rod, or coordinates set up in advance based on a specified rod.

3. Neither spatial separation nor simultaneity is invariant in SR. (see Taylor and Wheeler, Spacetime Physics, page 56: "not the same: space separations, time separations")

4. Because simultaneity and spatial separation are not invariant in SR, spatial length is not invariant in SR.

5. Any measure of the length of an object (or distance between two objects) in one reference frame is equally valid as a measure of the length of the object (or distance between the two objects) in any other reference frame.

Can you see why this . . . would yield a different length for the rod . . . because of the relative motion between O′ and O? O′ simply measures the rod to be a different (in fact shorter) length from its length as measured by O (the "rest" length since O is carrying the rod); however neither measurement is any more correct than the other i.e. they are both equally valid after being attributed to the respective observers.

WannabeNewton at https://www.physicsforums.com/showpost.php?p=4625309&postcount=6

These seem to be fundamental to any further analysis. But I have read a number of pieces that seem to take a different view. For example, sometimes it seems that people think of proper length (spatial separation of ends of an object measured at the same time in the object's rest frame) as essentially real and invariant, and length contracted length as being coordinate dependent and not real. An example is:

Time dilation and length contraction are artifacts of remote observers. That is, in your own frame of reference, you are always the same size and your clock always ticks at one second per second. Some other observer, from a different frame of reference, SEES you as being length contracted and with a slow moving clock.

You, right now as you read this, are moving at almost the speed of light from the reference frame of an accelerated particle at CERN. Do you feel any different?

phinds at https://www.physicsforums.com/showpost.php?p=4625297&postcount=5

But your length is also invariant in the other inertial reference frame -- it is just shorter there than in your frame. As WannabeNewton says, each length is equally valid. The muons in the classic experiment do not merely "see" the height of the mountain to be smaller than we do; the height of the mountain actually is smaller for the muons than it is for us. And neither measure of height is more valid or real than the other.

Another example imagines a construct of an odometer to determine invariant length or distance that is not dependent upon any coordinates:

The aim is not to involve frames per se, at all. Instead, generalize the idea of a 'road' going by. You reel out tape measure matching the speed of road as it passes. When done, you have a measure of how much road has gone by . . . The definition is coordinate and frame independent. The only thing specified is a congruence of world lines of 'reference objects'. This is much less than a coordinate system. Given this, only invariants are computed.

https://www.physicsforums.com/showpost.php?p=4621820&postcount=3

But again, there is no invariant length or distance in SR.

2. Jan 10, 2014

WannabeNewton

I agree with you on all your points. Measurements of "length" and "spatial distance" between events as made by one inertial observer can differ from similar measurements of "length" and "spatial distance" between the same events as made by another inertial observer but neither is any more correct than the other.

Even something like the "shape" of a rigid object has no observer independent meaning because of the relativity of simultaneity. If we have a rod parallel to the $x$ axis of $O$'s frame while accelerating uniformly along the $y$ axis then $O$ judges the rod to be a straight horizontal line at each "instant of time" (simultaneity line) relative to $O$ but an observer $O'$ moving uniformly along the $x$ axis will judge the rod to be parabolic because $O'$ has a simultaneity line that's tilted relative to that of $O$ and this slices the worldlines of all the points of the rod in a different way, yielding a parabolic shape.

That isn't to say however that one is not more preferred than the other. For example, often when dealing with general relativistic fluids we choose to work solely with properties of the fluid as measured by observers comoving with the fluid elements. The 4-velocity field of the fluid picks out a preferred family of observers and a preferred set of measurements of fluid properties. In particular, whenever defining quantities that depend on the "spatial distances" between fluid elements, such as the vorticity of the fluid, we use the spatial distances as measured by observers comoving with the fluid elements.

However "preferred" measurements should not be confused with "correct" measurements.

EDIT: as a side note, spatio-temporal intervals measured by an observer can easily be expressed in a coordinate-independent and frame-independent way even in the most general setting possible. See here: https://www.physicsforums.com/showpost.php?p=4559456&postcount=10

Last edited: Jan 10, 2014
3. Jan 10, 2014

Staff: Mentor

I would not use the word "invariant" here. In the context of relativity theory, "invariant" has the specific meaning "has the same value in every inertial reference frame."

4. Jan 10, 2014

Staff: Mentor

Hi JVNY. I looked up your profile and see you are a non-scientist with a profession in finance. It is much easier to explain what you are asking about if we could use a little math. What is the extent of your math background? Any analytic geometry? Any calculus?

Chet

5. Jan 10, 2014

Staff: Mentor

I don't know about the label "real". I think it is a rather useless philosophical label. However, it is correct that length is frame variant while proper length in frame invariant.

If you have an inertial coordinate system and if you have two events on opposite ends of the object then the length is $\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$ iff $\Delta t =0$. Since different frames will disagree about $\Delta t$ they will disagree about whether or not $\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$ fulfills the requirement to be a length, and therefore length is frame variant.

If you have an inertial coordinate system and if you have two events on opposite ends of the object then the proper length is $\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2 - c^2 \Delta t^2}$ iff $\Delta t =0$ in the rest frame. Since all frames will agree whether or not $\Delta t=0$ in the rest frame and since all frames will agree on the numerical value of $\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2 - c^2 \Delta t^2}$ the proper length is invariant.

6. Jan 10, 2014

JVNY

Chet, just the algebra used in texts like Taylor and Wheeler's Spacetime Physics; also I can use the computer to solve for the standard formulas for acceleration that are used for example in the relativistic rocket explanation at http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

Last edited by a moderator: May 6, 2017
7. Jan 10, 2014

JVNY

It seems that there is a disagreement between DaleSpam on one hand and WannabeNewton and jtbell on the other. jtbell advises that the word "invariant" means "has the same value in every inertial reference frame." The length of an object does not have the same value in every inertial reference frame. And WannabeNewton agreed with all of the points in the original post. But DaleSpam considers the proper length to be "invariant."

Is this a real disagreement? Or is it just that the proper length is a "preferred" measurement that should not be confused with a "correct" measurement (in WannabeNewton's description)? I have studied the Taylor and Wheeler text, and it asserts that there is no measure of length that is invariant.

8. Jan 10, 2014

PAllen

Proper length of an object is invariant, and is the same in any reference frame. Measured length of an object moving relative to you is differs from the length of the object measured in the rest frame. Proper length is only directly measured in the object's rest frame. In other frames it is computed (from several measurements).

Similarly, distance between two objects varies with the relative motion of who is doing the measurement. Proper distance between two specific events is invariant, and computed to be the same in any frame (but is only measured as a distance in a frame in which the events are simultaneous).

I don't think there are any real disagreements.

9. Jan 10, 2014

Staff: Mentor

There is no disagreement. I think you may be slightly confused. I agree with WBN that the points in the OP are correct and I agree with jtbell that invariant means has the same value in every inertial frame.

I am not sure why you think there is disagreement in the answers you have received.

"Proper length" is not "length". They are different quantities defined differently. "Proper length" is invariant, it has the same value in every frame. "Length" is variant, it has different values in different frames. I thought that I was very explicit on that point, even including formulas.

10. Jan 10, 2014

JVNY

OK, thanks, I may have misunderstood, so I will leave it to jtbell and WBN to reply if they have any disagreement. The next question involves radar distance. I think that I am misunderstanding something that WBN wrote on this point.

WBN gives an example of radar distance in the observer's rest frame as follows:

O attaches a mirror to the front end of the rod (he holds the back end of the rod). He then emits a beam of light towards the mirror which bounces off instantaneously from the front mirror and arrives back to him. He uses a clock to record the time for the round-trip and says it is Δt. Well if the speed of light in the forward direction is the same as the speed of light in the backwards direction, like we assumed, then surely the light beam reached the mirror when O's clock read Δt/2 and hence the distance traveled would obviously be L0=cΔt/2. This distance spans the length of the rod as measured by O.

https://www.physicsforums.com/showpost.php?p=4625309&postcount=6

I understand this. If the rod has 100 proper length, then in the observer's rest frame the light takes proper time 100 out and 100 back, for a total of Δt = 200, and proper length of 200/2, or 100.

WBN then states:

Can you see why this "radar echo" method would yield a different length for the rod when a similar measurement is made by O′ because of the relative motion between O′ and O? O′ simply measures the rod to be a different (in fact shorter) length from its length as measured by O (the "rest" length since O is carrying the rod); however neither measurement is any more correct than the other i.e. they are both equally valid after being attributed to the respective observers. [same link as above]​

I cannot see how this method yields a shorter rod length for O'. For example, in an O' frame in relative 0.8c motion, the rod has length 60. The O' time for the forward flash to reach the mirror is 300 [being 60 / (1 - 0.8)], and for the return flash is 33.333 [being 60 / (1 + 0.8)], for a total Δt = 333.333, and Δt/2 = 166.67 (a value greater than 100). So if O' uses this method, he gets a radar method length greater (not lesser) than the proper length.

How should one interpret WBN's post? I have not seen this method used except in the rest frame. In other frames typically writers use other methods, like Einstein's of having lightning strike at both ends of the object at the same time in the O' frame (e.g., the platform frame in the train and platform example).

11. Jan 10, 2014

WannabeNewton

It's infinitely more cumbersome to do it explicitly using radar signals but you would have to have $O'$ send a light signal at some instant of his clock aimed at the farther end of the rod and then at a later instant send out another light signal now aimed at the closer end of the rod such that the two signals reach the respective ends of the rod simultaneously relative to $O'$. By calculating the radar distance to each of the two simultaneous events, $O'$ can then subtract the two to get the length of the rod in his rest frame.

If you draw a space-time diagram representing the rest frame of $O'$ then the two ends of the rod would have two parallel but slanted worldlines. The proper length of the rod is the length of the line segment formed by the intersection of the $t = 0$ simultaneity line of $O$ with the worldlines of the two ends of the rod. The length of the rod as measured by $O'$ is simply the intersection of the $t' = 0$ simultaneity line of $O'$ with the worldlines of the two ends of the rod which, in the space-time diagram, is simply the intersection of said worldlines with the $x'$ axis. Clearly $O'$ will measure a shorter length than $O$ will.

The reason I mentioned radar signals is that these $t = \text{const.}$ and $t' = \text{const.}$ simultaneity lines are established by the respective observers using radar sets that they each carry. In their respective rest frames they use their radar sets to establish what they consider to be lines of simultaneity by applying the convention $t_B = \frac{1}{2}(t_A + t'_A)$ alluded to earlier.

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12. Jan 11, 2014

JVNY

OK, WBN, thanks. The radar method is certainly not cumbersome in the rest frame, so it seems useful there.

So, I think that I am down to the final measure, which is "coordinate." In inertial frames in SR, is there a distinct measure of "coordinate" length or "coordinate" distance? There does not seem to be any difference between (a) setting up coordinates covering a station platform and determining a passing train's length by reference to those coordinates, and (b) making marks on the platform at the train's front and rear at the same time in the platform's frame, then using a meter stick to measure the distance between the marks.

13. Jan 11, 2014

Staff: Mentor

Yes, it's the same thing if you can make the two marks at the same time in the platform's frame.

14. Jan 11, 2014

WannabeNewton

If it helps, imagine at first $O$ only has the clock he's carrying with him and a radar set. Furthermore imagine there's a clock at rest with respect to $O$ at each point in space. Right now $O$ can only label events with a time coordinate if they lie on his worldline but by using his radar set, $O$ can synchronize all the aforementioned clocks with his own and in doing so constructs what we call a global time coordinate. Whenever his clock reads some time $t$ so will all the other clocks in space because they have been synchronized with his so he can now assign a time coordinate to each point in space using his own clock.

By synchronizing each comoving clock in space with his own, $O$ has also defined what it means for events to be simultaneous because if two synchronized clocks $A$ and $B$ have their hands both positioned at $t$ at events $p_A$ and $p_B$ on their respective worldlines then $p_A$ and $p_B$ are deemed simultaneous. Einstein synchronization mathematically leads to the simultaneity planes $t = \text{const.}$

Imagine now a rigid rod connecting each such clock with another; this forms what one might call a "rigid coordinate lattice". By combining these rigid rods with his simultaneity planes, $O$ can now determine lengths of objects. This "rigid coordinate lattice" of rods and clocks is his coordinate system.

15. Jan 11, 2014

JVNY

Agreed. And this seems to be consistent with Chestermiller's response above: there is no separate "coordinate" length or distance in an inertial frame. The length of an object measured by using a ruler is exactly the same as the length measured by coordinates of the object's ends along the rigid lattice when noted at the same time (time being the time shown on the lattice clocks). Indeed, this is the same length as determined using the radar method as long as the object is at rest in the lattice frame (which will also equal the object's proper length). For an object at rest, proper length = length = radar length = coordinate length = ruler length. Or for the distance between two points at rest, proper distance = distance = radar distance = coordinate distance = ruler distance.

This takes us to back to Born rigid motion. If you have a set of points that are at rest with respect to each other then accelerate Born rigidly, they maintain their proper distance at all times. See http://www.mathpages.com/home/kmath422/kmath422.htm.

Put another way, they maintain their rest distance at all times. See Franklin, "Lorentz contraction, Bell’s spaceships, and rigid body motion in special relativity," page 8, at http://arxiv.org/abs/0906.1919 (referring to rest length of an object that is accelerated Born rigidly).

This makes sense, because they are at rest with respect to each other at all times.

They also maintain their ruler distance at all times. Observers riding along with them can slowly pass a ruler and measure the distance between the points throughout the acceleration, and they will agree that the points maintain the same ruler distance. See http://en.wikipedia.org/wiki/Rindler_coordinates

So, we have points that are at all times at rest with respect to each other and maintain their same proper, ruler and rest distances -- just like points that are at rest in an inertial frame.

However, the radar distance is not the same. Consider three ships, rear center and front, with equal proper distance between rear and center as between center and front. The ships begin at rest, then accelerate Born rigidly. Flashes of light sent simultaneously from the center toward the front and rear ships reflect off of those ships and return to the center. The forward flash returns before the rearward flash. This is true even though the proper distance between rear and center remains the same as the proper distance between center and front (and despite the fact that the rest distances and ruler distances remain the same also).

So, question 1: what is this radar distance? When we discussed the issue in the earlier thread, posters used examples of curves (such as the distance along a curve from one city to another along the surface of the earth). However, this cannot explain the radar distance here for two reasons. First, SR does not used curved spacetime; SR uses flat spacetime.

Second, the curved line along the surface of the earth is the ruler distance between the cities. If you lay a series of rulers between the cities you will measure the longer curved distance (rather than the shorter straight line distance that you would get in a tunnel bored straight between the cities underground). But we have confirmed that the ruler distance between center and rear remains the same as the ruler distance between center and front, yet the forward light flash returns to the center before the rearward one does.

Question 2: is it clear what the coordinate distance is between rear and center (and between front and center)? The lattice of rods should remain unchanged in the Born rigid acceleration, and so the distances as measured by the lattice rods should remain the same. In order to ensure simultaneous measurements you have to adjust the clock rates so that each clock ticks at the same rate (choosing a single clock for the Rindler coordinate time rate). But, that does not seem to be enough. The Einstein method of synchronizing the clocks as WannabeNewton describes above depends on radar signals, and so there seems to be a circularity. You need to use radar to synchronize the clocks, but the radar distance is different from the ruler, rest and proper distance.

It is conceptually easier in GR, where I understand that the ruler distance between concentric rings around a black hole is longer than the difference in radii calculated using Euclidean geometry, due to curved spacetime.

16. Jan 11, 2014

WannabeNewton

In the case of a rod being accelerated in a Born rigid fashion this is true but keep in mind that Born rigidity of a macroscopic object does not require all the microscopic constituent particles to be at rest with respect to one another because we can have Born rigid rotation wherein the constituent particles rotate relative to one another but spatial distances between neighboring constituent particles will remain constant as measured in the instantaneous rest frames of the neighboring constituent particles.

Indeed, so as long as this ruler distance is measured in the rest frames of the constituent particles making up the Born rigid object (see above).

Indeed this is true in general and it is precisely due to the reason you stated: if we imagine a rod that is accelerated in a Born rigid fashion from back to front then the closer to the front we get the harder the constituent particles have to accelerate in order to maintain Born rigidity; relative to a background global inertial frame this means that the particles closer to the front of the rod have greater instantaneous velocities than do the particles closer to the back of the rod. However note that radar distance agrees with ruler distance if the constituent particles we are using for measurement are infinitesimally close to one another.

Could you rephrase this question? I'm not sure if you're looking for some explicit calculation of radar distance, a conceptual explanation of radar distance, or something else entirely.

Einstein synchronization doesn't work for a rigid coordinate lattice carried by a uniformly accelerating observer. If a clock in the lattice is Einstein synchronized with the local clock of the observer at some initial local time then the clocks will become desynchronized immediately after. Because Einstein synchronization fails to be consistent for the rigid coordinate lattice carried by this observer, in the sense that initially synchronized clocks become desynchronized soon after, we cannot use Einstein synchronization to define simultaneity for this observer because a simultaneity convention derivative of a synchronization convention requires the synchronization convention to be consistent.

This is why radar distance doesn't agree to all orders with ruler distance for a uniformly accelerating observer: ruler distance automatically takes into account the clock desynchronization since ruler distance is calculated using the geometrical definition of simultaneity whereas radar distance relies on local and distant clocks being Einstein synchronized hence if we have the uniformly accelerating observer use radar distance we are assuming that simultaneous events have the same clock readings which will give us over-estimates and under-estimates of the ruler distance.

However for events infinitesimally close to events on the world line of the uniformly accelerating observer we will find that ruler distance and radar distance agree and this is because the clock desynchronization factor is negligible for clocks in the observer's rigid coordinate lattice that are infinitesimally close to the observer's own clock.

17. Jan 11, 2014

pervect

Staff Emeritus
I would say that the invariant interval is independent of the observer, is defined between any two events in space-time regardless of whether or not they are part of an object, and is given in special relativity by the formula given by DaleSpam (the square root of dx^2 + dy^2 + dz^2 - dt^2 for a spacelike interval).

Proper distance is most rigorously understood as applying to a curve with endpoints. If you integrate the invariant integral (as defined above) along the specified curve between the specified endpoints you get the proper distance.

Often times it is assumed that given the endpoints, one "knows" the right curve to integrate along, in which case specifying the endpoints is _assumed_ to specify the curve. This may or may not cause confusion depending on how explicit the author was about the assumptions.

In the flat space-time of Special relativity, the length of an object (which may or may not be rigid) at some time t is equal to the proper distance (the integral defined above) between the ends of the object along a curve of constant time - which curve this is is observer dependent, it depends on the observer's notion of simultaneity.

As a consequence of the above definitions, the length of an object at time t in a frame in which it as rest in the flat space-time of SR is also equal to the invariant interval between the endpoints at time t.

In general relativity or in accelerated coordinate systems, one needs to introduce a metric. The invariant interval then becomes defined only for "nearby" points, and is equal to:

$$ds^2 = \Sum_{i=0..3, j=0..3} g_{ij} dx^i dx^j$$

where $g_{ij}$ are the metric coefficients. If g_00 = -1 and g_ii = +1 for i = 1..3, one recovers the usual SR formula.

Proper distance is still defined by the integral of the invariant interval using the new definition, and the length of an object is still defined by the proper distance between its endpoints. The potential confusion over which curve to integrate over is greater, however. Usually it's correct to assume a curve of constant coordinate time when trying to figure out which curve to integrate along.

It is in general no longer true that one can simply take the invariant interval between the endpoints to get the length of an object in GR, one has to use the procedure above.

18. Jan 12, 2014

WannabeNewton

I missed the latter comment the first time around so let me address it now. While what you said about non-euclidean geometry in GR is true, keep in mind that the aforementioned issues regarding (and stemming from) clock synchronization exist in GR for the same reasons as in SR.

Imagine we have a rotating star (or rotating black hole if you prefer but it doesn't matter). Consider an observer outside the rotating star who is fixed with respect to the distant stars at some radius $R$ as well as a ring of clocks at $R$ (in the same plane as the observer) that are also fixed with respect to the distant stars. The clocks are therefore also all at rest with respect to one another and with respect to the observer; the ring formed by this observer+clocks system is Born rigid and all constituents of this system have the same proper acceleration. Can the observer Einstein synchronize all these clocks with his own?

Imagine further that at some given instant of the observer's local time he emits a light signals in the prograde direction around the ring and a light signal in the retrograde direction around the ring (each clock is equipped with a mirror angled appropriately in order to allow light beams to circulate around the ring). If the observer is to have any hope of Einstein synchronizing all the clocks in the ring then the retrograde signal and prograde signal should arrive back at the observer at the same time (as measured by his clock). It will turn out however that for the observer+clocks ring that we have constructed fixed with respect to the distant stars outside of a rotating star, the two signals will not arrive back at the same instant of local time. This is formally analogous to the failure of global Einstein clock synchronization for a rotating ring in flat space-time. This effect is called the Sagnac effect.

There is another method of clock synchronization that for a rigid lattice of inertial clocks is mathematically equivalent to Einstein synchronization but doesn't use light signals; this synchronization method is known as slow clock transport.* So you may think that we can instead try using slow clock transport for the ring described above because the Sagnac effect only manifests itself when we have counter-propagating signals (of any kind not necessarily electromagnetic) but it turns out that even using slow clock transport an observer in the ring fails to achieve synchronization with all the clocks in the ring.

*See section 7 of the following paper: http://arxiv.org/pdf/1002.0130v1.pdf

19. Jan 12, 2014

stevendaryl

Staff Emeritus
Just a pet peeve, but it annoys me that physicists use the words "philosophy" or "philosophical" pejoratively in this way. A real philosopher is not going to use a useless label, any more than a physicist is. One of the main occupations of philosophy is to try to figure which labels are useful and meaningful, and which ones are not.

Almost all scientists engage in philosophy, but only use the word "philosophy" to describe other people's useless thoughts.

20. Jan 12, 2014

stevendaryl

Staff Emeritus
That's not a disagreement. Proper length has the same value in every inertial reference frame.

21. Jan 12, 2014

JVNY

Thanks for a lot of good points to think about. Perhaps best to start with what seems most fundamental. Is spacetime different in an accelerating reference frame in SR than in an inertial reference frame in SR? I had learned that SR uses flat spacetime. But see the following from pervect:

This suggests that spacetime is different in SR in an accelerating frame (which is grouped with GR, and uses a metric) than in an inertial frame in SR (which is considered separately, and uses the interval).

And if it is different, is spacetime curved in an accelerating frame in SR?

22. Jan 12, 2014

stevendaryl

Staff Emeritus
There are two different concepts: (1) A manifold (space, or spacetime) with curvature, and (2) curvilinear coordinates. If your manifold is curved, then you are forced to use curvilinear coordinates to describe a large enough region. You can use curvilinear coordinates in flat spacetime, but that doesn't make it curved.

An accelerating frame is just using curvilinear coordinates to make a set of accelerating observers "stationary" (that is, have zero spatial velocity).

23. Jan 12, 2014

stevendaryl

Staff Emeritus
Just another point: flat spacetime has a metric, just as curved spacetime does. The significance of an "inertial reference frame" is that, as expressed in those frames, the metric is so trivial that you can pretty much get away without ever mentioning it explicitly (although it is involved implicitly whenever you see expressions such as "$x^2 + y^2 + z^2 - c^2 t^2$" and "$E^2 - p^2 c^2$). In curved spacetime, or in noninertial, curvilinear coordinates, it becomes more important to explicitly talk about the metric.

24. Jan 12, 2014

WannabeNewton

Yes. For one, there can be global anisotropies in the speed of light. Take, for example, an observer riding on a rotating ring in flat space-time. As mentioned earlier if the observer emits prograde and retrograde light signals that circulate around the ring (by means of mirrors placed around the ring) then the observer will record different local times of reception of the counter-propagating signals. In this example this is directly related to the fact that the tangent vector field $\vec{\xi}$ to the world tube of the rotating ring has a non-vanishing curl $\vec{\nabla}\times \vec{\xi} \neq 0$ in the instantaneous rest frame of the observer. However it is important to keep in mind that this anisotropy in the speed of light relative to the observer riding on the rotating ring is global i.e. it only manifests itself after the prograde and retrograde signals complete an entire circuit. Locally the speed of light will still be isotropic regardless of the Sagnac effect.

EDIT: Also, aside from the obvious presence of inertial forces, accelerating frames of reference also have finite extents in space-time. Since you have access to MTW, see section 6.3 for a discussion of this.

No.

Last edited: Jan 12, 2014
25. Jan 12, 2014

WannabeNewton

A frame is not the same thing as a coordinate system.