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## Large rotating disk.

 Quote by A.T. The rotating frame is rotating around some axis at some angular velocity wrt to an inertial frame. The rulers are rotating around the same axis at the same angular velocity wrt to that inertial frame.
Did you read the Wiki page on Born coordinates that I linked to, and my comments in post #48? I'm not sure the "rotating frame" you have just defined has quite the properties you think it has.

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 Quote by yuiop This way the space measured with rulers agrees with the idea of "space as a surface of simultaneity" and so there is no conflict or discarding of that notion.
Actually, I should clarify my comment on this: the notion of "surface of simultaneity" that you have defined does *not* agree with "the space measured with rulers", which is the notion of "space" that A.T. and DrGreg have described. That's part of the problem that we have been discussing.

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This thread has been moving along fast. I haven't had time to reply. I'll post few replies now. I'll start with pervect and the hypersurface of simultaneity spirals.

 Quote by pervect If you draw the set of points that are Einstein-synchronized on a rotating cylinder, you get a non-closed spiral. See for instance http://en.wikipedia.org/wiki/File:La...ronization.png Talking about the spatial geometry of such a non-closed surface (which I gather can be thought of also as a quotient manifold) is definitely odd and tends to cause confusion. Specifically, it's generally assumed the circumference of a disk is a closed curve, and we can clearly see from the diagram that this is not the case if one uses Einstein clock synchronization.
I'm not so sure that this spiral surface has the same metric as the quotient manifold. I haven't tried to work out the details, but I would expect them to be different. If I'm right, then the geometry of this spiral surface is not what people are referring to as "the spatial geometry".

Note by the way that while the spiral curves (like the blue line in the picture you linked to) can be thought of as simultaneity lines of component parts of the disk, other curves in the surface aren't simultaneity lines of the same observer. So the surface isn't really a surface of simultaneity.

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 Quote by A.T. Change? I for my part have always understood "spatial geometry" as "what rulers at rest measure".
And I have always understood "geometry" as a mathematical term. In this case, I'd say that it's a synonym of "metric". So the term "spatial geometry" should refer to the metric of a manifold that we can call "space". In pre-relativistic physics, "space" was a slice of spacetime. It's obvious that SR and GR needs to generalize the term, but it's very far from obvious that we're going to need a definition that's so general that "space" doesn't even need to be a subset of spacetime.

If there's a good argument for why the word "space" should refer to a congruence of timelike curves rather than a spacelike 3-dimensional hypersurface of spacetime, then I would have to say that "spacetime" was inappropriately named from the beginning.

 Quote by A.T. the concept of "surface of simultaneity" is useless to define "space" in a rotating reference frame.
If the term "rotating reference frame" refers to the orthonormal frame field* associated with the congruence of curves that represent the motion of the disk, then this is true. But the term "reference frame" often refers to an orthonormal frame field associated with a single observer, and is sometimes used informally as a synonym for "coordinate system". If it refers to an orthonormal frame field associated with (only) the point at the center, or a coordinate system that's rotating with the disk but has its spatial origin at the center, then there certainly is a hypersurface of simultaneity associated with it (and it's flat).

*) A frame field can be defined as a function that takes each member p of some subset of spacetime to a basis for the tangent space at p.

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 Quote by A.T. To me, physical quantities are defined by stating how you measure them: - time is what clocks measure - space is what rulers measure
This is how most physicists think about these things. I assume that you would also use the term "time" for the mathematical thing that corresponds to time, because that's what physicists usually do. I dislike this approach, because there is certainly a better way to specify how to interpret the mathematics as predictions about results of experiments than to simply use the same term for two very different things. I think this approach makes it unnecessarily hard for students to understand the difference between physics and mathematics.

I prefer to define terms (only) mathematically, and then explicitly state the rules that tell us how to interpret the mathematics as predictions about results of experiments. These correspondence rules are what turn a piece of mathematics into a theory of physics. No theory of physics is fully defined without a set of correspondence rules.

Note that the same term can have different definitions in different theories of physics. For example, in classical electrodynamics, "light" is an electromagnetic wave. In QED, it's a state that involves photons.

To deal with "what clocks measure", I would first define "proper time" as a coordinate-independent property of a curve given by an integral that I'm not going to write down here, and then I would state the correspondence rule that says that the difference between the numbers displayed by a clock at two events A and B, is the proper time of the curve that represents its motion from A to B. This correspondence rule is an essential part of the definitions of both SR and GR.

I like this approach better because it makes it easy to understand that a) mathematics doesn't say anything about reality, b) a theory of physics consists of a purely mathematical part and a set of correspondence rules (that do say something about reality), and c) how the specific theory we're talking about is defined.

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 Quote by A.T. - time is what clocks measure - space is what rulers measure
Initially I liked this definition, mainly because of its simplicity, practicality and the apparent dismissal of the requirement to consider definitions of simultaneity. However it is obvious from the posts in this thread that even with this approach, that simultaneity issues refuse to go away quietly. Another physical problem that is present here that is not present when purely inertial motion is considered, is the stress that the rulers or rods are under in a non inertial rotating reference frame. When we rotate a measuring rod from a tangential to a radial position in the rotating frame, we have to be sure that the rod is not distorted by purely Newtonian forces, resulting in a distortion of what we are trying to measure. One way to achieve this, is to check the radar length of the measuring rods in the two orientations. When we do this, the first thing we might notice is that radar length of a radial rod depends upon whether the clock used to measure the radar length of the rod is located at the end nearer the centre or the end nearer the disc perimeter. This can be minimised to a certain extent by using infinitesimal measuring rods.

I would now like to propose a method to measure the geometry of the disc that is indisputably independent of simultaneity issues. We use a single clock that is fixed to a point on the rim of the rotating disc to make all measurements. First we measure the circumference of the disc by sending a signal all the way around the disc back to the single clock and then all the way back again in the opposite direction, using suitably placed mirrors, to obtain the radar circumference of the disc. Doing this we find the circumference is is gamma times longer than the circumference measured in the inertial reference frame at rest with the centre of the disc. Next we measure the radar radius of the disc, by sending a signal from the rim to a mirror at the centre and back out to the clock on the rim again. This time we obtain that the result that the radius of the disc is gamma times shorter than the radius measured in the non rotating inertial reference frame. The end result is that in the rotating reference frame, the ratio of the circumference to the radius is 2*pi*gamma^2.

This method uses the same clock that is at rest in the rotating reference frame to measure both radar circumference and radius, so that the two measurements can be compared in a consistent way without any simultaneity issues or concerns about physical distortions of measuring rods due to "centrifugal forces". Interestingly, this alternative analysis obtains a different result from the usual interpretations, that the radius measured in the rotating frame is the same as the radius measured in the non rotating reference frame at rest with the spin axis of the disc.

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 Quote by yuiop First we measure the circumference of the disc by sending a signal all the way around the disc back to the single clock and then all the way back again in the opposite direction, using suitably placed mirrors, to obtain the radar circumference of the disc. Doing this we find the circumference is is gamma times longer than the circumference measured in the inertial reference frame at rest with the centre of the disc.
You have to assume a speed of light to transform this "round-trip" travel time into a distance, of course. What speed of light are you assuming?

 Quote by yuiop Next we measure the radar radius of the disc, by sending a signal from the rim to a mirror at the centre and back out to the clock on the rim again.
Again, you need to use an assumed speed of light to transform this into a distance measurement. Do you use the same speed as you used for the circumference measurement?

Also, because the clock on the rim is moving, relative to an inertial observer's clock, the travel time it measures will be *smaller* than the travel time an inertial observer would measure. This will result in a *smaller* result for the radius of the disk for the moving observer, *not* the same result as the inertial observer.

 Quote by PeterDonis You have to assume a speed of light to transform this "round-trip" travel time into a distance, of course. What speed of light are you assuming? Again, you need to use an assumed speed of light to transform this into a distance measurement. Do you use the same speed as you used for the circumference measurement? Also, because the clock on the rim is moving, relative to an inertial observer's clock, the travel time it measures will be *smaller* than the travel time an inertial observer would measure. This will result in a *smaller* result for the radius of the disk for the moving observer, *not* the same result as the inertial observer.
Is there some reason not to use c??

Is there some reason to assume a different speed for the radial measurement??

As I read it, yuiop said exactly what you are stating here. The radial evaluation taken from the rim would be smaller than the measurement by an inertial observer.

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 Quote by Austin0 Is there some reason not to use c??
I don't think so, but be aware that if you measured the two travel times separately for the beams going around the circumference (with the rotation, then back opposite to the rotation), the individual times would be *different*, because of the Sagnac effect:

http://en.wikipedia.org/wiki/Sagnac_effect

Some people interpret this as the speed of light being different for light going with the rotation and light going opposite to the rotation. (I don't agree with this interpretation, btw.)

There's also the more general point that the coordinate speed of light in non-inertial reference frames may not be c when evaluated over significant distances. See below.

 Quote by Austin0 Is there some reason to assume a different speed for the radial measurement??
Again, in this case I don't think so, but in general, in non-inertial reference frames, the coordinate speed of light may not be c, and may not be isotropic, when evaluated over significant distances. The *local* speed of light (evaluated using proper distances and proper times in a local inertial frame) is always c, but that doesn't automatically allow you to assume that it will always be c for a non-local measurement in a non-inertial frame, such as those yuiop has proposed.

 Quote by Austin0 As I read it, yuiop said exactly what you are stating here. The radial evaluation taken from the rim would be smaller than the measurement by an inertial observer.
Hm, you're right. I must have misread his post somehow.

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