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grav
2008-Feb-29, 01:47 AM
Let's say that we are all riding on a large merry-go-round with horses. Sound like fun? I'm on one and the rest of you are spaced out on the ride (but not too spaced out, I hope :) ). Even though we are all rotating at the same rate, always in the same positions relative to each other, and therefore observing no relative speed to each other, can we still say that we are in the same frame of reference? To an observer on the ground, if I am moving forward at some speed, then the rest of you are moving at different angles and perpendicular to me, and one in completely the opposite direction, doubling the relative speed between us according to the observer.

publius
2008-Feb-29, 02:19 AM
Grav,

First of all, everyone there is in a non-inertial (accelerated) frame of reference. Even in Newton, a simple rotating observer at the center of the merry-go-round is different from observers riding out at some radius r. The radius riders are experiencing translational accelerations as well as pure rotations ("revolving" as well as rotating about their own center). That would add a wee bit more complication to the fictitious forces, even in Newton-Galileo.

When one uses a rotating frame (Coriolis Frame, it is sometimes called), one pegs oneself at the center, r = 0. r > 0 isn't done much unless one just has to.

Now, in Relativity, this all gets very, very, very, very complex indeed. At the center, r = 0, one can adopt what is sometimes called the Born coordinates, and one can do relativistic physics in those just fine and dandy. The centrifugal and coriolis forces add pseudo-gravitational effects on clock rates and all that good stuff, with the coriolis part giving you frame dragging looking terms.

While that's not Romper Room math by a long shot, it's still fine and dandy. For an observer at r = 0.

Radius riders here, relativistically, become a absolute mess. This is part of something known as Ehrenfest's Paradox.

There is no real paradox of course, but resolving things is not trivial at all. It turns out the while radius riders certainly have their own local ruler and clock, those local coordinates just don't extend globally very well. The trouble is, in the language of the high powered 'rithmetic of differential geometry that governs this, that the radius rider's world lines have "vorticity". That means their spatial hyperslices of the space-time just aren't kosher, and don't work globally so well. That's about all I can tell you -- this is far from Romper Room 'Rithmetic, indeed.

Now, locally, in a region small enough to make those global problems vanishingly small, a radius rider sees a curved, hyperbolic spatial geometry. Just try to extend that geometry globally and it fails because of the vorticity. Apparently you can still work with it if you don't mind a very crazy non-Riemannian geometry (which is to Riemann as Riemann itself is to Euclid). Crap goes on there, IIRC, like the "distance" between two points being non symmetrical, ie the distance from A to B is different from B to A.

Nothing is wrong with the space-time mind you -- indeed it is flat Minkowski, we're just playing games with curved coordinates -- and nothing is wrong with any physics one does there. One just needs to use kosher global coordinates if one is going to try do global physics.

You will see a lot of misconecptions about Ehrenfest and rotating frames in relativity, and by a lot of physicists and others who should know better. That is common lament apparently by the real "high priests" of GR and differential geometry -- that part about the failure of spatial hyperslices is the root of all the problems there.

-Richard

grav
2008-Feb-29, 03:17 AM
That was quick. Thanks, Richard. Good to hear from you again. Where was all that when I was asking if rotation was absolute and that thing about an object orbitting at the same rate as the rotation of a gravitating body? :)

I guess that means that observers that aren't moving relative to each other can still be in different frames of reference. If a gravitating body were set to spin at the same rate that an observer orbits it (in freefall), the situation would appear stationary as well, but it can't be because then the observer should begin to fall straight toward the gravitating body from his point of view, which would happen if they were actually stationary, but that doesn't happen.

So I guess it all has to do with acceleration, then, right? It's not just who is moving relative to whom. We would always have to figure acceleration in as well to get the relative speed. If you are riding the horse directly opposite of me, then you are accelerating outward, away from the central axis, and I am accelerating in the opposite direction, so even though we are not moving relative to each other, we would still have an effective relative speed, other than being stationary to each other, adding equal and opposite speeds, same as with the acceleration, relative to the central axis, I think. That is fascinating. The riders perpendicular from me are accelerating perpendicularly relative to me, so their relative speed is perpendicular as well, it would seem. It would still all be relative to the central axis, then, but in exactly what way, I'm still not sure.

publius
2008-Feb-29, 03:52 AM
Grav,

The geosynchronous satellite situation in Newton is simple enough. The gravitational acceleration on the satellite, g(r), is exactly cancelled out by the centrifugual force, w^2*r. Thus the rotating observer concludes the satellite should remain stationary. The gravitational force is cancelled by the inertial psuedo forces of the non-inertial frame of reference. There is no problem there at all, it's just these psuedo forces can trip people up sometimes.

In General Relativity, the situation is exactly the same, albeit a bit more mathematically involved. The rotating observer sees the psuedo forces of the rotating frame as type of global gravitational field, which exactly cancels out the local gravitational field of the planet.

It's all in the metric -- while it gets complex, the rotating Schwarzschild metric (that is, Schwarzschild transformed to a rotating observer) is used in the GPS calculations, or at least to derive the calculations used. And that metric tells that a satellite at the geosynchronous radius would stay put and not fall from our rotating POV. Again, no big deal at all.

So I guess it all has to do with acceleration, then, right? It's not just who is moving relative to whom. We would always have to figure acceleration in as well to get the relative speed.

Grav, the most general way to put it is it depends on the metric and the worldlines against that metric. Relative motion means, in your coordinates, something is moving with time. That is, its position coordinates change with your time variable. :) Stationary means it doesn't change position with your clock.

In flat space-time, using inertial (Minkowski) coordinates, then the familiar rules of SR apply where the "magic factor" depends only on that relative velocity.

In non-inertial coordinates and/or curved space-time, then "other stuff", beyond the relative velocity (according to your coordinates) which is all encoded in your metric, will apply and govern what you "see". Succinctly, what you "see" depends on the space-time you're in, and your own path (worldine) through that space-time.

SR/Minkowski is the province of a straight time-like worldline in flat space-time.

-Richard

Jens
2008-Feb-29, 08:58 AM
I guess that means that observers that aren't moving relative to each other can still be in different frames of reference.

I wonder if that's really true. Are they moving relative to each other or not? It depends on whether you take your own (spinning) perspective or the fixed stars as guides. If you take stars as guides, then the person who was north of you at one point will be west and then south. Obviously, the distance isn't changing. But then, is the moon moving relative to the earth? If the earth were tidally locked to the moon, would the moon be moving relative to the earth? In a sense, if we are both on a merry-go-round, I could perceive you as orbiting around me, with me being tidally locked to you.

hhEb09'1
2008-Feb-29, 12:02 PM
Where was all that when I was asking if rotation was absolute and that thing about an object orbitting at the same rate as the rotation of a gravitating body?It was there, behind the calculus :)

I guess that means that observers that aren't moving relative to each other can still be in different frames of reference.I wonder if that's really true. It's absolutely true. :)

In fact, just sitting here this morining, I placed myself in four different reference frames, without moving from this chair. It doesn't have to be two different observers that aren't moving relative to each other--it can be a single observer, who, of course, is not moving relative to themself ("themself"? I like it...)

grav
2008-Feb-29, 10:18 PM
Thanks everyone. So, who's up for some cotton candy? My treat. :)

Jens
2008-Mar-03, 03:45 AM
In fact, just sitting here this morining, I placed myself in four different reference frames, without moving from this chair. It doesn't have to be two different observers that aren't moving relative to each other--it can be a single observer, who, of course, is not moving relative to themself ("themself"? I like it...)

What I meant to question wasn't about the reference frames, but rather about the "moving relative" part.

I think the people on a merry-go-round are moving relative to each other.

Aren't they?

hhEb09'1
2008-Mar-03, 04:10 AM
What I meant to question wasn't about the reference frames, but rather about the "moving relative" part.

I think the people on a merry-go-round are moving relative to each other.

Aren't they?Depends upon what you mean, I guess. They're staying the same distance apart, right? If you were at one, looking at the other, you wouldn't have to move, relative to your local horse, to keep the other in sight?

I think that's what grav meant.

Jens
2008-Mar-03, 04:47 AM
Depends upon what you mean, I guess. They're staying the same distance apart, right? If you were at one, looking at the other, you wouldn't have to move, relative to your local horse, to keep the other in sight?

I think that's what grav meant.

Yes, that's true. But where I wanted to go from there is, suppose you have a moon orbiting a planet in a perfectly circular orbit. Then they always maintain the same distance. And if the planet is tidally locked to the moon, like Pluto, then you would have (I think) precisely the same spatial relationship. In that case, would we say that the moon is moving relative to the planet? I would say yes.

Ken G
2008-Mar-03, 04:49 AM
I think the people on a merry-go-round are moving relative to each other.

Aren't they?I would say that they are in different reference frames, although whether or not they attribute the differences to relative motion depends on what reference frames they choose to analyze the situation. That's the key point physically in all this (publius is right that the key point mathematically is the metric), all that has to agree is the wavelength shift-- the reasons given for the shift do not have to be the same for different observers (one observer says it's "relative motion" in the special relativity sense, another says it is some sort of pseudo-gravity in the equivalence principle sense). The same thing happens with cosmological redshifts, how you attribute them depends on your chosen coordinatization (i.e., how you connect your local reference frame to the physics in the distant reference frame of interest).

Thus it is important to use empirical, not philosophical, determinations of this sort of thing. Two people on the same merry go round will experience relativistic Doppler shifts if they are not at the same radius on the merry go round and are exchanging light signals, even though if they have no points of reference they will not know whether to attribute those changes to gravity or time dilation of motion. But that's the beauty of the equivalence principle-- it doesn't matter. The shifts they will see are attributed to transformations between reference frames, as is characteristic of second-order (in v/c) shifts. First order (v/c) shifts are attributable to changing time-of-flight effects and are not relativistic in nature.

grav
2008-Mar-05, 12:08 AM
Looks like I still have a couple of questions. Maybe more. For one, let's say there is a "stationary" observer at the center of the merry-go-round. How should he view the riders? In this other thread (http://www.bautforum.com/questions-answers/71035-another-relativity-question.html), the time dilation and relativistic Doppler worked out to the same thing, but that was because, as Jeff pointed out, the actual time of observation was shorter for a ship travelling toward us, so it evened out to the same thing either way. In the case of a merry-go-round, however, we would consider the time of observation to be continuous, for a constant rotation, wouldn't we? So on the one hand, we have time dilation telling us that the stationary observer at the center will see the riders' clocks ticking at sqrt[1-(v/c)^2] normal speed. But on the other hand, the riders each ride tangent to the center observer, so regular Doppler says he will observe the same time, but divided by sqrt[1-(v/c)^2], so sees the riders' clocks ticking this much faster instead. Of course, each of the riders can also be said to be constantly accelerating away from the central observer due to the centrifugal force as well, so I'm sure that figures in somehow, but in what way exactly?

Ken G
2008-Mar-05, 12:29 AM
Of course, each of the riders can also be said to be constantly accelerating away from the central observer due to the centrifugal force as well, so I'm sure that figures in somehow, but in what way exactly?When both observers are inertial, they can each think the other's clock is slow because they are separating, so are continuously desynchronizing because of that. In the noninertial merry-go-round, they are not separating, but the continuous desychronization is obtained from the pseudo-gravity of the rotation.

grav
2008-Mar-05, 12:47 AM
When both observers are inertial, they can each think the other's clock is slow because they are separating, so are continuously desynchronizing because of that. In the noninertial merry-go-round, they are not separating, but the continuous desychronization is obtained from the pseudo-gravity of the rotation.You know, I'm not sure if I follow what you said, but it made me think of something. If one of the riders is continuously observing another, then if the one observes the other's clock ticking slower or faster, either way, then if the merry-go-round were to stop, they would both have to "catch up" to each other's times, because each underwent the same effects as the other, so their times should match. If they were to ride on the merry-go-round for minutes or for years, the effect of stopping the merry-go-round would be the same in either case, right? But if the time dilation between them is constant, then there should be a difference between stopping it short or after a very long ride, that the stopping part wouldn't account for, since it is the same. I am thinking that this means that whatever effects take place due to a constant outward acceleration exactly cancel the time dilation, then, and each would always observe the others' clocks ticking the same as their own. That should also be true regardless of the radius, then, if that is right. So that would mean that each is stationary relative to the others after all, anywhere on the ride, at least in respect to each other, but not in respect to a stationary (non-rotating) observer. Does that sound like it makes sense?

grav
2008-Mar-05, 12:55 AM
What was the formula for the time dilation of an acceleration relative to a stationary observer again? I'll try to find it.

grav
2008-Mar-05, 01:20 AM
So dT/dt (where T is the proper time) goes as sqrt(1 + gx/c^2).
I found this in another thread. If we figure the riders are rotating at v to an observer on the ground, then the relative speed between observers opposite each other would be 2v, making the time dilation sqrt[1 - (2v)^2/c^2] = sqrt[1 - 4v^2/c^2]. For the acceleration, we have g = 2 * (-v^2/r), and x = 2r, so sqrt[1 + gx/c^2] = sqrt[1 - 4v^2/c^2] also. Does this mean the time dilation cancels out? Of course, these two formulas are only right to the second order, because velocities do not add, so if we were to say that the ride is rotating at close to c on the perimeter, we wouldn't say the riders are travelling at close to 2c relative to each other. And the riders and the ride itself should contract, I think, even though distances themselves don't, but the riders and the portion of the ride they travel upon, as well as the rest of the way around would, right? So the entire ride should seem a little closer to each observing rider at a high rate of rotation. But the difference with these last two effects should also cancel out, then, shouldn't they?

publius
2008-Mar-05, 05:54 AM
Grav,

The 1 + gx/c^2 applies only to the Rindler frame, straight line constant proper acceleration (feeling constant 'g'). Don't even worry about trying to apply that to a rotating frame. [Actually, something close, but with a time varying 'g' in the metric would there for a *NON-ROTATING* radius rider -- that is one who was going around in a circle, but keeping his own axes inertial. The rotating radius rider has something more complex, much more complex. The observer at the center, the Born observer, is actually fine becuase the vorticity cancels out at r = 0]

But anyway, you're right about two radius riders at the same 'r', as can trivially be seen from an inertial frame watching them. If the merry-go-round goes for a while and then stops, both of them had the same gamma factor and their proper times should be equal (and less than ours).

How that would play out in the local coordinates of each is not a trivial thing, but it certainly has to work out. As I mentioned before, the local coordinates of the rotating radius riders are not kosher globally because of the vorticity of their world lines. All that is well beyond my understanding, and we'd need a "high priest", a real one, to explain it. If he could. :)

And while I think I mentioned this before, this is a good place to mention it again. A while back, I got to rambling about some papers by a Klauber about relativistic rotation. Klauber had it in his head that rotating radius riders were naturally in a non-time orthogonal metric, and should, among other things, seen tangential anisotropy in light speed. At the time he wrote the first papers, the lastest MM type experiments weren't sensitive enough to rule that out for the earth's rotation (1 part in 10^13 or something like that). But a later experiment was, and no such effect was found.

Well, later, I got to reading some commentary by some real "high priest" types on rotating frames, and well, Klauber was just getting screwed up by the vorticity, essentially, and doing something invalid because of the non-kosher problems with the coordinates. :)

But don't think Klauber some piker -- a *lot* of people get screwed up by relativistic rotation. And no wonder, it ain't romper room by a long shot. But what gets the goat of those who do understand this is how the same errors keep getting made, and somehow this "non kosherness of vorticity" and how this worksjust doesn't seem to catch on wider in the field.

-Richard

Ken G
2008-Mar-05, 06:51 AM
What I don't understand is, it seems we can always figure out what is happening to clocks for the rotating riders by considering their motion from an inertial frame. Surely the spinless riders and the rotating riders must have the same clock readings from the point of view of the inertial frame? And as they are at the same location (let's say), they cannot have any dispute between them about what time their clocks read either, can they? So why wouldn't the rotating riders have all the same time effects as spinless riders? Does spin at a point affect a clock comparison?

hhEb09'1
2008-Mar-05, 02:25 PM
What I don't understand is, it seems we can always figure out what is happening to clocks for the rotating riders by considering their motion from an inertial frame. Surely the spinless riders and the rotating riders must have the same clock readings from the point of view of the inertial frame? And as they are at the same location (let's say), they cannot have any dispute between them about what time their clocks read either, can they? So why wouldn't the rotating riders have all the same time effects as spinless riders? Does spin at a point affect a clock comparison?Which ones are the spinless riders? The ones whose horses are fixed?

Ken G
2008-Mar-05, 02:27 PM
The spinless ones always face the same distant star, so their horses are on frictionless pivots. They experience no centrifugal tides from one side of their bodies to the other, so that's how you can tell they are not spinning. I guess that must be related to my question above-- any real clock must have a lateral dimension, so centrifugal tides will affect its function. Perhaps a spinning clock does read time differently, but surely the effect could be mitigated in the limit of a clock the size of an atom, bringing me back to my question...

John Mendenhall
2008-Mar-05, 06:02 PM
The spinless ones always face the same distant star, so their horses are on frictionless pivots. They experience no centrifugal tides from one side of their bodies to the other, so that's how you can tell they are not spinning. I guess that must be related to my question above-- any real clock must have a lateral dimension, so centrifugal tides will affect its function. Perhaps a spinning clock does read time differently, but surely the effect could be mitigated in the limit of a clock the size of an atom, bringing me back to my question...

One for frictionless pivots, and one for infinitismally small clocks.

More seriously, read Richard's posts again. About once a month here or on ATM, somebody brings up some version of the relativistic wheel (in this thread a carousel), and the answers are always the same. It's one bear of a problem, it only yields to GR, and over the years some very good math people have gotten it wrong, as Richard says. Einstein treated the problem at least once in one of his early papers, I don't have the reference at hand.

But there are similar things in the (maybe) real world. Consider bunches of particles separated by fixed intervals and circulating around a cyclotron. Or material falling into the equator of a black hole at regular intervals.

Or how much math it takes to do the GR analysis.

Best regards, John M.

publius
2008-Mar-05, 08:34 PM
What I don't understand is, it seems we can always figure out what is happening to clocks for the rotating riders by considering their motion from an inertial frame. Surely the spinless riders and the rotating riders must have the same clock readings from the point of view of the inertial frame? And as they are at the same location (let's say), they cannot have any dispute between them about what time their clocks read either, can they? So why wouldn't the rotating riders have all the same time effects as spinless riders? Does spin at a point affect a clock comparison?

Ken,

Yes, as I understand things, from the POV of an inertial frame, the spinning and non-spinning clock rates are the same, given by the simple gamma factor.

The problem is trying to use the spinning observers's own local ruler and clock to globally coordinatize the space-time.

Further, as I understand it, in a small enough local region, where the globally problematic terms are small enough to be of no consequence, the notion of space will be hyperbolic, curved (just as Einstein himself concluded years ago, although some disagreed -- it's a long story).

That is, if you can imagine a group of local co-spinning observers on the rotating disc seeing each other as stationary and close enough together to elimainte the global prolems, they could do geometric experiments with their own little rulers and conclude they were in a non-Euclidean, hyperbolic space.

Again, the problem is trying to extend those local frames with "vorticity" globally.

And finally, all this is just about coordinates in flat space-time. Now, add "torsion" to space-time, ala spin and Einstein-Cartan, and you can get these vorticity effects in the space-time itself. Now, if that ever means a spinning clock rate might be different than a non spinning clock when you've got torsion afoot, I don't have a clue. I can only assume it gets very complex indeed. :)

-Richard

publius
2008-Mar-05, 08:50 PM
Spinless can be defined "easily" enough in GR -- something called the Fermi-Walker derivatives of the local "frame field", along the world line are zero. That defines what you can call proper rotation.

In flat space-time, that pretty much agrees with Newtonian expectations and means no inertial observer sees anybody's axes rotating according to his coodinates.

However, in curved space-time, that goes out the window, and you can have "coordinate rotation". This happens for orbiting observers relative to stationary observers in Schwarzschild, and is responsible for the "geodetic precession" (sometimes called deSitter precession, I believe) of gyroscopes.

That was one of the effects Gravity Probe B was looking for, and has confirmed. The geodetic effect is much larger than the frame-dragging effect, the latter they're having problems pulling out due to some problems.

However, according to Nordtvedt, both the geodetic *and* frame dragging precessional effects can be seen in the motion of the earth-moon system around the sun in the years of precision lunar laser ranging data. I don't know enough of the details of what the exact effects on the motion are, but Nordtvedt and others say they've confirmed it.

-Richard

Ken G
2008-Mar-05, 09:12 PM
Yes, as I understand things, from the POV of an inertial frame, the spinning and non-spinning clock rates are the same, given by the simple gamma factor.

The problem is trying to use the spinning observers's own local ruler and clock to globally coordinatize the space-time.

But this what I'm confused about. I can easily believe that trying to start with a neighborhood of a synchronously rotating and revolving observer and extending that spacetime outward across the center of rotation might encounter problems, like the problems of going north past the north pole.

But it seems to me that we can know the answer anyway by simply referencing to the inertial frame in which the whole operation is embedded, if spinning doesn't change local rulers and clocks (that are the size of atoms, say). In other words, if we have two observers A and B in synchronous rotation/revolution, and we consider an inertial observer I, then I can just use their own ruler and clock, along with special relativity, to transform I --> A or I --> B. Then if we wanted to go A --> B, could we not just use
[I-->A]-1 followed by I --> B,
thereby establishing a working algorithm A --> B? So it might not follow any global coordinatization rules, but it would seem to function just like a global coordinatization nevertheless. It's not guaranteed to give a result in all cases, A's spacetime might not be one-to-one and onto B's, which may be what you mean by "non-kosher", but when it works it should be what you need, and when it doesn't work you can say "you can't get there from here". That's why I say "when there is an inertial frame available that can be meaningfully used as a reference, you will get me into one of the noninertial frames only kicking and screaming".

grav
2008-Mar-05, 09:14 PM
Okay, the time dilation and acceleration parts might cancel out, but that still leaves relativistic Doppler. I'm thinking the subtly here is that of the difference of thinking about the carousel as a rotating disk, with all of the points upon it rotating together, so always stationary to each other, and that of thinking about all of the points separately travelling at their own inertial speeds in straight lines at different angles to each other at any particular moment during the rotation. We can easily see that riders opposite to each other will have no regular Doppler effect between them, since they are travelling parallel to each other. But what about to other points on the carousel, like those travelling perpendicularly to a co-rotating observer? So I guess the first step would be to find out what regular Doppler says about that, and then deal with the relativistic part of it after, since we would then simply divide the result by gamma.

For this, I am treating the vectors separately to make things easier, finding first for the speed directly toward or away from a point on a rotating disk to a stationary observer on the ground, and then for the speed directly toward or away from a point on a stationary disk to an observer travelling tangent to it. I have diagrams below to demonstrate. After we find the speed in each case for directly toward or away from an observer to any point on the disk, we can find the total speed directly toward or away for a observer moving tangent to the disk while the disk itself is rotating, by using just the sum of relative speeds along the same line of sight, in order to find the Doppler shift.

In the first diagram, the center of the disk is the origin, and we have a point on the disk at (x,y) travelling at v1. A stationary observer is outside of the disk at (R,0). The distance between the point and the observer is d, and g is the distance from the point on the disk to the intersection of the line of motion of the point and a line from the observer that is perpendicular to d, intersecting at (x',y'). The reason for the line perpendicular to d is that we are trying to find for the ratio of the speed between the observer and the point, and that along the line of motion of the point. Therefore, d must be a side of a right triangle where the line of motion of the point would be the hypotenuse.

So we can see in the diagram that since the line of motion is perpendicular to the axis at (0,0), then we can find the distance from the origin to (x,y) and (x',y') easily enough, so have the relationship [(x-0)^2 + (y-0)^2] + g^2 = [(x'-0)^2 + (y'-0)^2], so g^2 = x'^2 + y'^2 - x^2 - y^2. Since d is a side of a right triangle where g is the hypotenuse, we also have g^2 = d^2 + [(R-x')^2 + (0-y')^2] = d^2 + (R-x')^2 + y'^2. Finally, we can also directly find the square of the distance along g to be g^2 = (x'-x)^2 + (y'-y)^2. That is all we need for this, so we combine them to get

d^2 + [(R-x')^2 + y'^2] = x'^2 + y'^2 - x^2 - y^2

d^2 + R^2 - 2Rx' + x'^2 + y'^2 = x'^2 + y'^2 - x^2 - y^2

d^2 + R^2 - 2Rx' = -x^2 - y^2

x' = (d^2 + R^2 + x^2 + y^2) / (2R)

where d^2 = (R-x)^2 + y^2, so

x' = (R^2 - Rx + x^2 + y^2) / R

Then x'^2 + y'^2 - x^2 - y^2 = (x'-x)^2 + (y'-y)^2

x'^2 + y'^2 -x^2 - y^2 = x'^2 - 2x'x + x^2 + y'^2 - 2y'y + y^2

x'x + y'y = x^2 + y^2

y' = (x^2 + y^2 - x'x) / y

y' = [x^2 + y^2 - (R^2 - Rx + x^2 + y^2) x / R] / y

Finally g^2 = (x'-x)^2 + (y'-y)^2

g^2 = [(R^2 - Rx + x^2 + y^2) / R - x]^2 + [(x^2 + y^2 -(R^2 - Rx + x^2 + y^2) x / R ) / y - y]^2

= [(R^2 - 2Rx + x^2 +y^2) / R]^2 + [(x^2 - (R^2 - Rx + x^2 +y^2) x / R) / y]^2

= [((R-x)^2 + y^2) / R]^2 + [[(R^2x - 2Rx^2 + x^3 +y^2x) / (Ry)]^2

(gR)^2 = [(R-x)^2 + y^2)]^2 +[((R-x)^2 + y^2) x / y]^2

where again, d^2 = (R-x)^2 + y^2, so

(gR)^2 = d^4 (1 + x^2 / y^2)

(gR)^2 = d^4 (x^2 +y^2) / y^2

(d/g)^2 = R^2 y^2 / (d^2 (x^2 + y^2))

d/g = R y / ( d sqrt(x^2 + y^2))

That is the result for the first diagram. The speed on the perimeter of the disk is vp, and v1 = vp sqrt(x^2 + y^2) / r for the speed of the point, so the velocity directly between the point and the observer, then, is v1 (d/g) = [vp sqrt(x^2 + y^2) / r] [R y / (d sqrt(x^2 + y^2))] = vp R y / (d r).

grav
2008-Mar-05, 09:45 PM
Sorry, I had to take a break after that last post. There was a lot more to it than I intended. Anyway, the second diagram is simpler. We just have an observer travelling tangent to a stationary disk. For the right triangle with side d and hypotenuse g, that gives us

d^2 + [(R-x)^2 + (y'-y)^2] = y'^2

d^2 + R^2 - 2Rx + x^2 - 2y'y + y^2 = 0

y' = (d^2 + R^2 - 2Rx + x^2 + y^2) / (2y)

and g = y' and d^2 = (R-x)^2 + y^2 = R^2 - 2Rx + x^2 + y^2, so

g = d^2 / y

d/g = y/d

That is the result for the second diagram. The velocity directly between the point and the observer, then, is v2y / d. So in conjuction with the first diagram, if the disk is rotating counter-clockwise and we have a negative speed for the observer along the y axis tangent to the disk, then we would add the results of the velocities between the point and observer. For a positive speed along the y axis for the observer, moving in the same direction as the rotation of the disk, as I have set it up, then we would subtract them. We find, rather surprisingly, to me at least, that there is absolutely no relative motion between the observer and any point on the disk, even when only considering their instantaneous inertial motions, when v2 (y/d) = vp (R/r) (y/d), or when v2 = vp (R/r). Now, for an observer moving tangent to the disk travelling at v2 = vp (R/r), that observer is travelling at the same speed that would be required to co-rotate with the disk at that distance. So for the riders on the carousel, there is absolutely no Doppler shift with any of the other riders anywhere on the ride, as if stationary with them even with considering their inertial vector velocities in different directions. Now, that is the regular Doppler part. There is still the relativistic part for relativistic Doppler, where we just divide by sqrt(1-B^2) (I think I said gamma before, which I believe is 1/sqrt(1-B^2) for that term, so it would be multiplied in that case), which uses only scalar relative speed, not vectors that depend upon direction. So all of this little endeavor still doesn't really answer the question, unfortunately. But it seems to me that the only way, that I can see so far anyway, to make everything work out kosher, is to figure that the riders are always to be considered stationary to each other in the first place.

publius
2008-Mar-06, 12:02 AM
But this what I'm confused about.
....

But it seems to me that we can know the answer anyway by simply referencing to the inertial frame in which the whole operation is embedded, if spinning doesn't change local rulers and clocks (that are the size of atoms, say). In other words, if we have two observers A and B in synchronous rotation/revolution, and we consider an inertial observer I, then I can just use their own ruler and clock, along with special relativity, to transform I --> A or I --> B. Then if we wanted to go A --> B, could we not just use
[I-->A]-1 followed by I --> B,
thereby establishing a working algorithm A --> B?

Ken, understand this is well beyond my understanding. That is, I know what I've read, but it hasn't "clicked" in my own mind. :)

But, yeah, I would think that we could easily do an A --> I --> B, from the I frame. We would be "transporting" local measurements from A to B, using the I frame's simultaneity. And we could do things like if A fired a pulse of light which hit B at some event, we'd could easily calculate what frequency B would measure.

So it might not follow any global coordinatization rules, but it would seem to function just like a global coordinatization nevertheless. It's not guaranteed to give a result in all cases, A's spacetime might not be one-to-one and onto B's, which may be what you mean by "non-kosher", but when it works it should be what you need, and when it doesn't work you can say "you can't get there from here". That's why I say "when there is an inertial frame available that can be meaningfully used as a reference, you will get me into one of the noninertial frames only kicking and screaming".

I can add much more, save that by "non-kosher", the problem is the rotating, revolving, observers spatial hyperslices "don't work" globally because of the vorticity problem. So, you don't even want to go in them kicking and screaming.

Again, this is well beyond me. :)

-Richard