Does a relativistic rolling ball wobble?

[JesseM]

The difficulty is that simultaneity of the points on the rim is continuously changing.

How is that a difficulty? If you know the movement of the point in the frame of the center, the Lorentz transformation will tell you the time of any event on that point's worldline in another frame. It's really just a problem of the equations being difficult or impossible to solve exactly. For example, if the point on the rim is described by x'(t') = cos(t') and y'(t') = sin(t') in the rest frame of the center, applying the Lorentz transformation would give the equations:

gamma*(x - vt) = cos(gamma*(t - vx/c^2))
y = sin(gamma*(t - vx/c^2)

The difficulty is just in solving the first equation for x (if you could do that, you could plug the answer into the second equation to get y(t)). There may not be an exact solution, I don't know. But if you're just interested in a visual depiction of the path, it's an easy enough matter to just pick a bunch of different t' coordinates, find the corresponding x' and y' coordinates for each one, then convert each (x', y', t') to an (x, y, t) using the Lorentz transform.

 

[1effect]

How is that a difficulty? If you know the movement of the point in the frame of the center, the Lorentz transformation will tell you the time of any event on that point's worldline in another frame. It's really just a problem of the equations being difficult or impossible to solve exactly. For example, if the point on the rim is described by x'(t') = cos(t') and y'(t') = sin(t') in the rest frame of the center, applying the Lorentz transformation would give the equations:

gamma*(x - vt) = cos(gamma*(t - vx/c^2))
y = sin(gamma*(t - vx/c^2)

The difficulty is just in solving the first equation for x (if you could do that, you could plug the answer into the second equation to get y(t)). There may not be an exact solution, I don't know. But if you're just interested in a visual depiction of the path, it's an easy enough matter to just pick a bunch of different t' coordinates, find the corresponding x' and y' coordinates for each one, then convert each (x', y', t') to an (x, y, t) using the Lorentz transform.

Actually, it is possible to get a parametric solution, avoiding the nasty transcendental equations. Here is the trick:

$$x'^2+y'^2=1$$

$$\gamma^2 (x-vt)^2+y^2=1$$

$$y=sin(t)$$
$$\gamma (x-vt)=cos(t)$$
so:
$$x=\gamma^{-1} cos(t)+vt$$

 

[JesseM]

Actually, it is possible to get a parametric solution, avoiding the nasty transcendental equations. Here is the trick:

$$x'^2+y'^2=1$$

$$\gamma^2 (x-vt)^2+y^2=1$$

$$y=sin(t)$$

How did you get y=sin(t)? Wasn't the equation in the primed frame y'=sin(t'), which means it should be y=sin(gamma*(t - vx/c^2))

 

[1effect]

How did you get y=sin(t)? Wasn't the equation in the primed frame y'=sin(t'), which means it should be y=sin(gamma*(t - vx/c^2))

I parametrized :

$$\gamma^2 (x-vt)^2+y^2=1$$

There is an indefinite number of such parametrizations, they all produce the same curve.

For example, I could have chosen $$\frac{1-t^2}{1+t^2}$$ and $$\frac{2t}{1+t^2}$$

 

[JesseM]

I re-parametrized

$$\gamma^2 (x-vt)^2+y^2=1$$

I don't understand what you mean by "re-parametrized" here. Can you fill in the missing steps that allow you to go from that equation to y=sin(t)? Unless you mean that y=sin(t) is meant to be the definition of a new parameter t, but then you should have given the parameter a different name than the time coordinate that appears in (x-vt).

 

[JesseM]

I parametrized :

$$\gamma^2 (x-vt)^2+y^2=1$$

There is an indefinite number of such parametrizations, they all produce the same curve.

For example, I could have chosen $$\frac{1-t^2}{1+t^2}$$ and $$\frac{2t}{1+t^2}$$

But if you're just parametrizing y with a newly-defined parameter, you can't use the t-coordinate as the parameter! The t-coordinate already has a specific meaning--if you have some y' and t' which lie on the worldline of the point in the primed frame, you can't substitute in y=sin(t) unless it's true that when you do the Lorentz transform on that y' and t' on the point's worldline, the resulting y and t actually satisfy y=sin(t). If you want to define a new parameter you should give it a different name, like y=sin(p).

 

[1effect]

I don't understand what you mean by "re-parametrized" here. Can you fill in the missing steps that allow you to go from that equation to y=sin(t)? Unless you mean that y=sin(t) is meant to be the definition of a new parameter t, but then you should have given the parameter a different name than the time coordinate that appears in (x-vt).

I am not sure about that , I will need to think about it.
This would result into :

$$y=sin(\tau)$$

$$x=\gamma^{-1} cos(\tau)+vt$$

 

[JesseM]

I am not sure about that , I will need to think about it.
This would result into :

$$y=sin(\tau)$$

$$x=\gamma^{-1} cos(\tau)+vt$$

But that isn't very helpful, because you don't know what value of t corresponds to a given value of $$\tau$$.

 

[1effect]

But that isn't very helpful, because you don't know what value of t corresponds to a given value of $$\tau$$.

Yes, this wouldn't work. Too bad, I guess we are stuck with the unsolvable system of transcendental equations. :-(

 

[Antenna Guy]

Actually, it is possible to get a parametric solution, avoiding the nasty transcendental equations.

When you guys are done "simplifying" the problem, I'd like to see what you get when your observer is not at infinity.

Consider that the sphere traveling along x at relativistic speed with respect to an inertial observer will be described by:

$$\frac{\gamma^2 x'^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1$$

[note: sorry about leaving out the velocity term - I'd have probably screwed that up anyway ]

Where r is the radius of the sphere in its' rest frame.

If the inertial observer is not at infinity, the "shape" of the object will not be a projection on a plane - and that "shape" will vary with distance/aspect angle.

If you get that far, you can let the surface rotate about its' volume...

Regards,

Bill

 

[JesseM]

When you guys are done "simplifying" the problem, I'd like to see what you get when your observer is not at infinity.

At infinity? I wasn't trying to calculate what would be seen visually, just the coordinate path of a point on the rim in a frame where the rolling object was moving.

Consider that the sphere traveling along x at relativistic speed with respect to an inertial observer will be described by:

$$\frac{\gamma^2 x'^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1$$

I guess this would be the equation for the coordinates of the surface in the frame where the sphere is moving along the x' axis, at the moment the sphere was centered at the origin. But this is different from the shape the sphere would appear to be as seen by an observer at rest in this frame.

If the inertial observer is not at infinity, the "shape" of the object will not be a projection on a plane - and that "shape" will vary with distance/aspect angle.

What do you mean by "projection on a plane"?

 

[Phrak]

Did I miss it, or hasn't it been pointed out? It's not at all unambiguous as to what meant by a 'ball'?

First, is it a problem in dynamics, or simply kinematics? If only kinematics, the spinning ball has a circumference to diameter ratio of gamma*pi in the plane of angular momentum. So in this case, it's fair to ask 'was it a ball before spinning? (If it was a ball before spinning then its a ball of constant radius, otherwise constant circumference.)

 

[Antenna Guy]

Did I miss it, or hasn't it been pointed out? It's not at all unambiguous as to what meant by a 'ball'?

Are you aware that this relativistic 'ball' does not necessarily have an instantaneous circumference (per se) in the frame of the observer?

First, is it a problem in dynamics, or simply kinematics? If only kinematics, the spinning ball has a circumference to diameter ratio of gamma*pi in the plane of angular momentum. So in this case, it's fair to ask 'was it a ball before spinning? (If it was a ball before spinning then its a ball of constant radius, otherwise constant circumference.)

Let's say that the observed circumference of this ball spans a range of time that approaches $\frac{2r}{c}.$. What would you integrate over to verify that the rest length of the observed circumference is still $2\pi r$?

Regards,

Bill

 

[1effect]

When you guys are done "simplifying" the problem, I'd like to see what you get when your observer is not at infinity.

Consider that the sphere traveling along x at relativistic speed with respect to an inertial observer will be described by:

$$\frac{\gamma^2 x'^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1$$

[note: sorry about leaving out the velocity term - I'd have probably screwed that up anyway ]

The above is not the equation of the moving sphere. You need to apply to Lorentz transform correctly.

 

[JesseM]

The above is not the equation of the moving sphere. You need to apply to Lorentz transform correctly.

I think it is the equation for the sphere's surface at a given instant when it is centered at the origin and moving on the x'-axis. The equation describes an ellipsoid with radius r along the y' and z' axis, but Lorentz-contracted radius $$r/\gamma$$ along the x'-axis...is this not correct?

 

[1effect]

I think it is the equation for the sphere's surface at a given instant when it is centered at the origin and moving on the x'-axis. The equation describes an ellipsoid with radius r along the y' and z' axis, but Lorentz-contracted radius $$r/\gamma$$ along the x'-axis...is this not correct?

"moving" was key in my answer.
His equation is the snapshot for t=0, so we agree.

The correct equation is:

$$\frac{\gamma^2 (x'-vt')^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1$$

 

[JesseM]

Ah, OK. But one point that occurred to me is that if we're just looking at the total set of points occupied by the surface at any given instant, rather than trying to track an individual point on its surface, then in the rest frame of the sphere's center the coordinates of the surface won't be changing with time, so it will always occupy exactly the same coordinates as a certain nonrotating sphere centered at the same position. And if the surface of the nonrotating sphere occupies the same set of coordinates in this frame, its surface must occupy the same set of coordinates as the rotating sphere in every frame. So, this is one way of seeing that the rolling ball won't "wobble". The difficulty is just in figuring out how a given point on the rotating sphere will move around this ellipsoid surface in a frame where it's in motion.

 

[1effect]

Ah, OK. But one point that occurred to me is that if we're just looking at the total set of points occupied by the surface at any given instant, rather than trying to track an individual point on its surface, then in the rest frame of the sphere's center the coordinates of the surface won't be changing with time, so it will always occupy exactly the same coordinates as a certain nonrotating sphere centered at the same position. And if the surface of the nonrotating sphere occupies the same set of coordinates in this frame, its surface must occupy the same set of coordinates as the rotating sphere in every frame. So, this is one way of seeing that the rolling ball won't "wobble". The difficulty is just in figuring out how a given point on the rotating sphere will move around this ellipsoid surface in a frame where it's in motion.

I think that we can do the "tracking" by converting the problem to cylindrical coordinates and by describing the angle $$\phi=\phi(t)$$ as a function of time.

 

[1effect]

I am aware that that there is no length contraction perpendicular to the linear motion of the rolling sphere. I can guarantee that if the centre of mass is in the centre of the ball in its rest frame then it is not in the geometrical centre of the ball when it is rolling with respect to the observer. The centre of mass is further away from the point of rolling contact than the geometrical midpoint between the top of the rolling ball and the point of contact.

I think you need to have a look at http://www.spacetimetravel.org/filme/radv_t_0.93/radv_t_0.93-xe-320x240.mpg , there is no wobble. You can see more of the same.

 

[Phrak]

Are you aware that this relativistic 'ball' does not necessarily have an instantaneous circumference (per se) in the frame of the observer?

I'm afraid I haven't been clear. The ball rolls without slipping. Therefore it spins was well as translates.

In the inertial frame of the ball's center of mass the circumference is contracted from it's length at rest. The various radii are not contracted, because their extent is perpendicular to the instantaneous velocity develped by rotation.

Knowing this, the ball cannot be considered a rigid object. In fact relativity and rigid objects are contradictory in principle. This is why most answers given to pushing a long, relativistic stick down a short hole are flawed.

Until anyone trying to answer the original question addresses the shape the ball in a given, non-translating inerital frame, and what parts of the ball remain invariant under rotation, they're blowin' smoke.

 

[yuiop]

I think you need to have a look at http://www.spacetimetravel.org/filme/radv_t_0.93/radv_t_0.93-xe-320x240.mpg , there is no wobble. You can see more of the same.

First of all, when discussing relativity we normally discount purely visual effects to get objective measurements. The links you provided have animations that include the visual effect of Terrell rotatation that obscure the physics as far a relativity is concerned unless of course we were specifically investigating Terrell rotation.

Having said that, I have taken the liberty of modifying an image from one of the pages you linked to, to illustrate the point I was making in the post you responded to. The white blobs on the rims of the attached image are equal masses placed at the end of each spoke. (Assume the spokes, hub and perimeter have insignificant mass compared to the white test masses). The red wheel is not rotating and joining opposite pairs of masses shows the centre of mass is at the physical axle of the wheel. Joining opposite pairs of test masses on the green wheel, which is moving at relativistic speed to the observer, shows that the centre of mass (the light blue dot) is now much higher than the physical axle of the wheel. I think you will agree that this vertical shift of the centre of mass is not significantly affected by the Terrell rotation that causes an optical horizontal shift of parts of the object.

...
I am aware that that there is no length contraction perpendicular to the linear motion of the rolling sphere. I can guarantee that if the centre of mass is in the centre of the ball in its rest frame then it is not in the geometrical centre of the ball when it is rolling with respect to the observer. The centre of mass is further away from the point of rolling contact than the geometrical midpoint between the top of the rolling ball and the point of contact.

If you look at the bold part of my statement above my guarantee was that the centre of mass is not in the geometrical centre of the ball or wheel. I did not guarantee that the centre of mass wobbles. However, I will try and show later that is some wobble of the centre of mass, using geometrical software. I suspect this wobble is corrected by a counter rotating torque, generated by torque reactions that are result of force not always being parallel to the acceleration, in a transformed reference frame.

 

[Dale]

OK, I am not working today so I decided to start this. Here is an equation for the worldline of a particle on a rotating hoop in the reference frame where the axis of rotation is stationary.
$$s=(t,\cos (\phi +t \omega ),\sin (\phi +t \omega ),0)$$
in units where c=1 and r=1
Which can be differentiated to obtain the four-velocity
$$u=\left(\frac{1}{\sqrt{1-\omega ^2}},-\frac{\omega \sin (\phi +t \omega )}{\sqrt{1-\omega ^2}},\frac{\omega \cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},0\right)$$
Which can be differentiated to obtain the four-acceleration
$$a=\left(0,-\frac{\omega ^2 \cos (\phi +t \omega )}{1-\omega ^2},-\frac{\omega ^2 \sin (\phi +t \omega )}{1-\omega ^2},0\right)$$
A quick check shows u.u=1 and u.a=0

I then transformed the hoop into the frame where it rolls without friction
$$s'=L.s=\left(\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},\frac{t \omega +\cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},\sin (\phi +t \omega ),0\right)$$
and then differentiated to obtain u' and a'. I verified that u'=L.u and a'=L.a and that u'.u'=1 and u'.a'=0. So, kinematically everything works out just fine. I didn't actually work out the four-forces nor the tension on a differential element.

So then, to answer the question about the shape of the hoop in the primed frame I need to look at all of the points for a single value of t'.
$$t'=\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega ^2}}$$
I could not solve this equation for t as described above, but I could solve it for phi to obtain
$$\phi =-t \omega \pm \cos ^{-1}\left(\frac{t' \sqrt{-(\omega -1) (\omega +1)}-t}{\omega }\right)$$

Substituting this back into the expression for s' gives an expression for the shape of the hoop at a given t' in the frame where it is rolling

$$\left(t',\frac{t'-t \sqrt{1-\omega ^2}}{\omega },\pm \sqrt{1-\frac{\left(t-t' \sqrt{1-\omega ^2}\right)^2}{\omega ^2}},0\right)$$

I plotted this for a couple of different values of t' and it appears to be a "grape" that does not wobble.

 

[yuiop]

I'm afraid I haven't been clear. The ball rolls without slipping. Therefore it spins was well as translates.

In the inertial frame of the ball's center of mass the circumference is contracted from it's length at rest. The various radii are not contracted, because their extent is perpendicular to the instantaneous velocity develped by rotation...

The simple assumption that the spokes of rotating wheel without linear translation are not length contracted is in fact an over simplification because the principle of equivalence tells us that the acceleration due to rotation is similar to being in a gravitational field and there will be length contraction of all the spokes, with the greatest radial length contraction of the spokes occurring near the rim because that is where the acceleration is greatest. That is drifting into GR teritory and we probably don't need to go to that extent at this point.

We should be clear that the radii parallel to the linear translation of the rotating ball will be length contracted due to the linear motion superimposed on the rotation.

 

[1effect]

OK, I am not working today so I decided to start this. Here is an equation for the worldline of a particle on a rotating hoop in the reference frame where the axis of rotation is stationary.
$$s=(t,\cos (\phi +t \omega ),\sin (\phi +t \omega ),0)$$
in units where c=1 and r=1
Which can be differentiated to obtain the four-velocity
$$u=\left(\frac{1}{\sqrt{1-\omega ^2}},-\frac{\omega \sin (\phi +t \omega )}{\sqrt{1-\omega ^2}},\frac{\omega \cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},0\right)$$
Which can be differentiated to obtain the four-acceleration
$$a=\left(0,-\frac{\omega ^2 \cos (\phi +t \omega )}{1-\omega ^2},-\frac{\omega ^2 \sin (\phi +t \omega )}{1-\omega ^2},0\right)$$
A quick check shows u.u=1 and u.a=0

I then transformed the hoop into the frame where it rolls without friction
$$s'=L.s=\left(\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},\frac{t \omega +\cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},\sin (\phi +t \omega ),0\right)$$
and then differentiated to obtain u' and a'. I verified that u'=L.u and a'=L.a and that u'.u'=1 and u'.a'=0. So, kinematically everything works out just fine. I didn't actually work out the four-forces nor the tension on a differential element.

So then, to answer the question about the shape of the hoop in the primed frame I need to look at all of the points for a single value of t'.
$$t'=\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega ^2}}$$
I could not solve this equation for t as described above, but I could solve it for phi to obtain
$$\phi =-t \omega \pm \cos ^{-1}\left(\frac{t' \sqrt{-(\omega -1) (\omega +1)}-t}{\omega }\right)$$

Substituting this back into the expression for s' gives an expression for the shape of the hoop at a given t' in the frame where it is rolling

$$\left(t',\frac{t'-t \sqrt{1-\omega ^2}}{\omega },\pm \sqrt{1-\frac{\left(t-t' \sqrt{1-\omega ^2}\right)^2}{\omega ^2}},0\right)$$

I plotted this for a couple of different values of t' and it appears to be a "grape" that does not wobble.

Very nice, thank you !

 

[yuiop]

..

I plotted this for a couple of different values of t' and it appears to be a "grape" that does not wobble.

The shape of the Lorentz transformed hoop (rotating or not) is simply and formally a geometrical ellipse. (I can show you that the theory of relativity is falsified if that is not the case). I think we are agreed that the transformed outline of the hoop or ball does not change with time. The question, I think is more about the location of the centre of mass. If it does not remain constant over time, then you would expect the hoop or ball to "bounce" or wobble with an oscillating centre of mass rather than move in a straight line. The simplest way to analyse this, is to take just two test "point masses "on opposite sides of the hoop and find their combined centre of mass at any instant in the transformed frame.

 

[1effect]

First of all, when discussing relativity we normally discount purely visual effects to get objective measurements. The links you provided have animations that include the visual effect of Terrell rotatation that obscure the physics as far a relativity is concerned unless of course we were specifically investigating Terrell rotation.

This is not correct. The animations provided use the exact equations (very similar to the ones just derived by DaleSpam). They use rays of light with simultaneous arrival time in order to answer the question "how does the perimeter of the sphere (or spoke wheel, arbitrary object, etc.) look in the observer 's frame at the same time t'?" .

So, contrary to what you think, in this particular application the Terrell effect is not just an illusion, it is the mechanism of determining the dimensions of a moving object by marking its extremities simultaneously. It so happens that the raytracing programs are using rays of light for this purpose.

 

[1effect]

The shape of the Lorentz transformed hoop (rotating or not) is simply and formally a geometrical ellipse. (I can show you that the theory of relativity is falsified if that is not the case). I think we are agreed that the transformed outline of the hoop or ball does not change with time. The question, I think is more about the location of the centre of mass. If it does not remain constant over time, then you would expect the hoop or ball to "bounce" or wobble with an oscillating centre of mass rather than move in a straight line. The simplest way to analyse this, is to take just two test "point masses "on opposite sides of the hoop and find their combined centre of mass at any instant in the transformed frame.

Yes, you seem to have proven that (at least your images seem to indicate that to be the case since the spokes appear to be bent upwards). Since the spokes appear permanently bent upwards you will also note that the distance between the center of gravity and the rail doesn't change, so, there is no reason for wobbling.
In addition to the above, the fact is that the center of rotation (the wheel axle) remains at constant distance from the supporting rail. So, the wheel will not wobble. This is supported by all the linked animations where the wheel remains in permanent contact with the rail. So, again, no reason for any wobble.

 

[JesseM]

This is not correct. The animations provided (please note that they are done by professional phycists) use the exact equations (very similar to the ones just derived by DaleSpam). They use rays of light that arrive simultaneously in order to answer the question "how does the perimeter of the sphere (spoke wheel, arbitrary object, etc.) look in the observer frame at the same time t'?" .

So, contrary to what you think, the Terrell effect is not just an illusion, it is the mechanism of determining the dimensions of a moving object by marking its extremities simultaneously. It so happens that the raytracing programs are using rays of light for this purpose.

I'm not sure what you mean by "determining the dimensions of a moving object"--the Terrell effect will distort the length of objects from their "correct" length in the observer's coordinate system, making the visual length longer than the Lorentz-contracted length the object actually has in the observer's coordinate system (see this thread). And the animations you show are about visual appearance, not about the spatial coordinates occupied by different points on the object at a single coordinate time.

 

[1effect]

I'm not sure what you mean by "determining the dimensions of a moving object"--the Terrell effect will distort the length of objects from their "correct" length in the observer's coordinate system, making the visual length longer than the Lorentz-contracted length the object actually has in the observer's coordinate system (see this thread). .

What I mean is that this is raytracing , using only the rays of light that arrive simultaneously from the object endpoints. Thus, these rays act as the standard way of marking the object endpoints. The Terrell effect is just a byproduct of the method used for rendering (collecting simultaneously arriving rays). It is intereresting to read the notes and the paper included in the website. In fact, the authors are giving a superset of the Terrell paper.

And the animations you show are about visual appearance, not about the spatial coordinates occupied by different points on the object at a single coordinate time

I think that in these particular cases, the two coincide, the rays of light act as ways of marking the points on the object's periphery simultaneously, exacly like in DaleSpam's equations posted here.

 

[yuiop]

This is not correct. The animations provided (please note that they are done by professional phycists) use the exact equations (very similar to the ones just derived by DaleSpam). They use rays of light that arrive simultaneously in order to answer the question "how does the perimeter of the sphere (spoke wheel, arbitrary object, etc.) look in the observer frame at the same time t'?" .

So, contrary to what you think, the Terrell effect is not just an illusion, it is the mechanism of determining the dimensions of a moving object by marking its extremities simultaneously. It so happens that the raytracing programs are using rays of light for this purpose.

The ray tracing program uses rays of light to show what one one observer would visually see. A reference frame has an infinty of observers and there would be one observer at the back of the wheel/ball and one at the front of the ball to elliminate light travel times and take meaurements with previously syncronised clocks and compare notes later to get the measurements of the moving ball or wheel.

They might well be using exact equations, but the image is the point of view of an observer to one side and slightly above the table with a bit of distance perspective thrown in which is not good for us to make exact measurements from. The vertical displacement of the centre of mass is so great it is obvious despite the optical embelishments.

A rod with constant velocity relative to an observer appears to be longer when it is approaching a single observer and shorter as it goes away from the observer which is what a ray tracing program would show, but that obscures the fact that the length contraction is constant whether the rod is approaching or receding. Ray tracing is not a good method for objectively analysing Lorentz transformations.

[EDIT] Anyway, is anyone is interested here is a sketch of a beach ball progressively rolling faster with respect to the observer.

 

[JesseM]

What I mean is that this is raytracing , using only the rays of light that arrive simultaneously from the object endpoints. Thus, these rays act as the standard way of marking the object endpoints.

Yes, that's what the Terrell effect is about too. But just because they arrive simultaneously doesn't mean they were emitted simultaneously. Again, see the thread I linked to--when you calculate the visual size of an object based on rays from different ends that reach the observer's eyes simultaneously, she will actually see the object as being no different than its visual size at rest when at the same distance, despite the fact that the object is "really" Lorentz-contracted in her frame.

 

[1effect]

Yes, that's what the Terrell effect is about too. But just because they arrive simultaneously doesn't mean they were emitted simultaneously.

True. You don't want them emitted simultaneously. You want them simultaneous in the observer's frame (you want to mark the object's ends simultaneously in the observer's frame, the same way DaleSpam did it) . This is exactly what the authors do.

 

[1effect]

The ray tracing program uses rays of light to show what one one observer would visually see. A reference frame has an infinty of observers and there would be one observer at the back of the wheel/ball and one at the front of the ball to elliminate light travel times and take meaurements with previously syncronised clocks and compare notes later to get the measurements of the moving ball or wheel.

They might well be using exact equations, but the image is the point of view of an observer to one side and slightly above the table with a bit of distance perspective thrown in which is not good for us to make exact measurements from. The vertical displacement of the centre of mass is so great it is obvious despite the optical embelishments.

True. You keep missing the point that both the center of mass and the center of rotation do not change their distance to the supporting rail, therefore there is no reason for any wobbling. This is what the OP is all about :-)

 

[yuiop]

Yes, you seem to have proven that (at least your images seem to indicate that to be the case since the spokes appear to be bent upwards). Since the spokes appear permanently bent upwards you will also note that the distance between the center of gravity and the rail doesn't change, so, there is no reason for wobbling.
In addition to the above, the fact is that the center of rotation (the wheel axle) remains at constant distance from the supporting rail. So, the wheel will not wobble. This is supported by all the linked animations where the wheel remains in permanent contact with the rail. So, again, no reason for any wobble.

I am not going to specifically address the issue of wobble until someone can give an exact equation for the height of the centre of mass of a rolling wheel (without slipping) with a linear velocity of say 0.8c and two opposite point masses of 1M each at any point point during the rotation. I am not going to base the argument on a ray tracing program designed to entertain the masses, rather than come to any analytical conclusions.

 

[Dale]

The shape of the Lorentz transformed hoop (rotating or not) is simply and formally a geometrical ellipse.

Visually it certainly looked like an ellipse, but I didn't go through the effort to prove it so I thought it best to just call it a "grape".

The question, I think is more about the location of the centre of mass. If it does not remain constant over time, then you would expect the hoop or ball to "bounce" or wobble with an oscillating centre of mass rather than move in a straight line. The simplest way to analyse this, is to take just two test "point masses "on opposite sides of the hoop and find their combined centre of mass at any instant in the transformed frame.

I didn't cover that in my analysis. I don't think you can simply find the center of mass of two opposing points, I think you will need to integrate around the whole hoop.

I am not going to specifically address the issue of wobble until someone can give an exact equation for the height of the centre of mass of a rolling wheel (without slipping) with a linear velocity of say 0.8c and two opposite point masses of 1M each at any point point during the rotation.

I would know how to do this easily if I could have solved for t instead of phi, but as it is I am not sure how to proceed. I would have to think about this a bit.