# Does a relativistic rolling ball 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.
The argument is not based on raytracing, it is based on the fact that in the frame comoving with the sphere center there is no wobble. In this particular frame we know that the center of mass coincides with the center of rotation and the sphere (or wheel) doesn't wobble.
So, in the observer frame, the sphere (or wheel) cannot wobble either. You can't have physically different results in the two frames.
The raytracing program simply gives a visual confirmation for the above.

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The argument is not based on raytracing, it is based on the fact that in the frame comoving with the sphere center there is no wobble. In this particular frame we know that the center of mass coincides with the center of rotation and the sphere (or wheel) doesn't wobble.
So, in the observer frame, the sphere (or wheel) cannot wobble either. You can't have physically different results in the two frames.
The raytracing program simply gives a visual confirmation for the above.

Your argument is what bwr6 called a principled argument. We know relativity predicts that the rolling ball does not wobble in the comoving frame, but that tells us little about the physics that is going on and you are not necessarily correct to jump to the conclusion that the centre of mass stays in one place as the ball rolls with constant velocity.

It is possible that centripetal forces of the masses on the rim cause accelerations that are not parallel to the the force vectors and do not act through the instaneous centre of mass. In fact I believe the centre of mass has to move to compensate for the anisotropic centripetal forces, to prevent the "wobble".

I have attached a screendump from geometrical software that I used to simulate the motion of the particles under transformation. A line joining the red and blue particle shows the instantaneous centre of gravity and the two parallel black lines in the lower part of the diagram show the heights of the centre of mass at different points during the rotation of the ball. I have taken reasonable care to ensure an accurate simulation but I can not deny that the time dependent vertical harmonic motion of the centre of mass of the horizontally moving ball, may be an artifact of the resolution and accuracy of the geometrical software.

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Your argument is what bwr6 called a principled argument. We know relativity predicts that the rolling ball does not wobble in the comoving frame, but that tells us little about the physics that is going on and you are not necessarily correct to jump to the conclusion that the centre of mass stays in one place as the ball rolls with constant velocity.

It is possible that centripetal forces of the masses on the rim cause accelerations that are not parallel to the the force vectors and do not act through the instaneous centre of mass. In fact I believe the centre of mass has to move to compensate for the anisotropic centripetal forces, to prevent the "wobble".

I have attached a screendump from geometrical software that I used to simulate the motion of the particles under transformation. A line joining the red and blue particle shows the instantaneous centre of gravity and the two parallel black lines in the lower part of the diagram show the heights of the centre of mass at different points during the rotation of the ball. I have taken reasonable care to ensure an accurate simulation but I can not deny that the time dependent vertical harmonic motion of the centre of mass of the horizontally moving ball, may be an artifact of the resolution and accuracy of the geometrical software.

The results of your simulation depend on the equations you used. I would be very interested in seeing them.

It is possible that centripetal forces of the masses on the rim cause accelerations that are not parallel to the the force vectors and do not act through the instaneous centre of mass. In fact I believe the centre of mass has to move to compensate for the anisotropic centripetal forces, to prevent the "wobble".
Since the point-trails follow elliptical paths, they seem to be "rotating" about two centers of mass displaced in time. By "rotating", I mean a string of length $\frac{2r}{\gamma}$ tied off to two foci (centers of mass) displaced in time by $2r\sqrt{1-\frac{1}{\gamma^2 }}$ seconds should transcribe an ellipse (when held taut while rotating about the midpoint of the two foci) half the size of your point trails (analogous to the 2r radius hoops in the non-relativistic case).

For the time displacement, I rotated what I calculated for the focal point separation of the Lorentz-contracted ellipsoid 90 deg.

Regards,

Bill

P.S. I'm having a heck of a time with LaTeX today...

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Dale
Mentor
Well, I couldn't figure out how to determine the center of mass, but I did figure out how to color my Mathematica plots according to phi. So, if you look at the attached plot this is
$$\left(t',\frac{t'-t \sqrt{1-\omega ^2}}{\omega },\pm \sqrt{1-\frac{\left(t-t' \sqrt{1-\omega ^2}\right)^2}{\omega ^2}},0\right)$$
plotted for w=0.9c with t'=0 on the left and t'=1 on the right. In the coloring, each repetition of a given color represents a 30º chunk of hoop in the frame where the axis of rotation is stationary. So, for example, look at the left picture where the amount of hoop between the two blue spots next to the positive y axis is equal in mass to the amount of hoop between the two blue spots closest to the negative y axis. This is despite the fact that the distance is much greater on the bottom in this frame. In other words, the hoop is more "dense" on the top than the bottom.

This picture doesn't give you an analytical equation for the center of mass, but visually you can see that the center of mass will be somewhere on the positive y axis for both hoops. I cannot guarantee or prove it, but it looks to me like the center of mass doesn't change it's y-coordinate, although the y-coordinate of the center of mass is not 0.

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Easy. The particles in the edge of ball are rotating. At angular speed C*R (Hoping R<1 because if not Einstein gets mad). Then the contraction it's in the direction of the velocity (whicks keeps changin beacuse of the rotation). Then there's another movement seen form the center of the mass also at speed C. Using analytical mechanics for rigid bodies with the realtivistic vectorial sum of speeds theorem you'll get the right awnser. (So ofrcourse it wont roll as a grape, thinking that is just a consequence of treating the rigid body as a point particle)..
Hope it hepls. :)

I think there's a video around http://www.tubepolis.com [Broken] of a numerical simulation of this. Look for rolling relativistic ball. :)

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Well, I couldn't figure out how to determine the center of mass, but I did figure out how to color my Mathematica plots according to phi. So, if you look at the attached plot this is
$$\left(t',\frac{t'-t \sqrt{1-\omega ^2}}{\omega },\pm \sqrt{1-\frac{\left(t-t' \sqrt{1-\omega ^2}\right)^2}{\omega ^2}},0\right)$$
plotted for w=0.9c with t'=0 on the left and t'=1 on the right. In the coloring, each repetition of a given color represents a 30º chunk of hoop in the frame where the axis of rotation is stationary. So, for example, look at the left picture where the amount of hoop between the two blue spots next to the positive y axis is equal in mass to the amount of hoop between the two blue spots closest to the negative y axis. This is despite the fact that the distance is much greater on the bottom in this frame. In other words, the hoop is more "dense" on the top than the bottom.

This picture doesn't give you an analytical equation for the center of mass, but visually you can see that the center of mass will be somewhere on the positive y axis for both hoops. I cannot guarantee or prove it, but it looks to me like the center of mass doesn't change it's y-coordinate, although the y-coordinate of the center of mass is not 0.

It is not very difficult to show that , indeed, the y coordinate of the center of mass is constant. So, the hoop doesn't wobble.

It is not very difficult to show that , indeed, the y coordinate of the center of mass is constant. So, the hoop doesn't wobble.
Since the result you claim is important to the the whole point of this thread please show this explicitly.

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.
It is not clear to me how Dalespam was able to plot the last equation that contains both t' and t when he has stated it is not possible to solve for t in terms of t'.

Let's take the equation he derived that equation from and insert some numerical values.
Say time = 0 in both frames and and R=1 and w=0.8c in the rest frame.
We take the the positions of 4 elements a,b,c and d at the 4 points of the compass where phi = 0, 90, 180 and 270 degrees. Now insert these values into the equation for the rest frame (s):

$$s=(t,\cos (\phi +t \omega ),\sin (\phi +t \omega ),0)$$

and we get (x,y) coordinates a=(1,0), b=(0,1), c=(-1,0) and d=(0-1) in frame s.

Now if we insert the same values into the transformed equation:

$$s \single-quote=L.s=\left(\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},\frac{t \omega +\cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},\sin (\phi +t \omega ),0\right)$$

we get (x',y') coordinates a'=(1.67,0), b'=(0,1), c'=(-1.67,0) and d'=(0,-1) in frame s'.

It can be seen that the transformed ball is wider than it is high by this equation which is a false result. This is due to the "catch 22" situation of not being able to calculate phi without knowing how the proper time of each element transforms and it is not possible to calculate how the the proper time of each element transforms without knowing how phi transforms. On the face of it, the equation is not taking the simultanity issues into account between the transformed frames.

This can be seen when we do the calculations for the transformed time and space coordinates (t',x',y') to get a'=(1.33,1.67,0), b'=(0,0,1), c'=(-1.33,-1.67,0) and d'=(0,0,-1) in frame s'. The time t' is not simultaneous in the s' frame. I do not think Dalespams equation are wrong, but it does not seem that they resolved the issues of how the dimensions of the ball would be measured simultaneously in the transformed frame without doing numerical aproximations.

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The results of your simulation depend on the equations you used. I would be very interested in seeing them.
Imagine we have a rotating wheel with the axis of rotation stationary if frame S with a pen attached to the rim of the wheel and a moving paper chart behind the wheel on which the pen draws a graph. The path of the the pen in this frame is a circle and the graph is a cycloid. The equation for the circle and cycloid (in parametric form) are well known. In frame S' the observer is co-moving with the paper chart.The paper chart was length contracted in frame S because it was moving in that frame, so in frame S' the paper (and graph) is "stretched" in the direction parallel to the relative motions of the frames, so the equation for the transformed cycloid is the conventional parametric equation with x*gamma substituted for x. The transformed circle path is simply an ellipse with semi-minor axis (b) and semi-major axis (a) with the relationship b=a/gamma. The exact location of the pen at any time t' in the transformed frame is the intersection of the elliptical path and stretched cyloid graph. Finding the intersection mathematically is very difficult (although it might be possible using the Lambert W function which I think is implemented in Mathematica). Fortunately the free geometrical software I use, called "CaR" for Compass and Ruler can compute the intersection of two locii but I am pretty sure it uses numerical techniques to achieve this. Most commercial geometrical software is unable to find the intersection of two locii.

Imagine we have a rotating wheel with the axis of rotation stationary if frame S with a pen attached to the rim of the wheel and a moving paper chart behind the wheel on which the pen draws a graph. The path of the the pen in this frame is a circle and the graph is a cycloid. The equation for the circle and cycloid (in parametric form) are well known. In frame S' the observer is co-moving with the paper chart.The paper chart was length contracted in frame S because it was moving in that frame, so in frame S' the paper (and graph) is "stretched" in the direction parallel to the relative motions of the frames, so the equation for the transformed cycloid is the conventional parametric equation with x*gamma substituted for x. The transformed circle path is simply an ellipse with semi-minor axis (b) and semi-major axis (a) with the relationship b=a/gamma. The exact location of the pen at any time t' in the transformed frame is the intersection of the elliptical path and stretched cyloid graph. Finding the intersection mathematically is very difficult (although it might be possible using the Lambert W function which I think is implemented in Mathematica). Fortunately the free geometrical software I use, called "CaR" for Compass and Ruler can compute the intersection of two locii but I am pretty sure it uses numerical techniques to achieve this. Most commercial geometrical software is unable to find the intersection of two locii.
So, do you have any equations to drive your drawings or not ?

The transformed circle path is simply an ellipse with semi-minor axis (b) and semi-major axis (a) with the relationship b=a/gamma. The exact location of the pen at any time t' in the transformed frame is the intersection of the elliptical path and stretched cyloid graph. Finding the intersection mathematically is very difficult
Isn't your "stretched cycloid" an elliptical path twice the size of the transformed circle (albeit rotated 90deg, and translated such that its' center lies on the x-axis)?

Regards,

Bill

In the coloring, each repetition of a given color represents a 30º chunk of hoop in the frame where the axis of rotation is stationary.
Unless I'm not reading your picture correctly, the "chunks of hoop" appear to be rotating in the wrong direction.

Regards,

Bill

Dale
Mentor
Hmm, I should have numbered my equations to make talking about them easier. There are 7 equations in total, so eq1 is the first equation and eq7 is the last.
It is not clear to me how Dalespam was able to plot the last equation that contains both t' and t when he has stated it is not possible to solve for t in terms of t'.
It was not possible to solve eq5 for t. But I could determine the range of t directly from eq7, I just left it out for brevity, but here it is explicitly:

The y-coordinate in eq7 is imaginary unless
$$1-\frac{\left(t-t' \sqrt{1-\omega ^2}\right)^2}{\omega ^2}\geq 0$$ eq8

which gives:
$$t' \sqrt{1-\omega ^2}-\omega \leq t\leq t' \sqrt{1-\omega ^2} +\omega$$ eq9

Let's take the equation he derived that equation from and insert some numerical values.
Say time = 0 in both frames and and R=1 and w=0.8c in the rest frame.
We take the the positions of 4 elements a,b,c and d at the 4 points of the compass where phi = 0, 90, 180 and 270 degrees. Now insert these values into the equation for the rest frame (s):

$$s=(t,\cos (\phi +t \omega ),\sin (\phi +t \omega ),0)$$

and we get (x,y) coordinates a=(1,0), b=(0,1), c=(-1,0) and d=(0-1) in frame s.

Now if we insert the same values into the transformed equation:

$$s \single-quote=L.s=\left(\frac{t+\omega \cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},\frac{t \omega +\cos (\phi +t \omega )}{\sqrt{1-\omega ^2}},\sin (\phi +t \omega ),0\right)$$

we get (x',y') coordinates a'=(1.67,0), b'=(0,1), c'=(-1.67,0) and d'=(0,-1) in frame s'.

It can be seen that the transformed ball is wider than it is high by this equation which is a false result.
Actually, that is correct. If you take a set of events which are simultaneous in the axis frame then they are not simultaneous in the rolling frame, and their transformed locations are "smeared" out such that it is wider than it is high.

This is due to the "catch 22" situation of not being able to calculate phi without knowing how the proper time of each element transforms and it is not possible to calculate how the the proper time of each element transforms without knowing how phi transforms. On the face of it, the equation is not taking the simultanity issues into account between the transformed frames.
Eq5 takes simultaneity into account in the rolling frame, and therefore eq7 describes events which are simultaneous in the rolling frame.

This can be seen when we do the calculations for the transformed time and space coordinates (t',x',y') to get a'=(1.33,1.67,0), b'=(0,0,1), c'=(-1.33,-1.67,0) and d'=(0,0,-1) in frame s'. The time t' is not simultaneous in the s' frame. I do not think Dalespams equation are wrong, but it does not seem that they resolved the issues of how the dimensions of the ball would be measured simultaneously in the transformed frame without doing numerical aproximations.
Of course your events are not simultaneous in the rolling frame, they are simultaneous in the axis frame! That is why I went to the bother of obtaining eq7 instead of just stopping with eq4.

Unless I'm not reading your picture correctly, the "chunks of hoop" appear to be rotating in the wrong direction.

Regards,

Bill
Hi Bill,
I tend to agree with your observation. There appears to be a small mistake in DaleSpam's original equation that has the ball rotating counter clockwise while translating from left to right in the positive x direction which does not agree with the contact point being at the bottom. This is a non fatal error that is easily corrected by reversing the sign of the y coordinates.

Hmm, I should have numbered my equations to make talking about them easier. There are 7 equations in total, so eq1 is the first equation and eq7 is the last.
It was not possible to solve eq5 for t. But I could determine the range of t directly from eq7, I just left it out for brevity, but here it is explicitly:

The y-coordinate in eq7 is imaginary unless
$$1-\frac{\left(t-t' \sqrt{1-\omega ^2}\right)^2}{\omega ^2}\geq 0$$ eq8

which gives:
$$t' \sqrt{1-\omega ^2}-\omega \leq t\leq t' \sqrt{1-\omega ^2} +\omega$$ eq9

......

Eq5 takes simultaneity into account in the rolling frame, and therefore eq7 describes events which are simultaneous in the rolling frame.

Of course your events are not simultaneous in the rolling frame, they are simultaneous in the axis frame! That is why I went to the bother of obtaining eq7 instead of just stopping with eq4.
Hi Dalespam,
Of course I agree that events that are simultaneous in one frame are not simultaneous in the other. The point I was trying to make is that using eq7 we are still required to determine t for a given t' in eq7 by numerical aproximation using eq5. Do you agree?

By reversing the signs of the y coordinates (see above comment to Bill) to get the ball rotating in the same direction as the linear motion and by using the solve function of a spreadsheet to determine t in eq7 I get the following results for t'=0 in frame S' for four particles on the rim of the hoop with coordinates expressed as (t,x',y') where t is the proper time of each particle. (Assume w = v =0.8c where v is the relative velocity of frame S and S')

a' = -0.683390604 0.512542953 0.519881721
b' = 0.000000000 0.000000000 -1.000000000
c' = 0.683390604 -0.512542953 0.519881721
d' = 0.000000000 0.000000000 1.000000000

where in the frame at rest with axis of rotation the (x,y) coordinates for the particles are a = (1.0), b = (0,-1), c = (-1,0) and d = (0,1).

It can be seen that the y' coordinates of particles a and c are the same and above the centre of the axis rotation by a distance of 0.519881721 so we can assume the centre of mass for this pair of particles is at (x',y') = (0,0.519881721)

Now if we look at the pair of particles at the top and bottom of the ball (b and d), the particle at the contact point has a momentary velocity of zero in frame S' and the particle at the top has a velocity of 0.975609756c by velocity addition. If we assign a value of 1 to the rest mass of the particles then the particle b at the contact point has a inertial mass of 1.00000 and particle d has an inertial mass of 4.55555. The centre of mass of these two particles assuming radius (R=1) is found from 2*R*mb/(mb+md)-R = 0.64R or (x',y') = (0, 0.64) in terms of coordinates.

From the above it can be seen that the centre of mass when the particles are vertically alligned in S' is at y'=0.64 and moves to y'=0.519881721 after the particles on the hoop have completed a quarter rotation. Therefore the height of the centre of mass does not remain constant in time for a rolling object with constant linear and rotational velocity in the transformed frame. That is something we have to live with and it neccessary to find a compensating factor such as stress in the connecting rod or hoop joining the test masses, or a counter rotating ghost torque to sort things out (to justify no wobble observed in the rest frame of the rotation axis).

P.S. DAleSpam, I am not being critical of your work (I think it very good and the most positive and technical contribution to this thread so far ;)

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Dale
Mentor
By reversing the signs of the y coordinates ... and by using the solve function of a spreadsheet to determine t in eq7 I get the following results for t'=0 in frame S' for four particles on the rim of the hoop with coordinates expressed as (t,x',y') where t is the proper time of each particle. (Assume w = v =0.8c where v is the relative velocity of frame S and S')

a' = -0.683390604 0.512542953 0.519881721
b' = 0.000000000 0.000000000 -1.000000000
c' = 0.683390604 -0.512542953 0.519881721
d' = 0.000000000 0.000000000 1.000000000

where in the frame at rest with axis of rotation the (x,y) coordinates for the particles are a = (1.0), b = (0,-1), c = (-1,0) and d = (0,1).

It can be seen that the y' coordinates of particles a and c are the same and above the centre of the axis rotation by a distance of 0.519881721 so we can assume the centre of mass for this pair of particles is at (x',y') = (0,0.519881721)

Now if we look at the pair of particles at the top and bottom of the ball (b and d), the particle at the contact point has a momentary velocity of zero in frame S' and the particle at the top has a velocity of 0.975609756c by velocity addition. If we assign a value of 1 to the rest mass of the particles then the particle b at the contact point has a inertial mass of 1.00000 and particle d has an inertial mass of 4.55555. The centre of mass of these two particles assuming radius (R=1) is found from 2*R*mb/(mb+md)-R = 0.64R or (x',y') = (0, 0.64) in terms of coordinates.

From the above it can be seen that the centre of mass when the particles are vertically alligned in S' is at y'=0.64 and moves to y'=0.519881721 after the particles on the hoop have completed a quarter rotation. Therefore the height of the centre of mass does not remain constant in time for a rolling object with constant linear and rotational velocity in the transformed frame. That is something we have to live with and it neccessary to find a compensating factor such as stress in the connecting rod or hoop joining the test masses, or a counter rotating ghost torque to sort things out (to justify no wobble observed in the rest frame of the rotation axis).
I don't think that your conclusion follows from your analysis. It seems to me that, according to your analysis, when pair bd has center .64 pair ac has center .52 for an overall center of .58. Then, after a "quarter rotation" pair bd will have center .52 and pair ac will have center .64 for an overall center of .58. So the overall center of mass of all four points will be the same every "quarter rotation".

The hoop has more than two (or even four) particles. I think you need to integrate around the hoop, or at least do a numerical approximation to the integral.

Hi Bill,
I tend to agree with your observation. There appears to be a small mistake in DaleSpam's original equation that has the ball rotating counter clockwise while translating from left to right in the positive x direction which does not agree with the contact point being at the bottom. This is a non fatal error that is easily corrected by reversing the sign of the y coordinates.
Another way to correct it would be to start over with:

$$s=(t,\cos (\phi -t \omega ),\sin (\phi -t \omega ),0)$$

Regards,

Bill

I don't think that your conclusion follows from your analysis. It seems to me that, according to your analysis, when pair bd has center .64 pair ac has center .52 for an overall center of .58. Then, after a "quarter rotation" pair bd will have center .52 and pair ac will have center .64 for an overall center of .58. So the overall center of mass of all four points will be the same every "quarter rotation".

The hoop has more than two (or even four) particles. I think you need to integrate around the hoop, or at least do a numerical approximation to the integral.
I wondered if someone would raise that argument. The counter argument is this. A wheel with just two significant masses on opposite sides of the rotation axis will be balanced in the reference frame that is stationary with respect to the rotation axis. Therefore we should not have to rely on 4 masses to negate any wobble. In normal (Newtonian) physics a barbell with two main weights of equal mass on the ends of a bar when thrown through the air will rotate about its centre of mass. Relativity should be able to predict that the rotating barbell will not wobble when moving linearly and rotating at the same time without requiring four or more masses or a hoop to prevent wobble of the centre of mass.

Dale
Mentor
That's a good argument, although it really doesn't apply for the hoop geometry described by my equations. For a hoop I think the best conclusion is that the center of mass is stationary.

I would not recommend using my hoop equations to draw conclusions for the barbell geometry.

I would not recommend using my hoop equations to draw conclusions for the barbell geometry.
You started out with a single point on a given trajectory - why would that not hold for any number of points along the same trajectory?

Regards,

Bill

Dale
Mentor
You have to get rid of the variable phi and add a variable r.

I think it is an interesting problem, but it is a different problem than the one I worked.

Forgive me I just deleted this post and have to retype it so I am bound miss something I put before.

I have been studying the Dale's attachments on post 80 and they have been bothering me for some reason. Last night I think I finally realised why. To begin with some basic information about the drawings as I understand them.

1) The rings are in the roads view.
2) The colors represent 30 degrees in the starionary axis view.
3) c=1
4) r=1
5) v = .9c

I looked at the velocity of a particle in the roads view. Using the basic equation for velocity:
V = $$\frac{D}{T}$$​
I asked what if the wheel rolled forward pi/4. This yields the equation:
.9c = $$\frac{pi/4}{T}$$​
Then I looked at the distance the left most particle on the wheel would have traveled. To do this I first noticed that because of length contraction the bottem half of the wheel only contains 90 degrees. There for when the wheel turns pi/4 the left most particle will travel to the very bottem of the wheel. This is a distance of at least r (just look in the Y direction only). So using the same time as a pi/4 roll T, by particle has moved with velocity:
V' = $$\frac{1}{T}$$​
Dividing the V' equation by the V equation I get:
V'/.9c = $$\frac{1}{pi/4}$$​
Which yields the particles velocity V' to be 1.15c.

This obviously can not be correct though. I assume I am missing something. Can anyone help me understand what I did wrong?

By reversing the signs of the y coordinates (see above comment to Bill) to get the ball rotating in the same direction as the linear motion and by using the solve function of a spreadsheet to determine t in eq7 I get the following results for t'=0 in frame S' for four particles on the rim of the hoop with coordinates expressed as (t,x',y') where t is the proper time of each particle. (Assume w = v =0.8c where v is the relative velocity of frame S and S')

a' = -0.683390604 0.512542953 0.519881721
b' = 0.000000000 0.000000000 -1.000000000
c' = 0.683390604 -0.512542953 0.519881721
d' = 0.000000000 0.000000000 1.000000000

where in the frame at rest with axis of rotation the (x,y) coordinates for the particles are a = (1.0), b = (0,-1), c = (-1,0) and d = (0,1).

It can be seen that the y' coordinates of particles a and c are the same and above the centre of the axis rotation by a distance of 0.519881721 so we can assume the centre of mass for this pair of particles is at (x',y') = (0,0.519881721)
You can't draw any reasonable conclusion by using point particles attached to the circumference, you need to use the correct calculations, based on integrals.

In the frame S, comoving with the object, the center of mass (y-coordinate) is:

$$y_M=\frac{\int y \rho dV}{\int \rho dV}=\frac{\int y dV}{\int dV}$$

The center of mass coincides with the center of rotation because of the radial symmetry.

In the observer's frame S':

$$y'_M=\frac{\int y' \rho' dV'}{\int \rho' dV'}=\frac{\int y \gamma dV}{\int \gamma dV}= y_M$$

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You can't draw any reasonable conclusion by using point particles attached to the circumference, you need to use the correct calculations, based on integrals.

In the frame S, comoving with the object, the center of mass (y-coordinate) is:

$$y_M=\frac{\int y \rho dV}{\int \rho dV}=\frac{\int y dV}{\int dV}$$

The center of mass coincides with the center of rotation because of the radial symmetry.

In the observer's frame S':

$$y'_M=\frac{\int y' \rho' dV'}{\int \rho' dV'}=\frac{\int y \gamma dV}{\int \gamma dV}= y_M$$
Hi 1effect,
I am not sure how you got to your conclusions because you have not provided much detail. For example what is V? Is it volume? Anyway, you seem to conclude that y coordinate of the centre of mass in the S' frame is at same height as the centre of mass in the S frame. Dalespam reached the opposite conclusion that the height of the centre of mass of the rolling ball is higher in S' than in S and his conclusion is in agreement with the fairly well known fact that the centre of mass of a rotating object is not invariant under Lorentz transformation. It seems you have reached a false conclusion.