# Does a relativistic rolling ball wobble?

by bwr6
Tags: ball, relativistic, rolling, wobble
P: 3,791
 Quote by Antenna Guy 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.

 Quote by DaleSpam 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 ;)
Mentor
P: 15,624
 Quote by kev 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.
P: 303
 Quote by kev 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
P: 3,791
 Quote by DaleSpam 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.
 Mentor P: 15,624 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.
P: 303
 Quote by DaleSpam 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
 Mentor P: 15,624 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.
 P: 145 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?
P: 321
 Quote by kev 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':

$$y'_M=\frac{\int y' \rho' dV'}{\int \rho' dV'}=\frac{\int y \gamma dV}{\int \gamma dV}= y_M$$
P: 3,791
 Quote by 1effect 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.
Mentor
P: 15,624
 Quote by Antenna Guy Unless I'm not reading your picture correctly, the "chunks of hoop" appear to be rotating in the wrong direction.
 Quote by kev 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.
I looked into it, the hoop is rotating in the correct direction (clockwise) for the ground on the bottom and the hoop rolling to the right. Note that the vertical axis on the right-hand image crosses at x=0.6, so the hoop has rolled a decent distance to the right. Sorry about the confusing image.
P: 3,791
 Quote by Wizardsblade 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?
There is a minor mistake in DaleSpams drawing that shows the wheel rotating counter clockwise from while moving from left to right so that after one eighth of a turn (pi/4) in frame S the left most particle will move to the top and not to the bottom. Also, you do not seem have taken time dilation into account. If it takes t = pi/(4v) seconds to complete a quarter turn in frame S then it takes t' = pi/(sqrt(1-v^2)4v) seconds in frame S'. Also, it should be pointed out that the left most particle with ball moving from left to right will not rotate through 90 degrees in either frame after a rotation of pi/4 = (45 degrees) in frame S. So after taking everything into account including simultaneity and relativistic velocity addition I am sure you will not get v>c anywhere in the transformation of the rolling ball particles, using DaleSpam's equations.
P: 321
 Quote by kev 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?
It is volume, see the definition for center of mass.

 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.
Math says otherwise :-)
P: 3,791
 Quote by DaleSpam I looked into it, the hoop is rotating in the correct direction (clockwise) for the ground on the bottom and the hoop rolling to the right. Note that the vertical axis on the right-hand image crosses at x=0.6, so the hoop has rolled a decent distance to the right. Sorry about the confusing image.

Hi DaleSpam,

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

and substituting phi = 0 degrees, t=0 and w = 0.9c then the coordinates for the particle are (t,x,y,z) = (0,1,0,0). Staying with t= 0 and w =0.9 and using phi = 90 degrees the coordinates are (t,x,y,z) are (0,0,1,0) so that the convention you are using for phi is that a rotation from 0 to 90 degrees is counter clockwise in the rest frame of the axis of rotation. I think that is the cause of the confusion.

If I keep phi = zero degrees then when t=0.1 seconds the coordinates of our right most particle change from (t,x,y,z) = (0,1,0,0) to (0.1, 0.99, 0.09, 0) in frame S and the coordinates change from (t',x',y',z') = (2.06, 2.29, 0, 0) to (2.28, 2.49, 0.9) in frame S'. I am sure you will agree that is a counter clockwise rotation in both frames that is easily corrected by starting with:

$$s=(t,\cos (\phi +t \omega ),- \sin (\phi +t \omega ),0)$$
P: 3,791
 Quote by 1effect It is volume, see the definition for center of mass. Math says otherwise :-)
I think you have found the centre of volume which for an object of even density coincides with the centre of mass in Newtonian physics but not in relativity ;)
P: 321
 Quote by kev I think you have found the centre of volume which for an object of even density coincides with the centre of mass in Newtonian physics but not in relativity ;)
I would be very interested in your proving the above. The calculations I provided show the opposite. Could you provide yours?
P: 3,791
 Quote by DaleSpam 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 DaleSpam,

I have done the calculations for t = pi/4/w which is one eighth of a turn in the rest frame of the hoop axis which equates to t' = 1.636246174 seconds in frame S' when w and v is 0.8c. The transformed coordinates (t, x', y') of particles a, b, c and d are:

a' = 0.191000000 1.901914626 -0.152206102
b' = 1.772000000 0.715519375 -0.152597799
c' = 1.355638753 1.028609890 0.884074990
d' = 0.608000000 1.589744078 0.884021391

It is fairly easy (due to the symmetry) to work out that the centre of mass for the four particles at this posion is now y'=0.551

That means the centre of mass for the four particles is y' =0.58 every quarter of a turn (your calculation) and y' = 0.55 every eighth of a turn. It is reasonable to assume that as the number of particles on the rim increases the centre of mass will stabilise somewhere around y' = 0.56. This answers the question of the OP that a homogenous ball of even density will be grape shaped and move without wobbling. However, we have a new problem (perhaps needing a new thread?) that there appears to be a wobble of the centre of mass for a system of 4 or less particles with circular motion under Lorentz tranformation.
P: 321
 Quote by kev 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)
Thinking about the 4 particles is a bad way of thinking, there is no way that the center of mass for the complete wheel will be anywhere close to y'=0.519881721.

The correct way is to do the following:

$$dx'=-\gamma sin(\phi+t \omega) d \phi$$
$$dy'=cos (\phi+t \omega) d \phi$$
The arc element is :
$$ds'=\sqrt (dx'^2+dy'^2)$$

The center of mass is obtained by integrating from 0 to $$2 \pi$$ the expression:

$$\frac{\int y' ds'}{\int ds'}$$

It is very easy to show that $$\int y' ds'=0$$

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