Does a relativistic rolling ball wobble?

[yuiop]

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&#039; \sqrt{1-\omega ^2}-\omega \leq t\leq t&#039; \sqrt{1-\omega ^2}<br /> +\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 necessary 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 ;)

 

[Dale]

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 necessary 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.

 

[Antenna Guy]

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:

<br /> s=(t,\cos (\phi -t \omega ),\sin (\phi -t \omega ),0)<br />

Regards,

Bill

 

[yuiop]

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]

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.

 

[Antenna Guy]

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]

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.

 

[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 realized 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?

 

[1effect]

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&#039;_M=\frac{\int y&#039; \rho&#039; dV&#039;}{\int \rho&#039; dV&#039;}=\frac{\int y \gamma dV}{\int \gamma dV}= y_M

 

[yuiop]

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&#039;_M=\frac{\int y&#039; \rho&#039; dV&#039;}{\int \rho&#039; dV&#039;}=\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.

 

[Dale]

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

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.

 

[yuiop]

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 realized 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?

Hi WizardsBlade,
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.

 

[1effect]

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 :-)

 

[yuiop]

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,

Using your initial equation:

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)

 

[yuiop]

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 ;)

 

[1effect]

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?

 

[yuiop]

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.

 

[1effect]

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&#039;=-\gamma sin(\phi+t \omega) d \phi
dy&#039;=cos (\phi+t \omega) d \phi
The arc element is :
ds&#039;=\sqrt (dx&#039;^2+dy&#039;^2)

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

\frac{\int y&#039; ds&#039;}{\int ds&#039;}

It is very easy to show that \int y&#039; ds&#039;=0

 

[Dale]

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.

I agree, to calculate the actual center of mass requires an integral. I think that point particles attached to the circumference will represent a "rectangle rule" numerical approximation to the integral, which is known to be problematic.

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&#039;_M=\frac{\int y&#039; \rho&#039; dV&#039;}{\int \rho&#039; dV&#039;}=\frac{\int y \gamma dV}{\int \gamma dV}= y_M

Unfortunately, it is not radially symmetric in the primed frame. The density varies around the hoop, as you can see in the images with the funny colors around the rim that I posted earlier. The resulting asymmetry pushes the center of mass up.

On a tangentially related topic, the equation for the center of mass that you wrote down made me realize something. In the equation there is a volume V, but volume is well known to be a fundamentally problematic concept in relativity because of issues of simultaneity. Basically, the problems with volume are the reason that thermodynamics are difficult to do relativistically. In addition to the problems of simultaneity are problems of causality. The different events in a volume at a specific time are all spacelike separated so they cannot be causally connected.

How does this translate to the present discussion? Basically, I have some doubts about the physical significance of the center of mass. How can the position of a given particle at a time t' physically influence the center of mass at that same time t' when they are not causally connected? I think that a "wobble" must be caused by something different than the instantaneous center of mass, either that or there should be some definition of the center of mass with physical significance where the particles influencing the center of mass do so causally. It seems to me that "wobble" is fairly poorly defined, and I think it is not as simple as it may seem.

 

[1effect]

I agree, to calculate the actual center of mass requires an integral. I think that point particles attached to the circumference will represent a "rectangle rule" numerical approximation to the integral, which is known to be problematic.

Unfortunately, it is not radially symmetric in the primed frame. The density varies around the hoop, as you can see in the images with the funny colors around the rim that I posted earlier. The resulting asymmetry pushes the center of mass up.

I am confused, why would density vary? I looked back and I couldn't find any derivation of a variable density. In frame S, the density is uniform. Why would relative motion result into non uniform density? In my derivation, I assumed both \rho and \rho&#039; as being uniform, this is why they fall out from the final expression of center of mass.

On a tangentially related topic, the equation for the center of mass that you wrote down made me realize something. In the equation there is a volume V, but volume is well known to be a fundamentally problematic concept in relativity because of issues of simultaneity. Basically, the problems with volume are the reason that thermodynamics are difficult to do relativistically. In addition to the problems of simultaneity are problems of causality. The different events in a volume at a specific time are all spacelike separated so they cannot be causally connected.

Yes, :-) . I had a long exchange with kev on this, in the thread about pressure. There are also quite a few threads on temperature. This is a yet unresolved problem in relativity, a very interesting one indeed.

How does this translate to the present discussion? Basically, I have some doubts about the physical significance of the center of mass. How can the position of a given particle at a time t' physically influence the center of mass at that same time t' when they are not causally connected? I think that a "wobble" must be caused by something different than the instantaneous center of mass, either that or there should be some definition of the center of mass with physical significance where the particles influencing the center of mass do so causally. It seems to me that "wobble" is fairly poorly defined, and I think it is not as simple as it may seem.

I think the key to solving this problem is deciding if the density is uniform or not. If the density is uniform, then we have a solution to the problem. I need help in understanding how uniform motion can make density increase on the top of the object, especially since the object is symmetric about the x axis.

 

[1effect]

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.

How can this be? The hoop is symmetric about the x-axis in its comoving frame, why would it become asymmetric? Why would it "become denser" on the top and not on the bottom? Please help me understand this.

 

[JustinLevy]

How can this be? The hoop is symmetric about the x-axis in its comoving frame, why would it become asymmetric? Why would it "become denser" on the top and not on the bottom? Please help me understand this.

The situation is not symmetric about the x axis. The bottom point of the hoop (since it is rolling without slipping) is at rest, while the top point is moving even faster than the center point of the hoop.

It appears to me that this discussion is just an issue of semantics. Some people seem to be considering the center of "relativistic mass" and others to be considering the center of "proper mass".

If we sum over every particle, the "center of relativistic mass" will indeed be increased in the y direction.

I need to think about it some more to be sure, but I believe the center of "proper mass" however should not be changed. What I am not sure of is whether this is a valid concept at all (you can't just add proper masses of constituent particles to get the proper mass of a system).

I think we need to first define what is being discussed.

 

[Dale]

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)

Hmm, I think I have some bigger problem than just this. I have counter-clockwise rotation in the unprimed frame and clockwise rotation in the primed frame. Something in my transformation or plotting is wrong.

Unfortunately, I probably won't be able to spend the time to find and fix it.

 

[Dale]

I am confused, why would density vary? I looked back and I couldn't find any derivation of a variable density. In frame S, the density is uniform. Why would relative motion result into non uniform density? In my derivation, I assumed both \rho and \rho&#039; as being uniform, this is why they fall out from the final expression of center of mass.

The density varies because of length contraction. The bottom particle has velocity 0 so there is no length contraction while the top particle has velocity 2w/(1+w²) so there is length contraction. The length contraction implies that there are more particles in the same arc length and therefore a higher density.

I believe that kev, in addition, was also considering "relativistic mass" which further increases the density of the top. I was only considering proper mass, but since nothing was clearly defined and since I never derived a numerical or analytical result I didn't think it worth arguing over.

 

[1effect]

The density varies because of length contraction. The bottom particle has velocity 0 so there is no length contraction while the top particle has velocity 2w/(1+w²) so there is length contraction. The length contraction implies that there are more particles in the same arc length and therefore a higher density.

I believe that kev, in addition, was also considering "relativistic mass" which further increases the density of the top. I was only considering proper mass, but since nothing was clearly defined and since I never derived a numerical or analytical result I didn't think it worth arguing over.

OK,

JustinLevy pointed out the same thing. So, every point around the circumference moves at a different speed. Since the loop is not infinitely rigid, it means that it is getting stretched in a non-uniform way because different parts of the loop move at different speeds wrt the same reference. So, the problem cannot be treated as a purely kinematic one anymore, we need to really deal with the loop as a lattice. Most likely , the loop will not have axial symmetry anymore. This is getting really complicated, I think chapter 9 in MTW deals with this :-(

 

[Wizardsblade]

Hi WizardsBlade,
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.

I knew I would miss things ;).

I realize that the drawing have flaws in them but they do seem to give the general shape of a relativistic wheel so I am going to try to fix the errors I made and try to explain this a little better.

1) First off I meant the right most particle, not the left most. I wanted the particle that travels down.
2) I am doing everything in the road's view, I realize that I used primed variables but it was not for relativity but rather just to denote 2 different speeds and distances.

I also should have added this: We know that the ellipse keeps its upright shape as it moves (IE it does not topple over and over but looks like it translates.) Because it must keep this shape and because of the length contraction on top some funny things happen when you try to roll this wheel. Let's say I take the far right point and then roll the wheel a distance of pi. Where is the point now.. My first intuition is that is on the far left side of the wheel, but this can not be so. Because if this is so then if I pick a particle (or group of particles) at (or near) the top and roll it the same distance then the particle (or group of particles) will have moved to the bottom of the ellipse. If a group of the contracted particles have moved from the top to the bottom then the ellipse has not kept its contracted shape, IE the high density part will have moved to the bottom. From this I deduced that the right most particles would have to move pi/2 when the wheel has only moved a distance of pi/4. If this is not true then the wheel will have change its shape or its density configuration as it turns.

Now I have (hopefully) clarified what I ment and how I came about it. What did I do wrong?

 

[Dale]

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 realized 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?

Since either my graphics or my math has some errors I hesitate to post this, but according to my work the four-velocity in the primed frame is:
\left(\frac{\omega ^2 \sin (\phi +t \omega )-1}{\omega<br /> ^2-1},\frac{\omega (\sin (\phi +t \omega )-1)}{\omega<br /> ^2-1},\frac{\omega \cos (\phi +t \omega )}{\sqrt{1-\omega<br /> ^2}},0\right)
which you can verify results in a maximum speed
\frac{2 \omega }{\omega ^2+1}&lt;1

 

[Wizardsblade]

OK,

JustinLevy pointed out the same thing. So, every point around the circumference moves at a different speed. Since the loop is not infinitely rigid, it means that it is getting stretched in a non-uniform way because different parts of the loop move at different speeds wrt the same reference. So, the problem cannot be treated as a purely kinematic one anymore, we need to really deal with the loop as a lattice. Most likely , the loop will not have axial symmetry anymore. This is getting really complicated, I think chapter 9 in MTW deals with this :-(

It is the non-uniform stretching that I believe yields the results I was getting for why when you move the wheel pi/4 the particle moves pi/2 around the circumference.

 

[DrGreg]

Yesterday I bought a copy of Relativity: Special, General, and Cosmological by Wolfgang Rindler. One of the exercises in that book states that the "centre of mass" of a system must be calculated using energy instead of rest mass (or equivalently, using relativistic mass, if you like that sort of thing). It further states that different inertial observers may each calculate a different centre of mass. However all observers agree that their centre of mass is stationary relative to the centre of momentum frame (the frame in which the total momentum is zero). Or in other words, all of the different observers' centres of mass are all stationary relative to each other (and so the wobble in this question is impossible).

The book gave no further explanation, but I think one way of looking at this is via the conserved relativistic angular momentum tensor, whose 01, 02 and 03 components relate energy, linear momentum, time and position.

Finally, I would suggest that to model this as a system of a large number of particles might be easier than as a continuous density, because then you don't have the problem of deciding how the density should transform between frames.

 

[Dale]

Hmm, I think I have some bigger problem than just this. I have counter-clockwise rotation in the unprimed frame and clockwise rotation in the primed frame. Something in my transformation or plotting is wrong.

Unfortunately, I probably won't be able to spend the time to find and fix it.

I figured out my error. It was just a plotting mistake and I believe that all of the equations are still correct. Basically, I was coloring the hoop according to phi. My equation for phi has two roots, as you can see in eq6
\phi =-t \omega \pm \cos ^{-1}\left(\frac{t&#039; \sqrt{-(\omega -1) (\omega +1)}-t}{\omega }\right)

I simply got the order of the roots wrong and accidentally plotted the top root on the bottom and the bottom root on the top. Basically, just flip the colors about the horizontal axis to get the correct plots. This means that the hoop rotates counter-clockwise in both frames and is rolling to the right. Unfortunately, it also means that my "ground" is on the top, which is a little confusing.