Does a relativistic rolling ball wobble?

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  • #1
bwr6
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A spherical ball rolls without slipping along a surface, with uniform velocity v/c (close to 1) with respect to the lab frame. It is thus seen to contract by a factor of [tex]\gamma[/tex], forming a grape-like shape.

Why doesn't it wobble as it rolls, like a rolling grape?

One can certainly argue from the principle of relativity: a state of absolute rest could be determined by setting up a detector that beeps if the ball experiences a change in maximum distance from the surface, as it would if it rolled like a grape. But this is a principled argument. What is the mechanism?
 

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  • #2
Mentz114
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bwr6:
One can certainly argue from the principle of relativity: a state of absolute rest could be determined by setting up a detector that beeps if the ball experiences a change in maximum distance from the surface, as it would if it rolled like a grape.
How ? There is no absolute velocity, so different observers would see different things.

It is not possible to distinguish rest from uniform motion, because rest and motion are observer dependent.
 
  • #3
bwr6
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You were supposed to read that sentence as a reductio ad absurdum, Mentz114. Assume, *for the purpose of reductio*, that relativistic rolling balls wobble like grapes. Then absolute rest could be determined in the way in which I describe above. This contradicts the principle of relativiy; therefore, relativistic rolling balls do not wobble like grapes.

Look: it's clear that relativistic rolling balls don't wobble like grapes; my argument proves that much. But why is this the case? I'd prefer an answer that teaches me something about the kinematic of rolling balls in SR.
 
  • #4
Doc Al
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A spherical ball rolls without slipping along a surface, with uniform velocity v/c (close to 1) with respect to the lab frame. It is thus seen to contract by a factor of [tex]\gamma[/tex], forming a grape-like shape.
Interesting scenario (which I will not pretend to analyze in detail). But if it's rolling without slipping how can it have a uniform velocity with respect to the surface? For example, the speed at the point of contact is zero.
 
  • #5
Mentz114
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You were supposed to read that sentence as a reductio ad absurdum, Mentz114. Assume, *for the purpose of reductio*, that relativistic rolling balls wobble like grapes. Then absolute rest could be determined in the way in which I describe above. This contradicts the principle of relativiy; therefore, relativistic rolling balls do not wobble like grapes.

Look: it's clear that relativistic rolling balls don't wobble like grapes; my argument proves that much. But why is this the case? I'd prefer an answer that teaches me something about the kinematic of rolling balls in SR.
Apologies, I've become too accustomed to people seriously proposing scenarios like this as 'errors' in relativity.

The reason the ball does not wobble is that it appears to deform as it rotates, so it is more like a grape on tank-tracks.
 
  • #6
1effect
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A spherical ball rolls without slipping along a surface, with uniform velocity v/c (close to 1) with respect to the lab frame. It is thus seen to contract by a factor of [tex]\gamma[/tex], forming a grape-like shape.

Why doesn't it wobble as it rolls, like a rolling grape?

There are two effects at work , one is the well-known relativistic length contraction, the other one is the difference in arrival of light rays. When both of them are taken into consideration the result is very surprising: the object still looks spherical but the textures get distorted.

One can certainly argue from the principle of relativity: a state of absolute rest could be determined by setting up a detector that beeps if the ball experiences a change in maximum distance from the surface, as it would if it rolled like a grape. But this is a principled argument. What is the mechanism?

There is no such thing as a state of absolute rest . In the reference frame with the origin coinciding with the ball center there is no length contraction, so you will not be able to detect any of : "the ball experiences a change in maximum distance from the surface, as it would if it rolled like a grape."
 
  • #7
Doc Al
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There are two effects at work , one is the well-known relativistic length contraction, the other one is the difference in arrival of light rays. When both of them are taken into consideration the result is very surprising: the object still looks spherical but the textures get distorted.
Are you talking about a rolling ball or just a translating ball?
 
  • #8
1effect
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Are you talking about a rolling ball or just a translating ball?

Rolling about the vertical axis , as in http://www.spacetimetravel.org/filme/fussball_0.99/fussball_0.99-xe-320x240.mpg movie, so it is not exactly what the OP wanted. I am not sure if it makes any difference if the rolling happens about the horizontal axis.
This site arrives to the same result, see Orbit2.flc.
 
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  • #9
bwr6
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Interesting scenario (which I will not pretend to analyze in detail). But if it's rolling without slipping how can it have a uniform velocity with respect to the surface? For example, the speed at the point of contact is zero.
You raise a very good point. Here's an attempt to pose the problem more clearly:

Imagine that a homogeneous spherical mass rolls without slipping, such that its center of mass has uniform velocity v in the horizontal direction. Now take the limit as v/c approaches 1. (Perhaps even this is not very well posed, if the center of mass wobbles up and down.) My question is: How does the angular motion of the ball change?

I have the intuition that the ball will contract and that the motion, as Mentz114 declares, will look like a grape on tank-tracks. However, the situation need not be so simple. My worry is this: the group of lorentz transformations (as it is developed in the textbooks) is developed to model bodies with transverse velocity. But happens for *rotating bodies?* Is a rigid sphere that spins in empty space with relativistic angular velocity seen to contract, or does it remain a sphere?

Then, enter my question: if it does not remain a sphere, but rather contracts along the direction of its transverse velocity, then what determines the angular kinematics? The usual tools (moment and torque) seem to suggest that the ball will wobble, as in classical mechanics. But this can't be, by the principle of relativity!

I remain perplexed.
 
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  • #10
1effect
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You raise a very good point. Here's an attempt to pose the problem more clearly:

Imagine that a homogeneous spherical mass rolls without slipping, such that its center of mass has uniform velocity v in the horizontal direction. Now take the limit as v/c approaches 1. (Perhaps even this is not very well posed, if the center of mass wobbles up and down.)

It will not "wobble up and down". The distance of the sphere center to the supporting surface remains constant because it doesn't get Lorentz - contracted. So, it is like you were rolling a blob of jello :-)
 
  • #11
Doc Al
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Rolling about the vertical axis , as in http://www.spacetimetravel.org/filme/fussball_0.99/fussball_0.99-xe-320x240.mpg movie, so it is not exactly what the OP wanted. I am not sure if it makes any difference if the rolling happens about the horizontal axis.
This site arrives to the same result, see Orbit2.flc.
I think those sites just discuss the visual appearance of a translating (not rotating) ball. It appears to rotate. (The Penrose-Terell effect.)
 
  • #12
Antenna Guy
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Imagine that a homogeneous spherical mass rolls without slipping, such that its center of mass has uniform velocity v in the horizontal direction. Now take the limit as v/c approaches 1. (Perhaps even this is not very well posed, if the center of mass wobbles up and down.) My question is: How does the angular motion of the ball change?

I assume the radius of the ball is greater than zero - which creates an "interesting" situation for the top of the ball.

I'm picturing an ellipsoidal ball that leans backwards (center lags the contact patch).

Regards,

Bill
 
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  • #13
bwr6
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I've seen a lot of nice suggestions so far (and some irrelevant stuff). Some people seem convinced that the ball take on a contracted form "like jello" or "like a grape with tank-treads." That was my original intuition, but I don't see how one knows that this is true. What still has not been explained to me is the angular kinematics at relativistic speeds. The challenge remains:

Can anyone back that up their intuition with a proof or argument?

B.
 
  • #14
yuiop
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Interesting scenario (which I will not pretend to analyze in detail). But if it's rolling without slipping how can it have a uniform velocity with respect to the surface? For example, the speed at the point of contact is zero.

The wheels of a car moving along a road at normal non-relativistic speeds in everyday life, have velocity relative to the road, despite the speed of the contact patch of the wheels being zero. The wheels of a car are usually considered to roll without slipping, (but not when the roads are icy).

I assume the radius of the ball is greater than zero - which creates an "interesting" situation for the top of the ball.

I'm picturing an ellipsoidal ball that leans backwards (center lags the contact patch).

Regards,

Bill

Imagine the ball is enclosed by a non rotating shell with a small gap for the contact patch of the rolling ball with the road, a bit like the wheel arch of a car. The wheels of a car moving at relativistic speeds will not lean backwards, because if they do not not keep the same profile as the wheel arch this would provide a means to detect absolute motion. It is the same for the rolling ball enclosed in a non rotating shell. They must deform in the same way.


...
I have the intuition that the ball will contract and that the motion, as Mentz114 declares, will look like a grape on tank-tracks. However, the situation need not be so simple. My worry is this: the group of lorentz transformations (as it is developed in the textbooks) is developed to model bodies with transverse velocity. But happens for *rotating bodies?* Is a rigid sphere that spins in empty space with relativistic angular velocity seen to contract, or does it remain a sphere?...

The ball has to contract in exactly the same way as a non rotating spherical shell or you could detect absolute motion by comparing the shapes of a non rotating ball with a rotating ball.
 
  • #15
Doc Al
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The wheels of a car moving along a road at normal non-relativistic speeds in everyday life, have velocity relative to the road, despite the speed of the contact patch of the wheels being zero. The wheels of a car are usually considered to roll without slipping, (but not when the roads are icy).
The instantaneous velocity of the various parts of the wheel vary from zero (at the contact patch) to twice the speed of the center of mass. I'd think that would be relevant to the relativistic treatment of a rolling ball.
 
  • #16
bwr6
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The ball has to contract in exactly the same way as a non rotating spherical shell or you could detect absolute motion by comparing the shapes of a non rotating ball with a rotating ball.

I agree kev, that due to its transverse motion, the ball must contract in this way -- this follows from the principle of relativity, as I said in the beginning. However, what I would still like to know is 1) why the laws of rotation at relativistic speeds imply this (independent of the principle of relativity), and more generally 2) what the rules for a relativistically rotating body *actually are.*

I call arguments like this one (that the ball must do X because of the principle of relativity) a "principled argument," and generally don't find them very satisfying. They generally don't teach you anything about the mechanism at work. What we need is an argument that gets at the mechanics.

(For anyone who is still confused about the kind of argument I keep asking for, compare this to the two ways that one might argue for gravitational redshift. There is a principled argument from the equivalence principle, which derives the redshift by analyzing a beam of light in an elevator -- and this teaches you nothing about the mechanism behind the phenomenon. Or, you can derive the redshift by analyzing the way that frequencies get "stretched" along null-geodesics in curved spacetime, which is precisely the mechanism behind the redshift. What I'm interested in is the latter kind of argument -- in the case at hand, it would be an argument that explains to me the way that relativistic balls rotate.)
 
  • #17
yuiop
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The instantaneous velocity of the various parts of the wheel vary from zero (at the contact patch) to twice the speed of the center of mass. I'd think that would be relevant to the relativistic treatment of a rolling ball.

Fair comment, but it is also fair to assume that bwr6 meant the velocity of the centre of mass of the ball when he said "constant velocity". Of course your observation about the different velocities of the parts of the ball is correct and relevant. It raises the question about where exactly is the centre of gravity of a rolling ball (or wheel) is in the relativistic context and does it stay it stay parallel to the direction of motion (no wobble)?This is basically the question I posed in this thread https://www.physicsforums.com/showthread.php?t=230536 titled "Relativistic wobble?" coincidently about 5 hours before bwr6 posted this question.
 
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  • #18
Antenna Guy
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Imagine the ball is enclosed by a non rotating shell with a small gap for the contact patch of the rolling ball with the road, a bit like the wheel arch of a car. The wheels of a car moving at relativistic speeds will not lean backwards...

If r>>0, the top of the ball excedes c as v->c.

Neglecting rotation, the shape change is trivial. With rotation, something has to give so that no portion of the surface excedes c.

I'm looking at this as somehow analogous to the phase progression with polarization angle of an elliptically polarized wave (albeit with a 90deg rotation with respect to focal axes).

Regards,

Bill
 
  • #19
1effect
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I think those sites just discuss the visual appearance of a translating (not rotating) ball. It appears to rotate. (The Penrose-Terell effect.)

This one rotates, as I mentioned earlier. So, it is not the Terrell effect (only)
 
  • #20
yuiop
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If a rocket containing two spheres is moving relative to an observer and one ball is spun up while the other is not, then if the spinning ball is not exactly the same elliptical shape as the non spinning ball, with its longest dimensions exactly perpendicular to the linear motion, then relativity has a problem. In the rest frame there is no reason for the spinning ball not to be circular at the points furthest from the spinning axis.

Gravity Probe B had spherical gyros enclosed in spherical shells with quite narrow tolerances. If you imagine an exagerated version of the experiment with higher velocities then if the spinning spheres altered shape differently to the enclosing shells you could measure the difference in the sphere shell gap in variuos directions in the rest frame of the gyro and determine absolute velocity and direction (or the the gyros would jam).
 
  • #21
Antenna Guy
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If a rocket containing two spheres is moving relative to an observer and one ball is spun up while the other is not, then if the spinning ball is not exactly the same elliptical shape as the non spinning ball, with its longest dimensions exactly perpendicular to the linear motion, ...

I think three mutually orthogonal gyroscopes would be more telling:rolleyes:

Regards,

Bill
 
  • #22
bwr6
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If a rocket containing two spheres is moving relative to an observer and one ball is spun up while the other is not, then if the spinning ball is not exactly the same elliptical shape as the non spinning ball, with its longest dimensions exactly perpendicular to the linear motion, then relativity has a problem. In the rest frame there is no reason for the spinning ball not to be circular at the points furthest from the spinning axis.

Gravity Probe B had spherical gyros enclosed in spherical shells with quite narrow tolerances. If you imagine an exagerated version of the experiment with higher velocities then if the spinning spheres altered shape differently to the enclosing shells you could measure the difference in the sphere shell gap in variuos directions in the rest frame of the gyro and determine absolute velocity and direction (or the the gyros would jam).

To the first part: I like this analogy, but don't see how it can work. We want to understand the laws of motion of a relativistic rolling ball. In particular, the transverse motion of a rolling ball is determined in part by the "spinning" -- or more precisely (at least in the classical case) by its moment of inertia, which captures its "resistance" to spinning in all orientations. The transverse motion of your ball in a spaceship is not determined in this way.

To the second part: gravity probe B measured frame-dragging; I think it's best to neglect curvature effects as they can only complicate matters. The relativistic rolling ball problem is perfectly well posed in flat Minkowski spacetime.

However, I think you're exactly right to look to new experimental observations. What is becoming clear to me is that the motion of relativistic rolling bodies may not be well-defined by the textbook theory. Perhaps no matter what arguments are given (by way of the equivalence principle or something else), we must ultimately turn to empirical observation to learn how such systems behave.
 
  • #23
Antenna Guy
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The relativistic rolling ball problem is perfectly well posed in flat Minkowski spacetime.

If I understand this correctly, a family of Minkowski spaces would be needed to deal with it.

Regards,

Bill
 
  • #24
JesseM
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However, I think you're exactly right to look to new experimental observations. What is becoming clear to me is that the motion of relativistic rolling bodies may not be well-defined by the textbook theory. Perhaps no matter what arguments are given (by way of the equivalence principle or something else), we must ultimately turn to empirical observation to learn how such systems behave.
Why would it not be well-defined? The motion of a rotating disc is a well-understood problem in SR, so in the rest frame of the center of the rolling body, wouldn't you just have a rotating disc (or cylinder or sphere, both of which can presumably be sliced up into a series of rotating discs) with a floor that was moving inertially with the same velocity as the tangential velocity of the bottom of the disc? And if we know how things look in this frame, we can just use the Lorentz transform to figure out how it would look in other frames.
 
  • #25
Doc Al
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This one rotates, as I mentioned earlier. So, it is not the Terrell effect (only)
Which of the several animations on that page represents something rotating? The only mention of rotation that I see is apparent rotation.
 
  • #26
yuiop
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Which of the several animations on that page represents something rotating? The only mention of rotation that I see is apparent rotation.

Ieffect was referring to the animation with the file name "orbit2.flc" representmg a camera orbiting close to the Earth at 0.95c. (It is the one with a picture of the Eath from space). I am not sure if that means the camera is orbiting around the Earth which would give visual rotation of the Earth on top of the Penrose-Terell rotation.

Non of the animations are very helpful to the question posed by the OP because they hide the physical measurements behind visual effects (optical illusions).

In the Penrose Terell rotation the length contraction of a sphere is hidden by the optical rotation so it still appears like a sphere and 1effect likes to use this example to support his belief that length contraction is not a physical effect. His argument is weakened when it is noted that anything that is not an exact sphere will still be visually distorted and when light travel times are taken into account the length contraction is still there.
 
  • #27
yuiop
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To the first part: I like this analogy, but don't see how it can work. We want to understand the laws of motion of a relativistic rolling ball. In particular, the transverse motion of a rolling ball is determined in part by the "spinning" -- or more precisely (at least in the classical case) by its moment of inertia, which captures its "resistance" to spinning in all orientations. The transverse motion of your ball in a spaceship is not determined in this way.

I think as JesseM points out, if you want to understand the transverse motion of a rolling ball you should start with something simpler like a rolling disk or a rolling wheel and understand that first. JesseM is also right that we can simulate the rolling effect by taking a spinning ball or wheel and superimposing linear motion. The only stipulation is that the rim velocity of the ball or wheel in the rest frame of its axis is equal to the superimposed linear motion for rolling without slipping. The two are equivalent if we assume rolling without friction.

Why would it not be well-defined? The motion of a rotating disc is a well-understood problem in SR, so in the rest frame of the center of the rolling body, wouldn't you just have a rotating disc (or cylinder or sphere, both of which can presumably be sliced up into a series of rotating discs) with a floor that was moving inertially with the same velocity as the tangential velocity of the bottom of the disc? And if we know how things look in this frame, we can just use the Lorentz transform to figure out how it would look in other frames.

As I mentioned to brw6 I agree with most of what you say except for the fact that lorentz transformation of a rolling body is lot more complicated than it might first appear to be. The motion involves transcendental functions that can not be solved empirically and require iteration techniques to come to an approximate solution.

A geometrical solution of the motion of one particle on the rim of the rolling body can be arrived at as follows:

Observe that in the rest frame of the rolling body's rotation axis the particle follows a cycloid path. (The equation for a cycloid only has a parametric solution). In the moving frame the cycloid path of the rim particle is stretched by a factor of gamma. If we superimpose a moving ellipse (to represent the length contracted disk) on the cycloid path then the intersection of the elliptical locus and the stretched cycloid locus is the location of the rim particle at any instant in the transformed frame. The trouble is that the mathematical determination of the intersection is very difficult. Some of the difficulties stem from the fact that there is no exact equation for the perimeter length of an ellipse, there is no non parametric solution to a cycloid and the inverse of transcendental functions can not be found without iterative techniques.
 
  • #28
Dale
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This question should be simple but very tedious to answer. If I wanted the answer enough I would essentially take JesseM's approach, but I would start with a hoop to make the analysis simpler:

1) Start with a rotating hoop in the inertial reference frame where the axis of rotation is stationary.

2) Take a differential element of the hoop

3) Analyze the four-forces (tension only, ignore gravity etc.) acting on the element

4) Get an equation of (circular) motion for the element and an equation for the (centripetal) four-force

5) Lorentz transform to the inertial frame with velocity equal to the tangential velocity in the original frame.

6) Verify that the transformed four-forces still correctly describe the transformed motion for the differential element.
 
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  • #29
Antenna Guy
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Some of the difficulties stem from the fact that there is no exact equation for the perimeter length of an ellipse, there is no non parametric solution to a cycloid and the inverse of transcendental functions can not be found without iterative techniques.

Interesting. I came to a similar conclusion (iterative solution required) after several attempts at finding a closed form expression for the rim of (an ellipsoidal) Earth in spherical coordinates with respect to an arbitrary orbital frame...

Regards,

Bill
 
  • #30
yuiop
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Interesting. I came to a similar conclusion (iterative solution required) after several attempts at finding a closed form expression for the rim of (an ellipsoidal) Earth in spherical coordinates with respect to an arbitrary orbital frame...

Regards,

Bill

Thanks Bill :smile:

Maybe someone can make a breakthrough using the method suggested by Dalespam (below). With a bit of luck there may be some sort of cancellation of the unsolvable elements of the problem that I mentioned. If Richard Feynman was still alive he could probably sort this out. After all, he managed to get rid of infinities that were causing a problem in quantum mechanics ;)

This question should be simple but very tedious to answer. If I wanted the answer enough I would essentially take JesseM's approach, but I would start with a hoop to make the analysis simpler:

1) Start with a rotating hoop in the inertial reference frame where the axis of rotation is stationary.

2) Take a differential element of the hoop

3) Analyze the four-forces (tension only, ignore gravity etc.) acting on the element

4) Get an equation of (circular) motion for the element and an equation for the (centripetal) four-force

5) Lorentz transform to the inertial frame with velocity equal to the tangential velocity in the original frame.

6) Verify that the transformed four-forces still correctly describe the transformed motion for the differential element.
 
  • #31
1effect
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It raises the question about where exactly is the centre of gravity of a rolling ball (or wheel) is in the relativistic context and does it stay it stay parallel to the direction of motion (no wobble)?

This question has already been answered. The distance from the center of gravity to the surface is constant.
In the frame comoving with the sphere there is no length contraction.
In any moving frame there is no Lorentz contraction for directions perpendicular to the direction of motion of the sphere. :-)
 
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  • #32
JesseM
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As I mentioned to brw6 I agree with most of what you say except for the fact that lorentz transformation of a rolling body is lot more complicated than it might first appear to be. The motion involves transcendental functions that can not be solved empirically and require iteration techniques to come to an approximate solution.
Are you saying these transcendental functions would appear in the rest frame of the rolling body's center, or only when you try to transform into a frame where the center is moving? What was the # of the post where you mentioned this to brw6?
kev said:
A geometrical solution of the motion of one particle on the rim of the rolling body can be arrived at as follows:

Observe that in the rest frame of the rolling body's rotation axis the particle follows a cycloid path. (The equation for a cycloid only has a parametric solution). In the moving frame the cycloid path of the rim particle is stretched by a factor of gamma. If we superimpose a moving ellipse (to represent the length contracted disk) on the cycloid path then the intersection of the elliptical locus and the stretched cycloid locus is the location of the rim particle at any instant in the transformed frame. The trouble is that the mathematical determination of the intersection is very difficult. Some of the difficulties stem from the fact that there is no exact equation for the perimeter length of an ellipse, there is no non parametric solution to a cycloid and the inverse of transcendental functions can not be found without iterative techniques.
If you have parametrized the path of any point on the rim in terms of x(t), y(t) and z(t) in the rest frame of the center of the rolling object, then presumably you would then just use the Lorentz transformation on these equations, although the resulting equations might indeed be hard to solve exactly for x' in terms of t', y' in terms of t' and z' in terms of t' (but you could at least get an arbitrarily good numerical approximation of the path just by picking events on the path in the first coordinate system and then finding their coordinates in the second system).
 
  • #33
yuiop
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This question has already been answered. The distance from the center of gravity to the surface is constant.
In the frame comoving with the sphere there is no length contraction.
In any moving frame there is no Lorentz contraction for directions perpendicular to the direction of motion of the sphere. :-)

Your solution just says a rolling ball still looks like a sphere due to the optical illusion of Terrell rotation and due to light taking longer to reach the observer from the parts of the ball furthest from the observer. None of your example include rolling rotation (only rotation around the tranverse axis).

I am aware that that there is no length contraction perpendicular to the linear motion of the rolling sphere. I can guarantee that if the centre of mass is in the centre of the ball in its rest frame then it is not in the geometrical centre of the ball when it is rolling with respect to the observer. The centre of mass is further away from the point of rolling contact than the geometrical midpoint between the top of the rolling ball and the point of contact.
 
  • #34
yuiop
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Are you saying these transcendental functions would appear in the rest frame of the rolling body's center, or only when you try to transform into a frame where the center is moving?
The transcendental functions enter into the picture as soon as wheel starts moving at even at non relatavistic velocities and before any transformations are considered. It is do with there being no non parametric solution for the cycloid path of a particle on the rim of a wheel in terms of x and y without knowing the angle theta of the particle with respect to the wheel axis.

What was the # of the post where you mentioned this to brw6?

It was in post #27 when I just reiterating some things you mentioned earlier:

I think as JesseM points out, if you want to understand the transverse motion of a rolling ball you should start with something simpler like a rolling disk or a rolling wheel and understand that first. JesseM is also right that we can simulate the rolling effect by taking a spinning ball or wheel and superimposing linear motion. The only stipulation is that the rim velocity of the ball or wheel in the rest frame of its axis is equal to the superimposed linear motion for rolling without slipping. ...

If you have parametrized the path of any point on the rim in terms of x(t), y(t) and z(t) in the rest frame of the center of the rolling object, then presumably you would then just use the Lorentz transformation on these equations, although the resulting equations might indeed be hard to solve exactly for x' in terms of t', y' in terms of t' and z' in terms of t' (but you could at least get an arbitrarily good numerical approximation of the path just by picking events on the path in the first coordinate system and then finding their coordinates in the second system).

The difficulty is that simultaneity of the points on the rim is continuously changing. I did a simultion of the motion in geometry software a while ago, When I have more time I will try and post some pics. As far as I know there is no simple non iterative equation for the transformed rolling motion, but I would like to be proved wrong ;)
 
  • #35
1effect
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Your solution just says a rolling ball still looks like a sphere due to the optical illusion of Terrell rotation and due to light taking longer to reach the observer from the parts of the ball furthest from the observer.

You are misquoting me, see the correct quote here.

"There are two effects at work , one is the well-known relativistic length contraction, the other one is the difference in arrival of light rays. "

As an aside, "light taking longer to reach the observer from the parts of the ball furthest from the observer" and "the Terrell effect" are one and the same thing :-)

None of your example include rolling rotation (only rotation around the tranverse axis).

Yes, this is true, I pointed that out in one of my posts. I also mentioned that it is not clear why this is relevant.

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

Are you saying that the axis of rotation describes a cycloid? I would be very interested in the proof.
 
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