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Does a relativistic rolling ball wobble?

by bwr6
Tags: ball, relativistic, rolling, wobble
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1effect
#37
Apr23-08, 06:43 PM
P: 321
Quote Quote by JesseM View Post
How is that a difficulty? If you know the movement of the point in the frame of the center, the Lorentz transformation will tell you the time of any event on that point's worldline in another frame. It's really just a problem of the equations being difficult or impossible to solve exactly. For example, if the point on the rim is described by x'(t') = cos(t') and y'(t') = sin(t') in the rest frame of the center, applying the Lorentz transformation would give the equations:

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

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

[tex]x'^2+y'^2=1[/tex]

[tex]\gamma^2 (x-vt)^2+y^2=1[/tex]

[tex]y=sin(t)[/tex]
[tex]\gamma (x-vt)=cos(t)[/tex]
so:
[tex]x=\gamma^{-1} cos(t)+vt[/tex]
JesseM
#38
Apr23-08, 07:23 PM
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P: 8,470
Quote Quote by 1effect View Post
Actually, it is possible to get a parametric solution, avoiding the nasty transcendental equations. Here is the trick:

[tex]x'^2+y'^2=1[/tex]

[tex]\gamma^2 (x-vt)^2+y^2=1[/tex]

[tex]y=sin(t)[/tex]
How did you get y=sin(t)? Wasn't the equation in the primed frame y'=sin(t'), which means it should be y=sin(gamma*(t - vx/c^2))
1effect
#39
Apr23-08, 07:35 PM
P: 321
Quote Quote by JesseM View Post
How did you get y=sin(t)? Wasn't the equation in the primed frame y'=sin(t'), which means it should be y=sin(gamma*(t - vx/c^2))
I parametrized :

[tex]\gamma^2 (x-vt)^2+y^2=1[/tex]

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

For example, I could have chosen [tex]\frac{1-t^2}{1+t^2}[/tex] and [tex]\frac{2t}{1+t^2}[/tex]
JesseM
#40
Apr23-08, 07:45 PM
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Quote Quote by 1effect View Post
I re-parametrized

[tex]\gamma^2 (x-vt)^2+y^2=1[/tex]
I don't understand what you mean by "re-parametrized" here. Can you fill in the missing steps that allow you to go from that equation to y=sin(t)? Unless you mean that y=sin(t) is meant to be the definition of a new parameter t, but then you should have given the parameter a different name than the time coordinate that appears in (x-vt).
JesseM
#41
Apr23-08, 07:52 PM
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Quote Quote by 1effect View Post
I parametrized :

[tex]\gamma^2 (x-vt)^2+y^2=1[/tex]

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

For example, I could have chosen [tex]\frac{1-t^2}{1+t^2}[/tex] and [tex]\frac{2t}{1+t^2}[/tex]
But if you're just parametrizing y with a newly-defined parameter, you can't use the t-coordinate as the parameter! The t-coordinate already has a specific meaning--if you have some y' and t' which lie on the worldline of the point in the primed frame, you can't substitute in y=sin(t) unless it's true that when you do the Lorentz transform on that y' and t' on the point's worldline, the resulting y and t actually satisfy y=sin(t). If you want to define a new parameter you should give it a different name, like y=sin(p).
1effect
#42
Apr23-08, 07:52 PM
P: 321
Quote Quote by JesseM View Post
I don't understand what you mean by "re-parametrized" here. Can you fill in the missing steps that allow you to go from that equation to y=sin(t)? Unless you mean that y=sin(t) is meant to be the definition of a new parameter t, but then you should have given the parameter a different name than the time coordinate that appears in (x-vt).
I am not sure about that , I will need to think about it.
This would result into :

[tex]y=sin(\tau)[/tex]

[tex]x=\gamma^{-1} cos(\tau)+vt[/tex]
JesseM
#43
Apr23-08, 07:55 PM
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Quote Quote by 1effect View Post
I am not sure about that , I will need to think about it.
This would result into :

[tex]y=sin(\tau)[/tex]

[tex]x=\gamma^{-1} cos(\tau)+vt[/tex]
But that isn't very helpful, because you don't know what value of t corresponds to a given value of [tex]\tau[/tex].
1effect
#44
Apr23-08, 07:59 PM
P: 321
Quote Quote by JesseM View Post
But that isn't very helpful, because you don't know what value of t corresponds to a given value of [tex]\tau[/tex].
Yes, this wouldn't work. Too bad, I guess we are stuck with the unsolvable system of transcendental equations. :-(
Antenna Guy
#45
Apr23-08, 08:38 PM
P: 303
Quote Quote by 1effect View Post
Actually, it is possible to get a parametric solution, avoiding the nasty transcendental equations.
When you guys are done "simplifying" the problem, I'd like to see what you get when your observer is not at infinity.

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

[tex]\frac{\gamma^2 x'^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1[/tex]

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

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

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

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

Regards,

Bill
JesseM
#46
Apr23-08, 09:00 PM
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Quote Quote by Antenna Guy View Post
When you guys are done "simplifying" the problem, I'd like to see what you get when your observer is not at infinity.
At infinity? I wasn't trying to calculate what would be seen visually, just the coordinate path of a point on the rim in a frame where the rolling object was moving.
Quote Quote by Antenna Guy
Consider that the sphere travelling along x at relativistic speed with respect to an inertial observer will be described by:

[tex]\frac{\gamma^2 x'^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1[/tex]
I guess this would be the equation for the coordinates of the surface in the frame where the sphere is moving along the x' axis, at the moment the sphere was centered at the origin. But this is different from the shape the sphere would appear to be as seen by an observer at rest in this frame.
Quote Quote by Antenna Guy
If the inertial observer is not at infinity, the "shape" of the object will not be a projection on a plane - and that "shape" will vary with distance/aspect angle.
What do you mean by "projection on a plane"?
Phrak
#47
Apr23-08, 10:56 PM
P: 4,513
Did I miss it, or hasn't it been pointed out? It's not at all unambiguous as to what meant by a 'ball'?

First, is it a problem in dynamics, or simply kinematics? If only kinematics, the spinning ball has a circumference to diameter ratio of gamma*pi in the plane of angular momentum. So in this case, it's fair to ask 'was it a ball before spinning? (If it was a ball before spinning then its a ball of constant radius, otherwise constant circumference.)
Antenna Guy
#48
Apr24-08, 06:42 AM
P: 303
Quote Quote by Phrak View Post
Did I miss it, or hasn't it been pointed out? It's not at all unambiguous as to what meant by a 'ball'?
Are you aware that this relativistic 'ball' does not necessarily have an instantaneous circumference (per se) in the frame of the observer?

First, is it a problem in dynamics, or simply kinematics? If only kinematics, the spinning ball has a circumference to diameter ratio of gamma*pi in the plane of angular momentum. So in this case, it's fair to ask 'was it a ball before spinning? (If it was a ball before spinning then its a ball of constant radius, otherwise constant circumference.)
Let's say that the observed circumference of this ball spans a range of time that approaches [itex]\frac{2r}{c}.[/itex]. What would you integrate over to verify that the rest length of the observed circumference is still [itex]2\pi r[/itex]?

Regards,

Bill
1effect
#49
Apr24-08, 12:14 PM
P: 321
Quote Quote by Antenna Guy View Post
When you guys are done "simplifying" the problem, I'd like to see what you get when your observer is not at infinity.

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

[tex]\frac{\gamma^2 x'^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1[/tex]

[note: sorry about leaving out the velocity term - I'd have probably screwed that up anyway ]
The above is not the equation of the moving sphere. You need to apply to Lorentz transform correctly.
JesseM
#50
Apr24-08, 12:29 PM
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P: 8,470
Quote Quote by 1effect View Post
The above is not the equation of the moving sphere. You need to apply to Lorentz transform correctly.
I think it is the equation for the sphere's surface at a given instant when it is centered at the origin and moving on the x'-axis. The equation describes an ellipsoid with radius r along the y' and z' axis, but Lorentz-contracted radius [tex]r/\gamma[/tex] along the x'-axis...is this not correct?
1effect
#51
Apr24-08, 12:35 PM
P: 321
Quote Quote by JesseM View Post
I think it is the equation for the sphere's surface at a given instant when it is centered at the origin and moving on the x'-axis. The equation describes an ellipsoid with radius r along the y' and z' axis, but Lorentz-contracted radius [tex]r/\gamma[/tex] along the x'-axis...is this not correct?
"moving" was key in my answer.
His equation is the snapshot for t=0, so we agree.

The correct equation is:

[tex]\frac{\gamma^2 (x'-vt')^2}{r^2}+\frac{y'^2}{r^2}+\frac{z'^2}{r^2}=1[/tex]
JesseM
#52
Apr24-08, 12:43 PM
Sci Advisor
P: 8,470
Ah, OK. But one point that occurred to me is that if we're just looking at the total set of points occupied by the surface at any given instant, rather than trying to track an individual point on its surface, then in the rest frame of the sphere's center the coordinates of the surface won't be changing with time, so it will always occupy exactly the same coordinates as a certain nonrotating sphere centered at the same position. And if the surface of the nonrotating sphere occupies the same set of coordinates in this frame, its surface must occupy the same set of coordinates as the rotating sphere in every frame. So, this is one way of seeing that the rolling ball won't "wobble". The difficulty is just in figuring out how a given point on the rotating sphere will move around this ellipsoid surface in a frame where it's in motion.
1effect
#53
Apr24-08, 12:48 PM
P: 321
Quote Quote by JesseM View Post
Ah, OK. But one point that occurred to me is that if we're just looking at the total set of points occupied by the surface at any given instant, rather than trying to track an individual point on its surface, then in the rest frame of the sphere's center the coordinates of the surface won't be changing with time, so it will always occupy exactly the same coordinates as a certain nonrotating sphere centered at the same position. And if the surface of the nonrotating sphere occupies the same set of coordinates in this frame, its surface must occupy the same set of coordinates as the rotating sphere in every frame. So, this is one way of seeing that the rolling ball won't "wobble". The difficulty is just in figuring out how a given point on the rotating sphere will move around this ellipsoid surface in a frame where it's in motion.
I think that we can do the "tracking" by converting the problem to cylindrical coordinates and by describing the angle [tex]\phi=\phi(t)[/tex] as a function of time.
1effect
#54
Apr24-08, 01:56 PM
P: 321
Quote Quote by kev View Post

I am aware that that there is no length contraction perpendicular to the linear motion of the rolling sphere. I can guarantee that if the centre of mass is in the centre of the ball in its rest frame then it is not in the geometrical centre of the ball when it is rolling with respect to the observer. The centre of mass is further away from the point of rolling contact than the geometrical midpoint between the top of the rolling ball and the point of contact.
I think you need to have a look at this , there is no wobble. You can see more of the same.


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