Register to reply 
Does a relativistic rolling ball wobble? 
Share this thread: 
#37
Apr2308, 06:43 PM

P: 321

[tex]x'^2+y'^2=1[/tex] [tex]\gamma^2 (xvt)^2+y^2=1[/tex] [tex]y=sin(t)[/tex] [tex]\gamma (xvt)=cos(t)[/tex] so: [tex]x=\gamma^{1} cos(t)+vt[/tex] 


#38
Apr2308, 07:23 PM

Sci Advisor
P: 8,470




#39
Apr2308, 07:35 PM

P: 321

[tex]\gamma^2 (xvt)^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{1t^2}{1+t^2}[/tex] and [tex]\frac{2t}{1+t^2}[/tex] 


#40
Apr2308, 07:45 PM

Sci Advisor
P: 8,470




#41
Apr2308, 07:52 PM

Sci Advisor
P: 8,470




#42
Apr2308, 07:52 PM

P: 321

This would result into : [tex]y=sin(\tau)[/tex] [tex]x=\gamma^{1} cos(\tau)+vt[/tex] 


#43
Apr2308, 07:55 PM

Sci Advisor
P: 8,470




#44
Apr2308, 07:59 PM

P: 321




#45
Apr2308, 08:38 PM

P: 303

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 


#46
Apr2308, 09:00 PM

Sci Advisor
P: 8,470




#47
Apr2308, 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.) 


#48
Apr2408, 06:42 AM

P: 303

Regards, Bill 


#49
Apr2408, 12:14 PM

P: 321




#50
Apr2408, 12:29 PM

Sci Advisor
P: 8,470




#51
Apr2408, 12:35 PM

P: 321

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] 


#52
Apr2408, 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.



#53
Apr2408, 12:48 PM

P: 321




Register to reply 
Related Discussions  
Relativistic wobble?  Special & General Relativity  9  
Rolling ball problem  Introductory Physics Homework  2  
Rolling ball  Introductory Physics Homework  2  
Rolling ball  General Physics  7  
Rolling Ball Analysis  General Physics  0 