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Particle speed on a relativistic wheel?

  1. May 5, 2008 #1
    There have been some threads on relativistic wheels/balls from the view of a road the wheel/ball is rolling on, with out slipping. So I decided to research them. I found some nice movies and saw some nice drawings that help explained the movement/shape of the wheel. It is an ellipse that is length contracted such that (at high speed close to c) 7 of the 8 spokes are near the top (say in the top quarter of the wheel) while one spoke slides around the wheel and when that spoke gets close to the top the next spoke in turn starts to move around the wheel. This was reinforced when I thought about the fact that the top of the wheel can not move faster the c, i.e. if the axle is moving at .9 c then the top of the wheel can not move at 1.8c but rather a velocity very close to c. This also supports the idea of the spokes bunching up near the top.
    But now here is my beef. I noticed that in order for the wheel to work the spoke that travels from the front of the bunch to the back of the bunch must travel faster then c. Here is how I came to this conclusion, doing all my calculations from the roads view.
    First of all the radius R of the wheel is contracted along the x-axis, but not the y-axis, so that when a particle at the end of the spoke travels from the front of the spoke line to the back it must travel a distance of at least 2R, from the front most point to the ground to the back most point. Admittedly, the spoke’s distance traveled is greater then 2R, but 2R will suffice for this example.
    Now, in order for the wheel to remain in equilibrium When the axel moves forward 1/8th of the circumference (2 pi R) a spoke must travel from the front of the bunch to the back, so that when the wheel has moved like this 8 times every spoke is back to where it started and the wheel has turned 1 revolution after moving forward the circumference of the wheel.
    Here in lies the problem, if speed is distance/time and we set R = 1. Then while equation for the speed of the wheel is (1/8 2R pi)/t or (pi/4)/t, were t is some time interval in the roads frame of reference. The equation for the speed of the particle is (2R)/t or 2/t, where t is the same as in the prior equation because they derived from the same frame of reference. Now if you evaluate these 2 equations you find that the speed of the particle is at least 8/pi faster then the speed of the axel which we already said was moving close to c. This of course leads to the particle moving faster then c.
    I might have a geometric solution to this problem, but it is late and I need sleep. I am also perfectly willing to accept that I have made an error somewhere, please point it out to me. I have been told before that the spokes do not move so that when the axel moves forward 1 circumference the spokes will not have made 1 revolution, but I can not see this because it would make the wheel slip.
     
  2. jcsd
  3. May 6, 2008 #2

    Dale

    Staff: Mentor

    Do you have access to Mathematica? If so, I can send you my code from the wheel thread and you can play around with it. Otherwise, I am traveling this week and can do the derivation myself after I get back.
     
  4. May 6, 2008 #3
    Unfortnatly I do not have access to mathematica. I am really intrested to track a single particle as the wheel moves =).
     
  5. May 7, 2008 #4
    Hi Wizardblade,

    I have managed to "deploy" a java simulation of the relativistic wheel that I was working on, that is based on DaleSpam's equations and may be just what you are looking for: Relativistic Wheel Simulation applet.:

    Please let me know if the applet works OK for you (or anyone else) as I have not put a working applet onto the web before. :)
     
  6. May 8, 2008 #5
    Thanks guys tha applet opens for me I have not had time to play with it yet though =)
     
  7. May 8, 2008 #6
    Ok guys I want you to look at something on this applet. Put the speed at .99c and you just need 2 points. Now watch the points, specificaly when one point is at the top of the elipse. At the top of the elipse the particle is moving at nearly C, just below. And now watch the speed the other particle has. It looks to be greater then the top particles speed. This is most apparent in the top picture, but can be seen in the bottem one as well. Let me know what you guys think?
     
  8. May 8, 2008 #7
    Hi Wizard,

    I have modified the applet a bit so you can analyse it better with more clocks and a grid background. You might have to reload your browser to see the difference.

    Maybe if DaleSpam transforms the four velocity equations for us us I might be able to put a real time recording of one particle in the primed frame which will answer your question.
     
  9. May 8, 2008 #8
    Kev,

    Thanks that would be awsome. Ill look at it again =)
     
  10. May 9, 2008 #9

    Dale

    Staff: Mentor

    Nice applet. I like the way of streaming the paper behind the stationary wheel to get the cycloid.
     
  11. May 9, 2008 #10

    Thanks :-)

    I'm inspired to create some more relativity simulations now I know where to start ;)
     
  12. May 12, 2008 #11

    Dale

    Staff: Mentor

    I made my previous formulas a little more general. So, an arbitrary particle on a rolling wheel can be described in the axle frame by the worldline:
    [tex]s = \left(c t,r \cos (\phi +t \omega ),r \sin (\phi +t \omega ),0\right)[/tex] eq1
    In this expression the axle is at the origin, and the wheel is rotating in the counter-clockwise direction with an angular frequency w. We also have that R>r>0 where R is the radius of the wheel and r is the distance of the particle from the axle. Also, -pi<phi<pi.

    The four-velocity of this particle in the axle frame is:
    [tex]u=\frac{ds}{d\tau } = \left(\frac{c}{\sqrt{1-\frac{r^2 \omega ^2}{c^2}}},-\frac{r \omega
    \sin (\phi +t \omega )}{\sqrt{1-\frac{r^2 \omega
    ^2}{c^2}}},\frac{r \omega \cos (\phi +t \omega
    )}{\sqrt{1-\frac{r^2 \omega ^2}{c^2}}},0\right)[/tex] eq2

    Now, the velocity of the frame where the wheel is rolling without slipping is Rw where it is rolling to the left for positive w. So, using the Lorentz transform for a boost of Rw we have the particle's worldline in the rolling frame:
    [tex]s' = L(R \omega).s = \left(\frac{c t}{\sqrt{1-\frac{R^2 \omega ^2}{c^2}}}-\frac{r R
    \omega \cos (\phi +t \omega )}{c \sqrt{1-\frac{R^2 \omega
    ^2}{c^2}}},\frac{r \cos (\phi +t \omega )}{\sqrt{1-\frac{R^2
    \omega ^2}{c^2}}}-\frac{R t \omega }{\sqrt{1-\frac{R^2 \omega
    ^2}{c^2}}},r \sin (\phi +t \omega ),0\right)[/tex] eq3

    The four-velocity of this particle in the rolling frame is:
    [tex]u'=\frac{ds'}{d\tau } = \left(\frac{c^2+r R \omega ^2 \sin (\phi +t \omega )}{c
    \sqrt{1-\frac{r^2 \omega ^2}{c^2}} \sqrt{1-\frac{R^2 \omega
    ^2}{c^2}}},-\frac{\omega (R+r \sin (\phi +t \omega
    ))}{\sqrt{1-\frac{r^2 \omega ^2}{c^2}} \sqrt{1-\frac{R^2 \omega
    ^2}{c^2}}},\frac{r \omega \cos (\phi +t \omega
    )}{\sqrt{1-\frac{r^2 \omega ^2}{c^2}}},0\right)[/tex] eq4

    Upon examination of eq4 we note that it is real and finite for [tex]c^2>R^2 \omega ^2\geq r^2 \omega ^2[/tex] which means that the speed is less than c for the same conditions.
     
  13. May 12, 2008 #12
    Hi DaleSpam,

    Nice work! Thanks for doing the analysis for us. I have not checked it all out yet, but I am sure it will be to the high standard of your previous contributions :)
     
  14. May 12, 2008 #13
    Not to beat a dead horse, but consider a particle that sits on the rim at [itex]\phi=0[/itex] when t=0. With some small positive increase in t, shouldn't the y-component of that particle go negative with respect to the axle frame?

    I realize that the answer is "no" now that you have re-defined the ground to be in the +y direction of the axle frame, but I notice that the rolling frame appears to have a y-axis pointing in the opposite direction (i.e [itex]\phi[/itex] does not appear to be consistent between the two frames).

    [addendum: I suppose this is moot if you have changed the convention for the rolling frame as well]

    Regards,

    Bill
     
    Last edited: May 12, 2008
  15. May 12, 2008 #14
    I am probably wrong here but is not (x1 and x2) just the velocity in the x and y direction and to find the total velocity you need to add the vectors? ie (x1^2+x2^2)^.5?
     
  16. May 13, 2008 #15

    Dale

    Staff: Mentor

    Hi Bill,

    I knew this would come up so I was very careful and explicit with the sign conventions this time and I used the usual convention of putting the ground on the bottom. As I said above, the wheel is rotating counter-clockwise and rolling to the left for positive w. If you want a wheel rotating clockwise and rolling to the right simply use a negative value for w.
     
  17. May 13, 2008 #16
    In u and u'.

    If I understand correctly:

    [tex]u=\frac{ds}{dt}\times\frac{dt}{d\tau}[/tex]

    Regards,

    Bill
     
  18. May 13, 2008 #17

    Dale

    Staff: Mentor

    I am using four-vectors here, so the components of s are (ct,x,y,z). The four-velocity is the proper-time derivative of the four-vector, so it is γ(c,vx,vy,vz). Since γ increases without bound as |v| approaches c, the four-velocity is also unbounded, and for any finite value represents a speed less than c.

    The reason that I left it in terms of the four-velocity is that it is easier to look at a complicated expression and determine if it is real and finite than it is to look at a similarly complicated expression and determine if it is less than c. But a real and finite four-velocity is the same as a speed less than c.
     
  19. May 13, 2008 #18

    Dale

    Staff: Mentor

    Yes, that is correct.

    [tex]u=\frac{ds}{dt} \frac{dt}{d\tau}=\frac{d(ct,x,y,z)}{dt} \gamma=(c,dx/dt,dy/dt,dz/dt) \gamma=\gamma (c,vx,vy,vz)[/tex]
     
  20. May 13, 2008 #19
    Ok I think I see now thank you so much =).
     
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