Re: Transverse versus parallel torques -- did I miss something here?

From: Harry (harald.vanlintel_at_epfl.ch)
Date: 08/03/04 

Date: 3 Aug 2004 05:06:43 -0700

Sorry for the late reply, I only read it one hour ago...

sal <believer@nospam.org> wrote in message news:<pan.2004.07.07.16.26.02.134685@nospam.org>...
> Retic periodically posts something about the "rotating lever
> paradox". I got curious and tried to work it out. I'd like to hear
> what people think of this approach.
>
> We have two levers, one parallel to the line of motion, and one
> perpendicular, as in this picture:
>
> http://physicsinsights.org/images/rotating_levers.png
>
> F1 and F2 are forces and L1 and L2 are lengths.
>
> In the rest frame of the levers, F1 = F2 = F and L1 = L2 = L, and of
> course
>
> F1*L1 = F2*L2 = F*L
>
> In the rest frame of an observer who sees the whole thing going by at
> velocity V, lever L2 will be contracted, but lever L1 will not.
>
> We need to find the forces, and then we'll look at the torques.
>
> A transverse force of F in the MCRF of an object "O" with mass M will
> accelerate it at a rate of A = M/F. After time dTau, the object will
> be moving at velocity V1 = A*dTau.
>
> In the rest frame of an observer "S" who sees the object moving at V,
> while dTau time passes in O's MCRF, dT = dTau*gamma time will pass.
> The object's velocity at that point will be measured to be V1/gamma,
> since transverse lengths don't change but time is passing faster in
> S's frame. The velocity change S saw was 1/gamma times as large as
> the change O saw, and the time which passed was gamma times longer, so
> S sees the object accelerating 1/(gamma^2) times slower than it
> appears in the MCRF. But its mass, as measured by S, is gamma times
> larger as O measures it, so the transverse force as seen by S, "Fst",
> compared to the transverse force measured by O, "Fot", must actually
> be:
>
> Transverse force: Fst = Fot / gamma
>
> When a longitudinal force is applied instead, the above argument still
> holds

No, that's were you were mistaken. See below.

> _except_ that the longitudinal length seen by S is 1/gamma times
> the length seen in the MCRF of the object. So, the velocity change
> measured by S is reduced again by another factor of gamma, to obtain:
>
> Parallel force: Fsp = Fop / (gamma^2)
>
> So in the frame of S, where F1 is parallel to the motion and F2 is
> transverse, we must have
>
> F1s = F1/(gamma^2)
> F2s = F2/gamma
>
> Now, L1's length will be the same for S as it is for O, since it's
> transverse to the motion. But L2 is parallel to the line of motion,
> so it will be contracted by a factor of gamma. So:
>
> L1s = L1
> L2s = L2/gamma
>
> And the torques will be:
>
> F1s * L1s = F1/(gamma^2) * L1 = F*L / (gamma^2)
> F2s * L2s = F2/gamma * L2/gamma = F*L / (gamma^2)
>
> So the torques are identical from the point of view of S, just as they
> are from the point of view of O, and there is no paradox.
>
> Comments?
>
> Is something left out?

No doubt.

> Did I blow the analysis someplace?

Yes you did blow it! (I just checked, but it doesn't seem anyone
noticed!)
It would have been no real paradox if it were so simple.
You made the very common mistake to overlook the difference in
simultaneity. Consequently, you implied that dx/dt = dx'/dt'/gamma^2.
However, in relativity v_x = v_x'.
As all text books will show you, F_y = F_y'/gamma, but F_x = F_x'.

(I used more standard notation in which positive x is the direction of
motion, and primed symbols indicate observations in the moving
frame)).

Regards,
Harald

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