
#1
Feb2705, 06:36 AM

P: 37

i didn't quite understand the resolution to this paradox as explained by my modern physics book. in case anyone hasn't heard of this, let me explain the setup. there is a right angled lever (with arms of equal length) which is constrained to rotate about its bend point. in addition, there are two transverse forces (of equal magnitude) acting at the tips of each arm. thus the net torque is zero and the lever remains at rest. but how does this look like to an observer traveling at speed v = 0.866c (gamma = 2) along an axis parallel to one of the arms? for the observer, both the arm parallel to the motion and its respective force are reduced by onehalf. the other arm and force are unaffected. the result is that there now appears to be a net torque on the system. the solution is that the from the observer's frame work is being done by one of the forces and this energy increase is invested as a rise in mass that exactly counters the torque.
a couple of points i don't understand. (1) there must be a force at the center of the lever that constrains it to rotations. doesn't this force contribute to the work done? (2) how can the mass of the lever be rising if the natural frame measures no such change? isn't rest mass supposed to be an invariant? 



#2
Feb2705, 10:26 AM

P: 698

I’m not sure about this situation, but it might be that because in the moving frame, one of the levers is compressed while density is the same, its mass increases creating a larger moment of inertia, decreasing Net torque to 0. I’m not sure though, I’m assuming that the measured mass changes for the person in the moving frame. I might be wrong.
It’s an interesting question. I'm also interested in the answer. Regards, Nenad 



#3
Feb2705, 02:22 PM

P: 1,545

That means the "changing" distance of deflection is not on the same arm that is "reduced by onehalf". 



#4
Feb2705, 03:55 PM

P: 37

lever paradox 



#5
Feb2705, 06:54 PM

Emeritus
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P: 7,445

Maybe I'm screwing this up  but it seems to me that one arm shrinks and it's force does not shrink, will the other arm does not shrink, but it's force does. 



#6
Feb2705, 09:38 PM

P: 37





#7
Feb2705, 11:00 PM

Sci Advisor
PF Gold
P: 9,186

Think about your direction with respect to the lengths and forces acting upon both arms. You are travelling parallel to one arm, but perpendicular to the force acting upon it. The opposite is true for the other arm. Does that help?




#8
Feb2705, 11:10 PM

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P: 7,445

OK, the 4force in the rest frame should be (0, fx, fy, fz) Boosting in the x direction, the Lorentz transform should be (gamma*v*fx/c^2, gamma*fx,fy,fz) OK, that seems to say that you're right, the force in the direction of boost should increase 



#9
Feb2705, 11:17 PM

Sci Advisor
PF Gold
P: 9,186

Don't worry about the 'force' in the center of the lever. It is irrelevant.




#10
Feb2805, 12:39 AM

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P: 7,445

I've been thinking that maybe the angular momentum and torque need to be reformulated as bivectors (using the wedge product, not the cross product), but I'm not quite sure how to do it. [add] I think I see how to do it, and I think it solves the problem. The bivector for torque can be represented by (force) ^ (distance), where (force) is the oneform representing the force, and (distance) is the oneform representing the distance. (You can pick (force)^(distance) or the negative of that, (distance)^(force), but you have to be consistent  the wedge product is anticommutative). The wedge product ^ generates a second rank antisymmetric tensor from the two one forms, and the two torques generated by the problem statement are equal and opposite. So torque isn't a vector in GR,  it's a second rank antisymmetric tensor, a bivector. Thus you don't get the right results when you think of torque as a vector and transform it as such, but you do when you use bivectors, which are geometric objects. the following URL discusses the wedge product a little bit, but it may not be enough to explain it to someone who hasn't seen it before http://icl.pku.edu.cn/yujs/MathWorld/math/w/w054.htm This may also help http://www.av8n.com/physics/areavolume.htm I think one has to work with oneforms rather than vectors, but with a Minkowski metric the two are very closely related  one simply takes g_ab u^a to convert u^a to the one form u_b, and with a Minkowski flat spacetime (a Lorentzian metric), this means that one simply inverts the sign of the first component of the 4vector. That's about as clearly as I can describe it, anyway. I suspect there's a way to do the wedge product with 4vectors rather than oneforms, but it seems that the standard defintion uses oneforms. 



#11
Feb2805, 01:55 AM

Sci Advisor
PF Gold
P: 9,186

Length and forces are vectors. The original question was directionally confused. The applied forces are vectors, not masses [that is a mixed reference frame]. They are aligned 90 degrees apart from one another and rotate symmetrically. The vector products are not important. Your first intuition was essentially correct.




#12
Feb2805, 03:10 AM

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P: 7,445

These ideas all best described by Clifford algebras, which I learned (as much as I did learn) from the WWW, and not from a textbook, giving me a "seat of the pants" approach to the whole topic. Writing out the components of the bivector (which is just an antisymmetric tensor) might (or might not) help illustrate how it works the bivector in the original frame for fx^ry looks like this (f=force, L=length) [tex] \begin {array}{cccc} 0&0&0&0\\\noalign{\medskip}0&0&1/2\,fL&0\\\noalign{\medskip}0&1/2\,fL&0&0 \\\noalign{\medskip}0&0&0&0\end {array} [/tex] And when it's boosted (beta=v/c, applied in the x direction) it looks like this (this is just the usual tensor transformation rule) [tex] \begin {array}{cccc} 0&0&1/2\,{\frac {fL\beta}{\sqrt {1{\beta}^{2}}}}&0\\\noalign{\medskip}0&0&1/2\,{\frac {fL}{\sqrt {1{\beta}^{ 2}}}}&0\\\noalign{\medskip}1/2\,{\frac {fL\beta}{\sqrt {1{\beta}^{2} }}}&1/2\,{\frac {fL}{\sqrt {1{\beta}^{2}}}}&0&0\\\noalign{\medskip}0&0 &0&0\end {array} [/tex] (The two bivectors for fx^ry and rx^fy are the same except for the sign  they are additive inverses). 



#13
Feb2805, 10:23 AM

P: 1,545

the deflection of the still long arm will be shorter. How much more info is needed? RB 



#14
Feb2805, 02:52 PM

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P: 7,445

I think it's best to stick with the formalism. The 4force is defined as dP/dtau, where P is the energymomentum 4vector. IF we know the total momentum of the system P summed over some spacelike hypersurface representing a time T we can find the momentum of the system P' on another spacelike hypersurface represnting the time T' by summing up the 4forces F = dP/dtau multipled by the time interval dtau for all particles representing the system Note that we sum over dtau, not dt  dtau is the proper time it takes each particle to move from the hypersurface representing the system at T to the hypersurface representing T'. For the angular momentum of the system, the calculation is similar The angular momentum bivector L should be given by summing up P^r, where P is the energymomentum 4 oneform, and R is the radius oneform (and ^ is the wedge product). The angular momentum of the system is performed by doing this sum, again, over some spacelike hypersurface representing "the system" at some time T. The total change in angular momentum L over the time interval from T to T' is then the intergal of F^r dtau, just as the linear momentum change was the intergal of F dtau. That happens because F*dtau = dP, by defintion, and tau is a scalar. All we need to argue now is that internal forces can't cause the momentum or the angular momentum of the system to change. 



#15
Feb2805, 05:22 PM

P: 37

to be honest, i'm rather lost in all this 4space formalism. if you manage to figure it out, could you guys explain it in terms of 3dim in each frame?
http://scitation.aip.org/getabs/serv...cvips&gifs=yes http://panda.unm.edu/Courses/finley/.../relangmom.pdf (page 17) 



#16
Feb2805, 10:21 PM

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P: 7,445

http://phy1139862.rit.edu/blog/mathematics/index.php http://www.mrao.cam.ac.uk/~clifford/...00000000000000 http://planetmath.org/encyclopedia/Bivector.html They will be well worth your time. I can't read the first URL above without a subscription, alas. The second URL seems to contain mostly a much clearer exposition than mine of angular momentum as a bivector aka a 2form. I do see that it mentions your point about internal torque at the end of the paper, but it doesn't really explain it very well. 



#17
Mar105, 02:52 AM

P: 37

of course, this solution contradicts that given by my modern physics text and that is always a little disturbing. especially considering the book explicitly states its solution was first given by Max von Laue, a nobel laureate. 



#18
Mar105, 10:28 AM

P: 1,545

SO  If with a construct like a 4space “formalism” you can track all the measurements through “hyperspace” great! But at the end of the day once you have the “4space formalism” right. Please check your work by using it to, as genxhis requested, describe in 3dimensions within the time frame of the traveler's reference frame, all the observations there of force and deflections. 


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