1. Feb 27, 2005

### genxhis

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 one-half. 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. Feb 27, 2005

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,

3. Feb 27, 2005

### RandallB

How does traveling observer measure the force? Wouldn't it be by the deflection of the arms as they bend ever so slightly.
That means the "changing" distance of deflection is not on the same arm that is "reduced by one-half".

4. Feb 27, 2005

### genxhis

the change in the forces by changing reference frames is a given. actually, its presumbly based on the lorentz transformations (where F = dp/dt) and shouldn't be too difficult to derive.

5. Feb 27, 2005

### pervect

Staff Emeritus

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. Feb 27, 2005

### genxhis

it looks like your thinking under the false presumption that the transformation simply contracts those components of force parallel to the motion.

7. Feb 27, 2005

### Chronos

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?

Last edited: Feb 27, 2005
8. Feb 27, 2005

### pervect

Staff Emeritus

OK, the 4-force 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. Feb 27, 2005

### Chronos

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

10. Feb 28, 2005

### pervect

Staff Emeritus
I'm not sure I see how that helps.

I've been thinking that maybe the angular momentum and torque need to be reformulated as bi-vectors (using the wedge product, not the cross product), but I'm not quite sure how to do it.

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 one-form representing the force, and (distance) is the one-form 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 anti-commutative).

The wedge product ^ generates a second rank anti-symmetric 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 anti-symmetric tensor, a bi-vector. 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 bi-vectors, 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/area-volume.htm

I think one has to work with one-forms 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 space-time (a Lorentzian metric), this means that one simply inverts the sign of the first component of the 4-vector.

That's about as clearly as I can describe it, anyway. I suspect there's a way to do the wedge product with 4-vectors rather than one-forms, but it seems that the standard defintion uses one-forms.

Last edited: Feb 28, 2005
11. Feb 28, 2005

### Chronos

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.

Last edited: Feb 28, 2005
12. Feb 28, 2005

### pervect

Staff Emeritus
I really think the ultimate problem is that the cross product is only defined in 3 dimensions. The relevance of this fact is, of course, that torque = force (cross) distance. The mathematical equivalent of the 3-dimensional cross product in 4 dimensions is the wedge product/ bi-vector. The bi-vector approach works in three dimesnsions, too - but in 3d, one can take the dual (Hodges dual) of the bi-vector to get an ordinary vector. In 4d, the dual of a bivector is another bivector, so one can't reduce the bi-vector to an ordinary vector.

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 bi-vector (which is just an anti-symmetric tensor) might (or might not) help illustrate how it works

the bi-vector in the original frame for fx^ry looks like this (f=force, L=length)

\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}

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)

\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}

(The two bi-vectors for fx^ry and rx^fy are the same except for the sign - they are additive inverses).

Last edited: Feb 28, 2005
13. Feb 28, 2005

### RandallB

As I said in post 3 - assume you can measure the force by the deflection of the Arms. The deflection of the "shorter" arm will be unchanged.

the deflection of the still long arm will be shorter.

RB

14. Feb 28, 2005

### pervect

Staff Emeritus
I'm really not sure if you can define the force by the deflection of the arm.

I think it's best to stick with the formalism. The 4-force is defined as dP/dtau, where P is the energy-momentum 4-vector.

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 4-forces 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 bi-vector L should be given by summing up P^r, where P is the energy-momentum 4 one-form, and R is the radius one-form (and ^ is the wedge product). The angular momentum of the system is performed by doing this sum, again, over some space-like 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. Feb 28, 2005

### genxhis

to be honest, i'm rather lost in all this 4-space formalism. if you manage to figure it out, could you guys explain it in terms of 3-dim in each frame?

the following sources suggest that net internal torques can exist and are in fact neccesary to the resolution of this paradox:

http://scitation.aip.org/getabs/ser...00043000007000615000001&idtype=cvips&gifs=yes
http://panda.unm.edu/Courses/finley/P495/TermPapers/relangmom.pdf (page 17)

16. Feb 28, 2005

### pervect

Staff Emeritus
Bi-vectors aka two-forms are really very useful. They can be intuitively thought of as directed areas. A few sources that talk about them online are

http://phy1139-862.rit.edu/blog/mathematics/index.php
http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node4.html#SECTION00022000000000000000
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 bi-vector aka a 2-form. 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. Mar 1, 2005

### genxhis

thanks. the middle website gave a nice, rough overview of generalized (?) vectors.

i went to the library and read the first article (in the American Journal of Physics). it purports that the solution to the problem is in fact very simple. any partition of the lever into small pieces means each piece must have a balanced set of forces acting on it to remain at rest. but a set of forces balanced in one inertial frame is also balanced in any other (though in general their orientations and magnitudes may change). to resolve the paradox, just transform the internal forces along with the external ones. in addition to the net external torque, there will be a net internal torque which will precisely cancel the former. classically however we rarely observe a net internal torque and i would presume this arises from the low speeds of the bodies involved. still, that begs the question precisely why that is?

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.

Last edited: Mar 1, 2005
18. Mar 1, 2005

### RandallB

But the force is already well defined within it own reference frame, and those forces will directly account for 100% of the deflections - within that frame.

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

19. Mar 1, 2005

### pervect

Staff Emeritus
Well, I was having some problems even visualizing the internal forces on the system, unitl I hit on the expedient of adding a "brace" from one end of the L shaped rod to the other, making it a triangle, and making the all the hinge points a pviot joint.

I think we can get rid of anything but tension and compression forces this way, i.e. we can analyze the problem as beams in compression and tension, and get rid of all the torsion forces at the joints (which can rotate freely). A real beam might have a tendency to buckle, but I think we can idealize that out of the problem.

So we should wind up with a nice loop of forces. I haven't really worked the problem out in detail, though.

20. Mar 1, 2005

### pervect

Staff Emeritus
Here's the basic issue, as I see it. In the rest frame, none of the forces are doing any work - since the velocity is zero, force*velocity is zero. In the moving frame, this is not the case - some of the forces ARE doing work. (There is no net work done on the system as a whole - but the forces that point in the direction that the system is moving are doing work on parts of the system, work that is cancelled out by the opposing forces doing negative work on other parts of the system).

Energy and momentum interconvert, just like time and space do. So the work that these forces are doing in the moving frame shows up as momentum in the rest frame, and vica-versa. This is why I think the 4-space formalism is needed.

Handwaving, since work is being done on one part of the system, and is being extracted from another part, there must be a flow of energy through the system in the moving frame. This "flow" has to be accounted for when calculating the total system momentum / angular momentum.

Last edited: Mar 1, 2005