# A new angle on the Right Angle Lever Paradox.

1. Jan 5, 2008

### yuiop

This paradox has long been a thorn in the side of relativity. In 1909 Lewis and Tolman analysed the behaviour of a right angled lever with balanced forces with relative motion. They noticed there would be a potential imbalance as the arm parallel to the direction of motion would length contract. They proposed that the force transverse to the direction of motion should increase to compensate for the length contracting arm. Later it was shown that the transformation of force does exactly the opposite and the force actually decreases, making it difficult to explain why if the lever does not rotate in the rest frame, why does it not rotate in the moving frame? So the lever paradox was born.

Proposed solutions to date include a energy flow into the lever that counteracts the apparent imbalance of a right angle lever with relative motion (The Von Laue energy current) and an “internal torque” that exactly balances the external torque.

http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000043000007000615000001&idtype=cvips&gifs=yes [Broken]

http://panda.unm.edu/Courses/Finley/P495/TermPapers/relangmom.pdf [Broken]

Another proposed solution is that static forces transform in a different way to dynamic forces. http://www.springerlink.com/content/n6573281634614q2/

Einstein did not help when he stated relativity is not about statics but about time and motion.

To the layperson these solutions are non intuitive and appear a bit too convenient and ad hoc. In fact a search on the internet reveals that this paradox has lead many to believe there is something wrong with the force transformations or with relativity itself.

An example is this paper http://www.wbabin.net/science/ricker12.pdf that concludes that the accepted Lorentz transformation for transverse force is wrong.

Early in the paper the author substitutes x=0 in the Lorentz transformation $$t\prime =(t-vx/c^2)/ \sqrt{1-v^2/c^2}$$ to obtain $$t =t\prime\sqrt{1-v^2/c^2}$$. Clearly a big mistake. If x is set to zero, then there is no motion in the unprimed frame and v can only take the value zero. His result is also at odds with time dilation measured in experiments.

I would like to propose a new solution to the Right Angle Lever paradox that I feel is more intuitive and odes not require tampering with any of the classical force transformations of Special Relativity.

The new solution requires an introduction of a concept I call a geared lever.

The geared lever is here defined as a gear box with two equal length levers connected to the input and output shafts of the gear box. The gears are hidden in a “black box” that cannot be opened and we observe the behaviour of the input and output levers when forces are applied.

Li = length of input arm.
Lo = Length of output arm.
Ai = Angular displacement of input arm.
Ao = Angular displacement of output arm.
Di = Tangential displacement of input arm.
Do = Tangential displacement of output arm.

MA = Mechanical advantage (Output force/input force)

Classic lever

A) Li/Lo = MA
B) Ai/Ao = 1
C) Di/Do = MA

Geared lever

A) Li/Lo = 1
B) Ai/Ao = MA
C) Di/Do = MA

From the above it can be seen that only the ratio Di/Do can be applied generally to both classical levers and geared levers to obtain the correct mechanical advantage. When it is realised that the right angle lever in relative motion behaves as a geared lever, the toques are balanced and there is no paradox. When the right angle lever is rotated by a small angle in the rest frame of the lever, the tip of the transverse arm moves less than the tip of the arm parallel to the motion. It turns out that the distance moved by the tip of the transverse arm is less by $$\sqrt{1-v^2/c^2}$$ and the mechanical advantage of the lever due to the ratio of the tip displacements exactly compensates for the reduced transverse force applied to the parallel lever. The distance moved by the transverse tip is effectively length contracted. (By definition of the geared lever law, the lengths of the arms is irrelevant as far as mechanical advantage is concerned.)

In the relativistic right angle lever the reduced displacement of the transverse arm and the reduced force acting on the parallel arm means energy is conserved since one definition of energy is force multiplied by the distance the force acts through.

It should also be noted that the when the right angle lever is rotated by some small angle in the rest frame that the two arms rotate through different angles in the moving frame and the angle between the arms does not remain a right angle. I hope I have shown that the relativistic right angle lever has more in common with a "geared lever" than a classical lever and that the ratio of the tip displacements is a more generally applicable “law of the lever” than the simple ratio of the lever arm lengths.

Last edited by a moderator: May 3, 2017
2. Jan 6, 2008

### Mentz114

Hi kev,
You are not being totally ignored. I've read this and I'm thinking about it.

M

3. Jan 6, 2008

### yuiop

Thanks Mentz

I am aware of some potential objections and I'm waiting for someone to raise them

4. Jan 7, 2008

### pervect

Staff Emeritus

My end conclusion was that internal forces can contribute to the angular momentum of a system if the forces are not central forces. When you do the Lorentz transform, forces that are central forces in one frame may not be central in another, and in such frames where the forces are non-central, they contribute a term to the angular momentum.

See in particular https://www.physicsforums.com/showpost.php?p=477970&postcount=23

5. Jan 7, 2008

### Staff: Mentor

Is there a Minkowski "cross product"? I would think that such a construct would be pretty important for doing problems like this. Also, is there a Lorentz-invariant angular momentum and torque?

Last edited: Jan 7, 2008
6. Jan 7, 2008

### yuiop

Hi Pervect,

I had a look at the image you posted in the other thread. The direction of motion is not clear from the image and the two atom get further apart in the moving frame but reading between the lines the moving frame is going down the page rather than the usual left to right. I have taken the liberty to add in the lever outline and diagonal cross beam to your diagram and indeed it appears that the reaction force inside the cross beam appears to create an internal torque in the moving frame compensating for the imbalanced external torque. Unfortunately that is not the whole picture and the diagram fails to take account of all the forces and glosses over the fact that the compression forces (that are not shown) also rotate when transformed and the end result there is no net torque in the diagonal cross beam that can compensate for the inbalance of the external torque.

I have created a second diagram showing all the forces involved. The dashed vectors are forces within the springs connected to the right angle lever. These forces are resolved into compression forces on the cross beam and tension forces on the lever arms (blue vectors) and torque forces (red vectors). The reaction forces are represented by the green vectors. When all these forces are transformed in the moving frame (moving left to right in the diagram) it can be seen that all the compression and tension forces remain in balance with their respective reaction forces and the the main torque forces (red vectors) are still out of balance in the moving frame. I have adhered to the guidelines you mentioned that spatial dimensions contract in the direction parallel to the motion of the frame while forces reduce in the plane tangential to the motion of the frame. When all the forces are taken into account, internal torque seems to fail as an explanation as to why the lever does not rotate in the moving frame.

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7. Jan 8, 2008

### pervect

Staff Emeritus
I think you're missing the point I'm trying to make, or maybe I'm just not explaining it well.

If you look at the conservation law for the angular momentum of a system of particles in a standard mechanics text (for instance, Goldstein, Classical Mechanics, pg 6-7) for the conservation laws of angular momentum of a system of particles, you'll see that it is necessary for the forces between particles to lie along the line joining them.

Specifically, *assuming* the above condition is true (that the forces are directed in a line between the particles) and assuming also that the forces are equal and opposite (Newton's first law), it can be shown that the time derivative of the angular momentum of the system of particles is equal to the moment of the external force. This is done in the textbook I mentioned, for instance.

The necessary pre-condition of the force lying along the line connecting the particles is, however, not met in the moving frame. The theorem above fails, and there is now a contribution to the expression of the time rate of change of angular momentum of the system due to internal forces, as well as external forces.

8. Jan 8, 2008

### yuiop

I guess we are missing each others points. ;)

If we take two charged particles with the same sign of charge then they repel each other. If we take the two particles in your drawing the like charged particles go away from each other and follow a straight path that I have shown as as two parallel lines in the rest frame on the left of the attached drawing. The drawing on the right of the image is moving frame showing the transformed path and the transformed forces. It can be seen that the transformed forces would cause the particles to follow a curved path (shown in red) in the moving frame. Obviously that does not happen or this would be a method to detect absolute motion.

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9. Jan 8, 2008

### Mentz114

A simple take on it

I've tried to model this with real forces, provided by tension springs attached to the lever. All the materials are assumed to be rigid enough not to deform under the forces, and the pivot is frictionless. Spring constants K are the same in all frames (hmmm).

The lever is blue, attached to the pink material at the pivot. The green lines are tension springs with equations $$f_i = K_id_i$$.
In equilibrium we have -

$$K_1d_1r_1 = K_2d_2r_2$$

If we boost in the direction of the arrow, we have,

$$K_1\gamma d_1r_1 = K_2d_2\gamma r_2$$

and there's no problem.

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10. Jan 8, 2008

### yuiop

In the moving frame we tend to ignore d as length contraction happens both to the spring and whatever its mounted on. d in classic spring equations is displacement of the spring (tension or compression) from its rest length (No force). When a spring with no force on it in its own rest frame, is observed in a frame moving parallel to its length, it is seen as shorter due to length contraction but it still has no force on it. Basically d is the displacement of the spring from its length with no forces acting on it.

11. Jan 8, 2008

### Mentz114

Hi Kev,
OK, so we can't model it with springs. It sure doesn't work. And my spring equation for the compression case is wrong. But, if we have a spring-based clock, the frequency is given by,

$$\omega^2 = K/m$$. From a boosted frame, the clock is slower so K must decrease - or what ? My assumption that K is the same in all frames is suspect.

12. Jan 8, 2008

### yuiop

It can be quite difficult to visualise why spring forces transform as they do, but it can be quite easy to see how hydraulic forces transform. Assume we have a vessel that consists of two cylinders joined at right angles to each other to form two arms. The far ends of the cylinders have pistons of equal areas attached to springs. THe vessel is filled with water and put under pressure. In the absence of a gravitational field it is known that the pressure everywhere inside the vessel is equal in all directions. The pressure within the cylinder is the same in the rest and moving frames as otherwise we could detect absolute motion simply by attaching pressure gauges to a pressurised vessel.

The force on the pistons is simple the area multiplied by the pressure. It is easy to see that the area of the piston in the arm parallel to the motion is unchanged (A1) and so the force there is unchanged. The area of the piston in the transverse cylinder (A2) is reduced by gamma. The force on that piston is also reduced by a factor of gamma. To maintain equilibrium we have to assume the force in the transverse spring (FT) reduces by gamma and the force in the longitudinal spring (FL) remains unchanged.

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13. Jan 8, 2008

### Mentz114

Kev,
in your hydraulic vs springs, isn't there a decrease in the volume of the container, and therefore an increase in pressure ?

This goes against intuition ( but I'm not denying it might be correct).

Back to my springs - if I write my equations with a proper spring equation,

$$K_1(d_1-d_{01})r_1 = K_2(d_2-d_{02})r_2$$

where $$d_{0i}$$ is the unstressed length of the spring, then the boost leaves the equilibrium if

$$d_2 - d_{02} \rightarrow \gamma d_2 - \gamma d_{02}$$.

So it works if the gap and the unstretched length contracts.

I know, it looks too simple and I expect to get unseated on this. Nice problem, though.

M

14. Jan 8, 2008

### yuiop

This paper mentioned by Pervect in an earlier post may address the points you mention but its certainly non trivial. http://panda.unm.edu/Courses/Finley/P495/TermPapers/relangmom.pdf [Broken]

I like to think the angular momentum is invarient in the transformations from frame to frame in accordance with the principle of conservation of angular momentum. Reading around the topic it not clear that angular momentum is conserved in General Relativity but maybe I am reading the wrong sources.

A crude analysis: A very simple flywheel (two point particles connected by a massless rod) rotating around an axis parallel to the direction of a moving frame. Using the classic formula for angular momentum = m W r^2 the radius(r) is unchanged, the mass (m) increases by gamma but the angular velocity (W) reduces by gamma and so the quantity m W r^2 is conserved. (This glosses over the the fact that the mass is not its rest mass in the rest frame because of the speed of rotation) Similarly the transverse force in the connecting rod reduces by gamma. The classic formula for centripetal acceleration is m W^2 r. Again the radius is unchanged, the mass increases by gamma and the angular velocity is reduced by gamma giving the transformation of centripetal force (in simple terms) as m W^2 r (1-(v/c)^2)^(1/2) which is consistent with the reduced radial force in the connecting rod. The rotational energy in broad terms reduces by aproximately gamma due to the reduced rotation speed. This gives rise to the idle thought that maybe if every elementary particle has an intrinsic angular momentum, then the sum of all the reduced rotational energies in a moving block compensates for the increased energy of the block due to its linear kinetic energy when it has relative motion to conserve energy. As I said, just an idle thought and as far as I know energy is not assumed to be conserved between inertial frames.

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15. Jan 8, 2008

### yuiop

The volume of the container decreases, but so does does the volume of the water by exactly the same amount so there should be no change in pressure.

I'll get back to the rest of your post later (my bed time) but thanks ;)

16. Jan 9, 2008

### Staff: Mentor

Thanks for the paper. Unfortunately I don't know anything about forms (apparently a generalization of the cross product), so I didn't follow the math. However, the discussion was reasonably clear. In appears that the angular momentum is not generalized to a 4-vector, but rather to a 4x4 tensor. This tensor unifies the non-relativistic conservation of angular momentum and the uniformity of motion of the centroid. The angular momentum and the centroid, as components of this tensor, are not invariant. The fact that the centroid is not invariant may be related to the other geometrical concerns brought up by pervect. Interestingly, I gathered that the different hyperplanes of simultaneity in different frames was not a problem, although I didn't understand the reasoning.

Last edited by a moderator: May 3, 2017
17. Jan 9, 2008

### yuiop

Nice observation. Your suspicion that K is not the same in all frames turns out to be correct. In the transverse case the relativistic frequency in the moving frame should be $$\omega^2 = (Km)/y^2$$ where $$y= 1/\sqrt{1-v^2/c^2}$$.

If the transformation of m is m y then the transformation of K(transverse) must be K/y.

It seems you are right here again and the transformation $$d_2 - d_{02} \rightarrow \gamma d_2 - \gamma d_{02}$$ is valid for a spring parallel to the motion and the force seems to reduce by an amount equal to the length contraction of the spring. This bothered me at first but it turns out there is a solution.

First we have to look closer at the definition of the spring constant. The spring constant is defined as K= E A where E is the elastic modulus of the material the spring is made of and A is the cross sectional area of the spring. The spring constant transformations are then:

K(parallel) $$= E A = K_1$$
K(transverse) $$= E A/y = K_2/y$$

because the cross sectional area of the transverse spring is reduced by gamma. This agrees with our observation K must transform as K/y from the frequency calculation.

So if we put it all together we get

$$Force(parallel) = (E A_1) (d_1-d_{01})/y = K_1 (\Delta d_1)/y = F_1/y$$

$$Force(transverse) = (E A_2/y) (d_2-d_{02}) = K_2/y (\Delta d_2) = F_2/y$$ which is not good!

Up to now I have assumed that that the elastic modulus of the spring material (E) is the same in tranverse and parallel directions, but is that true? We have to look again at the harmonic spring equation again, but this time in the direction parallel to the motion. The inertial mass parallel to the direction of motion resists acelleration as if it had a mass of y^3. (Einstein's original derivation of relativistic mass comes in useful here ;) The cross sectional area remains unchanged so E(parallel) must transform to E_1y so that the frequency of spring oscillation reduces by y in both directions (as it must). Now the transformed spring equations are:

$$Force(parallel) = (E_1y A_1) (d_1-d_{01})/y = K_1y (\Delta d_1)/y = F_1$$

$$Force(transverse) = (E_2 A_2/y) (d_2-d_{02}) = K_2/y (\Delta d_2) = F_2/y$$

...which is in agreement with accepted lorentz transformations of force and with the hydraulic example I gave earlier. Thanks to your observations, we now have have a good idea how the classical concepts of spring constant, elastic modulus, cross sectional area and natural frequency of a spring relate to Special Relativity.

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18. Jan 9, 2008

### Mentz114

Hi kev,

that's very satisfactory. So the bulk properties of materials can look different from different frames. I suppose that is to be expected because those properties depend on the configuration of the EM fields holding it together.

But that leaves the 'paradox' untamed. I'll have to think some more ( and if all else fails, read the proposed solutions).

Thanks for running with this, it's a clear piece of analysis.

M

 later. I think there may be a problem. Have a look at the diagram. Here we have no lever but a piece of string. In the rest frame we have equilibrium when f1=f2, but in the boosted frame, the transverse force is reduced, and the pointer will turn anti-clockwise. The shortening of the parallel part of the string is matched by the overall contraction, so that won't affect things. The deformation of the pulley won't help ( I think ).

I leave it to you to point out why I'm wrong.

Looks right to me. Especially the bit I've emphasised. The trouble is I can't think of experiment to test any of this.

Regarding the conservation of angular momentum -
translational momentum is not invariant so I'd guess AM is not. I'll read the paper cited above.

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19. Jan 13, 2008

### yuiop

Hi Mentz, I see what you are getting at.

In the case of a pulley system (say a block sytem with lots of pulleys as seen on sailing ships or cranes) the overall mechanical advantage is measured as the ratio of the distance the input end of the rope moves relative to the distance the output end moves. In line with conservation of energy laws the force(input) * displacement(input) = force(output) * displacement(output). In the relativistic moving frame this relationship is maintained. The reduced tranverse force is always compensated for, by the length contracted displacement of the parallel force.

Any mechanical system made up of a mixture of levers, pulleys and gears always obeys the more generally applicable law (based on energy conservation) that force(input) * displacement(input) = force(output) * displacement(output) in both the classic and relativistic cases.

20. Jan 13, 2008

### Mentz114

Hi Kev,
OK, I'll stop looking for problems with the accepted view. I just set up the potential and field tensor for two repelling charges and did the Lorentz transformation at right angles to the line of the charges. The electric field in the transverse direction is reduced by gamma. Boosting along the line leaves the parallel component unchanged. I'm convinced.

Back to the drawing board.

M