# Right-angle lever paradox in special relativity

I remember to have read somewhere an interesting special relativity "paradox" considering two perpendicular rods $A$ and $B$ of equal proper length $L$ fixed at point $O$.

In the "rest" frame equal (in magnitude) forces are applied on the edges of the rods, perpendicular to them, thus giving zero net angular force: $$-F_AL_A + F_BL_B = -FL + FL = 0$$ (assuming $z$-axis is towards us).

In the frame moving parallel to rod $B$ with the velocity $v$ it's length is diminished due to Lorentz-contraction $$L_B' = L \sqrt{1-v^2/c^2}$$ and perpendicular component of the force is diminished also $$F_B' = F\sqrt{1-v^2/c^2}$$ while $L_A' = L$ and $F_A'=F$.

So the net angular force in this frame is not zero: $$-F_A'L_A' + F_B'L_B' = -FL + FL (1-v^2/c^2) = -\frac{v^2}{c^2} FL$$ and the system should be rotating clockwise with angular acceleration.

I don't remember the explaination of the paradox as I did not understand it at that time, but now it seems to be very simple: even in Newtonian Mechanics total angular force would not be zero in "moving" frame given the fact that the net force is not zero. If we take into account that rods are fixed relative to $O$, we should assume that in the junction point $O$ there is an additional force which compensates the forces on the edges of rods.

The same argument seems to be valid in the relativistic case too.

However, I remember that the explaination was not that simple. Am I wrong with my explaination, or the statement of the paradox is wrong? Does anyone know similar "paradox"?

• – user4552
Commented May 26, 2013 at 15:54
• – user4552
Commented May 26, 2013 at 16:06
• As the answer by @dmkeee pints out, it is not the force $F_B$ that gets contracted by the force $F_A$, this is what resolves the paradox. Commented Aug 16, 2017 at 9:41
• @MikaelFremling No, $F_B$ is contracted. $(F_x, F_y, F_z)$ are not spacial components of a 4-vector.
– xaxa
Commented Aug 16, 2017 at 14:14

To begin with, we should calculate the torque relative to the origin of the observer frame. We first need to add the diagonal force that keeps the lever in force balance down in the corner. The moment arm of the green force is then $$tv + \gamma^{-1}L$$ so the torque from the green force is $$\gamma^{-1}F (tv + \gamma^{-1}L)$$. But the $$tv$$ part will be cancelled by the moment from the balancing force in the corner, $$-\gamma^{-1}F \, tv$$. Therefore the total torque on the lever will still be $$\tau = -FLv^2/c^2.$$

Now to the solution of the paradox.

The red force acting in the direction of movement generates a power (energy per time unit) $$P = Fv.$$ This energy then flows along the red beam down to the corner where it will be "absorbed" by the force that keeps the lever in force balance.

According to the well-known equation $$E=mc^2$$ the flowing energy corresponds to a flow of mass, $$\mu = P/c^2 = Fv/c^2.$$ This flow corresponds to a momentum $$p = \mu L = FvL/c^2$$ (to be motivated later). Therefore there is an angular momentum $$J = -p \, tv = -Fv^2tL/c^2.$$ The time derivative of this is $$dJ/dt = -Fv^2L/c^2 = \tau.$$

So why does a mass flow $$\mu = dm/dt$$ along a line of length $$L$$ correspond to a momentum $$p=\mu L$$? Think of a tube of length $$L$$ and section area $$A,$$ containing a fluid of density $$\rho$$ flowing with speed $$u$$ such that through any section we have a mass flow $$\mu$$. Then $$\rho A u = \mu$$. But the total mass in the tube is $$m = \rho A L$$ and this mass has momentum $$p = mu = (\rho A L) u = (\rho A u) L = \mu L.$$

This answer is partly due to Fremlin (1969):

First, $$F_x'= F_x$$ $$F_y'= F_y/\gamma$$ $$L_x'=L_x/\gamma$$ $$L_y'=L_y$$ $$\delta\theta'=\mathrm{cos}^{-1}\left[ \frac{\mathrm{cos}\ \delta\theta -\beta}{1-\beta \ \mathrm{cos}\ \delta\theta }\right]$$

Apply the principle of virtual work to the lever, that is, imagine a virtual angular displacement $$\delta\theta$$.

In the lever's rest frame (Fig. a), work $$FL\delta\theta$$ is done on the lever at B. At C, work $$FL\delta\theta$$ is done by the lever. The net virtual work is zero, and the lever is in equilibrium.

In the moving frame (Fig. b) work $$FL\delta\theta'/\gamma$$ is done on the lever at B. At C, work $$FL\delta\theta'/\gamma$$ is done by the lever. The net virtual work is zero, and the lever is also in equilibrium.

The question is whether the "forces applied on the edges of the rods" are the ones that cause the relativistic movement?

If these rods are in a frame whose movement is caused by yet another force, than your explanation is not correct. So the two forces acting on the two rods can be in balance, and still there can be a third force that produces the movement of the whole frame.

However, if the movement of the frame is caused by the imbalance of the two forces (one being larger than the other) acting on the rods, than you are right - the net force is also not zero in the rest frame as well.

Hmmm, my relativity is weak, but...

In the usual Newtonian formulation of physics forces are vector quantities which suggests that in the relativistic formulation they should be the spacial part of the Lorentz four-vector and should transform as such. Further looking at Newton's second law we see force expressed as a time-derivative of momentum, which is to say as a Lorentz vector.1

So---sticking to the mixed formulation in which the the questions is framed---while the length of the arm draw green is contracted by the relative motion (reducing the torque by a factor of gamma), the strength of the force acting on the red arm should also be contracted (again reducing the torque by a factor of gamma). The result is then no net torque and no tendency to rotate. "Paradox" resolved.

It would be better to re-set the problem in a fully covariant way, but the math is typically a bit intimidating for beginners.

• This isn’t correct, the force transformation is already accounted for in the question, and it makes the paradox worse, not better. Commented Jul 19, 2021 at 18:44