In classical mechanics, the period $T$ of a pendulum is given by $$ T = 2\pi\sqrt{\frac{l}{g}},$$ where $g$ is the gravitational field and $l$ the length of the rope attaching the bob to the pivot. The formula is only valid for small angles, I know. But let's ignore that for the time being.

Since the period only depends on $l$, its velocity will adjust itself such that it can cover the amplitude $A$ (taken as the total arc length covered in one cycle) in $T$. In other words, wherever the pendulum is dropped from, the velocity $v$ will be such that $$\int_0^T v(t)\,dt = A.$$

Now: surely I can find an $l$ and an $A$ such that the speed of the pendulum, at some $t$, is greater than $c$, the speed of light.

I guess the solution to this 'paradox' is that we are only dealing with classical mechanics. But how would we go on about setting up the full, relativistic solution?

  • 1
    $\begingroup$ Use the relativistic Lagrangian $\endgroup$ – Phoenix87 Jul 14 '15 at 22:57
  • 5
    $\begingroup$ You want to ignore the fact that your using an approximation in a already approximate theory (classical mechanics)... $\endgroup$ – Zach466920 Jul 14 '15 at 23:01
  • $\begingroup$ @Phoenix87 How do I find the period though? $\endgroup$ – SuperCiocia Jul 14 '15 at 23:49

To simplify things a bit assume small oscillations and a point-like mass. The relativistic Lagrangian for the 1-dimensional case is $$L = -\frac{mc^2}\gamma - \frac12kx^2.$$ The equation of motion turns out to be $$\ddot x + \frac13\frac1{c^2-\dot x^2}\frac{\text d}{\text dt}\dot x^3 + \frac{\omega^2}\gamma x=0,$$ where $\omega^2 = \tfrac km$. I haven't really tried to solve this (I'm not even sure this can be solved analytically), but one can give an interpretation of the terms involved. The first together with the third are reminiscent of the classical harmonic motion, only now the frequence, hence the period, depend on the velocity through $\gamma$. The middle term can be interpreted as a damping term. As the speed approaches that of light this damping term diverges and we can make sense of this because of the postulate that a massive body cannot travel at the speed of light or faster. The non relativistic limit is achieved by requiring $|\dot x|\ll c$, whence the damping term becomes negligible, $\gamma\approx 1$ and therefore the equation reduces to $$\ddot x + \omega^2x=0,$$ i.e. the classical SHM.

The two plots on Wolfram Alpha with $c=\omega = 1$ are those of a pendulum at rest with displacement 1 from the equilibrium and that of a pendulum at equilibrium with initial velocity that of light respectively. In the former case oscillations appears, while in the latter the motion is of constant speed ($\dot x(t)=1$ at any time $t$) as expected.

  • $\begingroup$ I am not so sure small oscillation approximation works here, for a high speed means high momentum, its angle then naturally gets larger, the approximation probably will break. $\endgroup$ – Shing Jul 15 '15 at 12:18
  • $\begingroup$ What if the pendulum is very long? $\endgroup$ – Phoenix87 Jul 15 '15 at 12:24
  • $\begingroup$ I see your point now, but since there is no rigid body. I think a better and more interesting question is what happens if we do the same experiment on optics, rather than a Pendulum. $\endgroup$ – Shing Jul 15 '15 at 12:29
  • $\begingroup$ How about a massless spring? $\endgroup$ – Phoenix87 Jul 15 '15 at 12:31
  • $\begingroup$ Personally, I doubt if it is a good approximation for a very long spring to be massless. $\endgroup$ – Shing Jul 15 '15 at 12:33
  • There is no such a thing as rigid body in real world as well as in the theory of special relativity. Hence a very large pendulum is impossible.
  • For a string, Given the momentum of the pendulum has, small oscillation is impossible.
  • For very large string, depends on how large it is, special relativity is defined on small scale, in a very large (~edge of observable universe) scale, it is not well defined, one should not be surprised even if it is faster than speed of light.

To sum it up: in a nice scale (very roughly from the size of atom to galaxy, I am not exactly sure), your argument of faster than light does not stand, as there is no sounding reasons to support the period $T\propto l^{\frac {1}{2}}$

1.) Even if it is hang by a string, then it period surely will look completely different as well, as high speed means high momentum, then the motion must come with a larger angle, hence the small angle approximation breaks down.

2.) However, one way to solve it "logically" is to assume the $l$ will become shorter as it oscillates, but personally I don't see any feasible way to test it experimentally (maybe you will have to do the e optics, if you want to do experiment about it).

3.) Personally, I think it is more interesting to ask what will happens if we do experiment with a beam of light instead of pendulum .


There are two important reasons why this is impossible. For one, it is impractical. The gravitational pull would have to be so extreme that it would break the material used for the rope, or the length would be so long that it is under a weaker and weaker force of gravity as it gets farther away.

Even if you had an unbreakable rope and an enormous mass very close by that doesn't destroy the pendulum with its gravity, there is still a problem. The mass on the end of the string would never move faster than the speed of light. It would get closer and closer to the speed of light as you observe its motion. But, instead of accelerating at the rate you would expect, instead its mass begins to grow and so the acceleration of that mass only makes it go a little bit faster. In fact as it approaches the speed of light its mass goes toward infinity.


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