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- Homework Statement
- How to calculate the period of a simple pendulum in lab reference frame and the pendulum frame if special relativistic effect is considered? This is a very silly question, it is just for the purpose of understanding. Imagine we have a simple pendulum with a watch as pendulum bob. There is another watch initially synchronized with the watch on the pendulum. The rest mass of the watch is ##m## and the rest length of the pendulum is ##l##. The question is in which frame (lab versus pendulum) the reading is ##2 \pi \sqrt{ \frac{l}{g}}## for one oscillation, or neither frame will have that reading? Can we find the reading on each watch?

- Relevant Equations
- $$2 \pi \sqrt{ \frac{l}{g}}$$

While not exactly correct, we will continue to use Newtonian gravitational force and tension force in the lab frame. We will not concern ourselves with GR, besides the approximation is reasonable for low velocity and small mass.

In the lab frame, the forces acting on the pendulum is weight and tension, in an instantaneous tangential/radial frame,

$$\gamma m g \sin \theta = \gamma^3 m a_t $$

where ##a_t## is the tangential acceleration in the rest frame.

It's not trivial to find the speed of the pendulum in the lab frame because the mass of the pendulum keeps changing as it falls, note that the pendulum lengths remains ##l##,

$$ \gamma mg l \cos \theta - (\gamma - d (\gamma) ) m g l \cos (\theta + d \theta) = d \gamma m c^2 $$

It can be seen that neither frame can have the ##2 \pi \sqrt{ \frac{l}{g}}## result. Let me know if I made any mistake in the arguments or equations. Thanks,

In the lab frame, the forces acting on the pendulum is weight and tension, in an instantaneous tangential/radial frame,

$$\gamma m g \sin \theta = \gamma^3 m a_t $$

where ##a_t## is the tangential acceleration in the rest frame.

It's not trivial to find the speed of the pendulum in the lab frame because the mass of the pendulum keeps changing as it falls, note that the pendulum lengths remains ##l##,

$$ \gamma mg l \cos \theta - (\gamma - d (\gamma) ) m g l \cos (\theta + d \theta) = d \gamma m c^2 $$

It can be seen that neither frame can have the ##2 \pi \sqrt{ \frac{l}{g}}## result. Let me know if I made any mistake in the arguments or equations. Thanks,

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