Does a pendulum violate general relativity?

Question

General relativity asserts that gravity slows down clocks. The greater the gravitational field, the slower the clock. But taking into account the formula of the pendulum, i.e. $$T=2π\times \sqrt{L/g}$$ we see that if gravity gets stronger the pendulum (which is the simplest clock) gets faster. We consider the formula in its domain with g≠0.

Answer 1

This question is sort of interesting. There are two competing phenomena. One is the increased frequency of a pendulum with gravity and the other is the time dilation of such a pendulum as seen by a distant observer.

Let us look at the surface gravity. The Schwarzschild metric has the Killing vector $\xi_t~=~\sqrt{1~-~2m/r}$, $m~=~GM/c^2$. The gravity on a stationary surface held at a fixed radius is given by $$ g^2~=~(\nabla_r\xi_t)(\nabla^r\xi^t)~=~\frac{1}{4}\frac{1}{(1~-~2m/r)^2}\frac{4m^2}{r^4} $$ or $$ g~=~\frac{1}{1~-~2m/r}\frac{m}{r^2}~=~\frac{m}{r^2~-~2mr}. $$ What this says is that for $r$ large this is $g~=~GM/r^2$, which is the Newtonian result. For $r~\rightarrow~2m$, up to the event horizon of a black hole, the surface gravity $g~\rightarrow~\infty$.

Using this in a standard formula for a pendulum of length $\ell$ it indicates that the periodicity is $$ T~=~2\pi\sqrt{\frac{\ell(r^2~-~2mr)}{m}}. $$ The periodicity $T~\rightarrow~0$ as $r~\rightarrow~2m$. This means the frequency diverges. So a pendulum held fixed on a constant surface just above the event horizon has a huge frequency.

Now let us ponder what a distant observer witnesses. The time dilation of the periodicity will be simply $$ T'~=~\frac{1}{1~-~2m/r}T~=~2\pi\sqrt{\frac{\ell r^3}{m(r~-~2m)}}T. $$ The time dilation then means the distant observer will witness the pendulum with a huge periodicity very close to the event horizon. So the frequency $\nu~\rightarrow~0$. This means the pendulum will appear frozen when held fixed near the event horizon. Again, this holds of course for the pendulum on some fixed radius above the horizon and not for a freely falling one.

Answer 2

You are wrongly confusing two separate effects, namely the curvature of spacetime and the effect that curvature has on the performance of a device that measures time.

Answer 3

You are confusing two notions which are totally different :

• the gravitational potential : the deeper into a (negative) gravitational potential, the slower the clock. Whatever method you use to measure the time, all clocks there will measure the same time. You are right in as much as all clocks will be slower there than clocks very far from the bottom of the potential, but this will be the case for all clocks, whether you use a pendulum, an old fashioned spring-based watch, or a quartz watch. The unit of the gravitational potential is $m^2/s^2$, namely the square of a velocity.

• the gravitational field which is the gradient of the gravitational potential, and is usually noted $g$. The unit of the gravitational field is $m/s^2$, an acceleration.

And if you use a pendulum as your clock, indeed, it will oscillate faster in a stronger field. Note that the frequency also depends on its length so anyway it is not just the number of oscillations that determines the time that has elapsed. You get the time by multiplying the number of oscillations by the correct period. So a longer pendulum oscillates more slowly than a short one, but they measure the same time, if you do it correctly. So the fact that the period is shorter in a stronger field has nothing to do with the measure of time. You take both the length of the pendulum and the strength of the field into account when using your oscillating pendulum as a time measuring apparatus.

There is no direct relationship between gravitational potential and gravitational field. For instance the gravitational potential due to the Sun to an object on Earth orbit is about 14 times the one due to the Earth on an object on Earth surface. That means that an object on Earth surface has in fact a total negative gravitational potential 15 times "deeper". The "escape velocity" from gravitational potential of Earth is 11,2 km/s to overcome a potential of about $-125 (km/s)^2$: ignoring the friction of the atmosphere, a projectile sent at that speed will not fall back on Earth. But it will fall back on the Sun. To escape from the solar system one should send it at about 43 km/s to overcome a potential 15 times deeper (the square of 43 being about 15 times that of 11.2).

Contrariwise, the gravitational field of the Sun on the surface of the Earth is a paltry 0,06% of that of the Earth ! The reason is that the Sun being so far away compared to the center of the Earth, its gravitational potential varies very little with the position. Therefore, its gradient (the field) is much less than the gradient of the smaller gravitational potential of the Earth which varies much more rapidly with the distance from center of the Earth.

Answer 4

A rod is a pendulum. Is a rod a clock? No, I don't think so.

A pendulum clock is a clock. Do I need to justify that claim?

From the above we conclude that a pendulum is not a pendulum clock, and a pendulum clock is not a pendulum.

Now the question arises what is a pendulum clock?

Well, I would say that its a rod and a gravitating mass. And some support structure that keeps the rod and the mass separated.

This pendulum clock is such that it gets seriously messed up by nearby extra masses, it speeds up a lot or slows down a lot.

But if extra masses are placed symmetrically around the pendulum clock, then the clock slows down slightly as as predicted by general relativity. General relativity says that masses have that effect on clocks. That by the way is more correct statement than the "gravity has effect on clocks".

Answer 5

I believe you could see it as such, that the pendulum is capable of measuring 'time' in the configuration space, but is not a valid timekeeper in the spacetime manifold, which has 'time' as the fourth dimension in its tensors. These two 'times' are not congruent and contradicting, as you rightfully identified. This is exactly the problem between quantum mechanics (configuration space/ state space to be more precise) and the spacetime manifold. They haven't solved this, as this gets near the problem of quantum gravity.