0
$\begingroup$

I was wondering exactly how one would go about deriving an equation to describe the movement of a pendulum from the Schwarzschild metric. I have no real mathematical tools at the moment to actually go about this but even then, how would one deal with this problem.

I happened to have this thought with me for a while as I have known about solutions to a pendulums movement under newtonian gravitational fields where the force or acceleration is constant. As well as trying to investigate now, through very rough approximations through geometry and position versus time graphs, what a pendulums movement would be like from a point mass considering the changes in acceleration. All of this without calculus, I have not gotten to there yet, another year i'm afraid and a little bit more persistance on my part.

But what noticeable or un-noticeable difference would come about from applying the Schwarzschild metric to this situation?

$\endgroup$
1
$\begingroup$

Before we can even think of trying to derive it from the Schwarzschild metric, why not first have a go at it with Newton's law of universal gravitation in its full form, that is, using classical mechanics to its fullest extend before we pull in modern physics? I'd think one is rather hastily skipping a step here ... That will already be much closer to the Schwarzschild metric than the usual pendulum equation is which is based on the idea of a uniform gravitational field, and thus while it doesn't go all the way there, it nonetheless goes part way and so will provide a good deal of the insights.

So this won't directly answer your question as phrased, but I think it will, much better, answer to the gist of what you're after. Remember that in the Newtonian mechanics you're thinking of, the gravity field is being approximated as uniform - the immediate next step up from there is still to use Newtonian mechanics, but to now drop the uniformity, and switch for the full Newton's law of universal gravitation, since it seems what you're really interested in is the behavior in a more realistic gravitational field taking into account that the field of a real gravitator like the Earth is not uniform: you say "what a pendulums movement would be like from a point mass considering the changes in acceleration." In that case, general relativity is quite overkill if we're talking anything close to an everyday situation.

In that case, we would start with that the pendulum, when hanging straight down, where the pivot is at some distance $h$ from a gravitating center of mass $M$ and the pendulum has length $L$. Then the force on the pendulum is

$$\mathbf{F} = -\frac{GMm}{||\mathbf{r}||^2} \hat{\mathbf{r}}$$

where $m$ is the mass of the pendulum. However, analyzing this in this form is difficult, so to make the problem easier we will use the Lagrangian formalism. The potential energy is

$$U = -\frac{GMm}{||\mathbf{r}||}$$

and the kinetic energy is

$$K = \frac{1}{2} m \dot{\mathbf{r}}^2$$

so

$$\mathfrak{L} = K - U = \frac{1}{2} m \dot{\mathbf{r}}^2 + \frac{GMm}{||\mathbf{r}||}$$

It is possible to determine with geometry that

$$||\mathbf{r}|| = (L^2 + h^2) - 2Lh \cos(\theta)$$

and furthermore we have

$$\begin{align} \dot{\mathbf{r}}^2 &= [L \cos(\theta) \dot{\theta}]^2 + [L \sin(\theta) \dot{\theta}]^2\\ &= L^2 \dot{\theta}^2 \end{align}$$

Thus

$$\mathfrak{L} = \frac{1}{2} m L^2 \dot{\theta}^2 + \frac{GMm}{\sqrt{(L^2 + h^2) - 2Lh \cos(\theta)}}$$

is the system Lagrangian in the single angular coordinate $\theta$. Now we can set up the Euler-Lagrange equation

$$\frac{\partial \mathfrak{L}}{\partial \theta} = \frac{d}{dt} \frac{\partial \mathfrak{L}}{\partial \dot{\theta}}$$

And we get

$$\frac{\partial \mathfrak{L}}{\partial \theta} = -\frac{GMm}{2} \left[(L^2 + h^2) - 2Lh \cos(\theta)\right]^{-3/2} \cdot [2Lh \sin(\theta)]$$

$$\frac{\partial \mathfrak{L}}{\partial \dot{\theta}} = mL^2 \dot{\theta}$$

with the last equation being recognized as the angular momentum $I \omega$ of the point-mass bob. Thus the full equation of motion is


$$mL^2 \ddot{\theta} = -\frac{GMm}{2} \frac{2Lh \sin(\theta)}{[(L^2 + h^2) - 2Lh \cos(\theta)]^{3/2}}$$


This is the equation of motion of a pendulum in a Newtonian spherically symmetric gravitational potential. As you can see, it is considerably worse than the equation for the pendulum in a uniform field, and is definitely not solvable analytically. We should, however, check to see if it makes sense. If we take the pendulum very very small, that is, $L << h$, we get that $L^2 \approx 0$ and $2Lh \approx 0$ as well so the denominator is practically just $h^3$ and thus

$$mL^2 \ddot{\theta} = -\frac{GMm}{2} \frac{2L}{h^2} \sin(\theta)$$

$$mL^2 \ddot{\theta} = -\frac{GMm}{h^2} L \sin(\theta)$$

If you eliminate the mass from both sides, divide by $L$, and recognize that $g(h) = -\frac{GM}{h^2}$, you get

$$L \ddot{\theta} = -g \sin(\theta)$$

or

$$\ddot{\theta} = -\frac{g}{L} \sin(\theta)$$

which is the usual equation for a pendulum; thus we can be confident in our derivation.

To attempt to understand the equation, as said, we cannot solve it analytically any more than the simpler pendulum one (arguably in some sense this should be "even worse"!), but we can nonetheless do the same trick where we consider the small-angle approximation with $\theta \approx 0$, and thus $\sin(\theta) \approx \theta$ and $\cos(\theta) \approx 1$, which causes the equation to simplify to

$$mL^2 \ddot{\theta} = -\frac{GMm}{2} \frac{2Lh \theta}{[L^2 - 2Lh + h^2]^{3/2}}$$

which then becomes, recognizing that $(L^2 - 2Lh + h^2) = (h - L)^2$ (if we use $(L - h)^2$ we will get bad results below in that the square root will have a negative input in realistic situations, so that is not an accident to make this (equivalent) choice),

$$mL^2 \ddot{\theta} = -\frac{GMm}{2} \frac{2Lh}{(h - L)^3} \theta$$

and cancelling and collecting everything, we get

$$m \ddot{\theta} = -\left[\frac{GMm}{L} \frac{h}{(h - L)^3}\right] \theta$$

and thus taking this in the form $m\ddot{\theta} = -k\theta$ for the simple harmonic oscillator we see we again have simple harmonic motion but now with angular frequency

$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{GMh}{L(h - L)^3}}$$

versus the usual

$$\omega = \sqrt{\frac{g}{L}}$$.

Again, note that if $L << h$ you get $(h - L)^3 \approx h^3$ and the latter equation is recovered just same with recognition that $g = \frac{GM}{h^2}$. Indeed, using this relation for $g$ at the altitude $h$ above the gravitator we can get the more comparatively useful form

$$\omega = \sqrt{\frac{g}{L}} \left[\frac{h}{h - L}\right]^{3/2}$$

For $h > L$ the term on the right is easily seen to be larger than $1$, thus the effect to first order is that for very small oscillations the non-uniform Newtonian gravitational field causes the pendulum's frequency of oscillation to increase. This should make intuitive sense: near the bottom the gravitational force on the bob is stronger as it's closer to the gravitating mass, so there is more force "preferring" it point down and thus will want to wiggle around more vigorously near there. You can think of it as being "pinched" by the gravitational field lines near the bottom of its swing.

As a benchmark of the effect, consider near the Earth's surface with a pendulum of length $L = 1\ \mathrm{m}$ (was perhaps the original inspiration for the metre unit) and $h = 6371\ \mathrm{km}$. With these values you can easily figure the deviational term $\left[\frac{h}{h - L}\right]^{3/2}$ as about 1.000000235. So the pendulum's frequency (about, but not quite, 1 Hz - note that while we gave the above in angular frequency $\omega$, this is directly proportional to frequency, so we need not convert anything for the amplification term) is only slightly higher by about 235 parts per billion due to this correction. Arguably, we could expect that even more complex inhomogeneities in the field due to the Earth not being a uniform sphere but a whole, complex planet, will be even worse. These will, however, require a supercomputer loaded with a very accurate model of the Earth, to calculate to any precision, and will depend strongly on precisely where on Earth you are seeking to envision a pendulum (and it's for this reason that the pendulum was not used as a standard to define the metre, but instead the rather coincidentally similar 1/10,000,000 of a meridian arc from the Equator to the North Pole was used instead, as this has rather less deviation.).

I do not have enough expertise yet to do a full mathematical workup in general relativity as you ask for, but I figured I'd post this answer because it seems what you'd be after and at least answers it part-way, inasmuch as Newton's law is a better approximation of the real situation. But intuitively I'd guess from the non-linear effect that it will show an even greater worsening of the angular frequency near the equilibrium point.

$\endgroup$
  • $\begingroup$ Thank you, thank you, I can follow it almost hand in hand. . . almost. But this was extremely well put together and exactly what I was looking for with respect to part of the question. The reason for this interest in the "over kill" that is using the Schwarzschild metric is mainly that interest has intensely sparked within me regarding differences between what newtonian mechanics predicts and what relativistic mechanics predicts instead. $\endgroup$ – The victorious truther Apr 3 '18 at 11:26
  • $\begingroup$ Even then, a long time ago, I had come across a video of a relativistic crackpot who thought that relativity was nonsense, you know. . . that type, and he was talking about gravitational time dilation where on he thought he was clever in trying to disprove it with a pendulum and sand clocks. I know, this sounds terrible, but it did get me thinking as to whether you could actually test the existence of gravitational time dilation by noting experimental differences between what a newtonian pendulum clock would predict even in un-ideal situations as you had laid out and what the Schwarzschild. . $\endgroup$ – The victorious truther Apr 3 '18 at 11:31
  • $\begingroup$ . . . metric as well as other assorted solutions to general relativity, Kerr, Kerr-Newman, and Reissner-Nordström, would have in terms of subtle, unnoticeable, minute, or actually feasible measurable changes. $\endgroup$ – The victorious truther Apr 3 '18 at 11:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.