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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*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.

Qmechanic
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mfc
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  • Gravitational time dilation results from accelerating reference frame near a massive object. The period of oscillation of the pendulum decreases with more gravity, but that does not mean the pendulums clock will run differently with respect to some observer in such frame. See: gravitational time dilation on wikipedia – bleuofblue Jan 20 '17 at 19:39
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    @mfc: A pendulum clock doesn't define the passage of time... it measures the passage of time. A pendulum clock is designed to measure time semi-accurately when placed in a certain gravitational field. A pendulum clock will stop when in free fall, but that doesn't mean that time stops. – James Jan 20 '17 at 20:32
  • @James: yes, the pendulum measures the passage of time. We know it works as a clock. If it is not so accurate it doesn't matter. If we cut power also your electronic clock will stop and this doesn't mean time stops. – mfc Jan 20 '17 at 20:37
  • @blueofblue: we can't treat a pendulum differently of another kind of clock. The Pound-Rebka-experiment shows clocks are faster at the highest level of a building since gravity is weaker and for this reason a gravitational redshift was measured. But the pendulum would be slower and would measure a blueshift. So the question remains unanswered. – mfc Jan 20 '17 at 20:46
  • Possible duplicate of Is time dilation based on the formula for period of a pendulum? – James Jan 20 '17 at 20:48
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    A pendulum doesn't actually measure time. It measures the number of cycles it has completed, which we then calculate time from. If you assume $g = 9.81 m/s^2$ and shorthand the equation to a simple x-number-of-cycles-per-minute form for convenience, the shorthand equation no longer holds when $g$ increases. – Devsman Jan 20 '17 at 20:56
  • @James: no, with that question they wondered whether Einstein thought of time dilation by starting from the observation of a pendulum. And they do the same error. We cannot consider g=0 in the pendulum's formula, we can't have denominator = 0, it's basic mathematics. That formula has a domain, within which falls my question. – mfc Jan 20 '17 at 20:59
  • @Devsman: of course no clock directly measures time but ALL clocks perform cycles, movements etc. which we interpret as time. I am not interested here in discussing what time is or not. Pendulum has nothing special about time measuring in comparison with other clocks. Also: what do you mean with "the shorthand equation"? Post it if you wish, thanks: n cycles per minute, as you say, is rather the "result" of the equation not the equation itself. – mfc Jan 20 '17 at 21:05
  • @mfc Yes it has, in that the equation for the period of a pendulum depends on gravity in Newtonian physics. All I meant by a "shorthand equation" (there's no such thing officially) is a rule of thumb calculation like "This pendulum's period is 2 seconds." It depends on the value of $g$ at which it was calculated. – Devsman Jan 20 '17 at 21:09
  • @Devsman: again, you speak of the result given by a formula not of the formula. And all clocks are built according to Newtonian mechanics, as the pendulum... but all are subject to Einstein's time dilation due to a stronger (or weaker) gravitational field. Moreover, there aren't 2 different gravities... only 2 ways to describe it. Newton quantitatively describes it and Einstein adds a qualitative explanation as curved space-time. His field equations start from Newton, you see the constant G. And the stress-energy tensor T_μυ represents, among other things, the action of the quantity "mass". – mfc Jan 20 '17 at 21:24
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    I think your assertion that "we can't treat a pendulum differently of another kind of clock" is wrong. I can make a clock that tells time by counting the number of oxygen molecules that have bounced off it. Maybe it works perfectly well at sea level, but then slows down when I take it up a mountain or put it out in the cold. That doesn't mean time slowed down. Similarly, you chose to measure time in a bad way (with a pendulum) instead of in a good way (with cesium atoms), so you can't expect much. – Jahan Claes Jan 20 '17 at 22:25

2 Answers2

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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.

Lawrence B. Crowell
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  • I don't understand what you want to demonstrate. We know the parallelism with the Schwarzschild solution. Also: general relativity doesn't need event horizons to work and we don't know yet what black holes actually are. ANY clock's period diverges in such situation. So, let's consider the pendulum formula within its domain (g≠0) and without a diverging gravity. A local observer, if the planet increased its density, would measure a faster period (violation!) and a distant observer would measure a period difference inversely proportional to the difference (if there's) in gravity (violation!) – mfc Jan 20 '17 at 21:51
  • I meant (not characters left) the distant observer measures a difference in the period according to the difference in gravity of his reference system with respect to the other one. If ∆g is negative he should measure a slower clock. But for the pendulum this doesn't occur (violation), since the greater is g, the faster is the pendulum. No theory is perfect (e.g. GR doesn't fit in QFT). Gravity changes the weight of a clock's components and "depending on how this fact acts on the specific mechanics of the clock-system", its period can get faster or slower as the pendulum proves. – mfc Jan 20 '17 at 22:10
  • This is a rather canonical general relativity problem. The Killing vector field for the metric predicts the periodicity of oscillation will shorted due to increased gravity. However, a distant observer witnesses this slowed by time dilation. – Lawrence B. Crowell Jan 20 '17 at 22:42
  • we have special relativity where the discourse on two different reference systems comes into play if they move the one with respect to the other. This is not the case. Also, the distance between two reference systems is not important (there may be a time delay for signals from a system to the other, but time delay is not time dilation). So, we come to "general" relativity. Where the distance still doesn't matter. Gravity matter. You have to consider the ∆g. And the problem with the pendulum is still there. If not maybe you want to be clearer with equations thanks. – mfc Jan 20 '17 at 22:52
  • Also consider the Pound-Rebka experiment. A lower gravity accelerates clocks and a redshift is measured. This is also general relativity, i.e. the same issue (gravitational time dilation). Now, imagine to measure the frequency of the emitted light with a pendulum-based mechanism. You would measure a blueshift (violation of GR). What about this? – mfc Jan 20 '17 at 23:00
  • and a last experiment: imagine to put a pendulum on a heavier planet and to record its motion with a wi-fi videocamera. The signal (which of course should be strong enough) will need some time to arrive to the Earth, where you have a similar pendulum for comparison. When the signal arrives, you will observe that the other pendulum is faster despite the stronger gravity of the other planet, this according to the pendulum formula and confuting GR. – mfc Jan 20 '17 at 23:50
  • @mfc you are way off. Clearly as others have said the pendulum is a mechanical contraption, and it'll move as Lawrence Crowell explains. And a clock that is not mechanical like the cesium atom will measure it right. So you're barking at an invalid point. Nobody has to convince you, you have a right to fool yourself all you want. But you are simply wrong, and the others explained it. – Bob Bee Jan 21 '17 at 05:32
  • @Bob Bee: so, why GR doesn't specify that gravity slows down only non-mechanical clocks? That's contradictory. I never read that time dilation only works with atomic clocks. I don't bark, I'm not a dog, physics is not a religion. We have to discuss it. And if something doesn't work it doesn't work, we have to onestly solve that. – mfc Jan 21 '17 at 12:56
  • Won, it slows down all motion. How it affects a device made to try to determine how long an oscillation takes depends on the mechanism that defines the oscillation. That mechanism you have to analyze. The g in the pendulum gets affected but g is simply a gravity term, and the equation of motion may also be affected by the strong GR field. You think thru it and figure out the major or next major effects. That answers above did that and you are wrong. – Bob Bee Jan 21 '17 at 20:54
  • Suppose I had some physical effect or field that increased the periodicity of oscillators as $r~\rightarrow~2m$. Clearly in this case I worked the periodicity increases for a pendulum, but appears to slow down for the distant observer. However, with this bizarre field the oscillator increases its frequency enormously to the point that the asymptotic observer still sees it speed up. What would be wrong with this? Why can't this happen and is there some physical principle that would prevent the existence of this bizarre field? – Lawrence B. Crowell Jan 22 '17 at 00:52
  • @Lawrence: let's come back to your posted equations. You show how the period of the pendulum tends to zero as the Schwarzschild radium (R_s) is approached. OK. Then you show how it tends, on the contrary, to infinity by multiplying the T equation for 1/(1-R_s/r) = a. Well, what I call here "a" for brevity, is mathematically built to make everything diverge! So it's not a demonstration, but it is as to say: "the period diverges because I multiply it for a diverging factor! GR "assumes" that if g→∞ then T'→∞. But the pendulum says it's a "wrong" assumption. It depends on the clock type! – mfc Jan 25 '17 at 21:13
  • The period diverges for the pendulum, but the time dilation over takes that for the distant inertial observer. – Lawrence B. Crowell Jan 26 '17 at 21:37
  • @Lawrence: GR equations are made to obtain the result they want. You can multiply a cat, an apple and whatever equation by the factor 1/1-(Rs/r) and the result is what Einstein (and you) says! Experimental evidences have to confirm a theory and if we used a pendulum in time dilation tests, we wouldn't confirm what Einstein says. The dilation of time depends (as also Bob said) by the clock type. You can't say "the theory is right since I use its equations and it seems so". That factor gives a compulsory outcome. You can't verify a theory through its own formulas! It'd be self-referential – mfc Jan 28 '17 at 12:03
  • the other is the time dilation of such a pendulum as seen by a distant observer. – Ben Jul 14 '21 at 02:14
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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".

stuffu
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