0

Imagine a planet with the same properties as Earth, this time moving in an elliptical orbit around a heavy star of a large number of solar masses. Also imagine that the surface of this planet is as massive as that of the Earth and that you can therefore experience a normal force on it as on the Earth's surface. At a speed of 30 km/s at the apoapsis, the Earth has a radius of 6500 km and a mass of 6*10^24 kg.

Imagine that the planet is currently at the apoapsis (the farthest point) of an elliptical orbit around the heavy star. Now place a pendulum clock and an atomic clock on the planet's surface and somehow get them in phase. This depends on the length of the pendulum for the frequency that must be in step with the frequency of the atomic clock. The pendulum motion depends on the (Newtonian) field strength g on the surface of the planet according to Huygens formula - as a pendulum is fundamentally only active in an active gravitational field - while the rate of the atomic clock depends on the gravitational time dilation on the planet's surface according to general relativity - as the atom clock is sensitive to 'proper time' relativity effects. The pendulum however intrinsically cannot measure proper time. This is the crux of this system's setup according to my reasoning.

The planet varies in velocity in its elliptical orbit around the black hole: according to Kepler, it will absolutely move faster in its orbit at the periapsis (i.e. closer to the heavy star) of the ellipse than at the apoapsis (farthest from the black hole). According to special relativity, this change in velocity should increase the magnitude of time dilation affecting the atomic clock on the planet’s surface.

Imagine that the velocity increases to 30,000 km/s at the periapsis; this heavy star is very heavy indeed, and the planet’s orbit curves very closely near its surface. According to special relativity, time dilation should increase on the planet’s surface and affect the atomic clocks ticking rate, making it tick slower. The local gravitational field g of the planet stays unchanged however as the planet's mass is unchanged, which means the pendulum still ticks at the same original rate as at the apoapsis. The pendulum clock and the atomic clock will therefore run out of phase on the faster parts of the planet's elliptical orbit around the heavy star. Which implies an objective change in its celestial velocity can be detected due to this time dilation effect affecting the pendulum and atomic clock differently on the planet's surface.

Is this conclusion correct? If not, I'd like to hear where the error lies in the assumptions made above. Thanks

Apsteronaldo
  • 109

2 Answers2

3

You went wrong at the point where you assumed that atomic and pendulum clocks are affected differently by time dilation. All temporal processes are affected in the same. That's why it's called "time" dilation and not "clock" dilation. Another way to look at it is if the atomic clock and pendulum clock are synchronized in one reference frame, they are synchronized in all of them (this is the principle of relativity).

The best way to think about time dilation is as a geometric effect, similar to rotation in Euclidean space. If you take a set of coordinate axes $X', Y'$ that are rotated with respect to "standard" coordinates $X, Y$, you'll find that a point $(0, y')$ on the $Y'$ axis has coordinates $(x, y)$ on the original $Y$ axis, with $|x| > 0$ and $y < y'$. You could call this effect "x-dilation" and "y-contraction", but we don't usually bother because it's "obvious" to us what is happening. Motion in spacetime is basically the same thing: the "moving" object's coordinates are rotated relative to the "stationary" object's coordinates, so that the time coordinate is stretched out and the length coordinate is contracted. This coordinate change is independent of how the coordinates are measured (whether with a light clock, atomic clock, pendulum clock, or biological process).

Eric Smith
  • 9,064
  • Thanks for your comment and your time. My reply to you would be the following: how how would you explain the fact that a pendulum will slow down its ticking at altitude, while an atomic clock increases its ticking; and vice versa, if they relate to time dilation identically (which is the same as 'gravity' in the weak field limit of general relativity)? It is thus clear that they manifest different behaviour under identical circumstances of gravity, which in the weak field limit case is just this time dilation. Even if space curvature was involved, the conclusion would be the same in my opinion – Apsteronaldo Jan 23 '24 at 23:03
  • This means I don't agree with your statement that a pendulum clock would respond the same to this 'coordinate change' like a light clock, atomic clock or biological process. These last three are fully quantum mechanical systems, while the pendulum clock is a fully gravitational one. Even if time dilation is fully due to coordinate change, the fact is still the pendulums manifest behaviour differs in equipotential surfaces identical to the gravitational ones as the time dilation ones. Even if it isn't caused by time dilation, it directly correlates with it if you get what I mean. – Apsteronaldo Jan 23 '24 at 23:07
  • I guess my point is that the time measured by the pendulum is a different time than the one measured by an atomic clock or a light clock: they are two different kind of times! That is my whole point.

    Would love to hear your comment, thanks

     – Apsteronaldo Jan 23 '24 at 23:17
  • @Apsteronaldo Of course a pendulum clock will be affected by gravitational time dilation, but the effect is very small compared to the pendulum precision, so it's not detectable. The pendulum clock will also be affected by various other things, like thermal expansion. A very high precision pendulum clock can detect variations in gravity due to the tidal force of the Moon & the Sun, but you need a higher precision time measurement to compare the pendulum with. See http://leapsecond.com/hsn2006/ – PM 2Ring Jan 24 '24 at 00:52
  • @PM2Ring Thank you for your reply and your interest. I would like to investigate your statement that a pendulum clock would be affected by gravitatonal time dilation. Here it goes: would a falling grain of sand take more time to fall due to time dilation? – Apsteronaldo Jan 24 '24 at 01:30
  • 1
    @Apsteronaldo Sure! But you don't notice time dilation locally. You can only measure it when you can make a comparison, eg between clocks at different altitudes. Over the course of a year, clocks speed up & slow down due to the changing speed of the Earth, and the changing gravitational potential of the Sun, relative to a clock in flat spacetime at rest relative to the Sun. We don't worry about that in relation to Earth's atomic clock network. But we would have to take that into account if we set up atomic clocks on Mars and wanted to convert between Earth & Mars time. – PM 2Ring Jan 24 '24 at 01:42
  • @PM2Ring Also great to discover your webpage on pendulum clocks and timekeeping! Thanks for the link, I will surely investigate it thoroughly tomorrow when it is not midnight like now :-) It seems nice to (finally) meet somebody with a similar obsession with the concept of time as I have – Apsteronaldo Jan 24 '24 at 01:43
  • @PM2Ring If that is the case, that a grain of sand would take more time falling down in strong time dilation, as you say is only observed from an external observer, does this imply that for a local observer the space is shrunken, as we don't notice the movement slowing down? The grain has more time to cover the same distance (as it slows down due to time dilation), yet we cannot notice locally: this implies that locally the distance must have shortened as well to compensate for that! Thus space shrinks locally if we cannot measure a grain of sand falling slowly – Apsteronaldo Jan 24 '24 at 01:47
  • @Apsteronaldo Sorry, but it's not really appropriate to get into protracted discussions here in the comments. But we have a chat room: https://chat.stackexchange.com/rooms/71/the-h-bar Maybe we discuss this tomorrow... – PM 2Ring Jan 24 '24 at 01:51
1

As you said, the period of a pendulum clock depends on the local $g$, so it is not suitable to measure gravitational time dilation.

In reality, that effects are normally very tiny, and only atomic clocks are accurate enough to the task.

Claudio Saspinski
  • 16,319