Does a clock oscillating in a friction-free hole through the center of a planet run slower than a stationary clock on the surface?

Question

Assume a clock is dropped into a friction-free hole through the center of a symmetric, non-rotating planet, far from any other massive object. Clearly, the clock oscillates from one end of the hole to the other forever, and, as it is subject to time dilation due to relative speed and acceleration, it runs more slowly than a stationary clock at the center of the planet. But, does it run more slowly than a stationary clock at the surface, which is subject only to the acceleration of gravity, or does it keep exact time with the surface clock? I think it runs exactly the same as the surface clock. Am I correct?

[Added 10/20/2015 4PM] Thanks for the initial answers, but they indicate I need to clarify my question and explain my reasoning.

1) My "planet" (unlike the Earth) is "symmetric" (perfect sphere of constant density), "non-rotating" and "far from any other massive object" (not in orbit around a star).

2) At the instant the clock is dropped into the friction-free hole, it is adjacent to the surface clock and they are both set to zero. When it first passes the center clock, that clock is set to whatever time is on the oscillating clock.

3) After many cycles, as the oscillating clock passes each of the other clocks, the times are compared and, if the oscillating clock time is substantially lower, it is said to have been subject to more time-dilation than the center and/or surface clock, and vice-versa. If the times are equal, or nearly so considering the accuracy and precision of the clocks, it is said to have the same time-dilation. I believe the oscillating clock will have run more slowly than the center clock (more time-dilation) and equal to the surface clock, for the reasons in the following points. (Please correct me if I am wrong.)

4) Let the origin of the reference inertial frame be the center of the planet. The center clock is "at rest" in that frame, at zero speed and zero gravity. The surface clock is not moving, but it is subject to gravity, and therefore more time-dilation than the center clock.

5) The oscillating clock is subject to varying amounts of speed and/or gravity. When adjacent to the surface clock, it stops momentarily and reverses direction. At that instant, it is at zero speed and the same gravity as the surface clock, and so running at the same rate (same time-dilation) as the surface clock.

6) When the oscillating clock is adjacent to the center clock, it is at maximum speed and zero gravity. What is that maximum speed? By conservation of energy, the kinetic energy of the clock, as it zips past the center, is exactly the same as the gravitational potential energy the clock had when dropped from the surface. Therefore, the maximum speed just happens to be the escape velocity at the surface, since escape velocity perpendicular to the surface is defined as the speed where kinetic energy is exactly equal in magnitude to gravitational potential energy, and (assuming no atmosphere) a rocket will continue its journey into space forever.

7) Time-dilation due to gravitational potential energy of a stationary clock on the surface of a planet is exactly equal to the time-dilation due to kinetic energy of a rocket in deep space moving at the escape velocity corresponding to the surface of that planet. By conservation of energy, as the oscillating clock shuttles back and forth between the ends of the friction-free hole, its total energy (combination of kinetic and gravitational potential energy) remains constant, and exactly equal to the gravitational potential energy of the surface clock. Therefore, I believe the oscillating clock will experience exactly the same time-dilation as the surface clock.

Answer 1

There are two effects that cause time dilation:

1. the velocity of the falling clock

2. the gravitational time dilation

The implicit argument in your question is that these two effects might cancel out to make the falling clock run at the same speed as a clock on the surface.

Actually calculating the time dilation is a hard problem as it requires solving the geodesic equation for the Schwarzschild interior metric. Even that is only an approximation because the Schwarzschild interior metric only applies to a sphere of uniform density while the Earth is neither a sphere nor of uniform density. In practice the calculation would need to be done numerically.

However we can answer your question very easily because both effects $1$ and $2$ make the falling clock run more slowly, so overall the falling clock must run more slowly that the clock on the surface.

It's tempting to think that because the gravitational acceleration falls to zero as we approach the centre of the Earth then the time dilation must also fall to zero. However the time dilation is dependant on the gravitational potential not the gravitational acceleration. To a first approximation the time dilation is given by the weak field expression:

$$ \frac{dt_{falling}}{dt_{surface}} = \sqrt{1 + \frac{2\Delta\Phi}{c^2}} \tag{1} $$

where $\Delta\Phi$ is the difference between the gravitational potential of the falling clock and the gravitational potential of the clock on the surface. Obviously the potential energy decreases as we descend towards the centre, which is why things fall downwards. That means $\Delta\Phi \le 0$, and therefore the right hand side of equation (1) must be less than one:

$$ \frac{dt_{falling}}{dt_{surface}} \le 1 $$

In other words the time dilation is greater not less, as we descend into the Earth. So the falling clock must run more slowly than a clock on the surface.

The details:

Since Ira asks about the details here they are - the uninitiated may wish to run away screaming.

Let's start by looking at a moving clock in flat spacetime i.e. no gravity. We'll start with this because it's simpler, then we'll extend the calculation to include gravity. In flat spacetime the metric is the Minkowski metric:

$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 \tag{2} $$

The quantity $\tau$ is the time in the rest frame of the clock, while $t$ is the time measured in our frame as we are watching the clock. Likewise $x$, $y$ and $z$ are the clock's spatial coordinates in our frame. The metric tells us that if we observe the clock to move an infinitesimal distance through spacetime of $(dt, dx, dy, dz)$ then the metric tells us how to calculate the corresponding elapsed time $d\tau$ for the clock.

In this case the clock is moving radially, so we'll take the $x$ axis to be the $r$ axis and there is no motion in the $y$ and $z$ axes. Our equation (2) simplifies to:

$$ c^2d\tau^2 = c^2dt^2 - dr^2 \tag{3} $$

In our frame the velocity of the clock is just $v = dr/dt$ so $dr = vdt$. If we substitute this into equation (3) we get:

$$ c^2d\tau^2 = c^2dt^2 - v^2dt^2 $$

and a quick rearrangement gives:

$$ d\tau = dt \sqrt{1 - \frac{v^2}{c^2}} = \frac{dt}{\gamma} $$

Where $\gamma$ is the Lorentz factor. You should immediately recognise this as the usual formula for time dilation in special relativity. Because the Lorentz factor $\gamma \gt 1$ we find the clock's time $d\tau$ is less than our time $dt$ i.e. the moving clock is running slow. So far so good.

Where the going gets tough is that the curvature of spacetime caused by the mass of the Earth also affects the time dilation, and it does this by changing the metric i.e. equation (2) above. If we approximate the Earth by a sphere of uniform density then the geometry of spacetime inside the Earth is given by the Schwarzschild interior metric:

$$ c^2d\tau^2 = \left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}\right]^2c^2dt^2 - \frac{dr^2}{\left(1-\frac{2Mr^2}{R^3}\right)} - r^2 (d\theta^2 + sin^2\theta d\phi^2) $$

For radial motion $d\theta = d\phi = 0$, and as before the radial velocity is $v(r) = dr/dt$ so $dr = v(r)dt$, where the velocity $v(r)$ is now a function of $r$. Making these substitutions and rearranging gives:

$$ \frac{d\tau}{dt} = \sqrt{\left(\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}\right)^2 - \frac{v^2(r)}{c^2}\frac{1}{\left(1-\frac{2Mr^2}{R^3}\right)}} $$

And this is the equation you asked for i.e. the equation for the time dilation factor $d\tau/dt$, though note that this equation gives the time dialtion relative to an observer at infinity not at the surface of the Earth.

Note that the right hand side contains both $r$ and the function for the velocity $v(r)$. This function $v(r)$ is the velocity of the clock as it falls through the Earth, and to calculate it precisely would require solving the geodesic equation. However given that the time dilation would be small it would be a good approximation to use the $v(r)$ calculated using Newtonian mechanics. I'll leave this as an exercise for the reader.

To get the time dilation for a stationary clock just set $v(r) = 0$.

Answer 2

Assume a clock is dropped into a friction-free hole through the center of a symmetric, non-rotating planet, far from any other massive object. Clearly, the clock oscillates from one end of the hole to the other forever,

I'd expect the oscillation amplitude to decrease gradually; but perhaps only very gradually, in comparison to the (initial) oscillation period.

as it is subject to time dilation due to relative speed and acceleration, it runs more slowly than a stationary clock at the center of the planet.

That's incorrect. No: In order to compare durations $\Delta \tau$ of (these) two clocks to each other we must account for their geometric relations (relative speed, accelerations, ...) while they were separate from each other.

But in order to compare rates $\frac{\Delta t}{\Delta \tau}$ between clocks, such as to determine which one had "run more slowly" and which one had "run more quickly", or whether both had "run equally fast" in a trial under consideration, we'd also have to know and account for how the readings $t$ were assigned to either clock.

What can be said instead is that the duration of the dropping and oscillating clock from having encountered the planet surface once until having encountered the planet surface the next time is larger than the (corresponding) duration of a clock stationary on the planet surface from having encountered the oscillating clock once until having encountered the oscillating clock the next time. (Again: this alone doesn't allow any conclusions about a comparison of their "running rates" because we don't know how values (readings) $t$ are assigned to the described encounter indications of one clock, or the other.)

Looking instead at a stationary clock at the center of the planet, I'd expect that its durations from having been passed by the oscillating clock once until having been passed by the oscillating clock again is equal to the (corresponding) durations of oscillating clock from having been passed by the (clock at the) center of the planet once until having been passed by the (clock at the) center of the planet again. (Again: this alone doesn't allow any conclusions about a comparison of their "running rates" because we don't know how values (readings) $t$ are assigned to the described encounter indications of one clock, or the other.)