Is time dilation due to relative velocity equivalent in principle to time dilation due to relative gravitational strength?

Question

Is time dilation due to relative velocity and relative gravitational strength equivalent? That is, similar to Einstein's thought experiment where an observer in an enclosed capsule with no windows cannot tell the difference between acceleration due to gravity or due to some applied force, is there any observable difference between time dilation due to velocity versus that due to gravity?

Answer 1

Not really. The dilation due to the relative velocity is reciprocal, which means, both will see each other's clocks go slower. This time dilation is explained because moving faster in space implies moving slower in time so total velocity through spacetime is constant.

However, the gravitational dilation is not reciprocal, which means that the closest observer to the planet will be his clock going slower than the further observer and the further observer will see that his clock goes faster than the closer observer's clock, which is more logical. So, this dilation is caused by bending of space-time caused by the presence of the planet's mass.

Answer 2

The equivalence principle means that it is possible to have local coordinates, so that the spacetime can be regarded Minkowskian. Bob can that way estimate the time dilation of Alice, who passes by with a given velocity, using the Lorentz transformation.

But in order to know the time dilation after some time, when Alice is not so close, he must know the global spacetime metric to do the calculation.

One interesting case is if both have circular orbits with the same height, but opposing directions. For each meeting event it is valid what is written in the first paragraph. But nevertheless, due to the symmetry of the situation, their clocks will display the same time at that events. So, there are other points in the orbit where each one measures a time contraction. And between these extremes, points where there is no relative time dilation.

Answer 3

One example of that equivalence is the following:

Take a large ring-shaped space station. The structure is rotating so that at all points away from the axis of rotation the motion is pulling G's.

It is straigthforward to calculate for a given angular velocity of the space station at what radial distance the acceleration is pulling 1 G of acceleration.

Let's say there are several habitat levels, at various distances to the axis of rotation

Today atomic clocks are available that can measure time accurate enough such that they can detect that at the various habitat levels the amount of proper time that elapses is different.

In terms of GR that measured difference of amount of proper time that elapses can be accounted for in terms of either velocity time dilation or gravitational time dilation.

As a matter of principle: GR asserts that locally you cannot tell the difference

A habitat level at greater radial distance is, due to the circumnavigating motion, moving along a larger spatial distance than a habitat level at a lesser radial distance. Hence there will be difference in amount of proper time elapsing.

The G-load in a habitat level is directed away from the axis of rotation. That is, habitat levels at lesser radial distance are at a higher gravitational potential than levels at greater radial distance. The difference in proper time elapsing corresponds to the difference in gravitational potential.

Satellite orbit and proper time

A more striking example of that is the case of the amount of proper time that elapses for a satellite in orbit, compared to the amount of proper time that elapses on the surface of the primary.

In low Earth orbit the amount of proper time that elapses for the satellite is smaller than on the surface of the primary. This can be understood as follows: the difference in gravitational potential is small, and the orbital motion is very fast.

At high orbit (such as geostationary orbit) the difference in gravitational potential is large, and the velocity is far slower than in low Earth orbit, and the amount of proper time that elapses for the satellite is larger than on the surface of the primary.

In the case of satelites orbiting the Earth the cross-over point is at an orbital altitude of about 3000 kilometers above the surface of the Earth. For those satellites the amount of proper time that elapses is the same as on the Earth's surface.

It is tempting to think of that cross-over point in terms of two effects dropping away against each other, velocity-time-dilation, and gravitational-time-dilation.

The point is: making a distinction between velocity time dilation and gravitational time dilation is artificial.

Historically velocity time dilation was recognized first, but that does not mean that velocity time dilation and gravitational time dilation are distinct effects.

There is a very interesting newsgroup post in a thread titled relativistic time on satellites, posted in 1997 by Kevin Brown.

I need to explain:
Today Google is the custodian of those old newsgroup archives. Unfortunately, the content is somewhat munged. At the time the only way to represent mathematical equations at all was ASCII art. In ASCII art elements are lined up by using spaces. By the looks of it: Google's archiving software has deleted the spaces. That is why the equations are munged.

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Below is a reproduction of the derivation that Kevin Brown posted in the sci.physics newsgroup in 1997

I want to be clear that all below the horizontal line is quote, it's just that because of the length it is unpractical to format it as a blockquote

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[...] just for fun here's a pure GR derivation of orbital proper time. It's actually simpler (in my opinion) than trying to split up the problem into gravitational and non-gravitational effects.

$$ ds^2 = A dt^2 - B dr^2 - C d\theta^2 - D d\phi^2 $$

Where $A = (1 - 2\tfrac{M}{r})$, $B = \tfrac{1}{A}$, $C = r^2$ , $D = r^2\sin^2(\theta)$, $\phi = latitude$ and $\theta = longitude$

Let's say our radial position r and our latitude phi are constant for each path in question (treating r as the "radius" in the weak field approximation). Then A and D are both constants, and the metric reduces to

$$ ds^2 = A dt^2 - D d\phi^2 $$

If I'm sitting on the Earth's surface at the North Pole, I have D=0, so it follows that $ds = \sqrt{(1-2\tfrac{M}{r})} dt$ where $r$ is the radius of the Earth.

On the other hand, if I'm in an equatorial orbit with radius $R$ then we have $\theta=\tfrac{1}{2}\pi$, $\sin^2(\theta)=1$, and so $D$ is simply $R^2$. Now, recall Kepler's law $\omega^2 R^3 = M$, which also happens to hold exactly in GR (where $\omega$, $M$, and $R$ have their GR meanings). Since $\omega = \frac{d\phi}{dt}$ we have $D = R^2 = \frac{M}{\omega^2 R} = (\frac{dt}{d\phi})^2 (\frac{M}{R})$. Thus the path of the orbiting particle satisfies

$$ ds^2 = (1 - 2\tfrac{M}{R}) dt^2 - \tfrac{M}{R} dt^2 = (1 - 3\tfrac{M}{R} dt^2 $$

Now for each test particle, one sitting at the North Pole and one in a circular orbit of radius R, the path parameter s is the local proper time, so the ratio of the orbital proper time to the North Pole's proper time is

$$ \frac{d s_{orbit}}{d s_{earth}} = \sqrt{\frac{1-3\tfrac{M}{R}}{1-2\tfrac{M}{r}}} $$

To isolate the difference in the two proper times, we can expand the above function into a power series in $\frac{M}{r}$ to give

$$ \frac{d s_{orbit}}{d s_{earth}} = 1 + \left( 1 - \frac{3r}{2R} \right)\frac{M}{r} + ... $$

The mass of the earth, represented in geometrical units by half the Schwartzschild radius, is about 0.00443 meters, and the radius of the earth is about 6.38 $10^6$ meters, so this gives

$$ ds_{orbit} = ds_{earth} + 6.9^{-10} (1 - \tfrac{3k}{2}) ds_{earth} $$

which shows that the discrepancy in the orbit's lapse of proper time during a given lapse delta_T of proper time measured on Earth is

$$ (\Delta T) \ 6.9^{-10} \ \left( 1 - \frac{3r}{2R} \right) $$

Consequently, for an orbit at the radius R=3r/2 (about 2000 miles up) there is no difference in the lapses of proper time.