Does time move slower at the equator?

Question

While answering the question GPS Satellite - Special Relativity it occurred to me that time would run more slowly at the equator than at the North Pole, because the surface of the Earth is moving at about 464 m/s compared to the North Pole. The difference should be given by:

$$ \frac{1}{\gamma} \approx 1 - \frac{1}{2}\frac{v^2}{c^2} $$

and at $v = 464\ \mathrm{m/s}$ we get:

$$ \frac{1}{\gamma} \approx 1 - 1.2 \times 10^{-12} $$

This is a tiny difference – about 4 days over the 13.7 billion year lifetime of the universe – but according to Wikipedia the accuracy of current atomic clocks is about $1$ part in $10^{14}$, so the difference should be measurable. However, I have never heard of any measurements of the difference. Is there a flaw in my reasoning or have I simply not been reading the right journals?

I seem to recall someone doing this for LEO astronauts once. If there aren't any papers on this topic, you should hurry up and publish -- or at least open up a "Life Extension Resort" in an equatorial zone! — Carl Witthoft, Jul 17 '14 at 11:58
@michael: yes, but both the observer at the North Pole and at the equator are moving round the Sun/round the Milky way/along with the Milky Way at the same speed, so these motions do not contribute to their relative time dilation. — John Rennie, Jul 17 '14 at 15:47
Related: Understanding the "$\pi$" of a rotating disk. Also, the present title of this question, "Does time move slower at the equator?" appears utterly improper. Consider instead asking properly "How to compare the (proper) rates of a clock placed at the North Pole to a clock at the Equator (which rotates around the North Pole once a day)?" — user12262, Jul 17 '14 at 16:03
@user12262 we'd really rather you not use line breaks in comments. I've edited your comment accordingly. — David Z, Jul 17 '14 at 16:45
Thanks to everyone who responded. As always Luboš's answer nails it :-) — John Rennie, Jul 18 '14 at 15:24
You might be interested to know that someone took an atomic clock with them camping in the mountains, then compared it with the atomic clock they left at home: http://www.leapsecond.com/great2005/ — Adam Davis, Jul 18 '14 at 17:22

Luboš Motl · Accepted Answer · 2014-07-17T13:16:05.380

The difference would indeed be measurable with state-of-the-art atomic clocks but it's not there: it cancels. The reasons actually boil down to the very first thought experiments that Einstein went through when he realized the importance of the equivalence principle for general relativity – it was in Prague around 1911-1912. See e.g. the end of

http://motls.blogspot.com/2012/09/albert-einstein-1911-12-1922-23.html?m=1

to be reminded about Einstein's original derivation of the gravitational red shift involving the carousel.

The arguments for John's setup may be seen e.g. in this paper:

http://arxiv.org/abs/gr-qc/0501034

There is a sense in which the "geocentric" reference frame rotating along with the Earth every 24 hours is more inertial than the frame in which the Earth is spinning.

Consider one liter of water somewhere – near the poles or the equator – at the sea level. Keep its speed relatively to the (rotating) Earth's surface tiny, just like what is easy to get in practice.

Now, let's check the energy conservation in the Earth's rotating frame. The energy is conserved because this background – even in the "seemingly non-inertial" rotating coordinates – is asymptotically static, invariant under translations in time.

The energy is conserved but the potential energy of one static (in this frame) liter of water may be calculated as $$m_c^2 \sqrt{|g_{00}|}. $$ Because the $00$-component of the metric tensor is essentially the gravitational (which is normally called "gravitational plus centrifugal" in the "naive inertial" frame where the Earth is spinning) potential and it is constant at the sea level across the globe, $g_{00}$ which encodes the gravitational slow down as a function of the place in the gravitational field must be constant in everywhere at the sea level, too.

In the "normal inertial" frame where the Earth is spinning, the special relativistic time dilation is compensated by the fact that the Earth isn't spherical, and the gravitational potential is therefore less negative i.e. "less bound" at the sea level near the equator.

Some calculations involving the ellipsoid shape of the Earth may yield an inaccurate cancellation. (That error may be attributed to not quite correct assumptions that the Earth's mass density is uniform etc., assumptions that are usually made to make the problem tractable.) But a more conceptual argument shows that the non-spherical shape of the Earth is a consequence of the centrifugal force. Quantitatively, this force is derived from the centrifugal potential, and this centrifugal potential must therefore be naturally added to the normal gravitational potential to calculate the full special-relativistic-plus-gravitational time dilation. That makes it clear why this particular calculation is easier to do in the frame that rotates along with the Earth's surface and the effect cancels exactly.

Let me mention that the spacetime metric in the frame rotating along with the Earth isn't the flat Minkowski metric. If we allow the frame to rotate with the Earth, we just "maximally" get rid of the effects linked to the centrifugal force and the corresponding corrections to the red shift. However, in this frame spinning along with the Earth, there is still the Coriolis force. In the language of the general relativistic metric, the Coriolis acceleration adds some nontrivial off-diagonal elements to the metric tensor. These deviations from the flatness are responsible for the geodetic effect as well as frame dragging.

Every argument showing the exact cancellation of the special relativistic effect must use the equivalence principle at one point or another; any argument avoiding this principle – or anything else from general relativity – is guaranteed to be incorrect because separately (without gravity and its effects), the special relativistic effect is certainly there.

score 19 · Answer 2 · answered Jul 17 '14 at 16:09

The current answers by Luboš and David do a good job of explaining why it is essential to include general relativity in the picture. In fact, this is even more of an issue because the irregularities in the shape of the Earth do matter.

It's fairly easy to understand why this is the case: it's been known since 2010 that atomic clocks are sensitive to height differences as small as one foot (NIST press release, paper, doi). With this sort of sensitivity, how can one synchronize atomic clocks on altitudes over 1km apart? Because of this, the International Atomic Time includes corrections to rescale each contributing clock's frequency back to the mean sea level, and has done so since the seventies.

This is made worse by the fact that the 'relative altitude' is not even locally measurable, and it is in a sense a global property of the Earth's gravitational field. This is because what really matters is the difference in gravitational potential between the two clocks, which can be affected by changes the mass distribution between the two clocks but far from each of them. What really matters, then, is the shape of the geoid, and the relative height of each lab with respect to it.

This is bad for two reasons. The first is that the geoid can indeed change measurably (example), driven by things like earthquakes, plate tectonics, and even tides and water cycles. The second one is that these changes are not locally detectable, because changes in the geoid affect the local gravitational potential (relative to a point at infinity) but do not affect the local gravitational field, which is essentially uniform locally.

One thing you can do, though, is to turn this synchronization problem around, and see your atomic clocks as a way of measuring the geoid, and this has indeed been proposed (Phys.org piece, preprint, doi).

With all this in mind, then, it's clear that we do have the capacity to measure the special-relativistic effect you mentioned, even if it wasn't cancelled out exactly in the way pointed out by Luboš. However, there are a lot of other effects that need to be taken into account, and that's where the important and current science is at.

score 15 · Answer 3 · answered Jul 17 '14 at 13:24

Is there a flaw in my reasoning or have I simply not been reading the right journals?

Yes. The flaw is that you are ignoring general relativity. The poles are closer to the center of the Earth and are thus deeper in the Earth's gravity well than is the equator. The combined effects of gravitational and special relativistic time mean that clocks at sea level tick at the same rate. More precisely, clocks at the surface of the geoid tick at the same rate.

score 8 · Answer 4 · answered Jul 17 '14 at 23:49

If you look at the Earth's geoid, you can see that there isn't a particular "band" of gravitational variation along the equator. So while time would move slower / faster at varying locations around the globe, it is not correlated with the equator.

Here is, I believe, the latest model, in 2D:

enter image description here

And there is also a very nice 3D animation here.

Regarding the ability to measure these differences with atomic clocks; yes they are noticeable. For example, from an NIST experiment demonstrating that time moves faster at your head than at your feet:

In one set of experiments, scientists raised one of the clocks by jacking up the laser table to a height one-third of a meter (about a foot) above the second clock. Sure enough, the higher clock ran at a slightly faster rate than the lower clock, exactly as predicted.

Does time move slower at the equator?

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