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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 464m/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 $v$ = 464m/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?

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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 at 11:58
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In this paper, chapter "Relativity on the Earth’s surface", there is an interesting discussion, where it seems that the gravitationnal effect (+ $108$ ns by $24$ h) compensates the special relativity effect (- $104$ ns by $24$ h). –  Trimok Jul 17 at 12:12
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@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 at 15:47
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@user12262 we'd really rather you not use line breaks in comments. I've edited your comment accordingly. –  David Z Jul 17 at 16:45
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@Trimok consider placing your comment as an answer. –  ja72 Jul 17 at 18:29

4 Answers 4

up vote 27 down vote accepted

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.

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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.

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Emilio Pisanty: "the geoid can indeed change" -- Should the SI "second" definition consequently refer to a ceasium atom unperturbed by changes of the geoid ? (Cmp. "Does the definition of the SI unit “second” require that possible perturbation of primary frequency standards should be measured?", http://p.se.com/q/123563). "see your atomic clocks as a way of measuring the geoid," -- How should be determined whether (or to which accuracy) a given atomic clock had been "good" before and without having "measured the geoid" already? –  user12262 Jul 17 at 21:54
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No, that's not really the right way to see it. Each atomic clock gives (a reasonable enough implementation of) the SI second - with respect to the local time. The problem comes when comparing two clocks at different locations, whose local time itself runs at different rates, and trying to come up with an average time. In essence, this is the difference between measurements of frequency - like the SI second - and measurements of time like UTC and TAI. Because of this, the SI second itself remains unaffected. –  Emilio Pisanty Jul 18 at 0:11
    
Regarding your second question, that is a question for metrology and it won't be easy to solve. I haven't read the proposal paper in detail, but it presumably contains an in-depth discussion, so it makes a good starting point. –  Emilio Pisanty Jul 18 at 0:16
    
Emilio Pisanti: "Each atomic clock gives (a reasonable enough implementation of) the SI second - wrt. the local time." -- Well, thanks for providing a concrete justification to submit a question about that notion which I find so utterly mysterious: "local". (I plan it along the lines of: "Is a Cs133 atom defined as local?; Are changes of the geoid on scales typical for a Cs133 atom ruled out by definition?"). "comparing two clocks at different locations, whose local time itself runs at different rates" -- That seems a bit improper. (Does "1 Hz" not mean "equal rate, proper"?) –  user12262 Jul 18 at 4:36
    
Emilio Pisanty: "I haven't read the proposal paper [ arxiv.org/pdf/1209.2889v1.pdf ] in detail, but it presumably contains an in-depth discussion" -- Well, they (Bondarescu, et al.) seem to uncritically accept claims about "accuracy level in clock rate" from secondary literature. Now, I haven't seen the possibility of "changes of the geoid" and their impact on the determination of "accuracy" (or "systematic error") addressed anywhere else... (But at least MTW § 16.4.) –  user12262 Jul 18 at 4:50

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.

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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.

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