John Rennie – who is going to jump above me at Physics Stack Exchange in a few months unless I will find motivation to prevent him from doing so – has asked a very nice question:
Has someone measured this effect that should be measurable?
The answer is actually funny (especially if you say more than just the word "No"). Let me just repost mine.
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
The arguments for John's setup may be seen e.g. in this paper:
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 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 ellipsoidal 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.
A technical detail: note that I argued that the effects cancel in the frame where the Earth is spinning. But shouldn't it be cancelling in a frame where the Earth's surface is moving "slowly" because it subtracts the periodicity 23 hours 56 minutes 04 seconds instead of the 24 hours? Or vice versa? Well, these two frames differ by the effect of the Sun's gravitational field. You may only distinguish them if the solar contribution to the tides (and the equipotential surfaces on the Earth) and the associated tiny corrections to the red shift are carefully considered. The difference between the dilations in these two frames is about 366 times smaller than the difference considered in the main part of this article (it's no coincidence that it's the number of days in a year plus one) and such a small difference sits at the very boundary of the state-of-the-art atomic clocks' resolution.
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.
Does time move slower at the equator?The Earth is spinning so all people living at the equator are apparently moving at 464 m/s relatively to what seems like a "better inertial frame". By the special relativistic time dilation, this should slow their time by one part per trillion. That would be a 100 larger relative effect than the accuracy you may achieve with state-of-the-art atomic clocks.
Has someone measured this effect that should be measurable?
The answer is actually funny (especially if you say more than just the word "No"). Let me just repost mine.
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
Albert Einstein 1911-12, 1922-23 (TRF)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:
gr-qc/0501034 (arXiv, by S.P. Drake)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 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 ellipsoidal 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.
A technical detail: note that I argued that the effects cancel in the frame where the Earth is spinning. But shouldn't it be cancelling in a frame where the Earth's surface is moving "slowly" because it subtracts the periodicity 23 hours 56 minutes 04 seconds instead of the 24 hours? Or vice versa? Well, these two frames differ by the effect of the Sun's gravitational field. You may only distinguish them if the solar contribution to the tides (and the equipotential surfaces on the Earth) and the associated tiny corrections to the red shift are carefully considered. The difference between the dilations in these two frames is about 366 times smaller than the difference considered in the main part of this article (it's no coincidence that it's the number of days in a year plus one) and such a small difference sits at the very boundary of the state-of-the-art atomic clocks' resolution.
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.
Is time going slowly near the equator?
![Is time going slowly near the equator?]()
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July 17, 2014
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