Relativity of rotating system(s)


by S.R.Wilton
Tags: relativity, rotating, systems
S.R.Wilton
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#1
Oct10-09, 10:28 PM
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I was just playing around with thoughts while laying in bed, and I need confirmation or rejection of a random idea that popped up in my head. I have a mild idea of relativity from various readings. I'm an undergraduate student studying engineering physics, so I have a decent background on Newtonian mechanics and basic quantum mechanical principles, but not so much of relativistic effects. I always thought of relativity and the speed of light in a strictly linear sense because it was straightforward (haha, pun intended >_<), but today was the first day I thought from a different perspective.

Do the atoms on the surface of earth decay slower relative to the atoms in the core of the earth due to the velocity difference?

If so, wouldn't that mean there is a "gradient of time" that exists on the planet? (i.e, aquatic animals on the bottom of the ocean age more quickly relative to animals on the surface of the planet; Other animals/matter move through time in between the speeds of everything on the outside of the planet and everything toward the center.)

If so, wouldn't that mean that our entire galaxy rotating about it's center has a space-time gradient in which all matter on the outside edge of the galaxy is aging slower relative to the matter on the inside of the galaxy toward the center?

If not, why not?

Maybe I'm just tired and not thinking clearly.


Edit: Back awake again, just realized I completely neglected the gravitational effects of being at the center of the earth as opposed to being on the surface of earth. Now I'm just confused. Matter at the center feels (approximately) the same amount of gravitational attraction from all sides of the planet as opposed to being drawn toward the center of mass of the planet (yeah? no?)... but it's still attracted to the sun and everything else anyway.... so.... I just realized I don't know anything about general relativity at all. How does all this factor into relativity? How would all this factor into the rate at which time moves relative to us? I'm just confusing myself.

Hopefully my confusion will lul me to sleep, and I can have dreams space-time makes sense to me.
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Cleonis
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Oct11-09, 05:04 PM
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Quote Quote by S.R.Wilton View Post
Do the atoms on the surface of earth decay slower relative to the atoms in the core of the earth due to the velocity difference?
As you remark later on in your posting, in such a case the overall relativistic effect has both a gravitational nature and a velocity difference nature.

There is actually a more interesting scenario: if you have atomic clocks on the poles, and on the equator, which at some point in time have been synchronized, will they remain in sync?

In 1905 Einstein wrote they wouldn't. The 1905 special relativity predicted a velocity time dilation. Clocks on the equator are circumnavigating the Earth's axis, and clocks on the poles are stationary with respect to the Earth's axis.

But later the Equivalence principle came into view. There is an equatorial bulge, and clocks on the poles are closer to the Earth's geometrical center; they are deeper in the gravitational well.

The Earth's equatorial bulge is an equilibrium shape. Over its entire lifetime the Earth's rotation rate has decreased, and its equatorial bulge has readjusted accordingly.

The counterpart of that equilibrium in terms of General Relativity is that in equilibrium state the same amount of proper time elapses everywhere.

Matter tends to move towards a region where a larger amount of proper time elapses. If the amount of proper time at the poles would differ from the amount of proper time at the equator, then the shape of the Earth will deform until equilbrium state is reached.

Everywhere on Earth clocks located at sealevel count the same amount of lapse of proper time. That is: at every point on the surface of the Earth the contributions of velocity time dilation and gravitational time dilation add up to the same amount. (If they would not add up to the same amount then the principle of equivalence would be violated.)

So on completion of General Relativity it turned out that the 1905 prediction was wrong. Clocks all over the Earth (at sea level) count the same amount of proper time.


Actually, it's a bit more complicated than that. Spacetime curvature involves both space and time. The path of light through spacetime is affected about fifty-fifty. (In his very first exploration of a theory that implements the Equivalence principle, in 1907, Einstein predicted half the light bending of what the correct theory predicted.)

For things moving slower than light the contribution of space curvature is accordingly smaller. For motion of planets a theory that would fail to incorporate space curvature would only fail to predict Mercury's orbital precession.

Cleonis
S.R.Wilton
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Oct11-09, 06:53 PM
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Woah, that's pretty awesome. Thanks for the info! I really didn't know or understand anything about relativity at all until yesterday. I'm still having trouble conceptually grasping everything, but everything you said seems logical. What do you mean by "proper time" though?

Cleonis
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Oct11-09, 07:01 PM
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Relativity of rotating system(s)


Quote Quote by S.R.Wilton View Post
What do you mean by "proper time" though?
I suppose that expression may sound as if I just invented it. But it's a standard expression in relativistic physics. Think of the word 'property'. For any traveller, the traveller's proper time is his own time, the amount of time that is measured by a clock that is co-moving with that traveller.

For instance, the satellites of the GPS system have clocks onboard that are so accurate that relativistic effects are clearly noticable. The 'proper time' of those satellites is the time as measured by those onboard clocks.

Cleonis
JesseM
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Oct11-09, 07:36 PM
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Quote Quote by Cleonis View Post
The Earth's equatorial bulge is an equilibrium shape. Over its entire lifetime the Earth's rotation rate has decreased, and its equatorial bulge has readjusted accordingly.

The counterpart of that equilibrium in terms of General Relativity is that in equilibrium state the same amount of proper time elapses everywhere.
Is there any simple way to explain why this is? I don't understand why the equivalence principle should imply this about the equilibrium shape of a rotating planet.
Quote Quote by Cleonis
Matter tends to move towards a region where a larger amount of proper time elapses. If the amount of proper time at the poles would differ from the amount of proper time at the equator, then the shape of the Earth will deform until equilbrium state is reached.
Why would matter tend to move towards a region with greater proper time? I suppose the inertia of matter makes it "want" to move on a geodesic path which is where proper time is really maximized, but no point on the surface of the Earth is moving on a geodesic. Why wouldn't the equilibrium shape an object assumes depend on the nature of whatever non-gravitational forces (in this case electromagnetism) are preventing the particles from moving on geodesics? Can the equilibrium shape be defined as a minimum of potential or some other quantity which would depend on the non-gravitational force? If the electromagnetic force between atoms worked differently--if it obeyed an inverse-cube law or something--I wonder if it would it still be true that a solid object would naturally assume a shape where clocks on the surface all tick at the same rate.

If these kinds of questions aren't easy to answer without doing a complicated mathematical analysis, don't worry about it, I was just curious.
PeterDonis
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Oct11-09, 11:33 PM
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A quick calculation shows that the proper time rates are *not* the same at the equator and the poles. Define the following:

c = speed of light = 3.00 x 10^8 m/s

v = equatorial velocity at Earth's surface = 465 m/s

M = mass of Earth = 5.97 x 10^24 kg

G = Newton's constant = 6.673 x 10^-11

r_eq = equatorial radius of Earth = 6.378 x 10^6 m

r_po = polar radius of Earth = 6.357 x 10^6 m

Then we have

[tex]\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]

[tex]\phi_{eq} = \sqrt{1 - \frac{2 G M}{c^2 r_{eq}}}[/tex]

[tex]\phi_{po} = \sqrt{1 - \frac{2 G M}{c^2 r_{po}}}[/tex]

and the ratio

[tex]\frac{\phi_{eq}}{\phi_{po}}[/tex]

is what we should compare to [itex]\gamma[/itex] to see whether the effects on proper time rate of increased altitude and velocity cancel out. When I plug in numbers, I get:

[tex]\gamma = 1 + 1.2 * 10^{-12}[/tex]

[tex]\frac{\phi_{eq}}{\phi_{po}} = 1 + 2.3 * 10^{-12}[/tex]

So the speed-up due to increased altitude is almost twice the slow-down due to the equatorial velocity. Clocks will run faster at the equator than at the poles.
A.T.
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Oct12-09, 02:54 AM
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Quote Quote by JesseM View Post
Why would matter tend to move towards a region with greater proper time? I suppose the inertia of matter makes it "want" to move on a geodesic path which is where proper time is really maximized,
This is a model of how geodesics deviate towards greater gravitational time dilation:
http://www.physics.ucla.edu/demoweb/...urved_time.gif
But you could also think of space time having a density gradient. And the geodesics behave like light rays in a medium with varying optical density: they deviate towards the denser region.

More Generally on the topic of gravity and rotation: I guess you would have to combine the Schwarzschild-Mertic:
http://en.wikipedia.org/wiki/Schwarzschild_metric
with the metric for rotating frames of reference:
http://books.google.de/books?id=DH7j...metric&f=false
Ich
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Oct12-09, 03:38 AM
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Is there any simple way to explain why this is? I don't understand why the equivalence principle should imply this about the equilibrium shape of a rotating planet.
Gravity is a fictitious force just like centrifugal force. They are on equal footing, except that the latter is easier to transform away. In a rotating frame, you add both potentials to get the total time dilatation. Fictitious force potentials are the weak field approximation of GR when you can express everything as a change in the time-time component only.
It's not a coincidence that potential energy goes with time dilatation.
Crono
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Oct12-09, 06:26 AM
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I've been wondering a few things for a while and this looks like the place to find some answers and more questions, so here's another:

An object moving relative to another at content velocity has equal claims to it being stationary and the other moving (at least in otherwise empty space).

Nothing (nothing in our three brane) can (truly) exceeded light speed as (not because) the velocity along ANY (maybe all, but at least the 3 extended) spatial dimension is shared with the time dimension (if it exists :o), thus the closer you get to light speed the slower time travels relative to the rest of the universe).

So if an object, a sphere, is rotating at close to light speed, and moving through space at close to light speed what would be the effects of time dilation relative to a stationary object.

Thanks
Cleonis
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Oct12-09, 04:50 PM
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Quote Quote by PeterDonis View Post
Clocks will run faster at the equator than at the poles.
Hi Peter,

I need to check: have you taken the Earth's equatorial bulge into account?

What I mean by that is the following: in the case of an oblate spheroid the center of gravitational attraction does not coincide with the geometrical center.

Here, with 'center of gravitational attraction' I mean the following: the point where all of the Earth's mass would have to be concentrated, so that it would exert exactly the same gravitational force as the entire Earth does.

The Earth's equatorial bulge is such that the equator is about 20 kilometers further away from the Earth geometrical center than the poles.
In the case of the Earth, with its equatorial bulge: for an object located at the equator, the center of gravitational attraction does not lie on the Earth's axis, but about 10 kilometers away from it.

(Actually, I have that figure of 10 kilometers on authority. I would love to be able to carry out the integration to compute the location of the Earth's gravitational center (as a function of latitudinal position) but my math ability isn't up to that task.)

Cleonis
PeterDonis
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Oct12-09, 05:04 PM
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Quote Quote by Cleonis View Post
Hi Peter,

I need to check: have you taken the Earth's equatorial bulge into account?
Of course I did; look at the different values I used for the equatorial vs. the polar radius of the Earth.

Quote Quote by Cleonis View Post
What I mean by that is the following: in the case of an oblate spheroid the center of gravitational attraction does not coincide with the geometrical center.

Here, with 'center of gravitational attraction' I mean the following: the point where all of the Earth's mass would have to be concentrated, so that it would exert exactly the same gravitational force as the entire Earth does.

The Earth's equatorial bulge is such that the equator is about 20 kilometers further away from the Earth geometrical center than the poles.
In the case of the Earth, with its equatorial bulge: for an object located at the equator, the center of gravitational attraction does not lie on the Earth's axis, but about 10 kilometers away from it.
This is all wrong. An oblate spheroid is still symmetrical about one axis; in the case of the Earth that's the axis that runs through the North and South poles. The gravitational center of the Earth is the point on that axis that's equidistant between the poles. (This is all assuming, of course, that the Earth is a perfect oblate spheroid, which it isn't; but the variations are too small to matter here.) Also, the location of that center is not observer-dependent; it's at the same point for all observers, whether they're on the equator, at the poles, or orbiting far out in space.
Cleonis
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Oct12-09, 05:24 PM
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Quote Quote by PeterDonis View Post
An oblate spheroid is still symmetrical about one axis; in the case of the Earth that's the axis that runs through the North and South poles. The gravitational center of the Earth is the point on that axis that's equidistant between the poles. (This is all assuming, of course, that the Earth is a perfect oblate spheroid, which it isn't; but the variations are too small to matter here.) Also, the location of that center is not observer-dependent; it's at the same point for all observers, whether they're on the equator, at the poles, or orbiting far out in space.
Peter,

your comment above is very problematic.

The sphere is the only shape with the property that the center of gravitational attraction coincides with the geometrical center.

In the case of the Earth the fact that the center of gravitational attraction does not coincide with the geometrical center is what gives rise to the precession of the equinoxes. The precession of the equinoxes is a case of gyroscopic precession.

Let me discuss the example of a pole in a gravitational field.
Imagine a long pole in a gravitational field, aligned with the gradient in the field. Consider two halves of the pole; the first one closest to the source of the gravitational field, and the second one furthest away from the source. Evaluate the total gravitational force upon each section. Since gravity falls off with the square of the distance the half section closest to the source will be subject to stronger gravity than the half section furthest away from the source. The center of mass of the pole, however, will be at the geometrical center. Therefore in a gravitational field a pole will be subject to a torque.

The Earth's gyroscopic precession is due to the fact that the Earth is subject to a torque, arising from the Sun's gravity and the Moon's gravity.

Cleonis
PeterDonis
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Oct12-09, 06:02 PM
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Quote Quote by Cleonis View Post
Peter,

your comment above is very problematic.

The sphere is the only shape with the property that the center of gravitational attraction coincides with the geometrical center. ...
Not with the definition of "center of gravitational attraction" you've given. The "source point" of the Newtonian gravitational force of an object is the geometric center for any shape that has an axis of symmetry. For shapes that aren't spheres, there are additional multipole moments that come into play for things like precession, as you note, but those don't change the Newtonian force exerted on a small test object like a person standing on the Earth's surface, and that is the only factor that comes into play for determining the rate of lapse of proper time, which was the OP's original question.

Quote Quote by Cleonis View Post
Let me discuss the example of a pole in a gravitational field.
Here the pole isn't a small test object--it has enough extension to see the gradient in the gravitational field, so of course it's going to see effects that a small test object wouldn't see. But again, this is irrelevant to the OP's original question.
Cleonis
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Oct12-09, 06:42 PM
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Quote Quote by PeterDonis View Post
The "source point" of the Newtonian gravitational force of an object is the geometric center for any shape that has an axis of symmetry. For shapes that aren't spheres, there are additional multipole moments that come into play for things like precession, as you note, but those don't change the Newtonian force exerted on a small test object like a person standing on the Earth's surface, [...]
This issue will require further discussion, but it no longer fits/suits this relativistic physics thread. Tomorrow I will start a new thread in 'Classical Physics', where I will submit this issue to the forum. (As a name for that thread I think 'Center of gravitational attraction' is a good candidate, but I'm undecided yet.)

Needless to say, I'm convinced I made no errors.

Cleonis
PeterDonis
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Oct12-09, 07:10 PM
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Hmm... looking at various web pages discussing quadrupole moments, I'm no longer sure I was correct that they have *no* effect on the force on a small test particle. However, I haven't been able to find any definite formulas. Does anyone know of any online references?
PeterDonis
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Oct12-09, 07:14 PM
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Quote Quote by Cleonis View Post
This issue will require further discussion, but it no longer fits/suits this relativistic physics thread. Tomorrow I will start a new thread in 'Classical Physics', where I will submit this issue to the forum
I agree that this particular question belongs better in Classical Physics; however, the effects on proper time lapse still belong here, IMHO.
yuiop
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Oct12-09, 07:22 PM
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Quote Quote by Cleonis View Post
Matter tends to move towards a region where a larger amount of proper time elapses.
I think you may have this the wrong way round, if by "a region where a larger amount of proper time elapses", you mean a region where proper time is faster or a region where the least time dilation occurs.

For example imagine a mine shaft at one of the poles with clocks at the top and bottom of the shaft. The upper clock will be running faster than the lower clock and observers at the top and bottom of the shaft would both agree this was the case. The lower observer would see more tick pulses from the upper clock compared to his local clock and vice versa. Since matter would fall from the top of the mine towards the bottom, it seems that saying matter tends to move towards a region where a lesser amount of proper time elapses, would be more accurate. Talking in terms of coordinate time might be even better, because technically the rate of proper time is always one second per second.

Now imagine a very rigid perfectly spherical spinning Earth. The clock at equator would be running slower at than the clock at the pole because they both experience the same amount of gravitational time dilation and the equatorial clock experiences an additonal amount of time dilation due to its velocity. The velocity component subtracts from the gravitational component to make the effective gravitational potential at the equator lower than the gravitational potential at the poles. Now if liquid is added to the surface of this rigid spinning sphere at one of the poles, it will move from the higher effective gravitational potential to the lower effective gravitational potential at the solid surface of the equator. If enough liquid is added the effective gravitational potential will be the same everywhere at the surface of the liquid (sea level) creating the classic bulged shape of a rotating planet. Clocks anywhere on the liquid surface will run at the same rate because they are all at the same effective gravitational potential.

Quote Quote by S.R.Wilton View Post
If so, wouldn't that mean there is a "gradient of time" that exists on the planet? (i.e, aquatic animals on the bottom of the ocean age more quickly relative to animals on the surface of the planet;
Taking the statement by PeterDonnis "So the speed-up due to increased altitude is almost twice the slow-down due to the equatorial velocity." at face value, it follows that animals at the bottom of ocean (at the poles or at the equator) age slower relative to animals on the surface.

Quote Quote by S.R.Wilton View Post
Now I'm just confused. Matter at the center feels (approximately) the same amount of gravitational attraction from all sides of the planet as opposed to being drawn toward the center of mass of the planet (yeah? no?)...
The important thing to remember in these sort of considerations is that gravitational time dilation is more closely related to the gravitational potential, than to the attraction or force of gravity. For example a clock at the centre of a Moon sized object will be running faster than a clock at the centre of an Earth sized object, because although the force of gravity is zero in both cases, the gravitational potential at the centre of the Moon sized object is greater than the gravitational potential at the centre of the Earth sized object.
PeterDonis
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Oct12-09, 10:30 PM
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After some further noodling with this, I find that I made some errors in my previous posts. The effect of Earth's oblateness does make enough of a difference to matter, as Cleonis says. Here's a modified calculation, properly accounting for the Earth's oblateness and consequent quadrupole moment.

I'll use the same definitions of variables as above, plus the following additional ones:

J_2 = Earth's dimensionless quadrupole moment = approximately 1.35 * 10^-3. (This is an upper bound of 2/5 of the Earth's oblateness of 1/297; the upper bound is 2/5 of that because 2/5 is the correct coefficient for the moment of inertia of a spheroid. I haven't been able to find a more accurate value, but the real value should be smaller than this upper bound--see below for comments on this.)

Let me now re-write things in terms of the Newtonian gravitational potential, [itex]\phi[/itex]. (I apologize for changing the use of the variable [itex]\phi[/itex] from my previous post, but the use I'm giving here is a standard one.) What I was using previously was the value of [itex]\phi[/itex] for a spherically symmetric source, namely

[tex]\phi = - \frac{G M}{r}[/tex]

However, for an oblate spheroid, like the Earth, the potential gains an additional term, and looks like this:

[tex]\phi = - \frac{G M}{r} + \frac{G M r_{eq}^2}{2 r^3} J_2 \left( 3 cos^2 \theta - 1 \right)[/tex]

where [itex]\theta[/itex] is zero at the poles and [itex]\pi / 2[/itex] at the equator. (The formula comes from the book Classical Mechanics, by Kibble & Berkshire, which can be found on Google Books using this link.) Note that the effect will be to make the potential *less* negative at the poles and *more* negative at the equator.

For a test object at rest in the field (such as one at the poles), the proper time lapse rate is given by

[tex]\frac{d \tau}{dt} = T = \sqrt{1 + \frac{2 \phi}{c^2}}[/tex]

However, for a test object moving in the field (which an observer on the equator is, since he's rotating with the Earth), the lapse rate is

[tex]\frac{d \tau}{dt} = T = \sqrt{1 + \frac{2 \phi}{c^2} - \frac{v^2}{c^2}}[/tex]

(This is actually an approximate formula; for very weak fields and small velocities, it's more than good enough.)

So the two values we need to calculate are:

[tex]T_{pole} = \sqrt{1 + \frac{2 \phi_{pole}}{c^2}} = \sqrt{1 - \frac{2 G M}{c^2 r_{pole}} \left( 1 - \frac{r_{eq}^2}{r_{pole}^2} J_2
\right)}[/tex]

[tex]T_{eq} = \sqrt{1 + \frac{2 \phi_{eq}}{c^2} - \frac{v^2}{c^2}} = \sqrt{1 - \frac{2 G M}{c^2 r_{eq}} \left( 1 + \frac{1}{2} J_2 \right) - \frac{v^2}{c^2}}[/tex]

Again plugging in numbers, I get:

T_pole = 1 - 6.963 * 10^-10

T_eq = 1 - 6.966 * 10^-10

Since T_pole is closer to 1, clocks will run faster at the *poles* than at the equator, by about 3 parts in 10^-13, *if* we use the value of J_2 given above. However, if the value of J_2 is somewhat smaller (as it probably is), T_pole and T_eq will in fact be equal (since the effect of *decreasing* J_2 will be to move T_eq *closer* to 1 and T_pole *further* from 1). And the actual value of J_2 *is* smaller than the upper bound I used above. Kibble & Berkshire, in their textbook, discuss how the "equipotential surface" of the Earth can be determined by including the effect of centrifugal force; their calculation is basically equivalent to the one I did above, where I included the term in v^2 in T_eq. (That v^2 term can be considered as a "potential due to centrifugal force" if you want to look at it that way.) Their contention is basically the same as what Cleonis was asserting in an earlier post: the surface of the Earth *is* an "equipotential surface" when the effects of centrifugal force are included. The value of J_2 they calculate that would support that conclusion is about 1.1 * 10^-3, which is indeed somewhat smaller than the one I used above, and is a perfectly reasonable value.

So, Cleonis, mea culpa: you were correct that, assuming the real value of the Earth's J_2 is the one that Kibble & Berkshire calculate, the proper time rates at the poles and the equator *will* be the same.


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