2
$\begingroup$

Are the polar and equatorial radiuses of a rotation fluid ellipsoid approximately depended of the rotation velocity by this equation $$\frac{{GM}}{{{r_{{\text{pole}}}}}} - \frac{{GM}}{{{r_{{\text{equator}}}}}} \cong v_{{\text{equator}}}^2 = {\left( {\omega \cdot {r_{{\text{equator}}}}} \right)^2}{\text{ ?}}$$ This equation is performed with accuracy 2% for Earth for the polar radius 6356752 m, equatorial radius 6378137 m and the speed of rotation π/43200.

In small gravity the time dilation is $$1 - \sqrt {1 - \frac{{2GM}}{{r \cdot {c^2}}}} \approx \frac{{GM}}{{r \cdot {c^2}}}$$ At small rotation velocity the time dilation is $$1 - \sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} \approx \frac{{{v^2}}}{{2{c^2}}}$$ As a result, does the speed of the clock slow down on the pole relatively to the equator by this equation $$\frac{{GM}}{{{r_{{\text{pole}}}} \cdot {c^2}}} - \left( {\frac{{GM}}{{{r_{{\text{equator}}}} \cdot {c^2}}} + \frac{{v_{{\text{equator}}}^2}}{{2{c^2}}}} \right) \approx \frac{{{{\left( {\omega \cdot {r_{{\text{equator}}}}} \right)}^2}}}{{2{c^2}}}{\text{ ?}}$$ For Earth is it $\frac{1}{2}{\left( {\frac{{\frac{\pi }{{43200}} \cdot 6378137}}{{299792458}}} \right)^2} \approx 1.2 \cdot {10^{ - 12}}$ ?

On the web, I could only find that this is an equipotential surface, therefor there is not time dilation between pole and equator, because the time dilation from gravity is compensated by the time dilation from the movement. But the above calculation have showed that the movement compensates only a half of the change from gravity between pole and equator.

$\endgroup$

3 Answers 3

1
$\begingroup$

You can analyze this either in the rotating frame or in an inertial frame.

The analysis in the rotating frame is simple. In this frame, nothing is moving, so the surface of the earth, which is in hydrodynamic equilibrium, is an equipotential. The gravitational time dilation depends on the difference in potential, so it's zero.

You're doing the analysis in the inertial frame and trying to analyze the structure of the earth explicitly, which makes it quite a bit more complicated. In this frame, it is not true that the earth's surface is an equipotential. (Not only that, but it is not true that when you take a small mass from the pole to the equator, the sum of PE+KE for that mass is conserved -- in fact, both increase. This is balanced only by the reduction in KE of the rest of the earth, whose rotation slows down by conservation of angular momentum.)

The expression you're starting with for the equatorial-polar flattering is actually wrong for a self-gravitating, incompressible fluid. There is a nice treatment of the topic by Fitzpatrick here: https://farside.ph.utexas.edu/teaching/336k/Newton/node109.html One thing to watch out for is that you can't find the difference in potential between the pole and equator by using the potential of a sphere. In fact the potential of a uniform ellipse is different from that of a sphere, and the effect is of leading order. This produces a factor of 5/4, which is missing from your analysis.

On top of this is the fact that the earth's density is nonuniform. Therefore the polar flattening comes out to be different from the incompressible fluid result by about a factor of 1.28.

These two errors in your calculation almost cancel out, but they result in a net error of 2% in your calculation relative to what is actually observed for the earth.

So then you take this result for the flattening, which is fortuitously approximately right, and plug it in to the equation for the potential of a uniform sphere. But the potential isn't actually that of a uniform sphere, so you end up with a result that is wrong.

So although it is exactly true that the time dilation cancels, it's somewhat complicated to prove that by explicit calculation in the inertial frame, for a self-gravitating, rotating, incompressible fluid -- and the complication would be even greater if you wanted to model the actual earth, which is nonuniform.

If you want to see a little more explicitly that the cancellation of the relativistic effect always works to leading order and is independent of assumptions like incompressibility, a nice way to do it is the following. In Fitzpatrick's notation, let $\Phi$ be the Newtonian potential in the inertial frame. The field is minus the gradient of this. When we switch to the rotating frame, the potential changes by a term $\chi=-(1/2)\Omega^2 r^2\sin^2\theta=-(1/2)v^2$, whose gradient is the centrifugal acceleration. In the rotating frame, hydrodynamic equilibrium requires that $\Phi+\chi$ be constant for the surface of the earth. But this is equivalent to $\Phi-(1/2)v^2$ being constant, so that the gravitational and kinematic time dilations cancel to leading order, and this cancellation is exact in the sense of being independent of the detailed structure of the earth.

$\endgroup$
0
$\begingroup$

About the difference in gravitational potential between the poles and the equator:

When it comes to gravitational potential there is a difference between a spherical celestial body and a celestial body with an equatorial bulge.

As we know, in the case of a perfectly spherical gravitating body the center of gravitational attraction coincides with the geometrical center.

With a celestial body with an equatorial bulge (ellipsoid) that is not the case. Depending on the position of a test object relative to the ellipsoid the center of gravitational attraction is displaced from the geometric center in some direction. The evaluation of difference in amount of proper time between Equator and poles must take that into account.



There is a quick way of assssing the difference in gravitational potential between the Equator and the poles.

(The calculation is short; it is the explanation why it is valid that is not short.)

A rotating celestial body is in a state of dynamic equilibrium. (Over geologic time scale the solid mass of the Earth is sufficiently ductile to settle into equilibrium state.)

At each latitude the degree of equatorial bulge provides the required centripetal force to match the rotation rate. That is why the water of the oceans is not accumulating around the equator.

To give you an idea:
At 45 degrees latitude the corresponding slope is 0.1 degree. That is: at 45 degrees latitude the amount of required centripetal force is such that a slope of 0.1 degree provides that.

The ratio is about 1:580
Take your own body weight, and divide that by 580. At 45 degrees latitude, in the direction parallel to the local surface, that is the amount of required centripetal force.



The state of a fluid in hydrostatic equilibrium is called 'solid body rotation'. A property of solid body rotation is that the provided centripetal force is described with Hooke's law: the force increases linear with distance to the center of rotation.

With Hooke's law we have that the potential energy increases with the square of the distance to the center of rotation.

We have that kinetic energy is proportional to the square of velocity. In the case of solid body rotation: the velocity increases linear with the distance to the center of rotation.

Hence:
In the case of solid body rotation:
At each distance to the axis of rotation the amount of kinetic energy and the amount of potential energy are equal to each other.

The velocity of co-moving with the Earth along the Equator has a corresponding kinetic energy. The difference in gravitational potential energy between equator and poles is equal to that.

That difference in gravitational potential energy has a corresponding difference in height. That height difference is what you must use in your calculation to estimate the amount of difference in elapsed proper time between the Equator and the poles.

$\endgroup$
3
  • $\begingroup$ That height difference is used in calculation to estimate the amount of difference in elapsed proper time between the Equator and the poles. But was not written detailed. $\frac{{GM}}{{{r_{{\text{pole}}}} \cdot {c^2}}} - \frac{{GM}}{{{r_{{\text{equator}}}} \cdot {c^2}}} = \frac{{3.98600435436 \cdot {{10}^{14}}}}{{{{299792458}^2}}}\left( {\frac{1}{{{\text{6356752}}}} - \frac{1}{{{\text{6378137}}}}} \right) \approx 2.34 \cdot {10^{ - 12}}$ $\endgroup$
    – Imyaf
    Commented Aug 21, 2022 at 14:08
  • $\begingroup$ $\frac{{v_{{\text{equator}}}^2}}{{2{c^2}}} = \frac{{{{\left( {\omega \cdot {r_{{\text{equator}}}}} \right)}^2}}}{{2{c^2}}} = \frac{{{{\left( {\frac{\pi }{{43200}} \cdot 6378137} \right)}^2}}}{{2 \cdot {{299792458}^2}}} \approx 1.20 \cdot {10^{ - 12}}$ $\frac{{GM}}{{{r_{{\text{pole}}}} \cdot {c^2}}} - \frac{{GM}}{{{r_{{\text{equator}}}} \cdot {c^2}}} - \frac{{v_{{\text{equator}}}^2}}{{2{c^2}}} = 2.34 \cdot {10^{ - 12}} - 1.20 \cdot {10^{ - 12}} = 1.14 \cdot {10^{ - 12}} \approx \frac{{{{\left( {\omega \cdot {r_{{\text{equator}}}}} \right)}^2}}}{{2{c^2}}}$ $\endgroup$
    – Imyaf
    Commented Aug 21, 2022 at 14:12
  • 1
    $\begingroup$ @Imyaf You are using the geometric values for $r_{pole}$ and $r_{equator}$. However, in this case that introduces an error. The point is: the geopotential of a celestial body with an equatorial bulge is not spherically symmetrical. On Earth the difference in gravitational potential energy between Equator and poles corresponds to a height difference of about 11 kilometers, about half the difference between $r_{equator}$ and $r_{pole}$. I quote from the other answer: " [...] you can't find the difference in potential between the pole and equator by using the potential of a sphere." $\endgroup$
    – Cleonis
    Commented Aug 21, 2022 at 16:25
0
$\begingroup$

There is a problem with the expression that you are using.

To be clear: this post is not an answer, This is a comment. The only reason for using answer space: the comment space is too small to communicate what I need to communicate.

The expressions for time dilation that you used as your source express a ratio, not a difference.

Yet in your expression that contains both $r_{pole}$ and $r_{equator}$ you specify a subtraction; there is a minus sign in your expression.

But then: in the end you drop the difference in geopotential height altogether.

So in your expression there is an accumulation of two errors, and each is an error that renders the calculation invalid.

The hyperphysics page about gravitational time dilation gives the following expression for time dilation ratio betweeen a point at distance $R$ to a gravitating body, and infinity:

$$ T = \frac{T_0}{\sqrt{1-\frac{2gR}{c^2}}} \tag{1} $$

Then the following binomial expression is given:

$$ \frac{1}{\sqrt{1-x}} = 1 + \frac{x}{2} + \frac{3}{8}x^2 + \dots \tag{2} $$

Using (2) the following approximation is given:

$$ T = T_0(1 + \frac{gR}{c^2} + \dots ) = T_0(1 + 6.95\times 10^{-10} + \dots) \tag{3} $$

You want to calculate the ratio:

$$ \frac{T_{equator}}{T_{pole}} \tag{4} $$

That means you have to apply (3) in such a way that the difference in geopotential height between equator and poles is expressed. (For emphasis: what counts is difference in geopotential height, not difference in geometric height. The distinction between the two is described in the answer I submitted yesterday.)

$\endgroup$
3
  • $\begingroup$ Second order of binomial expression is $<10^{−18}$ , that does not matter. First order of binomial expression is enough in this task, because it gives relative tolerance $<10^{−6}$ after subtraction. $\endgroup$
    – Imyaf
    Commented Aug 21, 2022 at 14:17
  • $\begingroup$ Instead of division (4), subtraction was used $\frac{{1 + \Delta {T_1}}}{{1 + \Delta {T_2}}} - 1 \approx \Delta {T_1} - \Delta {T_2}$ , that introduces an insignificant error in this task. $\endgroup$
    – Imyaf
    Commented Aug 21, 2022 at 14:34
  • $\begingroup$ $$\frac{{\sqrt {1 - \frac{GM}{{{r_{\text{equator}}} \cdot c^2}} - \frac{{{{\left( {\omega \cdot {r_{{\text{equator}}}}} \right)}^2}}}{{2{c^2}}}} }}{{\sqrt {1 - \frac{{2GM}}{{{r_{pole}} \cdot {c^2}}}} }} = \frac{{\sqrt {1- \frac{{2 \cdot 3.98600435436 \cdot 10^{14}}}{{\text{6378137} \cdot {299792458^2}}}- \frac{{{{\left( {\frac{\pi }{{43200}} \cdot 6378137} \right)}^2}}}{{2 \cdot {299792458^2}}}} }}{{\sqrt {1 - \frac{{2 \cdot 3.98600435436 \cdot {{10}^{14}}}}{{6356752 \cdot {{299792458}^2}}}} }} = \frac{{1 - 6.965454 \cdot 10^{-10}}}{{1-6.976878 \cdot 10^{-10}}} = 1 + 1.1424 \cdot 10^{ - 12}$$ $\endgroup$
    – Imyaf
    Commented Aug 21, 2022 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.