2
$\begingroup$

Does General Relativity theory correctly explain the ellipsoidal shape of the earth?


It seems it does not because the Thirring expression¹ for the force of a spherical shell—of mass $M$, radius $R$, and spinning at $\vec{\Omega}$—on an internal test particle of mass $m$ at $\vec{r}$, $$\vec{F}=-\frac{4GM}{15Rc^2}\left[\color{red}{2m\left(\vec{\Omega}\cdot\vec{r}\right)\vec{\Omega}}+\color{green}{m\vec{\Omega}\times\left(\vec{\Omega}\times\vec{r}\right)}+5\cdot\color{blue}{2m\left(\vec{v}\times\vec{\Omega}\right)}\right],$$ his terms differing from those of the most general Newtonian expression for fictitious forces acting on a particle moving in a rotating, non-inertial frame: $$\vec{F}_f=\color{green}{m\vec{\omega}\times\left(\vec{\omega}\times\vec{r'}\right)}-\color{blue}{2m\left(\vec{\omega}\times\vec{v'}\right)}-m\frac{d\vec{\omega}}{dt}\times\vec{r'}-m\frac{d^2\vec{h}}{dt^2},$$$\vec{h}$ being the displacement between the inertial and non-inertial frames, the third term being Euler's force, and the last term being the fictitious "force of inertia."³

Thus, Thirring predicts:

  1. an additional fictitious force that has no equivalent in Newton's theory: $\color{red}{m\left(\vec{\Omega}\cdot\vec{r}\right)\vec{\Omega}}$;
  2. the coefficient of the centrifugal term, $\color{green}{m\vec{\Omega}\times\left(\vec{\Omega}\times\vec{r}\right)}$, is 5× smaller than that of the Coriolis term, $\color{blue}{2m\left(\vec{v}\times\vec{\Omega}\right)}$; but in Newton's theory, the coefficients are equal.

This would seem to greatly affect how GR explains the ellipsoidal shape of the earth.


References:

  1. Mashhoon, Bahram, Friedrich W. Hehl, and Dietmar S. Theiss. “On the Gravitational Effects of Rotating Masses: The Thirring-Lense Papers.” General Relativity and Gravitation 16, no. 8 (August 1, 1984): 711–50. doi:10.1007/BF00762913.

  2. Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Apeiron, 2014. p. 293 (PDF p. 313), equations (16.16) and (16.17). For the application of this to the earth's flattening, see §16.5.2, pp. 298-300 (PDF pp. 318-20).

  3. Symon, Keith R. Mechanics. Reading, Mass.: Addison-Wesley Pub. Co., 1971. ch. 7.
    Lanczos, Cornelius. The Variational Principles of Mechanics. Toronto: University of Toronto Press, 1970. ch. IV, §§4-5.

$\endgroup$
6
  • 6
    $\begingroup$ Yes. But then again, so does Newtonian mechanics. There's no need to invoke general relativity here. By design, general relativity agrees with Newtonian mechanics agree in the limit of small masses, small velocities. $\endgroup$ – David Hammen Sep 26 '14 at 23:41
  • $\begingroup$ To lowest orders essentially answered by physics.stackexchange.com/q/8074/2451 in combination with physics.stackexchange.com/q/89/2451. $\endgroup$ – Qmechanic Sep 27 '14 at 0:11
  • $\begingroup$ @Qmechanic: I added much background to the question. An answer should address Thirring. $\endgroup$ – Geremia Sep 27 '14 at 18:25
  • $\begingroup$ You need to revert all of those changes. You have completely changed the nature of the question with this stealth question editing. Ask your new question in a different question. Your new question is fundamentally wrong. The Lense-Thirring effect does not predict the forces you wrote. If it did, this would suffice to immediately falsify general relativity. $\endgroup$ – David Hammen Sep 28 '14 at 14:57
  • $\begingroup$ @DavidHammen: I simply added background to the question, the question remaining the same. Also, what is the correct expression for the Thirring force, then? $\endgroup$ – Geremia Sep 28 '14 at 22:00
2
$\begingroup$

Yes, the shape of the Earth is consistent with general relativity. Thirring's equations are not an indication that general relativity predicts results that are inconsistent with the observed shape of the Earth, in large part because Thirring's equations applied general relativity incorrectly.

For clarity, this question and answer are purely about Thirring's application of general relativity to the inside of a rotating spherical mass shell, that Thirring came up with on his own. They do not pertain to the application of general relativity to the outside of a solid rotating object that Thirring collaborated with Lense on, the Lense–Thirring effect, for which Thirring is primarily known.

Even if Thirring's equations were correct, they wouldn't have a significant implication for the Earth's shape, because Thirring's equations were only intended to be applicable for $r\ll R$. Thirring intentionally dropped terms of order $M\omega^{2}r^{2}/R^{2}$. The analysis was for a rotating shell of distant masses.

The problems with Thirring's analysis were corrected very slowly by a set of at least nine physicists over the course of about two-thirds of a century. One of the biggest problems with Thirring's model is that it violates the conservation of the stress energy tensor, ${T^{\mu\nu}}_{;\nu}=0$. One of the last two problems to be addressed is that any physically realistic rotating mass shell will experience a centrifugal deformation, rather than remain spherical.

The final remaining problems with the analysis of general relativity inside a rotating shell were resolved by Pfister and Braun's analysis in 1985. According to Pfister and Braun's analysis, the interior of the shell experiences Coriolis and centrifugal forces, but no other forces. Furthermore, Pfister and Braun's analysis is applicable for all $r<R$, not just for $r\ll R$.

Of your references, it's understandable if Symon presents the analysis of the inside of a rotating shell as being an ongoing problem, because that book predates Pfister and Braun's paper by about 14 years. Assis, on the other hand, does not have that excuse.

For a history of the general relativistic analysis of the inside of a rotating shell, see the section "Thirring's Work on the Rotating Mass Shell, and the Problem of a Correct Centrifugal Acceleration" on pages 498-500 of the book "General Relativity and John Archibald Wheeler" by Ciufolini and Matzner.

$\endgroup$
1
  • $\begingroup$ I'm considering the earth's inertia as due to the gravitational influence of the spinning shells (Mach's principle), so the earth is inside Thirring's spinning shells, and comparing that to the Newtonian expression, where, in Newton's theory, inertia is a property intrinsic to the earth's matter. $\endgroup$ – Geremia Sep 29 '14 at 7:09
6
$\begingroup$

(The following answer was written to address the original version of this question, which was simply "Does General Relativity theory correctly explain the ellipsoidal shape of the earth?")

Yes. General relativity predicts that the equator will bulge out just enough such that the reduction in gravitational time dilatation at the equator relative to the North pole due to the equator being further from the Earth's center will just balance out with the special relativistic time dilation at the equator due to the Earth's rotation. This prediction is consistent with observation; the Earth bulges out just enough such that atomic clocks run at essentially the same speed at sea level (or more precisely, on the geoid) no matter where they are on the Earth's surface. And a difference in elevation of only a meter will result in a measurable difference in how fast an atomic clock will run, so that's a rather precise description of how much general relativity predicts that the Earth will bulge out.

The basic reason why the Earth assumes the shape that it does can be described in the same way for general relativity as for Newtonian mechanics: The Earth assumes a shape such that points attached to the Earth's surface form an equipotential surface, such that the sum of the gravitational force and the centrifugal force form a combined force that has the same magnitude at any point on the Earth's surface, and that combined force everywhere points directly toward the surface (idealized as being a smooth surface).

It's easy to understand why it makes sense for the Earth to behave that way if you picture the Earth's surface as being smooth, and think about a loose spherical rock on that smooth surface. If the total force on the rock didn't point directly toward the surface, the rock would roll "downhill". The final stable configuration of the Earth's surface will occur after everything on the surface has "rolled downhill" as much as possible, at which point the surface will no longer have a "downhill", i.e., the sum of the gravitational force and the centrifugal force will point directly toward the surface.

One key guiding principle in general relativity is the equivalence principle, which says that the gravitational "force" at a point is locally indistinguishable from a pseudoforce that only appears to be a force due to using an accelerating frame of reference. A frame of reference attached to the Earth's surface is actually an accelerating frame of reference, whose origin is accelerating upward. A frame of reference centered on an object in free fall isn't actually accelerating; it's an inertial frame of reference.

The point is that due to the equivalence principle, it's impossible to locally distinguish between the "centrifugal force" and the "gravitational force" at a point on the Earth's surface. From the point of view of a frame of reference attached to the Earth's surface, there locally is just one pseudoforce involved, and that one pseudoforce is what's responsible for the time dilation relative to a nearby inertial frame of reference. Since the magnitude of the pseudoforce is constant at all points on the equipotential surface, the amount of time dilation is the same at all points on the equipotential surface, and only depends on the effective potential at the surface, so that's one way of describing exactly where that equipotential surface is.

The equipotential surface as calculated with general relativity is in a different location than where the equipotential surface would be according to Newtonian physics, but it's only a very tiny difference; the general relativistic effective potential only differs from the Newtonian effective potential by about one part in $10^{11}$.

For a more in-depth, mathematical exploration of how the equivalence principle and time dilation relate to the equatorial bulge, see this paper.

$\endgroup$
4
  • $\begingroup$ Nice answer, +1, and nice paper at the end, except for one thing. In the paper, Sam asks "Taking our model of Earth as a rotating perfect fluid, is the Earth an equipotential surface?" Kim's answer is "Because if it wasn’t, the sea water would feel a force $\vec F = −m \nabla \phi$ and would move until $\nabla \phi = 0$ everywhere on the surface." That is so very, very wrong! How did that pass peer review? The only way $\nabla \phi$ can be zero for an equipotential surface is if that surface is a plane. It obviously isn't a plane. It's an ellipsoid. Instead, $-\nabla \phi = g$. (Continued) $\endgroup$ – David Hammen Sep 27 '14 at 11:07
  • $\begingroup$ The right answer must invoke the second law of thermodynamics. A hanging chain follows a catenary, a ball comes to rest at the bottom of a bowl, a fluid in a tube forms a meniscus, and the Earth's surface is very close to an equipotential surface all for the same reason, which is the second law of thermodynamics. Minimizing the total potential energy maximizes entropy. If there's a pathway that takes a system toward the configuration that minimizes total potential energy / maximizes entropy, the system will try very hard to follow that path. $\endgroup$ – David Hammen Sep 27 '14 at 11:12
  • $\begingroup$ If you discuss the Thirring objection (see the edited question above), I'll consider accepting this answer. thanks $\endgroup$ – Geremia Sep 27 '14 at 19:47
  • 1
    $\begingroup$ @DavidHammen This answer essentially does address the second law of thermodynamics, it just does so at a simplified level. It says that the Earth's final stable state occurs after everything has "rolled downhill" as much as possible. "Rolling downhill" in classical terms is reducing potential energy, and "as much as possible" is when the entropy has been increased as much as it can be easily. The answer assumes that there's a final non-oscillatory state, i.e. it doesn't pretend that friction is zero and hence pretend that entropy isn't increased in the process of "rolling down hill". $\endgroup$ – Red Act Sep 28 '14 at 1:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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