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Re: [Phys-L] field of an ellipsoidal distribution



Yikes! For some reason my the thetas in my previous post showed up on the list as acccented e's. So to make sense of what I wrote below whenever you see a "θ" in my previous post it supposed to be a [theta]. This is confusing in the Somigliana equation because it also has the ecentricity e, in it, and it is easy to confuse an e with an accent with one without the accent.

David Bowman

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From: Phys-l <phys-l-bounces@www.phys-l.org> on behalf of David Bowman <David_Bowman@georgetowncollege.edu>
Sent: Saturday, August 8, 2015 5:13 PM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] field of an ellipsoidal distribution

Regarding JD's comment:

Consider the gravitational field of a slightly idealized
planet. We assume uniform density, but just for fun we
allow it to be neither static nor spherical. If it's
rotating, if it's in mechanical equilibrium it will be
an ellipsoid.

BTW, The result of a ellipsoid of revolution (spheroid) shape is more general than simply the constant density requirement. Other internal equations of state than a simple constant density independent of pressure are OK as well. All that is required for a spheroid surface shape is:

1) Newtonian physics (i.e. we neglect relativistic effects)
2) Perfectly flexible fluid-like behavior in mechanical equilibrium
3) The external universe is a vacuum with no other gravitational sources present.
4) The mass distribution at the surface all rotates together at the same angular velocity (i.e. there is no differential rotation or shearing on the surface between various latitude regions).

This remarkable result is sort of a rotational generalization of the non-rotational result that a non-rotating fluidic mass distribution assumes a spherical shape in equilibrium regardless of how compressible it is internally.

The general spheroidal shape result is a consequence of matching the external boundary condition at the surface for an external vacuum solution with the requirement that the surface be an equipotential surface in a common rotating frame with a common centrifugal field. The internal solution (whatever it is) never appears in the calculation. The result is (for me) easiest to show using confocal coordinates (with axial symmetry) because then the relevant surface is a surface of constant value of one of the coordinates independent of the other two coordinates. This makes the boundary condition matching easier. The mathematical reason for the result is that the confocal coordinate analog of the quadrupole moment field combined with the monopole field completely accounts for the anisotropy of the centrifugal field at the surface, so no higher order multipoles are needed to maintain the equipotential condition on the surface. The centrifugal field simply doesn't have any higher order multipole summetry than the lowest order quadrupole anisotropy.

This result means the external gravitational field of the mass distribution is completely characterized (in a frame is which it is at rest as a whole) by the specification of the distribution's total mass and its quadrupole moment strength (a function of the mass and spin angular momentum magnitude) and the orientation of its symmetry axis. It's sort of a Newtonian version of the fact that an external vacuum Kerr spacetime is completely characterized by the total mass and the total internal angular momentum.

Another consequence of this result is that the surface gravity g as a function of geodetic latitude, è must acquire the form:

g(è) = g_0*(1 + k*sin^2(è))/sqrt(1 - e^2 *sin^2(è))

where g_0 is the surface gravity on the equator, e is the spheroid's eccentricity, and k is a nonnegative constant that depends on how the mass is distributed internally. If the internal mass distribution has a constant density then k is zero. For Earth k = 0.00193185 indicating that the Earth is denser in its inside than on its outside. Also Earth's values for the other two parameters are e = 0.081819191 & g_0 = 9.7803268 m/s^2. I believe the the above equation for g(è) is called the Somigliana equation.

David Bowman
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