Re: Alternative to Kerr metric, (kst).

sal wrote:
We don't need high precision here to show the point. Let's just take
the linearized theory metric for Newtonian gravity. Let "p" be the
Newtonian gravitational potential ("p" for "\phi").
Then the metric is:
| -1-2p 0 0 0 |
g = | 0 1-2p 0 0 |
| 0 0 1-2p 0 |
| 0 0 0 1-2p |
[...]
Now we'll look at a rotating frame, with coordinates (t', x', y',
z'). [...]

That misses the point. You have a static physical situation, and you're going to look at it in rotating coordinates. Nothing _physical_ is rotating (relative to locally-inertial frames), just your coordinates, and you'll not get an equatorial bulge, an IFOG will detect no rotation, an artillery shell will display no "Coriolis force", etc. No _physical_phenomena_ related to rotation will be present; but if you use those coordinates for Newtonian mechanics you will need "centrifugal and Coriolis forces" (in GR they are merely aspects of the connection in these coordinates, so there's nothing special to do since you use the connection anyway -- GR automatically "compensates" for your coordinate choice. But NM does not and those fictitious "forces" are manual corrections).


Let's do this correctly:
Pick up a GR textbook and look up the Kerr metric (too complicated for ASCII). The key point is that it is not diagonal in _any_ coordinate system. That means the Kerr manifold is not static, because a static manifold is one in which there is a timelike Killing vector _and_ there exists a foliation of spacetime into space and time in which the spacelike 3-surfaces are orthogonal to the timelike Killing vector. As the Kerr metric is not diagonal in any coordinates, there is no such foliation. There is, however, a timelike Killing vector, and we call such a manifold stationary.

[Technical note: the above only applies to the region outside
the ergosphere.]

In Tucker's construction with a sphere and a hula hoop all at rest in some inertial coordinate system, the manifold is _static_, and therefore his construction cannot possibly be equivalent to the Kerr manifold -- it does not rotate (duh!).


Tom Roberts tjroberts@xxxxxxxxxx
.