On the equatorial plane of the Kerr geometry, there are three
quantities that are identical to the equivalent quantities in the
Schwarzschild geometry:

(1) The transverse tidal stretching within the plane for a stationary
observer is M/r^3 (where by "stationary" I mean someone whose 4-
velocity is parallel to the Kerr geometry's timelike Killing vector),
just as in the Schwarzschild case.

(2) The gravitational blueshift relative to infinity is 1/sqrt(1-2M/
r), just as in the Schwarzschild case. Equivalently, g_{tt} = - (1-2M/
r), where t is the coordinate corresponding to the timelike Killing
vector.

(3) The covariant derivative of the unit spatial vector pointing
outwards from the hole, taken in the time direction (wrt proper time
for a stationary observer) has squared magnitude:

-M^2 / [ r^3 (r - 2M) ]

In the Schwarzschild case the covariant derivative represents a pure
acceleration; in the Kerr case it includes a component of rotation due
to frame dragging. But regardless of the hole's angular momentum, the
overall *magnitude* of the covariant derivative in question takes
precisely the same form.

Now, one of these three conditions could be considered mere gauge-
fixing; for example, we could take (1) as being the *definition* of
the r coordinate in the equatorial plane.

My question is, does anybody know a simple proof of (2) -- that is, a
way to derive it without actually knowing the Kerr metric? The result
seems completely unsurprising to me; blueshift should measure the
hole's mass and be impervious to its angular momentum. That's
certainly true for the distant weak-field metric. But is there some
basic principle at work that *demands* this for the precise, near-
field metric? If there is, I can't seem to figure it out!

If anyone's curious, the reason I'm asking is because I thought it
would be fun to try to derive some aspects of the Kerr geometry with
an absolute minimum of high-powered mathematics. (The derivation of
the complete solution is quite difficult.) It turns out that you can
derive almost everything in the equatorial plane, if you take (1) as a
gauge condition, and (2) and (3) as assumptions. But (2) seems so
"obvious" that I'm wondering if there's some independent reason why it
must be true.