Re: opposite of relativistic time contraction

On Mar 17, 12:55=A0pm, Erik Max Francis <m...@[EMAIL PROTECTED]
> wrote:
> Tim Little wrote:
> > On 2008-03-16, Erik Max Francis <m...@[EMAIL PROTECTED]
> wrote:
> >> This doesn't sound right to me. =A0Remember that geodesic
> >> gravitational time dilations are actually very minor effects; the
> >> maximal time dilation effect you can get from orbiting close to a
> >> black hole is only about a factor of two.
>
> > It can be arbitrarily great, depending upon how close your orbit is to
> > the photon sphere.
>
> Not according to every source I could find. =A0Gravitational time
dilation=

> for an object in a circular orbit in a Schwarzschild geometry is
>
> =A0 =A0 =A0 =A0 tau'/tau =3D (1 - R/r)^(1/2),
>
> where tau' is the proper time experienced by the orbiting clock, tau is
> the proper time experienced by a far away clock, r is the Schwarzschild
> coordinate, and R is the Schwarzschild radius. =A0Orbits by massive
> particles are unstable within the photon sphere (r =3D (3/2) R), so the
> maximum time dilation that's possible is a ratio of about 58%.
>
> If you hover at constant radial r, rather than stay in orbit, and then
> get arbitrarily close the _event horizon_ -- not the photon sphere --
> then you can achieve arbitrarily high time dilations. =A0But this
requires=

> =A0 arbitrarily high thrusts.

I think Erik and Tim are both right, just talking about different
things.  If you're orbiting a black hole close to the photon sphere,
then you'll be moving very rapidly and there will be time dilation due
to your orbital velocity, in addition to the gravitational red****ft.
This orbital velocity approaches lightspeed, so as Tim says the total
dilation effect can be arbitrarily high.  But I guess Erik's point is
that when you're talking about voids versus galactic clusters, the
appropriate comparison is with purely gravitational red****fts, which
aren't that high even quite close to a black hole (e.g. sqrt(1/3) at
the photon sphere).