Re: does light travel forever?
- From: bdbryant@xxxxxxxxxxxxxxx (Bobby D. Bryant)
- Date: Tue, 17 Jan 2006 22:07:06 +0000 (UTC)
On Tue, 17 Jan 2006, "David Ewan Kahana" <dek@xxxxxxx> wrote:
> Bobby D. Bryant wrote:
>> On Tue, 17 Jan 2006, John Vreeland <vreejack@xxxxxxxxxxx> wrote:
>>
>> > 2. it climbs out of a gravity well (though it usually loses exactly
>> > what it gained when it fell into the well). You could argue that
>> > all light leaving a black hole (the ultimate gravity well) loses all
>> > its energy but there are other ways to interpret it as well.
>>
>> Is it legitimate to think of a black hole as an object where the
>> escape velocity exceeds the speed of light, and the event horizon
>> is the highest orbit you can reach if you're capable of light speed?
>
> It's more complicated. This is basically a non-relativistic
> way of thinking about a black hole.
>
> Essentially, there are no stable orbits at the radius of the
> event horizon of a Schwarzchild black hole, and you could
> never reach the horizon if you start from inside the horizon.
> The lowest stable circular orbit for a photon I believe is actually
> at twice the Schwarzchild radius. All initially light-like orbits at
> the horizon fall in, except for one special case. That's the case
> of a photon which is emitted directly radially outward exactly at
> the horizon ... such a photon hovers at the event horizon forever,
> moving neither outward nor inward.
>
> To maintain an orbit below the minimum stable radius (2R)
> for photons would require a continuous burn of fuel.
>
> The event horizon is what is called a closed, trapped, null
> surface. All future directed geodesics originating inside such
> a surface remaine inside that surface forever.
Thanks.
Do we have any idea what photons do inside a black hole? (When you
say "fall in", does that mean head toward the center?)
--
Bobby Bryant
Austin, Texas
.
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