A photon comes in from infinity and performs a hyperbolic curve-- the
tightest part of the curve touching part of a photon orbit and escaping to
infinity again. On a small section of the tightest part of the curve, what
will its linear speed on the curve be relative to c. ie its circumferential
speed?- to an outside observer. c or less than c.--?
regards Dave R.

What does "the tightest part" mean? You mean greatest curvature? What does
"touching part of a photon orbit" mean? Indeed, what the **** is a "photon
orbit"? Escaping to infinity? Where the hell is that?

"Ruth & Dave" <[Only registered users see links. ].nz> wrote in message
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Presumably you are referring to black holes and the reference to
"photon orbit" is the photon sphere?

If it is a black hole, a hyperbolic orbit with closest approach
*at* the photon sphere is impossible. Slightly outside the
photon sphere, the "path" would be *very* rotated (would not
appear to be hyperbolic), but escape to infinity would still be
possible.

For a freefalling observer, all local light speed is c. The last
stable orbit for matter is something like 6m, so this is the last
freefalling observer that you can poll for very long. The photon
sphere is 3m, with the event horizon at 2m. Beyond this limited
case, I cannot say. Perhaps you should ask over on
sci.physics.relativity?

And if you didn't mean black holes, I second odin's question-set.

Thanks David
You were right re photon orbit being photon sphere- my mistake.
Going out a little from the sphere is fine for the example but I am
interested in the apparent speed of the photon to an outside observer. I am
after knowing what the linear speed is on the radius line of the photons
path. Scribe a line at r- what speed will it be going along this line? Less
than c perhaps.
Example== would I be right in saying the linear speed of a photon along a
radius when the photon is on the event horizon for a short time would be
absolutely zero. Its going out speed is c- matched by gravity in at c it
can only head out vertically on or near the event horizon. Just slight
lateral movement and it is in.
If we come out a little to as you say just outside the photon sphere-
gravity will allow lateral movement but because there is an inward vector
of gravity- the radial speed will be less than c. True or false?
I hope this is a little clearer re my question.
regards Dave R

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"Ruth & Dave" <[Only registered users see links. ].nz> wrote in message
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c is just a conversion constant. You already know that
gravitational time dilation has observable effects.

No. You can say that a photon released, and directed, precisely
out of the hole will not leave the environs of the event horizon
in finite time. But photons with any other direction, will go
precisely into the hole. And any freefalling observers will
always measure all these photons to be propagating at c.

.... or ingestion of additional mass/energy.

I'm not trying to duck your question. I have not gotten to crack
_Gravitation_ by Misner, Thorne, and Wheeler yet... perhaps a
direct answer is in there. But I would suspect not.

As to light moving outside the photon sphere, witness "Einstein
ring" and "Einstein cross". These are photons that dive into a
gravity well, and are deflected by it.

Hello and Thanks David
re
out of the hole will not leave the environs of the event horizon
in finite time. But photons with any other direction, will go
precisely into the hole. And any freefalling observers will
always measure all these photons to be propagating at c.
This is where I can see something happening. I absolutely agree with
this statement for the example on the event horizon-- any other direction
and it is going in-
If we come out a little and the photon is as you say -directed precisely
out of the hole
we now have a photon which can escape easily if somewhat slowly. Right here
is my point of interest because it is about here that we can introduce just
the slightest angle away from "precisely out of the hole" and if that angle
were taken out to its maximum without the photon going in-- it would begin
to move a very slow orbital vector-- for sure very unstable. This is the
radial speed less than c I am on about.
I hope you can see what I am on about and the shoe fits.
regards Dave R

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"Ruth & Dave" <[Only registered users see links. ].nz> wrote in message
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I am leary of applying the term "orbital" to such trajectories.
The only orbital a photon can be in is the photon sphere at 3m.
You are discussing (presumably) the vicinity of 2m to 3m... At
3m a tangentially directed photon enters orbit. Beneath this,
there are angles between normal and 90 deg that will, perhaps,
also never quite get out.

But then black holes do get larger, with other matter nearby, and
such "fairy tale" trajectories are doomed.

It is a shame that no externally sourced (say outside 6m) photons
could be stored...

Hi David
At this moment in time I am clear about the movement of a photon just
outside the event horizon- yes I may be wrong about this "fairy tale" orbit
but to me it is non fiction and is clearly an orbit.
I used to think that the photon was either out or in. I have changed my
mind.
Could you try this---
Draw a horizontal straight line at the bottom of a sheet of paper and draw
another about three inches up and parallel. Draw a vertical line at 90
degrees right through both in the centre of the sheet.
The bottom intersection for now is a photons departure point going straight
up--vertically- in free space it would travel over a short amount of time to
the next intersection with the other horizontal line at c. Now lets make the
bottom line the event horizon and still its departure point and off we go
again. This time the photon stays absolutely still and the upper line comes
down at c. Now (at the risk of being too obvious) the photon is still
heading out at c but the gravitational environment it is in is going in at
c-- all unstable but accepted theory I presume. For this exercise it still
has just travelled between the lines.
Lets extend the vertical line slightly by say a quarter of an inch and
consider the two horizontal lines represent the gravitational force just a
little further out-- not quite enough to hold a photon from going out and so
it has managed to head out a little. Over the same period as before--the top
line has come down to the bottom line and left a photon a quarter of an inch
up the page. Now scribe a compass line through the lower line intersection
and through the photon just above the other line to intersect the top
horizontal line and draw a line from the compass point through the
intersection. Now what I am saying is a photon could start at this lower
point and head out along this intersecting line--Now bring the top line down
once more with the photon a quarter of an inch up and look at the ark and
the compass intersection. I am saying anything that leaves the lower line
will be able to get out as long as it leaves at a more vertical angle than
this new scribed line. The photon precisely on the scribed line ends upat
the scribed intersection--- the intersection is now at the bottom of the
page and the distance made good by the photon is the small straight between
the vertical intersection and the compass intersection. So here we have a
photon going at c but travels that short distance along the top line which
is a lot less than c.This top line is now the bottom line and the photon has
progressed orbitaly and will continue to do so as long as it heads out on
that same angle I suppose this is a little like a boat on a river.
Simple calculations for orbital progress in any position could be used as
all examples for a non rotating black hole would be right angle triangles-
including any point in space which curved a photons path-- all seem to slow
the photon down but leave it at c. Sorry if my explanation is confusing.
This would be a lot easier on a white board.
regards Dave R

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"Ruth & Dave" <[Only registered users see links. ].nz> wrote in message
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Try this modification. The lower line label "event horizon".
Upper line label "photon sphere". Take a compass, and sweep two
arcs tangent to the 90 degree bifurcator at the event horizon,
and tangent to the photon sphere to the left and right of the
bifurcator. It should look something like a "V" (or cleavage).

EH = event horizon; BH = black hole

The area below the V should be labelled all photons cross the EH
in finite time. The area above the V should be labelled all
photons with departure angle less than the arcs will leave the BH
in finite time. The inifintely thin surface represented by the V
are photons that will never leave the environs of the BH, unless
the BH consumes more mass, or the BH evaporates. The BH is a
dynamic system.

....

But I think it is merely an intellectual exercise. The only
photon donors to these trajectories are infalling at nearly c, so
any visible light would be horribly red shifted.

Hello David
Yes I do like the cleavage diagram and hope you believe I have been there.
It is from there that we can further modify the cleavage diagram and draw in
another horizontal line just above EH and we now have the cleavage lines
permitting slightly less than vertical angles for an escaping photon.
If we go step by step from totally vertical and see where the photon ends
up in each case. it will end up starting its journey from this new line at
precisely the maximum permitted angle on the cleavage line, and it is this
angle which can only leave it in a perfect if unstable orbit. Not quite
going in-- not quite going out. It will stay on this radius line from the
singularity as infinate an amount of time as a photon on the event horizon.
In all cases and especially this one- the radial progression of the
photon is less than c but it is travelling always at c Any curve or bend of
a photons path due to a gravitational source will result in its progress
along the path being less than c.
I originally compared your cleavage diagram to an apple surface by
its stalk- I am sure you will see the comparison- but I like the cleavage
diagram better.
regards Dave R

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