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View Full Version : A simple but abstruse black hole question

Jeff Root
2008-Nov-08, 11:51 AM
For a nonrotating black hole...

What is the minimum angle at which a photon that crosses from
outside the photon sphere to inside can cross the event horizon?

-- Jeff, in Minneapolis

hhEb09'1
2008-Nov-08, 12:06 PM
For a nonrotating black hole...

What is the minimum angle at which a photon that crosses from
outside the photon sphere to inside can cross the event horizon?cross? :)

AndreasJ
2008-Nov-08, 12:21 PM
Any angle greater than zero will do.

Edit: Seems I misread the question. Given the specification the photon starts outside the photon sphere, I'm not sure, offhand.

timb
2008-Nov-08, 12:32 PM
Is the photon sphere the distance at which the exit cone becomes a hemisphere?

Jeff Root
2008-Nov-08, 04:12 PM
Yes, it is. The photon sphere is located at 1.5 Schwarzschild radii,
and is where light could theoretically orbit a non-rotating black hole.

-- Jeff, in Minneapolis

RussT
2008-Nov-08, 11:57 PM
Abstruse: Great word Jeff, But what is even more Abstruse, is why would anyone even consider an 'event horizon' as 'spherical'/'hemisphere of an exit cone', when there is absolutely 0 evidence of a Non-rotating black hole anywhere in our universe?

And. in north's thread, Phil's black hole 10 most interesting things, was linked where he starts out with...

3) They’re spheres. And they’re definitely not funnel shaped.

Rotating Black Holes do NOT have a 'spherical event horizon'!!!

cjameshuff
2008-Nov-09, 01:31 AM
Rotating Black Holes do NOT have a 'spherical event horizon'!!!

In any black hole likely to occur in nature, it's close enough that for almost all purposes it is perfectly spherical.

Why do you keep bringing this up? It's an approximation. Non rotating black holes have practically zero probability of actually forming, but they are simpler to analyze and in many cases the rotation makes no significant difference. Earth, the moon, and the sun are not perfect spheres either...would you object to someone approximating them as spheres for the purpose of illustrating an eclipse, or lunar phases and the day/night cycle? For computing the trajectory of a thrown ball here on Earth, do you insist that the Coriolis effects of Earth's rotation to be taken into account?

hhEb09'1
2008-Nov-09, 01:31 AM
Abstruse: Great word Jeff, But what is even more Abstruse, is why would anyone even consider an 'event horizon' as 'spherical'/'hemisphere of an exit cone', when there is absolutely 0 evidence of a Non-rotating black hole anywhere in our universe?Same reason they studied black holes for forty years before there was any evidence of them? :)

PS: And, what cjameshuff said

Jeff Root
2008-Nov-09, 04:43 AM
Yes, what cjameshuff said was right on!

-- Jeff, in Minneapolis

timb
2008-Nov-09, 10:23 AM
Yes, it is. The photon sphere is located at 1.5 Schwarzschild radii,
and is where light could theoretically orbit a non-rotating black hole.

-- Jeff, in Minneapolis

That sort of answers your question, doesn't it? if the photon trajectory is below the plane tangent to the photon sphere then it is outside the exit cone and should be headed for a date with the singularity. This implies that, barring a source of acceleration, light (and anything else) that crosses the photon sphere is doomed.

RussT
2008-Nov-09, 10:27 AM
In any black hole likely to occur in nature, it's close enough that for almost all purposes it is perfectly spherical.

Why do you keep bringing this up? It's an approximation. Non rotating black holes have practically zero probability of actually forming, but they are simpler to analyze and in many cases the rotation makes no significant difference. Earth, the moon, and the sun are not perfect spheres either...would you object to someone approximating them as spheres for the purpose of illustrating an eclipse, or lunar phases and the day/night cycle? For computing the trajectory of a thrown ball here on Earth, do you insist that the Coriolis effects of Earth's rotation to be taken into account?

Why does everyone keep throwing "Irrelevencies" at this???

All the 'spheres' you are talking about have NOT collapsed and more importantly Imploded, to cause an Event Horizon to form!!!

None of the spheres you are talking about had 0 angular momentum when they began their collapse that ends in "Implosion"! There is NO linear collapse!

So, it is impossible for a 'spherical' event horizon to have ever even had a chance to form! Just because it is 'easier', does NOT mean the maths are modeling anything "REAL"!!!

And, if anyone thinks that equating a balls trajectory at the earths surface is 'nearly' equivilent to what is happening at the 'Boundary' of a rotating black hole, stellar or Massive, something is seriously wrong.

WaxRubiks
2008-Nov-09, 02:05 PM
I know what you mean RussT, there's a quantum difference in the models.
And by quantum difference, I mean they are completely different models.

Any way, I think light entered the theorised event horizon, at 90 degrees to the tangent of the event horizon; well with a non-rotating liquorice star(BH).

Jeff Root
2008-Nov-09, 02:43 PM
Russ, as the original poster in this thread, who has asked a question
You've made it clear that you disagree with premises of the question.
Fine. Take the rest to ATM. No more here, please.

-- Jeff, in Minneapolis

Jeff Root
2008-Nov-09, 03:04 PM
That sort of answers your question, doesn't it? if the photon trajectory
is below the plane tangent to the photon sphere then it is outside the
exit cone and should be headed for a date with the singularity. This
implies that, barring a source of acceleration, light (and anything else)
that crosses the photon sphere is doomed.
It doesn't answer my question at all. I knew that any photon which
crosses from outside the photon sphere to inside is doomed to also
cross the event horizon. I asked for the minimum possible angle of
that crossing.

Does it always cross at 90 degrees, as Frog March apparently thinks?
Or are shallower angles possible? It seems to me that if light crosses
the photon sphere (r=1.5) at an angle only slightly higher than zero
degrees, it should reach the event horizon (r=1) at some angle less
than 90 degrees.

Also, if light behaves at all like other particles, it seems to me that
it would have to fall in with some sideways speed in order for any
black hole to have rotation. If everything always fell straight in, no
black hole could rotate.

-- Jeff, in Minneapolis

hhEb09'1
2008-Nov-09, 04:32 PM
Also, if light behaves at all like other particles, it seems to me that
it would have to fall in with some sideways speed in order for any
black hole to have rotation. If everything always fell straight in, no
black hole could rotate.What if the mass were rotating before it bacome a black hole?

The path of a photon is problematic here, too. As the discussion in the two-slit thread reveals, the path of the photon could go through both slits. Each slit would result in a different angle from the photons final point of impact at the screen. Light photons don't behave like other particles, but we can assume they follow the geodesics of spacetime--which is probably where Frog March is coming from, right?

Jeff Root
2008-Nov-09, 04:40 PM
I just had to use the word "photon" in the OP, huh? Pretend that I
said "a beam of light" instead of " a photon". We can ignore QM here.

-- Jeff, in Minneapolis

hhEb09'1
2008-Nov-09, 04:51 PM
We can ignore QM here.TG! Still, a beam of light would follow a null geodesic, in spacetime, right? What do those look like around a non-rotating black hole?

timb
2008-Nov-09, 09:45 PM
It doesn't answer my question at all. I knew that any photon which
crosses from outside the photon sphere to inside is doomed to also
cross the event horizon. I asked for the minimum possible angle of
that crossing.

Does it always cross at 90 degrees, as Frog March apparently thinks?
Or are shallower angles possible? It seems to me that if light crosses
the photon sphere (r=1.5) at an angle only slightly higher than zero
degrees, it should reach the event horizon (r=1) at some angle less
than 90 degrees.

Also, if light behaves at all like other particles, it seems to me that
it would have to fall in with some sideways speed in order for any
black hole to have rotation. If everything always fell straight in, no
black hole could rotate.

-- Jeff, in Minneapolis

Sorry, I misunderstood the question. I don't think 90o could be right. Otherwise one has to wonder where the angular momentum went. If I understood this paper (http://th-www.if.uj.edu.pl/acta/vol20/pdf/v20p1015.pdf) I could probably answer the question. :sad:

hhEb09'1
2008-Nov-09, 09:58 PM
If I understood this paper (http://th-www.if.uj.edu.pl/acta/vol20/pdf/v20p1015.pdf) I could probably answer the question. :sad:The last four words of the paper are "of freely falling photons". Try that five times fast.

But, in what situations are photons not freely falling? How much deviation is there?

tdvance
2008-Nov-09, 10:26 PM
well, if you magnetize the photons...oh, wait a minute.

The only way I can think of a photon not being in free fall would be if you ramp up the microscope enough that quantum effects are important.

As far as I know, while photons are produced or absorbed according to electromagnetic, strong, or weak fields, only gravity affects them in flight.

Maybe it means, "but not at the point a photon is emitted or absorbed".

tdvance
2008-Nov-09, 10:29 PM
Actually, I just thought of something--since the subject at hand is photons affected by a black hole, perhaps "freely falling" is a bit of a misnomer for "affected by the Black Hole's gravity, and no other".

loglo
2008-Nov-12, 03:40 AM
Hi Jeff
Interesting question. I think what you are looking for is part of the mechanics of Plunge orbits, defined briefly here (http://www.uwec.edu/Math/Research/posters/2007_LightCone.pdf). Photon trajectories are analysed in this paper (http://adsabs.harvard.edu/abs/1976PhRvD..14.3281Y) but I can only see the abstract and it is for Kerr-Newman type holes. There might be something in there you could use though.

The paper is concerned with a study of parabolic orbits around a Kerr-Newman black hole, with special emphasis on the cross section for plunge orbits and the periastron of escape orbits. Photon orbits are also considered, and again the cross sections and accreted angular momenta for various types are calculated.

This stuff mostly seems to be used for studying black hole mergers and accretion. Wolfram (http://demonstrations.wolfram.com/OrbitsAroundASpinningBlackHole/)even has a demo up for such where you can fiddle with the starting parameters.

My intuition is that different sized BH's would have different photon entry angles at the EH simply because the speed remains constant but the distance travelled to the EH is larger for larger BH's and the gradients are shallower. But black holes have a habit of chewing up intuition and spitting it out, or not, as the case may be. :)

grant hutchison
2008-Nov-12, 11:14 AM
Jeff:
The answer is observer-dependent. Observers who are stationary in the Schwarzschild metric will measure one angle; free-falling observers will measure something completely different.

Grant Hutchison

Jeff Root
2008-Nov-12, 01:05 PM
The answer is observer-dependent. Observers who are stationary
in the Schwarzschild metric will measure one angle; free-falling
observers will measure something completely different.
I should have expected that.

Given that angular momentum would always disappear if anything
coming in initially at a low angle crossed the event horizon straight
down, it seems to me that all observers must see such light cross
at some angle. A high angle for the stationary observer and a low

I'm interested in what a distant, stationary observer would see,
as well as an observer in a stable orbit. But since I also want to
know the path of the light after it crosses the event horizon, the
view of a free-falling observer is the only one that would work for
the whole distance.

This is what I've got: Black hole animation (http://www.freemars.org/jeff2/BH3c.htm)

-- Jeff, in Minneapolis

grant hutchison
2008-Nov-12, 02:27 PM
This is what I've got: Black hole animation (http://www.freemars.org/jeff2/BH3c.htm)I assume this is a plot of the Schwarzschild metric. What are you using to calculate the light path?

Grant Hutchison

Jeff Root
2008-Nov-12, 04:10 PM
My brain.

It's a plot of "That looks about right..."

-- Jeff, in Minneapolis

grant hutchison
2008-Nov-12, 04:13 PM
My brain.

It's a plot of "That looks about right..."Ah.
Have you considered letting your brain use formulae and a calculator?

Grant Hutchison

Jeff Root
2008-Nov-12, 04:33 PM
Have you considered letting your brain use formulae and a calculator?
Yes, but that's pie in the sky. It would be nice...

-- Jeff, in Minneapolis

grant hutchison
2008-Nov-12, 05:03 PM
Yes, but that's pie in the sky. It would be nice...See if you can find a copy of Taylor and Wheeler's Exploring Black Holes: Introduction to General Relativity. They devote a chapter to deriving light trajectories, and deal with the difference between what a Schwarzschild observer plots, and what local stationary and free-falling observers see.
They express the answers in terms of an "impact parameter", b, which is exactly the parameter you're varying in your animation: the initial separation (at infinity) between the light beam and a parallel radial beam.

Grant Hutchison

thorkil2
2008-Nov-12, 06:15 PM
Rotating Black Holes do NOT have a 'spherical event horizon'!!!

The merits of this statement are questionable, but if there are any, they would be frame-dependent. Your statements about rotating/non-rotating black holes and donut-shaped event horizons have been addressed recently in another thread in ATM where they belong.