# Thread: Light Inside a Black Hole

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## Light Inside a Black Hole

Let's pretend that I'm inside the event horizon of a black hole, and can't be spaghettified. I have a flashlight that I point directly out and turn it on.

Now, as I understand it the light cannot change speed, only frequency (i.e., become redshifted or blueshifted). So, shouldn't the light have to get out in spite of gravity? Am i missing something simple? Do I get to count the inward pull of gravity against the speed of light? In that case, couldn't I get a "perfect" spot where the inward pull was equal to the outward motion of the light, so that the light would effectively be standing still?

I realize that in a real black hole, the force of gravity would not be a neat, simple, inward pull. But, as a thought experiment what am I getting wrong??

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Originally Posted by Yagdrasil
Let's pretend that I'm inside the event horizon of a black hole, and can't be spaghettified. I have a flashlight that I point directly out and turn it on.

Now, as I understand it the light cannot change speed, only frequency (i.e., become redshifted or blueshifted). So, shouldn't the light have to get out in spite of gravity? Am i missing something simple? Do I get to count the inward pull of gravity against the speed of light? In that case, couldn't I get a "perfect" spot where the inward pull was equal to the outward motion of the light, so that the light would effectively be standing still?

I realize that in a real black hole, the force of gravity would not be a neat, simple, inward pull. But, as a thought experiment what am I getting wrong??
the first question that I have is why can you not be " spaghettified " ?

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Originally Posted by Yagdrasil
I realize that in a real black hole, the force of gravity would not be a neat, simple, inward pull. But, as a thought experiment what am I getting wrong??
Well, inside the event horizon all sorts of weirdness takes place, with time and space coordinates swapping over. Once in there, you might well find it difficult to find any direction that actually pointed outwards.
But we could place you momentarily at the event horizon, shining your torch bravely outwards. The photons would travel outwards then, but would lose energy continuously as they climbed out of the gravitational field. So a distant observer would find the photons to be infinitely red-shifted (that is, lacking all energy, and therefore undetectible).

Originally Posted by north
the first question that I have is why can you not be " spaghettified " ?
Easy enough to avoid spaghettification, at least initially. If you choose a big enough black hole, the tidal forces at the event horizon will be very small. Fall into the black hole at the centre of the galaxy, for instance, and you won't even notice the tides as you cross the event horizon.

Grant Hutchison

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Originally Posted by north
the first question that I have is why can you not be " spaghettified " ?
I made a suit out of some bubble gum, paper clips, and SCUBA gear. I'll post the schematics later.

Ahh, good point Grant, I hadn't considered the time effects when thinking about this. I guess the short answer is that our physics simply break down, or that I'm a few college degrees short in the right area.

5. Originally Posted by grant hutchison
Easy enough to avoid spaghettification, at least initially. If you choose a big enough black hole, the tidal forces at the event horizon will be very small. Fall into the black hole at the centre of the galaxy, for instance, and you won't even notice the tides as you cross the event horizon.
You'd have other distractions to occupy your mind at that moment, no doubt.

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Originally Posted by north
the first question that I have is why can you not be " spaghettified " ?
Easy enough to avoid spaghettification, at least initially. If you choose a big enough black hole, the tidal forces at the event horizon will be very small. Fall into the black hole at the centre of the galaxy, for instance, and you won't even notice the tides as you cross the event horizon.

Grant Hutchison
but what do these " tidal forces " ride on ?

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Originally Posted by north
but what do these " tidal forces " ride on ?
I don't understand the question.
The tidal force on your body is the difference in gravitational force between your head and your feet. You'll get that in any gravitational field, because the force of gravity falls off with increasing distance from the centre of mass.
If that difference is very high, you "spaghettify", as if someone hung a huge weight on your feet. If that difference is very small (as it is right here on Earth), then you don't notice it.

Grant Hutchison

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Originally Posted by grant hutchison
I don't understand the question.
The tidal force on your body is the difference in gravitational force between your head and your feet. You'll get that in any gravitational field, because the force of gravity falls off with increasing distance from the centre of mass.
If that difference is very high, you "spaghettify", as if someone hung a huge weight on your feet. If that difference is very small (as it is right here on Earth), then you don't notice it.

Grant Hutchison
I see

9. Originally Posted by grant hutchison
If that difference is very small (as it is right here on Earth), then you don't notice it.
But the oceans, covering a much larger distance, do. Hence the reason why Lunar tidal forces affect the sea level.
Last edited by Noclevername; 2007-Dec-04 at 04:29 AM. Reason: D'oh! I can't believe I forgot to say Lunar!

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Originally Posted by Noclevername
But the oceans, covering a much larger distance, do. Hence the reason why tidal forces affect the sea level.
Yeah, I was actually thinking of the tidal force generated by the Earth at its surface, and its feebleness in the spaghettification stakes.
But of course, as you say, the Moon also has trouble spaghettifying a person, while nevertheless being able to make its tidal force evident across the width of the Earth.

Grant Hutchison

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Originally Posted by grant hutchison
Yeah, I was actually thinking of the tidal force generated by the Earth at its surface, and its feebleness in the spaghettification stakes.
But of course, as you say, the Moon also has trouble spaghettifying a person, while nevertheless being able to make its tidal force evident across the width of the Earth.

Grant Hutchison
interesting

12. There is a cute coordinate trick we can do to see what happens here. In Schwarzschild coordinates (radial ruler and clock of a far away observer stationary to the mass), everything just slows down to zero as it approaches the horizon and nothing ever gets there in finite time. They don't do too well.

But rather that using that clock, use the clock of a radial free-falller, but continue to use the 'r' of the Schwarzschild observer. When you do that an interesting result appears. r(tau), the resulting trajectory is exactly Newton. The speed, dr/tau = v = sqrt(2GM/r). But note this "speed" is our far away notion of distance per the clock of the free faller. He thinks distance is something else per his own ruler.

Anyway, the proper time to hit the singularity is given exactly by the Newtonian expression. That is a consequence of the high spherical symmetry, and does not hold in general for arbitrary mass distributions, but it is more than a conincidence.

Anway, at the horizon, that v = c, and goes straight to infinty at r = 0.
There's nothing wrong with this. This is a pure coordinate speed, an a hybrid one at that, and doesn't correspond to anything any local observer could conceivably measure.

One can see this as the "speed" that "inertial space" is being dragged into the black hole. To escape at some point r you must throw something upwards at a speed greater than that 'v' above. Think of it as a little radial conveyor belt going straight to the singularity at r = 0. The closer you get, the faster it goes. The velocity you throw something is relative to the conveyor.

To get out, you've got to run against the conveyor faster than it is moving inwards, but you can't go any faster against it than c. Once you get to the horizon, you're SOL, because it's going faster than c and is going to drag you in.

In this view, if you shot a light beam outwards right at the horizon, it would just "sit there", it's motion relative to the conveyor just cancelling the speed of the belt itself. Inside, an outward beam of light is dragged right in. It is desparately trying to get out, going as fast as possible, but to no avail.

So that gives you a picture of what happens to light inside the horizon -- dragged by the conveyor to the singularity There is some currency to the "balance point" in the OP. However, that is an exact point. GR is a classical theory, and classical EM fits together with it like a hand in a glove. The glove is a very weird one, but it's a perfect fit. An EM wave is not a point, but spread out, and so if you centered a wave train exactly at the horizon, one side would go out and the other would be dragged in.

You can't have a point wave, so you don't really worry about that exact balance point. It's a razor's edge that cuts waves in two so to speak.

But in the quantum picture, you can think of a photon, which is more point like and so you might worry about a photon sitting exactly on that razor's edge.

I don't understand the details, but with the Uncertainty Principle, that razor is blurred, and there is spot where "last photon has a snowball's chance of getting out" that is just *outside the horizon*.

-Richard

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Richard,

when I do.

At what point (so to speak...) does a mass become dense enough
to make the conveyor belt start moving? I suspect that a neutron
star isn't enough. But maybe falling matter at *any* density makes
the conveyor belt move at least a tiny bit. So a star collapsing to
become a neutron star would cause spacetime to 'move' in the way
you describe as a moving conveyor belt, but when the collapse
halts, the conveyor stops, too. In a black hole, the collapse never
ends, so the conveyor keeps on going forever.

But... but... It seems to me that only the spacetime where the
collapsing matter is actually located would be involved in the
conveyor belt action. Once the entire mass of the collapsing star
is inside the event horizon, the part of the conveyor *at* the
event horizon would stop moving. Only farther inside the black
hole would the conveyor continue to move.

So have I got it all wrong? Did my finger get caught in the analogy?

The way I look at it, as the matter collapses, becoming more and
more dense, spacetime is stretched farther and farther in the time
direction. The big rubber sheet has a dent in it that gets deeper
and deeper, faster and faster as the matter becomes denser. The
distance across the black hole measured by God's stiff, unbending
ruler doesn't change, but the distance down to the surface of the
collapsing star is increasing rapidly, and there, the rubber sheet
is stretching.

-- Jeff, in Minneapolis

14. Jeff,

The conveyor is a just a picture based on what (a restricted class of) geodesics are doing. I actually don't like it all that much because I like to restrict the notion of "moving space" to frame dragging, which gives a different sort of conveyor idea. But nonetheless it works pretty good to show why nothing can escape.

You don't have to have an event horizon/collapsing mass to have that conveyor. Every mass would have that conveyor action in this picture, it just wouldn't ever reach light speed. Only at an horizon does it reach light speed.

You don't even need real gravity to have this conveyor picture. A Rindler observer could define a conveyor picture as well. And at his horizon, the speed would reach c as well.

This is just a coordinate picture -- other observers would have a different conveyor picture in the same space-time.

-Richard

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Originally Posted by publius
The conveyor is a just a picture based on what (a restricted class of)
geodesics are doing. I actually don't like it all that much because I like
to restrict the notion of "moving space" to frame dragging, which gives
a different sort of conveyor idea. But nonetheless it works pretty good
to show why nothing can escape.

You don't have to have an event horizon/collapsing mass to have that
conveyor. Every mass would have that conveyor action in this picture,
it just wouldn't ever reach light speed. Only at an horizon does it
reach light speed.
I was thinking of the forever-stretching rubber sheet as a star
collapses as causing a kind of frame dragging, in the radial direction
for a non-rotating black hole, and thinking that that was your
conveyor belt. But it looks like your conveyor belt describes all
gravity fields all the time.

I must say that a conveyor belt with differential speed along its
length -- as opposed to across its width -- makes for a pretty
awful analogy.

This is just a coordinate picture -- other observers would have a
different conveyor picture in the same space-time.
I might as well ask a question I wrote down a few weeks ago:
How far away from a black hole does an observer need to be in
order for the event horizon to appear spherical? I suspect that
the correct answer might be that it approaches sphericity
asymptotically with increasing distance. But I don't know how
to define how one "sees" the event horizon.

-- Jeff, in Minneapolis

16. Originally Posted by Yagdrasil
I guess the short answer is that our physics simply break down, or that I'm a few college degrees short in the right area.
Actually, you don't need the degrees - just get ahold of Black Holes and Time Warps by Kip Thorne. You can check out parts of it here. Try the library.

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If black holes evaporate in a finite amount of time when seen from far away and it takes you infinite time to reach the event horizon as seen from far away then you'll never get there. It will evaporate out from under you as you approach.

18. Originally Posted by Jeff Root
I was thinking of the forever-stretching rubber sheet as a star
collapses as causing a kind of frame dragging, in the radial direction
for a non-rotating black hole, and thinking that that was your
conveyor belt. But it looks like your conveyor belt describes all
gravity fields all the time.

I must say that a conveyor belt with differential speed along its
length -- as opposed to across its width -- makes for a pretty
awful analogy.
I don't like "moving space" in this sense all that great either, but then I don't like the rubber sheet all that much either. All these things are 1) coordinate dependent, and 2) involve taking some mathematical entity, and usually an incomplete subset of it, and using that to construct a picture, usually literally to explain something.

Sometimes that picture is good to see something, but for other things, it's not good or meaningless.

Let's look at how the rubber sheet is done. We take the *spatial part* of the metric and construct some curved surface that would have that spatial metric against a flat embedding space. In the case of the Schwarzschild, the spherical symmetry makes that very easy, allowing you to supress a spatial dimension (you'd need four in the general case or even more in some cases I think) and come up with a 2D curved surface in 3D space.

Now, that "thing" blows up at the horizon. The slope of the surface goes becomes vertical, and so there's a "hole" at the horizon. If you think at that surface as some z = f(x, y), z --> -infinity at the horizon. And then space and time flip inside, and so your spatial metric doesn't make any sense inside unless you just flip things and let your time coordinate be spatial.

But that sort of shows the "hole in space" idea, the bottomless pit and all that. But when it comes to imagining orbits as little balls rolling around the sides of that thing, well you're missing information, the time part of the metric.

And it's completely arbitrary because other observers, falling or orbiting split space and time in a different way altogether and their spatial metric doesn't look anything like it does for stationary observers.

I might as well ask a question I wrote down a few weeks ago:
How far away from a black hole does an observer need to be in
order for the event horizon to appear spherical? I suspect that
the correct answer might be that it approaches sphericity
asymptotically with increasing distance. But I don't know how
to define how one "sees" the event horizon.
If by "appear" you meant how things *look*, optically, from an observer seeing light coming at him, I don't have a clue. That gets very complex indeed. There are some visualizations to be found on the web that try to show you what your sky would look like close to a black hole. But I remember one screwed something up and the author had to fix it after someone recognized an error. Which goes to show you how tedious that can be.

But roughly, for Schwarzschild, the magic factor is R/r, the ratio of the horizon to your radius. If that is small, the geometry is close to flat and so nothing should be terribly weird. You probably get "fringe effects" for light passing close that bends around, but you should otherwise see a black circle I think.

-Richard

19. Originally Posted by Chuck
If black holes evaporate in a finite amount of time when seen from far away and it takes you infinite time to reach the event horizon as seen from far away then you'll never get there. It will evaporate out from under you as you approach.
That is a whole 'nother kettle of fish indeed. It may well actually work like that. But even then, rather than thinking of something falling in to a pre-existing black hole, that is, falling in after the stuff that makes the thing in the first place, you consider it from a co-collapsing frame describing the collapse itself.

It may be that as the local curvature starts getting very high, the Hawking radiation (term to describe whatever quantum vacuum effects are at play) just "throws it back out" as radiation as it tries to collapse and essentially converts all the mass to radiation before a singularity forms. Because of the extreme time dilation relative to far away observers, that process would take eons for large original masses and so you've got a very dense object with a super strong local gravity field that meets the practical notion of "black hole".

The question is does a true "null surface" ever form during the collapse. If it does, then you've got a true horizon. If a null surface doesn't really form, then no problem, really.

But that is an open question, well beyond my understanding, and only will be resolved when a real quantum theory of gravity, the real McCoy comes along.

-Richard

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Richard,

What I had in mind was the fact that the event horizon is always
between the singularity and an observer, so the black hole must in
some sense appear to get a dimple in it as you approach it, and the
dimple becomes a hole when you get too close. A hole in a hole.

-- Jeff, in Minneapolis

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Since I don't understand the math, and don't know how to determine
what the slope of the rubber sheet is supposed to be at any given
place, my rubber sheet doesn't have infinite slope anywhere, although
it very quickly approaches infinite slope close to the singularity, and
is forever becoming steeper and steeper with time. In my conception
of it, the singularity is not actually quite a singularity, but is the place
where a singularity is forming, but would only be completed in infinite
time. So I say that the slope of the sheet, which I take to represent
gravitational field strength, is steeper close to the center than at the
event horizon. Come to think of it, by convention, if nothing else, it
seems to me that the slope would be exactly 45 degrees at the event
horizon. Since the horizon is at different locations for different
observers, the slope at any location depends on where you are
observing from. But my rubber sheet model uses the convention of
being viewed from an infinite distance. (Which isn't much different
from the view far enough from the black hole to not have any worries

Can I get my rubber sheet analogy vulcanized?

-- Jeff, in Minneapolis

22. Originally Posted by Jeff Root
So I say that the slope of the sheet, which I take to represent
gravitational field strength, is steeper close to the center than at the
event horizon. Come to think of it, by convention, if nothing else, it
seems to me that the slope would be exactly 45 degrees at the event
horizon. Since the horizon is at different locations for different
observers, the slope at any location depends on where you are
observing from. But my rubber sheet model uses the convention of
being viewed from an infinite distance. (Which isn't much different
from the view far enough from the black hole to not have any worries

-- Jeff, in Minneapolis
Just forget about these sheets. If you want the slope to be the field strength, you need to represent the potential as the sheet. But the problem is that potential is a tensor (the metric) and not a simple scalar that you could imagine being the height of some surface. Far away, when the spatial part is negligible, then the time part dominates and you could do something like that. That would be Newton, basically. However, close in, in the strong field, all the other terms are important.

What the standard rubber sheet is looking at the spatial part, ignoring the time part.

That sheet is called Flamm's Paraboloid for Schwarzschild, actually, and z(r) = 2sqrt[R(r - R) ] is one way to write it. (For anyone interested, the negative square root would correspond to the "white hole", time reversed metric, which is the other spherical solution for a point mass)

The slope goes to infinity at r = R, and it is undefined for r < R. There's your "hole" (and all that is because our notion of space doesn't mean anything inside the horizon).

And again, that sheet is all about the spatial curvature -- it isn't even looking at the time part.

-Ricahrd

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Bold mine:

Originally Posted by publius
Just forget about these sheets. If you want the slope to be the field strength, you need to represent the potential as the sheet. But the problem is that potential is a tensor (the metric) and not a simple scalar that you could imagine being the height of some surface. Far away, when the spatial part is negligible, then the time part dominates and you could do something like that. That would be Newton, basically. However, close in, in the strong field, all the other terms are important.

What the standard rubber sheet is looking at the spatial part, ignoring the time part.

That sheet is called Flamm's Paraboloid for Schwarzschild, actually, and f(z) = 2sqrt[R(r - R) ] is one way to write it. (For anyone interested, the negative square root would correspond to the "white hole", time reversed metric, which is the other spherical solution for a point mass)

The slope goes to infinity at r = R, and it is undefined for r < R. There's your "hole" (and all that is because our notion of space doesn't mean anything inside the horizon).

And again, that sheet is all about the spatial curvature -- it isn't even looking at the time part.

-Ricahrd
I confess that your post is usual for you (you're in rare form) yet the thing that fascinated me the most was that you actually managed to typo your own signiture/name.

All that math and I wanna call you Ricardo...

ETA: Usually it is the substance of your posts that is fascinating. I must be suffering from more sleep dep.

24. Originally Posted by Neverfly

I confess that your post is usual for you (you're in rare form) yet the thing that fascinated me the most was that you actually managed to typo your own signiture/name.

All that math and I wanna call you Ricardo...
Maybe I'll marry a Lucy, get some neighbors named Fred and Ethel and life'll be a hoot.

When I get to typing fast, stuff happens. Actually when I get to speaking stuff happens to. When I get to going, trying to speak while thinking, people tell me I have a very annoying tendency to just not finish sentences and start a new one. The trouble is that it takes forever to say a thought compared to how fast it takes to think it. The queue gets full and overwritten apparently. Typing is another layer on top of that. It's a wonder I do as well as I sgsheng. JSjoutn nglsk ebbgus. bvvmdsml.

All that math? My boy, that little bit is baby stuff. If I could find it quickly, I'd link to this paper showing what "all that math" actually looks like. It was paper about PPN, basically the GR "N-body". The authors had managed to greatly simplify things. Each equation only took a few .pdf pages each.

Always remember this, I'm but a mere piker in all this. I know just enough to be dangerous. When I compare what I don't know about GR to what I do know, I feel very small and just want to forget about it sometimes.

-Richard Ricardo Baba Loo

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Eh.. It isn't often I can find a fault in one of yur posts Sieze the moment...

However, the thing that got me was that for those posters who, like yourself, or danscope, or others that have their name on a closing line- I had actually thought it was automated or something.
To see a typo in one after all this time was like seeing a purple gorilla- and it overrode the rest of the post

26. Originally Posted by publius
(snip)
All that math? My boy, that little bit is baby stuff. If I could find it quickly, I'd link to this paper showing what "all that math" actually looks like. It was paper about PPN, basically the GR "N-body". The authors had managed to greatly simplify things. Each equation only took a few .pdf pages each.

Always remember this, I'm but a mere piker in all this. I know just enough to be dangerous. When I compare what I don't know about GR to what I do know, I feel very small and just want to forget about it sometimes.

-Richard Ricardo Baba Loo
Wow, I just don't have the gift for the mathematics but love reading and trying to get it. All I did was have a picture of sitting in a deck chair while going over a large enough event horizon. I figured the light would start relative to the position of the person and they could play some neat tricks with the beam.

Probably want a bit of smoke because it works so well for laser displays. Again this is just a visual but the light pointed into the black hole would blue shift the further in it went until it 'crossed the event horizon relative to your accelerating position going in. By shining it outwards it would look like a straight beam due to bent space time, then redshift and then even out and then blue shift and disappear giving the impression it did escape, but that is by the time the beam has returned to the black hole reference where you were ... you are way gone.

There should be some neat tricks like shine it sideways and see a straight beam of light with just red shift, as it got back to you time and time again it would be more and more red shifted as you received it in tighter and tighter space time. Aim up a bit and see first a red shift and then the light curve up and outwards as if it was leaving, then normal, then blue and gone in a curve.

Since a black hole isn't black at any given point and the 'event horizon' would be determined by the position of the viewer at each separate reference point on the way down a really bright light would give you the chance to see space time curvature for real but only for yourself and only as long as you stayed unspaghettified. You might gain a bit of extra time due to frame dragging ... it could even seem like the death of a thousand years boredom.

One thing for certain it would be a unique display you could have all to yourself because no matter what you did by yourself you could not relay that information back. However given a suitably long enough chain of transmitting stations there is a chance the information could be boosted to a number of sacrificial probes all travelling in to destruction because the event horizon would need to be based upon the position of each observer in the chain unless it is assumed the nice even flow of gravity takes some weird right angle turn instead of flowing over the event horizon.

Now that is something as the flow is smooth ... the event horizon would need to be defined by the co-ordinate position of each observer dependant on their own local position ... shouldn't it?

I should have included that with the previous post ... please forgive me. I have sat down to a few drinks and was going to annoy some poor conspiracy theorists with a human nature reference from cradle to grave and the dangers of suffering an apprentiship in any field of endevour what ever it may be.

So I'm most likely wrong for more reasons than I can imagine and in the spirit of being wrong and fair game to better logic from others will leave that mess where it is. Appologies to the readers and I beg forgiveness from the moderators. But the debunking may add some valuable stuff to any new member wishing to learn more about black holes.

My idea was not to retrieve the light but via a series of probes able to read the information (very red shifted) might be able to pass the information back via a long enough chain to in effect achieve a return of information although all the probes and the comfy armchair would be lost. First order of the day would be to requisition a handy drinks cooler for the ride then ... what do you say .... cheers (hic) pardon.

Thank goodness fror spiel check

28. ## Yup

Three bourbons a quick nap and now the coffee is kickin' in.

I fair got that lot messed up. The horizontal light is blue shifted in the smoke as it leaves you because you are seeing the reflection as it falls into the tighter curvature at the same time you are. So it left you and travelled in more dense space but returns the same way red shifting back to white light.

Since from the time it emitted, reflected and was received the time increase component was the same for the light and yourself time does not become an issue. Well it does if the black hole is rotating and then there is a difference of time for a pulse of light to travel one direction as there is in the opposite direction. But it is received back red-shifted as it was first emitted as the space it was originally travelling in was larger, again time is not an issue as the beam and yourself stayed in the same frame. It is a lot like a bullet hits the ground at the same time if dropped or fired horizontally.

So you get a rainbow of blue shift inwards to white horizontal to red shifting to blue on the outer before it appears to disappear outwards which of course it doesn't. All light from the guy in the armchair further in is red shifted as received by yourself and if the third guy behind you hasn't decided to chicken out his beam will be red shifted further out. Now if he has a super rocket pack he can get his light to change colour by catching you so it starts off red shifted and becomes normal at some nearer point and then as he is closer to you the light is blue; a flash of off-white when he is level. Depending how tight space is you may or may not see redshift from his light, it just depends how sensitive your eyes or instruments are that close.

One thing for certain ... because space is more compressed even though you are level and in the same time frame you will be more attracted to the other traveller the further you go.

Hope that is a neat little wrap up but in all fairness you should wait to get the thoughts of the other members of the forum here, cheers

29. Originally Posted by Chuck
If black holes evaporate in a finite amount of time when seen from far away and it takes you infinite time to reach the event horizon as seen from far away then you'll never get there. It will evaporate out from under you as you approach.
No, you'll get there all right, in short order. It just won't look like that to the observer from far away, who is in a different frame.

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If a distant observer sees the black hole evaporate before I reach the event horizon then I must have never crossed it. I don't see how I could be said to have crossed it in my own frame of reference. Either I crossed the event horizon or I didn't no matter what the point of view. If the black hole is gone and I'm still there from a distant point of view it's hard to believe that I'm actually not there from my own point of view.

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