[Ross Walsh]

Ross Walsh  asked the Naked Scientists:
   Hi Team,
 
My concern/question is in regard to the oft quoted statement that an observer posted outside a blackhole who watches an object descending into the abyss never actually loses sight of the unfortunate object because as it approaches the speed of light (due to the enormous gravitational force) it's time slows to infinity - I presume that this statement only applies if you maintain your gaze on the object and, if you looked away then back again (after the object had fallen into the blackhole), the object would now have disappeared.
 
I presume this because otherwise every thing that had been sucked into the hole would still be visible!  Obviously an absurdity.  I would be most appreciative if you could settle my mind on this matter.
 
Regards,
 
Ross Walsh
 
Sydney, Australia
 
P.S. What a newbielink:http://www.thenakedscientists.com/HTML/podcasts/ [nonactive] you guys/gals are delivering! Keep up the good work.
What do you think?

[imatfaal]

Hi Ross

It gets complicated and seemingly paradoxical when trying to get cosmological ideas into a human perspective.  A famous pair of intrepid explorers are usually talked about - Alice and Bob. 

It starts with both astronauts a long way away from the blackhole and maintaining their position with rockets.  Alice draws the short straw switched off her rockets and starts to free fall - Bob is allowed to keep burning.  A and B are now in different reference frames and communicate by sending a message via light/raditation; whilst they are a few miles apart everything seems normal.  But as A falls towards the blackhole she notices that the message from B is speeded up (blue shifted) and B notices that the message from A is slowed down (redshifted) - the same would apply to the actual image if they looked at each other through telescopes.   From B's perspective as A gets closer and closer to the event horizon A's message gets more and more slowed down/redshifted - till B calculates that A should almost have reached the event horizon at which point the message is so red-shifted that B can no longer even receive it properly and A seems to have stopped in time and be smeared out in space  (the wavelength is so large that B cannot tell which part of the blackhole it comes from.  So B observes A slow down in time and spread out in space.  BUT from A's  perspective she accelerates towards the black hole and passes through the event horizon without even noticing.  A will however notice that B's messages get increasing fast and high energy.  B can look away, go and make a cup of tea, have a family, and his grandchildren (for as many generations as you care to mention) will still be able to observe poor old Alice almost infinitely still, spread out over the event horizon and looking as if time has slowed almost to a stop for her.  From an external accelerated reference from (which is what B is in and you must be in to avoid being sucked in yourself) everything that has ever fallen into the black hole is smeared out over the even horizon and stuck in time.  when you try and mix human intuition and relativity you often get what seems absurd but is in fact true.   

What we have is a situation in which what is observed from a different frame of reference does not seem to match what is happening.  this is all due to the fact that simultaneity is a very tricky concept when distances are large - communication is not instantaneous and even our best method of passing information (light/electromagnetic radiation) is at finite speed and is affected by gravity.  B's picture of what is happening to A is slowed down; by the point A is almost crossing the even horizon the picture is slowed down so much that it is stopped for all practical purposes.  But B's picture is not A's reality - A continues through the event horizon and dies.  (damn - bit of a down beat ending)

[raw]

Thanks heaps imatfaal,

BUT I'm afraid that your explanation leaves me even more incredulous because, if the 'frozen in time' image/s of all the victims of the 'hole' were smeared-out at the Event Horizon then the 'hole' would be visible to an outside observer ... and that is surely not the common perception?

This enigma has annoyed me for a long time now ... and I would really like to put it to bed ... help!

Regards,

Raw

[yor_on]

It doesn't work that way Ross. The event horizon is the last 'place' that has contact with you. After that things disappear. So you watch something slow down as it comes closer to that Event Horizon, where it as defined by your frame of reference will 'hang' there forever. Why that is has to do with way your frame perceive all other 'clocks' existing in the universe. If you imagine that gravity can slow a clock, relative your clock measuring, then the gravity at the event horizon is so strong that any clock there, according to your observations, stops ticking. Gravity is coupled to mass, accelerations (and 'energy') and wherever there is gravity there will be a 'clock' ticking. It's a weird fact, but has been proven here on earth, by tests with atomic clocks.

But it is also so that if you were to be teleported to that black holes Event Horizon, materializing beside that thought up clock, you now would find it to work the same as your own wristwatch, showing the same durations. From that you can see that the solution to this enigma must rest between the frames of reference and how they present themselves to you. Distances and durations for what you observe outside your own frame of reference will change with relative motion, mass and energy. But inside that 'frame, loosely speaking here, your ruler will always give you the same measurements, and your wristwatch the same durations. And that's why that guy will pass the Event Horizon in his own 'time' and frame of reference.

[yor_on]

From this you might wonder if it isn't all an optical illusion, but as I said we have tested the way gravity 'slows' a clock here on earth relative the observer. And the most elegant experiment I know of is the one where we, by using two synchronized atomic clocks on a table, then move one to the floor, watching it desynchronize as it starts to tick 'slower' than the one left on the table, where the gravity is lower. And this you can Google.

But where you to lay down at that floor your frame would fit the floor clock, standing you might be closer to the one on the table, and getting real close to any of those clocks, comparing their durations to your atomic wrist watch you will find that the both will fit those durations. And it doesn't matter if you live at the event horizon or on earth, is your lifespan a hundred years then it will be a hundred years no matter where you measure your heartbeats to your wrist watch. So all clocks are of the same durations locally, but when separated by relative motion or placed in a different gravity they will to you start to differ from your clock. And the same goes for 'energy' which is a description of a thought up state from where all things we know come, maybe time too?

[raw]

I have struggled-with, but hopefully comprehended, Einstien's theory (now, of course, conclusively proven) but must reiterate that this is not the point of my query, which is that, if the following excerpt from imatfaal's post:
"everything that has ever fallen into the black hole is smeared out over the event horizon and stuck in time" is correct then an observer at a safe distance from said event horizon would have a visual indication that a black hole was present?

Unless I'm missing something (or the subject is beyond my intellectual abilities) and it's taken as a given that black holes are invisible ... then that statement must be wrong.

Sorry to be appearing stubborn here ... but a black hole cannot be invisible if 'frozen images in time' of all the debris that has disappeared into it is smeared-out over the event horizon ... surely?

Thanks for the feed-back,

Raw

[graham.d]

raw, I think the point is that these things that are stuck at the event horizon are really at a very, very low gravitational potential and the light (information) coming from them is very redshifted indeed. The energy per/second (of observer's time) is extremely low.

An interesting question might be that if object never can be seen to fall in to a BH no matter how long you wait, how do they ever get to be BH and grow their Schwartschild radii? I think the answer is that the mass that forms the BH starts off at a very low gravitational potential and not at a large distance like the observer. BHs are thought to be formed by the collapse of a large star and subsequently may grow through accretion of neaby matter. Having said this, I still find this slightly puzzling to think about.

[Soul Surfer]

The problem with this question is that the answer in the text books has been given by mathematicians.  true gravitational red shift and time dilation exist but it doesn't say how long things would take before they vanish from our sight which is pretty quick.

It is a bit like saying when I strike a bell the note continues decaying exponentially for ever and one strike will ensure that the bell goes on ringing for ever but we all know very well that after a few seconds the sound vanishes into the background noise. 

So in reality there is absolutely nothing special about this.  We are all familiar with exponential decays everywhere else in our lives so why worry about this one!

[imatfaal]

I have struggled-with, but hopefully comprehended, Einstien's theory (now, of course, conclusively proven) but must reiterate that this is not the point of my query, which is that, if the following excerpt from imatfaal's post:
"everything that has ever fallen into the black hole is smeared out over the event horizon and stuck in time" is correct then an observer at a safe distance from said event horizon would have a visual indication that a black hole was present?

Unless I'm missing something (or the subject is beyond my intellectual abilities) and it's taken as a given that black holes are invisible ... then that statement must be wrong.

Sorry to be appearing stubborn here ... but a black hole cannot be invisible if 'frozen images in time' of all the debris that has disappeared into it is smeared-out over the event horizon ... surely?

Thanks for the feed-back,

Raw

A black hole by itself is close to invisible (it might emit hawking radiation) - but a black hole with stuff falling into it is another case in point.  the stuff falling into it is what you see - and in real cases it would have angular momentum and spiral at vast velocities to form an accretion disk.  this is one of the most energetic things around and a gamma ray emitter - just from stuff hitting other stuff at very high speed.  (btw nothing in science is 'conclusively proven' - everything is up for debate, revision and tweaking).

The layer of smeared out stuff is equal to the increase in the schwarzchild radius - so the ev does seem to grow. 

SSurfer - not sure what you mean - we worry about this one because the physics is fundamentally different.  The maths and physics that we use to determine stuff breaks down near/at the event horizon.  we are still receiving signals from voyager from over 100AU - thats plently of distance to see how a signal would vary depending on what it was falling towards before losing the signal

[graham.d]

The layer of smeared out stuff is equal to the increase in the schwarzchild radius - so the ev does seem to grow. 

Matthew, this is an important point you make here but I am not 100% sure about it. On one hand you could say that all the mass around a BH can be viewed as contributing to the total mass of the BH when viewed from a distance. In this case you would expect, perhaps, that the Swartzchild Radius to be larger as a result, as you suggest. On the other hand, the SR for the earth is a few millimeters but light can still escape from the centre and, obviously, from the surrounding matter. Of course the field never gets to infinity in the notional SR of the earth so no EH. Looking at it another way, if a BH was close to the centre of our galaxy, there would be millions of stars around it much closer than us; I don't think it is thought that they would result in a larger SR for the BH as viewed from our position. What do you think?

[imatfaal]

Graham - remember that the increase in the EV through the smearedout remains of Alice is merely a crutch to support that fact that simultaneiety has broken down; from Alice's perspective she has already crossed horizon and whatever happens at the "singularity" has happened to her already.  But we are still observing Alice crossing - yet we know from her perspective she has already added to the mass of the black hole - so we have to find an explanation that fits both (it's a cludge but it works); bear in mind that whilst the picture we see of the event horizon is massively time dilated compared to our external accelerated reference frame, any changes affecting our measurements of the gravity (and thus eh etc) propagate  at c.

I am struggling to get my head round the second part of your question - I don't understand why the surrounding stars should make a difference to the measurement of the SR

[yor_on]

Raw, if you understand what frames of reference does to the observer, then there is the answer to that one. And it has to do with clocks and gravity.

From your first frame of reference (earth?) the object may well hang there forever and that is correct. But 'teleporting' to it you will find your frame to be the same as the Event horizons, and 'times arrow' acting in accordance with all biological chemical functions you have, nothing being different.

SoulSurfer discuss it in form a ringing that we all know will vanish. So, even though it from Earths point of view may hang there for a extremely long time, knowing relativity they too will now that it is a effect obeying the arrow. There are no places in the universe that I know of that is excepted from the arrow, not discussing the inside of the event horizon here. But if I were I would say that we have gravity there, that implies what I call 'clocks' and so a 'time'.

[yor_on]

There is one point more to be made when considering it. What carries information through SpaceTime is radiation. Radiation has one unvarying speed 'c', the same from any frame of reference thought up, be it a rocket or earth. That's also what is the ground for a time dilation.

If we think a time dilation to be real, then we accept the fact that even though all frames may have a same 'ground duration', as proven when 'teleported' to whatever frame you observed earlier. It first giving you a different duration of time than your own clock but, when teleported there, finding its durations to be the same as yours. Then we must accept the fact that 'time' is observer dependent. That means that the rocket hanging there from your first frame of reference actually hangs there, it is not an optical illusion.

The arrow intrinsically (locally), if i may use that word, is the same everywhere but all other 'frames' you observe, as that even horizon, will present you other durations relative your intrinsic clock defining your 'frame of reference'.

Remember that it is radiation that brings with it information, and that this radiation is proven to have only one, invariant, speed. That's what relativity is all about to me. So the universe you see is one painted by radiation. If information exchange is what defines reality then that spaceship must hang there for earth, whilst it in its own frame will pass that Event Horizon in a measurable time.

The other way to see it is to assume that 'time' isn't coupled to gravity mass energy and uniform and accelerated motion. That there is some 'cosmic clock' free from those influences, that assumption is expressed in a Lorentz transformation as I see it. But I don't think that is right myself. Your frame of reference will define SpaceTime to you, as mine will to me, that we can translate our observations when comparing them does not give us that 'cosmic clock'. You only need to introduce a frame more to break the model you've made by your first comparisons into another 'average' cosmic clock as I see it.

[graham.d]

Graham - remember that the increase in the EV through the smearedout remains of Alice is merely a crutch to support that fact that simultaneiety has broken down; from Alice's perspective she has already crossed horizon and whatever happens at the "singularity" has happened to her already.  But we are still observing Alice crossing - yet we know from her perspective she has already added to the mass of the black hole - so we have to find an explanation that fits both (it's a cludge but it works); bear in mind that whilst the picture we see of the event horizon is massively time dilated compared to our external accelerated reference frame, any changes affecting our measurements of the gravity (and thus eh etc) propagate  at c.

I am struggling to get my head round the second part of your question - I don't understand why the surrounding stars should make a difference to the measurement of the SR

As to the first part, it surely does not matter if simultaneity between the observer and Alice is broken down because, as you say, both the light and the gravitational effects of any change in the EH on a distant observer (if any) would arrive at the same time as far as the observer is concerned.

The second part is to do with the gravitational attraction of a sphere. The centre of gravity is at the centre. I think this still holds even for approximate spheres of galactic proportions and it may be a reasonable approximation to a certain density of material (in the form of stars say) that are positioned out to some large radius from a BH. Consider an approximately spherical galaxy. That would have a much lower gravitational potential at its centre than at its edge. Does this affect the SR of any BH at its centre? I think it does not because the SR is really only defined by the mass within the EH. This can be seen by the EH being the point where any particle (or photon) requires infinite energy to escape. The extra energy needed to get to the edge of the galaxy is obviously negligible compared to infinity so makes no difference to the SR.

I was using this as an example of how objects outside the BH should not affect its EH. However, this would lead to the impossibility of a BH ever be seen to grow. I have grown used to such apparent paradoxes but I would still like a clear resolution. I vaguely remember seeing a simulation of two BHs colliding and seem to remember that the EH of each one bulges out towards the other as they approach. But I don't remember from what viewpoint this was being observed; I suspect from a distant position.

So I think it is true that the EH grows, even to a distant observer, but I still cannot reconcile all the facts.

[yor_on]

Graham, what if we assume that gravity is the ocean constructing a SpaceTime?

That is is coupled to mass motion and energy is then a direct result of that. In that ocean we have those pinpricks containing singularities. They are where our calculations breaks down, presenting us infinities. So what happens when a mass is transferred inside the Event Horizon in such a scenario? SpaceTime lose some mass to the singularity, that mass is also equivalent to an amount of energy. We wont get it back, not until the very last of our universe at least even though we have the possible Hawking radiation. But the Event horizon is inside our SpaceTime and?

Shouldn't it grow?

I don't know, you might assume that what happens inside a Event Horizon not necessarily need to express itself outside? But we have those symmetries expressing themselves everywhere, and assuming that the inside of the Event Horizon obeys similar principles that 'expansion' of a event horizon should be able to take place?

Or maybe not in all circumstances, maybe there are both possibilities?

[graham.d]

The SR is directly proportional the mass within it according to theory:

r = 2GM/c^2

So it should grow. The mass is not exactly "removed" from the universe as it still has a gravitational field and still has properties of momentum, angular momentum and charge (as theory would have it). My comments were about "how" it grows and "when" as perceived from a distant observer. There seems a point when (as Matthew puts it) part of the "smeared out" falling object(s) become within the EH's expanding EH. 

[raw]

A most interesting debate but, whilst I have enjoyed it, it has not answered my original query i.e. I have read, more than once, in scientific texts that an object falling into a black hole would remain visible to a distant observer forever.

Obviously incorrect ... and this article I found is the first understandable explanation I've come across:

"To a distant observer, clocks near a black hole appear to tick more slowly than those further away from the black hole.[44] Due to this effect, known as gravitational time dilation, an object falling into a black hole appears to slow down as it approaches the event horizon, taking an infinite time to reach it. At the same time, all processes on this object slow down causing emitted light to appear redder and dimmer, an effect known as gravitational redshift.Eventually, at a point just before it reaches the event horizon, the falling object becomes so dim that it can no longer be seen." (1995 Matt McIrvin.)

Stealing a famous US of A electoral slogan "It's the photons stupid"! 

But the last sentence in the explanation was not in the science books I read in the past (an omission that ruins the authors credibility in my eyes ... and makes me wonder on their grasp of the topic) ... Thanks all ... I think I'm on top of it now!

raw

[imatfaal]

Becoming harder to detect ie dimmer through red-shifting is just not the same - there is a theoretical and a practical way of looking at this.  "just before it reaches the event horizon" so that's a moment before infinite time has elapsed, what's that in seconds?  Of course there exists a point at which an energetic gamma ray emitted by Alice will be redshifted to a wavelength greater than anything you care to mention - but that is firstly an experimental thing and secondly it's as close to forever as you are gonna get.

[imatfaal]

The SR is directly proportional the mass within it according to theory:

r = 2GM/c^2

So it should grow. The mass is not exactly "removed" from the universe as it still has a gravitational field and still has properties of momentum, angular momentum and charge (as theory would have it). My comments were about "how" it grows and "when" as perceived from a distant observer. There seems a point when (as Matthew puts it) part of the "smeared out" falling object(s) become within the EH's expanding EH. 

I have confused myself terribly - I think your post is correct.  At all times prior the EH the distant observer Bob and the in faller Alice must agree on the gravity and mass.  Surrounding stars cannot increase the SR
- the SR is the locus of points at which light can no longer escape
- the gradient needed for this is that of a blackhole,
- the surrounding volume of stars (no matter how densely packed) must have a lower mass than the equiv BH (otherwise it would be a blackhole)
- therefore light will escape and you are outside the EH. 

ie there is no shell around a blackhole which has a sufficient mass to create a gradient so great that light will not escape  otherwise it would be part of the blackhole. 

[graham.d]

It's a confusing subject. I think you are right that the BH does grow to encompass some of the matter very close to the EH (as perceived by a distant observer) though. This seems out of kilter with the formula, but then that is derived from a perfectly spherically symmetric Schwartzchild metric. I have to say I am not sure about this, but if this were not the case then you have to wonder how a BH can ever be formed because accreting matter would always take an infinite time as observed from a distance.

I must look for the animation/simluation of two BHs colliding. If I remember right, the EHs of each BH become distorted, not spherical, just prior to colliding.