Black Holes: An Issue For Discussion

Well, white holes are just a bar stool academic curiosity since no one has detected anything like them. I am not even aware of any remotely sound theory that would predict them. So, your schematic representation is as good as anyone else's.

It is comforting to think that there are wormholes and parallel universes that we could slip into and out of, but nothing we have to date really predicts them with any certainty. However, I would be very excited if such a discovery would ever be made.

Your description of the black holes was very good!

"Well, white holes are just a bar stool academic curiosity since no one has detected anything like them. I am not even aware of any remotely sound theory that would predict them."

Einstein's general relativity (gr) allows worm holes and white holes,^[1] but they do not have stable or even finite solutions in gr. The big bang is the closest thing to a white hole that (we think) we know of, but gr does not work for big bangs.

Jorrie

[1] Kip Thorne: Black Holes and Time Warps.

Could it be posible that a black hole just consumes matter until it becomes unstable. At which point it explodes. Don't think the theory of white holes will stand up. I would think that if a white hole was returning the matter gathered by a black hole. Then the new matter would have been noticed by now.

Interesting thought, but we haven't seen that kind of event. Then you have to ask the question what would make a black hole unstable and why?

Interesting theory and well described. But as far as I know white holes have not been noticed. Maybe the idea of a black hole connecting to a parrelel universe or there being a number of parrelel universes? I have a black hole in my washing machine that takes only 1 of each pair of sox. It would be great if there was a white hole in the drier bringing them back. But knowing my luck they would all be from a different universe where everyone has tiny little feet.

But doesn't the latest theories (string and membrane) have 11 dimensions? And are we to also consider alternate universes? Since Astronomy and Astrophysics is still in wonder of black holes and their cousins, I shall deffer to them until further notice. I have enough trouble understanding the world I can see, hear, touch, taste and smell. Things like the speed of light still keep me occupied.

Hello, Chtank... String Theory, indeed, claims that there are 11 dimensions... But they are all bended extremely in a very tiny space around each point of our 3 dimensional world... Only the 3 (well known) dimensions of the space (and the one of the time) are speaded out and are observable... The other 8 dimensions are shrunk and not observable...

And something else: I don't claim that there is, actually, a 4th dimension of space which is spreaded out... As you can see, in figures 4 & 10, I talk about a "4th dimension" and "hyperspace" (which are not perceptible) and I show that the wormholes are going through this "4 dimensional hyperspace"... but I put a lot of "???" as you see... I thing that there isn't, actually, such a thing as a "4 dimensional hyperspace"... Physicists says that there isn't such a need in order to explain the curvature of the Universe or the existence of a Black Hole... (Although it's very hard for me to conceive this... we speak about the curvature of the Universe but relative to what???... or the Black Holes exists into what???...)... Anyway, I tend to accept that this "hyperspace" is, more or less, an "imaginary structure" in order to try to understand how things work... A lot of "???" though...

Hi again George.

"... Anyway, I tend to accept that this "hyperspace" is, more or less, an "imaginary structure" in order to try to understand how things work... A lot of "???" though..."

Exactly.

For mathematical purposes we don't need 'hyperspace' at all; it's just so much easier to visualize, provided than one throws one or two space dimensions away...

Jorrie

Hi George.

I think your figures 3 and 8 are the more plausible representations. Worm holes, if they exist, can possibly connect two parts of our universe, but there may be no 'intersection' like in your fig. 7, I think.

BTW, what does the emphasized part mean: "The matter falls inside the event horizon from all directions around the BH, passes the event horizon (without any possibility of escape from now on) and is directed to the singularity (as it travels "deeper and deeper inside" the BH ) but never reach it." ??

Jorrie

Hello, Jorrie...

(as it travels "deeper and deeper inside" the BH ) but never reach it." ??

Well, this is a corresponding result as in figure 1... The question is: why something could never reach exactly the singularity... My point of view is like this: for an astronaut, who is falling in the centre of the BH, the time is retarded extremely (practically to zero)... So, for an external observer, it seems that he never, actually, reaches the singularity... For the astronaut, correspondingly, the time of "the world outside the BH" is accelerated extremely (practically to infinity)... So, from his point of view, also, he doesn't reach the singularity (or we could say that, when he "reaches" the singularity, the "outside world=universe" will have come to its "end" (whatever this means))... Hmmm...

Hi again George.

The fact that outside observers cannot view the astronaut inside the hole and 'see' her falling slower and slower, coming to an apparent standstill at the event horizon (not the central singularity), is just a coordinate singularity problem (infinite redshift).

Relativity theory says that the actual astronaut falls right through the event horizon and rapidly spirals into the central singularity (or the worm-hole), where she is crushed into pure energy. But then, relativity breaks down at the central singularity, so who can tell?

Seeing the outside world come to an end? I don't quite buy that one...

Jorrie

"Relativity theory says that the actual astronaut falls right through the event horizon and rapidly spirals into the central singularity (or the worm-hole), where she is crushed into pure energy..... Seeing the outside world come to an end? I don't quite buy that one..."

I agree that the astronaut (from his point of view) reach the singularity (and crush into)... But at the rest of the universe (and I mean outside the BH) an infinite time will have passed... This means the "infinite future" or the "eternity"... Am I wrong???...

Jorrie, could you, please, make some comments on the message #10... Thanks...

"But at the rest of the universe (and I mean outside the BH) an infinite time will have passed... This means the "infinite future" or the "eternity"... Am I wrong???..."

Yea, I'm afraid you're wrong. Sorry.

The fact that the distant observers cannot see what's happening at or inside the event horizon, makes it appear that they have to wait for all eternity and still do not see anything. In the meantime, the local free falling observer zips right past the horizon and before very long is crushed.

What does the in-falling observer see of the outside world? Definitely not an infinite blueshift/eternity. Remember that the in-falling observer reaches the speed of light as she passes the event horizon (with a BIG Doppler shift) and goes even faster inside the hole.

I have once calculated the redshift that the in-falling observer will measure when passing the event horizon, measuring a signal sent by a distant observer. The result was a redshift factor 2. (http://www.physicsforums.com/showthread.php?t=146912&page=5 post#60 and confirmed by a physics guru in #61.)

This means that the in-falling observer sees the outside universe slowing down.

I hope this makes sense.

Jorrie

"What does the in-falling observer see of the outside world? Definitely not an infinite blueshift/eternity."

I still think that happens this: "infinite blueshift/eternity"... General Relativity claims that inside a gravitational field (or "gravitational well") the time is retarded... (Such difference in time have been observed through experiments even in the very weak gravitatinal field of the Earth: the time is measured by two "precise clocks", one on the Earth's surphase and the other in an orbit (far away from the surphase)... The two clocks indicate a slightly different time, as is predicted by the General Relativity...)... So, the time of an observer, who is located inside a gravitational field, is "flowing" slower than the time of an observer, who is located fairly away from this gravitational field... (This is an essential difference with the special relativity which claims that when two observers move (relative to each other) with very high speeds, each one observe that the time of the other is retarded...)... The conclusion about these two observers is that the one, who is located inside the field, observe that the time of the other, who is located outside the field, is "flowing" faster... Am I right???...

Now go a step further: Imagine that you fall inside the extremely strong "gravitational well" of a BH... You'll see the time outside the BH to "flow" extremely fast... This means "infinite blueshift/eternity"... In fact he will be slushed by a radiation of infinite intensity (from the outside space)...

(Of course, I don't know what happens if you take into account that the in-falling observer obtain the speed of light inside the BH...)

So, what's your opinion about all these???...

"Remember that the in-falling observer reaches the speed of light as she passes the event horizon (with a BIG Doppler shift) and goes even faster inside the hole."

Is the in-falling observer going even faster inside the hole???... It doesn't seem logical...

Hi again George, your: "(Of course, I don't know what happens if you take into account that the in-falling observer obtain the speed of light inside the BH...)" is the crux of the issue.

The wavelength ratio λ_received / λ_emitted for a observer falling a escape velocity Ve=√[2M/r] towards a black hole M at distance r to the wavelength emitted by a distant observer is:

λ_received/λ_emitted = √[1+Ve]/√[1-Ve]*√[1-2M/r]= 1+2M/r.

This is the relativistic Doppler shift and the gravitational time dilation multiplied. At the event horizon 2M/r = 1 and the wavelength ratio becomes 2:1. It means that the Doppler shift wins all the way...

This means that the to the in-falling observer, time in the 'outside world' seems to run at 'half speed'. According to Prof. Chris Hillman, the equation is valid until just before it 'blows up' at the central singularity, where the Doppler ratio becomes infinite, but it's redshift, not blueshift.

"Is the in-falling observer going even faster inside the hole???... "

All material objects fall at Ve=√[2M/r] through the event horizon and the equation remains valid until it too 'blows up' at the central singularity.

Hope this helps to settle the 'eternity' issue...

Jorrie

Just a clarification that I neglected to put in:

The equations used 'geometric units', meaning c=G=1. In SI units, M must be replaced by GM/c² and Ve by Ve/c.

Jorrie

Thanks, Jorrie... You conviced me... I am somehow relieved and happy about all this "eternity" issue...

After this I have to reconsider my initial thought that an object which is falling inside a BH never "reaches" the singularity... So I have to apologize to all of you... And you must ignore the specific phrase of my presentation...

(Sorry, but I've read this idea (contraction of "outside time" & infinite blueshift) in a book of physics... It seemed very logical in accordance with the General Relativity and I extended this idea in my mind and I got the wrong conclusion that the object never "reaches" the singularity...)

"Is the in-falling observer going even faster inside the hole???... "

Do you mean that the object can obtain speed even larger than the speed of light (inside the BH)???... Is this possible???... Or I misunderstood something???...

No problem George, I understand why you held that view: "(Sorry, but I've read this idea (contraction of "outside time" & infinite blueshift) in a book of physics... "

It is perfectly true for a stationary observer just outside the event horizon. However, at the horizon there can be no stationary observer. If she attempts a 'controlled' powered descend to the horizon, she will still fall through it at the speed of light, unless she's got infinite power of propulsion... The reason is that the gravitational acceleration of a hypothetical 'stationary observer' at the horizon would diverge to infinity.^[1]

Once at the speed of light (when she crosses the horizon), the inward velocity will, according to general relativity, continue to increase under growing gravitational acceleration, which will cause her velocity to approach infinity as she reaches the central singularity.

Special relativity's speed limit does not apply inside a black hole. General relativity could of course be wrong when very close to the central singularity - that's where quantum gravity should take over.

Jorrie

[1] There's a lot more about this in Chapter 4: 'Static Gravity as Geometry' of Relativity 4 Engineers.

I wonder who is going to volunteer to check this out? Not I said the dinosaur. I ean one does have to go inside a black hole and take measurements does he not? Then he must report his findings.

If George's pictures are right, that Dino may not come back, but he may perhaps report his findings to some other universe, giving that civilization a huge technological advantage...

Jorrie

Thanks, Jorrie... This, also, makes sense...

This is all assuming you have figured out how to shield yourself and equipment from all the radiation. eg. cosmic, gamma, and x-ray.

I have not done the math but by the time you could see the event horizon with the naked eye, the naked eye will probably be cooked.

An event horizon that is cold from a lack of incoming matter could not be seen except by its effect on light passing close (relatively) before reaching the viewer as in Gravitational Lensing.

When the BH is feeding, for lack of a better term, the local area I would think would be a beautiful nightmare of energy. Best viewed from far away under high magnification.

Thanks, UV... I must not forget to take my sunglasses when I'll visit a BH...

Hi, Jorrie... After our discussion I had a second thought about the "2 dimensional model" in figure 1 of my presentation: As I thought that an object would never reach the singularity I had put the singularity at an infinite distance (as shown, also, at (A) of the above figure that I've drawn)... But, as the final result of our discussion was that the object would reach the singularity in a finite time (and, actually, very fast), maybe it is more appropriate to draw this model as in (B) of the above figure, where the singularity is placed at a specific point in finite distance... Afterall, the curvature of the space at the singularity is, also, infinite in (B) (as it is in (A))... Although the usual representation of a BH (that I usually see in books) is as (A) maybe the (B) is more correct...

What do you think???... I want your opinion...

Hmmm... But after a second thought (or I should say: a third thought) I thing that, maybe, the (A) could be, also, correct... The singularity can exist at an infinite distance... And that's because you've told me that the velocity of the "in-falling object" will become infinite at the singularity... So, t=S/u (where S=∞ & u=∞) gives any finite result... Am I right???...

I still want your opinion...

Note if ∞+1/∞-1 should enter the picture. In all cases, ∞ is not finite, the reason it is called infinite, and like the the singularity, very elusive. It is so elusive that one must say that ∞ = ∞, ∞ ≡ ∞, ∞ ≈ ∞ and ∞ ≠ ∞.

I'm not sure that I got your thought, Chtank... Clarify this, pls...

Afterall, my latter message may be nothing more than a nonsense...

The problem is that I participate to the CR4 forums through my PC at work (no PC at home yet, fortunately)... So I'll be back on Monday... I hope that we'll keep this conversation alive for a while...

Thanks...

my comment is that ∞ is not finite. It cannot be manipulated as the other mathematical expressions. When one reaches the ∞ he enters a new world. The light here is so bright that I must shade my eyes. I have difficulty imagining beyond infinity anyway. Guess I am just getting old or something.

Hi, Chtank... I agree that the ∞ is a very strange issue and we must use it carefully... The [∞ divided by ∞] could give anything (an indefinite result)... [Whatever is mutiplied by ∞] gives ∞... (even [∞ mutiplied by ∞] gives ∞)... So the result could be anything (indefinite value... so who can tell???...)...

Furthermore, I made another mistake: As Jorrie said, the in-falling object is accelerated inside the BH and its speed becomes infinite right on the singularity... As S=u.Δt and because we have u=∞ only at the "very last moment" we have the following at the "very last moment": dS=u.dt=∞.0= "no meaning"... (there is a wrong perception that x/0=∞ thus 0.∞=x, where x could be any number... this gives very fallacious results... I thought of this: for example one could say: 1/0=∞ and 2/0=∞ hence 1/0=2/0 hence 1=2... The fault is that, from the begining, the division of any number by zero is not permitted...)...

So I thing that what I said in message #27 was a nonsense...

The better is to assume that the singularity exists at a specific point in a finite distance from the event horizon... And that's makes sense...

Hi George. "Hmmm... But after a second thought (or I should say: a third thought) I thing that, maybe, the (A) could be, also, correct... "

The problem with these representations is the use of a fictitious fourth space dimension (non-existent from a general relativity (gr) point of view). So going to infinity in that dimension actually means nothing at all. All that is important is that the in-falling object reaches the vicinity of the singularity in a finite time. (I'm cautious here, because we expect gr to break down at or near the central singularity.)

Jorrie

Hi, Jorrie...

1) So, the figure (B), in the message 26, maybe is more correct (as a representation) than the figure (A), as figure (B) represents the singularity at a specific point in a finite distance from the event horizon... Do you agree???...

2) And another question: Is it sure that an in-falling object obtain the speed of light as it reaches the event horizon???... The truth is that I've never read this anywhere before... Here it's the first time that I read such a thing (written from you)... How did you come in such conclusion???... And if it's true for small BHs (as those which are produced from the "collapse" of a star), is it true, also, for the huge BHs (as those which exist in the centre of galaxies)???...

I need your opinion for these two issues... Thanks...

Morning George.

"So, the figure (B), in the message 26, maybe is more correct (as a representation) than the figure (A), as figure (B) represents the singularity at a specific point in a finite distance from the event horizon..."

I guess it depends a bit on the coordinate chart that you use. On a Schwarzschild chart, the A would be 'more correct', while on a 'free-fall' chart, B would be 'more correct', but both is wrong! There is no macroscopic dimension in that direction, only perhaps one of the tightly curled up dimensions of string theory.

"Is it sure that an in-falling object obtain the speed of light as it reaches the event horizon???..."

Any radially in-falling object's speed V_r approaches negative radial escape velocity, i.e. V_r = -Ve_loc = -√[2GM/r²], as it nears the event horizon of a non-rotating black hole. At the event horizon, r = r_h = 2GM/c², so V_r(hor) = -Ve_hor = -c. This is why light does not escape...

The equations are true for all non-rotating black holes, as you can find in any general relativity text, but differ for rotating black holes. There nothing can fall in precisely radially, so the local radial speed will be less than c. I think the resultant speed vector will still be c, however.

I reckon part of the reason for not finding this (falling at the speed of light) in books is the fact that one cannot have a stationary observer at the event horizon. So one can ask: falling in at the speed of light relative to what? One can however in principle have a stationary observer just outside (and arbitrarily close to the event horizon) who can observe the in-falling object's speed to approach the speed of light. The rest is extrapolation.

Jorrie

Oups... Jorrie, the above calculation gives V_r(hor)=-(c/√r_h) not -c... Did I misunderstand sth???... Clarify this, pls...

"There is no macroscopic dimension in that direction"... As I've already said, I agree with you... The "2 dimensional model" is just a tool for a kind of perception... It's not representing (actually) the reality as it is...

And what about the huge BHs that I refered in the previous message???... I mean is the [V_r=-c] valid???... Is this a general rule???...

Hi again George.

Oops, there's a typo in my formula (the square of r is wrong); should read:

Any radially in-falling object's speed V_r approaches negative radial escape velocity, i.e. V_r = -Ve_loc = -√[2GM/r].

"And what about the huge BHs that I referred in the previous message???... I mean is the [V_r=-c] valid???... Is this a general rule???..."

Yes, the horizon-crossing local velocity remains the same (V_r = -c); it's the same equation, so what else can it be? The only thing that changes is that the tidal gravity (differential force over the size of a spacecraft, human, etc.) is smaller for bigger BHs.

Jorrie

Thanks, Jorrie... I feel wiser right now...

About the tidal forces: I've read in an article that in huge BHs the tidal forces become more intense (and finally disrupts and kills someone) in the inside of BH (beyond the event horizon)... So someone has the opportunity to see what happens inside the BH before the tidal forces become significant (after a while) and strong enough, leading to his death... But in small BH's (like those after the death of a star) the tidal forces become strong and fatal before he succed to reach the event horizon... So, if someone decide to commit suicide by falling into a BH, a big one is suggested... At least, maybe, he will have some time to see what's happening in there...

Unfortunately, I didn't have a final answer to my main question: Is the representation of a white hole like in figure 3 or like in figure 5???... I have the sense that the geometry of a WH is the opposite than that of a BH, and figure 5 shows exactly this... But if so there must be intersections as the result (figure 7) (and I don't believe that such intersections exist either)... On the other hand, if the representation of a WH is like in figure 3 there isn't such a problem (but then, what is the difference between the geometry of a BH and a WH???)... If you don't have any comment to do on this I'll keep your opinion of your message #7 (that figure 3 represents a WH)...and I tend to agree with you)... Afterall, maybe the main property of the WH is that emits energy and matter and not whether is directed "up" or"down" in the "2 dimensional model"...

(Any comments on this are welcome...)

Jorrie (and all the others) thanks for your participation and your comments on this issue...

Lim _{x -> n} 1 / x - n = ∞ = Singularity