Black Holes, Ripples, and Warped Spacetime

my understanding is different but I'm not up on the latest. the "falling" part sounds wrong to me. its more of a massive gravitation that sucks in everything including light, hence the name black. we can observe it eating stars but the light disappears from the star once swallowed

Got this from Steven "the hawk" Hawking....

..."First, it's possible that the physical material (information) swallowed by the black hole never actually enters it at all. Instead, it's smashed into the point of no return and encoded as a two-dimensional hologram.

"The information is not stored in the interior of the black hole as one might expect, but in its boundary - the event horizon," he said. Working with Cambridge Professor Malcolm Perry (who spoke afterward) and Harvard Professor Andrew Stromberg, Hawking formulated the idea that information is stored in the form of what are known as super translations.

"The idea is the super translations are a hologram of the ingoing particles," Hawking said. "Thus they contain all the information that would otherwise be lost."

The information stored in these holograms is then emitted in the form of quantum fluctuations, though the data is so scrambled as to be useless for all intents and purposes. To return to our real-world analogy, imagine feeding a car through a crusher, industrial wood chipper, and coffee grinder. Even if you captured every bit of fluid, metal shavings, and tattered upholstery released at every stage of this process, there's no way to reconstitute two tons of finely-ground Volvo into a vehicle.

The advantage of this theory is that it doesn't violate quantum mechanics. The disadvantage is that it's rather boring.

Hawking's other proposed option is that black holes might serve as gateways into other universes. "The existence of alternative histories with black holes suggests this might be possible," Hawking said. "The hole would need to be large and if it was rotating it might have a passage to another universe. But you couldn't come back to our universe.

"So although I'm keen on space flight, I'm not going to try that.""...

http://www.extremetech.com/extreme/212968-stephen-hawking-may-have-finally-solved-the-black-hole-information-problem

My thinking, more of a feeling or intuition, is that theories like Hawking, string theory, etc are too complex and convoluted. That when we find the "true" theory it will be relatively simple.

Perhaps.

However, that analogy would posit that Newton's theory of motion and gravity, being simpler, are more correct than Einstein's general and special relativity.

There is a preponderance of evidence that suggests that the opposite is true.

I'd like to think it all condenses down to 42, but I suspect that the opposite is much more likely.

Google is your friend.

Gravity waves are a product of accelerating mass and two orbiting black holes do just that.

As mass accretes the warping of space-time increases and so does the radius (Schwarzchild Radius) of the event horizon (although the event horizon is being rethought of late it still is a viable tool). The formula is simple:

R = 2MG/C²

The reason that nothing escapes a black hole inside this radius is the escape velocity is greater than the speed of light.

The other statement about falling forever has to do with the intense gravity field distorting time (slowing proper time down) for the object near or entering the black hole.

As to the actual singularity, that's where the math breaks down (divide by zero) in the theoretical domain.

Dr Michio Kaku provides some interesting theories...

http://mkaku.org/home/articles/blackholes-wormholes-and-the-tenth-dimension/

Is it possible that we live on the edge of two dimensions..?

...""light" is nothing but vibrations rippling along the 5th dimension."...

...and to find the 'theory of everything' is so difficult because the quantum world is in fact another lower dimension??

The possible exception to your description might be at the event horizon where a huge energy flux is likely due to the black hole's accretion disks. That density of radiation is likely to fry you before you pass.

Ignoring the radiation, you are correct in that nothing special happens to the observer traveling across it.

May be in the other part of it, if one is willing to travel, they will find √-1

And if it were a stellar size black hole you would become considerably taller as the tidal forces stretched you in the radial direction and compressed you in the other two directions. A super massive black hole might be large enough that the tidal force, which falls off as the cube of distance, would not be fatal.

Either way I can think of better places to vacation.

Hi AH. You said: "... nothing special happens to the observer travelling across the event horizon." This is true only for huge, super-massive BH (i.e. very large event horizons). In the case of smaller BH the extreme tidal forces (differences of the gravity force on the several parts of a body) will stretch and dismember the observer as he is approaching the event horizon.

True, but that gravity gradient has nothing to do with the event horizon.

Either way you look at it, entering a black hole would have to be the last thing on your bucket list.

Also true, unless you pass through a wormhole and another, better universe waits for you on the other side...

"One thing that is very confusing is that gravity affects time.....As you fall into the black hole, your time appears to me to slow to a stop as you approach the event horizon, and from my perspective, you never reach the event horizon."

---------------------------

Confusing indeed! If (for an external observer) anything that falls into a BH never actually reaches the event horizon it means that the BH cannot actually grow in size (it would take an infinite amount of time to do that). But it means after all that it cannot form from the very beginning - as the initial matter is collapsing the time in the center of it slows down more and more until a fraction of a second there becomes longer than the age of the universe (for the perspective of an external observer) and this happens even before the event horizon (the BH) can appear.

So, do BH really exist? Or they are objects that will never have time to form (for the outside universe)?

Am I missing something here?

I think that the time dilation is not coming to a stop at the event horizon, but continues to slow down the closer you reach the singularity.

The Schwartzchild metric is used to compute the difference between proper time for the clock near a gravitational field and a clock outside the gravitational field.

t₀ = t_f √1- (2GM/rC²)

Where:

t₀ = ∆time for the clock inside the gravitational field

t_f = ∆time for the clock outside the gravitational field

G = Gravitational Constant

M = Mass of the black hole

r = Radius from black hole

C = Speed of light

If you graph the function you would see something similar to the Lorenz gamma factor. That is, until you get very close to the singularity the time dilation isn't as great as you might suspect.

Here is a graph of the gamma factor:

Hi AH. As far as I know, when an astronaut is falling towards a BH, an outside observer sees the astronaut slowing down as he is approaching the event horizon (time dilation) and he "freezes" exactly on the event horizon (not on the singularity). Of course, this doesn't mean that the observer can see the astronaut staying on the event horizon forever (as a frozen image). Actually, taking into account the redshift (gradual increase of the wavelength of the astronaut's image as he is approaching the event horizon) the astronaut's image is fading out. The total visual result should be something like the astronaut is slowing down and disappears.

I'd trust the math.

Most of these science magazines try to put things in terms that the reader is supposed to understand and many times the writer doesn't understand.

It's like the description of an expanding universe using a balloon. Most people think it's the volume of the balloon, but it's the surface of the balloon that is being used as an analogy.

From the singularity. That equation applies for non-black holes, too, as long as the other object is outside the radius of the larger mass.

Hi AH. Take a look at my post #35. (The equation gives an infinite time dilation for an in-falling object right on the event horizon -as observed by a distant observer- and not inside the BH or the singularity.)

Also, take a look at my post #37. (The redshift of an in-falling object becomes infinite right on the event horizon -as observed by a distant observer- which, also, corresponds to an infinite time dilation of the object. I.e. the only explanation that a distant observer can give for this infinite redshift is that the "local time" of the in-falling object has been expanded infinitely -i.e. infinite time dilation- as I, also, explain at my post #39.)

Take a look at this link:

http://www.phys.vt.edu/~jhs/faq/blackholes.html#q11

(Read the paragraph: "Will an observer falling into a black hole be able to witness all future events in the universe outside the black hole?" where it clearly depicts this infinite time dilation right on the event horizon.)

Regards

Well, no, an observer can't witness all future events while falling into a black hole because the information following him in is under the same laws of physics as he.

Think of it as a traffic jam. The photon behind you from the future remains behind you. In other words, it never reaches you any sooner than it would if you were a distant observer.

Unfortunately, you won't be around to know that since the black hole will shred you atom-by-atom. There are no loitering signs everywhere around a black hole.

I know the problem is confusing and I have difficulty wrapping my brain around it too, but things do fall into black holes.

Also, an event horizon is dynamic. That is, it will expand over time, so as an object reaches the event horizon its mass will contribute slightly and cause the event horizon to increase.

You will never witness anything reaching the event horizon because all the information in the form of radiation (light, etc.) redshifts to the point where it goes black at that threshold. Hence the name Black Hole.

AH, there is a misunderstanding. Don't try to convince me that an object actually falls into the BH. I know that perfectly well. (I have already replied to Alex. Roman about this issue.) Also, I know very well that an in-falling astronaut doesn't witness all the future events of the outside universe. (For God shake, he doesn't.) Right on the event horizon, he is experiencing a finite redshift (i.e. redshift=2) of the outside universe, so he actually observes the outside events running slower (not faster). (I have already mentioned this many times throughout this thread.) So, I strongly believe that you don't read my posts at all.

My objection was the following: You've claimed that an infinite gravitational time dilation of an in-falling object (as observed by a distant observer) occurs in the inside of the event horizon. I claimed that this occurs exactly on the event horizon. That's why I suggested to you to read my posts #35, #37 & #39 (please do). As you said, the maths are very clear. And I, also, trust maths.

Do a search about that and you'll see that I'm right. I just sent you this link:

http://www.phys.vt.edu/~jhs/faq/blackholes.html#q11

as an example. And I said read the specific paragraph #11 in this link (not the title of the paragraph).

Ah! I see your point and you are correct. My mistake.

The boundary is at the event horizon and not inside it.

The time dilation will come to a stop when 2GM/rc²=1 (when you reach the event horizon). So, the paradox remains.

Besides slowing down the time any gravitational field will also expand the space with the same factor (for a local observer). Similar to the Lorenz gamma factor there is a gravitational gamma factor that works in the same way.

So, there are in fact more paradoxes related to a BH:

1) For an external observer an object approaching the BH will appear to slow down (to a stop at the event horizon)

2) For the internal observer (falling to the BH) the closer it gets to the event horizon the greatest the space expansion (to infinity at the event horizon)

3) For any free falling object in a gravitational field the escape velocity in a certain place is equal to the velocity gained by the free fall (if the object falls from a very great distance, close to infinity). At the event horizon the object should have a velocity of c (because the escape velocity is c). But no object can be accelerated to c (because it will have an infinite mass)!!

The above paradoxes apply for any BH size, so from the very first moment the BH tends to appear. So (again), can a BH really exist?

I don't think that's what the math says. The term r is the radius from the center of the black hole, not the event horizon.

The equation you cited is actually for calculating the Schwartzchilde radius of a black hole, not the dilation of time.

The math for the proper time for an object within a gravity field is:

t₀ = t_f √1- (2GM/rC²)

Where:

t₀ = ∆time for the clock inside the gravitational field

t_f = ∆time for the clock outside the gravitational field

G = Gravitational Constant

M = Mass of the black hole

r = Radius from black hole

C = Speed of light

Here's the formula for escape velocity:

V_esc = √2GM / R

G = Gravitational Constant

M = Mass of body

R = Radius of the body

Note that there is no term for the initial velocity or even the change of velocity of an object.

Lastly, the appearance of an object falling into a black hole's event horizon doesn't actually freeze in time at that radius.

What you will observe is the photons from that object getting dimmer and dimmer as the energy is red shifted more and more. At the event horizon no energy or photons from the object return to the outside observer, so it looks like the falling object has gone black.

This is, in part, an optical effect by the bending of light rays.

You will pass through an event horizon if you fall towards it. Particularly if the black hole is greater than 8 solar masses where the tidal forces are not as violent until you are inside the event horizon.

I'm sorry that my reply was not clear enough - I thought the missing details will be easily grasped. As it's not the case, let me enter in more details.

I just took from your equation the term 1-(2GM/rC²) and made it equal to 0 - when this happens t₀=0 (the time stops). This is the moment the object reaches the event horizon. At that moment r is the radius of the BH, which is a sphere having the event horizon as its surface. The problem is that for an external observer (outside the influence of the time dilation caused by the gravitational field) the closer the object falls to the event horizon, the grater the time dilation. At a certain point a fraction of a second for the falling object will be seen as several years for the outside observer. Then, even closer, it will be seen as billions of years, and then even a Plank time interval will be seen as hundreds of billions of years, and so on.

Even it will be extremely red shifted and you will need to use other ways of detecting its presence (first infrared, then microwave, and so on) the object will still be there. It's the same case as for the red shift caused by very high speeds, close to the speed of light. The point is not if you can see the object or not, the point is that the object will need an infinite time (for the external observer - no matter how the observation is made) to reach the event horizon.

Now, regarding the escape velocity: what I tried to say is that if you have an object that is in free fall towards a very big mass (so that you can consider this big mass as stationary and the object has no other external influence), the theory says that if it started with a zero speed at infinite distance, while falling towards the big mass, in any point of its fall its velocity value will always be equal to the escape velocity corresponding to that point (just the direction will be opposite). The easiest way to prove this is that if it will suffer a perfectly elastic collision it will start going backwards with its velocity gradually decreasing until it will have a zero value at an infinite distance (so the velocity at the collision point was exactly the escape velocity).

So, since the escape velocity at the event horizon is c, it means that the falling object should have there a falling velocity of c - but this is not possible for any object that has m₀≠0.

Okay. I understand what you are saying.

The confusion is that many people think because objects at the event horizon are experiencing Tau_ZERO nothing actually enters a black hole because time stops. This is not quite true.

Now, you wrote, "the point is that the object will need an infinite time (for the external observer - no matter how the observation is made) to reach the event horizon."

No it actually doesn't take an infinite amount of time. If an object free falls toward the star or black hole it does so at an accelerated velocity. It still gets there, and it gets there very, very fast, but what you see as an outside observer is the light of the object, not the object itself. That return signature is delayed.

As the free falling object nears the event horizon its returning light is red shifted and appears to slow down, but you are only witnessing the energy signature of the object in the form of light that is disrupted by the black hole's gravity field.

Also, for an outside observer watching another approaching the event horizon there is an issue that there is no coordinate-independent way to say that two distant events are happening simultaneously in General Relativity.

You also cross the event horizon just a tiny fraction faster than what the math would say. This is because your mass adds to the black hole and the event horizon expands slightly. Inside the event horizon time is a spatial dimension.

I am heading out of town today, so I can't elaborate, but for those interested you can get more information from the following links:

Penrose Diagram

Minkowski Diagram

If AH will pardon me for jumping in late, and further to what he wrote, you must be careful of the "time stops at the event horizon" idea. This is just a coordinates time, which has no physical meaning other than you would "see" the infalling astronaut's clock tick slower and slower...

As far as the astronaut is concerned, his time ticks normally and if the BH is big enough, inside his ship he will not even notice that he is crossing the horizon - that is unless he looks outside.

Interestingly, if the remote observer is sending a signal towards the BH and the astronaut measures its wavelength shift, he will observe a signal that is (relativistically) red shifted by this relative speed AND blue shifted by his lower gravitational potential: (geometrically, where G=c=1)

λ/λo = √[(1+Ve)/(1-Ve)] * √(1-2M/r) = 1+Ve (because Ve = 2M/r)

At the horizon, where r=2M, Ve=1 and it gives λ/λo = 2, i.e. he will receive signal with twice the original wavelength (or in redshift terms, with z=1).

Since λ/λo = 1+Ve remains well behaved until the astronaut approaches the central singularity (r=0), he will continue to observe the original signal at higher and higher (diverging) redshift until he is almost "there". What happens then, we don't know...

-- Regards, Jorrie

Sorry I goofed in my qualifying remark "(because Ve = 2M/r)".

It should obviously be Ve = √[2M/r], which makes the multiplier √(1-2M/r) =
√(1-Ve²) = √[(1-Ve)(1+Ve)], causing the cancellations that lead to the simple result.

Hi Jorrie.

You said: "This is just a coordinates time, which has no physical meaning other than you would "see" the infalling astronaut's clock tick slower and slower..."

As I have also pointed out to Alex. Roman, the falling astronaut is actually falling inside the BH and doesn't stay forever on the event horizon. This is just the perception of a distant observer.

You said: "Interestingly, if the remote observer is sending a signal towards the BH and the astronaut measures its wavelength shift, he will observe a signal that is (relativistically) red shifted by this relative speed AND blue shifted by his lower gravitational potential: (geometrically, where G=c=1)"

This is what I mentioned in my post#26. There are two different causes for the wavelength shift of the outside universe as experienced by the falling astronaut: 1) a redshift due to his relativistic speed and 2) a blueshift due to the gravity.

You said: " At the horizon, where r=2M, Ve=1 and it gives λ/λo = 2, i.e. he will receive signal with twice the original wavelength (or in redshift terms, with z=1)."

This is what I said in my post#26. I remembered (from a very old thread of mine about the BHs) where you had mentioned that the overall shift of the wavelength (combination of shift due to gravity and relativistic speed) of the outside universe as observed by the falling astronaut has a finite value, but I couldn't recall this specific value. So, the value is 2. That's nice.

However, I have the same doubts as Alex. Roman has in his post#26 about where the astronaut's clock actually stops (as observed by a very distant observer).

So the distant observer observes that the astronaut's clock is gradually slowing down. This time dilation due to BH's gravity is:

t_r=t.√(1-(r_s/r)) [where r_s is the Schwarzschild radius (event horizon): r_s=2GM/C²]

According to this equation:

1) When r=∞ then t_r=t

2) As the astronaut is approaching the event horizon, the r becomes smaller and -as a result- the t_r becomes, also, smaller which means that the astronaut's clock runs slower (as observed by the distant observer).

3) When r=r_s (i.e. right on the event horizon) then t_r=0. This means that the astronaut's clock actually stops (as observed by the distant observer).

4) When r<r_s (i.e. inside the event horizon) then the t_r takes negative values and at r=0 (i.e. right on the singularity) we get t_r=-∞. I don't know what these negative values (or -∞) of the t_r actually mean.
So, as it seems from (3), the astronaut's clock stops on the event horizon. Am I wrong with that?

The problem is in the words "the astronaut's clock stops", because even with the qualifier "as seen by the distant observer", it is still just an interpretation, based on a coordinate time. It is so that clocks 'lower' in a gravitational well record less elapsed time than clocks 'higher' in the well, which can be tested by starting and ending the test with the clocks together at the same level. Relativistically extrapolating the equations to the horizon points towards zero elapsed time for a clock at the horizon.

The problem is that the same theory says that it a clock falling through the horizon does not physically stop ticking, because a falling observer at the same level (or somewhat deeper into the horizon) can still see the clock ticking! Do you see the potential for a paradox here?

Secondly, T_r does not go negative inside the horizon, but rather imaginary. Outside observers can receive no signal from there so the point is rather irrelevant. A different interpretation is that the space and time coordinates of spacetime swap places inside the metric, as AH has said. This is however also just an interpretation with no observable effects to justify it.

Relativists routinely analyze the insides of black holes using appropriate coordinates, like free-fall or Kruskal coordinates.

-J

Hi again. I still don't get it. It's not a matter of "interpretation". The equation itself says exactly that: The falling astronaut's clock (as seen by a very distant observer) actually stops right on the event horizon (not inside the event horizon or the center/singularity of the BH). I've also read exactly this on numerous physics books.

Here is again the maths:

The observed redshift due to gravity effect is:

1+z=1/√(1-r_s/r)

Hence, for r=r_s (i.e. right on the event horizon) we get 1+z=∞ (i.e. an infinite redshift, which corresponds to an infinite time dilation).

Here is what the Wikipedia says as a result:

"...The effect is very small but measurable on Earth using the Mössbauer effect and was first observed in the Pound-Rebka experiment.^[46] However, it is significant near a black hole, and as an object approaches the event horizon the red shift becomes infinite..."

I've never claimed that the astronaut's clock actually stops. The astronaut will observe his clock running normally. Also, if the astronaut could be static right above the event horizon (i.e. let's take into account only the gravity effect) he should see the distant observer's clock running extremely fast. In fact, right on the event horizon, he should see all the future time of the whole universe in just one moment (this is equivalent to an infinite blueshift). Of course, I know that this is not the real case, because if we also take into account his (unavoidable) relativistic speed, he is experiencing a final finite redshift of 2 (for the outside universe) as you said. (This is equivalent by seeing the clock of the distant observer "ticking" at half the rate of his clock).

From the distant observer's point of view, he should see an infinite redshift of the astronaut's image right on the event horizon. By taking into account only the gravity effect, he should observe an infinite redshift (as shown above). By taking into account only the astronaut's relativistic speed, he should see, again, an infinite redshift. Of course, the overall result must be an infinite redshift of the astronaut's image (which, of course, corresponds to an infinite time dilation).

George, I think there was never an argument about infinite redshift observed (from 'outside') at the event horizon, just that it does not equate to a "stopped clock", which I think has a well defined meaning.

Distant observers do not observe clock rates, just redshift.

-J

Hi Jorrie. You are right: Neither the infalling astronaut can actually see the distant observer's clock, nor the opposite. They only can observe a wavelength shift of the light. However, this can be interpreted as a time expansion (or contraction) of the time itself.

For example, let's suppose that the infalling astronaut is emitting a signal of a specific, well known frequency. As the astronaut is approaching the BH the distant observer observes an increasing redshift of this signal. As the frequency is f=1/T (and knowing that the signal's frequency is stable) he could conclude that the "local time" of the astronaut has been expanded. So, he could interpret this phenomenon as a time dilation (and this is exactly the time dilation due to gravity and relativistic speed). When the astronaut is exactly on the event horizon, the redshift of his signal becomes infinite (f=0 Hz) and it's like his "local time" has been expanded infinitely.

So, I think that (in a free way of thinking) we have the right to say that the astronaut's local time "stops" (as seen by the distant observer) exactly on the event horizon of the BH.

GK wrote: "So, I think that (in a free way of thinking) we have the right to say that the astronaut's local time "stops" (as seen by the distant observer) exactly on the event horizon of the BH."

I would avoid any interpretation that the astronaut's local time 'stops', because it is a coordinate dependent observation. And as we know, coordinate dependent time does not represent proper time, which can only be measured by the astronaut himself.

The distant observer observes redshift that tends to infinity, but we have no real 'right' to infer 'real clock rates' from that. It is a common practice, but I think it is a questionable inference.

A point to remember is that two inertial observers receding from each other at a relative speed of very close to 'c' (in flat spacetime), will each observe the others redshift to approach infinity. Do we interpret both clocks to have stopped?

-J

Hi Jorrie. You said: "I would avoid any interpretation that the astronaut's local time 'stops', because it is a coordinate dependent observation. And as we know, coordinate dependent time does not represent proper time, which can only be measured by the astronaut himself."

In a way, there is not a "proper time". Or (if you like) the "proper time" for each observer is his "local time". I.e. for the astronaut, the proper time is the time which is measured by his own clock. And he observes that his clock runs normally. I could agree that this is a kind of "real time" because -for the astronaut's coordinate- it is a fact (or, in other words, only this "time" matters to the astronaut). However, for a distant observer, the astronaut's time looks like it is gradually expanded (and infinitely expanded right on the event horizon). This is the gravitational time dilation. Of course, the only observable fact (as you said) is the astronaut's redshift. But the astronaut's "time dilation" is an indirect and unavoidable conclusion for the distant observer (in order to explain this redshift).

You said: "A point to remember is that two inertial observers receding from each other at a relative speed of very close to 'c' (in flat spacetime), will each observe the others redshift to approach infinity. Do we interpret both clocks to have stopped?"

Again, each observer interprets the other observer's extreme redshift as a "time dilation". It is like the other observer's "local time" has been expanded. For Special Relativity this time dilation (due to relativistic speeds) is, again, an indirect conclusion for each observer. It's a relative and subjective conclusion. (Of course, the only real time for each observer is his "local time".)

Another way to conclude this time dilation (due to gravity or relativistic speeds) is via light pulses. Let's suppose that a spacecraft emits a light pulse every second. As the spacecraft approaches a BH (or as it is accelerating at a speed close to c) we observe that these light pulses become more and more sparse. Again, the indirect conclusion for us is that the "local time" of the spacecraft has been expanded (relative to us). Right on the event horizon (or right at speed c) we receive no pulses at all. So, for us, it's like the "local time" of the spacecraft has been expanded infinitely (relative to us).

You wrote, "But the astronaut's "time dilation" is an indirect and unavoidable conclusion for the distant observer (in order to explain this redshift)."

Actually, the red shift is explained by the relativistic velocity of the in-falling object, not time dilation.

You wrote: "Actually, the red shift is explained by the relativistic velocity of the in-falling object, not time dilation."

So, do you mean that there is no red shift of the astronaut due to gravity??? Does this red shift occur only due to his relativistic speed??? I mean, if (somehow) the astronaut manages to overcome the BH's gravity (even for a while and it can be done if he is not so close to the event horizon and by applying enormous thrust) the distant observer will not experience any red shift of the astronaut image???

I think that there must be, also, a red shift of the astronaut (as seen by the distant observer) because of the gravity. For the same reason, the astronaut will observe a blue shift of the outside universe. Am I wrong???

According to Jorrie you are probably right. So, I am going to concede on this point.

AH, I'm afraid that I have to disagree with you here. We are looking at the remote observer in terms of the Schwarzschild metric and in those coordinates the radial speed of the infalling astronaut is zero at the event horizon. That's why it appears to be "frozen" at the horizon. In Schwarzschild coordinates, the infinite redshift comes purely from the gravitational time dilation:

t₀/t_f = √[1- 2GM/(rC²)] = 0 at the horizon.

Okay.

I have been treating the in-falling object and its return EM signature as two different things.

So, in summary, I think that we could say the followings:

1) What is observed by the distant observer (concerning the in-falling astronaut):

• Taking into account only the gravity, there is a redshift (which becomes infinite on the event horizon).
• Taking into account only the astronaut's relativistic speed, there is a redshift (which becomes infinite on the event horizon).

(Of course, the overall result must be an infinite redshift on the event horizon.)

2) What is observed by the in-falling astronaut (concerning the outside universe):

• Taking into account only the gravity, there is a blueshift.
• Taking into account only the astronaut's relativistic speed, there is a redshift.

The overall result is a finite redshift (=2) on the event horizon.

I hope that Jorrie would agree with this summary.

Hi GK, I can live with your summary, because it avoids the inference of unobservable relative clock rates.

I have a minor concern with the second part of point 1. We have treated the distant observer in Schwarzschild coordinates and as I wrote to AH before, the coordinate speed of the in-falling observer is zero at the horizon. In Schwarzschild coordinates the infinite redshift at the horizon comes purely from gravitational redshift. Otherwise we are mixing two reference frames in one argument.

This is however not too serious, because it makes no difference to the observable redshift.

Similarly, we are supposed to treat the in-falling observer in free-fall (or equivalent) coordinates, but it makes no difference to the prediction of the result.

-J

Thanks for your reply, Jorrie. Yes, I had read your reply to AH. Probably, it's not appropriate to mix this two approaches.

Thanks again.

George, I think we all have our own preferences, but I would never use the terms "clock stops" or even "clock running slower" in any relativistic context. The only terminology that is relativistically uncontroversial is that one clock records less elapsed time than another clock and that the difference depends on the respective spacetime paths that the two clocks took. It becomes quite complex in general relativity, but it is always possible to quantify that spacetime path and extract the elapsed time from it.

On this note, I would like to end the discussion, unless there are new perspectives that someone can offer.

-J

Hi Alex. When an object is falling onto a BH, the time dilation is observed only by an outside observer. This doesn't mean that the object never actually falls inside the BH. The BH is formed by the mass of a big, collapsing star and keeps on "swallowing" matter and energy for the rest of its life (getting bigger and bigger).

Let's assume that an astronaut is falling on a BH. An outside observer observes a redshift of the astronaut's image. The redshift occurs because of the gravity (time dilation) and, also, because the astronaut is moving at a very high speed. On the other side, the astronaut experiences a blueshift of the outside universe (due to the gravity) combined with a redshift (as the outside universe is "moving away" from him at a very high speed). As far as I can remember, the combined effect is a final, finite redshift (I'm not sure though). Anyway, when the astronaut passes the event horizon, there is no return. He will eventually end up on the BH's singularity (or its "wormhole", if such a thing exists).

My knowledge of this subject is not good. So I have an observation? Gravity is bent space and time, and gravity is normally shown graphical as a downward slope towards the mass, and a external smaller mass falling down that slope towards the larger mass. So my observation is, is gravity acting on the smaller mass driving it towards the larger mass, and secondly, gravity waves are shown radiating away from the mass of origin, so does gravity originate from outer space, or outward from the mass?

Remember, the bent sheet is only a 2D analogy of gravity's effect on space-time.

There is no actual force between the two masses, but mathematically you can predict their behavior as if there was (Newtonian physics).

Gravity simply impacts space-time and objects traveling through space-time follow a straight line geodesic. It's the warping of space-time that makes that path appear bent.

Gravity waves are simply oscillations in the space-time moving at C. From a 2D perspective they behave much like the ripples from a pebble dropped in the middle of a pond. However the amplitude is very small and that is why you need two large black holes circling each other at half the speed of light to make waves of enough amplitude to detect them.

If I remember correctly, LIGO detected a vacillation in space-time over a 4 km stretch that was akin to a change in distance less than the diameter of a proton.

Hi jdretired. Every mass (large or small mass or even an amount of energy) bends the space-time continuum. The larger the mass the more intense the bending. So, the larger mass -through its stronger bending- affects more the smaller mass than the opposite, forcing the smaller mass to fall onto (or orbit around) the larger mass. In Newtonian approach the gravity force is supposed to be the same on both objects but its effect (acceleration) is much stronger on the smaller mass than on the larger mass (γ=F/m) and so the smaller mass is moving towards the larger mass (more than the opposite). Of course, the General Relativity approach depicts the same results.

The gravity is not originated from the outer space. It is originated from the mass. The space (space-time continuum) is just the transmission medium for the gravity. As the mass produces a distortion (bending) of the "space fabric", this distortion is moving away from the mass through the space itself -affecting any other nearby mass (or energy)- at the speed of light. From the Quantum Physics perspective, the hypothetical carriers of the gravity force are the gravitons (which, of course, are travelling at the speed of light). It is, also, supposed that the gravity-waves are produced during very "violent" events. An example of such an event is a pair of BHs rotating relative to each other. Right before their collision and merging they produce gravitational ripples (waves) which travel away from the collapsing system.

We can only wish that old Volvo's were crushed, ground to a powder and sucked into massive black holes from which they never could escape. Instead, they are still running around spewing oil clouds and dripping oily slicks onto the roadways. Now let's get the black holes to tend to the VW TDI's if VW refuses to do a buyback or refund the purchases.

Hi HiTekRedNek. You said: "I always thought that when a black hole added matter.....the horizon became smaller" This is wrong. As the matter keeps on falling inside a BH (hence the mass of the BH is increased) the Schwarzschild radious (event horizon) is increased and the BH becomes bigger and bigger. The only way for the event horizon to become smaller is only due to the Hawking radiation. This procedure is equivalent to the loss of mass of the BH (it's like the BH radiates itself these particles of radiation, although these particles are created -in fact- just outside the event horizon of the BH).

I think my plumber had the TOE.

"Hot on the left,cold on the right and water don't flow uphill."

Heat flows from left to right(hot to cold),and everything takes the path of least resistance.

The universe is expanding because it is easier to expand than to resist whatever force is within.

Space time warps when it encounters matter because it encounters a different medium, like light striking the surface of water.

Space and time are both warped.

Every molecule has space within it,and it is this circuitous path around all of the subatomic structure that causes space time to bend.

If close enough to center of a large mass,relative time could actually stop.

A tip of the Hatlo Hat,and thanks to Joe The Plumber.