How does spacetime curvature become infinite?

I’m trying to understand black holes. Specifically I wish to understand--without knowing tensor calculus--how the escape velocity of a body can equal the speed of light or greater to form a black hole.

A book I have (Relativity Visualized) explains that mass curves spacetime, and the greater the mass the greater the curvature. I understand that. The book goes on to say that a black hole forms when the curvature becomes infinite (and notes that this does not require infinite mass). This I do not understand, because it seems to imply that as mass increases there is a particular moment at which the curvature transits from finite to infinite. I understand how a finite value can approach infinity, but how can a finite value become infinite?

Thank you.

 

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Good question

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It's my understanding that the curvature never truly becomes infinite. Rather, it perpetually grows steeper and steeper. At least, from GR standpoint. The curvature at the singularity itself is infinite, but then again the singularity is a mathematical concept that may have no counterpart in reality; whether a singularity can ever truly form is still a matter of debate. For example, string theorists prefer to think of the singularity as literally a ball of string.

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Note also that the curvature applies both to space and time (ergo, "spacetime") -- so both are stretched.

One "intiutive" way to visualize black holes is as whirlpools. If the current flows into the sink too fast, even light can't "swim" fast enough to overcome the current. You can picture spacetime "flowing" into the singularity, and dragging everything else with it.

Another "intuitive" analog would be to imagine that light (and everything else) beyond the event horizon suffers a total internal reflection of sorts back toward the black hole's center as it attempts to move away from the center.

(btw, welcome to the insane asylum!

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Excellent info. I’m enjoying the asylum!

That the curvature never truly becomes infinite agrees with my intuition. Along those lines it seems to me that a body could not truly become a singularity, because the body would have to transition from “having a size” to “having no size” at a particular moment, somehow completing the “approaching infinitely small” stage.

I’ve often thought of light swimming against a current as you suggest. For a black hole, the current (the escape velocity) would have to be c, the speed of light, or greater, and that is what I have read (sans an explanation not requiring tensor calculus).

Yet how can the escape velocity reach c? Light always gets to its destination in no time from its perspective. In a race to a nearby galaxy against a spaceship moving at .99c, the ship might take a century of proper time whereas light would take infinitely less proper time than a nanosecond and more correctly no time at all.

If I imagine light swimming directly against a current that is moving at .99c, I see the light getting to a destination across the universe in no proper time. (Although the light is moving against a current, the proper time would still be none, because if you had a formula to dilate the amount of proper time elapsed by the light, you’d be plugging “none”—not zero—into it, and the result of such input remains “none.”) I see the same for a current moving at .999c and .9(million 9’s)c. How then can I intuitively understand that there is a particular moment, in a region of ever-increasing curvature of spacetime never reaching infinity, precisely at current = c, at which there is a transition from “light may reach a destination across the universe in no proper time” to “light may not reach that destination”?

Or what is the flaw in my thinking?

 

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Most people assume that black holes would be catalysed by a gravitational body, but this might not necessary be the case. From what I understand a black hole could be created through a object or body suffering complete wave function collapses, causing the object/body to dent Spacetime and draw space into it's centre.

This of course would cause a compression of spacial folds towards it's centre, so would pretty much give the same effect as an Event Horizon.

Although it's said that the centre of a black hole would be giving more G's than light can elude. You have to remember that an Event Horizon is a like emminating circles from a drop of rain hitting a puddle.

The folds exist as normal space, with normal G that light can escape, but due to the Event Horizon causing a spacial stiffening, it just seems that the light can't ecape it's gravity.

With this in mind, light can escape, but it just has to be on the same level as the folded spacial plane, which means when it escapes and your looking from a different plane, it will have suffered a spectral shift.

Since most of the universe is filled with EM radation and Blackbody, it's safe to assume that they are the objects, bodies and energy that has transversed from a multiworlds plane into another dimension, either greater or lesser than the one you would be observing from.

Although Einstein made the Equation E=MC<SUP>2</SUP>, some people still argue over how a Theory can be taken for granted.
Well I drew a small descriptive passage in another post that explained, that light on paper is drawn as waves, and these waves are arc's.

If light was to try and pass it's own velocity, (namely it tries going too much faster than it can) then the arc's will try to form in a 90Deg angle so they won't be arc's, they would be just straight up and back down on the same spot.

This means light can't pass that threshold, otherwise it would INVERT its wavefunctions. (The arc would start moving the other way)

 

anyway

kewl explanation stryder

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the way I see black holes (slightly offtopic) is by starting to imagine a planet like Earth. We would be able to stand on it, and rockets are ale to get off it.
If we then take the Sun, that is qiute a bit heavier, we wouldn't be able to stand on it ~duh~, but neither will a rocket be able to get off it, as it cannot reach the speed necessary to escape.
If we then take a simple (but huge) cloud of gas and we let that slowly fall onto the Sun, the Sun will get heavier (and denser, hotter and so on) and the speed (watch that I don't write 'power') the rocket needs to escape will be even larger.
Eventually, after adding a lot of gas (and numerous collapses and im/explosions of the Sun), the escape speed will reach c. When that happens, the fastest things we know of, photons, won't even be able to escape! Then we've converted the Sun into a black hole.

At this point, spacetime (probably) won't be curved infinitely. What I could expect happening though is that the black hole will collapse, as more and more matter flows in. Personally I think there will be some point at which the black hole will 'fall through' spacetime. I would think that will happen when the black hole reaches a mass at which the quarks inside are pressed onto eachother so hard that they disintegrate (but if quarks happen to be made out of smaller particles it'll be when THEY disintegrate. Thus when the smallest possible particles disintegrate under the force applied on them).
What happens at such a point I cannot foresee. I suppose the particle will be 'gone'. leaving a 'hole in spacetime' or creating a wormhole to somewhere. Maybe even the moment at which it collapses will curve spacetime so that time stops there, making a 'collapsed black hole' the most stable thing there is

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In some interpretations black holes don't actually exist. Now before you get excited let me explain. As the hole starts to form the event horizon's time rate slows down. In fact it slows down so much that while it gets closer and closer to forming it never actually forms. Light gets slowed down so much it looks like it can never escape.

Not that it makes any real difference to what we see, as after forever when the hole has finished forming the light will have stopped and if it had not stopped it would not be able to escape anyway. But none the less the black hole never finishes forming so no singularity.

Dont ya just love infinities ? Now to head off the bright sparks out there, the matter thats collapsing is not at the black holes event hroizon so is not slowwd so can it form the black hole ?

Well firstly we aint sure what goes on inside the black hole, as the blinds are pulled down. And secondly if gravity has a speed and see other threads for that question, then it will take forever for the gravity "wave" to create the event horizon and another avoid the singularity.

As to singularities, there is a maths theory that says that if an infinity exists then the universe will be detroyed. (There is fine print in this, in that an infinite time to destroy is alllowed ) The argument goes :- We are here so there are no infinities in nature.

This theory is so strong that many physicists will look at a singularity/infinity as a sign that we have missed something.

For example of some recent finds.

As we go smaller the small the world fuzzes out (uncertainty principle) so no singularity.

For increasing speed mass/ and time gets fiddled with so no infinite speed.

The universe is a closed curve equals no edges equals no singularity.

 

Since the velocity of light is finite, no infinity or singularity is required to have an escape velocity greater than the velocity of light. Even classical physics calculations indicate that a black hole is a possibility. The Newtonian formula for escape velocity is a damn good approximation to the Relativity formula, and both predict that a large enough mass will result in an escape velocity greater than the velocity of light. The classical equations do not cope properly with other aspects of a black hole, but are okay for escape velocity.

The time contraction effects at the event horizon are deceptive. From one point of view, an object never gets past the event horizon in the time frame of a distant observer due to time contraction effects. Yet, a distant observer will see a black hole growing rapidly due to the matter falling into it. The following is overlooked.

Code:

Consider a black hole with event horizon radius [B]CurrentRadius.[/B]

Assume that the addition of [B]DeltaMass[/B] would result in an event horizon radius [B]NewRadius[/B]

If [B]DeltaMass[/B] is approaching the black hole and gets inside of [B]NewRadius,[/B]
the event horizon will expand to [B]NewRadius[/B] 
without any mass necessarily getting as far 
as the original event horizon.

Another effect of a black hole is the red shifting of light due to the intense gravity. Any object falling toward a black hole appears to get dimmer and dimmer due to the so called red shift. Note that red and blue shifts get their names due to the effect of gravity and/or motion on the appearance of visible light. The blue shift is actually a shift toward higher frequencies (wave theory) or higher photon energies (particle theory), while red shift refers to a shift toward lower frequencies/energies. As an object approaches the event horizon, the light is shifted way past the red part of the visible spectrum to extremely low frequencies/energies, making an object appear dimmer and dimmer. At some point before reaching the event horizon, it becomes too dim to be seen.

Oddly enough, there is no guarantee that the matter inside a black hole be at the center. The General Relativity equations deal with what happens if the matter is all in the center, and this is certainly the expected situation. However, extreme rotation, thermonuclear reactions, and/or other effects maintain some other configuration for a finite amount of time. Remember that there is no gravity inside a hollow spherical shell. Hence, there could be regions with little or no gravity inside a black hole

About 10-15 years ago, there was a fascinating article about the distant from a black hole at which the orbital velocity equals the speed of light. In a Torus shaped space station surrounding the black hole at that distance, an astronaut could see the back of his head. The torus would appear to be a cylinder to those inside it. In a torus shaped space station surrounding the black hole at a lessor distance, the torus would seem to curve away from the black hole, while to an observer farther out it would appear to curve toward the black hole. The article also mentioned some strange effects. For example: Inside a rotating torus surrounding the black hole closer than the light orbit distance, the rotational effects would cause you to move toward the black hole instead of away from it.

BTW: The article never suggested that such space stations could be built in the vicinity of a Black hole. I suppose they are theoretically possible around a planet or a star, although the engineering problems seem formidable, if not overwhelming.

Has anybody here read Ring World by Larry Niven?

 

Thank you all for sharing your interesting knowledge. I’ve given a lot of thought to everything posted.

It seems there are 2 conflicting camps here: Camp 1 says a black hole can never fully form, or that the formation takes forever. Camp 2 says that the black hole is fully formed when the escape velocity reaches c, the speed of light, which need not take forever.

I wish to know which camp is right, but the answer has to appeal to my intuition regardless of what theory says. Camp 1 does that for me now, but I’m flexible!

Regarding Camp 1, I’ve read that a black hole forms when the curvature of spacetime becomes infinite. I understand that intuitively. That the curvature can never truly become infinite, or that that takes forever, also makes sense to me. And if the curvature is finite, light has the possibility of escaping however slim the odds may be.

Regarding Camp 2, I will show that this camp has a subtle logic flaw. Let’s examine the concept of escape velocity.

If I throw a ball straight up, unless I throw it at more than 11.2 kilometers per second (earth’s escape velocity), it will eventually decelerate to a velocity of zero and fall back to earth. The gravity will relentlessly drag on the velocity, decreasing it with each passing moment.

Now, suppose I shine a flashlight straight up. If I measure the upward velocity of the light, it will always be c; gravity has no effect on the upward velocity. The photons that make up the light don’t decelerate (unlike the ball) so they’ll never fall back. They will escape.

Yet we know that the path of light, when moving on other than the axis of gravity, is deflected by gravity. If I shine the flashlight parallel to the ground, the light will arc toward the ground during its escape to space. As I move the flashlight toward the ground (keeping it parallel to the ground), at some particular height, still above zero, the light will arc to the ground rather than escape. Alternatively, if I keep the flashlight at its original height and the gravity is instead increased, the light will hit the ground at some particular gravity.

If the gravity increases further and I want to keep the flashlight at the same height and allow the light to escape, I’ll have to point it more upward than parallel. The light will still arc toward the ground, so as the gravity continues to increase the flashlight must be pointed more and more upward. Eventually I’ll have to point the flashlight at an 89-degree angle to the ground, and then 89.9 degrees, and then 89.99 degrees.

Is there a particularly strong gravity where a perfect 90-degree angle will contain the light? No, because as we’ve seen, gravity has no effect on the velocity of light that is moving solely along the axis of gravity, in this case straight up. Light doesn’t decelerate. Nor will the path of the light be deflected. If you doubt this, ask yourself, which way would the light be deflected?

Imagine a gravity well, as it is often illustrated in relativity books, like a funnel with a collapsing star at the small end. As the star collapses and gravity strengthens, the sides of the imaginary well steepen toward vertical and the funnel tube lengthens. For the light to escape, the flashlight must be pointed at the opening of the funnel. Otherwise, the light will eventually fall to the surface or at best orbit the star. But the angle of the well’s wall never reaches vertical, which is an infinite steepness, so the light from a flashlight pointed straight up, where an opening always exists, will escape.

Now I will show, both intuitively and mathematically using a simple formula, that an escape velocity = c cannot contain even a material object.

Suppose I wanted to travel to the Andromeda galaxy, some 3 million light years away, in one year of my life. I could do that if I had a very capable ship that could move at close to c, compliments of time distortion, as students of relativity know full well. But no observer may observe himself moving at c or greater, and here I am covering 3 million light years in 1 year, which is 3 million c. Something is wrong; if I will reach the galaxy in one year at less than c, the galaxy must be less than 1 light year away. And so it is, compliments of spatial distortion.

Here is a formula that tells me what my velocity must be to cover a specific number of light years in one year of my (proper) time:

LightVelocityPercent = SquareRoot((ApparentLightVelocityPercent ^ 2) / ((ApparentLightVelocityPercent ^ 2) + 1))

where LightVelocityPercent is the percentage of c you must travel, ApparentLightVelocityPercent is the multiple of c that you want to seemingly travel, and “^ 2” means “squared.”

For example, suppose I wanted to travel 10 light years in 1 proper year. That’s an apparent velocity of 10c. Plugging in ApparentLightVelocityPercent = 10 yields .995c. Moving at that velocity, distances along my axis of travel contract to 9.95% of their original distance (calculated using the special relativity formula for spatial distortion), reducing the distance to .995 light years, which I will cover in exactly 1 year at .995c.

Play with the formula to see that any distance may be traveled in any proper time at a velocity less than c. A gazillion light years in a nanosecond? No problem! If I can travel a gazillion light years in a nanosecond, can you begin to see intuitively that no amount of gravity may contain me? Can you see that any amount of gravity equates to an escape velocity less than c?

The ApparentLightVelocityPercent of light itself is infinity, not c, to represent the fact that light covers an infinite distance in no proper time. Plug infinity into the formula and you get LightVelocityPercent = c.

The logic flaw I mentioned above regarding escape velocity is this: in “escape velocity = c,” c is the ApparentLightVelocityPercent, not the advertised LightVelocityPercent. Plug 1 (that is, 1c) into the formula and you get .707c. That’s the minimum velocity needed to escape a “black” hole having an escape velocity = c. A true black hole would have an escape velocity = infinity.

I’ve shot my mouth off here, but I repeat that I’m flexible. If it’s my logic that’s flawed, please, I wanna know.

 

Zanket: Your analysis is interesting, and might have some (but not much) theoretical merit. Camp 1, which does not believe that Black Holes form, has little or no basis for their point of view.

From a practical point of view, astronomers have good observational evidence for the existence of Black Holes. General Relativity (not Special Relativity) provides a good theoretical basis for the existence of Black Holes. Even classical physics supports the notion of a Black Hole. Without a good alternative explanation for the observational evidence, it is hard to argue against the existence of Black Holes, especially when they are supported by General Relativity, which has a lot of evidence indicating that it is a valid theory.

As mentioned in a previous post, no singularities or infinite curvature effects are required to deal with what happens to light rays near the event horizon of a Black Hole. The singularities (if they exist) are beyond the event horizon.,

I do not know enough about the mathematics of General Relativity to deal with the light ray directed parallel to the gravitational force vector of a Black Hole. Even if I did, I doubt that anyone here would understand anyway. It could be that light ray speed is always c relative to some frame of reference, but I have no understanding of the time/distance contraction effects and reduced photon energy effects due to the red shift. Furthermore, General and Special Relativity treat light propagation and speed a bit differently. Maybe c is not an absolute constant and an absolute speed limit in General Relativity. In fact, I do not think it is an absolute speed limit in General Relativity.

Ignoring any possibilities hinted at by the above paragraph, there is a basic problem with your analysis of the parallel light ray. The problem is analogous to the mathematics and physics of pendulums. In classical theory, a pendulum could be balanced in a vertical position. Also in classical theory, a precise force acting on a pendulum could result in its swinging to a vertical position and staying there. Even classical thinkers did not expect either of these situations to occur in practice. Modern quantum theory refutes even the theoretical possibility of the balanced pendulum.

Remember that gravity has an effect on light rays, which was predicted by General Relativity and first verified in 1919 during a solar eclipse. In practice, there is no way that a light ray could remain exactly parallel to the gravitational force vector of a Black Hole. The least deviation of the light ray from the parallel direction or an almost infinitesimal deviation of the direction of the force vector would result in gravity curving the light ray away from the parallel path. Once some quantum effect caused it to curve ever so slightly, the deviation would get worse and worse, causing the light ray to curve back into the black hole. The parallel light ray argument is not valid.

As for denying that the Black Hole can form, consider the analysis from my previous post. Black Holes do not get bigger due to matter falling through the event horizon. They get bigger due to the event horizon getting larger, which a subtly different process. The mathematics of the time contraction effects suggest that nothing is ever observed to fall through an existing event horizon. Id est: In the refernce frame of a distant observer, it takes an infinite amount of time for a particle to reach the event horizon. This is likely to be a valid analysis, but it does not preclude an expanding event horizon engulfing nearby matter.

The article I mentioned in my previous post described the possibility of a light ray in a circular orbit around a Black Hole. Once again, this is a theoretical possibility never to be achieved in practice. I do not think that an elliptical orbit is a practical possibility for a light ray. Planets, binary star systems, satellites, asteroids, et cetera have remarkably stable orbits compared to a light ray in the vicinity of a Black Hole because such orbits are not critically sensitive to extremely small disturbing effects. If the Earth loses a bit of orbital speed or encounters a slightly stronger force due to irregularities in the Sun, it is accelerated toward the sun. This causes an increase in its speed, which tends to counter the disturbing effect and restore the original orbit. However, orbiting light rays and rays parallel to the gravitational force vector are situations for which there are no correcting forces to counter small destabilizing effects.

 

Dinosaur:

Great input!

I don’t think anything in my logic contradicts the observational evidence for black holes. From everything I’ve read, all we know is that there are dense dark objects out there. Even without an event horizon, massive objects would appear black due to the gravity redshifting the light waves to be beyond detectable. The jury is still out regarding the existence of an event horizon and hence a true black hole.

The strong point you make is that, how could I be right, given that general relativity is otherwise well confirmed? General relativity and I could both be right. The general relativity equations Schwarzchild used for calculating the event horizon radius could be right, but his application of those equations could be subtly wrong in the way I described.

In general relativity: The same formula is used to determine the deflection of a light ray as for any material object. Light can orbit just like anything else. Whereas light may be redshifted or blueshifted (which is superfluous to the existence of an event horizon), its velocity remains c.

Regarding the physics of pendulums you mention, are you saying that although light going straight up could escape, it probably wouldn’t because the odds of it remaining straight up are slim? Well, if there is a possibility of a photon escaping, that destroys the event horizon as the theory defines it, which is that light cannot escape. If the event horizon were statistical in nature, it would be defined in terms of odds not velocity.

My parallel light ray argument said the same thing you are saying. I said “If I shine the flashlight parallel to the ground, the light will arc toward the ground…” I agree with you.

Regarding your analysis of the event horizon, I agree with you if the event horizon exists. But I’m disputing its existence to begin with.

I think I gave a strong argument that when the escape velocity = c, light can easily escape, because the true escape velocity when properly converted is .707c. I showed that a true black hole needs an escape velocity = infinity, which cannot exist. I’d like someone to counter me on the logic or math I applied here.

 

Zanket: As suggested in my previous post, I do not know enough about Special Relativity mathematics to deal with Light rays in the vicinity of the event horizon of a Black Hole.

To the best of my knowledge, you must refute some basic General Relativity concepts to justify your denial of the the existence of Black Holes and/or to justify your claim that light can escape from a Black Hole.

By definition, the event horizon is the surface at which the escape velocity is exactly the velocity of light in the frame of reference of the event horizon itself. A ray of light just outside the event horizon can theoretically escape if directed parallel to the gravitational force vector. A ray of light at or beyond the event horizon cannot escape. At least this is what is claimed by General Relativity.

You are using a Special Relativity formula to calculate .707c as the true escape velocity at the event horizon. Special Relativity is only applicable in the absence of accelerated motion and in the absence of any gravitational effects.

BTW: Acceleration effects and gravitational effects are the same animal, which is one of the fundmental concepts of General Relativity.

As far as I know, Special Relativity formulae are not considered valid in the presence of gravitational fields. The stronger the field, the less accurate are the Special Relativity formulae.

General Relativity definitely indicates the existence of Black Holes. Therefore you are misapplying some Special Relativity formula, or the formula is not applicable at the event horizon, or something else is wrong with your analysis. I am sorry that I do not know enough to come up with the General Relativity equations supporting the existence of Black Holes and the inability of light rays to escape.

BTW: All of the formulae relating to time, distance, and velocity adjustments involve transforming equations valid in one inertial frame of reference to the corresponding equations valid in another frame of reference. Perhaps you are overlooking something related to reference frames in your application of the Special Relativity formula to escapre velocity. Also, I have never seen such formulae applied to light rays, although I see not reason why they should not be applied to light rays.

It is my understanding that at the event horizon, a photon has zero energy and no mass (particle view of light). Under other circumstances, the photon has finite energy and finite mass. From the wave view, light has has zero frequency coupled with an infinite wave length at the event horizon. I have no idea how all this applies to the velocity of light at the event horizon. It does not seem to me that your special relativity formula applies to these conditions.

When I described the instability of a light ray directed parallel to the force vector, I was not referring to some quantum probability phenomena. I was claiming that in practice, the perfect alignment cannot be maintained due to quantum fluctuations in the direction of the light ray and/or the direction of the force vector. This analysis suggests to me that in practice, a light ray cannot escape from near the event horizon, even though it is theoretically possible for it to escape.

BTW: Are you familiar with Hawking radiation? According to Hawking, quantum tunneling or some other quantum effect can cause a Black Hole to evaporate if not supplied with mass from an external source. For Stellar sized Black Holes, the evaporation time is many orders of magnitude longer than the life of the universe. For tiny Black Holes it happens fast enough to have disasterous effects in the immediate vicinity of the mini Black Hole.

BTW: I have read essays in which it was claimed that General Relativity allows for objects to move away from each other at speeds exceeding c, but have never read about the possibility of objects approaching each other faster than c.

 

Zanket,

Your "escape velocity" argument has a few basic flaws.

The first is gravitational redshift. If light tried to escape from a blackhole starting at a position infinitely close to the event horizon, by the time it does distance itself from the black hole to any appreciable degree it will be redshifted to a wavelength that is practically infinite (and, physically undetectable.)

The second can be best illustrated (IMHO

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) with a rather twisted thought experiment involving trains (what else did you expect?

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Relativity=trains, me=twisted

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) Imagine a train car at rest, with the track speeding along past it at relative speed v &gt; c. That's right, I do mean &gt;. Now, housed in the train car is a flashlight pointed in the direction opposite to the train's motion. Pulse the flashlight and observe the pulse's propagation away from the coordinate on the track at which it was flashed. Even though the distance between the lightfront and the train car will grow without bound, the lightfront will progressively fall behind the point along the track where it was originally emitted. In a similar fashion, space itself is perpetually "stretching" inside the event horizon, "flowing" toward the center, at a rate exceeding that of the light propagating through it. (this is the "sink" interpretation I've mentioned earlier.)

Mathematically (but no formulas) once you pass the event horizon of a black hole, concepts get blurred and the math becomes somewhat hard to interpret. Time starts behaving like space -- so with passing time the distance between you and the event horizon keeps growing even apart from any time-dependent movement through space. At the same time, space starts behaving like time, so moving in certain directions through space (i.e. toward event horizon) becomes equivalent to traveling back in time.

Of course, GR (just like SR) uses light to define space coordinates. So space is not something tangible but rather an abstract set of measurements that results as light propagates back and forth. So in reality the "sink" analogy is most probably deeply flawed, as no tangible thing called "space" is actually rushing into the black hole. However, mathematically gravity makes light behave exactly as if this were the case (and not just for black holes, but for any gravitational well.)

 

Dinosaur: About me using the special relativity formula, you may be right. Looks like I have to learn tensor calculus. As you say, gravity is equivalent to acceleration in GR, and acceleration is nothing but changing velocity. But alas, I may not know what I’m talking about. Interesting stuff about Hawking radiation.

Overdoze: I’m not sure what you intended about the redshift. I agree about the redshift making the light undetectable if it escapes. The hole may appear black, but I want to know if it is black.

About the train... I realize your twisted example is analogous to what GR says is happening, and it’s an excellent example. I just don’t buy it. I gave the reason before, but now I’ll apply it to the train. And the same logic refutes that an event horizon can exist.

Let’s pretend the train can indeed move at > c relative to the track, which is analogous to “stretching” space in GR (where “stretching” itself is an analogy). Rather than be a mere observer of the pulse, put yourself on the photon. You are riding the photon at the speed of light in the direction opposite to the train’s motion. Directly in front of you, a billion light years away, is a galaxy. Keeping in mind that the space between the galaxy and the train you just left is stretching at > c, how long will it take you to reach the galaxy? GR says you will never reach it, that the escape velocity is > c. I say you will reach it in no time at all. And the fact that you can reach it at all means that the escape velocity is < c.

Please forget for the time being what GR says about how space stretching affects light. Look solely at what I’m saying. If space were not stretching, then regardless of the train’s speed, you, riding the photon, would reach the galaxy in no time flat. That’s because light does not experience time. It travels only through space. I’m not making that up, SR says that. If you plug c into the spatial distortion formula, you see that, for light, space is contracted to zero. Light experiences no distance between locations, so it always gets to its destination in no time.

Now let’s add just a wee bit of space stretching back into the example. Say space is stretching at a rate of one meter per year between you and the galaxy. How much longer will it take you to get to the galaxy now? One second longer than nothing? One hour longer than nothing? Putting “nothing” into any formula returns “nothing.” Light cannot be made to experience time by stretching space. You will get to the galaxy in no time, regardless of the space stretching.

(Lest you think light actually takes zero time rather than no time, remember that light could travel a trillion times further in the same “time.” A thing that can move at the same “speed” to various distances in the same “time” is a no-time thing, not a zero-time thing.)

But won’t an outside observer see the light take longer due to the stretching? Sure. Then if space is stretching at > c, the light will never reach its destination (escape velocity > c), right? No, because that is incompatible with the conclusion that light always gets to its destination in no time. It cannot be that space can stretch >= c. If GR says it can, there must be a problem with it.

 

zanket

Much of what you've posted regarding reference frames is for the most part fairly accurate, however you are not relating the reference frames to anything.

Firstly, a photon traveling across the universe may not experience time in the traditional sense, but to a photon, the concept of time is meaningless. If you mean that a photon does not experience time, that would be in the reference frame of the photon only. A reference frame from an observer which may be considered a stationary frame could measure the time it takes for a photon to traverse space. The stationary observer will always measure the speed of light at c, but will measure the photons time and distance to destination relative to his own velocity. The photon will not experience time because time is a property of spacetime which in turn is the barrier to which the photons velocity is representative.

Secondly, in the reference frame of a stationary observer a safe distance away from a black hole will view certain aspects of the black hole relative to his frame. You submit the black hole does not form. To this stationary observer, that could be true relative to his frame. However, from the reference frame of the particles or matter caught up in the gravitational field of the black hole, these particles will pass through the event horizon and will impact the singularity in a finite amount of time. In fact, once the progentor star collapses, the matter is compressed to its end state in a finite amount of time. The black hole does form, however the information, ie. light, propagated from this event will not reach the outside stationary observer in the same amount of time. Time is dilated at the bottom of the gravity well at the event horizon and beyond.

Thirdly, in the space severely warped beyond the event horizon, due to the intense gravitational field, are the geodesics, or paths that the light travels. Similar to a roller coaster, the light is the car and the track is the geodesic. It is the severe gravitational field which warps the paths/geodesics (roller coaster track) back towards the center of the hole (singularity). The light (roller coaster car) will always travel in a straight line and will always travel along the path/geodesic. Therefore, no matter which direction you shine a light, it will always follow the paths/geodesics back towards the center of the hole.

 

Any light emitted by a black hole represents escaping energy. Light of infinite wavelength represents 0 energy, so "emitting" such light is the same thing as emitting nothing at all. And the point is that while light trying to escape from precisely the event horizon ends up not escaping at all means that light cannot escape the black hole from inside the event horizon.

The "conclusion" is incorrect. Sometimes, light does not get to a "destination" at all, ever. Like for example when it's flying away from the "destination". Stretching space in such a way that the distance between a photon and its destination grows faster than c has the effect of the photon traveling away from the destination (just calculate the net speed as seen by a "stationary" observer.) Not to mention that as you stretch space the photon itself gets stretched (this is used to explain cosmological redshift), so if this process goes on forever then the photon will become infinitely stretched and would disappear into the noise of the quantum background.

 

(Q): Thanks for sharing. I thought I was being clear about the reference frame by saying “...put yourself on the photon...you will reach [the galaxy] in no time at all.” When I say a photon does not experience time, that, to me, is the same thing as saying that, to a photon, the concept of time is meaningless. I prefer your terminology though.

When you say “You submit the black hole does not form ... However, from the reference frame of the particles or matter caught up in the gravitational field of the black hole, these particles will pass through the event horizon ...” I lose you there, because, as I understand it, a black hole is defined by the formation of an event horizon. If I’m submitting that an event horizon (hence a black hole) does not form, I can’t grasp a counter explanation that assumes the event horizon exists without first explaining how it must form, ideally by refuting the argument I’ve put forth. If indeed an event horizon forms, then I agree with all that you said about it.

Here’s how I’d summarize and regurgitate my argument, now that I’ve gotten some input from y’all (less your last post Overdoze, it's past my bedtime!):

This shows that an event horizon (hence a black hole) cannot form: A photon that leaves a body along the axis of the body’s gravity can neither be deflected nor decelerated by the body’s gravity (even in GR). That photon has no geodesic path back to the body, so it is free to escape. Escape = no event horizon.

I’ve been given some good arguments as to why the event horizon must form, but nothing yet that refutes the above, or perhaps it did refute it but I didn’t grasp it (a distinct possibility!). The train example is hardest for me to counter. I’d summarize my argument against it like this: If I can travel a gazillion light years in a nanosecond by my clock whilst moving at a real velocity < c (where a "real velocity" ranges from 0 to c, my terminology), which both SR and GR allow, meaning I can move at an apparent velocity approaching infinity, then no puny “black” hole with a finite gravity--a finite acceleration--is going to hold me back; I can always out-accelerate the body to escape. And light, which always travels at a flat-out infinite apparent velocity, would have no problem escaping. Given the intuitiveness of that, an escape velocity = c is most certainly an apparent velocity falsely advertised as a real velocity. Converted to a real velocity, it becomes < c, as would any apparent escape velocity < infinity. Applied to the train example, this means that space cannot “stretch” at a rate >= c, so no event horizon forms.

In a future post I hope to show the exact problem with Schwarzchild’s solution that defines the event horizon. I’ve got an idea of where the problem is. If I can do it at all, it won’t require tensor calculus to explain it.

 

I can’t grasp a counter explanation that assumes the event horizon exists without first explaining how it must form, ideally by refuting the argument I’ve put forth. If indeed an event horizon forms, then I agree with all that you said about it.

The event horizon forms as a result of a collapsing progenitor star. Every particle of matter has a gravitational field. The more matter in one place, the more gravity. Large objects will produce gravitational fields that will necessitate the need for high escape velocities, the velocities required to escape the gravitational field. As the mass of the object increases, so does the gravitational field, hence the need for higher escape velocities.

As well, if the mass of an object is squeezed smaller and smaller, as is the case with a collapsing star forming a black hole, the escape velocity increases. In other words, the escape velocity on the surface of the star before it collapses is less than the escape velocity on the surface of the star after it collapses, even though the mass itself did not increase. Therefore, light will escape the gravitational field of a star, yet will not escape the gravitational field of the resulting black hole formed once the star has collapsed. For example, if the mass is squeezed four times smaller, the escape velocity would need to be twice as large. (Laplace formula)

This shows that an event horizon (hence a black hole) cannot form: A photon that leaves a body along the axis of the body’s gravity can neither be deflected nor decelerated by the body’s gravity (even in GR). That photon has no geodesic path back to the body, so it is free to escape. Escape = no event horizon.

Mass and energy are interchangeable according to Einstein. Light is simply energy with no rest mass. Therefore, a gravitational field will affect light in the same way it affects particles of mass. There is no difference between a group of particles thrown in the direction of the axis of gravity and a beam of light shining along the axis of gravity. Both will succumb to the gravitational field and will fall back towards the center of the black hole.

In fact, escape velocity is best achieved if the escaping object begins to orbit the large object rather then trying to escape along the axis of gravity. This is how rockets are launched from Earth.

 

Some more thoughts occurred to me about the formation of an event Horizon.

BTW: Previous posts have mentioned that infinite curvature and a singularity are not required for the formation of a Black Hole with an Event Horizon The singularity (if any) is beyond the Event Horizon and cannot be observed. There are many experts who claim that the mathematics indicating a singularity provide good reason to assume that General Relativity is invalid at or near the center of a Black Hole. This is analogous to Classical Theory breaking down under less extreme conditions.

Note that a Black Hole could form without matter falling through the Event Horizon of an existing but smaller Black Hole. Consider CriticalMassX, to be the amount of mass required to form a Black Hole with an Event Horizon of RadiusX. The mathematics relating these quantities is very straight forward using Classical or General Relativity formulae. If CriticalMassX somehow finds its way inside a spherical volume of RadiusX, the Black Hole with RadiusX Event Horizon forms.

Now how can this happen? The following might not be how the experts describe the evolution of a Black Hole, but it seems reasonable to me.

Consider a massive star in a binary system. Massive stars use up their supply of fuel for the nuclear reactions which fight gravitational collapse. Before collapsing and during the collapse, the massive star pulls matter from the outer parts of its binary companion and from elsewhere in the solar system containing the binary pair. If the system is near the center of a galaxy, there are clouds of matter, stars, and other sources of matter which can be sucked in. Radiation is also received from the companion and from other sources. For most of its life cycle radiation leaves the surface of the collapsing star and escapes from the local solar system. .

It seems possible for the star to evolve to a situation for which the escape velocity is nearly the velocity of light within a spherical region surrounding the collapsing star. What happens to matter in the solar system containing such a star?

• It seems reasonable to believe that almost all matter within the critical region stays in this region and falls towards the surface of the collapsing star.
  
• Most of the matter from outside with velocity vectors directed toward the critical region will enter and stay in this region. Only a small percentage traveling at some fraction of light speed will enter this region and escape due to the direction of the velocity vector.
  
• A lot, but not all, of the matter from outside with velocity vectors initially directed close to but not directly toward this region will also be pulled in and never escape.

From the above, it seems reasonable to assume that the massive star gets more massive and the escape velocity increases.

What about radiation?

• If the escape velocity within a region is nearly the velocity of light, some radiation originating inside that region is not expected to escape. The extreme gravity will pull it back to the surface of the collapsing star and most will be absorbed, although some will be reflected or will produce secondary radiation which escapes. Any radiation escaping will be drastically red shifted and appear to be extremely dim to a distant observer.
  
• Some radiation from outside the critical region that comes close to or enters the region will be pulled to the surface of the star and fail to escape.
  
• Some radiation from outside will escape, but will be red shifted due to the extreme gravity and appear dim to a distant observer.
  
• A lot of radiation from outside will not come close enough to be dimmed significantly by red shifting. To a distant observer, only a tiny percentage of this escaping radiation will appear to be coming from the collapsing star. Almost all will escape in a direction away from the observer.

One might call the situation described above a dim gray hole.

From the above, it seems reasonable to assume that a massive star or other collection of a huge amount of matter will evolve to a dim object. While nova & super nova eject huge amounts of matter, if what remains is massive enough, the above circumstances seem likely. As more matter is sucked in and the escape velocity increases, the star will become dimmer. There seems to be no process that makes the star brighter or less massive. At some point, it becomes so dim that it becomes a Black Hole.

BTW: As mentioned in a previous post, a distant observer can see an Event Horizon growing, and it can grow rapidly if there is a nearby source of lots of matter. From the reference frame of a distant observer, it takes an infinite amount of time for an object to reach and pass through an Event Horizon. This paradox occurs because the Event Horizon expands to engulf nearby objects. Suppose a Black hole with RadiusX Event Horizon requires DeltaMass added to form a Black Hole with RadiusNewX. If DeltaMass gets inside RadiusNewX, the Event Horizon can expand to RadiusNewX without any matter passing through the original RadiusX Event Horizon.

 

Overdoze: Regarding the emitting of light of infinite wavelength being the same thing as emitting nothing at all, I agree and that’s a good point. I think the wavelength of light emitted from a “black” hole is of finite wavelength. More on that in a future post.

Regarding how, sometimes, light does not get to a “destination” at all, ever. Like for example when it's flying away from the “destination.” Of course I agree. By “destination,” I mean more specifically a point along the geodesic path of the light in the direction of the light’s motion.

Regarding space stretching: well said. I’ve given this concept a lot of thought. I intend to remove the mystery behind this in a future post to better show that space does not really stretch, it only apparently does.

(Q): I agree with your analysis regarding “As the mass of the object increases, so does the gravitational field, hence the need for higher escape velocities.” I think that the escape velocity never really reaches c, it only apparently does.

When you say “There is no difference between a group of particles thrown in the direction of the axis of gravity and a beam of light shining along the axis of gravity. Both will succumb to the gravitational field and will fall back towards the center of the black hole,” I disagree, and it’s a key point. The velocity of light is always c in its direction of motion; if the motion is along the axis of gravity, then gravity cannot decelerate it. Neither can gravity accelerate the light, if, for example, the photon is falling into the hole along the axis of gravity. The path of light may be deflected by gravity in the same way as a material object, but only along an axis other than the axis of its motion.

Another way to visualize this: Light from a galaxy that falls directly toward the earth always arrives exactly at c, but a so-falling baseball is accelerating. If the light were moving parallel to the ground (or on any other path not directly on the axis of gravity), then its path would be deflected to arc toward the ground.

The argument that still allows the light to fall back is the space stretching, e.g. the train example, in which the light pulse is moving away from the train (black hole) at c, but the track (space) is moving > c.

Dinosaur: You’d make a good author!

Regarding your earlier comment, “Special Relativity is only applicable in the absence of accelerated motion and in the absence of any gravitational effects,” I’ve given this some more thought. On the one hand, you are absolutely correct. On the other hand, gravity likens to acceleration, which is just a changing velocity. At each velocity, you can use SR. What SR will not easily give you, when applied to acceleration or gravity, is the right answer. That requires calculus or its equivalent to correctly apply SR. In another post I’ll present a formula I found in a book, SR-like in its tidiness, that gives the right answer for spacetime distortion for acceleration. Such a formula is applicable to our discussion here on the event horizon.

You can still use SR to draw correct logical conclusions about thought experiments involving acceleration when you don’t care about the exact numerical answer. For example, I can correctly state, using only SR, “If I accelerate from a point and return, my clock will have elapsed less time than a clock at that point.” How much less requires more than SR.

 

If you were to accept this assumption, then you must conclude that light does not get redshifted by gravity at all.

As it stands, the formula for gravitational redshift of light is (with light escaping to infinity):

f_inf/f_o = (1 - 2GM/Rc²)^1/2

where f_inf is frequency at infinity, f_o is frequency at the start of the trajectory, G is the gravitational constant, M is the (point) mass generating the gravitational well, R is the radius from center of the well where the trajectory begins and c is the speed of light.

As you can see, there exists a certain radius (called the Schwarzschild radius), namely R_s=2GM/c², at which f_inf is 0. Taking a radius smaller than Schwarzschild radius would give you an imaginary frequency, which doesn't make sense.

By the way, gravitational redshift of light has actually been confirmed experimentally and found to agree with theory to high accuracy, and is one of the few verifications of GR in existence.

Geodesics are curves of shortest distance connecting two distinct points, and light travels along geodesics. IOW, if you want to shoot light from point A to point B, you direct your beam tangentially to the geodesic between A and B (in flat spacetime, geodesics are simply straight lines.) Now, mathematically a geodesic is bidirectional, so that it is the same curve regardless of whether you go from A to B or from B to A. But physically this property doesn't hold. For example, you may be able to go from coordinates (x₁, y₁, z₁, t₁) to (x₂, y₂, z₂, t₂). However, if you wanted to go in the opposite direction you'd have to be moving back in time which is not allowed for obvious reasons.

When it comes to black holes, there is a certain "boundary" in spacetime (the event horizon, defined by the Schwarzschild radius), at which geodesics of spacetime as defined by light suddenly loose their bidirectionality. Going into the black hole light describes one geodesic, but trying to come out of the black hole it describes a different geodesic. It's as if light, upon entering the event horizon, passes through the looking glass into a different universe ... and there is no way back. In fact, if you define point A outside the black hole and point B inside the black hole, then there is a geodesic from A to B but there is no geodesic from B to A.

Intuitively, you can treat the entire volume of space inside the black hole as a one-way mirror. Light can travel in one direction (into the black hole), but it's instantly reflected back if it attempts to travel in the opposite direction (and actually, the closer it gets to the singularity, the harder it's "kicked" toward it, so that its frequency becomes ever greater.)

As to how this geodesic assymmetry can be represented in a static topological sculpture, don't ask. I have no idea, and I don't even know if it can be represented topologically. IMHO, the whole differential geometry approach of GR is fundamentally flawed or at least vastly misinterpreted, and this is just one of the cases where the flaws are glaring.