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.

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. :D

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! :D)

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?

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)

kewl explanation stryder :)

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 :D

Originally posted by zanket
I can understand how a finite value can approach infinity, but how can a finite value become infinite?

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.Consider a black hole with event horizon radius CurrentRadius.

Assume that the addition of DeltaMass would result in an event horizon radius NewRadius

If DeltaMass is approaching the black hole and gets inside of NewRadius,
the event horizon will expand to NewRadius
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 :)) with a rather twisted thought experiment involving trains (what else did you expect? :D Relativity=trains, me=twisted :p) 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.

Originally posted by zanket
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.


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.


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.


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.

Originally posted by zanket
Overdoze: I think the wavelength of light emitted from a “black” hole is of finite wavelength. More on that in a future post.


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.


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.


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.

zanket

I think that the escape velocity never really reaches c, it only apparently does.

What do you mean by "apparently?" Are you implying there is an illusion taking place ? The escape velocity of light is c because light travels at c. If the gravitational field of an object is strong enough, it will overtake the escape velocity of any object, including the escape velocity of light.

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

That makes no sense. In fact, the greatest effect of gravity on any object is when the object is attempting to escape a gravitational field along the axis of gravity. Again, I repeat, that is why rockets are launched to begin orbiting the Earth as soon as possible.

Iyour statement were true, there would be no redshifting or blueshifting of light. Gravity has an effect on all mass, and more specifically, energy, which is what light is comprised.

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

No one said anything about accelerating or decelerating of light except you. That is clearly not the case. The light is blueshifted as it falls towards an object, in other words, it picks up energy as it falls. In fact, the greatest effect of this will be if the light falls directly along the axis of gravity.

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 light will fall towards Earth at c no matter what trajectory it follows. The only difference is how much energy is picks up while falling and what geodesic it follows towards the ground. The baseball will accelerate until it reaches a velocity equivalent to the Earths gravitational influence. That same baseball will accelerate to a velocity equivalent to the gravitational field of any other planet or object which may be different then that of Earth's gravitational field.

Q:

What do you mean by "apparently?" Are you implying there is an illusion taking place ?

Yes, of sorts. I gave an example and a conversion formula earlier. If I am traveling at constant velocity from the earth to the Andromeda galaxy and will arrive in 1 proper year, then my apparent velocity is 3 million c. But my real velocity is less than c after taking into account spatial distortion that reduced the distance to less than 1 proper light year. I submit that "escape velocity = c" has not taken into account spatial distortion as it should, and when you do, escape velocity is < c.

In my thinking, no real velocity > c exists. Whenever I see that, I know it is an apparent velocity that requires conversion to a scale between 0 and < c.

In fact, the greatest effect of gravity on any object is when the object is attempting to escape a gravitational field along the axis of gravity.

That may be true, but the effect on light is different than the effect on a material object in this case, and the difference makes my point. I used acceleration as an example of that. If you throw up a beam of light, it rises at c and doesn't slow. If you throw up a ball, it decelerates. The light redshifts, the ball doesn't.

Because the light cannot slow and cannot be deflected, it must be able to escape (however redshifted it may be, the redshift being finite), so there is no event horizon. If ever the light could fall back along the axis of gravity, there would come a moment at which the body would measure the velocity of the light at < c.

That light is not slowed regardless of the escape velocity, and that I can travel a gazillion light years in a nanosecond while moving at < c, shows that light can be thought to move at an infinite velocity, which I call an apparent velocity. When converted using the formula I posted earlier, you get a real velocity = c and escape velocity < c.

Overdoze:

In the formula R_s=2GM/c², I submit that c should be an apparent velocity = infinity. If so, the formula always yields 0, the center of the body. If so, no finite mass yields a real escape velocity = c. If so, the light will still redshift (GR is still confirmed) as it escapes, but the redshift will be finite at any destination in the universe that the light reaches.

zanket

That may be true, but the effect on light is different than the effect on a material object in this case, and the difference makes my point. I used acceleration as an example of that. If you throw up a beam of light, it rises at c and doesn't slow. If you throw up a ball, it decelerates. The light redshifts, the ball doesn't.

No, no difference. The energy of the photon is diminishing due to the force of gravity. The photon is comprised entirely of energy, therefore the photon should disappear. However, because of the warping of spacetime, the geodesics of the photons are warped back towards the center of the mass. Regardless of their initial trajectory, the photon, which is traveling at c, winds up going back towards the surface of the mass before it can completely run out of energy. In fact, it will pick up energy as it travels back towards the mass.

The baseball is somewhat different because there is a rest mass present. The rest mass does not diminish, it is the energy of the thrown baseball which does diminish. Thus the baseballs rest mass will decelerate, but the energy will diminish just like the energy of the photon.

Because the light cannot slow and cannot be deflected, it must be able to escape

Why "must' the light escape ? The gravitational field diminishes the energy of the photon. The energy is eventually gone. The photon is comprised of energy. The photon will disappear. It will NOT escape if the gravitational field is strong enough AND if the photon is able to continue its ascent in a straight line. But it doesn't because the path of the photon is warped back towards the center of the mass. It cannot climb out of the gravity well. It does not have enough energy to do so.

Okay I've post this within this topic previously, I've also got it as a portion of another thread, But I thought I would re introduce it within the thread again.

Unconventional Blackhole Theory

A blackhole might not follow the previous conventional theory, it might not contain any significant mass in it's centre if one was to exist.

It's centre could actually be created from a wave function collapse of multiple universal layers. Universal Layer collapsing into layer to cause a spacial depression.

If it's done at the speed (Cx109) suggested with the Akimov, Shipov and Binghi: Torsion Field theory, it would cause spacetime to be shifted into the depression (A dip in space).

when it eventually reaches there it might still be pulling stiffened space along so it wrinkles up.

The wrinkled effect is what is known as an "Event Horizon", and because it's universal space and magnetism folded up, it gives the gravity that suggests the presence of mass.

Due the "Black hole" being made in this method, it might overtime suffer from the spacial folding being flattened out again (as it might unstiffen).

So eventually the black holes mass would lessen and then eventually allow it's gravity to disperse.

Q:

Regardless of their initial trajectory, the photon, which is traveling at c, winds up going back towards the surface of the mass before it can completely run out of energy.

If a photon traveling away from the surface at c along the axis of gravity could go back, c could not be constant.

If there is an event horizon, it would be analogous to the train example, wherein the photon continues to move away at c, never turning back, but space stretches past it at > c.

Why "must' the light escape ? The gravitational field diminishes the energy of the photon. The energy is eventually gone. The photon is comprised of energy. The photon will disappear.

Because no mass, being finite, could completely exhaust the photon's energy. In addition to the other reasons I gave.

[The photon] cannot climb out of the gravity well. It does not have enough energy to do so.

It does if the steepness of the well's sides is finite, which we ascertained is true at the beginning of this thread, to my satisfaction anyway, with this logic: The steepness of a gravity well's sides can approach infinity, but cannot become infinite.

Stryder

A blackhole might not follow the previous conventional theory, it might not contain any significant mass in it's centre if one was to exist.

What happens to the mass of the progenitor star ? Where does it go ?

The wrinkled effect is what is known as an "Event Horizon", and because it's universal space and magnetism folded up, it gives the gravity that suggests the presence of mass

Space and magnetism do not cause gravitational fields. Only the presence of mass can cause a gravitational field.

zanket

If a photon traveling away from the surface at c along the axis of gravity could go back, c could not be constant.

That makes no sense. And, I repeat again, the paths of light are warped back towards the center of the mass. Light follows these paths at c and remains constant. The path curves back towards the mass.

If there is an event horizon, it would be analogous to the train example, wherein the photon continues to move away at c, never turning back, but space stretches past it at > c.

Wrong. You are implying photons have infinite energy. Again, I repeat, for the last time I hope, the paths are warped back towards the center of the mass. This does not allow the photon to continue its journey away from the mass. Do you understand this concept ?

Because no mass, being finite, could completely exhaust the photon's energy.

It is the gravitational field which exhausts the photons energy. The mass, nor the gravitational field is infinite, it is finite. Photons DO NOT have infinite energy. Their energy is quite finite and can diminish in a gravitational field.

It does if the steepness of the well's sides is finite, which we ascertained is true at the beginning of this thread, to my satisfaction anyway, with this logic: The steepness of a gravity well's sides can approach infinity, but cannot become infinite.

A gravitational field is proportional to its mass. Neither are infinite nor will ever approach infinity. Your logic is seriously flawed.

Q,

I don't think I'm wrong in mentioning that nobody has yet seen a real black hole, let alone travelled to it's centre to suggest there is mass.

The only reason why the assumption of mass is drawn is due to the observation of gravity and gravitational bodies.

Just stars can "compress" to formulate dwarfs, you might say "thats due to gravity" but do you know what happens when you burn say the oxygen in a sealed container??? It causes a vacuum.

As for where a progenitor star's mass goes, I'm not to well versed on Cosmology for all portions of the universe. The only reason I've delved at blackholes has been over Parallel universes and multiworlds e ffects, not just what an astronomer would spot with a radio-telescope.

As for gravity not existing without mass, If I had explained it fully you would know that you don't need Mass, the idea is that space contains energy, this energy when measured on it's own doesn't weigh much and isn't that strong, but if it's universally folded with the stiffening of space caused by a reaction, it could compress together to form something that would act like gravity.

So I don't think just saying space and magnetism isn't going to create gravity, as I only termed it space because I didn't think you would think "complete vacuum, void of anything".

Again, I repeat, for the last time I hope, the paths are warped back towards the center of the mass. This does not allow the photon to continue its journey away from the mass. Do you understand this concept ?

Sorry, this concept is incompatible with my understanding of relativity, wherein the distance between a body and a photon moving directly away from it is always ct, where t is time. So I don't see how such a photon can fall back.

It is the gravitational field which exhausts the photons energy. The mass, nor the gravitational field is infinite, it is finite. Photons DO NOT have infinite energy.

They do seem to have enough energy to always move at c, even if the gravity of countless galaxies has been diminishing their energy for eons. That's good enough for me; to refute the existence of an event horizon, I need only know that the photon's velocity is forever constant.

Zanket: Nobody (including myself) has really given an explanation refuting your concept that a light ray traveling parallel to the gravitational force vector can escape and allow a distant observer to see the Black Hole. They have given explanations of how and/or why an Event Horizon forms and talked about related topics like there being no geodesic from inside to outside the Event Horizon.

Any argument about the escape velocity never equaling or exceeding the velocity of light has to be spurious. Both classical & GR gravitational formulae show escape velocity increasing with the square of inverse distance and the mass. There are various combinations of these quantities which result in a calculated escape velocity equal to the velocity of light, and other combinations resulting in a higher escape velocity. There is no current evidence or reasons to doubt the validity of the formulae, and I doubt that there ever will be any reason to doubt the validity of these formulae. At most, the precision of the GR formulae for certain conditions might be questioned due to some currently unknown effect, just as the classical formulae are no longer considered precise for all possible conditions.

The problem is whether or not the escape velocity concept is applicable to light itself. In particular, there is a question about a light ray directed parallel to the gravitational force vector and away from the Black Hole.

I am ignoring the concept of the steepness of a gravity well’s sides. Gravity Well is a cute phrase, but a poor analogy for use in a serious analysis of gravitation.

Following are two possible thoughts on why a light ray cannot escape from a Black Hole.

A pragmatic explanation. Light rays do not just happen. They are the result of electromagnetic processes involving atoms and/or the particles which make up atoms. If the process takes place external to the Event Horizon, it cannot produce a light ray that reaches the Event Horizon, and is directed parallel to the force vector, and is directed away from the Black Hole. A process external to the Event Horizon can, at best, create a light ray that tangentially touches the Event Horizon. Such a light ray is expected to be sucked into the Black Hole.

Any process taking place at the Event Horizon will at most generate a photon or two before the atomic particles are sucked inside the Horizon. After a short period of time, a distant observer will not see light rays produced by processes taking place at the Horizon.

The above indicates that only processes taking place inside the Event Horizon need be considered. If you assume that photons from inside can escape only if they are directed parallel to the force vector, a distant observer will be aligned to see very few of them, as most will be directed toward other observers.The above indicates that a distant observer can at most see an occasional blip from a Black Hole. For all practical purposes, he will not see a black hole, unless light rays not parallel to the force vector can escape. I do not think anybody is suggesting that such light rays can escape.

BTW: Another practical consideration: If (when) all the matter inside the Event Horizon becomes incredibly compressed near the center, the matter no longer exists as atoms & molecules. In such a compressed state, the processes that generate light do not occur. This reasoning is plausible (if not valid) even if we do not assume a singularity, which is a controversial concept likely to be abandoned if a good quantum description of gravity is ever developed. .

Hence, practical considerations indicate that a Black Hole is not seen by a distant observer. Id est: Light does not escape.

A GR theoretical explanation. For wave phenomena, the velocity of propagation is the product of the frequency and the wave length. The gravitational red shift at the Event Horizon results in an infinite (I suppose) wave length and a zero frequency. If the limit of this product is zero as you approach the Horizon, then the velocity is zero. This is not very convincing to me, but it is a thought.

From a particle point of view, photon energy decreases with increasing gravity. At the Event Horizon, the calculated energy is zero, resulting in no energy and no motion. Note that photon energy is related to frequency, making this view consistent with the red shift view.The above arguments are less convincing to me than acceptance of the concept that light is subject to gravity, and that the calculated escape velocity applies to it just as it applies to material objects. I have posted the above thoughts because they might be convincing to others.

Originally posted by zanket
Overdoze:

In the formula R_s=2GM/c², I submit that c should be an apparent velocity = infinity. If so, the formula always yields 0, the center of the body.


First of all, c is the speed of light which is a constant. The formula is unequivocal about the value of c as it is derived from the Schwarzschild solution of the GR equations for a uniform non-rotating spherical mass.

Secondly, if you assume c to be infinity then c² in the denomenator of the redshift formula would mean that there is no redshift no matter at what radius you start. (Are you really that bad at math?)

zanket

Sorry, this concept is incompatible with my understanding of relativity, wherein the distance between a body and a photon moving directly away from it is always ct, where t is time. So I don't see how such a photon can fall back.

I can only assume your understanding of relativity may be lacking or you are not understanding the explanations. You may be under the presumption the photon somehow slows down, perhaps stops, turns around and then falls back to towards the mass. If so, then you are certainly misunderstanding the concept or you're under the delusion the geodesic along the axis of gravity cannot be warped by the intense gravitational field.

Spacetime, in and around the black hole, is severely warped by the intense gravitational field. The warped spacetime is wrapped around the mass of the black hole such that all geodesics (straight paths) are curved back towards the center of this mass. If a photon were to escape the mass of the black hole, at any trajectory, including the axis of gravity, it will travel along the geodesic at c. The photon will also have a certain amount of energy.

The photon will follow the geodesic which has been warped and wrapped around the mass. The photon is still traveling at c and will follow this path back towards the center of the mass. However, as it travels, the gravitational field is diminishing its energy. As long as the photon has energy, it will still travel at c. If the gravitational field diminishes all the photons energy, the photon will simply disappear, it will cease to exist.

The photon will eventually impact the mass and get absorbed all the while traveling at c. The only difference is in the amount of energy the photon carries.

They do seem to have enough energy to always move at c, even if the gravity of countless galaxies has been diminishing their energy for eons.

Then you agree that gravitational fields can diminish the energy of a photon. Good. However, it still does not indicate the photon has infinite energy. If so, the energy of a photon would never diminish.

That's good enough for me; to refute the existence of an event horizon, I need only know that the photon's velocity is forever constant.

Your logic is flawed. You are confusing the velocity of a photon with its energy. The velocity of the photon is not a result of its energy. Photons always travel at c, however they may carry different energies. Photons follow paths in spacetime. If the path leads back towards the mass, the photon will follow this path. It will travel this path at c but will lose energy to the intense gravitational field.

Just because your logic is good enough for you does not make it correct. Your logic is flawed and your conclusions are wrong.

Dinosaur: Good stuff. My thinking is that if even an occasional blip from the hole may be seen by an observer outside the event horizon, then its definition, that nothing may escape, is at best incomplete. An occassional blip is a clue that something else might be happening.

The gravitational red shift at the Event Horizon results in an infinite (I suppose) wave length and a zero frequency.

I have a hard time believing that because it means that a once-finite value can become infinite. It seems to me that a once-finite value can only approach infinity.

From a particle point of view, photon energy decreases with increasing gravity. At the Event Horizon, the calculated energy is zero, resulting in no energy and no motion. Note that photon energy is related to frequency, making this view consistent with the red shift view.

I see the redshift or blueshift as a purely relative property. If I move toward a light source, the light is blueshifted. If I move away, a redshift occurs. If light is hitting me with a nearly infinite redshift, I need only move toward the light's source at sufficient velocity to make the light blueshift. I've read that redshift can approach infinity, but not that it can become infinity. Based on that, and if the photon's energy is indeed related to its frequency, then I have even less reason to believe that the photon's energy is either intrinisic (as opposed to a function of my velocity relative to the light source) or can drop to zero.

overdoze: You raise a good point about the redshift formula. Let me think about that over the weekend.

(Q):

The photon will follow the geodesic which has been warped and wrapped around the mass. The photon is still traveling at c and will follow this path back towards the center of the mass.

If the hole is not rotating, which way is the geodesic (of a photon moving against the direction of gravity) wrapped around the mass? To the south? East? How is that determined mathematically or intuitively? By random?

Suppose I am attempting to leave the hole in a ship that can move at any velocity < c and I can correct for any deviation in my trajectory. How am I prevented from passing the event horizon if I can prevent myself from falling back?

Then you agree that gravitational fields can diminish the energy of a photon. Good. However, it still does not indicate the photon has infinite energy. If so, the energy of a photon would never diminish.

No, I was saying that whether or not the photon's energy diminishes seems irrelevant. I neither agree nor disagree that it does. That's why I used "if." Please see my response to Dinosaur above.

To no one in particular:
Here's how I see the "space stretching"; that is, the train example.

Suppose I am passing by the earth on my way to the Andromeda galaxy, which is some 3 million light years distant from us on the earth. I'm moving at a velocity < c that will get me to the galaxy in 1 year of my time, so the galaxy is less than 1 light year away even though I can see the earth passing by in my starboard window.

Now suppose at that moment I begin a deceleration that reduces my velocity, relative to both the earth and the galaxy, to a full stop by the halfway point between them. At this point I measure the distance to either the earth or the galaxy at 1.5 million light years. During my deceleration, then, the distance between myself and the galaxy increased from 1 light year to 1.5 million light years even though I was moving toward the galaxy the whole time. If, say, the deceleration took me a tad less than 10 years, then the galaxy moved away from me at an average velocity of 150,000c.

That certainly sounds like space stretching to me. The big question is, did the galaxy disappear from my view during the deceleration? That is, did I stop receiving new light or information from it because it is moving away from me at > c?

The answer is no, because my velocity relative to the galaxy never really exceeded c. It only apparently did, compliments of spatial distortion. If you plug the apparent velocity of 150,000c into the conversion formula I gave earlier, it yields my average real velocity < c.

Nor did the photons stretch, or anything else stretch except in the way that spatial distortion apparently stretches or contracts things as a relativistic effect.

My take on the train example, then, is that the track is apparently moving at > c, but is really moving at < c.

If the hole is not rotating, which way is the geodesic (of a photon moving against the direction of gravity) wrapped around the mass? To the south? East? How is that determined mathematically or intuitively? By random?

Why would it matter what direction the geodesics take? What has that got to do with it ?

Suppose I am attempting to leave the hole in a ship that can move at any velocity < c and I can correct for any deviation in my trajectory. How am I prevented from passing the event horizon if I can prevent myself from falling back?

If you don't have enough velocity to escape the gravitational field, it wouldn't matter which direction you deviate, you'll eventually be pulled back to the mass.

That certainly sounds like space stretching to me.

No. It is a result of your misunderstanding of relativity and reference frames.

During my deceleration, then, the distance between myself and the galaxy increased from 1 light year to 1.5 million light years even though I was moving toward the galaxy the whole time.

Totally wrong. The distance always remained the same. You are confused because you did not take into account the reference frames, yours and a stationary observer.

(Q):

Why would it matter what direction the geodesics take? What has that got to do with it ?

I assume that an axis of gravity is straight and pierces the event horizon. If a photon leaves the surface along this axis but will eventually fall back to the surface, following a geodesic that curves back to the surface, then I assume that the geodesic path deviates from the gravity axis at some point to curve back, and that there are multiple directions of deviation available, like south and east. I'd like to know in which direction the photon deviates and more specifically how that direction is determined.

The equivalent of this question is, which direction does a compass point at the magnetic north pole?

If you don't have enough velocity to escape the gravitational field, it wouldn't matter which direction you deviate, you'll eventually be pulled back to the mass.

If I can leave the surface at all, then I can avoid being pulled back until I run out of fuel. If I didn't have that practical problem, say I use stray space hydrogen as fuel, then I could rise indefinitely. How am I prevented from passing the event horizon if I can prevent myself from falling back?

Totally wrong. The distance always remained the same. You are confused because you did not take into account the reference frames, yours and a stationary observer.

What distance is that? If I am moving toward the galaxy, I don't see how the distance I measure to it could remain the same.

I thought I was clear about the reference frame. I am moving relative to the earth and the galaxy. I measure my distance to them. Simple as that. How would you write the example?

(Q):

I realized my last post didn't answer your question.

Why would it matter what direction the geodesics take? What has that got to do with it ?

I'm asking because, if it cannot be determined which direction the geodesic takes other than random, then it probably doesn't work that way. A compass at the magnetic north pole, for example, is not attracted to any direction, so the arrow doesn't move. If a geodesic is not attracted to any direction then it continues to rise straight up.

Why must a photon curve back but a material object can double back, like a ball thrown straight up on earth? I thought gravity treated both the same.

zanket

I'd like to know in which direction the photon deviates and more specifically how that direction is determined.

The equivalent of this question is, which direction does a compass point at the magnetic north pole?

Photons propagate in all directions from their source. Some photons will travel along the axis of gravity. One could only speculate in what direction it would deviate, but be assured, it will not continue an upward journey for long.

A compass at the magnetic north pole, for example, is not attracted to any direction, so the arrow doesn't move. If a geodesic is not attracted to any direction then it continues to rise straight up.

You're talking apples and oranges. Magnetic fields have nothing to do with gravtitaional fields. Geodesics are not attracted to anything. They are folded over and over again in the space beyond the event horizon due to the gravtiational field. No geodesic will continue in a straight line away from the mass.

If I can leave the surface at all, then I can avoid being pulled back until I run out of fuel. If I didn't have that practical problem, say I use stray space hydrogen as fuel, then I could rise indefinitely. How am I prevented from passing the event horizon if I can prevent myself from falling back?

Unless you achieve the minimum velocity to escape the gravitational field of a massive object, you will be pulled back no matter how much fuel. The issue is not one of fuel but of velocity. For example, you need to achieve a minimum of 25,000 mph to escape the Earths gravitational field. You could have an endless supply of fuel, but unless you achieve this velocity, you will fall back to Earth.

What distance is that? If I am moving toward the galaxy, I don't see how the distance I measure to it could remain the same.[i]

From the reference frame of the moving ship, the distance is contracted relative to a stationary observer. Your measurement of the distance will differ from that of the stationary observer simply because you are moving at a substantial fraction of c. Once you decelerate and are not moving relative to the stationary observer, your measurement will be the same as the stationary observer.

The galaxy did not move away from you, it was your velocity relative to the galaxy that caused the change in measurements. The distance always remained the same, it was your velocity that changed your view of the measurement.

[i]I thought I was clear about the reference frame. I am moving relative to the earth and the galaxy. I measure my distance to them. Simple as that. How would you write the example?

You need to use the Lorentz transformation formula to discern your measurements between the different reference frames.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html#c2

Why must a photon curve back but a material object can double back, like a ball thrown straight up on earth? I thought gravity treated both the same.

Gravity does treat them both the same. The difference is that the ball has an effective rest mass and will follow Newtons law of Universal Gravitation. Photons have no rest mass because they are comprised entirely of energy. The ball, having an effective mass, may not necessarily follow a spacetime geodesic whereas the photon will.

The ball, having an effective mass, may not necessarily follow a spacetime geodesic whereas the photon will.The above is not correct. A material object follows a space time geodesic in a gravitational field if not acted on by other forces. That is one of the concepts of GR.

If you assume that the matter inside an Event Horizon is concentrated at the center, it seems to me that nothing can maintain a trajectory away from that central mass.

Apparently nobody here knows the equations of motion for objects inside the Event Horizon. If somebody posted them, it is likely that we would not understand them anyway.

I am in favor of accepting the opinion of the experts. They say nothing escapes. I am willing to accept it, especially since it seems reasonable based on the little that I do understand about Black Holes & GR.

If you assume certain possible but unlikely conditions inside the Event Horizon, perhaps light rays and/or matter can avoid being sucked into the central mass for some finite period of time.

For example, extremely fast rotation of the central mass combined with some nuclear reactions in the central mass, might create conditions allowing some matter and light to avoid being sucked in for a while.

It does not seem possible for anything to avoid being sucked in for any long period of time.

Oddly enough, if there was some weird phenomena which caused all the matter inside the Event Horizon to be concentrated in a hollow spherical shell, there would be no gravitational force inside that shell.

BTW: I suspect that our intuition about what goes inside the Event Horizon might be no good at all. I remember being surprised about gravity inside a hollow spherical shell, which does not seem as weird as a Black Hole.

From the outside, a hollow spherical shell of uniform thickness and density acts like a point mass at the center. Inside, there is no gravitational field at all due to the shell itself. If there were an atmosphere, you could swim in it. With some hand held fans, you could move around fairly easily and perhaps travel at 50mph (80 meters per hour) or more.

For example, I wonder if rotation of the central mass might result in a force enhancing gravity instead of fighting it.

I remember an article about a torus shaped space station surrounding a Black Hole. Impossible in practice, but an interesting thought experiment. There is a distance from the center of A Black Hole & external to the Event Horizon at which a light ray would orbit in a circle around the Black Hole.

A torus station at that distance would seem to be a cylinder to astronauts inside. They would be able to see the back of their heads. A torus station closer to the Black Hole would seem to curve away from the Event Horizon to an astronaut inside. I am not sure what either of these space stations would look like to an external observer.

Dinosaur

Please note I specifically said may not necessarily follow a spacetime geodesic. You've stated one circumstance; if not acted on by other forces.

In all likelyhood, matter would probably never move in any other direction than towards the center of the mass of a black hole.

(Q):

You're talking apples and oranges. Magnetic fields have nothing to do with gravtitaional fields. Geodesics are not attracted to anything.

Agreed. I was drawing a loose analogy. Perhaps I can do better: If I drilled a hole through the diameter of the earth and placed a ball at the center, the ball wouldn't move even though it is attracted by the mass that surrounds it. That is shown by attempting to answer the question, Which direction would the ball move?

Like the ball, it seems that a photon is attracted equally in all directions as it rises up from the surface along the axis of gravity, and so its geodesic path should not wrap back to the surface but rather pierce the event horizon.

Unless you achieve the minimum velocity to escape the gravitational field of a massive object, you will be pulled back no matter how much fuel. The issue is not one of fuel but of velocity. For example, you need to achieve a minimum of 25,000 mph to escape the Earths gravitational field. You could have an endless supply of fuel, but unless you achieve this velocity, you will fall back to Earth.

I can't buy that. A rocket that rises at 1 mph can keep rising as long as it has fuel. There is nothing to force it back. Once the rocket runs out of fuel, then it will begin to fall back to earth no matter how far away it got. How else do you explain that a hot air balloon can remain aloft for weeks?

The reason the space shuttle achieves escape velocity is so it can orbit with its engines off. A jet can orbit but its engines need to be running. With an endless supply of fuel, you can orbit or rise endlessly. That's why only the train example holds water for me as an argument for the event horizon's existence. Only an event horizon that is moving away from the body at >= c could prevent light or a capable ship from breaching it, assuming that the light or ship could leave the surface at all.

From the reference frame of the moving ship, the distance is contracted relative to a stationary observer. Your measurement of the distance will differ from that of the stationary observer simply because you are moving at a substantial fraction of c. Once you decelerate and are not moving relative to the stationary observer, your measurement will be the same as the stationary observer.

Agreed. It seems to me that the story accounted for all of that. How would you have written it?

The galaxy did not move away from you, it was your velocity relative to the galaxy that caused the change in measurements.

It did move away from me. My initial distance measurement to the galaxy was less than 1 light year. After the deceleration it was 1.5 million light years. Initially I would see the galaxy taking up half my field of vision. After the deceleration I would need a telescope to see it. That the change in measurements had something to do with my velocity relative to the galaxy is irrelevant.

You need to use the Lorentz transformation formula to discern your measurements between the different reference frames.

I did. In the first paragraph of the story I compared a 3 million light year distance that earthlings measure to the galaxy, to a less than 1 light year distance that I measured from effectively the same location. Thanks for the link, that's a good one.

Dinosaur:

I am in favor of accepting the opinion of the experts. They say nothing escapes.

Oh Dinosaur, what joy is there in accepting?! What if Mr. E. had done that? For me, I can accept only what makes sense. And making sense of something is how I learn.

There is a distance from the center of A Black Hole & external to the Event Horizon at which a light ray would orbit in a circle around the Black Hole.

This link says it's 1.5 times the Schwarzchild radius:

http://astsun.astro.virginia.edu/~jh8h/Foundations/chapter9.html

They would be able to see the back of their heads.

Interesting!

zanket

Like the ball, it seems that a photon is attracted equally in all directions as it rises up from the surface along the axis of gravity, and so its geodesic path should not wrap back to the surface but rather pierce the event horizon.

Wrong. The only direction the gravitational field attracts is in the direction towards the center of the massive object.

I can't buy that. A rocket that rises at 1 mph can keep rising as long as it has fuel. There is nothing to force it back. Once the rocket runs out of fuel, then it will begin to fall back to earth no matter how far away it got.

Now that I think about it, you're right, a rocket could escape at 1 mph with enough fuel. I'm used to thinking in terms of current technology. My bad.

How else do you explain that a hot air balloon can remain aloft for weeks?

This has nothing to do with gravitational fields. The balloon stays aloft because of the lighter gases inside the balloon compared to the atmosphere outside.

That's why only the train example holds water for me as an argument for the event horizon's existence. Only an event horizon that is moving away from the body at >= c could prevent light or a capable ship from breaching it, assuming that the light or ship could leave the surface at all

You're also making the assumption that light, or ship, or whatever is attempting to leave the gravitational field, has infinite energy.

It did move away from me. My initial distance measurement to the galaxy was less than 1 light year. After the deceleration it was 1.5 million light years.

While moving at near light speed, the length of your meter relative to the length of a meter for a stationary observer is less. That is why the distance to the galaxy appears to be more once you've decelerated. Your measurement for a meter then equals the measurement of a meter of the stationary observer. The galaxy did not move away, it was your relative measurement of the distance which changed.

Initially I would see the galaxy taking up half my field of vision. After the deceleration I would need a telescope to see it. That the change in measurements had something to do with my velocity relative to the galaxy is irrelevant.

Then you do not understand the concepts of relativity and length contraction. What you're implying is that an entire galaxy has moved away from you simply because you chose to decelerate. That is entirely ludicrous. Read up on length contraction and time dilation.

I did. In the first paragraph of the story I compared a 3 million light year distance that earthlings measure to the galaxy, to a less than 1 light year distance that I measured from effectively the same location.

Perhaps, but you did not understand why this occurs. It is based on time dilation and length contraction. Would you like me to provide a few links explaining these concepts ?

(Q):

Wrong. The only direction the gravitational field attracts is in the direction towards the center of the massive object.

Now I’m confused. You said before “And, I repeat again, the paths of light are warped back towards the center of the mass. Light follows these paths at c and remains constant. The path curves back towards the mass.” But why would light that left along the axis of gravity curve back if the only direction the gravitational field attracts is towards the center? It seems to me the light could only double back, and then only if it could stop and turn around like a ball, which it can’t.

This has nothing to do with gravitational fields. The balloon stays aloft because of the lighter gases inside the balloon compared to the atmosphere outside.

I meant that the balloon defies gravity as long as it has fuel.

You're also making the assumption that light, or ship, or whatever is attempting to leave the gravitational field, has infinite energy.

Only finite energy is needed. Like the rocket rising at 1 mph and attaining any distance, the ship need not attain c to reach and surpass the event horizon. The escape velocity may be c at the event horizon, but the ship does not intend to escape the gravitational field, only to breach the horizon.

While moving at near light speed, the length of your meter relative to the length of a meter for a stationary observer is less.

That’s what the observer measures along the axis of motion. I measure the opposite, that the observer’s meter is shorter than mine. Before deceleration, the space between the galaxy and me was contracted to a small percentage of the distance either of us would measure at relative rest. When I decelerated to relative rest, the space between us expanded to that distance. That is why the galaxy moved away from me.

Before deceleration an observer on either the earth or the galaxy would measure only me contracted, not the space between us. That’s because either observer was moving relative only to me. I was moving relative to both observers and the space between them (imagine that a rope connected them; I would be moving relative to the rope as well), so I measure them as well as the space between them contracted.

That is why the distance to the galaxy appears to be more once you've decelerated.

It doesn’t appear to be more. It is more. The way to determine distance is to measure it. If I measure that the distance increased, it did.

What you're implying is that an entire galaxy has moved away from you simply because you chose to decelerate. That is entirely ludicrous.

Isn’t relativity fascinating?

Perhaps, but you did not understand why this occurs. It is based on time dilation and length contraction.

All relativity books discuss time dilation and length contraction. Few mention that it works both ways. Time rate and length can both contract and expand in a range from approaching 0% of the time rate or length at relative rest, to 100%.

If I am at rest relative to something, I can contract the distance I measure between that something and me simply by increasing my velocity relative to it. Likewise, its time rate that I measure will decrease.

If I am moving relative to something, I can expand the distance I measure between that something and me simply by reducing my velocity relative to it. Likewise, its time rate that I measure will increase.

This contraction or expansion is quite independent of my movement. For example, if I could somehow survive an acceleration that brought me from rest to close to c relative to a galaxy in say 1 meter of my movement, I would measure the galaxy as having moved closer to me by perhaps billions of light years during the acceleration. The distance change would be real; if I brought the galaxy close enough I could read the aliens’ bumper stickers. If in the next meter of my movement I decelerated back to rest relative to the galaxy, it would move away from me, back to its original distance. I could have moved in any direction that changed my relative velocity as described.

A clarification on the above paragraphs: To contract or expand the space between something and me as I measure it, I have to change my velocity relative to the thing along the axis between the thing and me. To contract, I must increase this velocity. To expand, I must decrease this velocity if any.

zanket

But why would light that left along the axis of gravity curve back if the only direction the gravitational field attracts is towards the center? It seems to me the light could only double back, and then only if it could stop and turn around like a ball, which it can’t.

I explained that in earlier posts. But I need to ask. How is it possible for object to be attracted equally in all directions when the massive object propagating the gravitational field is in one direction ? If that were the case, I could fly, since the gravity of Earth is attracting me equally in all directions.

I meant that the balloon defies gravity as long as it has fuel.

Would the balloon, with its infinite supply of fuel, defy gravity if there were no atmosphere ?

The escape velocity may be c at the event horizon, but the ship does not intend to escape the gravitational field, only to breach the horizon.

That is basically one in the same. If the ship breaches the event horizon, it has essentially escaped the gravitational field.

That’s what the observer measures along the axis of motion. I measure the opposite, that the observer’s meter is shorter than mine. Before deceleration, the space between the galaxy and me was contracted to a small percentage of the distance either of us would measure at relative rest. When I decelerated to relative rest, the space between us expanded to that distance. That is why the galaxy moved away from me.

The space between you and the galaxy did not expand or contract. It is simply a matter of your measurement of that space which changed due to your increased velocity. The galaxy neither came closer to you nor moved away.

Before deceleration an observer on either the earth or the galaxy would measure only me contracted, not the space between us. That’s because either observer was moving relative only to me. I was moving relative to both observers and the space between them (imagine that a rope connected them; I would be moving relative to the rope as well), so I measure them as well as the space between them contracted.

Exactly. Your measurement of that space is what changed, not the space itself. Both observers on Earth and on the Galaxy will measure the distance the same regardless of your velocity.

It doesn’t appear to be more. It is more. The way to determine distance is to measure it. If I measure that the distance increased, it did.

You don't understand the concept of length contraction. Your measurement of the distance will differ to that of a stationary observer because of your velocity. It is not an inherent property of space to expand and contract simply because you are moving at a different velocity. It is your velocity which has changed your measurement of that distance.

Isn’t relativity fascinating?

It is. Why don't you take the time to fully understand it ?

All relativity books discuss time dilation and length contraction. Few mention that it works both ways. Time rate and length can both contract and expand in a range from approaching 0% of the time rate or length at relative rest, to 100%.

No. It is your measurement of time and length which changes.

If I am at rest relative to something, I can contract the distance I measure between that something and me simply by increasing my velocity relative to it. Likewise, its time rate that I measure will decrease.

Again, no. The contracting of distances is not an inherent property of distance. It is relative to velocity.

If I am moving relative to something, I can expand the distance I measure between that something and me simply by reducing my velocity relative to it. Likewise, its time rate that I measure will increase.

Wrong again.

This contraction or expansion is quite independent of my movement. For example, if I could somehow survive an acceleration that brought me from rest to close to c relative to a galaxy in say 1 meter of my movement, I would measure the galaxy as having moved closer to me by perhaps billions of light years during the acceleration. The distance change would be real; if I brought the galaxy close enough I could read the aliens’ bumper stickers. If in the next meter of my movement I decelerated back to rest relative to the galaxy, it would move away from me, back to its original distance. I could have moved in any direction that changed my relative velocity as described.

Totally wrong. One cannot manipulate the universe simply by accelerating and decelerating. Complete nonsense.

To contract or expand the space between something and me as I measure it, I have to change my velocity relative to the thing along the axis between the thing and me. To contract, I must increase this velocity. To expand, I must decrease this velocity if any.

Absolutely not. You cannot expand or contract space. Total gibberish.

(Q):

How is it possible for object to be attracted equally in all directions when the massive object propagating the gravitational field is in one direction? If that were the case, I could fly, since the gravity of Earth is attracting me equally in all directions.

You are right. I should have said "attracted equally in all directions in which there is mass." The earth is in more directions from me than just toward the center. But the attraction in all the other directions cancel each other out.

Would the balloon, with its infinite supply of fuel, defy gravity if there were no atmosphere ?

No.

That is basically one in the same. If the ship breaches the event horizon, it has essentially escaped the gravitational field.

Not as I understand it. The escape velocity is that needed to reach infinity with no additional force applied. If the ship rises past a point with less than the escape velocity for that radius, it must keep its engines running or else it will eventually fall back towards the hole. However, if the ship continues rising at 1 mph, it will eventually reach a radius where the escape velocity is only 1 mph. Then the ship will have escaped the gravitational field.

The space between you and the galaxy did not expand or contract. It is simply a matter of your measurement of that space which changed due to your increased velocity. The galaxy neither came closer to you nor moved away.

I'm not sure which you are debating: semantics, or whether the distance changed in a physical way. If semantics, whether "space contracted" or "my measurement changed," I can agree with your interpretation. But if you are denying that the galaxy shot away from my field of vision when I decelerated, or that I can read the aliens' bumper stickers, the evidence is against you.

Relativity does not allow me to accelerate to >= c. I am allowed, however, to traverse any finite distance in any proper time > 0. I cannot do that while moving at < c if space does not physically contract (or whatever your interpretation is that amounts to the same thing) when I accelerate.

One of the confirmations of special relativity involves the decay of a muon particle. Here is a link about it:

http://www-istp.gsfc.nasa.gov/stargaze/Lframes1.htm

In this link it says: "... the distance from the top of the atmosphere to the ground, which is 12 km in the Earth's frame, may be only 0.6 km in the frame of the muon."

Another link:

http://theory.uwinnipeg.ca/mod_tech/node136.html

In this link it says: "[The muon] must conclude that the atmosphere of the Earth is only about 600 meters deep, or about one seventh of what someone on Earth measures it to be."

If the space between the earth and the muon were not contracted to 0.6 km (or whatever your interpretation is that amounts to the same thing) in the frame of the muon, it would decay before it reached the earth. To the muon, the contraction is a physical reality.

zanket

I'm glad you're starting to understand the concepts however, you need to get over the hurdle regarding length contraction. I'm glad you brought up the case of the muon, a good example.

It is determined that the lifetime of a muon is 2 microseconds and travels near the speed of light. In the laboratories frame of reference, a clock will tick 2 microseconds, therefore the muon should travel 0.6 kilometers (300,000 x 2/1000,000 = 0.6 kilometers). According to relativity, clocks tick slower in moving frames. Therefore, the laboratory would view the clock on board the muon as ticking slower. In other words, 2 microseconds which ticked on the laboratories clock would be equal to about 0.1 microseconds on the muons clock. In the time taken for the laboratory to count 2 microseconds, the muon counted only 0.1 microseconds and in that time measured by both the laboratory and the muon, the muon traveled 0.6 kilometers.

We know that the muon has a lifetime of 2 microseconds, therefore according to the muons clock, it still has 1.9 microseconds of life before it decays. If it traveled 0.6 kilometers in 0.1 microseconds according to its onboard clock, it should be able to travel another 11.4 kilometers in the 1.9 microseconds it has available in its lifetime.

Now lets look at length contraction. The muon traveled 0.6 kilometers in 0.1 microseconds relative to its own reference frame, but the laboratory counted 2 microseconds. As you can see, the difference is in the way the time was measured in the separate reference frames. The muon is no where near that 12 kilometer destination and does not view that point either. Its view is that of 11.4 kilometers away from that 12 kilometer destination.

Your understanding is that the muon should be able to clearly view that 12 kilometer destination from its current position 11.4 kilometers away. That is not correct for it has only traveled 0.6 kilometers and it's view of that 12 kilometer destination is 11.4 kilometers away. In other words, the measurements for that 0.6 kilometer distance is viewed as traveled in 0.1 microseconds for the muon and 2 microseconds for the laboratory. If the muon stopped at this position, the distance would remain the same and would not contract or expand in any way simply because the muon has only traveled 0.6 kilometers.

The muon now continues on its merry way for another 1.9 microseconds relative to its onboard clock