Just wondering If there's any facts about what happens Inside a black hole with time Itself. Thank you for your comments.

Just wondering If there's any facts about what happens Inside a black hole with time Itself. Thank you for your comments.
Beyond the event horizon, all is speculation.
Facts likely exist, just beyond our reach.
what we can't see is seen by mathematics.however when the mathematical method fail to describe reality we modify it or even change it. but just inside a black hole we witness what we call singularity and just then our math stop talking and still we cannot observe.therefore we are both blind and dumb.
beyond sight, feeling, and math, we still exist if only for a few fleeting moments.
Our ignorance does not preclude our existance.
There will never be any "facts" about this, because the interior region is not causally connected to the rest of the universe. The only way to get facts would be to allow yourself to fall into that black hole.
The mathematics suggest that spacetime is smooth and regular in the interior, except at the singularity itself. There is, however, one other possibility, but for the sake of clarity I shall omit it here.
Why omit it, wouldn't it be better to let us know what it is than keeping us in the dark? Pun intended.The mathematics suggest that spacetime is smooth and regular in the interior, except at the singularity itself. There is, however, one other possibility, but for the sake of clarity I shall omit it here.
Well, it is possible to completely remove the interior of a black hole via a simple coordinate transformation, turning it into an EinsteinRosen bridge. In this scenario, the event horizon becomes the throat of a wormhole, and the interior region  including the singularity  quite simply doesn't exist. From an external point of view this is completely indistinguishable from a "normal" black hole  it would behave and appear exactly the same, just without any "inside" part. This particular solution is mathematically valid and rigorous, however, I know of no initial energymomentum configuration the gravitational collapse of which results in such a static, stable wormhole. So, even though this is a mathematically valid metric, it mightn't be a physically meaningful one.
The exact numbers depend on many factors, including the acceleration profile of the traveller, the size, mass and type of the black hole, and the actual trajectory the traveller took. This is not trivial. However, he will always have aged less than his counterparts back on Earth.
Let's say a 10 Solar mass black hole ()
a 75kg human.
Initial Distance? 5,000km?
That should be enough to determine the size of the event horizon, and the amount of time it takes to reach it, then calculate for time passing on the spaceship he jumped out of (or wherever). Magically teleport him back his ship, and compare clocks?
To make things simple we can assume he starts at rest relative to the black hole, the initial distance is his center of mass compared to the black hole's center of mass, and his trajectory is a straight line from his position toward the black hole. I mean, it's not like we're planning a real NASA mission here, I am just curious as to what effect the time dilation will have.
Ok, this is a really small black hole with a Schwarzschild radius of 29km; if we disregard any effects of motion and acceleration and just pretend the observer hovers at 5000km from the BH for a while, and then "teleports" back, his clock will be dilated by a factor of 0.997  a very small amount.
Nono, He starts at rest, and falls into the black hole, he doesn't teleport back to his ship until after he touches the event horizon.
I have another question which we can address next if you're up for it. If I am in a spaceship that is accelerating near the speed of light, but I don't have enough force to continue accelerating any faster, am I still considered to be accelerating for the duration that my engine is firing? Or does my time dilation stop when my engines can no longer push my ship faster than it is already going?
Yeah, starts at rest, 5000km from the center of mass of the black hole, is pulled into the black hole by it's gravity (as the only force acting on him), and when his center of mass is at the event horizon, he is teleported back to his ship (where he started from), and compares his clock to his ships clock. What is the actual time difference, given the scenario?
The question is not very meaningful and has no real answer, since you are comparing two observers that follow completely different world lines in a highly curved spacetime, and because the "teleporting" in the end is not physically possible. I can tell you only that, because the tidal forces for this small black hole are very large, the freely falling observer takes only 0.15 seconds of his own proper time to fall from 5000km to 29km ( the event horizon ). The far away stationary observer will never see him reach the event horizon; it would take an infinite amount of coordinate time to do so. He will see the in falling observer slow down and become dimmer and dimmer, eventually fading away.
Trying to figure out how much proper time in the rocket corresponds to the 0.15 seconds of proper time of the freely falling observer is quite nontrivial, and involves a very complicated integral which I don't immediately know how to evaluate. I will see what I can do for you, but give me some time for this.
Sounds fair, thanks for the effort. I'm in no real hurry, but it's a definite curiosity of mine.
Last edited by Ninja Pancakes; November 26th, 2013 at 03:03 AM.
The problem is that there is no common concept of "time" between these observers; you can only compare proper times ( i.e. what clocks physically measure in a frame ) for observers who follow smooth, differentiable paths between the same two fixed events. This is not the case in this scenario. I think it is best to not get motion involved at all, and just compare time dilation ( not actual proper times ) at different distances from the black hole. Even then it is problematic, because one observer uses proper time, and the other coordinate time ( for the in falling one ), which cannot be meaningfully compared. The only physically meaningful thing to do would be to send an observer on a round trip close to the event horizon, and then back to the stationary rocket; one could then compute the proper times for both world lines and compare. But again, you will then have movement through time and space for one of them, so the comparison really is still not very meaningful. Also, the calculation itself would be quite difficult and time consuming, since one would have to integrate over the entire world line, at least parts of which will involve acceleration ( to get back out ). It sounds like a simple scenario at first, but in GR it is actually quite nontrivial.
Proper time is the time measured by a clock that is taken with the observer on their journey. Coordinate time represents the family of threedimensional spaces defined by the notion of simultaneity in the frame of reference of the observer, parametised by the proper time of that observer.
Also, what is the difference between coordinate time and proper time, and how do they correspond to reality?
It means that fourdimensional spacetime is sliced up into a layering of threedimensional spaces where each threedimensional space has the same time in the frame of reference of an observer. The time of these slices is the proper time of the observer, but they are not themselves proper time because that is only defined at the observer's location.
To put this perhaps a little bit simpler  proper time is what a clock physically measures as it travels between two events. Coordinate time is the time that a distant observer measures between those events, using his own definition of time and distance ( i.e. his own coordinates ). In relativity, these notions do not generally coincide  for example, an observer falling into a black hole does so in a finite amount of his own proper time ( in his local frame ), but for a distant observer that takes an infinite amount of coordinate time.
Okay, I understand the concept of stacks of 3 dimensional events. I don't understand the last sentence though, and I don't understand how coordinate time differs from proper time, really, or how either of them relate to reality, i.e. what actually happens.
So, what is the reality of an infinite amount of coordinate time? If my own coordinate time, and proper time don't line up, what does that mean?
Proper time is the time measured by a physical clock. One can consider the threedimensional space that is simultaneous to the clock at a given proper time, and label this threedimensional space with the time value of the clock. This labelling is the coordinate time. But, a clock can only measure the proper time at the location of the clock itself. It can't measure the proper time at some remote location. Therefore, the proper time of another clock at the remote location need not correspond to the coordinate time defined by the proper time of the other clock.
Okay, so what you are saying is... proper time is "actual" real time. Coordinate time is time projected from your reference frame to a distant reference frame?
So, the only instance that coordinate time and proper time disagree then, should be when there is a difference in velocity between one frame and another. Right?
Yes.
No. In flat spacetime, two observers moving relative to one another have different threedimensional spaces representing their notion of simultaneity. I was actually referring more to accelerated frames of reference and in particular gravitation, where distant observers might share the notion of simultaneity, but their proper times between the same threedimensional spaces differ, giving rise to redshifts or blueshifts, even though there is no relative motion between the observers.
Proper time is what clocks measure for yourself, in your own frame of reference; coordinate time is what your clock measures for another frame of reference. Both of them are physically real; however, a particular measure of coordinate time is physically real only for the observer whose clock is used to make the distant measurement  it is observer dependent. Another observer somewhere else might measure a different coordinate time, even if the "same" event is observed. Proper time, on the other hand, is an invariant and all observers agree on it  it is defined as the length of an observer's world line.
That's as simple as I can put it without butchering the concept
We need to formulate it the other way around  the only instance when proper time and coordinate time coincide are when all clocks are brought together at rest and synchronised.So, the only instance that coordinate time and proper time disagree then, should be when there is a difference in velocity between one frame and another. Right?
I'm confused again. I thought that time dilation worked in such a way that if my proper time differs from your proper time, in reality, the coordinate time would also reflect this. If coordinate time is disconnected from reality, why is it even considered?
Let me see if I can rephrase this: I thought that coordinate time reflected the real difference between one proper time and another. Is this not the case?
You can't meaningfully compare proper times unless your world lines traverse the same two events in spacetime.
That depends on the observer. A particular coordinate time is, in general, valid only in a particular frame of reference.the coordinate time would also reflect this
Coordinate time is just as physically real as proper time, but only in a particular frame of reference. For a distant observer far from a black hole, from his own perspective, it is physical reality that no free falling object ever reaches the event horizon ( takes infinite coordinate time ). However, that reality is tied to that particular frame of reference  it would be wrong to conclude from that all other observers agree as to these realities  in particular, the freely falling observer himself will not agree, because on his own clock he reaches the horizon in a finite, well defined amount of ( his own proper ) time.If coordinate time is disconnected from reality, why is it even considered?
Time dilation is the comparison between a clock's proper time and the observer's coordinate time.
Sorry my last question seems to have been lost in the crowd. What does the observer falling into the black hole see when he looks at his spaceship and friends inside? Also, what does this mean?
unless your world lines traverse the same two events in spacetime.
A. Death. The gravity would crush anything before it could even get past the event horizon.Sorry my last question seems to have been lost in the crowd. What does the observer falling into the black hole see when he looks at his spaceship and friends inside? Also, what does this mean?
B. Hypothetically, I think the observer would just see his companions ever redshifted. Forever. and ever. and ever. and ever
That's not really answering the question, I mean, in regard to their time. He can already magically teleport back to his ship when he reaches the horizon, or if nothing else within say an atom of it, so he can magically not die as well, since the survivability isn't really part of the curiosity. If I recall correctly, he would see them blueshifted, not redshifted. They see him redshifted and he only has 0.15 seconds to see anythingSo, what does their time progression look like to him?the freely falling observer takes only 0.15 seconds of his own proper time to fall from 5000km to 29km ( the event horizon ).
This is not so easy to answer in general, because the answer will depend on both the state of relative motion as well as differences in radial position. Basically, as the observer falls into the black hole, he sees the stationary rocket receed from him at ever increasing speeds; if those speeds become relativistic then relative time dilation starts to become important, making the leftbehind clocks appear to go slower. At the same time, as the observer falls towards the black hole, his own proper time becomes gravitationally dilated as compared to a clock in the rocket, making the rocket clock appear to be speeding up. What the net effect will be, and how that effect develops over (proper) time for the infalling observer, depends on a number of variables such as mass of the black hole, the exact trajectory if the infall and so on. You may like to have a read through this :
What happens to you if you fall into a black holes
and also take a look at the videos and explanations here :
Journey into a Schwarzschild black hole
Ok, this isn't very easy to explain  let's see what I can do.unless your world lines traverse the same two events in spacetime.
Imagine two rockets  one of them ( rocket A ) travels from Mars to Jupiter, the other one ( rocket B ) travels from Earth to Venus. We'll put stop watches in both of them to record exactly how much proper time each one needs. At the end of their journeys the watches are stopped, and then they both return to earth and compare their clock readings. But what do those readings tell us ? All we can infer from them is that one of them needed more time than the other  but so what ? They travelled different trajectories, had different start and endpoints. The comparison between them is meaningless, because we are comparing two completely different world lines  it is like comparing apples and oranges.
However, let's say they both start at Earth, and end up on Venus ( = they travel between the same two events ), but on different trajectories  now we can meaningfully compare them, because they are connecting the same points in spacetime, just along different world lines.
It boils down to the fact that you can only meaningfully compare like things  you can't compare apples with oranges and expect anything meaningful to come out of that. Likewise, you can only meaningfully compare world lines and their lengths ( =proper time ) that connect the same two events, or else you are not comparing like things.
What this means is quite simple  to compare proper times, the observers must start together at rest, and they must end together at rest when the experiment is over. Only then can you make a physically meaningful comparison of proper time readings on clocks. Trying to compare a stationary rocket far away with the proper time of an observer falling into the black hole is meaningless, because they don't connect the same two events.
Am I making sense ?
Yeah, that makes sense. So what I should really be asking then is: What is the difference in the clocks when an astronaut falls from 5000km from his spaceship, toward a black hole, gets as close as he possibly can to the event horizon without passing it, and then returns to his rocket? What I really want is a definite figure. I've been getting a lot of abstracts, but those don't really tell me anything more than "the rocket will have recorded more time". How much more time?
Nothing special. If it is a very small black hole, there will be noticeable tidal force which will tear the spaceship (and even the people (eww)) apart. But otherwises, you will not be aware of falling through the event horizon; everything around you will look normal.
Yes, that makes sense, because you are comparing world lines between the same two events.
The problem here is that the actual calculation is anything but trivial, and  to be perfectly honest  I would take me some time to figure this out as I would have to go back over some old notes I made on this issue. I can however give you some sense of the magnitude of the effect if you willing to forego the figure of 5000km. I did, a few years back, an exercise in the context of numerical GR that involved putting together a numerical simulation of freefall orbits around Kerr black holes ( those are black holes which have angular momentum as well, i.e. rotating black holes ). There is a numerical algorithm in existence for such orbits ( it's too complicated to do analytically for Kerr black holes ), and the exercise was to implement the algorithm in C++ as a computer simulation.
Unfortunately I no longer have the source code, but I still have my notes as to the results  the black hole had 10 solar masses; at the apogee point the infalling spacecraft had a speed of 0.14c and an exentricity of 0.5 just above the event horizon; it recorded a proper time of 0.013s for one orbital revolution. The reference rocket was stationary at the farthest point of the orbit, and recorded a time of 0.046s for one orbital revolution ( after which they were back together ). The time dilation was thus of the order of approximately 350% per orbital revolution. Unfortunately I didn't note down how far away from the black hole the farthest orbit point was, or how much angular momentum the black hole had. But in any case, that should give you some idea of the magnitude of the effect; for a stationary black hole without angular momentum the time dilation effect would be slightly less, but probably still on the order of maybe 200250% percent per revolution.
Is this what you were looking for ?
I guess I am just confused as to how from the spacecraft it looks like it takes the astronaut an infinite amount of time, but the astronaut gets back in half of a tenth of a second. Shouldn't these two match up? If I understand time dilation correctly; if my time is dilated such that I experience 10 seconds for every 100 of your seconds, you should then experience 10 seconds for every 1 of my seconds, right? By this I guess I need to clarify real, actual time dilation. Not just apparent.
No, the infinite time is only for a straight infall along the radial direction, and even then only for Schwarzschild black holes ( no angular momentum ); Kerr black holes are more complicated. For elliptical orbits, the distant observer will see the ship go in and come out again in finite time, but they won't agree on the speeds or positions for any given time.
Gravitational time dilation ( gravity & acceleration induced ) is "actual" in the sense that it affects the accumulated time on a watch travelling along with the observer. That is why, in my simulation, the travelling ship accumulated less time than the stationary one farther away from the black hole. It is real, measurable effect, an actual recorded time on a clock.By this I guess I need to clarify real, actual time dilation. Not just apparent.
The reason why the amounts of time are so small in this example is that the black hole itself is very small ( about 29km event horizon radius ), so tidal forces and relativistic effects are very large, and the ship falls at a considerable speed along its orbit ( 14% of speed of light at apogee ! ). This is a pretty violent affair, and in reallife any object travelling on such a trajectory would be torn apart by the extreme tidal forces.
Can you explain why the times don't agree? It just doesn't make any sense to me at all. How does the angular momentum, or trajectory into the black hole cause such a radical difference?
Because spacetime is not flat, so there is no common, absolute notion of time between observers. How could there be ? Light has a finite speed and travels along complicated trajectories in a curved spacetime, so the notion of simultaneity cannot be easily defined. "Time" is something that really only makes sense in a small region ( an observer's frame of reference ).
That is an effect of General Relativity, related to the field equations being highly nonlinear. If you throw angular momentum into the mix, you get a completely different structure of spacetime as a result. This is not very easy to explain without going into the maths of the field equations, which are really quite complicated.How does the angular momentum, or trajectory into the black hole cause such a radical difference?
When I picture try to picture time in relation to time dilation, I sort of think of it like perpendicular lines. The perpendicular lines in proper time always have a specific distance between them, i.e. Any reference frame sees time as normal at their location. However, when one person's time is distorted and another person's is not, things change. If your time moves faster than my time, I see your lines as closer together, and I see my lines as normal. Meanwhile you see your lines as normal, but my lines appear farther apart. If there were an objective time, any time flowing faster would have lines closer together, and any time moving slower would have lines farther apart. If we place this objective time reference frame at an arbitrary, relatively empty point in space, and then look at the event horizon of a black hole, the closer in position to a black hole's event horizon, the farther the line spread in time... out to infinity at the horizon itself. So it should follow that if we place a person at this arbitrary, relatively empty point in space, and then move our objective time reference frame to a point as close to the event horizon as possible without touching it, the line spread for that person, from our new objective time reference frame would be so consensed that individual lines are barely separate at all. Time flows so fast there, compared to where our objective time reference frame is, if we were to just move it one step closer to the horizon, infinite time would happen in an instant at that person's location. So, please explain to me how this effect is cancelled?
This is actually precisely what happens for a faraway observer looking back onto a clock falling into a black hole  he will see that clock run ( and fall ! ) more and more slowly while getting dimmer and dimmer, and eventually it will fade away into nothingness. It would take infinitely long from an outside perspective for something to fall into a black hole  that is coordinate time. The problem is that spacetime between the faraway observer's location and the infalling clock is not flat; the infinite dilation is an artefact of light traversing that curved spacetime on null geodesics which are not straight, and which become increasingly complicated the closer the point of origin is to the event horizon. Of course the falling observer himself doesn't have that problem, hence the difference.
So you're implying that the time dilation, is caused by the trajectory of the light? I don't understand why that is the case. Obviously the person falling in falls in, in finite time. But he should see the real time of everything outside the event horizon increase faster and faster and faster until it reaches infinity when he reaches the event horizon. What does the trajectory of light have to do with that?
What I am implying is this : what a distant observer sees happening to an in falling object depends on how light gets back from that object to the distant observer.
Have a look at what null geodesics ( light ) look like in Schwarzschild spacetime :I don't understand why that is the case.
As you can see, light does not travel "straight out" to the distant observer, but it describes complicated curved trajectories. These trajectories become more and more spiral the closer the point of origin is to the event horizon  and hence light will take much longer ( as measured by the distant observer ) to get to him. That is the reason why coordinate time tends to infinity for someone watching something fall into a black hole.
Yes, indeed.Obviously the person falling in falls in, in finite time.
See the above rendering, that will make it clearer than I can explain it. Light has to travel farther and farther to escape the black hole and reach the distant observer; hence it takes longer and longer to get there.What does the trajectory of light have to do with that?
I don't understand how this explains time dilation. I understand that the velocity of light is relevant, but this doesn't make any sense to me. What if a person were to follow the same trajectory as the light in the diagram (they would have to be accelerating I imagine to escape). What if the person were to accelerate in a straight trajectory, from the edge of the horizon back to their ship? I thought time dilation was a distortion in spacetime, not just a convoluted journey of light. This diagram doesn't even look like it is representational of a person near an event horizon sending light signals toward a person on a spaceship farther away. Also, what if the person at the edge of the horizon is in a "black" box that emits no light, and blocks all incoming light? It would seem ridiculous to suggest that suddenly there would be no time dilation between that person and any other location in the universe. This seems to be confusing visual information about a distant subject with actual events happening. I mean, the clocks don't rely on an ability to detect light to tick.
I think there is a misunderstanding here; this was not meant to explain gravitational time dilation, but rather why coordinate time tends to infinity if you watch an object fall into a black hole. Gravitational time dilation happens because the geometry of spacetime isn't the same everywhere; curvature closer to a massive object is different than it is far out ( where it can be almost neglected and becomes Minkowskian for all intents and purposes ). It is this global geometry which leads to gravitational time dilation between observers; however, the very same geometry is also the reason why light travels the trajectories it does, so these phenomena are intrinsically linked. All relativistic effects are ultimately because of the geometry of spacetime itself.
In short  gravitational time dilation happens because spacetime is not globally Minkowskian. To put this in a less rigorous but perhaps more intuitive way : time is curved more the closer you get to a massive object, and that "curvature in time" manifests globally as gravitational time dilation, and as null geodesics that are no longer straight lines.
@Markus, Can you link me to the papers or studies that show the time dilation is caused by long trajectories? Because I'm not sure I can believe this, just reading it from you. No offense, but I need to see the basis of the statement you are making. It doesn't make any sense to me. Also, that diagram is only showing 4 trajectories. What about the trajectory of a photon emitted on the extact opposite side from where the event horizon is (going to the right, in the diagram, rather than straight up)? Wouldn't it go perfectly straight?
Okay... That doesn't answer any of my questions though =(
Please refer back to post #63  what I am saying is that gravitational time dilation and the nontrivial trajectories of light are manifestations of the same phenomenon, being the curvature of spacetime. I did not say that gravitational time dilation is due to light trajectories; what a distant observer ( coordinate time ! ) sees when he watches an object fall in, is. Time dilation happens because the geometry of spacetime looks something like this close to a massive object :
No offence taken, but you misunderstood me. I didn't equate these phenomena, all I said is that they have a common cause, being the geometry of spacetime.No offense, but I need to see the basis of the statement you are making.
In curved spacetime, "straight" does not mean the same as in Euclidean space; light always takes a "straight" path, but that path follows null geodesics in a curved spacetime.Wouldn't it go perfectly straight?
Hmm, I don't think my questions were answered. Null Geodesics, if google has served me properly, is a fancy term for light trajectories. That image showed four of them. One of them goes around the event horizon and toward a hypothetical ship farther out on a line drawn from horizon to emitter of these photons to ship. That light would never have reached the ship in the first place it would have gone perpendicular out to another destination. The diagram does not show the light emitted from side of the emitter which is directly opposite the black hole which would have been the light reaching the ship in the first place. The effect of these photons reaching the ship which never would have without the warping of space create a strange image for the ship, where they can see sides of the emitter that are not normally visible. This has nothing to do with time dilation, so why was it brought up? I don't follow that it explains why the coordinate time extends into infinity, because I see no way that it could do that? That's like saying that if I fire a bullet so that it makes a circuit around the earth and then strikes a target beside me, that it was time dilation that caused the bullet to take so long to hit the target. There was time dilation, surely, but not like that (it was so relativistically small in the bulletearthtarget example that it was ignored). The light emitted directly away from the horizon has no curvature at all. It follows a straight path, as in, the "null geodesic" aka "light trajectory" is straight. Light having only one speed is not at liberty to change it's trajectory by going faster or slower, so when I say the light is going straight I don't mean the light is trying to go straight but it's path is altered by gravity, I mean it's going straight. I wonder also, would the light that made a circuitous route around the black hole before reaching the ship be less redshifted than the light coming straight on?
Because it has the same underlying cause, the curvature of spacetime.
Do you want to see the maths ? It's straightforward to show that coordinate time will diverge, and just as straightforward to show the proper time is finite and well defined for a free fall into a black hole. The same goes for null geodesics  it is easy to show that they are all curved in the vicinity of a black hole.I don't follow that it explains why the coordinate time extends into infinity, because I see no way that it could do that?
This is where you are wrong. All null geodesics in the vicinity of a black hole are curved; there is no such thing as "straight" lines in such a curved spacetime. That's the whole point. Remember that in reality we are dealing with a 4dimensional manifold, not a flat diagram; even if you find a trajectory that looks straight in space, it will still be curved in spacetime. Hence time dilation.The light emitted directly away from the horizon has no curvature at all.
It isn't. A flat null geodesic would imply a flat spacetime, and hence no gravity. Clearly, that is not the case here.It follows a straight path, as in, the "null geodesic" aka "light trajectory" is straight.
Correct.Light having only one speed is not at liberty to change it's trajectory by going faster or slower
But it doesn't. Not in the vicinity of a black hole.so when I say the light is going straight I don't mean the light is trying to go straight but it's path is altered by gravity, I mean it's going straight.
The redshift is the exact same, since, at least in Schwarzschild spacetime, it depends only on the radial coordinate.I wonder also, would the light that made a circuitous route around the black hole before reaching the ship be less redshifted than the light coming straight on?
Here is a nice little interactive simulator to illustrate geodesics in a curved spacetime :
http://www.adamtoons.de/physics/gravitation.swf
Just set the initial velocity to 1 ( = speed of light ) to simulate a photon. Play around with it a bit, it is really quite instructive.
Here is a more explicit version, which shows the actual geodesics themselves :
Wolfram Demonstrations Project
I am not sure if you understand that, but it needs to be noted that null geodesics are what light always travels on; it cannot do anything else. It is the null geodesics in spacetime that determine the trajectories of light, and not the other way around, so if the null geodesics are curved, then that is how light will propagate.
P.S. I even found a version in 3D for you ( note : this is actually a Kerr black hole, hence the two horizons ) :
http://demonstrations.wolfram.com/3D...ackHoleOrbits/
Oh, I know that the proper time of the falling observer is finite. That's well established, and I understand how the null geodesics work. It's all very straightforward. Except for the part that I think you are saying, where the null geodesics are the cause of the divergence of the coordinate time. I don't really see the coordinate time as diverging so much as being complicated when compared to a very basic understanding of time (the old idea that time was the same everywhere), and I definitely see no causation in the correlation.
For your claim that all null geodesics are curved, I need to see some examples, some math, some diagrams and the like. From my understanding, all paths of light are affected, but I don't see how gravity can possibly curve a light trajectory that is moving directly away from it. I understood it to be that the light emitted directly away from the black hole travels the normal distance (and thus in the normal time), but the effect of gravity upon it causes the redshifting. Does this specific trajectory of light, directly away from the black hole, take longer to go from emitter to distant observer than it would without the presence of the black hole behind it?
How can this be represented in terms of the time dilation created between a person in standing in a box, being accelerated through space at a speed as near to the speed of light as can be imagined, and a person "at rest"?
Thanks for the links! I'll be messing around with these for a bit, and will likely have more questions. Also, thank you for keeping up this conversation. There's not much else in the forum catching my eye and reading the responses in this thread is a highlight of my day, or night as it were.
Argh, I wish the adamtoons model would allow for more than just 0.75 likeness of a black hole. I'm also unsure of the trajectory. It looks like my light ray travels past it along the outside. I can't seem to get a version that starts near the mass and moves directly away from it. It's a pretty neat model nonetheless. We'll see what the other two have to offer me.
Last edited by Ninja Pancakes; December 1st, 2013 at 05:18 AM.
This is a problem with General Relativity, and it's a semantic or pedagogical issue rather than with the theory itself. The fact of the matter is that all geodesics (null or not) are straight lines. To say that they are curved in the presence of spacetime curvature is nonsense, though it does provide some indication of the effect of spacetime curvature on how we view the world. However, because it is ultimately nonsense, one shouldn't take it too seriously, or else one may draw conclusions that simply aren't true.
With regards to gravitational time dilation, a feature of the Schwarzschild solution is that it is stationary (it is actually static, but stationary suffices for our purpose). This means that the spacetime can be sliced into metrically identical threedimensional spaces (the spacetime has translational symmetry along a timedirected Killing vector field). Therefore, suppose one has a lightlike trajectory from an emitter to an observer. Due to the translational symmetry, one can translate the entire lightlike trajectory to a future time, thus describing two lightpulses from the emitter to the observer. Time dilation means that the time between the pulses at the emitter is not the same as the time between the same pulses at the observer. The precise details of the path is irrelevant because we are only considering the time intervals at the two endpoints and the translational symmetry of the intervening spacetime.
Firstly, KJW is of course right when he says that the statement "geodesics are curved" is ultimately meaningless; I said that only to connect up with the diagrams I provided. The meaning was of course that if one was to embed or project such geodesic onto a flat plane in a suitable fashion, the result will not be a straight line. To be mathematically rigorous here : the proper definition of a geodesic is a curve that paralleltransports its tangent vector along itself. In other words, the covariant derivative of the tangent vector with respect to itself vanishes :
Written in a coordinate basis, this becomes
or, a little simplified :
This is precisely the geodesic equation. Now, in a flat spacetime, and only in a flat spacetime, do the Christoffel symbols vanish identically, and the equation simplifies to
the solutions to which are of course straight lines. If spacetime is not flat, the correction terms remain, and the resulting set of equations becomes pretty nontrivial  the solutions to those are not in general straight lines of the form y=ax+b, but parametrised curves which, if embedded into Euclidean space, aren't straight at all. I have linked to visualisation tools for all of this.
It is curved in spacetime, even if it appears straight in space. Be careful with the distinction between the two.but I don't see how gravity can possibly curve a light trajectory that is moving directly away from it.
Yes, because the trajectory is much "longer" than it would be in flat space, though a direct comparison is not very meaningful.Does this specific trajectory of light, directly away from the black hole, take longer to go from emitter to distant observer than it would without the presence of the black hole behind it?
The world line of the accelerated observer is curved in spacetime, the world line of the inertial observer isn't ( gravity aside ).How can this be represented in terms of the time dilation created between a person in standing in a box, being accelerated through space at a speed as near to the speed of light as can be imagined, and a person "at rest"?
Feel free to keep asking away.Thanks for the links! I'll be messing around with these for a bit, and will likely have more questions.
The time dilation between the person in the box, and the person "at rest" is still real though. The one at rest will have experienced more proper time than the one in the box. And this difference approaches infinity as the acceleration of the box causes it's speed to approach the speed of light, right?
Indeed it is.
Correct. Given two fixed events in spacetime, it is always the observer travelling inertially between those events who experiences the longest proper time.The one at rest will have experienced more proper time than the one in the box.
Technically, what happens is that, given a large enough acceleration, the "boxed" observer's proper time will asymptotically approach zero. Note that proper time dilation in this scenario is solely a function of acceleration, not relative speed.And this difference approaches infinity as the acceleration of the box causes it's speed to approach the speed of light, right?
It is true that if one projects a straight line in fourdimensional spacetime onto a threedimensional space, then the resulting curve in the threedimensional space will in general be curved in a real sense. For example, the orbits of planets around the sun, which are straight lines in fourdimensional spacetime, are genuinely curved as trajectories in threedimensional space. The curvature of the threedimensional space does not make these trajectories straight.
I forgot to mention that, in order to obtain specifically null geodesics for light, one needs to apply the additional constraint that
Physically this demand amounts to the norm of the 4velocity ( with respect to our affine parameter ) to remain constant  which of course makes sense, since photons can't accelerate.
What is the equivalence of acceleration to gravity at the event horizon of a black hole?
At first that was confusing, because I was thinking it would have to be moving FTL in order to escape the event horizon at all..... but then I remembered that "escape velocity" doesn't necessarily mean the velocity required to move above a certain point. It's just the velocity at which you never fall back down again.
So, really a space ship traveling near C could pass through the event horizon, and come back out again, and if it had super amazing rocket thrusters and fired them just in time while it is outside the event horizon to give itself a push as it's getting ready to fall back in ..... it might actually escape. Right?
Even failing that, it could transmit data to observers outside the black hole while it's outside the event horizon. Maybe in a billion years or so humanity will be able to get a probe to do that. Would be interesting to see what it tells us. Or.... maybe not. Maybe it would just say there's nothing special about the interior of the event horizon. :(
Another interesting issue, then, is that a beam of light traveling straight outward from a black hole could travel past the event horizon, right? It would curve around and fall back in soon after (or just get red shifted into oblivion at 0 hertz frequency). But, if an observer were traveling near enough to the event horizon, they might intercept that beam of light while it's still moving outward, and see the interior of the black hole.
This presents an interesting test for the whole theory of black holes. Some black holes should still be perceived to be emitting light if they're close enough to us, because the light may reach us before it manages to turn all the way around (or fade to zero hertz frequency  whichever it is.)
My understanding is that the event horizon is the point at which a photon, travelling directly away from the black hole neither escapes nor falls toward the singularity. Nothing passing the event horizon can ever go fast enough to come back out again. The photosphere is farther from the singularity, and is the point where light orbits a black hole.
That's the mistake in our thinking. The event horizon is the point at which the formula for escape velocity would yield C as its value.
Escape velocity  Wikipedia, the free encyclopedia
Schwarzschild radius  Wikipedia, the free encyclopedia
But you don't need to be traveling at escape velocity in order to move out/up. You will fall back down later, but you can temporarily move outside of the boundary. Or .... at least that's what it looks like. So light could leave the event horizon temporarily, just not permanently. But if it is detected/absorbed by an object while it's outside the event horizon, then maybe it doesn't have to fall back in anymore?
Best to wait for one of the actual physicists to confirm or deny that, before taking it to heart, though. I'm just speculating.
What do you mean by "you don't need to be traveling at escape velocity in order to move out/up"? Instead of thinking about the escape velocity, perhaps you should be thinking about the force vectors. "How much force is required to have a positive acceleration at the event horizon?" Should be the question you need answered here. In order to throw a ball upward from the earth, you need to exert more force upon the ball away from the earth, than gravity exerts on the ball toward the earth. So you have moved the ball away from the earth, and it fell back down, but that is because the force of gravity is constant, and the force you exerted was impulsory. Light either has a force which is constant, maintaining it's velocity, or is exempt from such a concept due to its unique characteristics. The latter would seem to be the case. A thing of mass however, would need infinite force to even approach the speed of light. This suggests you would need infinite force to even maintain at the event horizon once you have reached it, which is impossible.
Last edited by Ninja Pancakes; December 2nd, 2013 at 09:19 PM.
So now I'm curious why that is, if by chance the explanation isn't too complicated to state here.
Schwartzchild radius is the point at which the formula for escape velocity yields the value of C. Based solely on that, it would stand to reason that a beam of light emitted just inside that radius would be able to travel outward for a time before it turns around and falls back in. (And presumably an object moving at or near C would be able to do this also.)
But clearly there's more to it. In the first place, the formula for escape velocity is a Newtonian formula that doesn't necessarily take Relativity into account.
Since incremental changes in speed are observed differently by observers in different frames of motion, that certainly should affect things. I mean that one observer sees the object accelerated by 5 m/s, while perhaps another might think it was accelerated by 10 m/s. And they're both right, just looking at it from different frames. Or like how if a fast moving proton traveling (from Earth's perspective) at .999999998 C gets accelerated to .999999999 C  locally that would seem like an incredibly large increase in speed.
Is that where I should be looking in order to understand why this is so?
Don't attempt to apply Newtonian physics to a black hole; it's a recipe for disaster. Also, don't attempt to extract physical information from coordinate measures such as speed, because they are generally dependent on the observer ( as you rightly note ).
The answer is best understood in terms of the geometry of spacetime itself, specifically its geodesic structure. For any given event within the event horizon, all possible geodesics will always end at the central singularity, and never cross the event horizon  needless to say, this can be shown mathematically. If you need a more intuitive analogy, you could say that, once you cross the horizon, space and time switch places for you  no matter where you turn, you will always find a singularity in your future. No matter which direction you choose to move into, you will always get closer to the singularity, never further away. Therefore there are of course no trajectories which take you back out of the horizon, even just for short time  and good thing that is, because it would violate causality since that would enable you to bring information from beyond the horizon out with you.
Hrum, I thought my force vector explanation was pretty good.
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