# Thread: Black Hole Event Horizon

1. Two questions:

#1>
If at position A an object only experiences the net gravity.

Than does this follow from that?

If we say that position A is a tiny distance ( say a Planck Length ) inside the event horizon of a black hole. If we then add a high gravity object ( Neutron Star or Quark Star ) on the outside of the event horizon ( say a Planck Length ).

Shouldn't the two opposing gravity sources reduce the net gravity of the object at point A? Effectively shifting the location of the event horizon. The object at point A could then be able to move through and outside of the original location of the event horizon without a need for FTL?

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#2>
Part A:
If the escape velocity is defined as the velocity needed to escape a given gravity with no further propulsion.

Than does it follow from that?

If one has additional propulsion than then there is a different escape velocity. If I have a constant propulsion I don't need to be going as fast initially in order still be able to escape the same gravity.

Part B:
If inside the event horizon is inescapable because the escape velocity is greater than the speed of light, and you can not go faster than the speed of light.

Than does it follow from part A that?
If an object has additional propulsion it's effective event horizon will be shifted. It would be slightly different than the object with no additional propulsion.

2.

3. Gravity outside the event horizon will distort the event horizon, so the object will still either be inside the horizon or not. Now, whether or not this change could liberate material from the black hole is beyond me (and I suspect would probably require a theory of quantum gravity to make a prediction, as opposed to a guess).

As for escape velocity, that's why it's a velocity instead of an acceleration. Assuming you're not experiencing any significant drag forces (from an atmosphere or a solar sail, for example) it doesn't matter what your acceleration is. (The best current engines in terms of delta-v, ion engines, have very low accelerations.)

4. Originally Posted by IamIan
If at position A an object only experiences the net gravity.
There is no such notion as "net gravity" in GR, since we are not dealing with forces here; we are only dealing with the geometry of space-time, which is the same for all observers.

Effectively shifting the location of the event horizon. The object at point A could then be able to move through and outside of the original location of the event horizon without a need for FTL?
This is a very complex scenario, but you are correct in saying that the geometry of the event horizon will change; it will more than likely create a "bulge". The object itself, once inside the horizon, will always stay inside the horizon, even if the horizon itself changes.

If one has additional propulsion than then there is a different escape velocity.
No, escape velocity depends only on mass and radial distance ( in Schwarzschild space-time ).

If inside the event horizon is inescapable because the escape velocity is greater than the speed of light, and you can not go faster than the speed of light.
You can't escape because all time-like geodesics inside the event horizon terminate at the singularity. It is because of the geometry of space-time itself.

If an object has additional propulsion it's effective event horizon will be shifted.
No, the event horizon is the same for all observers.

5. Originally Posted by IamIan
Effectively shifting the location of the event horizon.
A lot of work has been done on the simulation of black hole mergers. As the black holes approach each other, the event horizons are distorted (pulled out towards the other black hole).

I can find the really simple animation I saw before, but there is one example here: News (there are many, many more).

If one has additional propulsion than then there is a different escape velocity. If I have a constant propulsion I don't need to be going as fast initially in order still be able to escape the same gravity.
It doesn't matter how or when you accelerate; you need to reach the escape velocity.

6. I am confused.
I will try and explain my points of confusion bellow.

Originally Posted by MagiMaster
As for escape velocity, that's why it's a velocity instead of an acceleration. Assuming you're not experiencing any significant drag forces (from an atmosphere or a solar sail, for example) it doesn't matter what your acceleration is.
Originally Posted by Strange
It doesn't matter how or when you accelerate; you need to reach the escape velocity.
Originally Posted by Markus Hanke
No, escape velocity depends only on mass and radial distance ( in Schwarzschild space-time ).
So what it seems is being described here by several people ( correct me if I misunderstood what was intended ) , is that the part I referenced about the escape velocity that normally assumes no additional propulsion, is being excluded. Propulsion or no Propulsion , either way makes no difference at all , not one additional Planck Length of distance will be traveled ( seems to be the claim ) ?????

Why bother putting that stipulation into the definition of escape velocity if it makes no difference if you have additional propulsion or not?

These claims also seem to contradict every rocket launch ( or ball thrown up ) . It goes up, it makes a length progress of distance. It does that even before it ever reaches escape velocity. Even if it never reaches escape velocity.

I also don't see why the following doesn't work. Say I traveled at a tiny velocity ( 1mm per year ) , but traveled in a direction away from the earth ( with just enough propulsion to maintain that velocity ). Why would that not escape if the propulsion is maintained ( even with a slower velocity )??

It also seems odd to me, because I always see that ( no additional propulsion ) stipulation included in text book definitions of 'escape velocity'.

So this is confusing.

Originally Posted by Markus Hanke
The object itself, once inside the horizon, will always stay inside the horizon, even if the horizon itself changes.
Originally Posted by Markus Hanke
I have no doubt though that it might be possible for particles to quantum mechanically tunnel through the event horizon, but that isn't the main mechanism of Hawking radiation.
I'm confused. Didn't you previously list that above as a viable method for objects on the inside to get out?

Which leads me to another question.
If enough objects closer to the singularity used your tunnel method to escape wouldn't that be able to reduce the event horizon radius of the whole black hole. If an object was further on the outside of the radius when the radius was reduced. Why wouldn't that object now find itself outside the now reduced radius of the event horizon?

Originally Posted by Markus Hanke
If inside the event horizon is inescapable because the escape velocity is greater than the speed of light, and you can not go faster than the speed of light.
You can't escape because all time-like geodesics inside the event horizon terminate at the singularity. It is because of the geometry of space-time itself.
I'm confused.

I thought you already said the escape velocity was set by the mass and the radius. Doesn't that mean , if I have greater than that escape velocity than I escape?

And despite the time-like geodesics you list now , you previously listed a viable method to escape?

Originally Posted by Markus Hanke
If an object has additional propulsion it's effective event horizon will be shifted.
No, the event horizon is the same for all observers.
That seems to be very different from what I've heard Leonard Susskind say about different observers and the event horizon. He wasn't talking about propulsion, but he did talk about the event horizon not being the same for all observers. I'll have to find that and go through it again ( maybe I missed something in his context ).

7. Ok, I'll leave the black hole details to others, but I can try and explain the escape velocity thing.

First, away from any massive objects, you just coast along at a fixed speed. You'll never slow down. The same is true near a massive body if you're in a circular orbit. So acceleration only controls how soon you get there, not whether or not you arrive. And in deep space, you only need the tiniest bit of acceleration to get anywhere (eventually).

In the presence of a gravity well though things are slightly different. The nature of rockets (due to Newton's laws of motion) means that what really limits you is how much you're able to change your velocity by. More acceleration doesn't change this, it just makes it quicker.

Escape velocity is based on this limit. If you fired an object from the center of the Earth outward, it would never return to Earth if it was faster than that. If it were slower, it eventually would. It doesn't matter how fast it was accelerated up to that speed, only that it was going that fast when it stopped accelerating. This is all based on some simplifications though like assuming that everything is a point mass and that there is no atmosphere or other bodies involved.

Now, in this simplification, the further away you start, the less gravity there is and the slower you'd need to be going to escape. In fact, if the hypothetical bullet were travelling at exactly the escape velocity, it would come to stop right at infinity. It's speed along the way would be the escape velocity from that particular distance.

Because of the changing gravity, travelling at any fixed speed would be difficult and would require less and less acceleration to maintain. Eventually, you'd get to the point where your fixed speed was greater than the remaining escape velocity and you could turn off your engines and coast out to infinity (though you would still start to slow down a little). If you summed up all the acceleration changes up to that point, it'd match the escape velocity, no matter what fixed speed you tried to keep. (To keep a fixed speed out to infinity would require an infinite amount of energy, but to any given distance would need less and less extra.) Being able to accelerate is why rockets don't need to suddenly jump right up to the escape velocity or even need to ever reach the calculated escape velocity really, which is why that's part of the definition of escape velocity.

Now things actually launched from Earth have an additional problem: drag from the atmosphere. If you fired a ball upwards at exactly the escape velocity, the atmospheric drag would sap enough speed that it might not even clear the atmosphere.

BTW, one definition of the event horizon of a black hole is the point where the escape velocity (from that point) exceeds the speed of light.

Edit: I'm not exactly sure, but it may be that the usual quoted escape velocity for Earth is calculated from sea level instead of from the center. Either way, it does depend on the starting altitude.

8. Originally Posted by IamIan
I'm confused. Didn't you previously list that above as a viable method for objects on the inside to get out?
Yes, you are right, I did say that. However, I took the current discussion to be about classical objects, so I didn't consider such tunnelling effects here. Bear in mind also that such tunnelling will happen only under a very specific set of circumstances ( sorry, can't find the paper at the moment ).

If enough objects closer to the singularity used your tunnel method to escape wouldn't that be able to reduce the event horizon radius of the whole black hole.
Tunnelling would happen only in the immediate vicinity of the event horizon; also, this would be a very rare event, so even over the entire lifetime of a black hole this wouldn't have any noticeable effect.

I thought you already said the escape velocity was set by the mass and the radius. Doesn't that mean , if I have greater than that escape velocity than I escape?
The question is moot, because you can't accelerate to superluminal speeds.

And despite the time-like geodesics you list now , you previously listed a viable method to escape?
It's not viable at all. It is a peculiar quantum effect which happens to particles under very specific circumstances very close to the event horizon. It is not a viable method for you or me to escape a black hole. Like I said, I took this discussion to be about classic, macroscopic objects, and those never escape; quantum objects are special in that one has to account for quantum effects, which isn't reflected in the notion of classical world lines and geodesics.

That seems to be very different from what I've heard Leonard Susskind say about different observers and the event horizon.
I think you might confuse that with Rindler horizons, which are event horizons for uniformly accelerated observers. Those are not the same as BH event horizons, and they are observer dependent. The Schwarzschild radius itself is a relativistic invariant.

9. One thing I forgot to mention about acceleration is that you do need enough to overcome the acceleration due to gravity during liftoff or you won't get anywhere in the first place. (Once in orbit, you can use a little bit of acceleration over a long time just as well.) After that, more acceleration only gets you there sooner, but won't actually get you any further.

10. Originally Posted by MagiMaster
Ok, I'll leave the black hole details to others, but I can try and explain the escape velocity thing.
Thanks.

Originally Posted by MagiMaster
Because of the changing gravity, traveling at any fixed speed would be difficult and would require less and less acceleration to maintain.
How is a fixed speed an acceleration in that direction of travel (away from the earth)?
Although when talking about both I guess velocity is the better term to use.

I would have thought no change in speed or velocity would mean no acceleration???

Originally Posted by MagiMaster
Now things actually launched from Earth have an additional problem: drag from the atmosphere. If you fired a ball upwards at exactly the escape velocity, the atmospheric drag would sap enough speed that it might not even clear the atmosphere.
But even without enough to escape ( out to infinity , etc ) it still travels some distance up before falling back.

Why would the escape velocity at that peak height ( further away from earth ) not be less than it is at the bottom ( before launch and closer to earth )????

Originally Posted by MagiMaster
BTW, one definition of the event horizon of a black hole is the point where the escape velocity (from that point) exceeds the speed of light.
??? Markus already objected against that when I wrote a version of that definition above ??? Then he seems to agree with it later ?? I don't understand. How is the same kind of definition both wrong and right at the same time??

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Originally Posted by Markus Hanke
Originally Posted by IamIan
Originally Posted by Markus Hanke
Originally Posted by IamIan
If inside the event horizon is inescapable because the escape velocity is greater than the speed of light, and you can not go faster than the speed of light.
You can't escape because all time-like geodesics inside the event horizon terminate at the singularity. It is because of the geometry of space-time itself.
I thought you already said the escape velocity was set by the mass and the radius. Doesn't that mean , if I have greater than that escape velocity than I escape?
The question is moot, because you can't accelerate to superluminal speeds.
I don't understand??
So what exactly are you then objecting to about my initial claim above?
It seems like , your last statement just agrees with my initial claim , and what you seemed to be objecting against??

Originally Posted by Markus Hanke

Originally Posted by IamIan
Originally Posted by Markus Hanke
No, the event horizon is the same for all observers.
That seems to be very different from what I've heard Leonard Susskind say about different observers and the event horizon.
I think you might confuse that with Rindler horizons, which are event horizons for uniformly accelerated observers. Those are not the same as BH event horizons, and they are observer dependent. The Schwarzschild radius itself is a relativistic invariant.
No , it was about BH event horizons.

Leonard Susskind on The Black Hole Wars - YouTube

Around ~18 minutes and onward he starts describing very significant differences different observers would have of that BH event horizon.

He even says in that video the person crossing the horizon experiences nothing unusual. That seems to mean to me that , they would feel their foot even if their foot is inside the event horizon and their head/brain had not yet passed through the event horizon. Not feeling your foot would count as unusual.???

11. Originally Posted by IamIan
So what exactly are you then objecting to about my initial claim above?
It seems like , your last statement just agrees with my initial claim , and what you seemed to be objecting against??
I'm a little confused what you are trying to say - I agree of course that they can't escape, it's just that I prefer to explain that geometrically instead of in terms of speed. The reason is simply that I am not convinced that a tachyon could escape the event horizon; this is often asserted, but I have yet to see an actual mathematical proof of that. I have a suspicion that even at superluminal speeds an escape might not be possible, though I haven't done the maths here yet because I don't know what the geodesic of a tachyon would look like. It is thus best not to try and explain this in terms of speed.

He even says in that video the person crossing the horizon experiences nothing unusual.
Of course not. The event horizon is not a physical object, and space-time there is smooth and regular.

That seems to mean to me that , they would feel their foot even if their foot is inside the event horizon
An observer in free fall would fully cross the event horizon long before the nerve impulses have time to travel all the way to the brain; this is point is thus moot. He would indeed not notice anything special. It is only an issue if you imagine an observer somehow "hovering" very close to the horizon ( ignore tidal forces for now ), and then letting his foot dangle across the horizon ( note - actually doing this would be close to impossible ). The nerve signals from his foot would in that case not be able to reach his brain, and he will no longer feel it or be able to pull it back out ( this experiment is definitely not recommended ! ).

Not feeling your foot would count as unusual.???
Like I said, so long as he is in free fall he won't experience anything unusual, he'll just fall through.

Around ~18 minutes and onward he starts describing very significant differences different observers would have of that BH event horizon.
The point is that all observers agree that there is a horizon, and they also agree as to where the horizon is, because those are invariants. I don't notice Prof Susskind saying any differently. The only major disagreement here would be time ( and possibly length ) measurements - proper times and coordinate time will not agree for different observers. Have a look here :

Journey into a Schwarzschild black hole

12. Originally Posted by IamIan
Originally Posted by MagiMaster
Ok, I'll leave the black hole details to others, but I can try and explain the escape velocity thing.
Thanks.

Originally Posted by MagiMaster
Because of the changing gravity, traveling at any fixed speed would be difficult and would require less and less acceleration to maintain.
How is a fixed speed an acceleration in that direction of travel (away from the earth)?
Although when talking about both I guess velocity is the better term to use.

I would have thought no change in speed or velocity would mean no acceleration???

Originally Posted by MagiMaster
Now things actually launched from Earth have an additional problem: drag from the atmosphere. If you fired a ball upwards at exactly the escape velocity, the atmospheric drag would sap enough speed that it might not even clear the atmosphere.
But even without enough to escape ( out to infinity , etc ) it still travels some distance up before falling back.

Why would the escape velocity at that peak height ( further away from earth ) not be less than it is at the bottom ( before launch and closer to earth )????

Originally Posted by MagiMaster
BTW, one definition of the event horizon of a black hole is the point where the escape velocity (from that point) exceeds the speed of light.
??? Markus already objected against that when I wrote a version of that definition above ??? Then he seems to agree with it later ?? I don't understand. How is the same kind of definition both wrong and right at the same time??
To maintain a fixed speed in deep space requires no effort. But if you're in the gravity well of a massive body, then you need to accelerate in your frame of reference enough to overcome the acceleration due to the gravity of the massive body. If those two exactly cancel out, you would maintain a fixed speed away from it. From the point of view of the massive body your net acceleration would then be 0. (BTW, if you include the rocket exhaust, you'd see the center of mass of the rocket and exhaust falling back to the planet at the normal rate, IIRC.)

And yes, if you hit a speed somewhere below escape velocity, you'll eventually fall back down (again, assuming no other bodies to interact with).

And I did say that escape velocity does depend on altitude. The commonly quoted escape velocity is calculated from sea level (looking it up, it couldn't be from the center as that'd be infinite under the given simplifying assumptions), but if you check the formula there is an r term in there, so the higher up you go, the lower it gets. The Wikipedia article on this is fairly good. So a ball fired upwards at escape velocity would lose some speed due to drag and from then on, it's speed would be lower than escape velocity everywhere along its path.

13. Two other new questions occurred to me out of curiosity:

#1>
Does the event horizon radius shrink if a moving BH crosses over a fixed position in space worm hole connecting two points in the universe? Even if it is only shrinking in location A as it expands in location B.

#2> If a BH grew large enough to encompass both points of a worm hole how would that contort the geometry of the BH's Event Horizon Shape?

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Originally Posted by Markus Hanke
I'm a little confused what you are trying to say -
I was trying to describe BH / Event horizon in terms of the needed ( but not attainable ) escape velocity/speed.

Originally Posted by Markus Hanke
I agree of course that they can't escape, it's just that I prefer to explain that geometrically instead of in terms of speed. The reason is simply that I am not convinced that a tachyon could escape the event horizon; this is often asserted, but I have yet to see an actual mathematical proof of that. I have a suspicion that even at superluminal speeds an escape might not be possible, though I haven't done the maths here yet because I don't know what the geodesic of a tachyon would look like. It is thus best not to try and explain this in terms of speed.
I'm confused. Doesn't the bold part above disagree with the prior claim of the escape velocity being determined by mass and radius?

Originally Posted by Markus Hanke
The point is that all observers agree that there is a horizon, and they also agree as to where the horizon is, because those are invariants. I don't notice Prof Susskind saying any differently.
If all observers agree where the horizon is, does that mean when a BH itself traveling closer to the speed of light is an exception and does not experience the changing of apparent length ( to some observers ) that other objects do in those conditions?

Originally Posted by Markus Hanke
The only major disagreement here would be time ( and possibly length ) measurements - proper times and coordinate time will not agree for different observers. Have a look here :

Although this comment of yours confuses me a bit. If there is a possible observer disagreement in length. Why would that not include the length of the BH radius or the length of the BH diameter?

And a claim in the link causes me a bit of confusion as well. It describes the space is not just bent or curved, but the space itself is moving / falling into the the black hole. Kind of almost like the exact opposite of dark energy expansion of space.

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Originally Posted by MagiMaster
But if you're in the gravity well of a massive body, then you need to accelerate in your frame of reference enough to overcome the acceleration due to the gravity of the massive body. If those two exactly cancel out, you would maintain a fixed speed away from it.
??? So it seems you are claiming acceleration means no change in speed ???
??? Same speed = Acceleration ???

This disagrees with every type of definition of acceleration I have ever previously read or been told.

All the definitions I've ever read of acceleration have it requiring a change in speed, and if there is no change in speed , than there is no acceleration.

Originally Posted by MagiMaster
And yes, if you hit a speed somewhere below escape velocity, you'll eventually fall back down (again, assuming no other bodies to interact with).
Why doesn't this also apply to the escape velocity of an event horizon / BH?

You can't do the escape out to infinity ( needed speed is too fast ). But one or two Planck Lengths is far less than out to infinity. Why doesn't it go like the ball that was propelled up , but didn't have enough to reach escape velocity ( go up, and then fall back )?

14. Do you think a ball moving directly away from the Earth will maintain a fixed speed on its own? No. Gravity will accelerate it back down. You need to apply a force to counteract that. In a rocket, that force comes from throwing exhaust gasses out the back really quickly. Since F=ma, this is still an acceleration. It's just that the net acceleration is zero (-g from gravity and +g from the rocket equals zero). Again though, if you take the rocket and exhaust together, their center of mass will accelerate downward due to gravity. It's only the rocket itself that's coasting along at a fixed speed, and then only if it's acceleration perfectly balances against the acceleration due to gravity. Any more or less and it will change velocity.

As for the event horizon, imagine you shot a beam of light upwards. If it got even a little out of the event horizon, it would continue to move at c and escape altogether. Since light can't get out of the event horizon, nothing going slower than light can either. But to get the full story, you need GR (if not some theory of quantum gravity) so I'll leave further explainations to others.

15. Originally Posted by MagiMaster
Do you think a ball moving directly away from the Earth will maintain a fixed speed on its own? No. Gravity will accelerate it back down. You need to apply a force to counteract that. In a rocket, that force comes from throwing exhaust gasses out the back really quickly. Since F=ma, this is still an acceleration. It's just that the net acceleration is zero (-g from gravity and +g from the rocket equals zero). Again though, if you take the rocket and exhaust together, their center of mass will accelerate downward due to gravity. It's only the rocket itself that's coasting along at a fixed speed, and then only if it's acceleration perfectly balances against the acceleration due to gravity. Any more or less and it will change velocity.
You are just using the term acceleration completely differently than any text book or physics professor or anyone else I have ever known , when you claim things like constant speed is acceleration.

As for what I think about the ball moving away from the earth. I think the ball moving up has a negative acceleration , aka a deceleration as it's speed changes ( slows ). At least that is how everyone else I've known has ever used the term acceleration.

I'm curious , is there any dictionary that defines acceleration this way you are using it? None that I've found do. All the one's I've found define acceleration like the following:

Originally Posted by merriam-webster
Acceleration: physics : the rate at which the speed of a moving object changes over time.
??If the speed is not changing I don't see how that fits the definition of the term acceleration??

Originally Posted by MagiMaster
As for the event horizon, imagine you shot a beam of light upwards. If it got even a little out of the event horizon, it would continue to move at c and escape altogether. Since light can't get out of the event horizon, nothing going slower than light can either. But to get the full story, you need GR (if not some theory of quantum gravity) so I'll leave further explainations to others.
Seems to me it shouldn't escape altogether , just like the ball that didn't have escape velocity wouldn't escape either.

Maybe I'm missing something. But if Escape velocity is the amount needed to reach infinitely far away, not having that much just means you can't reach infinity far away. Just like the ball that didn't have enough. It goes some finite maximum distance, and that's it. It falls back , etc.

In the specific case of light which doesn't slow down the same as other objects.

It also seem to me , athat in the case of light the Kinetic Energy it looses does not result in a loss of speed, but it is the kinetic energy that is equal to the potential energy at infinity that is needed for an escape velocity. So it wouldn't have enough KE. Even though it's speed isn't changing.

16. Originally Posted by IamIan
I was trying to describe BH / Event horizon in terms of the needed ( but not attainable ) escape velocity/speed.
You would just be making your life really, really hard, because speed is observer dependent. That is why in the literature the event horizon is given as a function of radius, which is an invariant.

I'm confused. Doesn't the bold part above disagree with the prior claim of the escape velocity being determined by mass and radius?
No, because the kinematics of a tachyon wouldn't work the same way as those of "normal" particles. You can't just plug superluminal speeds into standard formulas and expect something meaningful to come out of it. I remember having once read a paper about tachyonic kinematics, and things were very different from how they work for normal speeds.

If all observers agree where the horizon is, does that mean when a BH itself traveling closer to the speed of light is an exception and does not experience the changing of apparent length ( to some observers ) that other objects do in those conditions?
This is a very complicated scenario, because space-time is not flat in the vicinity of a black hole, so you cannot just use the normal formulas from Special Relativity. What you describe is modelled by a metric called the Aichelburg-Sexl ultraboost :

Aichelburg

The resulting space-time is highly non-trivial, and involves something called "axisymmetric vacuum pp-waves", which is a form of gravitational radiation. I am not very familiar with this particular solution to the field equations; there is still length contraction involved here, but only in one direction, and not in the same way as in Special Relativity.

If there is a possible observer disagreement in length. Why would that not include the length of the BH radius or the length of the BH diameter?
See above - it will, but only along one direction, and not in the same way as in SR.

And a claim in the link causes me a bit of confusion as well. It describes the space is not just bent or curved, but the space itself is moving / falling into the the black hole
What you are referring to is the "space-time waterfall" analogy, which results from describing space-time in and around a black hole with a set of coordinates called the Gullstrand-Painleve metric. Please do take note that this is just an analogy, just like the rubber sheet visualisation is just an analogy. You can find some good animations about this here :

Waterfall

Physically, space-time itself has no kinematic properties, so it doesn't actually "fall" into a black hole; the analogy just arises out of a certain choice of coordinates which "carry along" an orthonormal tetrad, thereby giving the illusion of motion. It's just a coordinate effect. The details are really quite complicated to explain, but the physical content is the exact same as when one uses the Schwarzschild metric to describe the space-time. This demonstrates one of the key features of GR very well - one is free to choose any suitable system of coordinates one likes without effecting the physical laws; a feature called general covariance.

17. Originally Posted by IamIan
You are just using the term acceleration completely differently than any text book or physics professor or anyone else I have ever known , when you claim things like constant speed is acceleration.
I think you need to remember that both acceleration and velocity are 4-vectors in space-time, and not scalar quantities. These two 4-vectors are always perpendicular to each other ( i.e. their inner product vanishes everywhere ). What this means is that an object "hovering" close to a black hole actually traces out a curved world line in space-time, which means it must necessarily be accelerated, even if its 4-velocity is constant everywhere.

Don't make the mistake of trying to interpret Minkowski 4-vectors in a globally curved space-time in terms of Newtonian physics, or else much unnecessary confusion will ensue.

18. The thing about light is that the moment a photon is created, it immediately travels at C without any acceleration. So maybe it would make more sense to you to think about this in terms of escape velocity. A photon can't decelerate, only wavelength shifted.

But in any case, thinking about event horizons in terms of escape velocity only gives you the radius of the event horizon of a non-rotating black hole. It doesn't represent what is really going on. Aside from that, event horizons represent the radius before which all vectors from a given particle point away from the event horizon, from an outside perspective. From that particle's point of view, there is no outside, no direction that point to the outside. So even light, which always follows geodesics from it's own perspective, would never near the event horizon again.

19. Originally Posted by Markus Hanke
What this means is that an object "hovering" close to a black hole actually traces out a curved world line in space-time, which means it must necessarily be accelerated, even if its 4-velocity is constant everywhere.
I don't see hovering describing the same thing as was described previously. Previously it was not hovering it was moving directly away from the earth at a fixed unchanging speed. If the speed is not changing I don't see how it satisfies the definition of the term acceleration.

Orbiting would be accelerating ( otherwise it would travel off in a straight line ). But orbiting , is not what was described previously.
Sense speed is not directional it can be constant in one direction while changing in another. But if it does not change in any direction that brings me back to the terms definition.

As for your description of hovering near the BH. I don't see how it can satisfy the definition of the word accelerate if the speed does not change in some direction?? My confusion on this issue ( use of the word accelerate ) is not about Newtonian physics. I'm confused because the definition of the word accelerate ( which requires a change in speed ). If there is no change in speed ( in any direction ) than I don't see how it satisfies the definition of the term accelerate???

20. Originally Posted by IamIan
Previously it was not hovering it was moving directly away from the earth at a fixed unchanging speed.
Fixed and unchanging with respect to what ? Remember that space-time is not flat in the vicinity of earth, so the 4-velocity is not constant, even if the scalar speed is.

Orbiting would be accelerating ( otherwise it would travel off in a straight line ).
Orbiting is free fall, so yes, there is acceleration involved.

Sense speed is not directional it can be constant in one direction while changing in another. But if it does not change in any direction that brings me back to the terms definition.
Once again, do not confuse "speed" with 4-velocity; they are not the same things ! A constant speed does not imply a constant velocity.

As for your description of hovering near the BH. I don't see how it can satisfy the definition of the word accelerate if the speed does not change in some direction??
I already told you that the world line of such an observer is curved in space-time; hence the 4-velocity vector does change, even if the Newtonian speed does not; the definition is hence satisfied. This is analogous to a satellite orbiting the earth - its linear speed is the same everywhere on its orbit, but its velocity vector is not. It works the same way for the hovering observer, only in 4 dimensions.

You need to forget about Newtonian mechanics in a 3-dimensional world here and think in terms of 4-vectors in a curved space-time.

21. Originally Posted by KALSTER
The thing about light is that the moment a photon is created, it immediately travels at C without any acceleration. So maybe it would make more sense to you to think about this in terms of escape velocity. A photon can't decelerate, only wavelength shifted.
Which is why I was thinking that in terms of light in particular even from smaller objects like the earth. It looses energy in the form of red shifting, not slowing down. Light from the inside of the BH would get redshifted as it tried to move away. From inside the BH it would get redshifted to virtually noting ( CMBR ). Even that red-shifting process takes some time/distance to redshift away the light's energy. Anything less than infinity means it doesn't escape, and it is less than infinity if it came from inside the event horizon.

An object that can speed up or slow down , like an neutron at 99% the speed of light would slow down. If it was at 99% the speed of light 1 Planck length inside the even horizon. It doesn't have the escape velocity, so it can't escape ( to infinity ). I would expect it to do the same thing the ball on earth did, go up some distance before falling back in.

Originally Posted by KALSTER
Aside from that, event horizons represent the radius before which all vectors from a given particle point away from the event horizon, from an outside perspective.
I don't understand this.
If at that point all vectors point away then there is no possible direction ( vector ) that would point in. Which would suggest there is no way to get inside. I don't understand??

Originally Posted by KALSTER
From that particle's point of view, there is no outside, no direction that point to the outside. So even light, which always follows geodesics from it's own perspective, would never near the event horizon again.
I don't understand this??
It occupies a finite volume, there is an outer edge. How can there be no direction that points to the outside?

22. Originally Posted by IamIan
It occupies a finite volume, there is an outer edge. How can there be no direction that points to the outside?
Because of space-time curvature: space is so curved that, inside the event horizon, there is no direction that is towards the event horizon. All paths lead to the singularity. (Oddly, I have been told that if you try and take a path away from the singularity you will actually get there quicker than free fall.)

Also, because of that space time curvature, I don't believe "volume" is well defined for the inside of the event horizon.

23. Originally Posted by Markus
it must necessarily be accelerated, even if its 4-velocity is constant everywhere.
Originally Posted by Markus Hanke
hence the 4-velocity vector does change,
?Confused?
How are you claiming the 4-velocity is constant and changing at the same time and not contradicting yourself???

Originally Posted by Markus Hanke
Once again, do not confuse "speed" with 4-velocity; they are not the same things ! A constant speed does not imply a constant velocity.
I already agreed with constant speed not imply constant velocity.

Originally Posted by Markus Hanke
even if the Newtonian speed does not; the definition is hence satisfied. This is analogous to a satellite orbiting the earth - its linear speed is the same everywhere on its orbit, but its velocity vector is not. It works the same way for the hovering observer, only in 4 dimensions.

You need to forget about Newtonian mechanics in a 3-dimensional world here and think in terms of 4-vectors in a curved space-time.
My confusion is not about Newtonian physics, or 3-dimentions. It's the definitional requirements of the term accelerate. You could have 100 dimensions, and you would still have the same issue.

The definition of the term accelerate requires a change in speed. That change could be in any dimension you like. It can be constant ( no acceleration ) in the X direction, while at the same time changing ( acceleration ) in the Y direction. Same for any number of dimensions you like. If it's speed is changing in the 78th dimension than it is in the 78th dimension that it is accelerating.

If there is no change in speed it does not satisfy the definition of the term acceleration.
Acceleration: physics : the rate at which the speed of a moving object changes over time.

24. Originally Posted by Strange
Originally Posted by IamIan
It occupies a finite volume, there is an outer edge. How can there be no direction that points to the outside?
Because of space-time curvature: space is so curved that, inside the event horizon, there is no direction that is towards the event horizon. All paths lead to the singularity. (Oddly, I have been told that if you try and take a path away from the singularity you will actually get there quicker than free fall.)

Also, because of that space time curvature, I don't believe "volume" is well defined for the inside of the event horizon.
Why would the direction have to follow the curvature?

25. That is the nature of General Relativity, it describes the universe as existing inside of a 3 + 1 dimensional Spacetime manifold. The crucial thing about this, is that our universe can have intrinsic curvature and that is what gravity is according to General Relativity.

What this means is that mass and energy warps spacetime in it's vicinity, so that from an outside perspective, the straightest path through such a portion of space follows a curve. One consequence of this is that we can look at the light of a star that is effectively behind the sun for instance, since the spacetime around the sun is curved and light then follows this path. It is because of this distortion that even light, which is massless, is affected by gravity. This experiment has been done and the predictions confirmed.

In the case of a black hole, the space-time is curved so much that from a point inside the event horizon, any direction you might choose to go in would all curve towards the centre of the black hole, even though for you, you are moving straight forward at each instance of time. If you were to imagine a 3D grid that represents the 3 spacial dimensions, then inside a black hole no degrees of freedom grid line would cross the event horizon again once you have crossed it.

26. Originally Posted by IamIan
Why would the direction have to follow the curvature?
Because that is what the curvature of space means. In our local, flat space we have up-down, left-right, forward-back. These are all straight lines and at right angles to one another. They all take you in different directions.

In curved space this is no longer true. So if you are find yourself heading towards the singularity, you might turn round and go the other way: tough luck, you are still heading towards the singularity. Space is so curved that forward, back, left and right all point towards the singularity.

27. Originally Posted by Strange
In curved space this is no longer true. So if you are find yourself heading towards the singularity, you might turn round and go the other way: tough luck, you are still heading towards the singularity. Space is so curved that forward, back, left and right all point towards the singularity.
Originally Posted by KALSTER
In the case of a black hole, the space-time is curved so much that from a point inside the event horizon, any direction you might choose to go in would all curve towards the centre of the black hole, even though for you, you are moving straight forward at each instance of time. If you were to imagine a 3D grid that represents the 3 spacial dimensions, then inside a black hole no degrees of freedom grid line would cross the event horizon again once you have crossed it.
Doesn't this mean the quantum tunneling can't escape either? There would be no direction it can go that leads out.

28. Originally Posted by IamIan
Originally Posted by Strange
In curved space this is no longer true. So if you are find yourself heading towards the singularity, you might turn round and go the other way: tough luck, you are still heading towards the singularity. Space is so curved that forward, back, left and right all point towards the singularity.
Originally Posted by KALSTER
In the case of a black hole, the space-time is curved so much that from a point inside the event horizon, any direction you might choose to go in would all curve towards the centre of the black hole, even though for you, you are moving straight forward at each instance of time. If you were to imagine a 3D grid that represents the 3 spacial dimensions, then inside a black hole no degrees of freedom grid line would cross the event horizon again once you have crossed it.
Doesn't this mean the quantum tunneling can't escape either? There would be no direction it can go that leads out.
Quantum tunnelling is one of those "weird" bits of quantum mechanics. There is not yet a full theory that can describe both quantum en relativistic effects, but quantum field theory comes close. It predicts that particle-antiparticle pairs just outside the event horizon is produced, with one with negative energy falls into the black hole and decreases it's mass. But it is quite a bit more complicated than that and certainly way above my understanding.

Hawking radiation - Wikipedia, the free encyclopedia

Physical insight into the process may be gained by imagining that particle-antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles.[11] As the particle-antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles takes away some of the mass of the black hole.[12]

A slightly more precise, but still much simplified, view of the process is that vacuum fluctuations cause a particle-antiparticle pair to appear close to the event horizon of a black hole. One of the pair falls into the black hole whilst the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole). By this process, the black hole loses mass, and, to an outside observer, it would appear that the black hole has just emitted a particle. In another model, the process is a quantum tunnelling effect, whereby particle-antiparticle pairs will form from the vacuum, and one will tunnel outside the event horizon

29. Originally Posted by IamIan
Doesn't this mean the quantum tunneling can't escape either? There would be no direction it can go that leads out.
I can't answer that. I doubt anyone can, definitively, because the mathematics of quantum mechanics in curved spacetime is hard (*).

But as tunnelling is about passing an insurmountable barrier then it seems plausible that a particle could tunnel out.

In quantum tunneling, it isn't clear that the particle actually travels between A and B. (And in fact, it couldn't because there is, by definition, a barrier in the way.) The "speed" of a tunneling particle can therefore appear to be faster than light.

(*) It is an area of very active research with several contradictory hypotheses...
Black Hole Firewalls Confound Theoretical Physicists: Scientific American
Astrophysics: Fire in the hole! : Nature News & Comment
Entanglement = Wormholes | Quantum Frontiers

There appear to be some sort of relationship between quantum effects and event horizons but as yet it isn't clear what they are....

30. Originally Posted by IamIan
As for your description of hovering near the BH. I don't see how it can satisfy the definition of the word accelerate if the speed does not change in some direction?? My confusion on this issue ( use of the word accelerate ) is not about Newtonian physics. I'm confused because the definition of the word accelerate ( which requires a change in speed ). If there is no change in speed ( in any direction ) than I don't see how it satisfies the definition of the term accelerate???
Welcome to general relativity. Acceleration may be considered as a non-constant four-dimensional velocity in a locally Minkowskian coordinate system. The reason why an observer hovering above the event horizon of a blackhole or standing on the surface of the earth is considered to be accelerated is that in these cases the coordinate system is not locally Minkowskian. Coordinate-transform to a locally Minkowskian coordinate system and the acceleration becomes evident.

31. In an arbitrary coordinate system, the four-dimensional acceleration is expressed as:

The quantity is known as the connection and "corrects" for the curvilinear nature of the coordinate system.

32. Originally Posted by KJW
The quantity is known as the connection and "corrects" for the curvilinear nature of the coordinate system.
I remember long ago you were describing the affine connection to me while I was trying to make sense through a thought experiment of the impacts of mapping various quantum states such as charge, mass etc. to a spacetime geometry. It was your description of the connection and the infintesimal changes attributed to Christoffel symbols that I think lay at the heart of attempting to map a variable such as mass which was embedded with the spacetime curvature and could not be transformed away. This was different to say a variable such as charge which appeared to me to be independent and floating freely in the spacetime context. Your description of the connection has stuck with me as a critical notion of GR. :-))

33. Originally Posted by KJW
Welcome to general relativity. Acceleration may be considered as a non-constant four-dimensional velocity in a locally Minkowskian coordinate system. The reason why an observer hovering above the event horizon of a blackhole or standing on the surface of the earth is considered to be accelerated is that in these cases the coordinate system is not locally Minkowskian. Coordinate-transform to a locally Minkowskian coordinate system and the acceleration becomes evident.
I don't see a problem with the change in speed being not in any one specific dimension, in Y but not Z, etc. It could be a change in speed in a 4th, or a 5th, or the 100th dimension.

If there is no change in speed in any dimension. that leaves only two options that I see.

#1> You're changing the definition of the term acceleration.
#2> You are not accelerating.

Also note, the source of my confusion over the use of the term acceleration was not used for a BH , but for traveling away from the Earth at a constant speed ( but called accelerating , with no change in speed ).

34. Originally Posted by IamIan
Originally Posted by KJW
Welcome to general relativity. Acceleration may be considered as a non-constant four-dimensional velocity in a locally Minkowskian coordinate system. The reason why an observer hovering above the event horizon of a blackhole or standing on the surface of the earth is considered to be accelerated is that in these cases the coordinate system is not locally Minkowskian. Coordinate-transform to a locally Minkowskian coordinate system and the acceleration becomes evident.
I don't see a problem with the change in speed being not in any one specific dimension, in Y but not Z, etc. It could be a change in speed in a 4th, or a 5th, or the 100th dimension.

If there is no change in speed in any dimension. that leaves only two options that I see.

#1> You're changing the definition of the term acceleration.
#2> You are not accelerating.

Also note, the source of my confusion over the use of the term acceleration was not used for a BH , but for traveling away from the Earth at a constant speed ( but called accelerating , with no change in speed ).
You're missing the main point. The key issue is the notion of change. To say that acceleration is a change in velocity implies a frame of reference relative to which the change is specified. The question then becomes: Which frame of reference is the change relative to? What general relativity does is to provide a uniform definition of the notion of change that can be applied to any coordinate system in any spacetime. But, in doing so, it has to account for reference frames that are themselves curved. That is why the definition of acceleration in general coordinates contains the connection term. It isn't really that the definition of acceleration has changed, but that the definition has been made more generally applicable.

35. Originally Posted by KJW
You're missing the main point. The key issue is the notion of change. To say that acceleration is a change in velocity implies a frame of reference relative to which the change is specified. The question then becomes: Which frame of reference is the change relative to? What general relativity does is to provide a uniform definition of the notion of change that can be applied to any coordinate system in any spacetime. But, in doing so, it has to account for reference frames that are themselves curved. That is why the definition of acceleration in general coordinates contains the connection term. It isn't really that the definition of acceleration has changed, but that the definition has been made more generally applicable.
It isn't just more generally applicable it gives an entirely different answer.

The dictionary definition for acceleration is for a change in speed ( not velocity ).

Originally Posted by Mirriam-Webster
Acceleration: physics : the rate at which the speed of a moving object changes over time.
Using this old ( un-modified ) dictionary version of the term.
If asked. "Does an object with zero change in speed have non-zero acceleration?"
The answer using the Mirriam definition would have been:
"No."

Using this new modified version you get an entirely different answer to the same question.
The modified definition would give the answer of , "Yes."

That is vastly different.

I don't see anything inherently wrong with changing or modifying the definition of the term. Under this new version the object with no change to it's speed in any dimension is still accelerating. But that is what is being done.

It should not have come as any surprise if someone who read the term acceleration thought of it as describing a change in speed. Which is how about 99% of the world uses the term. When asked about it, saying one was using a different modified definition for the term would have explained why it didn't fit.

36. Originally Posted by IamIan
The dictionary definition for acceleration is for a change in speed ( not velocity ).
Which is a great example of why you shouldn't rely on (non-specialist) dictionaries when discussing science.

37. Originally Posted by Implicate Order
Originally Posted by KJW
The quantity is known as the connection and "corrects" for the curvilinear nature of the coordinate system.
I remember long ago you were describing the affine connection to me while I was trying to make sense through a thought experiment of the impacts of mapping various quantum states such as charge, mass etc. to a spacetime geometry. It was your description of the connection and the infintesimal changes attributed to Christoffel symbols that I think lay at the heart of attempting to map a variable such as mass which was embedded with the spacetime curvature and could not be transformed away. This was different to say a variable such as charge which appeared to me to be independent and floating freely in the spacetime context. Your description of the connection has stuck with me as a critical notion of GR. :-))
I don't recall the specific conversation, but the thing that struck me when I first started to learn GR was how a simple rectangular coordinate system can be coordinate transformed in such a way that the metric becomes unrecognisable as a description of a flat space. That is, when compared to the metric of a curved space, there seemed to be no obvious difference between the two. Because of this, I saw no reason why a space ought to be constrained to being flat. It seemed a very small step to go from the set of all flat space metrics to the set of all metrics. Thus, any resistance I may have previously had to the notion of curved spacetime evaporated.

I think it's interesting to consider a metric on its own terms without trying to relate it to some standard coordinate system. I recall presenting to the other forum the metric of a spherical coordinate system but without the usual references to spherical coordinates (I think I used as the coordinates in order to divert attention away from obvious recognition). It wasn't recognised by people I would expect to recognise a spherical coordinate system, indicating to me how much we tend to rely on standard coordinates. My point at the time was how this is the way general relativity should be viewed. Not so long ago I pointed out to Markus Hanke how the spherical coordinate system has a big bang like singularity when considered without reference to spherical coordinates.

I think this is important to understanding the Einstein equation because there you do not get a coordinate system until you actually solve the equation. I'm inclined to think that this is what Mach's principle is really about.

38. Originally Posted by IamIan
How are you claiming the 4-velocity is constant and changing at the same time and not contradicting yourself???
Sorry, that was my mistake - the first of the quotes should have read "speed" instead of "4-velocity". My apologies.

I already agreed with constant speed not imply constant velocity.
Then what is the issue ?

My confusion is not about Newtonian physics, or 3-dimentions. It's the definitional requirements of the term accelerate.
Acceleration means a change in velocity - that change can be in magnitude ( speed ), direction, or both.

The definition of the term accelerate requires a change in speed.
No. Consider a satellite on a circular orbit - its speed will be constant everywhere, but its velocity changes of course. Obviously, there is acceleration since it is in free fall.

The dictionary definition for acceleration is for a change in speed ( not velocity ).
The dictionary definition is irrelevant here - the correct and precise definition has been given in post #30, and as you can see it clearly involves vector quantities.

I don't see anything inherently wrong with changing or modifying the definition of the term. Under this new version the object with no change to it's speed in any dimension is still accelerating. But that is what is being done.
What are you saying ? Of course there is a change in the 4-velocity vector; for an object that remains stationary in space that change is in the time-like part of the 4-vector. We are dealing with space-time here, not just space; the metric of a uniformly accelerating observer differs from the Minkowski metric in the time coefficient.

Also note, the source of my confusion over the use of the term acceleration was not used for a BH , but for traveling away from the Earth at a constant speed ( but called accelerating , with no change in speed ).
There is no difference, except in the acceleration required.

When asked about it, saying one was using a different modified definition for the term would have explained why it didn't fit.
We are not using anything but the standard definition of the 4-acceleration :

Four-acceleration - Wikipedia, the free encyclopedia

Of course that definition is more general than the Newtonian version, but we are dealing with General Relativity after all.

39. Originally Posted by IamIan
Originally Posted by KJW
You're missing the main point. The key issue is the notion of change. To say that acceleration is a change in velocity implies a frame of reference relative to which the change is specified. The question then becomes: Which frame of reference is the change relative to? What general relativity does is to provide a uniform definition of the notion of change that can be applied to any coordinate system in any spacetime. But, in doing so, it has to account for reference frames that are themselves curved. That is why the definition of acceleration in general coordinates contains the connection term. It isn't really that the definition of acceleration has changed, but that the definition has been made more generally applicable.
It isn't just more generally applicable it gives an entirely different answer.
It isn't a different answer, it's the correct answer. Suppose one is on a spinning merry-go-round and throws a ball towards the centre. Is the ball's trajectory curved? Regardless of what the observer on the merry-go-round sees, the answer is no. The equation I presented above will correctly deduce that even from the perspective of the observer on the merry-go-round.

40. Originally Posted by Markus Hanke
I already agreed with constant speed not imply constant velocity.
Then what is the issue ?
I have no idea what issue you had with me already agreeing with that.

Originally Posted by Markus Hanke
My confusion is not about Newtonian physics, or 3-dimentions. It's the definitional requirements of the term accelerate.
Acceleration means a change in velocity - that change can be in magnitude ( speed ), direction, or both.
Except what has been proposed ( by others that confused me ) , is that NO CHANGE is still Acceleration.

Originally Posted by Markus Hanke
The definition of the term accelerate requires a change in speed.
No.
Yes. That definition in mirriam I was referencing in that context there , does do that exactly.
Originally Posted by Mirriam-Webster
Acceleration: physics : the rate at which the speed of a moving object changes over time.
Originally Posted by Markus Hanke
The dictionary definition for acceleration is for a change in speed ( not velocity ).
The dictionary definition is irrelevant here
It is not irrelevant.
It is reasonable for a person to read a word and think it means the dictionary definition of that word.

Originally Posted by Markus Hanke
I don't see anything inherently wrong with changing or modifying the definition of the term. Under this new version the object with no change to it's speed in any dimension is still accelerating. But that is what is being done.
What are you saying ?
I'm saying many words have more than one definition.
There is no crime in using changing or modifying the definition of a term that one is going to use.
It is reasonable to expect some confusion about the use of words that have more than one definition.

Instead of this one sided attack Ian for thinking of terms as they are defined in a dictionary. It would have been far more helpful ( and faster , and easier ) if when I asked the question originally , the simple answer of , 'using a different definition for the word', had been given.

Originally Posted by Markus Hanke
Of course there is a change in the 4-velocity vector; for an object that remains stationary in space that change is in the time-like part of the 4-vector. We are dealing with space-time here, not just space; the metric of a uniformly accelerating observer differs from the Minkowski metric in the time coefficient.
I'm not the one claiming no change. I'm the one who got confused by other claims that no change is still acceleration.

Originally Posted by Markus Hanke
When asked about it, saying one was using a different modified definition for the term would have explained why it didn't fit.
We are not using anything but the standard definition of the 4-acceleration :

Four-acceleration - Wikipedia, the free encyclopedia

Of course that definition is more general than the Newtonian version, but we are dealing with General Relativity after all.
Originally Posted by Wikipedia
four-acceleration is defined as the change in four-velocity over the particle's proper time:
Notice the emphasis I put on the term change.
Even in this version of the definition.

I am the one who got confused by other people's claims of zero change has non-zero acceleration.

Also,
#1> I'd take Mirriam as a better source than Wikipedia.

#2> When someone ( me in this case ) points specifically to the dictionary definition they (O me in this case ) was thinking of. At that point just telling the person you are using a different definition is helpful and constructive. This much harder method of people arguing against the use of the dictionaries definition for a word, is not nearly as helpful.

41. Originally Posted by KJW
What????
Now you're claiming 'yes' is not a different answer from 'no'?????
How does that work?

42. Back in Newtonian mechanics with the rocket travelling away from Earth at a fixed velocity...

BTW, before anything else, there is no such thing as deceleration. Any change in velocity is an acceleration. Deceleration is not a scientific terms and roughly means a decrease in speed which is subsumed in any proper definition of acceleration.

Now, if the rocket had its engines off, it would accelerate towards the Earth due to gravity. Agreed? Therefore, doing nothing would not keep it travelling at a fixed velocity. You would have to apply a force to the rocket to cancel out the force from gravity. From the frame of reference of someone free falling alongside the rocket, the rocket would appear to accelerate away from them, even though from the point of view of the Earth it would now be coasting along at a fixed speed.

I do agree (and did say) that the net acceleration would be zero. You have an acceleration downward due to gravity cancelled out by an acceleration upward due to the rocket engine. So zero net acceleration.

Edit: In math and science, most words do not have multiple definitions. Things like acceleration get precise, quantifiable definitions and are typically used consistently. Even subtle differences often get separate names, like 4-acceleration.

43. Originally Posted by IamIan
Originally Posted by KJW
What????
Now you're claiming 'yes' is not a different answer from 'no'?????
How does that work?
No. I'm saying that general relativity gives the one and only correct answer to the question of whether or not an object is accelerating. Any answer that is different is quite simply the wrong answer. The point I was making is that general relativity doesn't define acceleration in a way that is different to the usual definition of acceleration. Any apparent difference is due to the frame of reference in which the object's trajectory is viewed. The error one makes is due to applying the wrong frame of reference to the situation.

The following may be taken as the definition of inertial motion (zero acceleration): An object in a flat spacetime with a rectangular coordinate system is in inertial motion if its spacetime trajectory is described by a straight line. This definition is always true, even in general relativity. It applies to any spacetime, curved or not. However, in curved spacetime, the definition is applied locally. To apply it in an arbitrary coordinate system, the coordinate system is coordinate-transformed to a (locally) rectangular coordinate system. The equation in post #30 is fully consistent with this definition. However, that equation can be applied directly to the coordinate system at hand.

44. Originally Posted by KJW
No. I'm saying that general relativity gives the one and only correct answer to the question of whether or not an object is accelerating. Any answer that is different is quite simply the wrong answer.
Then perhaps you should write to mirriam about the error in their dictionary.

If you use the definition they give (not the one you've listed), you can get a very different answer compared to that one you listed. Yes vs No.

It is that difference between the two definitions of the same term which lead me to the point that the (version you listed) is a significantly different definition of the same term.

Originally Posted by KJW
The point I was making is that general relativity doesn't define acceleration in a way that is different to the usual definition of acceleration. Any apparent difference is due to the frame of reference in which the object's trajectory is viewed. The error one makes is due to applying the wrong frame of reference to the situation.

The following may be taken as the definition of inertial motion (zero acceleration): An object in a flat spacetime with a rectangular coordinate system is in inertial motion if it's spacetime trajectory is described by a straight line. This definition is always true, even in general relativity. It applies to any spacetime, curved or not. However, in curved spacetime, the definition is applied locally. To apply it in an arbitrary coordinate system, the coordinate system is coordinate-transformed to a (locally) rectangular coordinate system. The equation in post #30 is fully consistent with this definition. However, that equation can be applied directly to the coordinate system at hand.
I'm have not been trying to argue against this version. I did point out that it is different ( and it is ) from the other definition for the same term.

Can I just drop it now?
I guess it was my bad for ever looking for a words definition in a dictionary ( or asking about the difference in usage I noticed ).

- - - - - -

Time to go eat some pie.
Pie for Everyone.

45. Originally Posted by IamIan
I'm have not been trying to argue against this version. I did point out that it is different ( and it is ) from the other definition for the same term.
In terms of the point I have been trying to make, the distinction between speed and velocity is a minor distinction, and the distinction between three-dimensional space and four-dimensional spacetime is a minor distinction. Apart from that, it is the same definition. You have simply been applying it incorrectly when you said that an object with a constant rate of change in the height above earth is not accelerating. It's an important issue because it is the difference between gravity being a force and gravity not being a force, with general relativity saying that gravity is not a force.

46. Just to clarify, I've been arguing from the Newtonian point of view because my understanding of GR isn't all that great. But GR is the more correct answer, so don't take the Newtonian answer as definitive if GR gives a slightly different one. Trying to apply Newtonian intuitions and results to GR leads to lots of headaches.

47. Originally Posted by IamIan
Except what has been proposed ( by others that confused me ) , is that NO CHANGE is still Acceleration.
I have explained that to you - it is about velocity and speed, and the difference between them.

It is reasonable for a person to read a word and think it means the dictionary definition of that word.
If you insist on giving preference to dictionary definitions over the appropriate mathematical definition, you will have a very hard time understanding physics. This discussion is the best example of what I mean. If you are looking for accurate definitions of physics terms, a dictionary is not a suitable source.

It would have been far more helpful ( and faster , and easier ) if when I asked the question originally , the simple answer of , 'using a different definition for the word', had been given.
I ( and others ) have all along trying to point out to you that acceleration in the context of relativity is defined in terms of 4-velocity, not speed.

Notice the emphasis I put on the term change.
Even in this version of the definition.
Did you also notice that the link uses the term 4-velocity, and not speed. That is the crucial bit, as I have tried to explain to you all along. Speed need not change, even in the presence of acceleration; 4-velocity will always change. That is why you use the latter to define the notion; the issue is not the concept of change, but it is about what changes.

I am the one who got confused by other people's claims of zero change has non-zero acceleration.
Ok, so to make it clear once and for all - in the presence of acceleration, it is 4-velocity which changes. Speed is the magnitude of that 4-vector, and that magnitude does not necessarily change, even if acceleration is present. Speed and velocity are not the same things.
This is really as clear as anyone can make it.

I'd take Mirriam as a better source than Wikipedia.
I suggest that in matters of accurate definitions you abandon both, and refer to a proper physics textbook; for example chapter 6 in Misner/Thorne/Wheeler "Gravitation" devotes an entire chapter to accelerated observers in the context of relativity, and how to mathematically describe them.

This much harder method of people arguing against the use of the dictionaries definition for a word, is not nearly as helpful.
Again, physics does not use dictionary definitions in many instances. Look it up in the appropriate textbooks, and pay attention to what is being explained in response to a question.

48. One thing worth mentioning with regards to acceleration: Suppose one has a spacetime trajectory described parametrically in terms of an arbitrary parameter:

where is the arbitrary parameter along the curve. Then, how can one determine if the acceleration along the curve is zero? General relativity usually describes acceleration in terms of an arc-length or proper time parameter, both of which are zero for a light-like trajectory. So, how can one establish whether a light-like trajectory has zero acceleration? The following applies to arbitrary parametisation of curves whether they are null or not:

If there exists a function of the parameter such that:

then there exists a reparametisation of the curve with parameter such that:

and therefore the acceleration is zero.

49. Originally Posted by IamIan
Originally Posted by KJW
No. I'm saying that general relativity gives the one and only correct answer to the question of whether or not an object is accelerating. Any answer that is different is quite simply the wrong answer.
Then perhaps you should write to mirriam about the error in their dictionary.

If you use the definition they give (not the one you've listed), you can get a very different answer compared to that one you listed. Yes vs No.
You learn about the difference between velocity and speed in high school science and similarly that acceleration does not necessarily imply a change in the magnitude portion of velocity (i.e. the moon is in an accelerating state in it's orbit around the earth). Mirriam Webster is simply not the place you should be looking for the meaning of scientific terms.

Time to go eat some pie.
Pie for Everyone.
Cake or death.

50. Originally Posted by IamIan
Can I just drop it now?
I guess it was my bad for ever looking for a words definition in a dictionary ( or asking about the difference in usage I noticed ).
Given the reponses. I'm guessing the answer is no. Request denied, and an attempted apology not accepted.

- - - - - - - - - -

Originally Posted by MagiMaster
Just to clarify, I've been arguing from the Newtonian point of view because my understanding of GR isn't all that great. But GR is the more correct answer, so don't take the Newtonian answer as definitive if GR gives a slightly different one. Trying to apply Newtonian intuitions and results to GR leads to lots of headaches.
At least someone else also remembers the context of this.
I have no problem to use the GR version of the definition of the term. I can't go back and time about what was. Why I was confused doesn't change ( it's in the past ).

- - - - - - - - - - -

Originally Posted by KJW
In terms of the point I have been trying to make, the distinction between speed and velocity is a minor distinction, and the distinction between three-dimensional space and four-dimensional spacetime is a minor distinction. Apart from that, it is the same definition. You have simply been applying it incorrectly when you said that an object with a constant rate of change in the height above earth is not accelerating. It's an important issue because it is the difference between gravity being a force and gravity not being a force, with general relativity saying that gravity is not a force.
It seems to me a point that seems to have been lost somewhere , is that the context of the original confusion did not involve GR ( but was Newtonian ). So rewind everything and put the no change in speed OR velocity in to the Newtonian model ( where gravity is a force ).

I said it before and I'll still say it. I don't mind using a different definition of the term acceleration. How different doesn't matter to me either. I can't go back in time and change the reasons I was confused back then. Any questions related to why I was confused will always get the same answer, my past is history and I can't change what happened.

- - - - - - - - - -

Originally Posted by Markus Hanke
I have explained that to you - it is about velocity and speed, and the difference between them.
And as I've tried multiple times to remind you this context involves a confusion caused by no change. Not to either one. ( no change in speed AND no change in velocity )

Reminder , here is the question you are arguing against , that is the root of this line of discussion ( back on post #9 )

Originally Posted by IamIan
I would have thought no change in speed or velocity would mean no acceleration???
And weather it's the mirriam with speed , or your 4-Acceleration both requires a change to have acceleration. No change = no acceleration.

Originally Posted by Markus Hanke
It is reasonable for a person to read a word and think it means the dictionary definition of that word.
If you insist on giving preference to dictionary definitions over the appropriate mathematical definition, you will have a very hard time understanding physics. This discussion is the best example of what I mean. If you are looking for accurate definitions of physics terms, a dictionary is not a suitable source.
Clarification:
I am not insisting. I've been trying to explain what had happened. I can't go back in time to change it now.

As I have posted many times. I don't mind using a different definition for the term. The original confusion was because of the definition I sited. 100 years from now that piece of history will still be the same, I can't change it now.

I do still think the definition of terms from a dictionary is a good place to start. Any term , be it Chair, North, Open, etc. When there is a difference , I think the easiest solution is to just say it's a different definition and be done with it. Fighting against the definition in the dictionary is less helpful to successful communication.

Originally Posted by Markus Hanke
It would have been far more helpful ( and faster , and easier ) if when I asked the question originally , the simple answer of , 'using a different definition for the word', had been given.
I ( and others ) have all along trying to point out to you that acceleration in the context of relativity is defined in terms of 4-velocity, not speed.
And as I've been trying to get through to you. I have no problem using the different definition for the term. It is different from the original one used.

Reminder this context of this issue with the term acceleration is not GR ( the context was newtonian ) , and it was both no change in speed and no change in velocity. NO CHANGE.

Originally Posted by Markus Hanke
Notice the emphasis I put on the term change.
Even in this version of the definition.
Did you also notice that the link uses the term 4-velocity, and not speed. That is the crucial bit, as I have tried to explain to you all along. Speed need not change, even in the presence of acceleration; 4-velocity will always change. That is why you use the latter to define the notion; the issue is not the concept of change, but it is about what changes.
Yes I did notice that.
But as I've tried to explain before, and I'll say again. As long as the questions are about what was, the history is set in the past. I can not change why I was confused in the past.

And that context was NOT GR.
And, that context ( of that confusion ) had NO CHANGE for both speed AND velocity.

So I tried to bring that change being needed to your attention. But you overlooked it again.

Originally Posted by Markus Hanke
I am the one who got confused by other people's claims of zero change has non-zero acceleration.
Ok, so to make it clear once and for all - in the presence of acceleration, it is 4-velocity which changes. Speed is the magnitude of that 4-vector, and that magnitude does not necessarily change, even if acceleration is present. Speed and velocity are not the same things.
This is really as clear as anyone can make it.
The past is set. I can't change why I was confused. That is in the past.

I still continue to have no problem using a different definition of the term acceleration. Like the one you list there.

I still continue to remind you the context was not GR ( was Newtonian ).

I still continue to remind you the original question I had of confusion , was that there was no change to BOTH speed and velocity.

Originally Posted by Markus Hanke
I'd take Mirriam as a better source than Wikipedia.
I suggest that in matters of accurate definitions you abandon both, and refer to a proper physics textbook; for example chapter 6 in Misner/Thorne/Wheeler "Gravitation" devotes an entire chapter to accelerated observers in the context of relativity, and how to mathematically describe them.

Until someone says they are using a different definition I think a dictionary is a good place to start.

As for a Physics text book. I don't think that would have helped here either. I think the problem was a break down in the communication process itself. People misunderstanding what each other was talking about.

The text book I have (Serway Physics for Scientists and Engineers 3rd Edition) page 44 defines acceleration as:
Originally Posted by Serway
When the velocity of a particle changes with time, the particle is said to be accelerating.
As long as I had the confusion I did that there was no change of both speed AND velocity. Even this physics text book would still have suggested no change was no acceleration. The back and forth would likely have been exactly the same.

Originally Posted by Markus Hanke
This much harder method of people arguing against the use of the dictionaries definition for a word, is not nearly as helpful.
Again, physics does not use dictionary definitions in many instances. Look it up in the appropriate textbooks, and pay attention to what is being explained in response to a question.
The context of the question is also useful.
Not GR ( it was Newtonian ).
No Change of BOTH speed AND velocity.

I still say arguing against the dictionary is the wrong ( less constructive ) method. Just start off saying it is a different definition and be done with it. When talking to someone it is a reasonable to assume the person will use words as defined in a dictionary. When they tell you that is the source, it should remove any and all doubt. Arguing against the dictionary wording is not helpful. Just say it's a different definition. Many words have more than one definition. I do not think this kind of hassle is needed or useful to a successful communication process..

= = = = = = = = = = = = = = = = = =

I'll repeat my request again.

Originally Posted by IamIan
Can I just drop it now?
I guess it was my bad for ever looking for a words definition in a dictionary ( or asking about the difference in usage I noticed ).

51. Originally Posted by IamIan
It seems to me a point that seems to have been lost somewhere , is that the context of the original confusion did not involve GR ( but was Newtonian ). So rewind everything and put the no change in speed OR velocity in to the Newtonian model ( where gravity is a force ).
Let's not forget that this is a thread about blackhole event horizons, where the GR definition of acceleration must be used. And the Newtonian concept of gravity being a force is just as wrong here on earth as it is hovering above the event horizon of a blackhole. It wasn't my intention to correct specific things that you said (where context may be important), but rather to provide you with the correct understanding of acceleration so that the topic of this thread is given the proper foundation.

52. GR gives an interesting resolution to this as well. Climbing out of a gravity well at a fixed 3-velocity apparently involves a changing 4-velocity and thus a non-zero 4-acceleration (if I've understood Markus's posts).

53. Originally Posted by KJW
Originally Posted by IamIan
It seems to me a point that seems to have been lost somewhere , is that the context of the original confusion did not involve GR ( but was Newtonian ). So rewind everything and put the no change in speed OR velocity in to the Newtonian model ( where gravity is a force ).
Let's not forget that this is a thread about blackhole event horizons, where the GR definition of acceleration must be used. And the Newtonian concept of gravity being a force is just as wrong here on earth as it is hovering above the event horizon of a blackhole. It wasn't my intention to correct specific things that you said (where context may be important), but rather to provide you with the correct understanding of acceleration so that the topic of this thread is given the proper foundation.
Which is one reason why I'd like to be allowed to move on past the confusion of the past ( about term usage ) back in post #9. If allowed to move on, then we can get on with other aspects of the overall thread itself.

I appreciate the efforts and information provided in the larger scale overall thread context. The different view GR gives is very helpful.

- - - - - - - - - -

They say a picture is worth a thousand words.
So bellow here I'll use a picture Link to try and help summarize the over all thread progress as of this point.

AFAIK there are two perspectives give to the thread overall concept of the black hole and the event horizon.

- - - - - - - - - - -

Perspective #1>
Escape velocity

Perspective #2>
Curvature of Space-Time

- - - - - - - - - - -

#2> This seems to result in there being no escape velocity at any magnitude , because there is no direction that leads out.

If we look at it from this perspective we get something like the following. Even if the escape velocity is 2x the speed of light ( SoL ) at some point , and you ( some how) were able to achieved 10xSoL you still can't get out. You are SOL ( shit out of luck ).

This perspective requires the BH and event horizon to be an exception and violate the more general ( less than SoL ) notion of escape velocity.

- - - - - - - - - - -

#1> This seems to paint a different picture.

Using the attached picture points as references.

At point C. If an object had a velocity > than the escape velocity at point C, then it could get to point E ( infinity ). This would move past point D and keep going. Ride never ends.

At point C. If an object had a velocity < the escape velocity at point C, then it will only get to point D ( anything less than infinity ). Once it reached point D it would then fall back down again. Rides over.

This perspective says an object at point C even one Planck Length inside a BH event horizon , can't get to point E (infinity) because it is not possible to reach a velocity or speed that is faster than the speed of light.

= = = = = = = = = = = = = = = = =

Is this crude simplification of the thread concept progress acceptable?

54. Originally Posted by IamIan
I think the problem was a break down in the communication process itself. People misunderstanding what each other was talking about.
Ok, I am happy enough to leave it at this; I just don't understand why you assumed this was about Newtonian physics, since this is a thread about black holes, where Newton's theory is not valid.

Perspective #1>
Escape velocity

Perspective #2>
Curvature of Space-Time
As mentioned before, I am unsure what happens if a superluminal observer is considered; if we exclude tachyons, then the only "direction" that leads back out of a black hole is the past. Obviously, going back in time is just as unphysical as tachyons ( they might even be the same things ), so the entire point is really largely academic.

55. Originally Posted by Markus Hanke
I just don't understand why you assumed this was about Newtonian physics, since this is a thread about black holes, where Newton's theory is not valid.
I don't mind explaining , but I also do not want to reopen that again either. I'd rather the conversation just move on, past the misunderstand/miscommunication.

I assumed that particular response from MagiMaster was about Newtonian physics because in that post he wrote.
Originally Posted by MagiMaster
(due to Newton's laws of motion)
Later reaffirming that it was about Newtonian physics when he wrote later:

Originally Posted by MagiMaster
Since F=ma,
Originally Posted by MagiMaster
Back in Newtonian mechanics with the rocket travelling away from Earth at a fixed velocity
Originally Posted by MagiMaster
Just to clarify, I've been arguing from the Newtonian point of view
That is basis for why I assumed that point in the past was from a Newtonian point of view. AFAIK that assumption was correct ( it was Newtonian Based ).

Now , I'd much rather if we buried it and move on please.

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

Originally Posted by Markus Hanke
Perspective #1>
Escape velocity

Perspective #2>
Curvature of Space-Time
As mentioned before, I am unsure what happens if a superluminal observer is considered; if we exclude tachyons, then the only "direction" that leads back out of a black hole is the past. Obviously, going back in time is just as unphysical as tachyons ( they might even be the same things ), so the entire point is really largely academic.
I'll agree it is mostly academic.

There is are two minor implication though that I don't think is entirely academic.

Implication A:
If Perspective #1 is true. Even if you can't reach the escape velocity to reach point E ( infinity ) , that doesn't mean a slower speed ( Sublight ) can't reach the shorter distance of point D ( less than infinity ). It would end up falling back in. But it would still not rule out initial travel to point D.

I think it would be important to recognize this would be for an object that is not light itself. Light climbing out of a gravity well even a non-BH one ( AFAIK ) does not slow down in speed or velocity it gets red shifted instead.

Implication B:
Of course if Perspective #2 is true , than it nullifies the Implication A. But seems to create a new one of it's own. If there is no direction which leads out. Then it should not seem to be possible even for quantum tunneling or quantum leaps ( etc ) to get out. Any direction it quantum tunneled would not lead to the outside ( as in Perspective #2 no direction points to the outside ).

56. I also said that if GR gave a different answer, take that one. I've been talking about Newtonian mechanics because that's what I'm familiar with and also because I feel that that gives a mostly satisfactory answer to your Earth-based escape velocity questions. But in this case, forget perspective #1. It's only an approximation. Don't try to assume it's true and then work from there. And really, if you want to talk about black holes and quantum tunneling, the Newtonian approximation isn't going to give you a very good answer anyway. (And even GR might not as GR and QM are known to not get along very well.)

57. Originally Posted by IamIan
Can I just drop it now?
Originally Posted by IamIan
I'll repeat my request again.
Originally Posted by IamIan
Can I just drop it now?
Originally Posted by IamIan
Which is one reason why I'd like to be allowed to move on
Originally Posted by IamIan
Now , I'd much rather if we buried it and move on please.
4th request denied.
Apparently MUST continue to beat the dead horse.

Originally Posted by MagiMaster
I also said that if GR gave a different answer, take that one.
Why can't we let this drop and move on?

And I multiple times said during the back and forth I didn't mind doing that. Yet, the dead horse continues to be beaten.

5th request now.
Can we now drop it and move on, please?

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

Originally Posted by MagiMaster
But in this case, forget perspective #1. It's only an approximation. Don't try to assume it's true and then work from there. And really, if you want to talk about black holes and quantum tunneling, the Newtonian approximation isn't going to give you a very good answer anyway. (And even GR might not as GR and QM are known to not get along very well.)
Perspective #1 being false is fine. I have no problem with that. It does seem to mean abandoning escape velocity being determined by mass and distance. Or at least there are exceptions to it.

That leaves me with perspective #2 , and implication B.

If our current model ( where GR and QM meet ) of implication B 'don't get along well'. Doesn't that put significant doubt and issue to the premise of quantum tunnel solution of implication B where there is no direction to tunnel that gets out of the BH???

It also occurs to me that there may also be a implication C & D:

C: Discontinuity
If the fabric of space-time is soo bent that there is no direction out from the inside. How can anything ( even space itself as the BH itself moves ) get inside from the outside, without there being a discontinuity of space and time to travel through??

D: Retained fabric behind it as BH moves forward
If a device was built that anchored an object to a point of the fabric of space time itself ( bent or not ). If that object is located in the rear of a BH volume. What happens as the BH itself travels forward??

58. Originally Posted by IamIan
If there is no direction which leads out. Then it should not seem to be possible even for quantum tunneling or quantum leaps ( etc ) to get out.
That "no direction leads out" bit was based purely on GR, which is a classical theory and does not make allowance for quantum effects. Hence the disagreement in this case.

59. We can't let GR drop because you're asking a GR question. Actually, you're asking a quantum gravity question which currently doesn't have a well established answer. (If you want to talk about escape velocity away from black hole event horizons, we can get back to Newtonian mechanics.) But no direction leading out doesn't mean there's no direction leading in because you have to include time in the bent space-time. (Actually, I'll let someone who knows better try and explain that.)

60. Surely the escape velocity model doesn't really work. It is possible to leave the surface of the earth at less than escape velocity but you will fall back again. You can't leave, even temporarily, the "surface" (event horizon) of a black hole.

61. Here is a new what if that occurred to me today that does not involve the border between GR / QM.

How big would a BH have to get for the event horizon to loose the fight against Dark Energy?

AFAIK they are both relativistically based concepts based on the same model of the cosmos.

As the BH gains more and more mass the event horizon grows larger and larger. It encompasses greater and greater volume of space-time.

Each bit of volume is undergoing dark energy expansion.

When the interior volume is small enough gravity can easily overcome the expansion.

Eventually the enclosed volume is large enough that the dark energy expansion of space itself is expanding faster that the speed of light.

Objects 1 Planck length inside the event horizon could not fall toward the singularity fast enough to overcome the FTL expansion. They would soon find themselves outside the event horizon. Etc.

Thoughts of concept?

62. Originally Posted by IamIan
How big would a BH have to get for the event horizon to loose the fight against Dark Energy?
I recently had an "epiphany" with regards to blackholes and dark energy.

It's my intention to post this in the New Hypotheses and Ideas section of the forum, so I won't reveal my thoughts until then.

63. Originally Posted by KJW
I recently had an "epiphany" with regards to blackholes and dark energy.

It's my intention to post this in the New Hypotheses and Ideas section of the forum, so I won't reveal my thoughts until then.
Looking forward to that

64. [QUOTE=KJW;496720]
Originally Posted by IamIan
It's my intention to post this in the New Hypotheses and Ideas section of the forum, so I won't reveal my thoughts until then.
"Attention"......"Battlestations"......"Prepar e for incoming"..../ TIC.

Looking forward to it KJW but I know it is going to be a brain twister for me :-))

65. In one of the science video about worm hole and black hole ,

Physicist uses the word INFLATION to explain big bang.

and adds , Universe may be just 20% bigger than we might have thought.

Narrator says , There might be space existing out there where inflation might have not yet happened

narrator : Morgan freeman (might be easy to search video)

Is it right what they stated in that video ?

66. Originally Posted by sciencestudy
Is it right what they stated in that video ?
As they said, it might be. There is a hypothesis that inflation happens continuously in different parts of space. For example: Eternal inflation - Wikipedia, the free encyclopedia

On the other hand, inflation may never have happened at all.

67. Two Things:

#A> For Personal Clarification, I'd like to confirm I have this picture / concept correct.
On the overall 'big picture' and each small bit that makes it up.
Bellow crude , it is not perfect , don't expect it to be perfect.

#1> Prior to a given region having sufficient density to establish an event horizon, say 1 Planck mass under.

#2> In (#1), if an object doesn't have sufficient Kinetic energy ( even with additional ongoing propulsion ) in order to escape 'out to infinity far away'.

#3> Object as (#2) would travel out some distance ( maybe even as far as several light years away ) before eventually falling back.

#4> If that region of #1 gains that last 1 Planck mass to reach the critical density for the formation of an event horizon in that region.

#5a> After #4 happens , zero outward distance can be traveled.
#5b> That last 1 Planck mass ( in the given Region #1 ) tips the scale so much , so as to reduce even the possibly multiple light year distances of object travel down to nothing ( less than 1 Planck Length ).

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

#B> In case I just missed this one previous question I asked answer.

#1> If a specific point of the fabric of space-time itself is inside the event horizon of a black hole.

#2> If the Black hole has some speed / velocity.

#3> Does that point in the fabric of space-time get out the tail end as the BH moves forward?

#4> If #3 is 'no' where does the fabric of space-time on the tail end of the moving forward BH come from ( if not from the bent fabric of space-time from inside the BH )?

#5> If #3 is 'yes' is that because the fabric of space-times itself is only bent and not torn/severed by gravity?

Originally I last proposed this in terms of a fictional piece of technology that were able to 'anchor' to a point of the fabric of space-time itself. However, the concept of curiosity if the passage of the fabric itself as the BH itself moves.

68. is that because the fabric of space-times itself is only bent and not torn/severed by gravity?
In GR, space-time only gets curved, but always remains a smooth, differentiable manifold - i.e. it never "tears".

69. #4> If #3 is 'no' where does the fabric of space-time on the tail end of the moving forward BH come from ( if not from the bent fabric of space-time from inside the BH )?
Space-time is not "stuff", it is just geometry; measurement of distances.

70. For A, yes, but it should be mentioned that the distance traveled by an object with a given kinetic energy will be a smooth curve towards 0 as the mass of the body it's travelling away from increases. (It'd be easier to model if you assume a sphere of constant radius and slowly increase its density.)

71. Originally Posted by Markus Hanke
is that because the fabric of space-times itself is only bent and not torn/severed by gravity?
In GR, space-time only gets curved, but always remains a smooth, differentiable manifold - i.e. it never "tears".
Thanks for confirming that part of what I wrote in #B#5.
Care to weigh in on the concept being asked in #B itself?

Originally Posted by Strange
#4> If #3 is 'no' where does the fabric of space-time on the tail end of the moving forward BH come from ( if not from the bent fabric of space-time from inside the BH )?
Space-time is not "stuff", it is just geometry; measurement of distances.
Who claimed it was "stuff" (whatever you mean by that) ?

Care to weigh in on the concept being asking in #B itself?

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

Originally Posted by MagiMaster
For A, yes, but it should be mentioned that the distance traveled by an object with a given kinetic energy will be a smooth curve towards 0 as the mass of the body it's travelling away from increases. (It'd be easier to model if you assume a sphere of constant radius and slowly increase its density.)
Bold = Easier? How is that any functionally different than what I already proposed in #A?

If at A#1 the object can travel as A#3 , but at A#4 it can not. And given that one can not add fractions of a Planck mass. Doesn't that combination suggest it only appears as a smooth curve , if looked at in much closer detail is actually small steps? (Even if we do not yet have a model of quantum gravity to accurately model those small steps.)

72. Originally Posted by IamIan
And given that one can not add fractions of a Planck mass.
Do you know what the Planck mass is?

73. Originally Posted by IamIan
Who claimed it was "stuff" (whatever you mean by that) ?
It isn't "fabric" and it doesn't need to "come from" anywhere.

Care to weigh in on the concept being asking in #B itself?
I'm not quite sure what is being asked, which is why I only addressed that one point. What happens to an object at a point in space-time depends on where it is and how it is moving relative to the black hole. It will either fall in never to be seen again, get into a stable orbit or have its path changed by the encounter.

74. Originally Posted by IamIan
And given that one can not add fractions of a Planck mass.
"Nature’s stable point-mass particles, such as electrons and quarks, are many, many orders of magnitude smaller than the Planck mass"
Planck mass - Wikipedia, the free encyclopedia

75. Originally Posted by IamIan
Care to weigh in on the concept being asked in #B itself?
Just go into the frame of reference of the black hole itself. Think about what happens then, and I am sure you can answer that question yourself

76. Originally Posted by KJW
Originally Posted by IamIan
And given that one can not add fractions of a Planck mass.
Do you know what the Planck mass is?
It was (previously) my impression that it (Planck Mass) was the smallest divisible unit of actual mass.

Looking further into it, that was a false impression of mine. My error.

Which leads to the question of if there a minimum mass? One that bellow one can not have a smaller unit of mass.

If there is no minimum amount of mass , than that would seems to follow that there would not be the small steps I described previously, for #A.

Such that (if truly no minimum) there can always have fractions of any amount, even say a tiny fraction of 10-xgrams. the x could be = a googolplex or googolplexgoogolplex , etc out to any arbitrarily small amount of mass to transfer at previously described #A#4.

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

Originally Posted by Strange
Originally Posted by IamIan
Who claimed it was "stuff" (whatever you mean by that) ?
It isn't "fabric" and it doesn't need to "come from" anywhere.
#a> I still don't see you explaining , who claimed it was "stuff"?

#b> "Fabric" is just a term I've sometimes heard used by some people when describing the 4Dimentional Space-Time that is bent by gravity. It is not actual "Fabric". No one that I know of is claiming that it is actual "Fabric" (like a quilt or Kevlar). If it helps I'll try and avoid the use of the term "Fabric" to try and reduce future confusion.

#c> I'm not sure what you are replying to with that bit about , it doesn't needed to "come from" anywhere? My only guess is that maybe, you are claiming and assuming the axiom listed of #B#4 , ( meaning that you would have to be claiming #B#3 is answered as 'no' ). If that is not the case, I don't understand what you are talking about. If that is the axiom you are asserting, than I need additional explanation please.

Originally Posted by Strange
Care to weigh in on the concept being asking in #B itself?
I'm not quite sure what is being asked, which is why I only addressed that one point. What happens to an object at a point in space-time depends on where it is and how it is moving relative to the black hole. It will either fall in never to be seen again, get into a stable orbit or have its path changed by the encounter.
Not an object.
A point of space-time itself (that 4D which Gravity bends in GR). Pick a point of that as the BH conmtinues to move.

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

Originally Posted by Markus Hanke
Originally Posted by IamIan
Care to weigh in on the concept being asked in #B itself?
Just go into the frame of reference of the black hole itself. Think about what happens then, and I am sure you can answer that question yourself
Maybe I'm missing something.
When I tried to think of it in that way, I came back to two different things that seem to be in conflict. Thus my #B question.

#I> I've been told the 4D of space time is so bent there is no out direction that points out (except via the the past on the 4th D of time). If this #I is true , than the 4D space-time itself on the inside has no direction to go to get out in the furture as the BH move forward.

#II> The conflict I see in #I above is the 4D space-time that is behind the BH as the BH moves forward. This seems to suggest to me that the 4D space-time itself is not bound by the BH event horizon as objects are. Bent by it but not completely prevented from passing through it.

#I and #II (that I thought of) seem to be in conflict to me. Thus I ask about #B.

77. Originally Posted by IamIan
#c> I'm not sure what you are replying to with that bit about , it doesn't needed to "come from" anywhere?
<sigh> This is why conversations with you are so frustrating. You say things and when people reply, you say "what are you talking about".

Just so you know, I was replying to this:
#4> If #3 is 'no' where does the fabric of space-time on the tail end of the moving forward BH come from ( if not from the bent fabric of space-time from inside the BH )?
I'm not sure what you meant by "come from" ...

78. Originally Posted by KJW
Originally Posted by IamIan
As for your description of hovering near the BH. I don't see how it can satisfy the definition of the word accelerate if the speed does not change in some direction?? My confusion on this issue ( use of the word accelerate ) is not about Newtonian physics. I'm confused because the definition of the word accelerate ( which requires a change in speed ). If there is no change in speed ( in any direction ) than I don't see how it satisfies the definition of the term accelerate???
Welcome to general relativity. Acceleration may be considered as a non-constant four-dimensional velocity in a locally Minkowskian coordinate system. The reason why an observer hovering above the event horizon of a blackhole or standing on the surface of the earth is considered to be accelerated is that in these cases the coordinate system is not locally Minkowskian. Coordinate-transform to a locally Minkowskian coordinate system and the acceleration becomes evident.
KJW has a good understanding of his own statements.

79. Originally Posted by IamIan
Originally Posted by MagiMaster
For A, yes, but it should be mentioned that the distance traveled by an object with a given kinetic energy will be a smooth curve towards 0 as the mass of the body it's travelling away from increases. (It'd be easier to model if you assume a sphere of constant radius and slowly increase its density.)
Bold = Easier? How is that any functionally different than what I already proposed in #A?

If at A#1 the object can travel as A#3 , but at A#4 it can not. And given that one can not add fractions of a Planck mass. Doesn't that combination suggest it only appears as a smooth curve , if looked at in much closer detail is actually small steps? (Even if we do not yet have a model of quantum gravity to accurately model those small steps.)
It's no different. It's just easier to model, that is, write down the equations that describe what's going on. You're asking GR questions, and in GR everything is assumed to be infinitely divisible. GR is extremely accurate within its domain which includes things just inside an event horizon (but not all the way to the singularity itself), at least AFAIK. QM says this isn't really true, but since we don't have a theory of quantum gravity, we can't yet provide any more accurate answers by assuming quantums of mass or space exist.

So in the mean time, the height a ball is thrown will be a smooth curve towards 0 as the density of the planet increases.

80. Originally Posted by IamIan

Maybe I'm missing something.
When I tried to think of it in that way, I came back to two different things that seem to be in conflict. Thus my #B question.

#I> I've been told the 4D of space time is so bent there is no out direction that points out (except via the the past on the 4th D of time). If this #I is true , than the 4D space-time itself on the inside has no direction to go to get out in the furture as the BH move forward.

#II> The conflict I see in #I above is the 4D space-time that is behind the BH as the BH moves forward. This seems to suggest to me that the 4D space-time itself is not bound by the BH event horizon as objects are. Bent by it but not completely prevented from passing through it.

#I and #II (that I thought of) seem to be in conflict to me. Thus I ask about #B.
I think you are thinking about this in too complicated a way. A point at some distance r from the centre of the BH will remain at the distance r regardless of whether the BH is moving with respect to the observer or not ( relativistic effects aside for now ); this is easy to see, because you can always put yourself in the same frame of reference as the BH, in which case it simply is at rest. Relative motion cannot change what the BH is, just as relative motion cannot turn an object into a BH through relativistic mass increase. The universe just doesn't work that way

81. Originally Posted by MagiMaster
You're asking GR questions, and in GR everything is assumed to be infinitely divisible. GR is extremely accurate within its domain which includes things just inside an event horizon (but not all the way to the singularity itself), at least AFAIK. QM says this isn't really true, but since we don't have a theory of quantum gravity, we can't yet provide any more accurate answers by assuming quantums of mass or space exist.
I was not asking GR exclusive questions.

That's why I included this statement:
Originally Posted by IamIan
(Even if we do not yet have a model of quantum gravity to accurately model those small steps.)
If the GR answers become less accurate as it approaches the domain of (yet to be completed) quantum gravity.
Then that leads me to 2 questions:

-1>
How small of mass steps does GR remain reasonably accurate for?

-2>
How does GR smoothly transition for an object in the #A concept above?

Or -2 said another way.
Where is the transitional phase (increasing regional mass density) between infinity and zero travel distance? Where light ( or other relativistic objects ) don't reach infinity, but do travel more than zero distance. ie, Light only gets 3,000 light years away ( not infinity ) , then only 300 light years away ( not infinity ) , etc. In order to have that smooth transition.

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

Originally Posted by Strange
Originally Posted by IamIan
#c> I'm not sure what you are replying to with that bit about , it doesn't needed to "come from" anywhere? My only guess is that maybe, you are claiming and assuming the axiom listed of #B#4 , ( meaning that you would have to be claiming #B#3 is answered as 'no' ). If that is not the case, I don't understand what you are talking about. If that is the axiom you are asserting, than I need additional explanation please.
<sigh> This is why conversations with you are so frustrating. You say things and when people reply, you say "what are you talking about".
I added pointer references this time (#A, #B, etc ) in an effort to prevent / reduce the type of confusion that happened previously from crossing different lines of conversation out of the context for a given line of conversation.

Originally Posted by Strange
Just so you know, I was replying to this:
#4> If #3 is 'no' where does the fabric of space-time on the tail end of the moving forward BH come from ( if not from the bent fabric of space-time from inside the BH )?
I'm not sure what you meant by "come from" ...
That would seem to fit in my best guess I listed as the rest of my #c answer your quoted part. In the above I re-included the rest of #c that includes that guess I was making.

The #B#4 you site there includes in it a Axiom ( see bold ).

that #B#3 is:

Originally Posted by IamIan
#3> Does that point in the fabric of space-time get out the tail end as the BH moves forward?
Only when that axiom is met does it create the question of #B#4:

Bellow I will try and elaborate on the concept of #B#4, in an effort to help clarify that #B#4 concept to you.

If and only if (#B#3 is no) , points (not objects) of the 4D (3Space+1time) itself that are inside the BH are not available to be the points of 4D(3Space+1Time) that trail behind the BH as the BH moved forward. Yet there is 4D (3Space+1ime) beyond the BH as it moves forward.

Does that help clarify the #B#4 question?

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

Originally Posted by Markus Hanke
I think you are thinking about this in too complicated a way. A point at some distance r from the centre of the BH will remain at the distance r regardless of whether the BH is moving with respect to the observer or not ( relativistic effects aside for now ); this is easy to see, because you can always put yourself in the same frame of reference as the BH, in which case it simply is at rest. Relative motion cannot change what the BH is, just as relative motion cannot turn an object into a BH through relativistic mass increase. The universe just doesn't work that way
Perhaps I am over thinking it.

It seems you are asserting a 'no' to #B#3.(maybe I'm wrong)
How do you reconcile there being points of 4D (3Space+1Time) trailing behind a BH if the points of 4D (3Space+1Time) inside the BH can not be those points of 4D (3Space+1Time)? Was more created (like but different from expansion)?

The only other option I currently see , seems to fall back to #II above. ie, There is always a point of 4D (3Space+1Time) at distance r from the center of the BH, but it is not the same point of 4D (3Space+1Time) as it was prior to the BH itself moving. Which can only be the case if ( and on if) the 4D itself is not bound by the event horizon like objects are.

Two separate question about the Bold
-3>
Does the relativistic mass increase still increase the 4D (3Space+1Time) curvature (ie Gravity) of the object that is on the edge of transitioning into a BH? One where that increase in mass ( if it were a rest mass , instead of relativistic mass ) would have been enough mass density in the region in order to form a BH.

-4>
Does that mean the concept of the Kugelblitz (a light based BH who only has relativistic mass) is false?

82. The #B#4 you site there includes in it a Axiom ( see bold ).
Predicate, not axiom.

#3> Does that point in the fabric of space-time get out the tail end as the BH moves forward?
I'm not even sure the question is meaningful, but if it is it almost certainly depends what coordinate system you are using. Although, whichever coordinates you use, I think the answer is no because nothing comes out of a black hole.

There is one coordinate system that describe space as falling into a black hole: Waterfall

If and only if (#B#3 is no) , points (not objects) of the 4D (3Space+1time) itself that are inside the BH are not available to be the points of 4D(3Space+1Time) that trail behind the BH as the BH moved forward. Yet there is 4D (3Space+1ime) beyond the BH as it moves forward.
I don't know if it is easy to transform the Schwarzschild metric (which I assume you are thinking of) to the frame of reference of an observer moving wrt the black hole. So there may not be a simple answer to this.

83. Originally Posted by IamIan
Perhaps I am over thinking it.
Yes, I think so too

How do you reconcile there being points of 4D (3Space+1Time) trailing behind a BH if the points of 4D (3Space+1Time) inside the BH can not be those points of 4D (3Space+1Time)? Was more created (like but different from expansion)?
I am sorry, but I really do not understand what you are trying to say here. Four dimensional space-time is completely static, so each event is unique and doesn't "change" in any way.

Does the relativistic mass increase still increase the 4D (3Space+1Time) curvature (ie Gravity) of the object that is on the edge of transitioning into a BH?
No. The source of gravity is not relativistic mass, but the energy-momentum tensor; the defining property of a tensor is that it is the same for all observers, so if you change to a different observer, the components of the tensor will vary accordingly, and the tensorial object itself remains the same. That is why you cannot push an object "over the edge" simply by moving relatively to it.

Does that mean the concept of the Kugelblitz (a light based BH who only has relativistic mass) is false?
Not at all. Like I said, the source of gravity is the energy-momentum tensor, not relativistic mass, or even just rest mass. This tensor describes all forms of energy, which includes mass and radiation, and also things like stresses, strains and fluxes - a Kugelblitz is an object that is purely radiation ( and associated dynamics ), and hence has a non-zero energy-momentum tensor, like any other more "normal" object. After the gravitational collapse occurs, the two are no longer distinguishable anyway ( no-hair theorem ), even though the original rest mass of the Kugelblitz is technically zero ( like that of photons, of which it is composed ).

84. Originally Posted by IamIan
Does that mean the concept of the Kugelblitz (a light based BH who only has relativistic mass) is false?
Interesting, I hadn't heard of that before. I'm not sure why you think it would be false, but energy and mass are equivalent so I see no reason why you couldn't create black hole from photons (in principle, it may not be possible in practice).

85. Originally Posted by Strange
Interesting, I hadn't heard of that before.
It is one of the lesser known, more arcane applications of relativity. In practice it would be extremely difficult to create the required energy densities to create such a Kugelblitz; it is definitely outside our capabilities by many, many orders of magnitude.

86. Originally Posted by Strange
Originally Posted by IamIan
Originally Posted by Markus Hanke
just as relative motion cannot turn an object into a BH through relativistic mass increase
Does that mean the concept of the Kugelblitz (a light based BH who only has relativistic mass) is false?
Interesting, I hadn't heard of that before. I'm not sure why you think it would be false, but energy and mass are equivalent so I see no reason why you couldn't create black hole from photons (in principle, it may not be possible in practice).
Bold = I never said I thought it was false.

If it matters any (probably not), here is an effort to describe what my thoughts were at that time referenced in the quote above. I also added the context back to the above, thinking it might help as well.

#1> Markus was making a claim about a relativistic mass increase not being able turn an object into a BH.
#2> I knew a Kugelbiltz is a case where relativistic mass creates a BH.
#3> Because , I saw a possible conflict between 1 & 2 , I asked a question about it.
#4> Markus gave an explanation , as to why there isn't a conflict with his statement. Question answered.

- - - - - - - - - -

Originally Posted by Markus Hanke
It is one of the lesser known, more arcane applications of relativity. In practice it would be extremely difficult to create the required energy densities to create such a Kugelblitz; it is definitely outside our capabilities by many, many orders of magnitude.
Still cool though.

87. Originally Posted by IamIan
Still cool though.
I agree

88. it was a bad joke it was as useless as junk

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