The Arrow of Time: The Laws of Physics and the Concept of Time Reversal

[Passionflower]

I appreciate the depth of the response but I already get all that and it doesn't have anything to do with what I'm trying to get across. I don't believe that an object can even accelerate into a black hole, but I'm being told that it doesn't even have to accelerate, it can free-fall in. I'm thinking tidal force could be the answer. Imagine a very long object free-falling towards a black hole. It's at an angle so that (if we ignore length contraction for a moment) if it's ten metres long then the back will be ten metres further from the black hole than the front. There has to be an easier way of explaining that. There's probably a single word that describes that angle. Anyway, the front will be pulling the back along, even though the front is just trying to free-fall. What's free-fall speed at the front isn't the same as what's free-fall speed at the back. The difference is caused by the fact that the front end is more time dilated/length contracted than the back. From the backs perspective the front is accelerating, so objects closer than whoever's observing them do accelerate from that observers perspective. In other words objects do accelerate when they free-fall. As they accelerate they become time dilated/length contracted and the rest of the universe (including the life span of the black hole) appears to speed up. Anything at the event horizon would, in effect, have accelerated to c. I don't see how a free-faller could gradually reach infinite time dilation/length contraction. It doesn't matter how long the black hole lasts because they journey inwards would always last longer. I think tidal force comes from the fact that you can't assign a set proper time or length in which anything can happen when acceleration's involved because you'd have to constantly recalculate as you go. You'd have to take into account that what's a metre or a second at the speed your traveling now won't be the same after you accelerate. I think we all know that's true in special relativity and I don't see how it works any differently with gravity.

You may think you get it but really I would urge you to take some advice, clearly your understanding is flawed.

A free falling observer will reach the event horizon and singularity in finite proper time (e.g. the time on his clock)."

How? Where does this postulate actually come from? If you work out how much time it would take from one position, then move closer and work it out again then you'll get a different answer. It takes more time the closer you get.

Again you are wrong, a free falling observer reaches the singularity in finite proper time, and the closer he gets, the less time he has left.

Do you perhaps want to discuss the formulas that show this?

 

[PeterDonis]

"You mean if their initial velocities are different then they will remain different, and if they both start off hovering at the same distance then it would be zero?" was right than. What's epsilon?

No, I mean that their initial velocities start out both being zero, but as the two objects fall, their velocities *become* different (the lower object falls faster than the higher object), even though they are both freely falling and both feel no force.

Epsilon is just the initial difference in radial coordinates of the two objects.

I'm not sure there's anything to resolve here. I've already stated that the only time tidal force wouldn't apply is to a point-like object.

And yet you keep on trying to bring in tidal force when I and others have told you multiple times that an object can free-fall through the horizon even in the case where tidal force is negligible (the object's length is so short compared to the size of the black hole that the tidal gravity of the hole at the horizon has no appreciable effect on the object's motion). See further comment below.

Same thing isn't it? If a light ray can reach the horizon then nothing can.

I never said a light ray can't (I assume that's what you meant, not "can") reach the horizon; I said the horizon *is* the path of a light ray--an outgoing light ray at r = 2M.

I'm thinking tidal force could be the answer.

No, tidal force is a red herring that's distracting you from the real physics involved. That's why we keep recommending that you consider a Rindler horizon, which works like a black hole horizon in key respects but occurs in flat spacetime, where there is *no* tidal gravity to confuse things.

 

[clitvin]

" Imagine a very long object free-falling towards a black hole. From the backs perspective the front is accelerating, so objects closer than whoever's observing them do accelerate from that observers perspective.

If the black hole is massive enough the front and the back will accelerate at the same speed.

Very large black holes will only experience tidal forces beyond the EH closer to the singularity.

I don't see how a free-faller could gradually reach infinite time dilation/length contraction.

You do not experience any time dilation. Time is always contant for yourself. Outside observers would see an image of you "get stuck" just beyond the event horizon but only because the light coming towards them would make it appear so. However even this won't last forever as eventually the light wavelenght would become so long it is no longer inside the visible spectrum.

 

[PeterDonis]

Imagine a very long object free-falling towards a black hole...What's free-fall speed at the front isn't the same as what's free-fall speed at the back.

This is true, but irrelevant. As I showed in a recent post, even for an extended object like this one, subject to tidal gravity, the center of mass of the object moves on a freely falling worldline--an accelerometer mounted at the center of mass of the object would read zero proper acceleration. This alone is enough to prove that the object will reach the horizon in a finite proper time, starting from a finite radius R > 2M. (The proof is what I posted earlier, since that proof applies to *any* freely falling worldline, including the one followed by the center of mass of an extended object moving solely under the influence of gravity.) This is why I said the whole issue of tidal gravity is a red herring--it has nothing to do with whether or not an object can reach the horizon.

 

[Passionflower]

This is why I said the whole issue of tidal gravity is a red herring--it has nothing to do with whether or not an object can reach the horizon.

I completely agree with Peter.

For a black hole the tidal force between the bottom and the ceiling of a free falling object traveling is directly related to the remaining time until the singularity is hit, the location of the EH, and thus the black hole's mass, is totally irrelevant to this.

 

[yuiop]

You clearly misunderstood; I posted that example as a reductio ad absurdum counterargument to your position. I am glad that you recognized that the argument is wrong in the case of a brick wall, but it is frustrating that even after more than 200 posts you don't realize that it is just as wrong in the case of an event horizon also.

Here are some incontrovertible mathematical facts:
1) the proper time along an infalling radial geodesic is finite in all coordinate systems
2) there are coordinate systems where the coordinate time is also finite
3) the Schwarzschild chart does not smoothly cover the event horizon
4) other coordinate charts do
5) the coordinate time is infinite in the Schwarzschild coordinates

Your position has been that because of 5) no observer can cross the event horizon despite 1)-4). In other words, your position is that an object cannot cross the horizon for no other reason than the fact that there exists some coordinate system where the coordinate time goes to infinity. If this logic were correct then it would apply to other coordinate systems in other spacetimes also, and therefore you would not be able to hit a brick wall because there exists some coordinate system where the coordinate time goes to infinity. This is your argument as you have described it here.

Wikipedia http://en.wikipedia.org/wiki/Hawking_radiation gives the time t_{ev} for a black hole to evaporate as:

t_{ev} = \frac{5120 \pi G^2 M^3}{\hbar c^4}

This evaporation time is proportional to the cube of the black hole mass and if finite for any finite mass black hole. The coordinate time for an observer to fall to the event horizon is infinite so the coordinate falling time is always greater than the coordinate evaporation time and so the observer never arrives at or passes through the event horizon in practice (if we give any credibility to Hawking or his evaporation prediction). Now coordinate measurements *exactly* to or at the event horizon are contentious because of the impossibility of having a stationary observer at that location so the event horizon is considered "off the Schwarzschild chart". This is not a problem, because we can always find a local stationary observer outside but arbitrarily close to the event horizon that will say that when the free falling observer arrived the black hole and event horizon had evaporated. In other words we can always find a point *outside* the event horizon where a local observer will confirm that the black hole evaporated before the free falling object's arrival. The proper time for a free falling observer that falls from a finite height less than infinite but greater than 2m is finite. Let us say that this finite free falling proper time is 16 seconds, then the black hole evaporates in less than 16 seconds of the free faller's proper time. This means, if we believe Hawking, that in practice no free falling particle has ever fallen to or through an event horizon, so points 1 to 4 in the list become academic because they never happened in nature and in principle never will.

 

[PeterDonis]

The coordinate time for an observer to fall to the event horizon is infinite so the coordinate falling time is always greater than the coordinate evaporation time and so the observer never arrives at or passes through the event horizon in practice (if we give any credibility to Hawking or his evaporation prediction).

This is not correct. The infinite coordinate time for an observer to free-fall to the event horizon assumes an "eternal" black hole, one that never evaporates; that is what the Schwarzschild solution to the EFE describes. When you add in evaporation a la Hawking, the spacetime as a whole is no longer the exact Schwarzschild spacetime, so it no longer takes infinite coordinate time for an observer to free-fall to the horizon. Instead, the coordinate time at which the hole finally evaporates (the formula you gave) is also the coordinate time at which any free-falling observer hits the horizon. However, you have to be very careful drawing implications from that statement; it is *not* the same as saying that all free-falling observers hit the horizon at the same event.

If you look at a Penrose diagram for an evaporating black hole spacetime, as shown for example on this page (towards the bottom of the page)...

http://nrumiano.free.fr/Estars/bh_thermo.html

...you will see that the horizon is still a line (a null line going up and to the right), just as it is for an "eternal" black hole, so there is room for an infinite number of "crossing" events at which a free-falling observer crosses the horizon. The difference is that now the horizon has a future endpoint; since the horizon is a null line, light rays from *all* events that took place on the horizon (such as all the "crossing" events) all pass through that endpoint and then radiate outward to be seen by distant observers.

So basically, at the event of final evaporation of the hole, a flash is emitted that contains images of all the free-falling observers that crossed the horizon, just as they were crossing it. (Unfortunately, the free-falling observers themselves have already fallen into the singularity, since that is still to the future of any timelike worldline that crosses the horizon.) That single flash therefore gets assigned a single coordinate time in the "approximately Schwarzschild" coordinate system used by a distant observer, which will be the coordinate time of the final evaporation *and* of all the "crossing events" whose images are contained in the flash.

 

[surajc0504]

that means will an isentropic process be always time reversible ??

 

[yuiop]

If you look at a Penrose diagram for an evaporating black hole spacetime, as shown for example on this page (towards the bottom of the page)...

http://nrumiano.free.fr/Estars/bh_thermo.html

...you will see that the horizon is still a line (a null line going up and to the right), just as it is for an "eternal" black hole, so there is room for an infinite number of "crossing" events at which a free-falling observer crosses the horizon. The difference is that now the horizon has a future endpoint; since the horizon is a null line, light rays from *all* events that took place on the horizon (such as all the "crossing" events) all pass through that endpoint and then radiate outward to be seen by distant observers.

Well, I am not completely convinced by the Penrose diagram but maybe that is just because I do not fully trust/understand the diagrams, which seem to demonstrate all sorts of unlikely things like nipping from one universe to another, which probably don't happen in reality. For example, take a look at this adaptation of the diagram you linked to:

The diagram shows light rays coming from the surface (orange curve) of a star collapsing to form the black hole. Examination of the path of observer A (green curve) shows he sees the star continue to shine all the way until he hits the singularity. Examination of the path of the second observer (light blue curve) shows he sees the star stop shining at event B but at that point the black hole has already ceased to exist. It seems that no observer sees a region of space that is gravitationally attractive that is not shining and only stop seeing the ghost of the star that formed the black hole after they hit the singularity or after the the black hole has evaporated, so its seems no one ever sees the black hole as black, although there will be a brief time when the light from the star that formed the black hole is highly redshifted and difficult to observe just before the black hole expires in a flash. Does that seem right?

 

[PeterDonis]

Well, I am not completely convinced by the Penrose diagram but maybe that is just because I do not fully trust/understand the diagrams, which seem to demonstrate all sorts of unlikely things like nipping from one universe to another, which probably don't happen in reality.

When you look at the Penrose diagrams for, e.g., the maximally extended Schwarzschild spacetime (which includes the "white hole" and the second exterior region), I agree the diagrams are showing things which, although they follow mathematically from the given solution to the EFE, are probably not physically reasonable. However, I don't think that's the case for this particular diagram; it presents a picture which seems, at least to me, to be consistent and in principle physically realizable, although perhaps counterintuitive.

Examination of the path of observer A shows he sees the star continue to shine all the way until he hits the singularity.

Agreed.

Examination of the path of the second observer (light blue curve) shows he sees the star stop shining at event B but at that point the black hole has already ceased to exist.

Yes, it has just evaporated--the second observer will see the final evaporation at the same time he sees the star stop shining, where "stop shining" means "just crossing the horizon" (since light from both events is coming to him along the same null worldline), and the term "stop shining" is justified by the fact that this light ray is the *last* one that will reach the second observer from any event on the star's worldline.

It seems that no observer sees a region of space that is gravitationally attractive that is not shining...

Not quite true, because there's a long period of time before all those light rays reach the second observer, during which he can definitely see the effects of the black hole's gravity but can't yet see any of the light coming from events close to the horizon. During that time he will indeed see a region of space that is gravitationally attractive but "not shining".

...and only stop seeing the ghost of the star that formed the black hole after they hit the singularity or after the the black hole has evaporated...

Yes.

...so its seems no one ever sees the black hole as black...

No, see above.

...although there will be a brief time when the light from the star that formed the black hole is highly redshifted and difficult to observe just before the black hole expires in a flash.

Yes.

 

[A-wal]

You may think you get it but really I would urge you to take some advice, clearly your understanding is flawed.

LOL. My understanding is flawed is it? I don't think I'm the one under a false impression of understanding to be honest. If I understood it perfectly they'd be no point in me being here. What don't I get now? I hate it when I'm told I don't get it but can't be told what it is I don't get for some reason. What's the advice then? My understanding is flawed? That's really more of an observation than advice. I'll give you some advice instead. Don't assume that you're the one that must be right when clearly you either don't even know what you're talking about or can't just be bothered to explain yourself. Which is it?

Again you are wrong, a free falling observer reaches the singularity in finite proper time, and the closer he gets, the less time he has left.

Again you're being a lot less than helpful. I don't think you even understood what I meant. You're just saying it is because that's the way it is. How does that sentence bring anything at all to the discussion. I'm beginning to think I've accidentally stumbled into a new age religious website

Do you perhaps want to discuss the formulas that show this?

I'm not sure it will mean much to me. We can try though. I just need to see if it's frame dependant really. If the formula for it shows that an object can reach an horizon then I can't be. At least I don't see how it can be.

No, I mean that their initial velocities start out both being zero, but as the two objects fall, their velocities *become* different (the lower object falls faster than the higher object), even though they are both freely falling and both feel no force.

Okay, that's twice now that you've said "No I meant this..." and then repeated what I've said but worded it slightly differently. I'm getting used to it but you've now done it twice to the same statement.

And yet you keep on trying to bring in tidal force when I and others have told you multiple times that an object can free-fall through the horizon even in the case where tidal force is negligible (the object's length is so short compared to the size of the black hole that the tidal gravity of the hole at the horizon has no appreciable effect on the object's motion). See further comment below.

Saying it multiple times doesn't make it right, and it definitely doesn't help me to understand why or how it's possible. It doesn't make sense.

I never said a light ray can't (I assume that's what you meant, not "can") reach the horizon; I said the horizon *is* the path of a light ray--an outgoing light ray at r = 2M.

Oops. Yes that's what I meant. But light can't reach the horizon because no object can be seen reaching the horizon.

No, tidal force is a red herring that's distracting you from the real physics involved. That's why we keep recommending that you consider a Rindler horizon, which works like a black hole horizon in key respects but occurs in flat spacetime, where there is *no* tidal gravity to confuse things.

But as I've already said twice: The equivalent to the Rindler horizon using gravity just places a limit on the distance an object can be in order to catch the free-faller before the black hole's gone. This distance is individual and depends on the distance of the free-faller to the other object and to the black hole. Maybe I've misunderstood the Rindler horizon?

If the black hole is massive enough the front and the back will accelerate at the same speed.

Very large black holes will only experience tidal forces beyond the EH closer to the singularity.

The front and the back will never accelerate at the same speed because that doesn't make sense. You can reduce the difference by increasing the mass, but it's still there. Very large black holes don't experience less tidal force. It's just more spread out.

You do not experience any time dilation. Time is always contant for yourself. Outside observers would see an image of you "get stuck" just beyond the event horizon but only because the light coming towards them would make it appear so. However even this won't last forever as eventually the light wavelenght would become so long it is no longer inside the visible spectrum.

You do not experience time dilation and time is always constant for yourself you say. Thanks for that. Yes I know time is always constant for yourself. It would be ridiculous if it wasn't. If you want it from the fallers perspective then the black hole will evaporate quicker and quicker as you get closer to it and you will never have time to reach it.

This is true, but irrelevant. As I showed in a recent post, even for an extended object like this one, subject to tidal gravity, the center of mass of the object moves on a freely falling worldline--an accelerometer mounted at the center of mass of the object would read zero proper acceleration. This alone is enough to prove that the object will reach the horizon in a finite proper time, starting from a finite radius R > 2M. (The proof is what I posted earlier, since that proof applies to *any* freely falling worldline, including the one followed by the center of mass of an extended object moving solely under the influence of gravity.) This is why I said the whole issue of tidal gravity is a red herring--it has nothing to do with whether or not an object can reach the horizon.

Maybe I wasn't clear enough. I'm going to explain this one more time as simply as I can. My thinking is that objects do accelerate in free-fall, kind of. That's what the Schwarzschild coordinates show. This acceleration has to be felt and I think it's felt as tidal force because tidal force is caused by the difference in velocity between objects large enough to notice the difference. Tidal force is just the difference in the rate of acceleration. Constant acceleration (as perceived by a distant observer) due to free-fall isn't felt as acceleration (so please no-one bother to point this out again, it's getting very frustrating). It's the equivalent to being at rest but an increase in the rate of acceleration is felt, as proper acceleration. This is all tidal force is. The rate of acceleration determines the value of length contraction and time dilation. The event horizon is the closest possible Rindler horizon to the singularity.

I completely agree with Peter.

For a black hole the tidal force between the bottom and the ceiling of a free falling object traveling is directly related to the remaining time until the singularity is hit, the location of the EH, and thus the black hole's mass, is totally irrelevant to this.

What? One of the factors is the location of the event horizon, which is determined by the black holes mass.

Wikipedia http://en.wikipedia.org/wiki/Hawking_radiation gives the time t_{ev} for a black hole to evaporate as:

t_{ev} = \frac{5120 \pi G^2 M^3}{\hbar c^4}

This evaporation time is proportional to the cube of the black hole mass and if finite for any finite mass black hole. The coordinate time for an observer to fall to the event horizon is infinite so the coordinate falling time is always greater than the coordinate evaporation time and so the observer never arrives at or passes through the event horizon in practice (if we give any credibility to Hawking or his evaporation prediction). Now coordinate measurements *exactly* to or at the event horizon are contentious because of the impossibility of having a stationary observer at that location so the event horizon is considered "off the Schwarzschild chart". This is not a problem, because we can always find a local stationary observer outside but arbitrarily close to the event horizon that will say that when the free falling observer arrived the black hole and event horizon had evaporated. In other words we can always find a point *outside* the event horizon where a local observer will confirm that the black hole evaporated before the free falling object's arrival. The proper time for a free falling observer that falls from a finite height less than infinite but greater than 2m is finite. Let us say that this finite free falling proper time is 16 seconds, then the black hole evaporates in less than 16 seconds of the free faller's proper time. This means, if we believe Hawking, that in practice no free falling particle has ever fallen to or through an event horizon, so points 1 to 4 in the list become academic because they never happened in nature and in principle never will.

I appreciate the backup but i don't think hawking radiation is needed to show that no object can possibly reach an EV, it's common sense. The fact that in-falling objects can reach a point when they can no longer be caught should be a big hint. Everything will reach the RH before reaching the EV. The EV is the closest any possible RH can be to the singularity.

This is not correct. The infinite coordinate time for an observer to free-fall to the event horizon assumes an "eternal" black hole, one that never evaporates; that is what the Schwarzschild solution to the EFE describes. When you add in evaporation a la Hawking, the spacetime as a whole is no longer the exact Schwarzschild spacetime, so it no longer takes infinite coordinate time for an observer to free-fall to the horizon. Instead, the coordinate time at which the hole finally evaporates (the formula you gave) is also the coordinate time at which any free-falling observer hits the horizon. However, you have to be very careful drawing implications from that statement; it is *not* the same as saying that all free-falling observers hit the horizon at the same event.

If you look at a Penrose diagram for an evaporating black hole spacetime, as shown for example on this page (towards the bottom of the page)...

http://nrumiano.free.fr/Estars/bh_thermo.html

...you will see that the horizon is still a line (a null line going up and to the right), just as it is for an "eternal" black hole, so there is room for an infinite number of "crossing" events at which a free-falling observer crosses the horizon. The difference is that now the horizon has a future endpoint; since the horizon is a null line, light rays from *all* events that took place on the horizon (such as all the "crossing" events) all pass through that endpoint and then radiate outward to be seen by distant observers.

So basically, at the event of final evaporation of the hole, a flash is emitted that contains images of all the free-falling observers that crossed the horizon, just as they were crossing it. (Unfortunately, the free-falling observers themselves have already fallen into the singularity, since that is still to the future of any timelike worldline that crosses the horizon.) That single flash therefore gets assigned a single coordinate time in the "approximately Schwarzschild" coordinate system used by a distant observer, which will be the coordinate time of the final evaporation *and* of all the "crossing events" whose images are contained in the flash.

What? How can the free-falling observers have already fallen into the singularity? If you're ten light minutes away from them I they don't cross in the next ten minutes then they hadn't already crossed the horizon. The fact that you'll never see them cross means they never cross the bloody horizon! I'll say it again: The reason the light from those objects slows as they approach the horizon is because time slows as they approach the horizon. It's not an illusion!

Well, I am not completely convinced by the Penrose diagram but maybe that is just because I do not fully trust/understand the diagrams, which seem to demonstrate all sorts of unlikely things like nipping from one universe to another, which probably don't happen in reality. For example, take a look at this adaptation of the diagram you linked to:

The diagram shows light rays coming from the surface (orange curve) of a star collapsing to form the black hole. Examination of the path of observer A (green curve) shows he sees the star continue to shine all the way until he hits the singularity. Examination of the path of the second observer (light blue curve) shows he sees the star stop shining at event B but at that point the black hole has already ceased to exist. It seems that no observer sees a region of space that is gravitationally attractive that is not shining and only stop seeing the ghost of the star that formed the black hole after they hit the singularity or after the the black hole has evaporated, so its seems no one ever sees the black hole as black, although there will be a brief time when the light from the star that formed the black hole is highly redshifted and difficult to observe just before the black hole expires in a flash. Does that seem right?

Time's frozen at the horizon so of course some light from the star will always take longer then the black hole's been there for as you get close to the horizon.

When you look at the Penrose diagrams for, e.g., the maximally extended Schwarzschild spacetime (which includes the "white hole" and the second exterior region), I agree the diagrams are showing things which, although they follow mathematically from the given solution to the EFE, are probably not physically reasonable. However, I don't think that's the case for this particular diagram; it presents a picture which seems, at least to me, to be consistent and in principle physically realizable, although perhaps counterintuitive.

That exact same logic is used to try to show that books like the bible make some kind of sense. "Well I like the final conclusion so I'll work backwards from there, even though this forced reverse engineering will lead to things that seem 'counterintuitive'."

 

[PeterDonis]

Okay, that's twice now that you've said "No I meant this..." and then repeated what I've said but worded it slightly differently. I'm getting used to it but you've now done it twice to the same statement.

But the "slightly different words" indicate a *very* different meaning, which you keep refusing to accept. In this case, the key item you refuse to accept is that the two objects are both freely falling and both feel no force even though their radial separation is changing with time. See further comment below.

But as I've already said twice: The equivalent to the Rindler horizon using gravity just places a limit on the distance an object can be in order to catch the free-faller before the black hole's gone. This distance is individual and depends on the distance of the free-faller to the other object and to the black hole. Maybe I've misunderstood the Rindler horizon?

Two things here: first, yes, you *are* misunderstanding the Rindler horizon (or at least you're mis-stating something), because the Rindler horizon (in the case of flat spacetime, no gravity) marks the limit of the distance an object can be in order to catch an *accelerating* object, meaning an object that is undergoing proper acceleration, due to rockets firing or some other physical mechanism that changes its 4-momentum. (More precisely, it's the limit of the distance an object can be and still emit a light ray that can catch an accelerating object).

Second, you keep on bringing in the evaporation of the black hole in the gravity case, and you still haven't answered the question I keep asking: in the case where the black hole is eternal--it never evaporates--do you agree that a freely falling observer *would* reach the horizon (and then the singularity after that)? If you do, then we can refocus this whole discussion on whether or not having the black hole evaporate eventually makes a difference. If you don't, then the evaporation of the black hole is irrelevant because you don't believe the horizon is reachable even when there is no evaporation.

Maybe I wasn't clear enough. I'm going to explain this one more time as simply as I can. My thinking is that objects do accelerate in free-fall, kind of. That's what the Schwarzschild coordinates show. This acceleration has to be felt and I think it's felt as tidal force because tidal force is caused by the difference in velocity between objects large enough to notice the difference. Tidal force is just the difference in the rate of acceleration. Constant acceleration (as perceived by a distant observer) due to free-fall isn't felt as acceleration (so please no-one bother to point this out again, it's getting very frustrating). It's the equivalent to being at rest but an increase in the rate of acceleration is felt, as proper acceleration. This is all tidal force is. The rate of acceleration determines the value of length contraction and time dilation. The event horizon is the closest possible Rindler horizon to the singularity.

You've been quite clear enough about this multiple times, and as I and others have told you multiple times, your thinking here is *wrong*. Period. (I've put the particular statement that zeroes in on the wrong idea in bold.) As long as you refuse to accept that, and keep on hanging on to this wrong idea, you will remain stuck and unable to understand what we're telling you about how the black hole horizon actually works.

 

[A-wal]

But the "slightly different words" indicate a *very* different meaning, which you keep refusing to accept. In this case, the key item you refuse to accept is that the two objects are both freely falling and both feel no force even though their radial separation is changing with time. See further comment below.

Yes, I definitely refuse to accept that (unless they're point like objects). If they're freely falling and separating as they do it then that same force that's separating them will also affect the objects themselves. It my be marginal but that's not the point. How could there possibly be a force acting on the space between them but not the space that they occupy?

Two things here: first, yes, you *are* misunderstanding the Rindler horizon (or at least you're mis-stating something), because the Rindler horizon (in the case of flat spacetime, no gravity) marks the limit of the distance an object can be in order to catch an *accelerating* object, meaning an object that is undergoing proper acceleration, due to rockets firing or some other physical mechanism that changes its 4-momentum. (More precisely, it's the limit of the distance an object can be and still emit a light ray that can catch an accelerating object).

Right! And in the case of a black hole it marks the limit of the distance an object can be in order to catch a *free-falling* object. The event horizon marks the tightest possible Rindler horizon.

Second, you keep on bringing in the evaporation of the black hole in the gravity case, and you still haven't answered the question I keep asking: in the case where the black hole is eternal--it never evaporates--do you agree that a freely falling observer *would* reach the horizon (and then the singularity after that)? If you do, then we can refocus this whole discussion on whether or not having the black hole evaporate eventually makes a difference. If you don't, then the evaporation of the black hole is irrelevant because you don't believe the horizon is reachable even when there is no evaporation.

I've answered that question more than once. If I was to accelerate away from Earth forever then I would still never reach c, just as it's impossible to reach an EV even if it lasts forever. Think about it. If it wasn't possible to reach any event horizon of a radiating BH then it can't be possible to reach the EV of an everlasting one because an object would have to cross at a given time (from any coordinate system, it doesn't matter). The length of time between the start of the black holes life and the time when the object crosses would obviously have to be finite so it's a nothing question.

You've been quite clear enough about this multiple times, and as I and others have told you multiple times, your thinking here is *wrong*. Period. (I've put the particular statement that zeroes in on the wrong idea in bold.) As long as you refuse to accept that, and keep on hanging on to this wrong idea, you will remain stuck and unable to understand what we're telling you about how the black hole horizon actually works.

I don't mind if I'm *wrong*. I'm just trying to make sure you know what I mean, and therefore know exactly what it is that you're refuting. And I don't see how hanging on to this idea will make me any less open minded to what I'm being told. I'll stop hanging on to it when it no longer makes sense to me for whatever reason.

 

[PeterDonis]

Yes, I definitely refuse to accept that (unless they're point like objects). If they're freely falling and separating as they do it then that same force that's separating them will also affect the objects themselves. It my be marginal but that's not the point. How could there possibly be a force acting on the space between them but not the space that they occupy?

You're assuming that there is a force acting on the space between them. There isn't. The *spacetime* in question does not have any "force" acting on it. It's just curved. The curvature of the spacetime is what causes the two freely falling objects to separate, even though they both feel no force. Your mental model of what's going on is giving you wrong answers because it uses the concept of "force" as a basic concept, instead of the concept of "curved spacetime". (In the curved spacetime view, the term "force", strictly speaking, should not be used for what we've been calling, informally, "tidal force"; that's why I've tried to use the term "tidal gravity" instead where possible.)

I realize that I've simply stated a contrary view, without *explaining* why you should prefer that view to yours. Nor does it help, apparently, to point out that my view gives correct answers to other questions (such as the current one, whether or not a black hole's event horizon is reachable) and yours doesn't, because you simply deny that your view's answer is incorrect--your view seems logically consistent to you, so you accept the answers it gives you even when the contradict the answers given by standard general relativity. Since we can't directly test the specific question about black holes experimentally, I'm not sure how to resolve the discrepancy. But see further comments below.

Right! And in the case of a black hole it marks the limit of the distance an object can be in order to catch a *free-falling* object. The event horizon marks the tightest possible Rindler horizon.

No, this is wrong. The analogy between a black hole horizon (which occurs in a curved spacetime, the Schwarzschild spacetime) and a Rindler horizon (which occurs in flat spacetime) is exact on this point: both represent the limit of the distance an object can be in order to catch (more precisely, send a light ray to catch) an *accelerating* object.

It *is* also true that an object inside a black hole's horizon can't send a light ray to catch outgoing free-falling objects *outside* the horizon (for example, an object moving at almost the speed of light directly radially outward, just outside the black hole horizon, could, if it were moving fast enough, escape to infinity; a light ray launched outward at or beneath the horizon can't ever get outside the horizon). But this feature of the black hole spacetime has nothing to do with the horizon being analogous to a Rindler horizon.

I've answered that question more than once. If I was to accelerate away from Earth forever then I would still never reach c, just as it's impossible to reach an EV even if it lasts forever. Think about it. If it wasn't possible to reach any event horizon of a radiating BH then it can't be possible to reach the EV of an everlasting one because an object would have to cross at a given time (from any coordinate system, it doesn't matter). The length of time between the start of the black holes life and the time when the object crosses would obviously have to be finite so it's a nothing question.

After reading back through the thread, you're correct, you have answered this before; but I think the review was helpful. Once again, your analysis is wrong, and I've highlighted the particular phrase that zeroes in on the point where you're wrong in bold. The length of time in question does *not* have to be finite in every coordinate system; in particular, in Schwarzschild coordinates, the "Schwarzschild time" t at which any object that crosses the horizon does so is plus infinity. But the region labeled "plus infinity" in a Schwarzschild coordinate diagram is *not* a single point (i.e., not a single event); it's an infinite line of events, all of which get labeled "plus infinity" by Schwarzschild coordinates. That's why Schwarzschild coordinates are not good ones to use to study what happens at the horizon. Other coordinate systems work better because they give all those events on the horizon separate labels, so that they are "visible" and can be analyzed. In Painleve coordinates, for example, the horizon corresponds to a line (r = 2M, T = minus infinity to plus infinity); each event on this line is a *different* event at which some object can cross the horizon. The region covered by the Schwarzschild coordinates (r = 2M, t = minus infinity to plus infinity) is all shoved down "below" T = minus infinity in Painleve coordinates.

Once again, I've just stated a contrary view, without really explaining why you should accept it. And again, in the absence of actual experimental data about black hole horizons, I'm not sure how to resolve the discrepancy (but see next comment). However, I do emphasize that what I've said above is not at all controversial; I've stated the position of standard general relativity, accepted and used by everyone that works in the field, as best I can (and others are welcome to correct or clarify if I've misstated something). I should also emphasize that, however unusual it sounds, what I've said above is perfectly consistent, logically and mathematically.

I don't mind if I'm *wrong*. I'm just trying to make sure you know what I mean, and therefore know exactly what it is that you're refuting. And I don't see how hanging on to this idea will make me any less open minded to what I'm being told. I'll stop hanging on to it when it no longer makes sense to me for whatever reason.

I think I have a decent understanding of two points on which you hold views that are inconsistent with the standard view of general relativity, and therefore lead you to answers to certain questions that are inconsistent with those of standard general relativity (to put it as neutrally as possible and avoid using the word "wrong"). Just to recap, the two points are:

(1) You believe that two freely falling objects separating due to tidal gravity must feel some force (because there must be a "force on the space between them"). In standard GR (and in Newtonian gravity as well), there is no such force; the two objects, separating due to tidal gravity, are both freely falling and feel no force. (In Newtonian gravity, both objects are responding to the "force" of gravity, and they separate because of the spatial variation in that force; but even in Newtonian physics, the "force" of gravity is not *felt* as a force--objects moving only under gravity are in free fall and feel no force, they are weightless. A person standing on the surface of the Earth feels weight because the Earth is pushing up on them, not because gravity is pulling them down, even in Newtonian physics.)

(2) You believe that any object that crosses a black hole horizon must do so at a finite value of the time coordinate in any coordinate system. In standard GR, this is false (I explained the standard GR view in my last comment just above).

As I said above, without experimental data on black hole horizons, it's impossible to directly test point (2). However, there is a lot of indirect evidence that GR holds just fine in strong gravity regimes fairly close to objects that we can't explain, on our current knowledge, unless they are black holes. And also, as I said, mathematically the standard GR view on point (2) is perfectly consistent, so if you're having trouble seeing how it can be so, looking further into how the various coordinate charts of Schwarzschild spacetime work might help.

As far as point (1) is concerned, I would think that there would be plenty of direct tests that freely falling objects separating due to tidal gravity are both weightless, but I haven't been able to find any decent treatment of such tests on the web. But, once again, mathematically, the curved spacetime model is perfectly consistent, so if you're having trouble understanding how it can be so, a better understanding of how it works might help.

 

[A-wal]

You're assuming that there is a force acting on the space between them. There isn't. The *spacetime* in question does not have any "force" acting on it. It's just curved. The curvature of the spacetime is what causes the two freely falling objects to separate, even though they both feel no force. Your mental model of what's going on is giving you wrong answers because it uses the concept of "force" as a basic concept, instead of the concept of "curved spacetime". (In the curved spacetime view, the term "force", strictly speaking, should not be used for what we've been calling, informally, "tidal force"; that's why I've tried to use the term "tidal gravity" instead where possible.)

Okay but if you look at it as curved space-time rather than a force then I'm not really sure how that changes anything. It's just semantics. I do normally look at it as curved space-time btw when there's more than two objects because it becomes easier. Looking at it in this way the space that the objects occupy is curved as well as the space between them so there'll still be a force acting on the objects themselves.

No, this is wrong. The analogy between a black hole horizon (which occurs in a curved spacetime, the Schwarzschild spacetime) and a Rindler horizon (which occurs in flat spacetime) is exact on this point: both represent the limit of the distance an object can be in order to catch (more precisely, send a light ray to catch) an *accelerating* object.

RIGHT! I didn't mean an actual Rindler horizon because that's already been defined. I'm saying you can use the same principle for gravity by substituting acceleration for free-fall because of tidal acceleration. An observer in flat space-time who sends a message from beyond the Rindler horizon to an accelerating observer shouldn't expect the message to ever reach the accelerating observer until their acceleration slows. The same thing happens if you send a message to a free-falling object. The free-faller will appear to slow from the distant observers perspective but so does the message.

After reading back through the thread, you're correct, you have answered this before; but I think the review was helpful. Once again, your analysis is wrong, and I've highlighted the particular phrase that zeroes in on the point where you're wrong in bold. The length of time in question does *not* have to be finite in every coordinate system; in particular, in Schwarzschild coordinates, the "Schwarzschild time" t at which any object that crosses the horizon does so is plus infinity. But the region labeled "plus infinity" in a Schwarzschild coordinate diagram is *not* a single point (i.e., not a single event); it's an infinite line of events, all of which get labeled "plus infinity" by Schwarzschild coordinates. That's why Schwarzschild coordinates are not good ones to use to study what happens at the horizon. Other coordinate systems work better because they give all those events on the horizon separate labels, so that they are "visible" and can be analyzed. In Painleve coordinates, for example, the horizon corresponds to a line (r = 2M, T = minus infinity to plus infinity); each event on this line is a *different* event at which some object can cross the horizon. The region covered by the Schwarzschild coordinates (r = 2M, t = minus infinity to plus infinity) is all shoved down "below" T = minus infinity in Painleve coordinates.

Okay, maybe not in any coordinate system then. I really just meant that if it's impossible to reach the EV of a black hole then it's impossible no matter how long it lasts because it would have to happen at a certain time and what if the black hole's life is finite but it last longer than this?

I think I have a decent understanding of two points on which you hold views that are inconsistent with the standard view of general relativity, and therefore lead you to answers to certain questions that are inconsistent with those of standard general relativity (to put it as neutrally as possible and avoid using the word "wrong"). Just to recap, the two points are:

(1) You believe that two freely falling objects separating due to tidal gravity must feel some force (because there must be a "force on the space between them"). In standard GR (and in Newtonian gravity as well), there is no such force; the two objects, separating due to tidal gravity, are both freely falling and feel no force. (In Newtonian gravity, both objects are responding to the "force" of gravity, and they separate because of the spatial variation in that force; but even in Newtonian physics, the "force" of gravity is not *felt* as a force--objects moving only under gravity are in free fall and feel no force, they are weightless. A person standing on the surface of the Earth feels weight because the Earth is pushing up on them, not because gravity is pulling them down, even in Newtonian physics.)

Yea that's how I look at it. The Earth accelerates outward pushing us downward. Length contraction means the Earth isn't expand as this happens. I got shot down for looking at it that way a few posts ago. I've already relied to the first bit.

(2) You believe that any object that crosses a black hole horizon must do so at a finite value of the time coordinate in any coordinate system. In standard GR, this is false (I explained the standard GR view in my last comment just above).

Not really, I think it's just that I worded it wrong. I shouldn't have said “in any coordinate system”. I should of said “has to happen at a given time regardless of the coordinate system you want to use”. If it is possible to pass the event horizon then you should be able to work out the time this happens according to the watch of the external observer even though they won't be able to see it yet. If that true then they'll be able to see the object reach the horizon if they wait long enough. It's not a self-consistent idea. Not to me anyway.

As I said above, without experimental data on black hole horizons, it's impossible to directly test point (2). However, there is a lot of indirect evidence that GR holds just fine in strong gravity regimes fairly close to objects that we can't explain, on our current knowledge, unless they are black holes. And also, as I said, mathematically the standard GR view on point (2) is perfectly consistent, so if you're having trouble seeing how it can be so, looking further into how the various coordinate charts of Schwarzschild spacetime work might help.

Coordinate systems aren't how I like to look at it. By definition they give an extremely limited perspective. I don't see how the standard view is self-consistent. Nobody's posted an answer to the paradox I posted ages ago yet.

As far as point (1) is concerned, I would think that there would be plenty of direct tests that freely falling objects separating due to tidal gravity are both weightless, but I haven't been able to find any decent treatment of such tests on the web. But, once again, mathematically, the curved spacetime model is perfectly consistent, so if you're having trouble understanding how it can be so, a better understanding of how it works might help.

Weightless? The very front would be but not the rest of the object. The rest is being dragged along at least slightly.

 

[PeterDonis]

Okay but if you look at it as curved space-time rather than a force then I'm not really sure how that changes anything. It's just semantics. I do normally look at it as curved space-time btw when there's more than two objects because it becomes easier. Looking at it in this way the space that the objects occupy is curved as well as the space between them so there'll still be a force acting on the objects themselves.

Nope. At least, if that's how you are using the concept "curved spacetime", then you are using it wrong. Or if you object to the word "wrong", then I'd say that the concept you are labeling with the term "curved spacetime" is *not* the same as the concept that standard GR labels with the term "curved spacetime". In standard GR, spacetime (flat *or* curved) does not exert any force on anything. All it does is define what states of motion are freely falling (geodesic) at each event and what states of motion are not. Which particular state of motion any actual object is actually *in* depends on the object and what it is doing (for example, is it firing rockets or not), not on "spacetime".

A general comment in view of what I've just said: as I see it, this discussion has gone on so long because every time someone tries to explain how standard GR treats a certain concept, you come back with your own personal version of that concept, instead of trying to understand the standard GR version.

RIGHT! I didn't mean an actual Rindler horizon because that's already been defined. I'm saying you can use the same principle for gravity by substituting acceleration for free-fall because of tidal acceleration. An observer in flat space-time who sends a message from beyond the Rindler horizon to an accelerating observer shouldn't expect the message to ever reach the accelerating observer until their acceleration slows. The same thing happens if you send a message to a free-falling object. The free-faller will appear to slow from the distant observers perspective but so does the message.

I can sort of see what you're describing here in the case of free fall towards a black hole, but it still has nothing to do with the black hole horizon being analogous to a Rindler horizon, and the idea of "substituting acceleration for free-fall because of tidal acceleration" makes no sense in standard GR.

Yea that's how I look at it. The Earth accelerates outward pushing us downward. Length contraction means the Earth isn't expand as this happens. I got shot down for looking at it that way a few posts ago.

"Length contraction" due to spacetime curvature has nothing to do with the Earth not expanding even though it's pushing objects outward. That's why you got shot down. (Also, the Earth pushes us *upward*--or outward--not *downward*.)

If it is possible to pass the event horizon then you should be able to work out the time this happens according to the watch of the external observer even though they won't be able to see it yet. If that true then they'll be able to see the object reach the horizon if they wait long enough. It's not a self-consistent idea. Not to me anyway.

Once again, your reasoning is based on a false premise (I've highlighted it in bold again in the quote above). In standard GR, the "time this happens" (i.e., the event of an observer crossing the horizon) is plus infinity according to the watch of any external observer (outside the horizon). They will *never* see the object crossing the horizon, even if they wait an infinite amount of time. This is all perfectly self-consistent; it just uses a model that does not satisfy your intuition.

I don't see how the standard view is self-consistent. Nobody's posted an answer to the paradox I posted ages ago yet.

Can you re-post it or point me to it? I don't remember what it was. (Maybe it wasn't in this thread?)

Weightless? The very front would be but not the rest of the object. The rest is being dragged along at least slightly.

I've asked several times already to restrict discussion to idealized, point-like "test objects" that have no internal parts, and therefore no internal forces between the parts. For such objects, there's no such thing as the front of the object "dragging the rest along", so what you've said above would not apply. As I've also said several times already, the reason this idealization is not unreasonable is that even for a non-idealized object, the motion of its *center of mass* will still be "weightless" motion; the object will feel no weight (no force, no acceleration) at its center of mass, and therefore will undergo no proper acceleration as a whole (though parts of it may accelerate and feel force relative to other parts). It's the center-of-mass motion that is being referred to when standard GR (and Newtonian physics too) say that bodies moving only under gravity (including tidal gravity) are weightless.

 

[PeterDonis]

I've asked several times already to restrict discussion to idealized, point-like "test objects"

Looking back at your post #263, I realized that you did include the qualifier "unless they're both point-like objects". So you *do* agree that two point-like objects with no internal parts will both be freely falling and feel no force even if their radial separation changes with time due to tidal gravity? If so, does that mean you agree that a point-like object *could* reach a black hole horizon?

 

[A-wal]

Nope. At least, if that's how you are using the concept "curved spacetime", then you are using it wrong. Or if you object to the word "wrong", then I'd say that the concept you are labeling with the term "curved spacetime" is *not* the same as the concept that standard GR labels with the term "curved spacetime". In standard GR, spacetime (flat *or* curved) does not exert any force on anything. All it does is define what states of motion are freely falling (geodesic) at each event and what states of motion are not. Which particular state of motion any actual object is actually *in* depends on the object and what it is doing (for example, is it firing rockets or not), not on "spacetime".

What the hell were you reading? I didn't think I was being that unclear. You're so eager to keep disagreeing with everything I say that you're taking everything I write to mean something different to what I've actually said. This is semantics, as I said in the last post. You can look at it as a force acting between two objects. If it works then it's right. That's what right means. It may be more accurate to think of it as curved space-time but that's the same thing here. Instead of a force acting on the objects you could say the same-time between them feels the force rather than the objects themselves. But if you want to take that view then the space occupied by the objects is curved as well as the space between them, effectively replacing the force acting on them.

A general comment in view of what I've just said: as I see it, this discussion has gone on so long because every time someone tries to explain how standard GR treats a certain concept, you come back with your own personal version of that concept, instead of trying to understand the standard GR version.

This discussion has gone on for so long because virtually everything I post gets misunderstood or taken out of context so I spend most of the time on here rewording posts that I've already written.

I can sort of see what you're describing here in the case of free fall towards a black hole, but it still has nothing to do with the black hole horizon being analogous to a Rindler horizon, and the idea of "substituting acceleration for free-fall because of tidal acceleration" makes no sense in standard GR.

If there was an object that we could see as being one millimetre away from the horizon then would it reach the horizon if it was to move two millimetres forward from its own perspective? No, because that's not what it would look like if you were the one approaching the horizon. You'd be in length contracted space so you could move forwards without crossing the horizon. The millimetre from the previous perspective could be a mile or a light year or whatever if you were actually there, depending on the mass of the black hole. Bigger ones last longer but there's more length contracted space to pass through before you get to the horizon, and you'd be more time dilated so it would take longer to get there. I don't see how any conversion formula that's accurate for moving between the two perspectives can possibly claim that you could cross the horizon if you were there but not from a distance.

"Length contraction" due to spacetime curvature has nothing to do with the Earth not expanding even though it's pushing objects outward. That's why you got shot down. (Also, the Earth pushes us *upward*--or outward--not *downward*.)

LOL. I'm fairly positive it does and it doesn't, respectively.

Once again, your reasoning is based on a false premise (I've highlighted it in bold again in the quote above). In standard GR, the "time this happens" (i.e., the event of an observer crossing the horizon) is plus infinity according to the watch of any external observer (outside the horizon). They will *never* see the object crossing the horizon, even if they wait an infinite amount of time. This is all perfectly self-consistent; it just uses a model that does not satisfy your intuition.

Well it's not self-consistent as far as I can see. See below.

Can you re-post it or point me to it? I don't remember what it was. (Maybe it wasn't in this thread?)

A ship is deliberately pulled into a black hole. It crosses the horizon when the black hole is a certain size and there's a second observer who follows close behind but doesn't allow themselves to be pulled in. There's a very strong rope linking the two. The second observer can never witness the first one reaching the event horizon so it can never be too late for them to find the energy to pull the first observer away from the black hole even after the it's shrunk to a smaller than when the first one crossed from it's own perspective. If the closer one always has the potential to escape the black hole under its own power from the further ones perspective then it should always be possible for the closer one to escape under the further ones power. So the first ship can't escape from it's own perspective but it does from the second ships perspective. Paradox!

I've asked several times already to restrict discussion to idealized, point-like "test objects" that have no internal parts, and therefore no internal forces between the parts. For such objects, there's no such thing as the front of the object "dragging the rest along", so what you've said above would not apply. As I've also said several times already, the reason this idealization is not unreasonable is that even for a non-idealized object, the motion of its *center of mass* will still be "weightless" motion; the object will feel no weight (no force, no acceleration) at its center of mass, and therefore will undergo no proper acceleration as a whole (though parts of it may accelerate and feel force relative to other parts). It's the center-of-mass motion that is being referred to when standard GR (and Newtonian physics too) say that bodies moving only under gravity (including tidal gravity) are weightless.

Looking back at your post #263, I realized that you did include the qualifier "unless they're both point-like objects". So you *do* agree that two point-like objects with no internal parts will both be freely falling and feel no force even if their radial separation changes with time due to tidal gravity? If so, does that mean you agree that a point-like object *could* reach a black hole horizon?

Yes, I agree that two point-like objects with no internal parts will both be freely falling and feel no force even if their radial separation changes with time. Of course this has no effect on whether a point like object can reach the horizon. I don't see how anything can reach the horizon including light. It's because of the curvature of space-time that no object can reach the horizon. There'll be no time and an infinite amount of potential space as the object heads into ever more length contracted/time dilated space.

 

[PeterDonis]

If it works then it's right. That's what right means.

And if it doesn't work, then it's not right. Your view of tidal gravity as a force doesn't work; it leads you to incorrect conclusions about what happens at the horizon of a black hole. Or at least, it does if your view of tidal gravity as a force is contributing to your claim that nothing can reach a black hole horizon. See further comment below on post #263 and your response to my question about it.

I should clarify that the view of tidal gravity as a force can work in more restricted domains; for example, it works fine for understanding how the Moon causes ocean tides on the Earth. But the curved spacetime view also works fine in those more restricted domains, *and* it works in domains where the view of tidal gravity as a force doesn't. It's because we're talking about the latter type of domain that I've been saying the view of tidal gravity as a force is wrong.

If there was an object that we could see as being one millimetre away from the horizon then would it reach the horizon if it was to move two millimetres forward from its own perspective? No, because that's not what it would look like if you were the one approaching the horizon. You'd be in length contracted space so you could move forwards without crossing the horizon. The millimetre from the previous perspective could be a mile or a light year or whatever if you were actually there, depending on the mass of the black hole. Bigger ones last longer but there's more length contracted space to pass through before you get to the horizon, and you'd be more time dilated so it would take longer to get there. I don't see how any conversion formula that's accurate for moving between the two perspectives can possibly claim that you could cross the horizon if you were there but not from a distance.

You don't see how, but it's true. I've posted the calculation already in this thread--both the calculation that the proper time for an object to fall to the horizon from a finite radius is finite, *and* the calculation that the proper distance to the horizon from an object at a finite radius is finite. Those calculations already take into account *all* of the "length contraction" and "time dilation" caused by gravity as the horizon is approached and reached; which means that length contraction and time dilation do *not* "go to infinity" as the horizon is approached and reached, even though your intuitive picture tells you that they do. You appear to prefer your intuitive picture to the actual math, but that doesn't make your intuitive picture right. I've tried in previous posts to give an alternate intuitive picture that helps to illustrate how the math arrives at the result it does, but the intuitive picture is not why physicists accept the result. They accept it because of the math, and because wherever we've checked the math against experiment, it's been confirmed.

LOL. I'm fairly positive it does and it doesn't, respectively.

So you believe that the force exerted on your feet by the ground is pushing you downward? Remember that I'm speaking from the GR spacetime perspective, where "gravity" is not a force, so the only force the Earth is exerting on you is the push from the ground on your feet.

A ship is deliberately pulled into a black hole. It crosses the horizon when the black hole is a certain size and there's a second observer who follows close behind but doesn't allow themselves to be pulled in. There's a very strong rope linking the two. The second observer can never witness the first one reaching the event horizon so it can never be too late for them to find the energy to pull the first observer away from the black hole even after the it's shrunk to a smaller than when the first one crossed from it's own perspective. If the closer one always has the potential to escape the black hole under its own power from the further ones perspective then it should always be possible for the closer one to escape under the further ones power. So the first ship can't escape from it's own perspective but it does from the second ships perspective. Paradox!

The answer to this has already been posted in this thread, but it's been a while and it was spread over several posts, so I'll recap it here. But first, you keep bringing in evaporation even though you've said (or at least I think you've said) that you believe the horizon of an eternal black hole, one that never evaporates and has the same mass for all time, is also unreachable. If that's the case, the evaporation is irrelevant and you should rephrase the paradox to apply to the case of an eternal black hole. In what follows I'll treat the paradox in that form.

The error in your argument is in bold above. The key point you're missing is that the ability of the second observer to see the first one reaching the horizon depends on the behavior of *outgoing* light rays, whereas the ability of the second observer to affect the trajectory of the first observer depends on the behavior of *ingoing* light rays (and other ingoing objects or causal influences, like infalling observers and pulls on ropes). Those two behaviors are different in the spacetime around a black hole. Outgoing light rays that start at the horizon, r = 2M, stay at r = 2M forever; but ingoing light rays (and other ingoing things) can reach the horizon from outside it just fine. So the fact that the second observer can never receive an *outgoing* light signal from the event of the first observer crossing the horizon, does not prevent the *ingoing* first observer from falling through the horizon; nor does it prevent there from being a "last point" on the second observer's worldline where an *ingoing* causal influence can be emitted that will reach the first observer before he crosses the horizon.

You may ask what happens if the second observer pulls on the rope *after* the "last point" has passed. The answer is that the rope will already have broken, so no causal influence transmitted down the rope can reach the first observer any more. The rope will break at some point *before* the "last point" is reached, from the second observer's perspective; a rope that breaks just before the "last point" is reached is an "idealized" rope with the maximum possible tensile strength allowed by relativity, a tensile strength so high that the speed of sound in the rope equals the speed of light. A real rope, of course, will break long before that.

And what force is it that breaks the rope? Not "tidal gravity". In fact it's just the force exerted by the second observer, who has to be exerting a force on himself and anything attached to him in order to hover at a constant radius R above the horizon. This force holds one end of the rope and forces the rope to stretch as the first observer free-falls; eventually (just before the first observer crosses the horizon, if the rope is the "idealized" rope--much earlier, for any real rope), that stretching force will overcome the rope's tensile strength and the rope will break.

Yes, I agree that two point-like objects with no internal parts will both be freely falling and feel no force even if their radial separation changes with time. Of course this has no effect on whether a point like object can reach the horizon.

Does this mean you agree that, if we limit the discussion to point-like objects, tidal gravity is irrelevant to the question of whether an object can reach the horizon? Or do you still think tidal gravity plays a role somehow? I'm not clear about the role that your view of tidal gravity as a force is playing in your logic.

I don't see how anything can reach the horizon including light. It's because of the curvature of space-time that no object can reach the horizon. There'll be no time and an infinite amount of potential space as the object heads into ever more length contracted/time dilated space.

See my comment above. The time is not zero and the space is not infinite.

 

[A-wal]

And if it doesn't work, then it's not right. Your view of tidal gravity as a force doesn't work; it leads you to incorrect conclusions about what happens at the horizon of a black hole. Or at least, it does if your view of tidal gravity as a force is contributing to your claim that nothing can reach a black hole horizon. See further comment below on post #263 and your response to my question about it.

I should clarify that the view of tidal gravity as a force can work in more restricted domains; for example, it works fine for understanding how the Moon causes ocean tides on the Earth. But the curved spacetime view also works fine in those more restricted domains, *and* it works in domains where the view of tidal gravity as a force doesn't. It's because we're talking about the latter type of domain that I've been saying the view of tidal gravity as a force is wrong.

I wasn't talking about tidal gravity, I was just talking about gravity. I don't view gravity as a force unless it's easier. You're arguing with yourself again. And the view of tidal gravity isn't contributing to my claim that nothing can reach a black hole horizon. It's the other way round. I'm trying to bridge the gap between what I'm being told and how I think of it.

You don't see how, but it's true. I've posted the calculation already in this thread--both the calculation that the proper time for an object to fall to the horizon from a finite radius is finite, *and* the calculation that the proper distance to the horizon from an object at a finite radius is finite. Those calculations already take into account *all* of the "length contraction" and "time dilation" caused by gravity as the horizon is approached and reached; which means that length contraction and time dilation do *not* "go to infinity" as the horizon is approached and reached, even though your intuitive picture tells you that they do. You appear to prefer your intuitive picture to the actual math, but that doesn't make your intuitive picture right. I've tried in previous posts to give an alternate intuitive picture that helps to illustrate how the math arrives at the result it does, but the intuitive picture is not why physicists accept the result. They accept it because of the math, and because wherever we've checked the math against experiment, it's been confirmed.

If the equations work then that doesn't necessarily tell the whole story. People thought Newtons equations worked until they were applied to the orbit of Mercury. It's what the equations represent that has to make sense. Equations can work but if what they represent doesn't than the they aren't complete. I don't know how anyone can be satisfied with a bunch of numbers and symbols if they can't understand what they're actually supposed to represent.

So you believe that the force exerted on your feet by the ground is pushing you downward? Remember that I'm speaking from the GR spacetime perspective, where "gravity" is not a force, so the only force the Earth is exerting on you is the push from the ground on your feet.

No I believe the force exerted on my feet by the ground is pushing me upward while length contraction stops me from going anywhere.

The answer to this has already been posted in this thread, but it's been a while and it was spread over several posts, so I'll recap it here. But first, you keep bringing in evaporation even though you've said (or at least I think you've said) that you believe the horizon of an eternal black hole, one that never evaporates and has the same mass for all time, is also unreachable. If that's the case, the evaporation is irrelevant and you should rephrase the paradox to apply to the case of an eternal black hole. In what follows I'll treat the paradox in that form.

The error in your argument is in bold above. The key point you're missing is that the ability of the second observer to see the first one reaching the horizon depends on the behavior of *outgoing* light rays, whereas the ability of the second observer to affect the trajectory of the first observer depends on the behavior of *ingoing* light rays (and other ingoing objects or causal influences, like infalling observers and pulls on ropes). Those two behaviors are different in the spacetime around a black hole. Outgoing light rays that start at the horizon, r = 2M, stay at r = 2M forever; but ingoing light rays (and other ingoing things) can reach the horizon from outside it just fine. So the fact that the second observer can never receive an *outgoing* light signal from the event of the first observer crossing the horizon, does not prevent the *ingoing* first observer from falling through the horizon; nor does it prevent there from being a "last point" on the second observer's worldline where an *ingoing* causal influence can be emitted that will reach the first observer before he crosses the horizon.

You may ask what happens if the second observer pulls on the rope *after* the "last point" has passed. The answer is that the rope will already have broken, so no causal influence transmitted down the rope can reach the first observer any more. The rope will break at some point *before* the "last point" is reached, from the second observer's perspective; a rope that breaks just before the "last point" is reached is an "idealized" rope with the maximum possible tensile strength allowed by relativity, a tensile strength so high that the speed of sound in the rope equals the speed of light. A real rope, of course, will break long before that.

And what force is it that breaks the rope? Not "tidal gravity". In fact it's just the force exerted by the second observer, who has to be exerting a force on himself and anything attached to him in order to hover at a constant radius R above the horizon. This force holds one end of the rope and forces the rope to stretch as the first observer free-falls; eventually (just before the first observer crosses the horizon, if the rope is the "idealized" rope--much earlier, for any real rope), that stretching force will overcome the rope's tensile strength and the rope will break.

I already answered that. From the perspective of the distant observer it will always be possible to pull the free-faller back because the free-faller will always be outside of the horizon from the perspective of the distant observer (as well as their own). So it is always possible to pull anyone away from a black hole in a finite amount of time, using a finite amount of energy and a finite strength rope.

Does this mean you agree that, if we limit the discussion to point-like objects, tidal gravity is irrelevant to the question of whether an object can reach the horizon? Or do you still think tidal gravity plays a role somehow? I'm not clear about the role that your view of tidal gravity as a force is playing in your logic.

I don't think tidal gravity is the reason why objects can't reach the horizon. I think it's a consequence of it.

See my comment above. The time is not zero and the space is not infinite.

But we know the time is zero and the space is infinite. Look at the outgoing light rays. At the horizon they're frozen. The only way light can be frozen like that is if time has come to a complete stop at the horizon. The speed of light is a measurement of the speed of time. I don't see how it could possibly work one way but not the other. The version you're pushing contradicts itself when it switches between ingoing light rays that aren't effected by time dilation as much as outgoing ones, and also when switching between the views of a distant observer who does witness infinite time dilation at the horizon and the free-faller who doesn't. Look at it in terms of light years/minutes etc. The free-faller would measure the distance between themselves and the horizon by the local speed of light. At the horizon light stops so there's always an infinite amount of light years between any object and any event horizon. So if there's an infinite amount of potential space between you and the horizon htf are you meant to reach it? I really do believe that it's the equivalent of trying to reach c. The reason you can't reach c is because c is effectively infinitely fast, so how are you supposed to go from a finite speed to an infinite one? You obviously can't, but it doesn't look infinite from any given rest frame. So even though the speed of light appears reachable it's not because it effectively changes as you change frame in order to remaining constant. The distance between a free-faller and the horizon always looks finite but it isn't.

 

[PeterDonis]

the view of tidal gravity isn't contributing to my claim that nothing can reach a black hole horizon. It's the other way round. I'm trying to bridge the gap between what I'm being told and how I think of it.

Ok, this helps me to understand where you're coming from.

If the equations work then that doesn't necessarily tell the whole story. People thought Newtons equations worked until they were applied to the orbit of Mercury. It's what the equations represent that has to make sense. Equations can work but if what they represent doesn't than the they aren't complete. I don't know how anyone can be satisfied with a bunch of numbers and symbols if they can't understand what they're actually supposed to represent.

I do understand what the numbers and symbols are supposed to represent. So do the physicists who accept the predictions of relativity regarding black holes. I apologize if I didn't make it clear that when I was talking about "math", I was including the fact that the math has to have meanings assigned to the symbols and equations that link them to actual experimental data, and that the data has to match the math under those assigned meanings.

Also, I've already agreed that we don't have any direct experimental evidence about what happens as black hole horizons are approached (although we do have indirect evidence and so far it bears out the predictions of relativity). If in the future we do get direct evidence that does not agree with the predictions of GR, physicists will look for a better theory, just as they did with Newtonian mechanics when there was evidence that that wasn't quite right.

No I believe the force exerted on my feet by the ground is pushing me upward while length contraction stops me from going anywhere.

How can "length contraction" stop you from moving? Please clarify the mental picture you have here. Are you using "length contraction" to refer to something like the "river model" of Schwarzschild spacetime (described in this paper http://arxiv.org/abs/gr-qc/0411060), where space itself is viewed as "flowing" inward towards the black hole?

I already answered that. From the perspective of the distant observer it will always be possible to pull the free-faller back because the free-faller will always be outside of the horizon from the perspective of the distant observer (as well as their own). So it is always possible to pull anyone away from a black hole in a finite amount of time, using a finite amount of energy and a finite strength rope.

And I already explained what is wrong with your reasoning here; you're assuming that the behavior of outgoing signals and causal influences must be the same as the behavior of ingoing signals and causal influences. They're not, and this invalidates the reasoning in the quote above. Try again.

But we know the time is zero and the space is infinite. Look at the outgoing light rays. At the horizon they're frozen. The only way light can be frozen like that is if time has come to a complete stop at the horizon.

No, they're not frozen. Outgoing light rays at r = 2M stay at r = 2M forever because spacetime is curved at r = 2M to the point where outgoing null lines are vertical ("vertical" meaning vertical as shown in a spacetime diagram that has time going up-down and radius going left-right). That doesn't mean outgoing null lines are "frozen"; outgoing light rays at r = 2M still "move" the same as light rays anywhere else.

The speed of light is a measurement of the speed of time.

Huh?

The version you're pushing contradicts itself when it switches between ingoing light rays that aren't effected by time dilation as much as outgoing ones...

Ingoing light rays behave differently from outgoing ones because they are ingoing, not outgoing. There's no physical requirement that the two *must* behave the same. In flat spacetime it so happens that light rays going in any direction behave the same, but that's because the spacetime is flat, not because of any physical requirement that light rays behave the same regardless of the direction they're traveling, in any spacetime whatsoever.

...and also when switching between the views of a distant observer who does witness infinite time dilation at the horizon and the free-faller who doesn't. Look at it in terms of light years/minutes etc. The free-faller would measure the distance between themselves and the horizon by the local speed of light. At the horizon light stops so there's always an infinite amount of light years between any object and any event horizon. So if there's an infinite amount of potential space between you and the horizon htf are you meant to reach it? I really do believe that it's the equivalent of trying to reach c. The reason you can't reach c is because c is effectively infinitely fast, so how are you supposed to go from a finite speed to an infinite one? You obviously can't, but it doesn't look infinite from any given rest frame. So even though the speed of light appears reachable it's not because it effectively changes as you change frame in order to remaining constant. The distance between a free-faller and the horizon always looks finite but it isn't.

False premise marked in bold above. All the rest of your argument here is based on that false premise, so it fails since the premise fails. Do you have any other argument that isn't based on a false premise?

 

[A-wal]

How can "length contraction" stop you from moving? Please clarify the mental picture you have here. Are you using "length contraction" to refer to something like the "river model" of Schwarzschild spacetime (described in this paper http://arxiv.org/abs/gr-qc/0411060), where space itself is viewed as "flowing" inward towards the black hole?

Length contraction and time dilation mean that my body is attracted to the mass at the centre of the Earth and acceleration in the opposite direction (due to the fact that atoms don't particularly like being squashed too close together) is in perfect balance with that length contraction/time dilation, meaning that I can stay stationary relative to the centre of the Earth. So if it wasn't for the length contraction/time dilation then we would be accelerating outwards. Of course in reality there couldn't be more acceleration than length contraction/time dilation because the acceleration is a reaction to them. But that misses the point I was trying to make.

And I already explained what is wrong with your reasoning here; you're assuming that the behavior of outgoing signals and causal influences must be the same as the behavior of ingoing signals and causal influences. They're not, and this invalidates the reasoning in the quote above. Try again.

That doesn't even make a little bit of sense, just like a black hole that lasts forever. How could it last forever? It's a finite sphere in three dimensions but not four? That's almost as silly as thinking you can get sucked into a hole to nowhere. So tell me Alice, how does the direction of the light have any effect on the flow of time, forcing the same point in space-time to exist in multiple states simultaneously? I didn't know it could do that.

No, they're not frozen. Outgoing light rays at r = 2M stay at r = 2M forever because spacetime is curved at r = 2M to the point where outgoing null lines are vertical ("vertical" meaning vertical as shown in a spacetime diagram that has time going up-down and radius going left-right). That doesn't mean outgoing null lines are "frozen"; outgoing light rays at r = 2M still "move" the same as light rays anywhere else.

I don't know what that meant. It seemed to me that you were trying to show that this graph would suggest that time is frozen at the horizon but this actually isn't the case. Why not?

Huh?

I could go deeper into how the speed of light also represents the speed of time but this will do. If I see light moving at .5c either because that light is coming from somewhere close to a gravitational source or because the light is the internal clock of something moving at a high velocity relative to me then I can work out what speed it's going or the mass of the object in the area the light is coming from.

Ingoing light rays behave differently from outgoing ones because they are ingoing, not outgoing. There's no physical requirement that the two *must* behave the same. In flat spacetime it so happens that light rays going in any direction behave the same, but that's because the spacetime is flat, not because of any physical requirement that light rays behave the same regardless of the direction they're traveling, in any spacetime whatsoever.

Huh? Ingoing light rays behave differently from outgoing ones because they are ingoing, not outgoing? That's not a reason!There may be no physical requirement that the light rays must behave the same but I'm not talking about the light rays themselves, I'm talking about the space-time that they're passing through. I don't see how the curvature of space-time can change the behavioUr of light depending on its direction.

False premise marked in bold above. All the rest of your argument here is based on that false premise, so it fails since the premise fails. Do you have any other argument that isn't based on a false premise?

No, I'm quite happy to argue that it isn't a false premise for now.

I think we've finally honed in on where the differences in perception are coming from.

 

[QuantumClue]

I definitely remember reading something official that said the laws of physics don't distinguish between the past and the future. I thinkit might have been A Brief HistoryOf Time. You could run it backwards and it would still work just as well. But now I've thought about it, there's something I can't resolve. Take two objects in space that are static relative to each other. They would gravitate towards each other. Now if time was running backwards then they would be moving away from each other. So gravity would be a repulsive force. But that doesn't work because if time was running backwards on Earth, we would still be pulled towards the planet, not pushed away. In other words it would work in freefall/at rest, but not when accelerating against gravity. How can it be both repulsive and attractive at the same distances?

http://www.motionmountain.net/download.html

"Time is a concept introduced specially to describe the flow of events around us; it does not itself flow, it describes flow. Time does not advance. Time is neither linear nor cyclic. The idea that time flows is as hindering to understanding nature as is the idea that mirrors Page 71 exchange right and left. The misleading use of the expression ‘flow of time’, propagated first by some flawed Ref. 36 Greek thinkers and then again by Newton, continues. Aristotle (384/3–322 bce), careful to think logically, pointed out its misconception, and many did so after him. Nevertheless, expressions such as ‘time reversal’, the ‘irreversibility of time’, and the much-abused ‘time’s arrow’ are still common. Just read a popular science magazine chosen at random''

For there to be an arrow of time, from past to future, there needs to be linearity involved in it. This does not exist, so an arrow time does not exist. There is a psychological arrow of time which appears to us as being linear in nature, that is, one which destinguishes our past from our future, but that is psychological, not a physical arrow.

 

[PeterDonis]

I think we've finally honed in on where the differences in perception are coming from.

I agree we're zeroing in on a key difference in our models, but it's not just a difference in perception. Your model makes a definite prediction that is different from standard GR, and could in principle be tested, as follows:

Hover at a given radial coordinate r > 2M above a black hole's horizon. Calculate, using standard GR, the proper time \tau_{crit} that will elapse to the "point of no return" for you with respect to a freely falling object dropped from that radial coordinate--in other words, the proper time that will elapse to the last possible moment at which a causal influence from you, hovering, can reach the freely falling object before it crosses the horizon, according to standard GR. Drop an object attached to a rope, let it free-fall, and wait for a time by your ship's clock significantly longer than \tau_{crit}. Then tug on the rope. If your model is correct, you should be able to pull the object back up; if standard GR is correct, every time you run this experiment you'll only get back a broken rope.

Of course, as I've noted before, we're not going to be able to run this experiment any time soon. Can you think of any other experiments where your model makes a different prediction from standard GR? If so, it might help me to understand your model, because from what you've described of it so far, I don't see any underlying structure that would allow me to derive predictions from it, other than the prediction for the experiment I just described. All I see is a flat statement of one proposition--"time stops at the horizon"--that is contrary to a theorem of GR, but you haven't derived it as a theorem in your model, you've just stated it. (You've also stated that the various ways I and others have tried to explain how GR shows that time doesn't stop at the horizon don't convince you, but that isn't the same as giving a positive chain of reasoning from your own model.)

 

[PeterDonis]

That doesn't even make a little bit of sense, just like a black hole that lasts forever. How could it last forever? It's a finite sphere in three dimensions but not four? That's almost as silly as thinking you can get sucked into a hole to nowhere. So tell me Alice, how does the direction of the light have any effect on the flow of time, forcing the same point in space-time to exist in multiple states simultaneously? I didn't know it could do that.

I never said the light affects the flow of time, or the "existence" of points in spacetime. All I said was that ingoing light behaves differently from outgoing light, because it's going in a different direction--towards the central mass instead of away from it. See next comment.

Huh? Ingoing light rays behave differently from outgoing ones because they are ingoing, not outgoing? That's not a reason!There may be no physical requirement that the light rays must behave the same but I'm not talking about the light rays themselves, I'm talking about the space-time that they're passing through. I don't see how the curvature of space-time can change the behavioUr of light depending on its direction.

The curvature of spacetime around a central mass has a directional asymmetry built into it because there's a central mass; going towards the central mass and going away from it are physically different. What's so difficult about that? We know this happens from experiments showing the gravitational redshift. Outgoing light is redshifted; ingoing light is blueshifted. Outgoing light behaves differently from ingoing light. What's the problem?

 

[PeterDonis]

How could it last forever? It's a finite sphere in three dimensions but not four?

In four dimensions, it's a hypercylinder, with infinite extension in the time direction. What's wrong with that? It's the same as mathematically modeling an infinitely long cylinder in three dimensions--the cylinder is a "finite circle" in two dimensions, but infinite in the third. Just add one dimension to the circle to make a sphere. What's the problem?

I should note, by the way, that I'm not sure the black hole actually *is* a "finite sphere" in three dimensions, because of the singularity at the center, r = 0. But I agree that the horizon is a 2-sphere--more precisely, the intersection of the horizon with a slice of constant time is a 2-sphere. So in the full spacetime, including the time dimension, the horizon (of an "eternal" black hole) will be an infinitely long hypercylinder, as I described above.

 

[A-wal]

http://www.motionmountain.net/download.html

"Time is a concept introduced specially to describe the flow of events around us; it does not itself flow, it describes flow. Time does not advance. Time is neither linear nor cyclic. The idea that time flows is as hindering to understanding nature as is the idea that mirrors Page 71 exchange right and left. The misleading use of the expression ‘flow of time’, propagated first by some flawed Ref. 36 Greek thinkers and then again by Newton, continues. Aristotle (384/3–322 bce), careful to think logically, pointed out its misconception, and many did so after him. Nevertheless, expressions such as ‘time reversal’, the ‘irreversibility of time’, and the much-abused ‘time’s arrow’ are still common. Just read a popular science magazine chosen at random''

My question was within the context of a perceived moving timeline. If you run the clock backwards then everything would obviously happen in reverse, but it wouldn't change the laws of physics. Gravity would still be attractive. If you run the clock backwards in the version of relativity that I'm being presented with then matter going into a black hole wouldn't come back out if you reversed the clock, because gravity is still attractive and nothing can escape from beyond the horizon. So it's completely self-contradictory to say that an object can cross the horizon according to gr because gr is supposed to be time reversible. It's no wonder general relativity breaks down at the horizon.

For there to be an arrow of time, from past to future, there needs to be linearity involved in it. This does not exist, so an arrow time does not exist. There is a psychological arrow of time which appears to us as being linear in nature, that is, one which destinguishes our past from our future, but that is psychological, not a physical arrow.

I agree that there's no real arrow of time. In the same way that the speed of light/time gets its value purely from how fast our brains work. The universe is static, but four-dimensional. The only difference between the past and the future is the fact that we remember one and not the other. We have to perceive it like that to have conciousness/intelligence. Otherwise we'd die the instant we were born. I wonder what would happen if we met aliens who's memories worked the opposite way? They'd already remember us, until we meet. Then we'd remember them and they'd forget us. That would be a interesting and very confusing conversation.

I agree we're zeroing in on a key difference in our models, but it's not just a difference in perception. Your model makes a definite prediction that is different from standard GR, and could in principle be tested, as follows:

Hover at a given radial coordinate r > 2M above a black hole's horizon. Calculate, using standard GR, the proper time \tau_{crit} that will elapse to the "point of no return" for you with respect to a freely falling object dropped from that radial coordinate--in other words, the proper time that will elapse to the last possible moment at which a causal influence from you, hovering, can reach the freely falling object before it crosses the horizon, according to standard GR. Drop an object attached to a rope, let it free-fall, and wait for a time by your ship's clock significantly longer than \tau_{crit}. Then tug on the rope. If your model is correct, you should be able to pull the object back up; if standard GR is correct, every time you run this experiment you'll only get back a broken rope.

I've been saying it's a slightly different model from the start, and that experiment looks exactly the same as the one I posted in the paradox. You saying the rope will break no matter how strong it is? Infinite energy? That's not possible! If if was then you could reach c.

Of course, as I've noted before, we're not going to be able to run this experiment any time soon. Can you think of any other experiments where your model makes a different prediction from standard GR? If so, it might help me to understand your model, because from what you've described of it so far, I don't see any underlying structure that would allow me to derive predictions from it, other than the prediction for the experiment I just described. All I see is a flat statement of one proposition--"time stops at the horizon"--that is contrary to a theorem of GR, but you haven't derived it as a theorem in your model, you've just stated it. (You've also stated that the various ways I and others have tried to explain how GR shows that time doesn't stop at the horizon don't convince you, but that isn't the same as giving a positive chain of reasoning from your own model.)

It's not really that time's frozen at the horizon. Time dilation/length contraction increase exponentially as you approach the event horizon, just as they do when approaching c. Describing a black hole at the horizon is like describing what a rainbow looks like from directly underneath. It's frozen in the same sense that time is frozen at c. It can't be reached no matter how much energy is involved. I don't see how you could gradually reach something that's effectively infinitely far away. When you approach the horizon it recedes (in comparison to flat linear space-time). In other words if an object is exactly five light years away and you move one light year closer then it will be at least slightly further away than four light years. You don't need a black hole for this. It applies to any amount of mass. The rate it recedes depends on the amount of length contraction/time dilation (time dilation would make it recede just as must as length contraction). The closer you get the more pronounced the difference is. What would happen if there was enough mass to cause so much length contraction/time dilation that even something moving at c (or as close to c as you like) would never be able to reach it? To me this is the obvious definition of an event horizon. The same thing happens with the Rindler horizon. I don't really think of a black hole as a physical object. Just define the event horizon as the closest point to the singularity that light can reach from the outside due to there being enough mass to produce time dilation/length contraction >c over a certain area of space-time. I don't see why it needs to be any more complicated than that.

I never said the light affects the flow of time, or the "existence" of points in spacetime. All I said was that ingoing light behaves differently from outgoing light, because it's going in a different direction--towards the central mass instead of away from it. See next comment.

You're saying that two light rays at the same point in space-time are effected by different amounts of time dilation/length contraction because of the direction they're traveling in. Doesn't seem to make any kind of sense. Presumably if this were true then velocity in that direction would also be a factor in gravitational time dilation. Velocity as cause of time dilation/length contraction has nothing to do with direction.

The curvature of spacetime around a central mass has a directional asymmetry built into it because there's a central mass; going towards the central mass and going away from it are physically different. What's so difficult about that? We know this happens from experiments showing the gravitational redshift. Outgoing light is redshifted; ingoing light is blueshifted. Outgoing light behaves differently from ingoing light. What's the problem?

What difference does it make if it's in-falling rather than out-going? The curvature of space-time doesn't depend on the direction of what's passing through it. Gravitational red/blue shift is the same as Doppler red/blue shift, or the equivalent to the effect of time dilation when viewing something that’s moving at a different relative velocity but not accelerating. In the twin paradox, that's actually not a paradox at all, it means both twins are the same age when they meet up. If you add acceleration then one will be older. If objects behave differently depending on the direction they're traveling through space-time (whether flat or not) then it would create a situation that would be the equivalent to each twin being older than the other one when they meet up.

In four dimensions, it's a hypercylinder, with infinite extension in the time direction. What's wrong with that? It's the same as mathematically modeling an infinitely long cylinder in three dimensions--the cylinder is a "finite circle" in two dimensions, but infinite in the third. Just add one dimension to the circle to make a sphere. What's the problem?

The problem is that you're okay with something being infinitely long in one dimension but presumably not in the other three. For one thing the idea that anything can be infinitely long is ridiculous, and for another, why would it be a different length in time? Time dilation and length contraction are equivalent to each other. You should be able to work out how much time it's got left based on it's size in the other dimensions. It should be a simple conversion.

I should note, by the way, that I'm not sure the black hole actually *is* a "finite sphere" in three dimensions, because of the singularity at the center, r = 0. But I agree that the horizon is a 2-sphere--more precisely, the intersection of the horizon with a slice of constant time is a 2-sphere. So in the full spacetime, including the time dimension, the horizon (of an "eternal" black hole) will be an infinitely long hypercylinder, as I described above.

As an actual object it's infinitely small because it's the singularity that actually exists as a physical object. A black hole is just an effect. An infinitely long lived black hole would have to be infinitely big as well. There's no such thing as infinity. Why is there an infinity symbol in the equation for black hole formation? What would happen if it were replaced with c? I bet the difference between the two is tidal force because it represents acceleration into a stronger gravitational field. The energy needed approaching c and the EV should even be the same. The Rindler horizon was presented to me as evidence that an object can reach the horizon, but it in fact suggests the opposite. The Rindler horizon actually shows that an object that accelerates away at a fast enough rate can't be caught by a given object/wave, even traveling at c. When an object accelerates into a higher gravitational field (tidal force) this acts exactly as an object accelerating away from another object in flat space-time. When this acceleration reaches a certain value then nothing further away from the black hole can catch the closer object. The event horizon is the point that nothing can reach, when everything goes beyond that Rindler horizon.

 

[PeterDonis]

I've been saying it's a slightly different model from the start, and that experiment looks exactly the same as the one I posted in the paradox.

Good, then I'm understanding you correctly. Since your model is physically different--it makes different experimental predictions--I don't see that further discussion is going to be very productive, since we're not talking about the same physical model. I'm still willing to try and explain how the GR model works, but I'm not sure how much progress we're going to make, since you basically keep on asserting "I can't see how..." based on the different model you have in your head. Since that model seems incoherent to me, I can't really see how to explain why GR works differently.

In view of what I've just said, I'll refrain from commenting on most of the rest of your post, since I would basically just have to keep on asserting things that are obvious facts in the GR model, but which seem obviously false to you based on your model. Maybe someday we'll be able to run the rope experiment and see who's right (but see next comment for a clarification on the GR view of that experiment).

You saying the rope will break no matter how strong it is? Infinite energy? That's not possible! If if was then you could reach c.

No, I'm saying that relativity imposes a limit on the strength of materials; they can't be infinitely strong. The simplest way to state the limit is that the speed of sound in the material must be less than the speed of light. This imposes a finite limit on the material properties that determine the speed of sound, among which is the tensile strength.

 

[PeterDonis]

The problem is that you're okay with something being infinitely long in one dimension but presumably not in the other three.

Presumption incorrect. Just to clarify, the "eternal" black hole spacetime is infinite in all four dimensions, one of time and three of space. Standard spherical coordinates may be misleading in this respect since the three infinite spatial dimensions are all encompassed in the radial coordinate r. However, if you transform to Cartesian coordinates you can clearly see the three infinite spatial dimensions. The reason that's not normally done is that the metric for a black hole spacetime looks a lot uglier in Cartesian coordinates.

 

[A-wal]

Good, then I'm understanding you correctly. Since your model is physically different--it makes different experimental predictions--I don't see that further discussion is going to be very productive, since we're not talking about the same physical model. I'm still willing to try and explain how the GR model works, but I'm not sure how much progress we're going to make, since you basically keep on asserting "I can't see how..." based on the different model you have in your head. Since that model seems incoherent to me, I can't really see how to explain why GR works differently.

In what way does it seem incoherent? The model is identical apart from the extra relative component of gravity, which to my mind tidies it up nicely. The standard black hole description is so messy. Normally the difference is marginal.

No, I'm saying that relativity imposes a limit on the strength of materials; they can't be infinitely strong. The simplest way to state the limit is that the speed of sound in the material must be less than the speed of light. This imposes a finite limit on the material properties that determine the speed of sound, among which is the tensile strength.

Can you clarify this please? I get the idea but are you saying there's a definite upper limit on the amount of energy that can be exerted onto an object? What happens if you use a thicker rope once this limit's been reached, or two ropes?

Presumption incorrect. Just to clarify, the "eternal" black hole spacetime is infinite in all four dimensions, one of time and three of space. Standard spherical coordinates may be misleading in this respect since the three infinite spatial dimensions are all encompassed in the radial coordinate r. However, if you transform to Cartesian coordinates you can clearly see the three infinite spatial dimensions. The reason that's not normally done is that the metric for a black hole spacetime looks a lot uglier in Cartesian coordinates.

"The three infinite spatial dimensions are all encompassed in the radial coordinate r" seems contradictory to me. Do you mean it's only infinite from the inside? Is that why you can't escape? I know the standard view is that every direction is facing the singularity which is why acceleration in any direction will speed up your approach to the singularity but I've never heard of the interior being infinite in size before. The Tardis hypothesis! I can't get my head round that one. Maybe that's not what you meant?

 

[PeterDonis]

In what way does it seem incoherent? The model is identical apart from the extra relative component of gravity, which to my mind tidies it up nicely. The standard black hole description is so messy. Normally the difference is marginal.

Your model makes definite predictions that are very different from the standard model. It predicts that the horizon can't be reached, and that the "rope experiment" will give different results. But I have seen no logical structure from you that leads to those predictions; all I see is your intuitive sense that they "make sense". You certainly haven't stated a coherent model that is "identical" to standard GR "apart from the extra relative component of gravity"; I'm not exactly sure what that means, but I do know that you can't arrive at a consistent model that makes the predictions you're making just by "adding in" some simple extra ingredient to standard GR. Standard GR is a very precise, specific logical structure, and adding anything to it like that would make it inconsistent. You may think you have stated a coherent model, but you haven't; all you've done is made some intuitive, hand-waving statements that don't form a coherent logical structure.

Can you clarify this please? I get the idea but are you saying there's a definite upper limit on the amount of energy that can be exerted onto an object? What happens if you use a thicker rope once this limit's been reached, or two ropes?

I'm saying there's a finite upper limit on the *breaking strength* of any material. There's no limit on the amount of force you can exert on the material (at least, not in principle), so any material will eventually break.

However, just saying "finite breaking strength" may not be the best way of stating the limit, because it suggests questions like you asked, about adding more ropes or making the rope thicker. Previously, I stated the limit as "the speed of sound in the material must be less than the speed of light", which should make it clear that in the case we've been discussing, adding more ropes or making the rope thicker won't help, because it won't change the fastest speed at which any force exerted on one end of the rope can propagate to the other end. That speed must be less than the speed of light, and as long as it is, the rope must break.

But even that way of stating it may still be misleading, because it makes it seem like it's the speed of *inward* propagation of the force through the rope that's critical. That's not quite right. The real problem is that, once the lower end of the rope drops below the horizon, *it* would have to move faster than light to keep up with the portion of the rope that's still above the horizon. It can't do that, so the rope has to break. The "critical time" I've been referring to is simply the last time when a force applied at the top end of the rope, by the "hovering" observer, can propagate down the rope, at the speed of light, and reach the free-falling observer at the lower end before he crosses the horizon--meaning, before he would have to move faster than light to keep up.

"The three infinite spatial dimensions are all encompassed in the radial coordinate r" seems contradictory to me.

Consider ordinarly 3-dimensional Euclidean space. It is infinite in all three spatial dimensions. If we describe it in Cartesian coordinates, x, y, z, then all three coordinates have infinite range, so it's "obvious" in these coordinates that all three spatial dimensions are infinite.

Now look at the same space in spherical coordinates, r, theta, phi. Only r has an infinite range; theta and phi are limited to a finite range (the normal convention is theta from 0 to pi, and phi from 0 to 2 pi). But it's the same space as before, so all three spatial dimensions are still infinite. It's just that the coordinates make it harder to see because the infinite range of all three spatial dimensions is "tied up" in the infinite range of the r coordinate alone. (A vector pointing in the direction of increasing r can point in *any* direction in the space, depending on the theta and phi coordinates.)

Describing spacetime around a black hole works the same way; you're just adding a t coordinate, so you have four infinite dimensions now instead of three, but only two coordinates (t and r) with infinite range. The r coordinate encompasses the infinity of three of the dimensions of the spacetime (all the spatial ones). Strictly speaking, this only refers to the exterior portion of the spacetime (outside the horizon); the portion inside works differently (see below).

Do you mean it's only infinite from the inside? <snip> Maybe that's not what you meant?

You're right, it's not what I meant. The spacetime as a whole is infinite in extent, but if you consider just the interior, the portion inside the horizon, I'm pretty sure that portion is finite. (I haven't actually done the calculation of its 4-volume to be certain.)

 

[A-wal]

Your model makes definite predictions that are very different from the standard model. It predicts that the horizon can't be reached, and that the "rope experiment" will give different results. But I have seen no logical structure from you that leads to those predictions; all I see is your intuitive sense that they "make sense". You certainly haven't stated a coherent model that is "identical" to standard GR "apart from the extra relative component of gravity"; I'm not exactly sure what that means, but I do know that you can't arrive at a consistent model that makes the predictions you're making just by "adding in" some simple extra ingredient to standard GR. Standard GR is a very precise, specific logical structure, and adding anything to it like that would make it inconsistent. You may think you have stated a coherent model, but you haven't; all you've done is made some intuitive, hand-waving statements that don't form a coherent logical structure.

I respectfully disagree. It’s based on the idea that the curvature created by matter is indistinguishable from the curvature created by energy accept that the curvature from energy is *c squared greater than the curvature from matter. That’s why gravity’s so weak.

I'm saying there's a finite upper limit on the *breaking strength* of any material. There's no limit on the amount of force you can exert on the material (at least, not in principle), so any material will eventually break.

However, just saying "finite breaking strength" may not be the best way of stating the limit, because it suggests questions like you asked, about adding more ropes or making the rope thicker. Previously, I stated the limit as "the speed of sound in the material must be less than the speed of light", which should make it clear that in the case we've been discussing, adding more ropes or making the rope thicker won't help, because it won't change the fastest speed at which any force exerted on one end of the rope can propagate to the other end. That speed must be less than the speed of light, and as long as it is, the rope must break.

But even that way of stating it may still be misleading, because it makes it seem like it's the speed of *inward* propagation of the force through the rope that's critical. That's not quite right. The real problem is that, once the lower end of the rope drops below the horizon, *it* would have to move faster than light to keep up with the portion of the rope that's still above the horizon. It can't do that, so the rope has to break. The "critical time" I've been referring to is simply the last time when a force applied at the top end of the rope, by the "hovering" observer, can propagate down the rope, at the speed of light, and reach the free-falling observer at the lower end before he crosses the horizon--meaning, before he would have to move faster than light to keep up.

Nice. Especially the third paragraph.

Consider ordinarly 3-dimensional Euclidean space. It is infinite in all three spatial dimensions. If we describe it in Cartesian coordinates, x, y, z, then all three coordinates have infinite range, so it's "obvious" in these coordinates that all three spatial dimensions are infinite.

Now look at the same space in spherical coordinates, r, theta, phi. Only r has an infinite range; theta and phi are limited to a finite range (the normal convention is theta from 0 to pi, and phi from 0 to 2 pi). But it's the same space as before, so all three spatial dimensions are still infinite. It's just that the coordinates make it harder to see because the infinite range of all three spatial dimensions is "tied up" in the infinite range of the r coordinate alone. (A vector pointing in the direction of increasing r can point in *any* direction in the space, depending on the theta and phi coordinates.)

Describing spacetime around a black hole works the same way; you're just adding a t coordinate, so you have four infinite dimensions now instead of three, but only two coordinates (t and r) with infinite range. The r coordinate encompasses the infinity of three of the dimensions of the spacetime (all the spatial ones). Strictly speaking, this only refers to the exterior portion of the spacetime (outside the horizon); the portion inside works differently (see below).

I think we’re getting further from the point that from the exterior it has a finite three dimensional spherical shape so I see absolutely no reason why it should be any different in the other one.

You're right, it's not what I meant. The spacetime as a whole is infinite in extent, but if you consider just the interior, the portion inside the horizon, I'm pretty sure that portion is finite. (I haven't actually done the calculation of its 4-volume to be certain.)

I would have thought it would have to be.

 

[PeterDonis]

I respectfully disagree. It’s based on the idea that the curvature created by matter is indistinguishable from the curvature created by energy accept that the curvature from energy is *c squared greater than the curvature from matter. That’s why gravity’s so weak.

I'm not sure what you're disagreeing with. In standard GR, curvature created by matter *is* indistinguishable from curvature generated by energy; in fact, "matter" and "energy" are really the same thing, just measured in different units, and the speed of light squared is just a conversion factor between the different units.

So if you mean that your own model is based on this idea as well, then your model should give the same predictions as standard GR, and it doesn't. So your model can't be "based on" this idea and still be consistent. If you mean something else, please clarify.

I think we’re getting further from the point that from the exterior it has a finite three dimensional spherical shape so I see absolutely no reason why it should be any different in the other one.

The *horizon* has a finite three dimensional spherical shape (more precisely, the intersection of the horizon with a slice of constant time has that shape). But the horizon is not the entire spacetime. When I was talking about dimensions being infinite in extent, I was talking about the entire spacetime (more precisely, the portion exterior to the horizon).

I would have thought it would have to be.

I think so too, but it's not entirely obvious because the portion of the spacetime inside the horizon still covers an infinite range of "time" coordinates. (This is true not only in the interior Schwarzschild coordinates, but in Painleve coordinates and in Kruskal-Szeres coordinates.) But I believe (though I haven't confirmed by explicit calculation--maybe one of the experts on these forums has) that the integral for the 4-volume of the interior portion of the spacetime converges fast enough as the "time" coordinate goes to infinity to make the total 4-volume finite, for reasons similar to the reasons that the integrals for proper time to fall to the horizon and proper distance to the horizon converge (as I showed much earlier in this thread).

(Also, I put "time" in scare-quotes because inside the horizon, the "time" coordinates of the first two coordinate systems I mentioned, Schwarzschild interior and Painleve, are not timelike! In Schwarzschild interior coordinates, the radial coordinate r is timelike inside the horizon. In Painleve coordinates, all four coordinates are spacelike inside the horizon! George Jones discusses that in this post from a thread where the subject came up:

https://www.physicsforums.com/showpost.php?p=3000266&postcount=121

Kruskal coordinates are more "normal" in this respect, there is one timelike coordinate and three spacelike--one "radial" and the two angular coordinates--and the timelike/spacelike nature of each coordinate remains the same throughout the full range.)

 

[A-wal]

I'm not sure what you're disagreeing with. In standard GR, curvature created by matter *is* indistinguishable from curvature generated by energy; in fact, "matter" and "energy" are really the same thing, just measured in different units, and the speed of light squared is just a conversion factor between the different units.

So if you mean that your own model is based on this idea as well, then your model should give the same predictions as standard GR, and it doesn't. So your model can't be "based on" this idea and still be consistent. If you mean something else, please clarify.

I know one of the principles of general relativity is that the curvature is the same, but it then goes on to describe a type of curvature that's different from the curvature in special relativity. It should be relative, just like velocity, which is why you can't feel the difference between being in gravitational fields of different strengths when in free-fall. In special relativity there is a limit to how fast you can go relative to anything else. You can keep on accelerating but you'll never reach c. I think it should work exactly the same for gravity. You shouldn't be able to curve it beyond 90 degrees whether the force is created by energy or matter. There should be an upper limit equivalent to c (event horizon) even though you can keep on accelerating just by free-falling (tidal force). One way of looking at it is that gravity can and does reach a value that exceeds c but that would be unreachable for any observer because the world-line between you and the horizon would be potentially infinite. Not actually infinite but a line that actually reaches the horizon would be infinite, so it's impossible. Its four-dimensional volume(?) would depend on its mass and the distance of the observer from the horizon. That's not new. In standard GR the size of the Earth in all four dimensions decreases the closer you get to it, but it's normally marginal. In the case of a black hole its volume in all four dimensions would always reach zero before anything could reach the horizon. The fact that the curvature is caused by matter rather than energy means the equivalent curvature from energy would be much greater. So the question of why gravity is so weak is really asking why E=mc squared, and we have special relativity to answer that.

The *horizon* has a finite three dimensional spherical shape (more precisely, the intersection of the horizon with a slice of constant time has that shape). But the horizon is not the entire spacetime. When I was talking about dimensions being infinite in extent, I was talking about the entire spacetime (more precisely, the portion exterior to the horizon).

Okay but I don't know how anything could be infinite in extent. You say the horizon has a finite three-dimensional spherical shape so why wouldn't it have a four-dimensional spherical shape?

I think so too, but it's not entirely obvious because the portion of the spacetime inside the horizon still covers an infinite range of "time" coordinates. (This is true not only in the interior Schwarzschild coordinates, but in Painleve coordinates and in Kruskal-Szeres coordinates.) But I believe (though I haven't confirmed by explicit calculation--maybe one of the experts on these forums has) that the integral for the 4-volume of the interior portion of the spacetime converges fast enough as the "time" coordinate goes to infinity to make the total 4-volume finite, for reasons similar to the reasons that the integrals for proper time to fall to the horizon and proper distance to the horizon converge (as I showed much earlier in this thread).

(Also, I put "time" in scare-quotes because inside the horizon, the "time" coordinates of the first two coordinate systems I mentioned, Schwarzschild interior and Painleve, are not timelike! In Schwarzschild interior coordinates, the radial coordinate r is timelike inside the horizon. In Painleve coordinates, all four coordinates are spacelike inside the horizon! George Jones discusses that in this post from a thread where the subject came up:

https://www.physicsforums.com/showpost.php?p=3000266&postcount=121

Kruskal coordinates are more "normal" in this respect, there is one timelike coordinate and three spacelike--one "radial" and the two angular coordinates--and the timelike/spacelike nature of each coordinate remains the same throughout the full range.)

See, messy!

 

[PeterDonis]

I know one of the principles of general relativity is that the curvature is the same, but it then goes on to describe a type of curvature that's different from the curvature in special relativity.

Huh? There is no curvature in special relativity; in SR, spacetime is flat.

It should be relative, just like velocity, which is why you can't feel the difference between being in gravitational fields of different strengths when in free-fall.

Yes, this is called the equivalence principle, and is one of the cornerstones of general relativity. It is perfectly consistent with being able to reach and pass through a black hole's horizon.

In special relativity there is a limit to how fast you can go relative to anything else. You can keep on accelerating but you'll never reach c.

Fine so far, but...

I think it should work exactly the same for gravity. You shouldn't be able to curve it beyond 90 degrees whether the force is created by energy or matter.

What does this have to do with velocity? And what does "curve it beyond 90 degrees" mean? Are you talking about the tilting of the light cones (see my next comment)? If so, you're wrong that they can't tilt beyond "90 degrees" (meaning "vertical" in the sense I describe below). They can.

There should be an upper limit equivalent to c (event horizon) even though you can keep on accelerating just by free-falling (tidal force).

There is an upper limit to velocity in GR, just as in SR; but it's expressed in a more general form that works when spacetime is curved (the form "you can't go faster than c relative to anything else" only works when spacetime is flat). In GR, we say that worldlines of objects can't go outside the light cones. That still works the same in flat spacetime, because all the light cones are aligned with each other (the "sides" of the light cones tilt upward at a 45 degree angle, to the left and right if you're looking at a spacetime diagram showing one time and one space coordinate). But it carries over to curved spacetime, where light cones at different events can be tilted with respect to each other. At the horizon of a black hole, the light cones are tilted over so that the radially outgoing sides of the cones are vertical--they stay at r = 2M forever. The ingoing sides of the cones, however, still tilt inward, just as they do far away from the horizon.

One way of looking at it is that gravity can and does reach a value that exceeds c but that would be unreachable for any observer because the world-line between you and the horizon would be potentially infinite.

Except that it isn't, as has already been shown in this thread.

The fact that the curvature is caused by matter rather than energy means the equivalent curvature from energy would be much greater.

Huh? Energy and matter are just different forms of the same thing; a given amount of energy causes exactly the same curvature as the equivalent amount of matter (with the formula E = mc^2 defining what is "equivalent").

Okay but I don't know how anything could be infinite in extent. You say the horizon has a finite three-dimensional spherical shape so why wouldn't it have a four-dimensional spherical shape?

Because it doesn't. That's not what the mathematics of a black hole solution to the Einstein Field Equation describes. There are other solutions that (sort of) describe a four-dimensional sphere (for example, the kind of "Euclidean" models that Stephen Hawking talks about in his "no boundary" proposal for the beginning of the universe), but they don't describe black holes.

See, messy!

How so? Different coordinate systems are useful for different purposes, so they have different properties. This is not something that's particular to GR; it happens in ordinary geometry too. Look at maps of the Earth in different projections--Mercator, stereographic, etc.--and compare to an actual globe. The different projections distort the globe in different ways in order to represent particular properties on a flat map. You can't make a flat map of a curved surface and not have it distorted, so "messiness", if you insist on calling it that, is unavoidable unless you want to carry a globe around with you all the time. And in the case of spacetime, we don't have that option; there's *no* way to represent a 4-dimensional manifold in three or fewer dimensions without distortion, and we choose maps that have different distortions for different purposes. What's the problem?

 

[A-wal]

Huh? There is no curvature in special relativity; in SR, spacetime is flat.

Space-time is flat. Objects can follow curved paths through it using energy to accelerate. The curvature from acceleration can never get you to c, and the curvature from gravity can never get you to c. You said that the rope would always break because anything beyond the event horizon has broken the light barrier. If you can’t use energy to do it then what makes you think you can do it with the curvature from matter? The strength of the gravitational field makes no difference, just like velocity. The only difference is that you accelerate harder in free-fall when in higher gravity, but you’ll never be able to accelerate hard enough to reach c.

Yes, this is called the equivalence principle, and is one of the cornerstones of general relativity. It is perfectly consistent with being able to reach and pass through a black hole's horizon.

Only if you choose to ignore or have the cheek to disagree with what I just said.

What does this have to do with velocity? And what does "curve it beyond 90 degrees" mean? Are you talking about the tilting of the light cones (see my next comment)? If so, you're wrong that they can't tilt beyond "90 degrees" (meaning "vertical" in the sense I describe below). They can.

A velocity of c as a right-angle on a space-time diagram like the one I described before. Gravitational curvature shouldn’t be able to move you to 90 degrees any more than energy curvature can. You say it can, but why when we know that they’re equivalent?

There is an upper limit to velocity in GR, just as in SR; but it's expressed in a more general form that works when spacetime is curved (the form "you can't go faster than c relative to anything else" only works when spacetime is flat). In GR, we say that worldlines of objects can't go outside the light cones. That still works the same in flat spacetime, because all the light cones are aligned with each other (the "sides" of the light cones tilt upward at a 45 degree angle, to the left and right if you're looking at a spacetime diagram showing one time and one space coordinate). But it carries over to curved spacetime, where light cones at different events can be tilted with respect to each other. At the horizon of a black hole, the light cones are tilted over so that the radially outgoing sides of the cones are vertical--they stay at r = 2M forever. The ingoing sides of the cones, however, still tilt inward, just as they do far away from the horizon.

Why would using the two different causes of curvature together be any different from using more of one?

Except that it isn't, as has already been shown in this thread.

Except that it is, as has also already been shown in this thread, in a slightly less technical and more hand-wavy way.

Huh? Energy and matter are just different forms of the same thing; a given amount of energy causes exactly the same curvature as the equivalent amount of matter (with the formula E = mc^2 defining what is "equivalent").

Right, so why is it such a mystery that the strength of electro-magnetism is so much greater than gravity when electro-magnetism is curvature from energy and gravity is from matter?

Because it doesn't. That's not what the mathematics of a black hole solution to the Einstein Field Equation describes. There are other solutions that (sort of) describe a four-dimensional sphere (for example, the kind of "Euclidean" models that Stephen Hawking talks about in his "no boundary" proposal for the beginning of the universe), but they don't describe black holes.

Perhaps they do. There’s no complex mechanism to consider when talking about the life span of a black hole, just time-dilation and length contraction which are equivalent. So it should have same length in all four dimensions.

How so? Different coordinate systems are useful for different purposes, so they have different properties. This is not something that's particular to GR; it happens in ordinary geometry too. Look at maps of the Earth in different projections--Mercator, stereographic, etc.--and compare to an actual globe. The different projections distort the globe in different ways in order to represent particular properties on a flat map. You can't make a flat map of a curved surface and not have it distorted, so "messiness", if you insist on calling it that, is unavoidable unless you want to carry a globe around with you all the time. And in the case of spacetime, we don't have that option; there's *no* way to represent a 4-dimensional manifold in three or fewer dimensions without distortion, and we choose maps that have different distortions for different purposes. What's the problem?

It’s messy because from what I’ve seen, there’s so many different ways of measuring it in the standard version that you can manipulate it to show anything you want.

 

[PeterDonis]

Space-time is flat. Objects can follow curved paths through it using energy to accelerate. The curvature from acceleration can never get you to c, and the curvature from gravity can never get you to c.

You are using very confusing terminology, which may help to explain why your thinking is confused. Specifically, you are using the same word, "curvature", to refer to two distinct concepts that should not be conflated. Let me try to restate what I think you mean by the two uses of the word "curvature" to see if I'm understanding you correctly:

(1) What you are calling "curvature due to matter" or "curvature from gravity" is curvature of *spacetime* itself. As I've pointed out, spacetime curvature can actually be caused by matter *or* energy; it can also be caused by pressure and stress in a material. When the word "curvature" is used in standard GR without qualification, it is this kind of curvature that is meant; and every time I've used the word "curvature", including the expression "curvature due to energy", I've been using it to mean this.

(2) What you are calling "curvature due to energy" or "curvature from acceleration" is curvature of the *worldline* of an object *within* a spacetime (which may itself be either flat or curved). The word "curvature" is *never* (as far as I've seen) used in standard GR to mean this; the word "acceleration" (or the term "proper acceleration" if more precision is needed) is used. You are correct that, in order to follow such a path, an object must normally expend energy, for example by firing a rocket engine. However, this is not always required; we don't have to expend any energy to remain stationary on the Earth's surface, but our worldlines are still accelerated; we are not in free fall.

The reason it's important to make the distinction I've just made is that, as far as standard GR is concerned, there is *no* useful analogy between the two things I've just described, curvature of spacetime and having an accelerated worldline. They are simply different things, and have to be analyzed and thought about separately. Your mental model appears to be based on drawing a close analogy between these two phenomena that simply is not justified.

You said that the rope would always break because anything beyond the event horizon has broken the light barrier.

No, that's not what I said. I said that the lower end of the rope, once it drops below the horizon, would have to move faster than light to *keep up* with the portion of the rope that's above the horizon. That's no different than saying that, in flat spacetime, an object that is beyond the Rindler horizon of a uniformly accelerating observer would have to move faster than light to keep up with that observer. It's simply a consequence of the way the light cones are laid out in the spacetime. You simply look at where the two objects are, and ask, "Would object A have to move outside the light cone in order to keep up with object B?" If the answer is yes, then object A can't keep up. So the object below the horizon has not "broken the light barrier"; quite the opposite. It's precisely the fact that it *can't* break the light barrier that prevents it from keeping up with objects that are hovering above the horizon.

If you can’t use energy to do it then what makes you think you can do it with the curvature from matter? The strength of the gravitational field makes no difference, just like velocity. The only difference is that you accelerate harder in free-fall when in higher gravity, but you’ll never be able to accelerate hard enough to reach c.

As I said above, acceleration of a particular worldline (for example, by firing a rocket engine, or standing on the surface of a planet) is simply a different thing from curvature of spacetime itself due to the presence of matter (more precisely, to the presence of stress-energy, which includes all the things I listed above that can cause spacetime curvature). An object in free fall is *not* accelerating (this is another confusing terminology you keep insisting on using, to use the word "acceleration" to refer to an object that is in free fall--avoiding this sort of confusion is a big reason why the word "acceleration" is *not* used in standard GR to refer to what is called "acceleration due to gravity" in Newtonian physics). Since the object is in free fall, there's no question of being able to "accelerate hard enough to reach c"; the object isn't accelerating at all. It's just following the straightest possible path in the spacetime it's in. If spacetime is flat, that path (worldline) will be a "straight line". If the spacetime is curved, the worldline of the object will be a geodesic of that curved spacetime--the analogue of a "straight line" in a curved manifold (where there are no "straight lines" in the strict Euclidean sense). If following such a geodesic worldline takes the object into portions of the spacetime where the light cones are tilted differently enough, then it may "look like" the object *is* moving "faster than light" if you insist on using coordinates that don't take proper account of the different tilting of the light cones. But that's a problem with your coordinates, not with the physics. The physics is simply that the free-falling object goes wherever the geodesic worldline takes it, just as it does in flat spacetime; it requires no more "energy" or "effort" than it does for an object in flat spacetime to remain in the same inertial frame for all time.

Only if you choose to ignore or have the cheek to disagree with what I just said.

I'll take option 2.

A velocity of c as a right-angle on a space-time diagram like the one I described before. Gravitational curvature shouldn’t be able to move you to 90 degrees any more than energy curvature can. You say it can, but why when we know that they’re equivalent?

They're not equivalent, once we clarify the terminology as I did above. What you have been calling "curvature due to energy" is *not* the same thing as what *I* have been calling "curvature due to energy". What you have been calling "curvature due to energy" is *not* equivalent to what you have been calling "gravitational curvature". I wasn't able to make that clear before because I didn't understand your non-standard use of the expression "curvature due to energy".

Why would using the two different causes of curvature together be any different from using more of one?

Because they're causing different kinds of "curvature", as I explained above--one of which should not even be described by the term "curvature".

Except that it is, as has also already been shown in this thread, in a slightly less technical and more hand-wavy way.

Except that your "hand-wavy" way is based on a false analogy and confusing terminology that obfuscates crucial distinctions.

Right, so why is it such a mystery that the strength of electro-magnetism is so much greater than gravity when electro-magnetism is curvature from energy and gravity is from matter?

Now you're using the term "curvature from energy" in yet *another* different way, which I don't understand, but which doesn't appear to me to be compatible with *either* of the uses of the term I described above.

Perhaps they do.

No, they don't. There is no event horizon in any of the cosmological models I referred to, so they can't describe a black hole.

There’s no complex mechanism to consider when talking about the life span of a black hole, just time-dilation and length contraction which are equivalent. So it should have same length in all four dimensions.

The spatial size of the horizon (i.e., the intersection of the horizon with a slice of constant time) is not finite because of "length contraction". It's finite because it's at a finite radius, r = 2M. But the horizon extends through an infinite range of "time" coordinates, so its behavior in time is *not* "equivalent" to its behavior in space.

It’s messy because from what I’ve seen, there’s so many different ways of measuring it in the standard version that you can manipulate it to show anything you want.

I don't understand where you're getting this from. The underlying geometry--either of the Earth's surface, or of a black hole spacetime--is the same *regardless* of what coordinate system you use to map it. Different coordinate systems are useful for different purposes, but they all describe the same underlying geometry, and they all give the same answers for all physical predictions within the portion of the geometry that they cover. You certainly can't "manipulate" the answers just by changing coordinates; if you get a different answer for a physical prediction in one coordinate system vs. another, then you've done something wrong.

 

[A-wal]

You are using very confusing terminology, which may help to explain why your thinking is confused. Specifically, you are using the same word, "curvature", to refer to two distinct concepts that should not be conflated.

Yes, that's intentional. I'm trying to make a point.

Let me try to restate what I think you mean by the two uses of the word "curvature" to see if I'm understanding you correctly:

(1) What you are calling "curvature due to matter" or "curvature from gravity" is curvature of *spacetime* itself. As I've pointed out, spacetime curvature can actually be caused by matter *or* energy; it can also be caused by pressure and stress in a material. When the word "curvature" is used in standard GR without qualification, it is this kind of curvature that is meant; and every time I've used the word "curvature", including the expression "curvature due to energy", I've been using it to mean this.

(2) What you are calling "curvature due to energy" or "curvature from acceleration" is curvature of the *worldline* of an object *within* a spacetime (which may itself be either flat or curved). The word "curvature" is *never* (as far as I've seen) used in standard GR to mean this; the word "acceleration" (or the term "proper acceleration" if more precision is needed) is used. You are correct that, in order to follow such a path, an object must normally expend energy, for example by firing a rocket engine. However, this is not always required; we don't have to expend any energy to remain stationary on the Earth's surface, but our worldlines are still accelerated; we are not in free fall.

Yes we do have to expend energy to remain stationary on the surface. It's called our sodding weight!

The reason it's important to make the distinction I've just made is that, as far as standard GR is concerned, there is *no* useful analogy between the two things I've just described, curvature of spacetime and having an accelerated worldline. They are simply different things, and have to be analyzed and thought about separately. Your mental model appears to be based on drawing a close analogy between these two phenomena that simply is not justified.

They're the same. Same effect, different (but equivalent) causes.

No, that's not what I said. I said that the lower end of the rope, once it drops below the horizon, would have to move faster than light to *keep up* with the portion of the rope that's above the horizon. That's no different than saying that, in flat spacetime, an object that is beyond the Rindler horizon of a uniformly accelerating observer would have to move faster than light to keep up with that observer. It's simply a consequence of the way the light cones are laid out in the spacetime. You simply look at where the two objects are, and ask, "Would object A have to move outside the light cone in order to keep up with object B?" If the answer is yes, then object A can't keep up. So the object below the horizon has not "broken the light barrier"; quite the opposite. It's precisely the fact that it *can't* break the light barrier that prevents it from keeping up with objects that are hovering above the horizon.

I know what you meant. You explained it very well but that description has the problem of there being an infinite (well c anyway) amount of difference between something fractionally outside the horizon and something fractionally inside it. The Rindler horizon in flat space time is the equivalent to an object in free-fall that can't be caught by an object (as long as it stays in free-fall) if it's a certain distance away - the Rindler horizon. An object doesn't have to cross the event horizon for this to happen. The event horizon is always the value of c more curved at the horizon than it is away from it. That's why the energy needed to escape at the horizon suddenly goes from finite to infinite. Any transition would have to be smooth. I know that it represents moving at c, so you could argue that it's a smooth transition but the fact remains that if it was a fixed event horizon you would need an finite amount of energy to escape from just outside the horizon and it would suddenly go up to infinity one Planc length later. How can you be in a position because one force of a certain strength has put you there, yet a different force can't move you away no matter how strong it is? You would need to travel faster than c to escape, which means gravity has effectively accelerated you past c. You should always be able to escape with a velocity of less than c because c is unreachable. It represents a right angle, and I don't think you can use curvature due to energy or gravity to get to that point, or a combination of both.

As I said above, acceleration of a particular worldline (for example, by firing a rocket engine, or standing on the surface of a planet) is simply a different thing from curvature of spacetime itself due to the presence of matter (more precisely, to the presence of stress-energy, which includes all the things I listed above that can cause spacetime curvature). An object in free fall is *not* accelerating (this is another confusing terminology you keep insisting on using, to use the word "acceleration" to refer to an object that is in free fall--avoiding this sort of confusion is a big reason why the word "acceleration" is *not* used in standard GR to refer to what is called "acceleration due to gravity" in Newtonian physics). Since the object is in free fall, there's no question of being able to "accelerate hard enough to reach c"; the object isn't accelerating at all. It's just following the straightest possible path in the spacetime it's in. If spacetime is flat, that path (worldline) will be a "straight line". If the spacetime is curved, the worldline of the object will be a geodesic of that curved spacetime--the analogue of a "straight line" in a curved manifold (where there are no "straight lines" in the strict Euclidean sense). If following such a geodesic worldline takes the object into portions of the spacetime where the light cones are tilted differently enough, then it may "look like" the object *is* moving "faster than light" if you insist on using coordinates that don't take proper account of the different tilting of the light cones. But that's a problem with your coordinates, not with the physics. The physics is simply that the free-falling object goes wherever the geodesic worldline takes it, just as it does in flat spacetime; it requires no more "energy" or "effort" than it does for an object in flat spacetime to remain in the same inertial frame for all time.

My definition of acceleration due to gravity is tidal force, not free-fall. I agree that being in free-fall is the same as being at rest. I also think there's an upper relative limit that no object can reach and it's the equivalent to c, and it's called an event horizon. I don't think you can gravity to break the light barrier either, if that's what you meant.

I'll take option 2.

I thought you might. Don't worry, I was prepared for that eventuality.

They're not equivalent, once we clarify the terminology as I did above. What you have been calling "curvature due to energy" is *not* the same thing as what *I* have been calling "curvature due to energy". What you have been calling "curvature due to energy" is *not* equivalent to what you have been calling "gravitational curvature". I wasn't able to make that clear before because I didn't understand your non-standard use of the expression "curvature due to energy".

In what way are they not equivalent?

Because they're causing different kinds of "curvature", as I explained above--one of which should not even be described by the term "curvature".

Apart from the fact that one curves outward (pushes you away from the source) and the other curves inward (pulls you towards the source), and their strength, what's the difference?

Except that your "hand-wavy" way is based on a false analogy and confusing terminology that obfuscates crucial distinctions.

My point is that I don't see why those distinctions actually exist.

Now you're using the term "curvature from energy" in yet *another* different way, which I don't understand, but which doesn't appear to me to be compatible with *either* of the uses of the term I described above.

No, same way. Energy curves your path through space-time. Matter curves your path through space-time. You can be at rest or you can accelerate. Whether the acceleration is coming from energy or from the tidal force of free-falling into a higher gravitational field doesn't make any difference. The effect is the same. Can accelerate in the opposite direction to the pull of gravity to cancel it out as we do on Earth. You can always match the acceleration from one with the other. If there's no such thing as absolute velocity then there's no such thing as absolute gravity. The fact that we're pulled down towards the ground is because of the difference in the strength of gravity between our heads and our feet. That's why it's more comfortable to lay down. There's no absolute strength of gravity which is what would be needed to cross an event horizon.

No, they don't. There is no event horizon in any of the cosmological models I referred to, so they can't describe a black hole.

I thought there was an event horizon in the standard description of the beginning of the universe. It's the edge of the universe. We're inside a big black hole from this way of looking at it. And guess what; you can't reach the horizon.

The spatial size of the horizon (i.e., the intersection of the horizon with a slice of constant time) is not finite because of "length contraction". It's finite because it's at a finite radius, r = 2M. But the horizon extends through an infinite range of "time" coordinates, so its behavior in time is *not* "equivalent" to its behavior in space.

Why? What makes it different? And what does r=2M actually mean? Two what? And why two? I didn't mean it's finite because of length contraction.

I don't understand where you're getting this from. The underlying geometry--either of the Earth's surface, or of a black hole spacetime--is the same *regardless* of what coordinate system you use to map it. Different coordinate systems are useful for different purposes, but they all describe the same underlying geometry, and they all give the same answers for all physical predictions within the portion of the geometry that they cover. You certainly can't "manipulate" the answers just by changing coordinates; if you get a different answer for a physical prediction in one coordinate system vs. another, then you've done something wrong.

So why is it okay to switch between Rindler coordinates where objects can cross the horizon and Schwarzschild coordinates where nothing ever can? Just use Schwarzschild coordinates. Rindler ones don't take into account the fact that gravity is relative. You're at rest when in free-fall so you'll never even be able to start approaching the point when no amount of energy would be enough to escape. And it's messy precisely because you have to use multiple coordinate systems to describe it properly, and even then it contradicts itself.

 

[Dmitry67]

So why is it okay to switch between Rindler coordinates where objects can cross the horizon and Schwarzschild coordinates where nothing ever can?

Here is a toy model for you (attached)
Red bug is moving conter-clockwise on a circle from the south pole to the north pole. It emits light every seconds.

I also have a second coordinate system - a projection from a center of a circle to the black line. There is a second, black bug moving at that line to the right. Somehow, their communication is posisble only via that projection line (it is a toy model).

For the black bug moving along the horisontal line, red bug becomes more and more 'dilated' in time: in gets the second splash at 1.2 seconds, thrird - at about 2 seconds, etc.

Now black bug claims that red bug would NEVER reaches the north pole, because for it (black bug) it takes forever to go to the end of the line, and still red bug will cross only a half of a distance.

 

[PeterDonis]

Yes, that's intentional. I'm trying to make a point.

If the point you are trying to make is that there somehow *is* a useful analogy between what you are calling "curvature due to acceleration" and what you are calling "curvature due to gravity", then your point is simply not valid. If you are trying to make some other point, please clarify what it is.

Yes we do have to expend energy to remain stationary on the surface. It's called our sodding weight!

Objects stationary on the surface of a planet, like the Earth, don't have to expend any energy just to remain stationary and have weight. Yes, the Earth exerts a force on the object, but the force is a static force (because the object is stationary) and no work is done, so no energy is expended. This isn't even particular to relativity; the analysis works the same way in simple Newtonian physics.

Perhaps you are referring to the fact that you and I, being bipedal living organisms, do have to "expend energy" to remain standing. But that's just because our bodies are unstable in a standing position. You can lie down and expend no energy to remain stationary. Or take the simplest possible example: a rock. It expends no energy--no chemical reactions take place in the rock, no nuclear reactions, no fuel being burned, no rocket engine being fired--yet it remains stationary on the Earth's surface just fine.

Or make it even simpler, since the Earth is an "active" planetary body with magnetic fields inside it, molten magma beneath the surface, tectonic plates moving about, etc. Consider a rock on the surface of the Moon. The Moon itself is just a big spherical rock, and is as static and non-reactive as the rock I described above. Yet a rock can sit on the Moon's surface, and have weight, when neither the rock or the Moon are expending any energy.

If it seems like I'm belaboring this point, it's because I am surprised that I need to make it at all. The fact that static forces do no work and expend no energy is very, very basic physics.

They're the same. Same effect, different (but equivalent) causes.

In standard GR, neither the causes nor the effects are equivalent. If you want to claim they are, you are going to have to give a detailed argument for why you think so. Just saying that it seems obvious to you won't do.

I know what you meant. You explained it very well but that description has the problem of there being an infinite (well c anyway) amount of difference between something fractionally outside the horizon and something fractionally inside it.

No, this is wrong. There's only a small amount of difference, because for an object that's just a little bit outside the horizon, the object has to move radially outward at a speed just a little bit less than c to avoid falling below the horizon. Any lesser speed, even if it's directly radially outward, will result in the object falling below the horizon.

(I've snipped the rest of your comments about the Rindler horizon because I address your misconception about that further below.)

My definition of acceleration due to gravity is tidal force, not free-fall. I agree that being in free-fall is the same as being at rest.

Fine so far.

I also think there's an upper relative limit that no object can reach and it's the equivalent to c, and it's called an event horizon.

If you mean by this that no objects can move outside the light cones, you are correct. But if you mean that there is a limit to how much light cones at a given event can be "tilted" relative to light cones somewhere else, that's wrong.

I don't think you can gravity to break the light barrier either, if that's what you meant.

It wasn't.

In what way are they not equivalent

...

Apart from the fact that one curves outward (pushes you away from the source) and the other curves inward (pulls you towards the source), and their strength, what's the difference?

...

My point is that I don't see why those distinctions actually exist.

Curvature of spacetime itself determines which paths in spacetime are geodesics (freely falling worldlines) and which are not. Being a geodesic or not a geodesic is an intrinsic property of a path in a particular spacetime geometry. The geometry itself depends on global specifications of the quantities that feed into the Einstein Field Equation: what (if any) matter or energy is present (globally), and what the boundary conditions are (for example, a black hole spacetime is asymptotically flat).

"Curvature due to acceleration" is just a way of saying the particular path you are following is not a geodesic. Specifying which path you are following is a property of the *object*; you figure it out by figuring out the object's physical situation and what it is doing (is it firing rockets, resting on the surface of a planet, etc. as opposed to floating freely in space).

These are two distinct concepts because you can "mix and match" them freely in any combination. You can have a geodesic path in flat spacetime, and an object moving on it (e.g., an inertial observer in Minkowski spacetime); you can have a non-geodesic path in a flat spacetime, and an object moving on it (e.g., a "Rindler observer" uniformly accelerating in Minkowski spacetime); you can have a geodesic path in curved spacetime, and an object moving on it (e.g., an observer freely falling into a black hole), and you can have a non-geodesic path in curved spacetime, and an object moving on it (e.g., an observer firing rockets to hover at a constant radius r > 2M above a black hole's horizon). So knowing just one thing doesn't impose any restrictions on the other thing.

No, same way. Energy curves your path through space-time. Matter curves your path through space-time. You can be at rest or you can accelerate. Whether the acceleration is coming from energy or from the tidal force of free-falling into a higher gravitational field doesn't make any difference. The effect is the same. Can accelerate in the opposite direction to the pull of gravity to cancel it out as we do on Earth. You can always match the acceleration from one with the other.

No, you can't. If you are below the horizon of a black hole, no amount of acceleration will prevent you from continuing to fall inward. It is true that *outside* the horizon, there is always some finite acceleration that will just "cancel" the inward "acceleration" due to the presence of the hole, but there is no requirement that that must be true throughout the spacetime. (And no, the fact that it seems "obvious" to you that you should always be able to "cancel gravity with acceleration" is *not* a proof that that *must* be true. If you think you can concoct an actual proof, please post one. But you can't just *assume* that it's true.)

The fact that we're pulled down towards the ground is because of the difference in the strength of gravity between our heads and our feet. That's why it's more comfortable to lay down.

Tidal gravity is *not* the same as the apparent "force of gravity" that pulls us towards the center of the Earth. Again, this isn't particular to relativity; the same is true in Newtonian physics. Tidal gravity is the *rate of change* of the apparent "force of gravity" (note that in relativity, the rate of change is in *spacetime*, not just space; there can be tidal gravity in the "time direction" as well as in the "space directions").

The case you give, of standing vs. lying down, actually illustrates the difference. Yes, the tidal gravity on your body is less when you're lying down, but you still weigh the same. The reason it's more comfortable to lie down is that your body is unstable standing up, and your muscles have to continually compensate to keep you upright, which tires you out.

I thought there was an event horizon in the standard description of the beginning of the universe. It's the edge of the universe. We're inside a big black hole from this way of looking at it. And guess what; you can't reach the horizon.

You are misunderstanding the standard cosmological model of the Big Bang; it does *not* say we are inside a big black hole. (And the universe does not have an "edge"; it is either finite but unbounded, like the surface of the Earth--that's if the universe is closed--or it is infinite in extent and has no edge, if it is open. The current "best fit" model has the universe being infinite, but the error bars are large enough that there is still an outside chance we will end up judging that it is closed when we have more data.)

It is true that, if you pick different worldlines coming out of the Big Bang, and go back in time close enough to the Big Bang (how close depends on how far apart the worldlines are), you will come to a point where there had not been enough time since the Big Bang for light to travel between the two worldlines. The term "horizon" is sometimes used to refer to this phenomenon, but it's not the same as a black hole event horizon, because as time goes on, more and more worldlines "come into contact" with each other, meaning there has been enough time for light to travel between them. A black hole's event horizon remains the same through time; the region inside can *never* send light signals to the region outside, while the region outside can send signals to the region inside at any time.

what does r=2M actually mean? Two what? And why two?

r = 2M is just the standard expression for the radius of a black hole's horizon in terms of its mass, where the mass is expressed in "geometric units"--i.e., mass and length have the same units. In conventional units, the expression is r = 2 G M / c^{2}, where G is Newton's gravitational constant and c is the speed of light. M is then the mass of the hole in conventional units.

I didn't mean it's finite because of length contraction.

You were trying to argue (as I understood you) that the black hole's "size in time" should be the same as its "size in space" because time dilation and length contraction are equivalent. This argument only makes sense if the hole's finite "size in space" is due to length contraction. Since it isn't, the argument fails; time dilation and length contraction are irrelevant.

So why is it okay to switch between Rindler coordinates where objects can cross the horizon and Schwarzschild coordinates where nothing ever can? Just use Schwarzschild coordinates.

Objects can't cross the Rindler horizon in Rindler coordinates. Those coordinates only cover the region "above" the horizon (where all the accelerating observers are) in flat spacetime. To "see" objects crossing the horizon in flat spacetime, you have to use Minkowski coordinates, since those cover the entire spacetime. So, can objects cross the Rindler horizon in flat spacetime? Of course they can. It's just that Rindler coordinates aren't good coordinates to analyze that case.

Similarly, in a black hole spacetime, to "see" objects crossing the horizon, you have to use, for example, Kruskal coordinates, which cover the entire spacetime in a way similar to the way Minkowski coordinates cover the entire spacetime when spacetime is flat. So once again, can objects cross the horizon in a black hole spacetime? Of course they can. It's just that Schwarzschild coordinates (which are analogous to Rindler coordinates) aren't good coordinates to analyze that case.

 

[PeterDonis]

No, this is wrong. There's only a small amount of difference, because for an object that's just a little bit outside the horizon, the object has to move radially outward at a speed just a little bit less than c to avoid falling below the horizon. Any lesser speed, even if it's directly radially outward, will result in the object falling below the horizon.

I realized after re-reading this that I should clarify: the above statement applies to objects that are moving inertially (i.e., they may have a very large velocity, approaching c, but their velocity is constant; they are not accelerating, in the invariant, coordinate-independent sense of "accelerating"). Such an object, just outside the horizon, must have an outward radial velocity just short of c to avoid falling through the horizon.

An *accelerating* object just outside the horizon can be stationary (i.e., can "hover" at a constant radial coordinate), as long as its acceleration is large enough--the required acceleration increases without bound ("approaches infinity") as the horizon is approached. However, as I noted elsewhere in my previous post, there is no requirement that one must be able to "hover" stationary at a finite acceleration everywhere in a black hole spacetime. It just happens to be the case that you can do so anywhere outside a black hole's horizon.

 

[A-wal]

Here is a toy model for you (attached)
Red bug is moving conter-clockwise on a circle from the south pole to the north pole. It emits light every seconds.

I also have a second coordinate system - a projection from a center of a circle to the black line. There is a second, black bug moving at that line to the right. Somehow, their communication is posisble only via that projection line (it is a toy model).

For the black bug moving along the horisontal line, red bug becomes more and more 'dilated' in time: in gets the second splash at 1.2 seconds, thrird - at about 2 seconds, etc.

Now black bug claims that red bug would NEVER reaches the north pole, because for it (black bug) it takes forever to go to the end of the line, and still red bug will cross only a half of a distance.

Isn't that describing a Rindler horizon rather than an event horizon?

 

[A-wal]

If the point you are trying to make is that there somehow *is* a useful analogy between what you are calling "curvature due to acceleration" and what you are calling "curvature due to gravity", then your point is simply not valid. If you are trying to make some other point, please clarify what it is.

No, I'm trying to make the point that you consider invalid.

Objects stationary on the surface of a planet, like the Earth, don't have to expend any energy just to remain stationary and have weight. Yes, the Earth exerts a force on the object, but the force is a static force (because the object is stationary) and no work is done, so no energy is expended. This isn't even particular to relativity; the analysis works the same way in simple Newtonian physics.

If the Earth exerts a force on the object then obviously the object exerts a force on the Earth. The energy we need to remain stationary is felt as our weight. Two opposing forces cancelling each other out doesn't mean there's no force. We only feel the acceleration, not the gravity pulling us the other way because gravity is relative, just like velocity. Is there a test that you can perform, even in principle to show that you're in a gravitational field? Is there a test you can perform to show that there's a moving time-line? Is there a test to see if you're moving at all? It's all the same thing.

Perhaps you are referring to the fact that you and I, being bipedal living organisms, do have to "expend energy" to remain standing. But that's just because our bodies are unstable in a standing position. You can lie down and expend no energy to remain stationary. Or take the simplest possible example: a rock. It expends no energy--no chemical reactions take place in the rock, no nuclear reactions, no fuel being burned, no rocket engine being fired--yet it remains stationary on the Earth's surface just fine.

You weigh the same whether you’re standing or not. The "expended energy" needed stays the same. I don't know what you mean by a static force, other than that we remain stationary relative to the Earth, and I really don't see how that's relevant. The only reason I can see for it to be considered what you call a static force is because gravity and electromagnetism are balanced. Or to put it another way the outward curvature from your acceleration matches the inward curvature from gravity.

Or make it even simpler, since the Earth is an "active" planetary body with magnetic fields inside it, molten magma beneath the surface, tectonic plates moving about, etc. Consider a rock on the surface of the Moon. The Moon itself is just a big spherical rock, and is as static and non-reactive as the rock I described above. Yet a rock can sit on the Moon's surface, and have weight, when neither the rock or the Moon are expending any energy.

Yes they are, through their weight, in the same way that our bodies and everything else on the planet needs to expend energy to stop it from being crushed.

If it seems like I'm belaboring this point, it's because I am surprised that I need to make it at all. The fact that static forces do no work and expend no energy is very, very basic physics.

Ah, that explains it. I know next to nothing about very, very basic physics. I've never even heard of a static force before, other than static electricity.

In standard GR, neither the causes nor the effects are equivalent. If you want to claim they are, you are going to have to give a detailed argument for why you think so. Just saying that it seems obvious to you won't do.

What the hell do you think I'm trying to do? Space-time can be curved in one of two ways. Energy can accelerate an object away from its source which causes time dilation and length contraction because everything has to move through space-time at c and c has to remain constant. You can use tidal force to accelerate you towards an object, but it's a much weaker force because it comes from matter rather than energy and E=mc^2. To me this is very, very basic physics. If you want to claim there's some fundamental difference between these two processes to make one behave differently from the other one then I think the onus should be on you to show that there is.

No, this is wrong. There's only a small amount of difference, because for an object that's just a little bit outside the horizon, the object has to move radially outward at a speed just a little bit less than c to avoid falling below the horizon. Any lesser speed, even if it's directly radially outward, will result in the object falling below the horizon.

What did I say just after that? The energy required to escape from the black hole jumps from finite to infinite in an instant. Surely that doesn't make sense to you?

I realized after re-reading this that I should clarify: the above statement applies to objects that are moving inertially (i.e., they may have a very large velocity, approaching c, but their velocity is constant; they are not accelerating, in the invariant, coordinate-independent sense of "accelerating"). Such an object, just outside the horizon, must have an outward radial velocity just short of c to avoid falling through the horizon.

An *accelerating* object just outside the horizon can be stationary (i.e., can "hover" at a constant radial coordinate), as long as its acceleration is large enough--the required acceleration increases without bound ("approaches infinity") as the horizon is approached. However, as I noted elsewhere in my previous post, there is no requirement that one must be able to "hover" stationary at a finite acceleration everywhere in a black hole spacetime. It just happens to be the case that you can do so anywhere outside a black hole's horizon.

Nothing just happens. You can always hover at a finite acceleration because nothing can produce infinite force. If you need infinite energy to escape from within the horizon then the black hole has infinite gravitational force past the horizon.

(I've snipped the rest of your comments about the Rindler horizon because I address your misconception about that further below.)

What misconception? If I'm understanding the concept correctly (I don't entirely trust my interpretation of the Rindler horizon because I had to look it up rather than figure it out for myself) then a free-falling object has crossed the Rindler horizon when it can't ever be caught by a specific more distant accelerating object, whereas a free-falling object has crossed the event horizon when it can't ever be caught by any more distant accelerating object. It would mean gravity accelerates the closer object faster than any amount of energy could accelerate the further object, no matter how close it is. That's very clever. It's so clever that I don't think it's exaggerating to call it magical.

Fine so far.

So you agree that tidal force is equivalent to acceleration in flat space-time? So why can tidal force accelerate you to the equivalent of c and beyond when energy can only accelerate you to other relative velocities?

If you mean by this that no objects can move outside the light cones, you are correct. But if you mean that there is a limit to how much light cones at a given event can be "tilted" relative to light cones somewhere else, that's wrong.

Is it? Any particular reason or is it just wrong? Why is it okay for gravity to tilt the light cone past 45 degrees but not energy? What makes gravity so special? If they can tilt past 45 degrees then would they be moving backwards through time?

It wasn't.

Okay, but that's basically what you said whether you meant it or not. You said that in order to escape from the event horizon you would have to move faster than c, which means that gravity has effectively moved you past c to get there in the first place.

Curvature of spacetime itself determines which paths in spacetime are geodesics (freely falling worldlines) and which are not. Being a geodesic or not a geodesic is an intrinsic property of a path in a particular spacetime geometry. The geometry itself depends on global specifications of the quantities that feed into the Einstein Field Equation: what (if any) matter or energy is present (globally), and what the boundary conditions are (for example, a black hole spacetime is asymptotically flat).

The only practical difference I can see other than their direction and strength is that energy doesn't tend to effect an object for long. It's there and then it's gone. Matter lasts long enough for us to map its effect.

"Curvature due to acceleration" is just a way of saying the particular path you are following is not a geodesic. Specifying which path you are following is a property of the *object*; you figure it out by figuring out the object's physical situation and what it is doing (is it firing rockets, resting on the surface of a planet, etc. as opposed to floating freely in space).

And gravity is also just a way of saying the particular path you are following is not a geodesic from a distance.

These are two distinct concepts because you can "mix and match" them freely in any combination. You can have a geodesic path in flat spacetime, and an object moving on it (e.g., an inertial observer in Minkowski spacetime); you can have a non-geodesic path in a flat spacetime, and an object moving on it (e.g., a "Rindler observer" uniformly accelerating in Minkowski spacetime); you can have a geodesic path in curved spacetime, and an object moving on it (e.g., an observer freely falling into a black hole), and you can have a non-geodesic path in curved spacetime, and an object moving on it (e.g., an observer firing rockets to hover at a constant radius r > 2M above a black hole's horizon). So knowing just one thing doesn't impose any restrictions on the other thing.

The fact that you can “mix and match” them doesn't prove they are distinct concepts. If anything the opposite is true.

No, you can't. If you are below the horizon of a black hole, no amount of acceleration will prevent you from continuing to fall inward. It is true that *outside* the horizon, there is always some finite acceleration that will just "cancel" the inward "acceleration" due to the presence of the hole, but there is no requirement that that must be true throughout the spacetime.

I think it is a requirement for the universe to make sense.

(And no, the fact that it seems "obvious" to you that you should always be able to "cancel gravity with acceleration" is *not* a proof that that *must* be true. If you think you can concoct an actual proof, please post one. But you can't just *assume* that it's true.)

Bugger!

Tidal gravity is *not* the same as the apparent "force of gravity" that pulls us towards the center of the Earth. Again, this isn't particular to relativity; the same is true in Newtonian physics. Tidal gravity is the *rate of change* of the apparent "force of gravity" (note that in relativity, the rate of change is in *spacetime*, not just space; there can be tidal gravity in the "time direction" as well as in the "space directions").

Tidal gravity as the "rate of change of the apparent force of gravity" is exactly what I've been saying forever.

The case you give, of standing vs. lying down, actually illustrates the difference. Yes, the tidal gravity on your body is less when you're lying down, but you still weigh the same. The reason it's more comfortable to lie down is that your body is unstable standing up, and your muscles have to continually compensate to keep you upright, which tires you out.

1). It was a joke. 2). There is a difference in tidal force as you said, but it's so marginal that you could never feel it. That was the joke. 3). I know you weigh the same standing up as laying down, and the reason it's more comfortable is because your weight's spread out. The question is why laying down spreads out your weight. There's nothing to determine the direction that gravity pulls you other than the direction in which it's strongest. In other words the difference between the gravity at your head and your feet. In other other words, tidal force. The reason your more comfortable laying down is because all the weight goes to your feet when you’re at a lengthwase angle to the direction of the tidal force. So in fact you are more comfortable when laying down because the gap between the top and bottom parts of your body are at their lowest.

You are misunderstanding the standard cosmological model of the Big Bang; it does *not* say we are inside a big black hole. (And the universe does not have an "edge"; it is either finite but unbounded, like the surface of the Earth--that's if the universe is closed--or it is infinite in extent and has no edge, if it is open. The current "best fit" model has the universe being infinite, but the error bars are large enough that there is still an outside chance we will end up judging that it is closed when we have more data.)

Why do you always assume I've misunderstood the principles before you assume that you misunderstood what I meant? I know it doesn't have that kind of an edge. How would that even work? I didn't actually say we were inside a black hole, although you could definitely look at it in that way. If it's closed then I don't think it's wrong as far as the standard model goes, just the same thing from the inside, accept you can’t even start to approach the horizon. And I'd say there's more than an outside chance that it's closed.

It is true that, if you pick different worldlines coming out of the Big Bang, and go back in time close enough to the Big Bang (how close depends on how far apart the worldlines are), you will come to a point where there had not been enough time since the Big Bang for light to travel between the two worldlines. The term "horizon" is sometimes used to refer to this phenomenon, but it's not the same as a black hole event horizon, because as time goes on, more and more worldlines "come into contact" with each other, meaning there has been enough time for light to travel between them. A black hole's event horizon remains the same through time; the region inside can *never* send light signals to the region outside, while the region outside can send signals to the region inside at any time.

That one-way road situation is one of the many reasons why it doesn't make sense to have a fixed crossable horizon. And if there wasn't enough time for light to travel between the two world-lines then how could there have been enough time for the two world lines to get that far apart?

r = 2M is just the standard expression for the radius of a black hole's horizon in terms of its mass, where the mass is expressed in "geometric units"--i.e., mass and length have the same units. In conventional units, the expression is r = 2 G M / c^{2}, where G is Newton's gravitational constant and c is the speed of light. M is then the mass of the hole in conventional units.

But not its length in time? What makes time so special?

You were trying to argue (as I understood you) that the black hole's "size in time" should be the same as its "size in space" because time dilation and length contraction are equivalent. This argument only makes sense if the hole's finite "size in space" is due to length contraction. Since it isn't, the argument fails; time dilation and length contraction are irrelevant.

I don't think its finite volume is due to length contraction. I think the whole thing is length contraction and time dilation, caused by the singularity. If you can't reach the black hole then what actually is it? It's just a four-dimensional bubble of nothingness that nothing can ever reach, because there's always too much space and time between it and you, just like the edge of the universe (it still has an edge, just not a fixed one).

Objects can't cross the Rindler horizon in Rindler coordinates. Those coordinates only cover the region "above" the horizon (where all the accelerating observers are) in flat spacetime. To "see" objects crossing the horizon in flat spacetime, you have to use Minkowski coordinates, since those cover the entire spacetime. So, can objects cross the Rindler horizon in flat spacetime? Of course they can. It's just that Rindler coordinates aren't good coordinates to analyze that case.

I thought objects could cross the event horizon using Rindler coordinates. It doesn't matter. The point was that you need to use two different points of reference to describe the same thing fully because one shows something the other doesn't, which in this case, is just another way of saying they contradict each other. Using that logic, whether or not you fall into a black hole is dependent on what coordinate system you choose to measure your decent. Why is it so acceptable to switch between two entirely views when one shows something that's automatically assumed wrong? It's very easy to say "of course they can" but to my mind it's obvious that they can't.

Similarly, in a black hole spacetime, to "see" objects crossing the horizon, you have to use, for example, Kruskal coordinates, which cover the entire spacetime in a way similar to the way Minkowski coordinates cover the entire spacetime when spacetime is flat. So once again, can objects cross the horizon in a black hole spacetime? Of course they can. It's just that Schwarzschild coordinates (which are analogous to Rindler coordinates) aren't good coordinates to analyze that case.

Schwarzschild coordinates cover the entire space-time external to the black hole. It's just that time and space are compressed to c/infinity at the horizon. It's different than the way it happens on a standard graph showing acceleration up to c. What that doesn't show is that it would look different if you were the one accelerating, just like it would look different if you were the one falling and accelerating through tidal force. Length contraction means you have to travel further and further, and time dilation means you have less and less time to do it the closer you get to the horizon. You think that you can use a different type of acceleration to outrun this process and reach the horizon but they're equivalent. You could use standard acceleration through energy to move towards it faster but than your left with exactly the same problem. If you accelerate in the opposite direction you can cancel it out or overcome it. There'll never be a point when one take somewhere the other can't take you away from. Gravity's just as relative as velocity, but you're trying to use it to break the light barrier. Not a chance in hell son.

 

[PeterDonis]

If the Earth exerts a force on the object then obviously the object exerts a force on the Earth.

Yes.

The energy we need to remain stationary is felt as our weight.

No, weight is a force, not energy. You feel the force of the Earth pushing on you. No energy is expended (see below).

Two opposing forces cancelling each other out doesn't mean there's no force.

True.

We only feel the acceleration, not the gravity pulling us the other way...

Not quite. We feel the *force* of the Earth pushing on us. The Earth, similarly, "feels" the force of us pushing on it. You're correct that there is no need to describe either of these forces as "gravity". They're not. However...

...because gravity is relative, just like velocity.

This is not quite right either, at least not in the way you use the term "gravity" (because you say tidal gravity and "acceleration due to gravity" are the same thing, which they're not). There is a sense in which "gravity is relative", meaning the "acceleration due to gravity" is relative, because you can always find a freely falling reference frame in which, locally, that acceleration disappears. But *tidal* gravity is *not* relative; it is a frame-invariant, genuine physical quantity.

Is there a test that you can perform, even in principle to show that you're in a gravitational field?

There is no *local* test you can perform to show that you're "being accelerated by gravity" if you are in free fall (i.e., not feeling any force). (Obviously you can use non-local information, such as the fact that you are above a large planetary body like the Earth, to infer that that body's mass is causing you to fall towards it. But that's not a local test; it requires information about distant objects.) But there *are* tests you can perform to show whether *tidal* gravity is present or absent between you and neighboring objects, even if you and all those other objects are all in free fall.

You weigh the same whether you’re standing or not. The "expended energy" needed stays the same.

Only in the sense that the "expended energy" is zero in both cases.

I don't know what you mean by a static force, other than that we remain stationary relative to the Earth, and I really don't see how that's relevant.

You have the correct definition of static force: it's a force which does not result in any relative motion of either body. It's relevant because at least one body has to move for any work to be done and any energy to be expended. For example, if you stand up from a prone position, you *do* expend energy, because you have to do work to raise your body's mass from a lower to a higher position (i.e., you move your body further away from the Earth--more precisely, your body's center of mass is further from the Earth's center of mass). Or if you are far out in space in a rocket and you fire the rocket's engine, the rocket does work by expelling exhaust out the back--the exhaust and the rocket are moving relative to each other, so work is done and energy is expended. There has to be relative motion.

The only reason I can see for it to be considered what you call a static force is because gravity and electromagnetism are balanced. Or to put it another way the outward curvature from your acceleration matches the inward curvature from gravity.

There is no "force of gravity" anywhere in the problem. The electromagnetic force of the atoms in your body pushes against the electromagnetic force of the atoms in the ground, and the ground pushes back.

Yes they are, through their weight, in the same way that our bodies and everything else on the planet needs to expend energy to stop it from being crushed.

The Moon and the rock on the Moon are not expending any energy. How can they? Where is it going to come from? There are no chemical reactions going on, no nuclear reactions going on, no fuel being burned. Where does the energy come from? The answer is that it doesn't have to come from anywhere because no energy is being expended. The Moon and the rock are in a state of static equilibrium, with no work being done, and they will stay that way indefinitely, with no work being done and no energy expended. By your logic, the rock and the Moon would eventually "run out" of energy and collapse somehow. They don't.

What the hell do you think I'm trying to do? Space-time can be curved in one of two ways.

No, spacetime can be curved in only one way, by the presence of stress-energy (matter, energy, pressure, and internal stresses of materials--the stuff that goes into the stress-energy tensor).

An object's *path* through spacetime can also be curved, in the frame-invariant sense of "curved", in one way: by an object having a force (again, in the frame-invariant sense of "force") exerted on it, causing it to accelerate (in the frame-invariant sense of "accelerate"). You can tell whether an object's path is curved in this sense by seeing whether it feels weight.

A path can "look" curved without being curved (in the frame-invariant sense), or be curved (in the frame-invariant sense) without "looking" curved, if you adopt a coordinate system that obscures the frame-invariant curvature. For example, in Schwarzschild exterior coordinates, the paths of objects "hovering" at constant radius above a black hole's horizon look straight, but they are curved in the frame-invariant sense. (The same is true of your path through spacetime when you're standing motionless on the Earth's surface.)

If you want to claim there's some fundamental difference between these two processes to make one behave differently from the other one then I think the onus should be on you to show that there is.

I've done this. I've explained how:

(a) Curvature of *spacetime* is different and distinct from curvature of an object's *path* through spacetime.

(b) Tidal gravity is different and distinct from "acceleration due to gravity".

What did I say just after that? The energy required to escape from the black hole jumps from finite to infinite in an instant. Surely that doesn't make sense to you?

There is no jump. The outward velocity required to escape (for an object that's in free fall, not accelerated) increases smoothly to c as the horizon is approached and reached, and the energy required to escape (again, for an object that's in free fall, not accelerated) increases smoothly to infinity. This works the same as an object's energy going to infinity as its speed approaches c (in a given reference frame) in flat spacetime. There's no jump.

Edit: I should clarify that by "works the same" I mean only that, mathematically, the limiting process works the same. I do *not* mean that the horizon can't be reached and passed. Let me know if I need to elaborate on this.

If you need infinite energy to escape from within the horizon then the black hole has infinite gravitational force past the horizon.

There is no "force" anywhere. It's just geometry. See below.

a free-falling object has crossed the Rindler horizon when it can't ever be caught by a specific more distant accelerating object, whereas a free-falling object has crossed the event horizon when it can't ever be caught by any more distant accelerating object.

No. You keep on getting this backwards. The Rindler horizon is the path of a light ray that just fails to catch up to a family of uniformly accelerating observers outside (or "above") it. Similarly, a black hole's horizon is the path of a light ray that just fails to catch up to a family of "hovering" observers outside it. Neither horizon has anything to do with whether an object outside the horizon can catch up with an object that free-falls *into* the horizon. That's a separate question.

So you agree that tidal force is equivalent to acceleration in flat space-time?

No. I apologize for a misstatement there; I should have snipped the quote from your previous post more carefully. The only statement of yours I meant to agree with was the statement that being in free fall is equivalent to being at rest. I did not mean to imply agreement with your statement about tidal force.

Why is it okay for gravity to tilt the light cone past 45 degrees but not energy? What makes gravity so special?

"Energy" in the sense you are using the term (meaning energy expended by an object to accelerate itself so it feels weight) has no effect on the light cones. It can't tilt them at all. Also, "gravity" is not a separate "thing" that somehow tilts the light-cones. Gravity *is* the tilting of the light cones. They're the same thing (spacetime curvature).

If they can tilt past 45 degrees then would they be moving backwards through time?

No. The tilt is not "past 45 degrees". The *ingoing* sides of the light cones remain pointed at 45 degrees inward. The *outgoing* sides gradually point more and more *upward* (i.e., *less* than 45 degrees) until they become vertical (0 degrees) at the horizon. Inside the horizon, they tilt over the other way, which simply means objects can no longer avoid moving inward, even if they move at the speed of light. It does not require going backward in time.

Okay, but that's basically what you said whether you meant it or not. You said that in order to escape from the event horizon you would have to move faster than c, which means that gravity has effectively moved you past c to get there in the first place.

No, that doesn't follow, because in order to escape, you have to move *outward* faster than light. You do not have to move *inward* faster than light to get inside the horizon. You can free-fall there, without "moving" at all (you yourself said being in free fall is the same as being at rest, so all you have to do to get inside the hole is to stay at rest).

And gravity is also just a way of saying the particular path you are following is not a geodesic from a distance.

No, "gravity" is a way of saying that *spacetime* is curved. It does not say anything about particular *paths* in that spacetime. You can follow a geodesic path (freely falling path, weightless path) in a curved spacetime, where gravity is present. They are separate, distinct concepts.

I think it is a requirement for the universe to make sense.

To make sense, yes. To make sense in a way that satisfies all your intuitions, no.

Tidal gravity as the "rate of change of the apparent force of gravity" is exactly what I've been saying forever.

And yet you keep trying to claim that tidal force is the *same* as "acceleration due to gravity" or "apparent force of gravity". How can something be the same as its rate of change?

I know you weigh the same standing up as laying down, and the reason it's more comfortable is because your weight's spread out. The question is why laying down spreads out your weight.

Um, because a larger area of you is in contact with the ground?

There's nothing to determine the direction that gravity pulls you other than the direction in which it's strongest. In other words the difference between the gravity at your head and your feet. In other other words, tidal force.

But there are tidal forces in multiple directions, not just one. There can even be tidal force in the time direction. You can't extract a unique direction for gravity to pull you from tidal force.

The reason your more comfortable laying down is because all the weight goes to your feet when you’re at a lengthwase angle to the direction of the tidal force. So in fact you are more comfortable when laying down because the gap between the top and bottom parts of your body are at their lowest.

No, it's because your weight is spread more evenly over a larger area of your body. See above.

Why do you always assume I've misunderstood the principles before you assume that you misunderstood what I meant?

I wasn't assuming anything. I was responding to the actual words you used. You said: "I thought there was an event horizon in the standard description of the beginning of the universe. It's the edge of the universe." You didn't elaborate on what you meant by "edge". How was I supposed to interpret it except in the obvious way?

That one-way road situation is one of the many reasons why it doesn't make sense to have a fixed crossable horizon.

Why not? I know it doesn't make sense in the mental model you have in your head. But can you give a logical argument, starting from premises we all accept, that shows that it *can't* make sense, in *any* consistent model? I don't think you can, but if you have such an argument, let's see it. Just claiming that it doesn't make sense won't do, because I have a consistent model where it does makes sense.

And if there wasn't enough time for light to travel between the two world-lines then how could there have been enough time for the two world lines to get that far apart?

Because the worldlines are expanding in different directions. Ned Wright's Cosmology FAQ, at the link below, briefly explains the kind of thing that's going on.

http://www.astro.ucla.edu/~wright/cosmology_faq.html#DN

But not its length in time? What makes time so special?

I'm not sure what you're asking here. I was just explaining what r = 2M means, since you asked that; that radius is the radius of the horizon. Radius is not a quantity in the time dimension.

If you can't reach the black hole then what actually is it?

Since you can reach the black hole, this question is meaningless.

I thought objects could cross the event horizon using Rindler coordinates.

They can't.

It doesn't matter.

Oh, yes, it does. It's crucial, because it grounds the analogy between Rindler coordinates and Schwarzschild coordinates (as compared with Minkowski coordinates and Kruskal coordinates). See below.

Using that logic, whether or not you fall into a black hole is dependent on what coordinate system you choose to measure your decent.

I have never said that, and in fact I've explicitly said several times that whether or not you fall into the hole does *not* depend on the coordinates you use. What *does* depend on the coordinates is whether particular events on your worldline--such as the real, actual events of you crossing the hole's horizon and moving inward beyond that point--are "visible" in those coordinates. It's no different than saying that something that happens on the surface of the Earth at a point beyond your horizon--where you can't see--isn't visible to you.

In Rindler coordinates, the portion of a free-falling object's worldline that lies at or below the Rindler horizon is not visible, but it *is* visible in Minkowski coordinates. That's all there is to it; one coordinate system can see events that aren't visible in the other. I really don't see why this is so hard to grasp.

Similarly, in Schwarzschild exterior coordinates, the portion of a free-falling object's worldline that lies at or below the black hole horizon is not visible, but it *is* visible in Kruskal coordinates. Again, one coordinate system can see events that aren't visible in the other. What's impossible about that?

Why is it so acceptable to switch between two entirely views when one shows something that's automatically assumed wrong?

The only one who's automatically assuming it's wrong is you. I make no such assumption.

Schwarzschild coordinates cover the entire space-time external to the black hole.

Yes.

It's just that time and space are compressed to c/infinity at the horizon.

No, they're not. Nor do you have to "break the light barrier" to fall into the hole, because you're moving inward, not outward. The fact that you can't comprehend the consistent model that explains all this does not mean that model is false or inconsistent; it just means you can't understand it.

 

[Dmitry67]

Isn't that describing a Rindler horizon rather than an event horizon?

It is a toy model

It just shows the idea that infinities in one coordinate system do not always have any physical meaning. Like infinities in S. coordinates are a problem of that coordinate system, while infalling objects can cross the horizon without any problems.

 

[A-wal]

No, weight is a force, not energy. You feel the force of the Earth pushing on you. No energy is expended (see below).

Not quite. We feel the *force* of the Earth pushing on us. The Earth, similarly, "feels" the force of us pushing on it. You're correct that there is no need to describe either of these forces as "gravity". They're not. However...

Only in the sense that the "expended energy" is zero in both cases.

You have the correct definition of static force: it's a force which does not result in any relative motion of either body. It's relevant because at least one body has to move for any work to be done and any energy to be expended. For example, if you stand up from a prone position, you *do* expend energy, because you have to do work to raise your body's mass from a lower to a higher position (i.e., you move your body further away from the Earth--more precisely, your body's center of mass is further from the Earth's center of mass). Or if you are far out in space in a rocket and you fire the rocket's engine, the rocket does work by expelling exhaust out the back--the exhaust and the rocket are moving relative to each other, so work is done and energy is expended. There has to be relative motion.

Okay then, I should have said force instead of energy.

This is not quite right either, at least not in the way you use the term "gravity" (because you say tidal gravity and "acceleration due to gravity" are the same thing, which they're not). There is a sense in which "gravity is relative", meaning the "acceleration due to gravity" is relative, because you can always find a freely falling reference frame in which, locally, that acceleration disappears. But *tidal* gravity is *not* relative; it is a frame-invariant, genuine physical quantity.

Okay, if "you can always find a freely falling reference frame in which, locally, that acceleration disappears" what about at the event horizon? I didn't say tidal gravity is relative. I said it's the equivalent to acceleration in flat space-time. If I was using relative velocities as a measurement of the strength of gravity then it wouldn't be relative and the same area of space-time would change its properties depending on which observer makes the measurements. One of the reasons I have a problem with standard GR is exactly that. The properties of the same area of space-time shouldn't change depending on which direction you travel through it.

There is no *local* test you can perform to show that you're "being accelerated by gravity" if you are in free fall (i.e., not feeling any force). (Obviously you can use non-local information, such as the fact that you are above a large planetary body like the Earth, to infer that that body's mass is causing you to fall towards it. But that's not a local test; it requires information about distant objects.) But there *are* tests you can perform to show whether *tidal* gravity is present or absent between you and neighboring objects, even if you and all those other objects are all in free fall.

I wonder why. Do you think maybe it's equivalent to acceleration? Tidal force is the only way, and there's always tidal force because it has infinite range. That's acceleration though because it's the difference, or the rate of change as you put it, which is also the definition of acceleration.

There is no "force of gravity" anywhere in the problem. The electromagnetic force of the atoms in your body pushes against the electromagnetic force of the atoms in the ground, and the ground pushes back.

What? But why do they push against each other? Both together are the reaction to the force of gravity. The pull of gravity is being countered by acceleration by the combination of the Earth and our bodies both being solid objects.

The Moon and the rock on the Moon are not expending any energy. How can they? Where is it going to come from? There are no chemical reactions going on, no nuclear reactions going on, no fuel being burned. Where does the energy come from? The answer is that it doesn't have to come from anywhere because no energy is being expended. The Moon and the rock are in a state of static equilibrium, with no work being done, and they will stay that way indefinitely, with no work being done and no energy expended. By your logic, the rock and the Moon would eventually "run out" of energy and collapse somehow. They don't.

If you were to wait long enough then the rock would be crushed under its own weight. Admittedly, I'd have to invent a whole new set of epically large numbers to describe how long that would take, but I'm sure it would happen, eventually. If there's too much gravity for the object to take then it will collapse. If the object is close to the limit then it will collapse after a time. If it's not close then it will collapse after a long time. This is the same principle as gradually reaching infinity. I don't think anything can be strong enough to resist anything forever.

No, spacetime can be curved in only one way, by the presence of stress-energy (matter, energy, pressure, and internal stresses of materials--the stuff that goes into the stress-energy tensor).

You've just listed four. Space-time can be curved BY one of two main causes.

An object's *path* through spacetime can also be curved, in the frame-invariant sense of "curved", in one way: by an object having a force (again, in the frame-invariant sense of "force") exerted on it, causing it to accelerate (in the frame-invariant sense of "accelerate"). You can tell whether an object's path is curved in this sense by seeing whether it feels weight.

You would feel your weight if you were in free-fall in a strong enough gravitational field. Remember, gravity's very weak. It looks stronger than it really is because of the change it causes in relative velocity (yes I do mean it like that this time), but that doesn't count because it's mostly cause by the acceleration of object resisting gravity. Only acceleration into a higher gravitational field counts, but the object's being accelerated from a constantly increasing relative speed, making it look stronger.

A path can "look" curved without being curved (in the frame-invariant sense), or be curved (in the frame-invariant sense) without "looking" curved, if you adopt a coordinate system that obscures the frame-invariant curvature. For example, in Schwarzschild exterior coordinates, the paths of objects "hovering" at constant radius above a black hole's horizon look straight, but they are curved in the frame-invariant sense. (The same is true of your path through spacetime when you're standing motionless on the Earth's surface.)

It's curved any time you feel acceleration. It's the same thing.

I've done this. I've explained how:

(a) Curvature of *spacetime* is different and distinct from curvature of an object's *path* through spacetime.

(b) Tidal gravity is different and distinct from "acceleration due to gravity".

(a) Not really.

(b) It's the “acceleration of gravity”.

There is no jump. The outward velocity required to escape (for an object that's in free fall, not accelerated) increases smoothly to c as the horizon is approached and reached, and the energy required to escape (again, for an object that's in free fall, not accelerated) increases smoothly to infinity. This works the same as an object's energy going to infinity as its speed approaches c (in a given reference frame) in flat spacetime. There's no jump.

Which proves my point. You can't accelerate smoothly up to c in flat space-time can you? So why do you insist on being able to do the exact same thing using gravity instead? Infinity can't be reached gradually, so it can't be reached.

There is no "force" anywhere. It's just geometry. See below.

If you're going to look at it like that then acceleration is also just geometry. You can't claim that they're different types of curvature because one affects matter and the other affects things in it. That doesn't make sense. You can't affect space-time on its own because space-time is nothing on its own. It's just a measurement of the relative distances between matter. If what you're saying about it being dependant on the direction you're moving through it and all those other stupid things them maybe it would seem as though it could have properties in its own right. This is what happens when you let equations direct your intuition rather than the other way round. Of course there's a force.

No. You keep on getting this backwards. The Rindler horizon is the path of a light ray that just fails to catch up to a family of uniformly accelerating observers outside (or "above") it. Similarly, a black hole's horizon is the path of a light ray that just fails to catch up to a family of "hovering" observers outside it. Neither horizon has anything to do with whether an object outside the horizon can catch up with an object that free-falls *into* the horizon. That's a separate question.

I KNOW they're different questions! Are you doing this on purpose? My point before was exactly that. I was asked why I thought an object couldn't cross an event horizon when it could obviously cross a Rindler horizon. Not having the slightest clue what a Rindler horizon was, I looked it up. It's the only time I've done that so it's not surprising that I got it slightly wrong. It's not how I'd normally do it. My point here was that the event horizon marks the point when nothing from the outside can catch the free-faller because they've broken the light barrier. Silly isn't it?

No. I apologize for a misstatement there; I should have snipped the quote from your previous post more carefully. The only statement of yours I meant to agree with was the statement that being in free fall is equivalent to being at rest. I did not mean to imply agreement with your statement about tidal force.

Well it's a start. How is tidal force in a gravitational field any different from acceleration in flat space-time?

"Energy" in the sense you are using the term (meaning energy expended by an object to accelerate itself so it feels weight) has no effect on the light cones. It can't tilt them at all. Also, "gravity" is not a separate "thing" that somehow tilts the light-cones. Gravity *is* the tilting of the light cones. They're the same thing (spacetime curvature).

That's not what you said before. Didn't you say that acceleration in flat space-time was equivalent, so you could tilt light cones but not all the to 45 degrees from this view?

No. The tilt is not "past 45 degrees". The *ingoing* sides of the light cones remain pointed at 45 degrees inward. The *outgoing* sides gradually point more and more *upward* (i.e., *less* than 45 degrees) until they become vertical (0 degrees) at the horizon. Inside the horizon, they tilt over the other way, which simply means objects can no longer avoid moving inward, even if they move at the speed of light. It does not require going backward in time.

The angle depends on which side you are? Right.

No, that doesn't follow, because in order to escape, you have to move *outward* faster than light. You do not have to move *inward* faster than light to get inside the horizon. You can free-fall there, without "moving" at all (you yourself said being in free fall is the same as being at rest, so all you have to do to get inside the hole is to stay at rest).

I think you may have misunderstood what I meant. You don't have to move to get in but you have to move to get out? Free-fall is the equivalent to being at rest in the sense that you don't feel any acceleration and you can't reach reach if you're always at rest.

No, "gravity" is a way of saying that *spacetime* is curved. It does not say anything about particular *paths* in that spacetime. You can follow a geodesic path (freely falling path, weightless path) in a curved spacetime, where gravity is present. They are separate, distinct concepts.

Of course they're not. What you call a geodesic path is one that goes with the curvature caused by gravity rather than trying to fight it. You could use an ongoing source of energy to replicate the exact same thing with outgoing rather than ingoing curvature.

To make sense, yes. To make sense in a way that satisfies all your intuitions, no.

We'll see.

And yet you keep trying to claim that tidal force is the *same* as "acceleration due to gravity" or "apparent force of gravity". How can something be the same as its rate of change?

Because being in free-fall is equivalent to being at rest, so you have to ignore the difference in relative velocities because that's all it is. The only thing that determines the direction of gravity is the direction it's strongest in, in other words the difference in gravity between the two end points of the object. The sharper the increase, the stronger the gravity. I'm not talking about the increase in velocity of an object free-falling in a gravitational field relative to something not free-falling.

Um, because a larger area of you is in contact with the ground.

But there are tidal forces in multiple directions, not just one. There can even be tidal force in the time direction. You can't extract a unique direction for gravity to pull you from tidal force.

No, it's because your weight is spread more evenly over a larger area of your body. See above.

If gravity is relative then the only thing that determines which way you are pulled is tidal force because that represent acceleration and it's that and only that that can really be classed as the force of gravity. That's all I was getting at. Stop wandering off.

I wasn't assuming anything. I was responding to the actual words you used. You said: "I thought there was an event horizon in the standard description of the beginning of the universe. It's the edge of the universe." You didn't elaborate on what you meant by "edge". How was I supposed to interpret it except in the obvious way?

My point is that's not the obvious way, especially not in a relativity forum.

Why not? I know it doesn't make sense in the mental model you have in your head. But can you give a logical argument, starting from premises we all accept, that shows that it *can't* make sense, in *any* consistent model? I don't think you can, but if you have such an argument, let's see it. Just claiming that it doesn't make sense won't do, because I have a consistent model where it does makes sense.

No it doesn't. To describe one thing it needs multiple coordinate systems that completely contradict each other, it effectively accelerates you past c, it breaks the arrow of time and it has areas of space-time that have different values for gravity depending on which direction you travel through them. Apart from that it might be a consistent model that makes sense.

Because the worldlines are expanding in different directions. Ned Wright's Cosmology FAQ, at the link below, briefly explains the kind of thing that's going on.

http://www.astro.ucla.edu/~wright/cosmology_faq.html#DN

Nothing can travel faster than light... accept when a physicist needs it to.

I'm not sure what you're asking here. I was just explaining what r = 2M means, since you asked that; that radius is the radius of the horizon. Radius is not a quantity in the time dimension.

Debatable. Length is though. A singularity has zero volume in all four dimensions, so it covers no space and no time. Why would the effect of the singularity be anything other than a four-dimensional sphere? Length contracts and time dilates, same thing.

Since you can reach the black hole, this question is meaningless.

Only by cheating and bending or ignoring altogether the rules that it contradicts. I don't think you can actually do it in real life. Doesn't it seem in the least bit silly to you?

They can't.

It still doesn't matter.

Oh, yes, it does. It's crucial, because it grounds the analogy between Rindler coordinates and Schwarzschild coordinates (as compared with Minkowski coordinates and Kruskal coordinates). See below.

I have never said that, and in fact I've explicitly said several times that whether or not you fall into the hole does *not* depend on the coordinates you use. What *does* depend on the coordinates is whether particular events on your worldline--such as the real, actual events of you crossing the hole's horizon and moving inward beyond that point--are "visible" in those coordinates. It's no different than saying that something that happens on the surface of the Earth at a point beyond your horizon--where you can't see--isn't visible to you.

Yes it is different. I can bring it into view by moving towards it. Nothing can witness an object crossing into black hole unless it crosses itself. But that object itself can't cross as far as any external observers can see, so no object crosses the event horizon from the outside

In Rindler coordinates, the portion of a free-falling object's worldline that lies at or below the Rindler horizon is not visible, but it *is* visible in Minkowski coordinates. That's all there is to it; one coordinate system can see events that aren't visible in the other. I really don't see why this is so hard to grasp.

It wouldn’t be a problem if that’s all there were to it. It stops working when you use infinity in one of them, because you can’t translate that into the other one without the two contradicting each other. I really don't see why that is so hard to grasp.

Similarly, in Schwarzschild exterior coordinates, the portion of a free-falling object's worldline that lies at or below the black hole horizon is not visible, but it *is* visible in Kruskal coordinates. Again, one coordinate system can see events that aren't visible in the other. What's impossible about that?

Nothing. Of course it's not impossible, provided they don't contradict each other. It's not okay two use two contradictory points of view and claim both are correct.

The only one who's automatically assuming it's wrong is you. I make no such assumption.

You serious?

Yes.

Schwarzschild coordinates cover the entire space-time external to the black hole and show that time dilates and length contracts to infinity at the horizon. At no point on any of that entire space-time external to the black hole does any object ever reach the horizon so I’d love to know why you think they can cross it.

No, they're not. Nor do you have to "break the light barrier" to fall into the hole, because you're moving inward, not outward. The fact that you can't comprehend the consistent model that explains all this does not mean that model is false or inconsistent; it just means you can't understand it.

Or maybe you just don't understand why it's not consistent?

 

[A-wal]

Use the river model, but add a river bed to act as a kind of ether. Not a real ether, just an imaginary one to compare our starting position with the horizon. We start far enough away for the gravitational pull of the black hole to be negligible enough to ignore. Our river bed provides us with an uncurvable surface that’s at rest relative to our starting position and the black hole. We wait a while and eventually notice we’re moving towards it as the river slowly moves us along relative to our starting position. We don't feel as though we're accelerating because we're still at rest relative to the river. That river moves relative to the river bed in exactly the same way any object using energy to accelerate would. It can’t reach c. The river changes in relation to the river bed as it moves faster, and the black hole changes with it. Time dilation and length contraction relative to the river bed mean that the its life span and size decrease exponentially as we approach the horizon. We would be traveling at the speed of light at the horizon if it made sense for anything to move that fast. You could try to use energy to accelerate you through the river to get you there faster, but that would just have the effect of adding very high velocities.

It is a toy model

It just shows the idea that infinities in one coordinate system do not always have any physical meaning. Like infinities in S. coordinates are a problem of that coordinate system, while infalling objects can cross the horizon without any problems.

I think it's a problem with the coordinate system you're using for the in-faller. They can't both be right.

 

[Dmitry67]

The example was supposed to illustrate that there is no objective 'mapping' of times in different coordinate systems. Only in nearly flat times you can use 'time dilation', which is like a low order correction. In BH you can'tuse that notion because it fails (you get infinities).

But if you draw light cones everything is simple and logical.

It can’t reach c.

In GR, speeds (for distant objects) faster than c are well know. Just an example - areas behind the cosmological horizon.

That thread is quite long, please let me know: do you deny GR or do you deny some consequences of GR?

 

[PeterDonis]

Okay, if "you can always find a freely falling reference frame in which, locally, that acceleration disappears" what about at the event horizon?

Sure, you can do it at the horizon. Take a freely falling observer and pick the event on that observer's worldline where he is just crossing the horizon. Make that event the origin of the local freely falling reference frame in which the observer's "acceleration" towards the black hole disappears. Call the (local) coordinates of that frame X and T; X increases in the outgoing radial direction, and T increases in the future time direction. Then the line X = T in that frame is the event horizon--more precisely, it's the little piece of the event horizon that lies within the range of the local freely falling frame.

I didn't say tidal gravity is relative. I said it's the equivalent to acceleration in flat space-time.

Which it isn't.

If I was using relative velocities as a measurement of the strength of gravity then it wouldn't be relative and the same area of space-time would change its properties depending on which observer makes the measurements. One of the reasons I have a problem with standard GR is exactly that. The properties of the same area of space-time shouldn't change depending on which direction you travel through it.

The properties of spacetime don't change depending on direction of travel. Outgoing and ingoing observers both see the same curvature of spacetime--meaning tidal gravity. They also both experience the same "acceleration due to gravity", inward towards the hole. The very fact that that "acceleration" is inward for all observers, regardless of their direction of travel, is *why* outgoing signals can't escape from any point on or inside the horizon, while ingoing signals can pass inward.

I wonder why. Do you think maybe it's equivalent to acceleration? Tidal force is the only way, and there's always tidal force because it has infinite range. That's acceleration though because it's the difference, or the rate of change as you put it, which is also the definition of acceleration.

No, "acceleration" is the rate of change of *velocity*, *not* the rate of change of acceleration, which is what "tidal force" is.

What? But why do they push against each other? Both together are the reaction to the force of gravity.

No, they aren't. There is no "force of gravity". A body that is moving solely under the influence of "gravity" feels no force at all; it's weightless, in free fall.

The pull of gravity is being countered by acceleration by the combination of the Earth and our bodies both being solid objects.

Once again, there is no "pull of gravity". It's simply that spacetime around the Earth is such that all the freely falling worldlines move inward, towards the Earth's center. You're correct that what prevents an object on the Earth's surface from following such a worldline is the fact that the object and the Earth are solid bodies; the Earth therefore pushes up on the object and keeps it from falling freely towards the center, so the object feels weight. The object also pushes down on the Earth--more precisely, it pushes down on the piece of the Earth's surface directly underneath it. But there's more Earth underneath that piece pushing back on it, and under that, and so on, so the Earth can't move inward in response to the object's push; its surface stays the same.

If you were to wait long enough then the rock would be crushed under its own weight. Admittedly, I'd have to invent a whole new set of epically large numbers to describe how long that would take, but I'm sure it would happen, eventually.

You're wrong. The rock will never be crushed under its own weight; it and the Moon can sit there in equilibrium indefinitely.

If there's too much gravity for the object to take then it will collapse. If the object is close to the limit then it will collapse after a time. If it's not close then it will collapse after a long time. This is the same principle as gradually reaching infinity. I don't think anything can be strong enough to resist anything forever.

Again, you're wrong. You're basically claiming that every object will eventually collapse into a black hole, regardless of its initial state. That's wrong, and has been known to be wrong since the 1950's, when John Wheeler and some students of his studied the possible end states of matter. Kip Thorne talks about it in Black Holes and Time Warps.

You've just listed four. Space-time can be curved BY one of two main causes.

No, I listed four different manifestations of one cause, the stress-energy tensor. "Matter" and "energy" are just different units for expressing the stress-energy tensor; "matter-energy" vs. "pressure" and "stress" are just different components of the tensor, and how the tensor breaks up into components like that is frame-dependent; different observers in different states of motion will break up the tensor into "energy" vs. "pressure" or "stress" in different ways, but they will find the same physical laws, which depend only on the tensor as a whole, a single geometric object. It's all one cause.

You would feel your weight if you were in free-fall in a strong enough gravitational field.

No, you wouldn't. A freely falling object is always weightless, regardless of how curved spacetime is (how strong "gravity" is). This is one of the most basic ideas in general relativity; if you don't understand that, then it's no wonder you're having trouble with the rest of it.

It's curved any time you feel acceleration. It's the same thing.

Only if you mean "curved" in the frame-invariant sense I gave. In that sense, the worldline of an object "hovering" at a constant radius over a black hole is curved. So is your worldline when you are standing motionless on the surface of the Earth. If you agree with both those statements, then we're OK.

Which proves my point. You can't accelerate smoothly up to c in flat space-time can you? So why do you insist on being able to do the exact same thing using gravity instead?

I'm not insisting on any such thing.

If you're going to look at it like that then acceleration is also just geometry.

No, it isn't. Again, this (the difference between curvature of spacetime itself and curvature of a path in spacetime) is one of the most basic concepts in relativity. If you don't understand that, it's no wonder you're having trouble.

You can't affect space-time on its own because space-time is nothing on its own. It's just a measurement of the relative distances between matter.

Not according to general relativity. In GR, spacetime is a dynamical entity of physics, right alongside matter-energy; its dynamics are contained in the Einstein Field Equation. That equation also includes matter-energy, so spacetime and matter-energy can affect each other.

My point here was that the event horizon marks the point when nothing from the outside can catch the free-faller because they've broken the light barrier.

And I keep on telling you that you've got this backwards, and you keep on saying it anyway. The event horizon marks the point where nothing *outgoing* can catch up with *accelerating* objects outside the horizon. It does not mean that nothing from outside can catch an object free-falling inward.

Well it's a start. How is tidal force in a gravitational field any different from acceleration in flat space-time?

Have you not been reading all the previous posts where I explained this? Objects can experience "tidal force" in free fall. Acceleration in flat spacetime (or in curved spacetime, for that matter) means that an object is not in free fall. They're different concepts.

That's not what you said before. Didn't you say that acceleration in flat space-time was equivalent, so you could tilt light cones but not all the to 45 degrees from this view?

No. Acceleration, in the invariant sense (an object is accelerating if it feels weight), has nothing to do with the orientation of the light cones, either in flat or in curved spacetime.

The angle depends on which side you are? Right.

Wrong. The "angle" of the light cones at a given event is an invariant geometric feature of the spacetime. As such, it's the same for all observers, regardless of their state of motion.

I think you may have misunderstood what I meant. You don't have to move to get in but you have to move to get out?

You can't get out (from inside the horizon) at all, no matter how you move. You can be accelerated instead of in free fall and still get in; just accelerate inward, or accelerate outward but not enough to "hover".

Of course they're not. What you call a geodesic path is one that goes with the curvature caused by gravity rather than trying to fight it. You could use an ongoing source of energy to replicate the exact same thing with outgoing rather than ingoing curvature.

No, you can't. They're different things. The "curvature" you're talking about using energy to "replicate" is curvature of a path--an object follows a curved path if it feels weight. The "curvature" of a geodesic path is curvature of spacetime itself; an object following a geodesic path feels no weight, and the *path* itself is as "straight" as it can be, in the spacetime it's in.

The only thing that determines the direction of gravity is the direction it's strongest in, in other words the difference in gravity between the two end points of the object.

No, it isn't. In GR, the "direction of gravity" (meaning the direction of the effective "force" an object feels) also depends on the motion of the object. For example, a planet in an elliptical orbit does not feel a force that points directly towards the Sun. That's why the perihelion of the orbit precesses; this has been measured for the planet Mercury and the measurement agrees with the GR prediction. In all the discussion in this thread, I've been talking only about purely radial motion, where this issue doesn't arise.

My point is that's not the obvious way, especially not in a relativity forum.

Then what is the "obvious" way in a relativity forum?

To describe one thing it needs multiple coordinate systems that completely contradict each other, it effectively accelerates you past c, it breaks the arrow of time and it has areas of space-time that have different values for gravity depending on which direction you travel through them.

None of these claims are true. The multiple coordinate systems are consistent with each other, there is no acceleration past c (objects always move within the light cones), the arrow of time is the same throughout the spacetime, and the "value of gravity" does not depend on your direction of travel (the curvature of spacetime is the same for all observers whatever their state of motion). This has been explained multiple times in this thread. If all you can do is keep repeating these false claims, there's not much point in continuing to discuss them, so I won't respond unless you actually can offer some substantive arguments. I'll only respond to correct outright false statements, or when I'm not sure I'm understanding something correctly.

Nothing can travel faster than light... accept when a physicist needs it to.

No object ever moves outside the light cones. That's all that "can't travel faster than light" means.

Debatable. Length is though.

Length is a quantity in the time dimension?

A singularity has zero volume in all four dimensions, so it covers no space and no time.

Wrong; that is not a necessary feature of a singularity. It happens to be true of the Big Bang singularity in the spacetimes used in cosmology, but it is *not* true of the singularity at the center of a black hole. That singularity is a spacelike line.

Doesn't it seem in the least bit silly to you?

No.

It wouldn't be a problem if that's all there were to it. It stops working when you use infinity in one of them, because you can't translate that into the other one without the two contradicting each other. I really don't see why that is so hard to grasp.

Because you don't understand the mathematics and logic that underlie the translation. It's perfectly straightforward and consistent. The description of Kruskal coordinates on the Wikipedia page shows how the translation is done. If you want more detail, all the major relativity textbooks describe the translation.

Schwarzschild coordinates cover the entire space-time external to the black hole and show that time dilates and length contracts to infinity at the horizon. At no point on any of that entire space-time external to the black hole does any object ever reach the horizon so I'd love to know why you think they can cross it.

Um, because the event of the crossing of the horizon is not in the spacetime external to the black hole, but another part of spacetime that those coordinates don't cover?

Or maybe you just don't understand why it's not consistent?

If it's not consistent, then it's not just me that doesn't understand. It's the entire relativity physics community, that has been using the theory to make correct predictions for almost a century now (starting from Einstein in 1915).

 

[PeterDonis]

Use the river model, but add a river bed to act as a kind of ether. Not a real ether, just an imaginary one to compare our starting position with the horizon.

You're describing the "river model" pretty well for the portion of spacetime outside the horizon; but adding the "river bed" to the model is gratuitous unless you can come up with a physical reason why it's necessary. The model predicts all the physics outside the horizon perfectly well without including the river bed at all, so Occam's razor says eliminate the river bed unless you can give a physical reason (some physical prediction that the model makes differently) why it's needed. And no, the "prediction" that nothing can cross the horizon is *not* a good physical reason, because that's precisely the point at issue. You have to find a reason *outside* the horizon why you need the river bed to make correct predictions. Otherwise you can just eliminate it and have the standard river model, which deals perfectly well with objects crossing the horizon and going inside.