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

[A-wal]

Three questions:

1). Using the river model; what speed would a free-faller have accelerated to relative to the hovering observers of the river bed when they hit the horizon?

2). Let’s do this the other way round. New four-dimensional shape. This one does expand and contract at a constant speed. What would that shape look like?

3). Why is the length contraction and time dilation from gravity considered as curvature when the length contraction and time dilation from acceleration isn’t? Seems a bit hypocritical.

 

[Dale]

2). Let’s do this the other way round. New four-dimensional shape. This one does expand and contract at a constant speed. What would that shape look like?

A cone.

3). Why is the length contraction and time dilation from gravity considered as curvature when the length contraction and time dilation from acceleration isn’t? Seems a bit hypocritical.

The Riemann curvature tensor is defined as:

$${R^\ell}_{ijk}= \frac{\partial}{\partial x^j} \Gamma_{ik}^\ell-\frac{\partial}{\partial x^k}\Gamma_{ij}^\ell +\sum^{n}_{s=1}(\Gamma_{js}^\ell\Gamma_{ik}^s-\Gamma_{ks}^\ell\Gamma_{ij}^s)$$

http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Curvature_tensors

Where the Christoffel symbols are given by:

$$\Gamma^m_{ij}=\frac12 g^{km} \left( \frac{\partial}{\partial x^i} g_{kj} +\frac{\partial}{\partial x^j} g_{ik} -\frac{\partial}{\partial x^k} g_{ij} \right)$$

The curvature tensor evaluates to 0 for a metric like the Rindler metric where there is acceleration. So non-gravitational acceleration is not due to curvature of the manifold.

There is a different concept, called the covariant derivative, which measures the proper acceleration felt by a particle on a given worldline. I.e. this is the acceleration which would be measured by an accelerometer.

$$\nabla_j v^i=\frac{\partial v^i}{\partial x^j}+\Gamma^i_{jk}v^k$$

http://en.wikipedia.org/wiki/List_o...y#Christoffel_symbols.2C_covariant_derivative

In the Rindler metric, a particle with a constant position coordinate has a constant and non-zero covariant derivative, despite the fact that the curvature tensor is 0. So it is clear that acceleration and curvature are different things.

 

[cosmik debris]

Three questions:

1). Using the river model; what speed would a free-faller have accelerated to relative to the hovering observers of the river bed when they hit the horizon?

2). Let’s do this the other way round. New four-dimensional shape. This one does expand and contract at a constant speed. What would that shape look like?

3). Why is the length contraction and time dilation from gravity considered as curvature when the length contraction and time dilation from acceleration isn’t? Seems a bit hypocritical.

1). The local speed of the event horizon is c.
2). I don't know what this means.
3). They are considered different because they are. GR and the acceleration thing are only equivalent to first order.

I really shouldn't wade in at this late stage but I have if only to reinforce what Dale and Peter have told you. You have so many misunderstandings about GR and SR and you keep repeating them. You don't seem to have learned anything in hundreds of posts. These are mathematical subjects and maths is used because it is the only unambiguous way of expressing the relationships. But not understanding the maths is only a small part of your problem as I see it, your main problem is your refusal to move forward.

Listen to Dale and Peter, I don't know how they have remained so patient.

 

[PeterDonis]

I posted a response earlier but somehow it seems to have vanished, at least when I view the site, so I'm posting again. Apologies if this ends up being a double post.

1). Using the river model; what speed would a free-faller have accelerated to relative to the hovering observers of the river bed when they hit the horizon?

There are two possible answers, depending on how the term "relative speed" is interpreted:

(1a) If "relative speed" means "the speed at which a hypothetical observer hovering at a constant radius r would see the free-faller falling, even if such an observer can't actually exist", then the answer at the horizon is the speed of light, c (and inside the horizon it's some speed greater than c). But if "relative speed" is interpreted this way, then the river model does *not* require that relative speed be less than c (for objects with nonzero rest mass). "Relative speed" interpreted this way is just a ratio of coordinates, and has no direct physical meaning. Sometimes it so happens that you can equate it to something that does have direct physical meaning, but there's no requirement that you must always be able to do so.

(1b) If "relative speed" means "the speed at which an actual hovering observer would observe the free-faller to be falling", then the question has no meaning at or inside the horizon since there are no such hovering observers in that region of spacetime. This is the only definition of "relative speed" that requires relative speed to be less than that of light (for objects with nonzero rest mass); at or inside the horizon, it's still true that actual physical objects with mass must move at less than the speed of light relative to each other; but it's also true that they must all be moving inward (i.e., decreasing radius with time), so none of them can be "hovering".

None of this contradicts the river model--at least not the "standard" river model, the one described in this paper:

http://arxiv.org/abs/gr-qc/0411060

If you have some other model in mind as the "river model", please give a reference.

3). Why is the length contraction and time dilation from gravity considered as curvature when the length contraction and time dilation from acceleration isn’t? Seems a bit hypocritical.

DaleSpam gave the mathematical answer to this; I just want to supplement a little bit what he said.

I gave the standard GR definitions of curvature of spacetime and curvature of a worldline (i.e., proper acceleration) in non-mathematical terms in an earlier post. (DaleSpam's math is a more precise way of saying what I said.) So it's not true that acceleration (if you mean proper acceleration) is not considered as curvature: it's just curvature of a worldline instead of curvature of spacetime. Those two types of curvature are different, independent fundamental concepts in GR. Length contraction and time dilation are not fundamental concepts in GR; they are just "side effects" of curvature, and the fact that both types of curvature happen to give rise to them does not indicate that both types of curvature are somehow "the same". They just both happen to have similar side effects in that particular way.

 

[Dale]

So it's not true that acceleration (if you mean proper acceleration) is not considered as curvature: it's just curvature of a worldline instead of curvature of spacetime. Those two types of curvature are different, independent fundamental concepts in GR.

good way of saying it.

 

[A-wal]

I'll explain this purely in the context of Schwarzschild coordinates. Apparently this system is incomplete and the entire space-time isn't covered in these coordinates, but I haven't got a clue what that means. The entire space-time is covered but it's hidden. Not very well hidden, but apparently well enough. A black hole is an area of space-time where time dilation and length contraction have gone beyond the point that they would if you had accelerated to c, beyond infinity. But the Schwarzschild coordinates don't actually allow this to happen. Instead, while time dilation and length contraction would shrink the size and lifespan of the black hole down to a single point in time and space (the singularity) if you get close enough, from a distance we "see" curvature. If we look at the inside we see an area where objects would be accelerated by gravity beyond the point of c. Objects obviously can't do that, but c becomes a "visible" place. In other words there is now a specific area of space-time that no amount of acceleration could ever take you to. This is because the time dilation and length contraction from the acceleration of gravity would be infinite at the horizon, in the same way they are at c. Everything outside is moving slower than c and everything inside would be moving at over c, which is why you're getting a black hole in the first place, but the horizon isn't always in the same place. It moves. It's relative, like c.

The space-time’s compressed from a distance so that objects closer to the singularity appear shorter and slower than the objects are in their own frames, just like an accelerator does in flat space-time. The horizon represents infinite time dilation and length contraction, just like you get at c. So there's a potentially infinite amount of space-time between you and the horizon (just like you get a potentially infinite amount of space-time between you and the c in flat space-time) which is determined by the life-span of the black hole. This means that it's not possible for an object to reach the horizon from the perspective of anything on the outside, and it's not possible for an object to reach the horizon from the perspective of anything on the inside either because of more than infinite curvature. There are apparently other coordinate systems that do allow an in-faller to reach and cross the horizon but they directly contradict Schwarzschild coordinates because they can't cross from the perspective of the outside, even from the perspective of the parts of the in-falling object that haven't crossed yet. How can you move smoothly between the Schwarzschild coordinates, where an object can't cross the horizon, to a coordinate system where it can? At what point does the changeover happen? It should be smooth, but it can't be. In your version an object can't ever cross one second, and then it does.

Anything at the horizon would be moving at c relative to everything in the universe outside the horizon, even the atom next to it. The rope isn't needed, it just makes it easier to visualise. There would be no way of observing an in-faller and knowing whether or not they made it to the horizon, which doesn't make sense. If an object hasn't crossed from the perspective of an observer then it hasn't crossed (ignoring the delay of the time the light takes to reach you of course). There’s no reason for this sudden jump to infinity at the horizon if you treat it the same way as c, and everything’s simpler and makes a lot more sense. It takes infinite energy to accelerate an object to c. You think you've got it with a black hole but it doesn't work.

A cone.

In three dimensions it’s a cone, but only because there's a spatial dimension missing. Extend that cone into the horizontal direction you're using for time and you have half a sphere. Now I'm going to describe what I think happens during the formation (and whole life span come to think of it) of a black hole. A gravity wave (a change in the strength of gravity which moves outwards at c) is released from the singularity. This is the horizon, so the sphere is limited to c. It can't "hit" anything as it expands because the space-time that makes up the black (and white) hole is being created behind it, like length contracting from the back in Rindler coordinates when an object accelerates in flat space-time. The gravity wave carries on moving outwards at c and does hit objects in its path but that's not the event horizon. This is the first you would notice that something has changed. When the horizon has reached its maximum size (when gravity has no longer curved space-time beyond c) it rushes inwards at c because it’s determined by how close an object could get to it in that amount of time. There's the other half of the sphere, and your white hole. We now have a four-dimensional bubble that nothing can ever reach from any direction (including of time) formed around the singularity, from a distance anyway. Because this is all happening at c it isn't actually experienced by the black/white hole at all, in the same way that light doesn't experience time. It's infinitely small from its own perspective but it’s bigger the more distance there is between you and it.

The Riemann curvature tensor is defined as:

$${R^\ell}_{ijk}= \frac{\partial}{\partial x^j} \Gamma_{ik}^\ell-\frac{\partial}{\partial x^k}\Gamma_{ij}^\ell +\sum^{n}_{s=1}(\Gamma_{js}^\ell\Gamma_{ik}^s-\Gamma_{ks}^\ell\Gamma_{ij}^s)$$

http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Curvature_tensors

Where the Christoffel symbols are given by:

$$\Gamma^m_{ij}=\frac12 g^{km} \left( \frac{\partial}{\partial x^i} g_{kj} +\frac{\partial}{\partial x^j} g_{ik} -\frac{\partial}{\partial x^k} g_{ij} \right)$$

The curvature tensor evaluates to 0 for a metric like the Rindler metric where there is acceleration. So non-gravitational acceleration is not due to curvature of the manifold.

There is a different concept, called the covariant derivative, which measures the proper acceleration felt by a particle on a given worldline. I.e. this is the acceleration which would be measured by an accelerometer.

$$\nabla_j v^i=\frac{\partial v^i}{\partial x^j}+\Gamma^i_{jk}v^k$$

http://en.wikipedia.org/wiki/List_o...y#Christoffel_symbols.2C_covariant_derivative

In the Rindler metric, a particle with a constant position coordinate has a constant and non-zero covariant derivative, despite the fact that the curvature tensor is 0. So it is clear that acceleration and curvature are different things.

Well it's not clear to me. I know you said you'd be using equations, and I shouldn't moan that I'm finally getting the kind of response that may actually answer my questions, but it'd be lovely to get it in English as well. I take it "In the Rindler metric" means using Rindler coordinates, and "constant and non-zero covariant derivative" means constantly accelerating at the same rate. So an object that is undergoing zero proper acceleration can still be accelerating in Rindler coordinates despite the fact that there’s no curvature? There’s no curvature because it’s being expresses as acceleration instead. You also said "there is a different concept, called the covariant derivative, which measures the proper acceleration felt by a particle on a given worldline." What's stopping that from being viewed as curvature? Whatever energy’s accelerating that particle will also accelerate every other object in the universe, a bit, so how’s that different from curvature? I'm still failing to see the distinction. Maybe I misunderstood what you said?

1). The local speed of the event horizon is c.
2). I don't know what this means.
3). They are considered different because they are. GR and the acceleration thing are only equivalent to first order.

1). Yes it is, and that's okay for an event horizon but not for matter approaching it.

2).That's okay. Apparently shapes can be hard.

3).They are considered different because they are? Why are GR and the acceleration thing only equivalent to first order?

I really shouldn't wade in at this late stage but I have if only to reinforce what Dale and Peter have told you. You have so many misunderstandings about GR and SR and you keep repeating them. You don't seem to have learned anything in hundreds of posts. These are mathematical subjects and maths is used because it is the only unambiguous way of expressing the relationships. But not understanding the maths is only a small part of your problem as I see it, your main problem is your refusal to move forward.

Listen to Dale and Peter, I don't know how they have remained so patient.

They're not misunderstandings about GR and SR. I know how they say it works. Don't assume I have misunderstandings from a few posts that you've read where I've used the concepts in a way you're not used to. Read the blogs I posted if you want a more conventional description, then you can tell me I have so many misunderstandings if you think there's some things I’ve gotten wrong. I have been listening to Dale and Peter, intently. I still haven’t had my questions properly answered. It may be that there's something I don't get but I don't think so. After this long I don't think I’m jumping to conclusions. They've probably stayed patient because we've only recently starting talking about it in this context. I was building up to it. I wanted to learn the official stance and see if I was missing something. It has been a while though. It would have been quicker to go to collage. I wouldn't be able to earn money at the same time though. These are not mathematical subjects. There’s only one mathematical subject and that’s mathematics. If that’s all you’re using then that’s all you’re doing because you have no conceptual understanding of what it is you're describing. In other words you don’t know what you’re talking about. But I do agree that maths is the only unambiguous way of expressing the relationships. It's not open to interpretation. It's a tool that should be based on understanding, not the other way round. I'll move on when someone explains to me why it doesn't work this way.

There are two possible answers, depending on how the term "relative speed" is interpreted:

(1a) If "relative speed" means "the speed at which a hypothetical observer hovering at a constant radius r would see the free-faller falling, even if such an observer can't actually exist", then the answer at the horizon is the speed of light, c (and inside the horizon it's some speed greater than c). But if "relative speed" is interpreted this way, then the river model does *not* require that relative speed be less than c (for objects with nonzero rest mass). "Relative speed" interpreted this way is just a ratio of coordinates, and has no direct physical meaning. Sometimes it so happens that you can equate it to something that does have direct physical meaning, but there's no requirement that you must always be able to do so.

(1b) If "relative speed" means "the speed at which an actual hovering observer would observe the free-faller to be falling", then the question has no meaning at or inside the horizon since there are no such hovering observers in that region of spacetime. This is the only definition of "relative speed" that requires relative speed to be less than that of light (for objects with nonzero rest mass); at or inside the horizon, it's still true that actual physical objects with mass must move at less than the speed of light relative to each other; but it's also true that they must all be moving inward (i.e., decreasing radius with time), so none of them can be "hovering".

None of them can be hovering on the other side of the event horizon, but you've got to reach it first. Remember the riverbed covers the entire "exterior" of the black hole, which is allowed. What speed would the free-faller be moving relative to the riverbed, or any object outside the horizon for that matter?

None of this contradicts the river model--at least not the "standard" river model, the one described in this paper:

http://arxiv.org/abs/gr-qc/0411060

If you have some other model in mind as the "river model", please give a reference.

The one I'm using might be a bit different, seeing as the first and only time I've heard of it was when somebody mentioned it in this thread and I thought it was a nice way of looking at it. Still do.

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.

An observer who is very, very far away would measure the distances as less than the starting frame of the in-faller and furthest hoverer, so let's have the book keeper in this frame. They see an equal distance between the hoverers, like I said. The first hoverer can be the bookkeeper.

The hoverers have to expend different amounts of energy to maintain a constant separation from the perspective of the bookkeeper.

Anything else you need clarified?

DaleSpam gave the mathematical answer to this; I just want to supplement a little bit what he said.

I gave the standard GR definitions of curvature of spacetime and curvature of a worldline (i.e., proper acceleration) in non-mathematical terms in an earlier post. (DaleSpam's math is a more precise way of saying what I said.) So it's not true that acceleration (if you mean proper acceleration) is not considered as curvature: it's just curvature of a worldline instead of curvature of spacetime. Those two types of curvature are different, independent fundamental concepts in GR. Length contraction and time dilation are not fundamental concepts in GR; they are just "side effects" of curvature, and the fact that both types of curvature happen to give rise to them does not indicate that both types of curvature are somehow "the same". They just both happen to have similar side effects in that particular way.

What a coincidence. They both have the exact same two consequences and they're both caused by curvature. What are you defining as curvature anyway? It just means you're viewing movement as a change in the space-time between objects rather than actual movement of the objects themselves. Tomato, tomato. You’re saying it’s okay that matter can reach the equivalent of c if it’s because of curved space-time, but it doesn’t make any difference. What makes them distinct concepts? The results shouldn’t be any different from viewing it as accelerating towards c in flat space-time. You could view acceleration as curved space-time as well to completely switch it round and it still wouldn’t make any difference. Energy doesn't just affect the object that's producing it. It spreads out to affect every object in the universe to some extent, but it's an inverse square so most objects are hardly affected at all. I can't see any reason why this can't be viewed as curvature of space-time, or why time dilation and length contraction aren’t fundamental concepts and curvature as just one way of looking at it in either case.

 

[PeterDonis]

A-wal, I see that you went to considerable trouble to "explain" your viewpoint, but your explanation (which is the first part of your post, before you responded to others) doesn't change my viewpoint at all. I still think you're making the same mistake I've thought you were making all along. So I still don't see much point in responding to it, but I do want to comment on a couple of things:

Anything at the horizon would be moving at c relative to everything in the universe outside the horizon, even the atom next to it.

By "anything at the horizon", do you mean an object free-falling inwards that is just crossing the horizon, relative to an object hovering just outside the horizon? If so, then there is an interpretation of "moving at c" that makes the above statement true, but that interpretation is *not* one that requires objects to move slower than c, as I explained in a previous post. See further comments below on the "river bed".

There would be no way of observing an in-faller and knowing whether or not they made it to the horizon, which doesn't make sense.

Why not? Why must the laws of physics ensure that an observer anywhere in the universe can always receive information from everywhere else in the universe? Why must it be impossible that there are some portions of the universe that simply can't send signals to other portions? I realize it's counterintuitive, but what iron law of physics or logic do you think makes it impossible? Just saying it doesn't make sense to you isn't enough.

Read the blogs I posted if you want a more conventional description, then you can tell me I have so many misunderstandings if you think there's some things I’ve gotten wrong.

Which blogs are you referring to? I don't remember the links, but it's been a long thread...

None of them can be hovering on the other side of the event horizon, but you've got to reach it first. Remember the riverbed covers the entire "exterior" of the black hole, which is allowed. What speed would the free-faller be moving relative to the riverbed, or any object outside the horizon for that matter?

You keep on assuming, without proof or argument, that there is some genuine physical restriction on the speed of objects relative to the "river bed". Why must there be such a restriction? The river bed itself is imaginary; it doesn't exist. Why must the speed of anything be restricted relative to something that doesn't exist?

The one I'm using might be a bit different

It is if you think objects are restricted to moving slower than c relative to the river bed. Read the paper I linked to; it makes perfectly clear that the "river model" works just fine inside the horizon, even though objects are moving "faster than light" relative to the "river bed" there, because the river itself is moving "faster than light" relative to the "river bed" there. But that's fine because the river bed doesn't actually exist.

What a coincidence. They both have the exact same two consequences

No. They both have two particular consequences in common, length contraction and time dilation. I did *not* say that they have *all* consequences in common. They don't. One very important consequence they do *not* have in common is that objects moving in curved spacetime do not have to feel weight; objects can free-fall in curved spacetime. But an object moving on a curved worldline *has* to feel weight; that's what "curved worldline" means.

You’re saying it’s okay that matter can reach the equivalent of c if it’s because of curved space-time

No. I'm saying that in curved spacetime, the "equivalent of c" is moving within the light cones at any given event. It does *not* require that objects have to move "slower than c" relative to objects that are in parts of spacetime where the light cones are tilted differently.

 

[Dale]

How can you move smoothly between the Schwarzschild coordinates, where an object can't cross the horizon, to a coordinate system where it can? At what point does the changeover happen? It should be smooth, but it can't be. In your version an object can't ever cross one second, and then it does.

I highly recommend Sean Carroll's lecture notes located here:
http://lanl.arxiv.org/abs/gr-qc/9712019v1

Your questions above are addressed in chapter 2 entitled "Manifolds". In particular, be sure to carefully study from 33-35 on maps and diffeomorphisms and 37-39 on coordinate charts. Note in particular the explanation about coordinate systems being defined on open subsets of the manifold and that some manifolds require multiple charts to span the entire manifold.

In three dimensions it’s a cone, but only because there's a spatial dimension missing.

You are correct. This is a good example of why math is preferable to English. A cone has the equation:
$$x^2+y^2-z^2=0$$

A sphere has the equation:
$$x^2+y^2+z^2=r^2$$

A hypercone has the equation:
$$w^2+x^2+y^2-z^2=0$$

A 4D hypersphere has the equation:
$$w^2+x^2+y^2+z^2=r^2$$

The figure you asked about is technically a hypercone. It is most definitely not a hypersphere.

Now I'm going to describe what I think happens during the formation (and whole life span come to think of it) of a black hole.

The math of black hole formation is much more complicated than the math for the static Schwarzschild spacetime. I think you need to learn to walk before you try to run. I would approach the topics in the following order:
1) Rindler spacetime - acceleration in flat spacetime, coordinate systems, event horizons
2) Schwarzschild spacetime - curvature, singularities, static spacetimes
3) black hole formation - non-static spacetimes, stress-energy, stellar structure

You simply aren't ready for it now, but when you are here is a good reference:
http://www.phys.uu.nl/~prokopec/MichielBouwhuis_bh2.pdf

Well it's not clear to me. I know you said you'd be using equations, and I shouldn't moan that I'm finally getting the kind of response that may actually answer my questions, but it'd be lovely to get it in English as well.

You have gotten it in English for more than 350 posts; I'm not saying anything new here with the math that hasn’t already been said multiple times before in English. The English approach didn't work and I don't want to repeat that; all that leads to is confusion on your part and nastiness on my part. I think you really need to tackle the math head-on. Otherwise you won't be able to learn GR.

I take it "In the Rindler metric" means using Rindler coordinates, and "constant and non-zero covariant derivative" means constantly accelerating at the same rate. So an object that is undergoing zero proper acceleration can still be accelerating in Rindler coordinates despite the fact that there’s no curvature?

All correct, yes. Let's go ahead and work this out in detail.

In units where c=1 the Rindler metric is given by:
$$ds^2 = -x^2 dt^2 + dx^2 + dy^2 + dz^2$$
or
$$g=\left( \begin{array}{cccc} -x^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

Using the expression that I gave above for the Christoffel symbols, we find that there are three which are non-zero:
$$\Gamma^{t}_{tx}=\Gamma^{t}_{xt}=\frac{1}{x}$$
$$\Gamma^{x}_{tt}=x$$

We can then use the expression above to determine the curvature tensor. For example, we have
$${R^x}_{txt}= \frac{\partial}{\partial x} \Gamma_{tt}^x-\frac{\partial}{\partial t}\Gamma_{tx}^x +\sum^{n}_{s=1}(\Gamma_{xs}^x\Gamma_{tt}^s-\Gamma_{ts}^x\Gamma_{tx}^s)= 1-0+(0-1) = 0$$

Similarly, all the other 64 elements of the curvature tensor also evaluate 0. So the curvature is 0, meaning that the Rindler metric describes a flat spacetime.

Now, on the other hand, if we have a particle at a constant x position in the Rindler metric then its tangent vector is given by:
$$v=\left(\frac{1}{\sqrt{x^2}}, 0, 0, 0 \right)$$

Plugging that into the equation above for the covariant derivative gives:
$$\nabla v = \left(0, \frac{1}{x} ,0 ,0 \right)$$

So the covariant derivative is non-zero, meaning that an accelerometer on a stationary particle (coordinate acceleration = 0) in the Rindler metric will measure a non-zero proper acceleration in the positive x direction whose magnitude is inversely proportional to x. This example shows the difference between curvature which is a rank 4 tensor, the covariant derivative of the tangent vector (proper acceleration) which is a rank 1 tensor, and the coordinate acceleration which is not a tensor.

 

[A-wal]

A-wal, I see that you went to considerable trouble to "explain" your viewpoint, but your explanation (which is the first part of your post, before you responded to others) doesn't change my viewpoint at all. I still think you're making the same mistake I've thought you were making all along.

I really just wanted to clarify the way I see it as much as possible. It wasn't really any trouble. I just waited until I knew what I wanted to say before I started writing it instead of forcing it. I was actually less effort than usual. It just takes a lot longer. What mistake do you think I'm still making? Thinking the entire space-time can be contained within just one coordinate system?

By "anything at the horizon", do you mean an object free-falling inwards that is just crossing the horizon, relative to an object hovering just outside the horizon? If so, then there is an interpretation of "moving at c" that makes the above statement true, but that interpretation is *not* one that requires objects to move slower than c, as I explained in a previous post. See further comments below on the "river bed".

I’m not talking about coordinate acceleration. I'm using proper acceleration, otherwise know as tidal force. Anything crossing the horizon (if that were possible) would have to be moving at c relative to every single other object in the universe that's not inside the horizon, including moving at c relative to the rest of the object that hasn't crossed yet. You said the rope would always break, well so would the object crossing the horizon. It would also be feeling infinite proper acceleration, so anything that gets pulled past the horizon would be ripped apart. That jump shouldn't be there, and why can objects survive it in GR?

Why not? Why must the laws of physics ensure that an observer anywhere in the universe can always receive information from everywhere else in the universe? Why must it be impossible that there are some portions of the universe that simply can't send signals to other portions? I realize it's counterintuitive, but what iron law of physics or logic do you think makes it impossible? Just saying it doesn't make sense to you isn't enough.

Because time can't be infinitely time dilated because nothing with mass can reach c. Whether c is an event horizon in curved space-time or light speed in flat space-time makes no difference because they're exactly the same thing. The information would reach you eventually. I know that the Rindler horizon means information wouldn’t reach a constantly accelerating observer, but you can’t accelerate forever because that would take infinite energy. It doesn’t make sense because you can see that there's enough time dilation and length contraction to stop anything reaching the horizon if you constantly compare the in-faller with a more distant observer. If the in-faller is moving through time at half speed relative to the distant observer then the distant observer is moving at double speed relative to the in-faller. If the black hole is about to die from the perspective of the distant observer and the in-faller hasn’t crossed yet then it’s about to die from the perspective of the in-faller and the in-faller hasn’t crossed yet (after taking into account the time it takes for the light to reach you).

You can see that there’s a contradiction if you measure proper time using the age/size of the black hole as a ruler. An in faller measures the size of the black hole when they supposedly cross the horizon and finds it’s about half way though its life. A distant observer sees that the black hole is near the end of its life and the in-faller still hasn’t crossed. An in-faller doesn't reach the event horizon in a finite amount of proper time using Schwarzschild coordinates, so why is it okay to switch to a coordinate system that contradicts this and assume this one is the right one for crossing the horizon, and if it is the right one then why doesn't it invalidate the Schwarzschild coordinates?

Which blogs are you referring to? I don't remember the links, but it's been a long thread...

Special Relativity made simple and General Relativity made simple. The links are by my name.

You keep on assuming, without proof or argument, that there is some genuine physical restriction on the speed of objects relative to the "river bed". Why must there be such a restriction? The river bed itself is imaginary; it doesn't exist. Why must the speed of anything be restricted relative to something that doesn't exist?

I forgot to quote this bit.

Don't think of it as one entity. Think of all the pebbles on the riverbed as individual 'hovering' observers. Now you can't move at c relative to them.

So it does exist. In fact it's everything that exists. Anything reaching the horizon would reach c relative to anything outside the horizon.

It is if you think objects are restricted to moving slower than c relative to the river bed. Read the paper I linked to; it makes perfectly clear that the "river model" works just fine inside the horizon, even though objects are moving "faster than light" relative to the "river bed" there, because the river itself is moving "faster than light" relative to the "river bed" there. But that's fine because the river bed doesn't actually exist.

It's not fine. It means that anything inside the horizon is moving faster than c relative to everything outside it. The fact that you wouldn't be able to see them doing it doesn't make it okay. I used the evenly spaced accelerating hoverers for the riverbed to remove the curvature and show the proper acceleration going up towards infinity and the relative velocity of the in-faller approaching c towards the horizon. The hoverers get further apart as you accelerate to keep you BELOW c. You could use the in-fallers velocity relative to the hoverers to work out the time dilation and length contraction just as you would in flat space-time, so it makes no sense to say there would be enough time dilation and length contraction to stop you from reaching the horizon.

No. They both have two particular consequences in common, length contraction and time dilation. I did *not* say that they have *all* consequences in common. They don't. One very important consequence they do *not* have in common is that objects moving in curved spacetime do not have to feel weight; objects can free-fall in curved spacetime. But an object moving on a curved worldline *has* to feel weight; that's what "curved worldline" means.

An object free-falling in curved space-time does always feel weight, but it’s called tidal force instead. That’s what “curved space-time” means. Free-falling is the equivalent to a different but constant relative velocity except that you’re always accelerating. You’re always moving into a higher gravitational field because you’re always getting closer to the source of the gravity, so you’re always feeling tidal force. Tidal force in curved space-time is the equivalent to acceleration in flat space-time. What else don’t they have in common?

No. I'm saying that in curved spacetime, the "equivalent of c" is moving within the light cones at any given event. It does *not* require that objects have to move "slower than c" relative to distant objects that have light cones tilted differently.

I would have thought you could view the light the cones as tilted when accelerating in flat space-time in exactly the same way as you can with curved space-time. The Rindler horizon would be when they tilt to 90 degrees?

I highly recommend Sean Carroll's lecture notes located here:
http://lanl.arxiv.org/abs/gr-qc/9712019v1

Your questions above are addressed in chapter 2 entitled "Manifolds". In particular, be sure to carefully study from 33-35 on maps and diffeomorphisms and 37-39 on coordinate charts. Note in particular the explanation about coordinate systems being defined on open subsets of the manifold and that some manifolds require multiple charts to span the entire manifold.

Thanks, but I’m still not sure why you would need multiple charts to span the entire manifold. It just says that you do. I still think you've gone off the edge of the map and said here there be monsters. Space-time curves round past 90 degrees at the horizon. That means all the space-time in the universe could fit in the gap between wherever you are and the horizon, just like all the velocity in the universe would fit between you and c.

You are correct. This is a good example of why math is preferable to English. A cone has the equation:
$$x^2+y^2-z^2=0$$

A sphere has the equation:
$$x^2+y^2+z^2=r^2$$

A hypercone has the equation:
$$w^2+x^2+y^2-z^2=0$$

A 4D hypersphere has the equation:
$$w^2+x^2+y^2+z^2=r^2$$

The figure you asked about is technically a hypercone. It is most definitely not a hypersphere.

I thought I was describing a four-dimensional circle. If a hypersphere is something else then how would that expand and contract? Why would it be a hypercone? It takes up the same amount of space in each dimension and the equivalent in time as well, so it should be a hypersphere. If it expands and then contracts at a constant rate so that the time it exists for is equivalent to how big it gets (which would happen if it always moves at c) then it's the same size in every dimension, so it's a sphere.

The math of black hole formation is much more complicated than the math for the static Schwarzschild spacetime. I think you need to learn to walk before you try to run. I would approach the topics in the following order:
1) Rindler spacetime - acceleration in flat spacetime, coordinate systems, event horizons
2) Schwarzschild spacetime - curvature, singularities, static spacetimes
3) black hole formation - non-static spacetimes, stress-energy, stellar structure

You simply aren't ready for it now, but when you are here is a good reference:
http://www.phys.uu.nl/~prokopec/MichielBouwhuis_bh2.pdf

But different coordinate systems should be just different ways of looking at the same thing, so it should be as simple as the static Schwarzschild space-time. I just don’t see how it could work any other way. A black hole would have no way of holding itself up. If you treat the event horizon in the same way as you treat c then you have an area that’s length contracted and time dilated beyond the point where it’s reachable and it would expand at c, and then recede at c because that’s how far you could reach with c as a speed limit. That way it doesn’t need to stay for any length of time in its own frame and doesn’t have to worry about holding itself up. c is a limit on how far you can go in certain amount of time. When G>C there's an area where you can't go because they'll never be enough time. It's nothing but an effect of time dilation and length contraction over the area of space time that they exceed c. It only looks like something from a distance. There's no space-time between the EV and the singularity, just like there's no space-time outside of the universe or before the big bang.

You have gotten it in English for more than 350 posts; I'm not saying anything new here with the math that hasn’t already been said multiple times before in English. The English approach didn't work and I don't want to repeat that; all that leads to is confusion on your part and nastiness on my part. I think you really need to tackle the math head-on. Otherwise you won't be able to learn GR.

I don’t think the equations will help me understand it because learning the equations won’t show me why those equations are the right ones. Besides, I don’t think in equations so learning all the maths in the world won’t help me to understand why it works in a way that seems to me to be completely self-contradictory and unnecessary. They would only help me to describe something that I don’t think can happen. I don’t think learning the equations are essential for learning GR, or anything other than the equations themselves.

All correct, yes. Let's go ahead and work this out in detail.

In units where c=1 the Rindler metric is given by:
$$ds^2 = -x^2 dt^2 + dx^2 + dy^2 + dz^2$$
or
$$g=\left( \begin{array}{cccc} -x^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

Using the expression that I gave above for the Christoffel symbols, we find that there are three which are non-zero:
$$\Gamma^{t}_{tx}=\Gamma^{t}_{xt}=\frac{1}{x}$$
$$\Gamma^{x}_{tt}=x$$

We can then use the expression above to determine the curvature tensor. For example, we have
$${R^x}_{txt}= \frac{\partial}{\partial x} \Gamma_{tt}^x-\frac{\partial}{\partial t}\Gamma_{tx}^x +\sum^{n}_{s=1}(\Gamma_{xs}^x\Gamma_{tt}^s-\Gamma_{ts}^x\Gamma_{tx}^s)= 1-0+(0-1) = 0$$

Similarly, all the other 64 elements of the curvature tensor also evaluate 0. So the curvature is 0, meaning that the Rindler metric describes a flat spacetime.

Now, on the other hand, if we have a particle at a constant x position in the Rindler metric then its tangent vector is given by:
$$v=\left(\frac{1}{\sqrt{x^2}}, 0, 0, 0 \right)$$

Plugging that into the equation above for the covariant derivative gives:
$$\nabla v = \left(0, \frac{1}{x} ,0 ,0 \right)$$

So the covariant derivative is non-zero, meaning that an accelerometer on a stationary particle (coordinate acceleration = 0) in the Rindler metric will measure a non-zero proper acceleration in the positive x direction whose magnitude is inversely proportional to x. This example shows the difference between curvature which is a rank 4 tensor, the covariant derivative of the tangent vector (proper acceleration) which is a rank 1 tensor, and the coordinate acceleration which is not a tensor.

But if there’s no curvature that would mean there’s no proper acceleration if they’re the same thing, so it’s fine. The fact that there is coordinate acceleration but no proper acceleration is irrelevant. In my riverbed example they are at rest despite the fact that the proper acceleration in different for each one. You're saying that you use a different process to work out the two types of curvature, but I don’t know why they would be different. Is it a different kind of tensor for curved space-time because you're describing what would happen in four dimensions rather than two? After you've plotted the path of the free-faller you could then uncurve that path and you should be left with acceleration in flat space-time. You could do the same thing the other way round using energy for mass, making acceleration a source of curvature.

This is a good coordinate system. You could view time as a spatial dimension if you look at space as expanding everywhere at c. In this view the past is inwards and the future is outwards. Everything is at the centre of its own bubble. Energy has no mass so it’s carried outwards as space expands making light expand at c in a sphere shape. You can use energy to accelerate making the light bubble warp and causing the expansion to pass through you more slowly in one spatial dimension so you get time dilation and length contraction. The expansion is also slowed through mass. Matter doesn’t expand outwards like energy but that movement has to go somewhere. The electrons are held in place by the strong nuclear force and are forced to orbit instead of expanding. Whether you could look at the space as curved by the strong nuclear force and the electrons moving in straight lines through curved space-time on a tiny scale I don't know but I suspect you probably could. The orbit of the electrons causes length contraction and time dilation because they have a different relative velocity, which radiates outwards from the source. So in these coordinates gravity is caused by the frame dragging of atoms. A black hole happens when the time dilation and length contraction are greater than the expansion over a certain area. You can’t reach that area because it’s obvious using these coordinates that you would always run out of time.

That coordinate system is only good for viewing one moment at a time though. If you want to view more than one time at once you can remove one spatial dimension to turn it into the ball analogy so that the surface of the ball represents all three spatial dimensions in the present with the past towards the centre and the future further out. The bubbles are then cones instead of spheres.

 

[PeterDonis]

What mistake do you think I'm still making? Thinking the entire space-time can be contained within just one coordinate system?

Not precisely; it is thinking that there is a flat background to the spacetime (what you call the "river bed") that constrains how things can move (or, equivalently, constrains how much the spacetime can curve). Any time you reason based on an analogy between the horizon and reaching light speed in flat spacetime, you're making this mistake. It shows itself in other ways as well, some of which I'll comment on below.

Anything crossing the horizon (if that were possible) would have to be moving at c relative to every single other object in the universe that's not inside the horizon, including moving at c relative to the rest of the object that hasn't crossed yet. You said the rope would always break, well so would the object crossing the horizon. It would also be feeling infinite proper acceleration, so anything that gets pulled past the horizon would be ripped apart.

There are at least two misstatements here:

(1) The freely falling object feels *no* proper acceleration. Zero. You keep on using the term "proper acceleration" incorrectly. If you're talking about freely falling objects, then just stop yourself whenever you want to use the term "proper acceleration", because it's always zero for a freely falling object.

(2) You are mixing up two scenarios. One (the "rope" scenario) involves a rope that is not freely falling--at least, one end of it is not (because it's attached to an observer that is hovering at a constant radius above the horizon). Such a rope will indeed always break when it crosses the horizon (actually, as I noted before, a real rope would break well before that, but in the absolute limiting case where the rope's tensile strength is the maximum allowed by relativity, it would break as it crossed the horizon). But that is because the top end of the rope is constrained; it is not freely falling.

The second scenario, which you are wrongly conflating with the "rope" scenario, is that of an object freely falling through the horizon--i.e., not attached to any other object, just falling by itself. Such an object *might* break as it crossed the horizon, if its extension in the radial direction was long enough compared to the strength of tidal gravity at the horizon. But if the freely falling object is small enough that tidal effects can be neglected over its length (by "neglected" I mean "compared to the tensile strength of the object", so obviously the material the object is made of can make a difference; an object made of styrofoam will have to be smaller than one made of steel for tidal effects to be negligible), then it will *not* break. I can set up a local inertial frame around the object as it crosses the horizon, and within that frame, the object is simply floating at rest, the force on it is negligible, and it holds together just fine.

(In that local inertial frame, by the way, the horizon is an outgoing light ray; for example, if I set up the local coordinates such that the origin is the point where the center of mass of the freely falling object crosses the horizon, and the positive x direction is radially outgoing, then the horizon is the line t = x. Since the object is at rest in this frame, the worldline of its center of mass is just the t axis; and the worldlines of other parts of the object are just vertical lines parallel to the t axis. So all the parts of the object are at rest relative to each other; even those that have not yet crossed the horizon--they are below the line t = x--are at rest relative to those that have crossed the horizon--they are above the line t = x. This should be obvious since the object is freely falling, and if it is short enough that tidal effects are negligible, all parts of the object are falling at the same rate.)

I know that the Rindler horizon means information wouldn’t reach a constantly accelerating observer, but you can’t accelerate forever because that would take infinite energy.

And if an accelerating observer in flat spacetime stops accelerating, they fall through the Rindler horizon and can now see the information beyond it. Similarly, if an observer hovering above a black hole stops accelerating, they are no longer hovering; they fall into the hole and can now see the information inside the horizon that was hidden from them before. None of this changes the fact that observers above the horizon can't receive information from below it *while they remain above the horizon*.

You can see that there’s a contradiction if you measure proper time using the age/size of the black hole as a ruler.

You keep on bringing in a finite lifetime for the black hole, when we've already agreed that's irrelevant; you said many posts ago that the horizon of even an "eternal" black hole, which lasts for an infinite time in the past and future, could not be reached. So reasoning based on a finite black hole lifetime is irrelevant. Please re-state your argument in a way that applies to an eternal black hole, so we can get that case nailed down first.

Special Relativity made simple and General Relativity made simple. The links are by my name.

Ok, found them. I'll take a look.

I forgot to quote this bit.So it does exist. In fact it's everything that exists. Anything reaching the horizon would reach c relative to anything outside the horizon.

You're assuming that the entire spacetime can be covered by a family of hovering observers, all at rest relative to each other. That's another form of the mistake I referred to at the start of this post.

It's not fine. It means that anything inside the horizon is moving faster than c relative to everything outside it.

And your continued refusal to accept this possibility in a curved spacetime is another form of the same mistake.

An object free-falling in curved space-time does always feel weight, but it’s called tidal force instead. That’s what “curved space-time” means.

No, it isn't. If you really think you're saying something relevant here, you will need to give a detailed scenario that illustrates why you think objects moving solely under the influence of tidal gravity feel weight. In standard GR (and in standard Newtonian gravity, for that matter), they don't.

I would have thought you could view the light the cones as tilted when accelerating in flat space-time in exactly the same way as you can with curved space-time. The Rindler horizon would be when they tilt to 90 degrees?

No. In flat spacetime the light cones are never tilted at all. All of the light cones throughout the spacetime line up exactly with each other. That's part of what "flat spacetime" means.

Also, the motion of an observer doesn't affect the light cones, in either flat or curved spacetime.

 

[PeterDonis]

A-wal, I've looked at your "special relativity made simple" blog post here:

https://www.physicsforums.com/blog.php?b=743

I have a request: can you post some actual images to illustrate how "tilting the V" works in the last paragraph? It looks to me like you're describing spacetime diagrams, but when you talk about "tilting the V" I can't follow what you're saying, because you don't seem to be describing how Lorentz transformations are normally represented in spacetime diagrams. Some pictures would help enormously in understanding what you're saying.

(I've posted this request in the blog post comments as well.)

 

[PeterDonis]

It means that anything inside the horizon is moving faster than c relative to everything outside it.

On re-reading, I realized I should comment further on this, since it's not even true as stated. What is true for all objects inside the horizon is that they would have to move faster than c to (a) "hover" at a constant radius, or (b) get back outside the horizon. It does *not* follow from that, however, that objects inside the horizon must be moving faster than c relative to *everything* [edited: changed from "anything"] outside the horizon. (In fact, I gave a counterexample in my last post but one: the portion of a free-falling object outside the horizon can be at rest relative to the portion inside the horizon, since the whole object can be at rest in a local inertial frame whose origin is the event where the object's center of mass crosses the horizon.)

There is a sense in which objects inside the horizon are moving "faster than c" relative to *some* objects outside the horizon (I've explained how this works in previous posts), but there is no prohibition against objects moving "faster than c" relative to other objects in that sense.

(EDIT: Changed "anything" to "everything" in first paragraph above.)

 

[Dale]

Thanks, but I’m still not sure why you would need multiple charts to span the entire manifold. It just says that you do.

That is true, it does not go into great detail about that, although some hints and examples are given. Recall (from page 37) that coordinate charts are smooth mappings of open subsets of the manifold to open subsets of R^n. So any closed manifold requires at least two charts simply because charts are open subsets. This is the case in the example given in the last paragraph on page 38 as well as the example on page 39. However, the point of the chapter isn't that some manifolds require multiple charts, but simply to give an understanding about what a coordinate chart is and its relationship to the manifold and to other charts.

If a hypersphere is something else then how would that expand and contract? ... If it expands and then contracts at a constant rate

Sorry, I didn't notice that you were saying "expand and contract", I only noticed the "constant rate" part. A hypersphere won't expand at a constant rate. It will expand infinitely fast at first, its expansion will slow, then stop, then contract, then contract faster, until it contracts infinitely fast. A hypercone will contract at a constant rate until it becomes a point and then expand at a constant rate. I don't know of any named figure that would first expand at a constant rate and then contract at a constant rate. I suppose that you could construct one by piecing together parts of a hypercone.

But different coordinate systems should be just different ways of looking at the same thing, so it should be as simple as the static Schwarzschild space-time.

They are completely different manifolds, not merely different coordinate charts on the same manifold.

I don’t think the equations will help me understand it because learning the equations won’t show me why those equations are the right ones.

These equations are right for the same reason that any equation is right in any branch of science: because it agrees with experiment.

Besides, I don’t think in equations so learning all the maths in the world won’t help me to understand why it works in a way that seems to me to be completely self-contradictory and unnecessary. They would only help me to describe something that I don’t think can happen. I don’t think learning the equations are essential for learning GR, or anything other than the equations themselves.

This is just an excuse to avoid some mental effort. If you do not wish to invest the effort to learn GR that is fine, there is probably no tangible benefit to your learning. But to think that you can really learn GR without the math is pure hubris, particularly given the history of this thread.

Is it a different kind of tensor for curved space-time because you're describing what would happen in four dimensions rather than two?

Yes. Curvature is a rank 4 tensor, but proper acceleration (covariant derivative of the tangent) is a rank 1 tensor, and coordinate acceleration is not even a tensor. I don't know how it can possibly be more clear that they are different things. Please go back and look at the math carefully again. I would be glad to work another example if you wish, or add more detail at any step which is unclear.

This is a good coordinate system. ... That coordinate system is only good for viewing one moment at a time though.

I agree it is a good coordinate system. You can learn a lot from Rindler coordinates. However, it is certainly not limited to a single moment in time. It covers 1/4 of the flat spacetime manifold, which is known as the Rindler wedge.

http://en.wikipedia.org/wiki/Rindler_coordinates
http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html

If I have time later I will work out the details for an object falling through the Rindler horizon.

 

[PeterDonis]

A-wal, I've looked at your "General Relativity Made Simple" blog post here:

https://www.physicsforums.com/blog.php?b=744

I don't think the following from your post is correct:

It’s not the gravity (velocity) you feel, it’s the relative difference in its strength (acceleration). On Earth that difference is very small but it’s enough to determine the direction of gravity.

Imagine a very tall building. It’s so tall that height and time on the top floors are noticeably different to the lower floors. Gravity strength is inversely proportional to the square of the distance to the mass. In zero dimensions (point) it’s infinite (singularity). In one spatial dimension (straight line) its strength would remain constant regardless of distance. In two dimensions (flat plane) it would be directly proportional to the distance. And in three the strength is divided by four if the distance is doubled and multiplied by four if the distance is halved. It’s always proportional to the volume it fills. The acceleration is determined by the difference in the gravitational field (how sharp the curve is), not its strength. That how objects know which way to fall.

You seem to be saying here that the "acceleration" is determined by the spatial derivative of the "field strength", *not* the "field strength" itself. Since you don't define your terms, I can't tell for sure what you mean by "gravitational field"; but you appear to be using it to refer to the field that causes objects to fall, for example a rock to fall towards the Earth. In that usage (which is a normal usage), what is quoted above is incorrect. The field in this sense, meaning the field that determines the acceleration (in Newtonian terms) with which objects fall, is *not* determined by "how sharp the curve is"--it is determined by the "strength" of the field itself. The difference in the "field" determines tidal gravity, but that is *not* the same as the "gravity" that causes objects to fall and determines the acceleration and direction of their fall.

I've posted this as a comment on the blog post as well.

 

[PeterDonis]

Just to clarify my last post a little further (since I talked about defining terms, I should do so myself).

The normal usage of "gravitational field" is to refer to the acceleration, as I said. In other words, the "field" is given by

$$a = \frac{G M}{r^2}$$

i.e., it's directly proportional to the mass and inversely proportional to the square of the distance. This seems to match up with what A-wal was describing as the "gravity strength", which he appears to be using as synonymous with the "gravitational field" in at least a part of his post.

Tidal gravity is then the rate of change of the field; in the radial direction (there are also rates of change in other directions--in general relativity there is even one in the time direction--so tidal gravity can't be described by a single number, you need a tensor), the rate of change is:

$$t = - \frac{da}{dr} = \frac{2 G M}{r^3} = \frac{2 a}{r}$$

(The minus sign in front of da/dr is because r increases outward, but tidal gravity causes radially falling objects to diverge--get farther apart--if they start from different radii and fall inward. So the positive sense of t, which corresponds to objects diverging tidally, corresponds to the negative sense of da/dr, with a incresing as r decreases.)

The above formula should make it obvious that a and t are different (even leaving out the fact that a is a spatial vector and t is a tensor--really what I'm calling "t" above is something like $t_{r r}$, the radial component of the radial rate of change of a). To see the difference more explicitly, consider two planets; one is the Earth, and the other has a hundred times the Earth's mass and ten times its radius (this, by the way, is by no means impossible; in fact, the planet Saturn comes pretty close to those numbers). You can see from the above formulas that a at the radius of each planet will be the same, but t on Earth at Earth's radius will be 10 times t on the other planet at its radius. So we can have the same a but very different t; objects can fall with the same acceleration but still be subject to very different tidal gravity.

There is one other quantity that is sometimes referred to as the "gravitational field", but is more properly called the "potential"; it is given by:

$$U = - \frac{G M}{r}$$

You can see then that a = dU / dr; the acceleration is the rate of change of the potential. (Actually, in full vector form, the acceleration vector is the spatial gradient of the potential, which is just a number--more precisely, a number at each point of space.) So if the word "field" is used to refer to the potential, then one can say that the acceleration is determined by how the "field" varies in space; but in that usage, the "acceleration" (the spatial gradient of the potential) is still *not* the same as tidal gravity (which is still the rate of change of the acceleration, *not* the potential). And also, of course, the "gravity strength" or "field" in the sense of the potential is *not* inversely proportional to the square of the distance; it's only inversely proportional to the distance.

Hopefully this clarifies the terms I was using and the substance of my objection to what I quoted from the blog post. I've added this to the blog post comments as well.

 

[A-wal]

Not precisely; it is thinking that there is a flat background to the spacetime (what you call the "river bed") that constrains how things can move (or, equivalently, constrains how much the spacetime can curve). Any time you reason based on an analogy between the horizon and reaching light speed in flat spacetime, you're making this mistake. It shows itself in other ways as well, some of which I'll comment on below.

The riverbed is an actual constraint on how things move. I was using the hovering observers as an example but just imagine space-time is completely full of matter, ignoring the extra gravity that would create. They're all moving at different velocities and in Gravitational fields of varying strengths (same thing). This is just to give you a constant comparison of an object right next to you. Do you think you would be able to reach c relative to any of them? If the answer is no then you wouldn’t be able to reach an event horizon.

There are at least two misstatements here:

(1) The freely falling object feels *no* proper acceleration. Zero. You keep on using the term "proper acceleration" incorrectly. If you're talking about freely falling objects, then just stop yourself whenever you want to use the term "proper acceleration", because it's always zero for a freely falling object.

I thought proper acceleration was the acceleration that you actually feel like proper time. You feel tidal force. I’ll stop calling it that so it’s clear whether I mean proper acceleration from energy or from mass.

(2) You are mixing up two scenarios. One (the "rope" scenario) involves a rope that is not freely falling--at least, one end of it is not (because it's attached to an observer that is hovering at a constant radius above the horizon). Such a rope will indeed always break when it crosses the horizon (actually, as I noted before, a real rope would break well before that, but in the absolute limiting case where the rope's tensile strength is the maximum allowed by relativity, it would break as it crossed the horizon). But that is because the top end of the rope is constrained; it is not freely falling.

The second scenario, which you are wrongly conflating with the "rope" scenario, is that of an object freely falling through the horizon--i.e., not attached to any other object, just falling by itself. Such an object *might* break as it crossed the horizon, if its extension in the radial direction was long enough compared to the strength of tidal gravity at the horizon. But if the freely falling object is small enough that tidal effects can be neglected over its length (by "neglected" I mean "compared to the tensile strength of the object", so obviously the material the object is made of can make a difference; an object made of styrofoam will have to be smaller than one made of steel for tidal effects to be negligible), then it will *not* break. I can set up a local inertial frame around the object as it crosses the horizon, and within that frame, the object is simply floating at rest, the force on it is negligible, and it holds together just fine.

(In that local inertial frame, by the way, the horizon is an outgoing light ray; for example, if I set up the local coordinates such that the origin is the point where the center of mass of the freely falling object crosses the horizon, and the positive x direction is radially outgoing, then the horizon is the line t = x. Since the object is at rest in this frame, the worldline of its center of mass is just the t axis; and the worldlines of other parts of the object are just vertical lines parallel to the t axis. So all the parts of the object are at rest relative to each other; even those that have not yet crossed the horizon--they are below the line t = x--are at rest relative to those that have crossed the horizon--they are above the line t = x. This should be obvious since the object is freely falling, and if it is short enough that tidal effects are negligible, all parts of the object are falling at the same rate.)

But tidal force can’t be negligible at an event horizon no matter how short the object is. It’s the point at which an infinite amount of proper acceleration would be needed to resist gravity, so the strength of gravity reaches infinity at the horizon relative to anything outside the horizon, including the back of the object that’s crossing it.

And if an accelerating observer in flat spacetime stops accelerating, they fall through the Rindler horizon and can now see the information beyond it. Similarly, if an observer hovering above a black hole stops accelerating, they are no longer hovering; they fall into the hole and can now see the information inside the horizon that was hidden from them before. None of this changes the fact that observers above the horizon can't receive information from below it *while they remain above the horizon*.

They’re no longer hovering but they’re still accelerating through tidal force. Replace tidal force with acceleration in flat space-time and you can see why you get a Rindler/event horizon and you can see why it can’t be crossed.

You keep on bringing in a finite lifetime for the black hole, when we've already agreed that's irrelevant; you said many posts ago that the horizon of even an "eternal" black hole, which lasts for an infinite time in the past and future, could not be reached. So reasoning based on a finite black hole lifetime is irrelevant. Please re-state your argument in a way that applies to an eternal black hole, so we can get that case nailed down first.

It's irrelevant to whether an object can reach the horizon but that doesn't mean it's size can't be used to compare what two observers see. Supressing common and all other forms of sense for the moment and assuming objects can, sigh, cross the event horizon; the first one waits until the last possible second to send a signal to the distant observer to tell them exactly how long the black hole had left to live just before they crossed. The second observer sees the black hole get to this age and beyond, but still the first observer is OUTSIDE of the horizon.

You're assuming that the entire spacetime can be covered by a family of hovering observers, all at rest relative to each other. That's another form of the mistake I referred to at the start of this post.

Why wouldn't a family of hovering observers be able to be at rest relative to each other if they're all outside the horizon?

And your continued refusal to accept this possibility in a curved spacetime is another form of the same mistake.

Curved space-time can make it look as though an object is traveling faster than light relative to an object where the gravity is different but that doesn't mean they're really moving faster than c. You still wouldn’t be able to catch up to it, and that’s exactly what you would need to be able to do if you wanted to reach an event horizon. Tidal force makes the speed of light slower to the in-faller in the same way it does for an accelerator in flat space-time and the Rindler horizon becomes an event horizon.

No, it isn't. If you really think you're saying something relevant here, you will need to give a detailed scenario that illustrates why you think objects moving solely under the influence of tidal gravity feel weight. In standard GR (and in standard Newtonian gravity, for that matter), they don't.

They would feel their weight in exactly the same way we’re feeling our weight now, through acceleration. Being in a stronger gravitational field doesn’t mean you weigh any more than you would in a weaker one (if we ignore tidal force for a second). It only means that it’s harder to resist the pull of gravity, so if you wanted to stay at rest relative to the source of the gravitation then you would feel more weight than you would in a lower gravitational field because you have to accelerate harder. Tidal force causes objects to accelerate as they fall, so it should affect an objects weight in exactly the same way.

No. In flat spacetime the light cones are never tilted at all. All of the light cones throughout the spacetime line up exactly with each other. That's part of what "flat spacetime" means.

Also, the motion of an observer doesn't affect the light cones, in either flat or curved spacetime.

But why couldn't you tilt the light cones when objects have a different relative velocity in either flat (or curved) space-time in the same way you do for gravity? In the riverbed example the light cones would be tilted further as you approach the horizon. They would be at 90 degrees if you could reach the horizon.

That is true, it does not go into great detail about that, although some hints and examples are given. Recall (from page 37) that coordinate charts are smooth mappings of open subsets of the manifold to open subsets of R^n. So any closed manifold requires at least two charts simply because charts are open subsets. This is the case in the example given in the last paragraph on page 38 as well as the example on page 39. However, the point of the chapter isn't that some manifolds require multiple charts, but simply to give an understanding about what a coordinate chart is and its relationship to the manifold and to other charts.

I'm still struggling understand why one manifold would need multiple coordinate charts to describe it fully. It seems like you’re saying the coordinate charts are incomplete, which would go along with what’s been repeatedly said about them not covering the entire space-time. I don’t see how a coordinate system doesn’t cover all of space-time because they take a frame of reference and map around it, telling you how things relate to each other from that perspective. The Rindler horizon in flat space-time shows that an accelerating observer wouldn’t be able to see things from beyond a certain point in space-time, but that is only if you ignore time. If you wait long enough then that observer will see everything because they can’t accelerate forever. It’s not the same as falling through an event horizon though because that would permanently remove matter from the perspective of anyone outside the horizon. You don’t fall through a Rindler horizon when you stop accelerating. Instead the Rindler horizon moves depending on how hard you’re accelerating.

Sorry, I didn't notice that you were saying "expand and contract", I only noticed the "constant rate" part. A hypersphere won't expand at a constant rate. It will expand infinitely fast at first, its expansion will slow, then stop, then contract, then contract faster, until it contracts infinitely fast. A hypercone will contract at a constant rate until it becomes a point and then expand at a constant rate. I don't know of any named figure that would first expand at a constant rate and then contract at a constant rate. I suppose that you could construct one by piecing together parts of a hypercone.

The singularity has no length in any direction including time, and the curvature caused by it is the same in all four dimensions, so you get a hypersphere. It's always a hypersphere but its length in all four dimensions changes with distance. It's smaller the closer you are and it wouldn't have any size at all if you actually reached it, and it obviously makes less difference the further you are away from it because it's an inverse square. Two cones put together at their bases makes a shape that expands outwards at a constant rate and then contracts at a constant rate, but that's looking at in three dimensions again. In four dimensions that shape is a hypersphere. It looks like a cone because of the limitation of c.

They are completely different manifolds, not merely different coordinate charts on the same manifold.

So not only does a manifold need multiple coordinate systems to fully describe it, but space-time has multiple manifolds (I thought I knew what that word meant is this context but now I'm not sure) and each of these need multiple coordinate systems to fully describe them? Work out the proper time using Rindler and Schwarzschild coordinates and you get exactly the same answer – it would take infinite proper time to reach an event horizon in exactly the same way it takes infinite proper time to accelerate to c. This is not dependent on the coordinate system that you use because the proper time it would take for an observer to do something can’t change.

These equations are right for the same reason that any equation is right in any branch of science: because it agrees with experiment.

But they don't always have to be exactly right to agree with experiment. If we could find a small black hole and a really strong rope then I doubt it would agree with experiment.

This is just an excuse to avoid some mental effort. If you do not wish to invest the effort to learn GR that is fine, there is probably no tangible benefit to your learning. But to think that you can really learn GR without the math is pure hubris, particularly given the history of this thread.

Not really. You don’t need to learn Hebrew to read the bible. I wish you did. There wouldn’t be as many people walking around with crap in their heads. The equations don’t tell you why things work the way they do. The energy released when an atom is split isn’t because E=mc^2. E=mc^ because of what happens when an atom is split. It's not the equations that let you understand something, it's knowing how things relate to each other. Maths can't help you with that. It's true that I would need to know the equations if I wanted to be able to work out exactly what would happen in a given situation, like for the amount of proper time someone would experience when they accelerate compared to someone who doesn't for example. I don't mind not knowing that I as long as I know why the accelerator experiences less proper time. You could know how to work out anything like that and still not have a clue why it works that way. They're completely separate things and I honestly believe you don't need to know any maths to understand any of it. I don’t mind you posting the equations with a description, like you have been, but to claim equations represent something that can't be understood without them is just silly, and a copout. And that’s just one way of interpreting the history of this thread.

Yes. Curvature is a rank 4 tensor, but proper acceleration (covariant derivative of the tangent) is a rank 1 tensor, and coordinate acceleration is not even a tensor. I don't know how it can possibly be more clear that they are different things. Please go back and look at the math carefully again. I would be glad to work another example if you wish, or add more detail at any step which is unclear.

So it's not because they're different things, it's just that you're viewing gravity as curvature of space-time and are therefore describing how it effects everything, but viewing acceleration as purely the curved worldline of one object without taking into account how energy curves space-time so that it's not only one object that's accelerated. So they're both rank 4 tensors. And you said: "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."

I agree it is a good coordinate system. You can learn a lot from Rindler coordinates. However, it is certainly not limited to a single moment in time. It covers 1/4 of the flat spacetime manifold, which is known as the Rindler wedge.

http://en.wikipedia.org/wiki/Rindler_coordinates
http://gregegan.customer.netspace.ne...erHorizon.html

If I have time later I will work out the details for an object falling through the Rindler horizon.

Start with one moment in time and view inwards and outwards as the past and the future with time expanding at c and taking light/energy with it. If you want to actually see things from a different time then this doesn't work because you're not really moving out into the future when you look outwards, you’re looking along a spatial dimension. If you want to be able to move along the timeline then you have to remove a spatial dimension. Then space-time makes a normal sphere with two spatial dimensions representing three and one representing time. The hypershpere now looks like a hypercone.

Rindler coordinates represent 1/4 of space-time? Hmm that's interesting, why a 1/4? Don't tell me, I'm going to try to figure it out for myself.


Edit: Haven't got time now to edit the blogs or reply to the more recent posts. I'll do it next time.

 

[PeterDonis]

A-wal, you continue to restate your beliefs without demonstrating why the assumptions underlying them must be true, and since those assumptions are false in GR, I continue to see little point in responding, other than to reiterate some specific instances that are central to the discussion. But you do say some other things that are worth commenting on.

The riverbed is an actual constraint on how things move.

Sorry, but you can't just assert this. You have to prove it. What you've given is not a proof; it's simply a restatement of the assertion in different words.

I thought proper acceleration was the acceleration that you actually feel like proper time.

It is. But...

You feel tidal force.

...is incorrect, as I've said repeatedly before. A body moving solely under the influence of tidal gravity feels no force and no proper acceleration. If you think it does, please describe a specific scenario that makes you think so. Just asserting it won't do; as above, you have to prove it, and you've never given a proof, you've just repeatedly asserted it. Bear in mind also that the phrase "moving *solely* under the influence of tidal gravity" is crucial. For bodies which have internal structure and therefore internal forces between their parts, the parts are *not* moving solely under the influence of tidal gravity, and the forces they feel are due to those internal forces, not tidal gravity.

But tidal force can’t be negligible at an event horizon no matter how short the object is.

Yes, it can. In my comments on your GR blog post I gave the Newtonian equations for acceleration and tidal gravity; but those equations are only approximations, which happen to work well enough when the spacetime curvature is small (which could be very, very far away from a black hole, or at the surface of an object like the Earth whose radius is much, much larger than its mass in geometric units--i.e., the radius R is much, much larger than 2GM/c^2). Now let me give the GR equations, which are correct even at the horizon of a black hole (and inside, all the way down to the central singularity at r = 0). As before, we're just considering the acceleration and tidal gravity in the radial direction.

$$a = \frac{G M}{r^2 \sqrt{1 - \frac{2 G M}{c^2 r}}}$$

$$t = \frac{2 G M}{r^3}$$

Notice that the equation for a differs from the Newtonian one by the extra square root term in the denominator. That is a reflection of the fact that, as you say, the acceleration goes to infinity as the black hole horizon (r = 2GM/c^2) is approached. (The square root term is negligible if r is much greater than 2GM/c^2, hence my statement above about where the Newtonian approximation is valid.) However, the equation for t is the *same* as the Newtonian equation! That seems surprising, but it's true; I can post a derivation of how this result arises if desired.

(Of course, this also means that in the full GR model, when we stop making the Newtonian approximation, the tidal gravity t is no longer exactly equal to the spatial rate of change of the acceleration a, and the disparity gets worse as we get closer to the horizon, ultimately becoming "infinite" at the horizon. Maybe this is a key point that we should go into further; at the moment I don't have a simple way of explaining physically how the result comes about. I'll have to think about this some more--or maybe one of the experts on the forum can provide one.)

So given the above equation for t, the tidal gravity at the horizon (r = 2GM/c^2) varies inversely as the square of the mass of the hole; the larger the hole, the smaller the tidal gravity at the horizon. So for any level of tidal gravity that you consider "negligible", there is some black hole with a mass large enough to have its tidal gravity negligible at the horizon.

Why wouldn't a family of hovering observers be able to be at rest relative to each other if they're all outside the horizon?

If they are all *outside* the horizon, they can. But if some are inside the horizon, the ones inside the horizon can never be at rest relative to the ones outside the horizon, since they would have to move faster than light in order to do so. That is *not* a proof that no observers can get inside the horizon, or that the portion of spacetime that is "inside the horizon" can't exist; it is simply an illustration that your implicit assumptions about how spacetime has to work are false.

They would feel their weight in exactly the same way we’re feeling our weight now, through acceleration.

You're getting careless again about the term "acceleration". We, sitting at rest on the surface of the Earth, feel weight because we are undergoing *proper* acceleration; we are not in free fall. A body moving *solely* under the influence of tidal gravity is in free fall; it is undergoing *no* proper acceleration and so feels no weight. (See my previous comments above.) When you say...

Tidal force causes objects to accelerate as they fall

...you are using the word "accelerate" in a different way; you mean two objects being separated by tidal gravity "accelerate" relative to each other because their separation increases with time, and the rate of increase itself increases with time. But in curved spacetime, that can happen even though both objects are in free fall and feel no weight and have no proper acceleration. That's the definition of curved spacetime.

But why couldn't you tilt the light cones when objects have a different relative velocity in either flat (or curved) space-time in the same way you do for gravity?

I'm not sure what you're asking. Are you asking (A) why, physically, relative velocity doesn't affect the tilting of the light cones? Or are you asking (B) why we can't just choose to draw a diagram with the light cones tilted differently when the relative velocity is different?

The answer to (B) is simple. The tilting of the light cones is a frame-invariant, physical phenomenon. You can't change it by changing coordinates or drawing your spacetime diagram differently. It's part of the invariant physical structure of spacetime. You may be able to change the way the light cones "look" on your spacetime diagram by drawing it differently, but that won't change the physics.

I can't answer (A) at this point, because I don't understand why you think relative velocity should have any effect on the curvature of spacetime (which is what the tilting of the light cones is a reflection of) in the first place.

 

[A-wal]

A-wal, you continue to restate your beliefs without demonstrating why the assumptions underlying them must be true, and since those assumptions are false in GR, I continue to see little point in responding, other than to reiterate some specific instances that are central to the discussion. But you do say some other things that are worth commenting on.

After re-reading my last post it came across a little more assertive than I intended. I said I was going to try to stop doing that but I’m just stating the way I see it. I’m not saying it has to be right. The same goes for this post. I’m not sure what else I can do in terms of demonstrating my assumptions. I’ve tried to apply them to real world examples as often as I can.

Sorry, but you can't just assert this. You have to prove it. What you've given is not a proof; it's simply a restatement of the assertion in different words.

You can’t move at c relative to another object in flat space-time. You don’t have to look at gravity as curvature if you don’t want to. The riverbed is simply SR in a gravitational field.

...is incorrect, as I've said repeatedly before. A body moving solely under the influence of tidal gravity feels no force and no proper acceleration. If you think it does, please describe a specific scenario that makes you think so. Just asserting it won't do; as above, you have to prove it, and you've never given a proof, you've just repeatedly asserted it. Bear in mind also that the phrase "moving *solely* under the influence of tidal gravity" is crucial. For bodies which have internal structure and therefore internal forces between their parts, the parts are *not* moving solely under the influence of tidal gravity, and the forces they feel are due to those internal forces, not tidal gravity.

You're getting careless again about the term "acceleration". We, sitting at rest on the surface of the Earth, feel weight because we are undergoing *proper* acceleration; we are not in free fall. A body moving *solely* under the influence of tidal gravity is in free fall; it is undergoing *no* proper acceleation and so feels no weight. (See my previous comments above.) When you say...

...you are using the word "accelerate" in a different way; you mean two objects being separated by tidal gravity "accelerate" relative to each other because their separation increases with time, and the rate of increase itself increases with time. But in curved spacetime, that can happen even though both objects are in free fall and feel no weight and have no proper acceleration. That's the definition of curved spacetime.

It takes energy to accelerate an object with mass. So anything with mass will create a drag. Tidal force means that the front end is pulling the back end along even though the object is in free-fall. It’s mass will create drag which would be felt as weight.

Yes, it can. In my comments on your GR blog post I gave the Newtonian equations for acceleration and tidal gravity; but those equations are only approximations, which happen to work well enough when the spacetime curvature is small (which could be very, very far away from a black hole, or at the surface of an object like the Earth whose radius is much, much larger than its mass in geometric units--i.e., the radius R is much, much larger than 2GM/c^2). Now let me give the GR equations, which are correct even at the horizon of a black hole (and inside, all the way down to the central singularity at r = 0). As before, we're just considering the acceleration and tidal gravity in the radial direction.

$$a = \frac{G M}{r^2 \sqrt{1 - \frac{2 G M}{c^2 r}}}$$

$$t = \frac{2 G M}{r^3}$$

Notice that the equation for a differs from the Newtonian one by the extra square root term in the denominator. That is a reflection of the fact that, as you say, the acceleration goes to infinity as the black hole horizon (r = 2GM/c^2) is approached. (The square root term is negligible if r is much greater than 2GM/c^2, hence my statement above about where the Newtonian approximation is valid.) However, the equation for t is the *same* as the Newtonian equation! That seems surprising, but it's true; I can post a derivation of how this result arises if desired.

(Of course, this also means that in the full GR model, when we stop making the Newtonian approximation, the tidal gravity t is no longer exactly equal to the spatial rate of change of the acceleration a, and the disparity gets worse as we get closer to the horizon, ultimately becoming "infinite" at the horizon. Maybe this is a key point that we should go into further; at the moment I don't have a simple way of explaining physically how the result comes about. I'll have to think about this some more--or maybe one of the experts on the forum can provide one.)

So given the above equation for t, the tidal gravity at the horizon (r = 2GM/c^2) varies inversely as the square of the mass of the hole; the larger the hole, the smaller the tidal gravity at the horizon. So for any level of tidal gravity that you consider "negligible", there is some black hole with a mass large enough to have its tidal gravity negligible at the horizon.

I thought tidal force is the spatial rate of change of the acceleration? I don’t think about it using equations, as I’m sure you’ve noticed, but this seems very wrong.

If they are all *outside* the horizon, they can. But if some are inside the horizon, the ones inside the horizon can never be at rest relative to the ones outside the horizon, since they would have to move faster than light in order to do so. That is *not* a proof that no observers can get inside the horizon, or that the portion of spacetime that is "inside the horizon" can't exist; it is simply an illustration that your implicit assumptions about how spacetime has to work are false.

I’ve already said that in this example they are all outside the horizon. They lead all the way up to the horizon and you would be traveling at c relative to the one at the horizon.

I'm not sure what you're asking. Are you asking (A) why, physically, relative velocity doesn't affect the tilting of the light cones? Or are you asking (B) why we can't just choose to draw a diagram with the light cones tilted differently when the relative velocity is different?

The answer to (B) is simple. The tilting of the light cones is a frame-invariant, physical phenomenon. You can't change it by changing coordinates or drawing your spacetime diagram differently. It's part of the invariant physical structure of spacetime. You may be able to change the way the light cones "look" on your spacetime diagram by drawing it differently, but that won't change the physics.

I can't answer (A) at this point, because I don't understand why you think relative velocity should have any effect on the curvature of spacetime (which is what the tilting of the light cones is a reflection of) in the first place.

Curvature of space-time is just another way of measuring acceleration. Something in a stronger gravitational field is the equivalent of something not at rest (accelerated) relative to you. Why can’t proper acceleration (as in acceleration from energy rather than mass) be thought of as curved space-time?

 

[PeterDonis]

You can’t move at c relative to another object in flat space-time.

True.

You don’t have to look at gravity as curvature if you don’t want to.

Technically true; you can simply avoid ever considering or utilizing the fact that tidal gravity is equivalent to spacetime curvature, and just work with tidal gravity directly without using any of the theoretical machinery of curvature. (It would be very difficult, but in principle you could do it.) But that won't change the physics. It will still be possible to reach and pass inside a black hole's horizon.

The riverbed is simply SR in a gravitational field.

But SR can't be used "as is" in the presence of tidal gravity. The normal way of saying that is to say that SR assumes that spacetime is flat, and in the presence of gravity spacetime is curved. But I just said you could avoid ever considering curvature. So instead I'll discuss the limitation of SR directly as it relates to tidal gravity.

Why can't SR be used "as is" in the presence of tidal gravity? Because SR assumes that, if two freely falling objects are at rest relative to each other at anyone instant of time, they will remain at rest relative to each other as long as they both remain freely falling. In the presence of tidal gravity, however, that assumption is false. In the presence of tidal gravity, you can have two objects, both freely falling, that start out at rest relative to each other but don't stay that way, even though they both remain freely falling. So SR is simply physically wrong in the presence of tidal gravity.

You appear to think that objects moving solely under tidal gravity will not be freely falling. I'll discuss that further below.

It takes energy to accelerate an object with mass.

Once again you're being careless about the word "acceleration", and also the word "energy", so I'm going to belabor those concepts some.

Suppose we drop a rock from a height, and consider the period of time during which the rock is freely falling. Assume air resistance is zero--we drop the rock in a very tall vacuum chamber, if you like. Assume also that the rock is small enough that tidal effects across it are negligible. Consider two questions: (1) What is the rock's acceleration? (2) What, if any, changes in energy are involved?

The standard GR answers to those questions are:

(1) The rock's *proper* acceleration is zero; it is freely falling. The rock's *coordinate* acceleration depends on the coordinates used; with respect to coordinates in which someone standing on the ground is at rest, the rock has a downward acceleration, but that acceleration is not felt by the rock.

(2) The rock's total energy is constant. Yes, the rock's kinetic energy increases as it falls; but its potential energy decreases by the same amount. So its total energy remains constant. (Edit: I should make clear that in the last sentence I am talking about kinetic and potential energy relative to coordinates in which the person standing on the ground is at rest. The point I was trying to make is that the individual "components" of the total energy can be frame-dependent--for example, the rock's kinetic energy does *not* change relative to an observer that is freely falling along with the rock. But the total energy, kinetic plus potential, is an invariant. I should also note that not all spacetimes even have an invariant energy in this sense; the spacetime surrounding a single gravitating body does because it has particular symmetry properties that not all spacetimes share.)

Now consider the case of a person standing on the ground watching the rock fall. Consider the same two questions. The standard GR answers for the person are:

(1) The person's proper acceleration is nonzero; the person is not freely falling, and feels the acceleration (more properly, the acceleration times his mass) as weight. The person's *coordinate* acceleration, relative to the ground, is zero, but that doesn't stop him from feeling weight.

(2) The person's energy is constant. He remains at the same height and doesn't move, so neither his kinetic nor his potential energy change.

So here we have two objects, one freely falling (zero proper acceleration) and one feeling weight (nonzero proper acceleration). But they both have constant energy. So how, again, does it require energy to accelerate something?

Tidal force means that the front end is pulling the back end along even though the object is in free-fall. It’s mass will create drag which would be felt as weight.

Did you read in my last post where I talked about bodies with internal structure and subject to internal forces? That's what you're talking about here. The force experienced by each end of an extended object, in response to the other end "pulling it along" or "holding it back", is *not* tidal gravity. It's the internal forces between the atoms in the object. Those atoms, individually, are *not* in free fall; they have a nonzero proper acceleration (because they're being pulled on by other atoms), and they feel weight.

If the object as a whole is in free fall (i.e., if the only *external* force acting on it is gravity), then its center of mass will move just as if the internal forces inside the object didn't exist; i.e., the center of mass will be in free fall. So the center of mass of the object can be considered to be moving solely under the influence of gravity. But the object as a whole cannot be if you insist on considering its internal structure.

The reason I insist on making this distinction is that tidal gravity can be observed with two neighboring objects that have no connection between them. For example, consider two rocks dropped from a height at the same instant of time (at that instant, they are mutually at rest), one slightly higher than the other. The rocks will separate--the lower one will fall faster than the higher one, even though they are both in free fall, and the distance between them will increase. But the lower rock is not pulling the higher one along. They are both in free fall and there is no force between them. Yet they still separate. *That* is tidal gravity. And since, as I noted above, such a situation, with two freely falling objects, initially at rest relative to each other, not staying at rest, simply cannot be modeled in SR. That's why we need GR to deal with gravity.

I thought tidal force is the spatial rate of change of the acceleration?

Only in the Newtonian approximation, when gravity is weak. Not in general. Yes, it's counterintuitive, but it's true.

I’ve already said that in this example they are all outside the horizon. They lead all the way up to the horizon and you would be traveling at c relative to the one at the horizon.

Two points. First, there is no "hovering" observer at the horizon (because such an observer would have to move at the speed of light to "hover"), so there is no actual observer that ever sees a freely falling object moving inward at c. And inside the horizon, there are also no "hovering" observers (since not even moving at c will allow you to "hover"; you would have to move faster). So there is no observer anywhere that ever sees a freely falling object moving past him at c or faster.

Second, yes, the speed at which successive "hovering" observers will see a freely falling object fall past them does approach c as the horizon is approached. But it never reaches c, because there are no hovering observers at or inside the horizon. So again, there is no observer anywhere that ever actually sees a freely falling object moving past him at c or faster.

In other words, there's actually no problem. You would like to claim there is, but the laws of physics don't say what you think they say. They only say that two objects can't move faster than c relative to each other *locally*--when they're in the same local piece of spacetime, so they can directly observe each other's relative velocity. The laws do *not* say, and never did say, that an object in one region of spacetime can't move "faster than c" relative to another object in a different region of spacetime.

Why can’t proper acceleration (as in acceleration from energy rather than mass) be thought of as curved space-time?

Let's continue what I started above, and stop talking about "curved spacetime" since that term seems to produce confusion. Since curved spacetime, in standard GR, is equivalent to tidal gravity, let's just substitute the term "tidal gravity" for "curved spacetime" and ask your question above again. The question then is, why can't proper acceleration be thought of as tidal gravity? And the answer is, as I've said a number of times, that tidal gravity can be observed with objects that are in free fall and have no proper acceleration. So they can't possibly be the same thing.

 

[Dale]

I don’t see how a coordinate system doesn’t cover all of space-time because they take a frame of reference and map around it, telling you how things relate to each other from that perspective.

In GR coordinate systems don't have anything directly to do with frames of reference or observers and it is not useful to think of them in that context. They are just mappings of numbers (points in R4) to events (points in spacetime manifold). Some coordinate systems don't even have timelike and spacelike basis vectors, but instead have null basis vectors. Other coordinate systems have basis vectors that are not orthonormal or that change character from place to place. Don't think of them as reference frames, think of them as mappings from events to numbers, as Carroll clearly explains on p 37 in words, equations, and pictures.

 

[Dale]

Work out the proper time using Rindler and Schwarzschild coordinates and you get exactly the same answer – it would take infinite proper time to reach an event horizon in exactly the same way it takes infinite proper time to accelerate to c. This is not dependent on the coordinate system that you use because the proper time it would take for an observer to do something can’t change.

Have you actually worked out the proper time using Rindler and Schwarzschild coordinates? If so, please post your math, if not, please stop making unfounded claims about what the math would work out to be.

I highly value math due to its being clear and unambiguous and you are completely misrepresenting the facts here. Your claim is not acceptable. You may freely claim what your opinion is, but if you attempt to give it a veneer of mathematical veracity or theoretical support then you should be prepared to back up your words with the claimed math.

 

[Dale]

You don’t need to learn Hebrew to read the bible. I wish you did. There wouldn’t be as many people walking around with crap in their heads.

Good analogy. You are "walking around with crap" in your head on the subject of GR because you are not reading GR in the original language, math. The translations from math to english are not very good.

I don’t mind you posting the equations with a description, like you have been, but to claim equations represent something that can't be understood without them is just silly, and a copout.

You are right. I should make a more specific claim and not generalize beyond what is warranted. The basic fundamental concepts of GR probably can be understood by some people without the equations, particularly people with a strong physics backgroud who can do the translation back to math in their heads. However, based on the past 350+ posts, you are not one of those people. Whether you are capable of understanding GR with math is not yet clear, particularly since you appear not to even be attempting it. I hope you do attempt it, because you appear to be interested in the subject, and learning completely new concepts is very enlightening and enjoyable.

 

[Dale]

So it's not because they're different things, it's just that you're viewing gravity as curvature of space-time and are therefore describing how it effects everything, but viewing acceleration as purely the curved worldline of one object without taking into account how energy curves space-time so that it's not only one object that's accelerated. So they're both rank 4 tensors.

Let me be clear:
$${R^{\mu}}_{\nu\eta\kappa} \ne \nabla_{\tau} v^{\mu}$$
They are not equal under any circumstances whatsoever. The Riemann curvature is a rank 4 tensor and the covariant derivative of the tangent vector is a rank 1 tensor. You may solve the EFE including the stress-energy tensor of the accelerating object and you will still get that the curvature is rank 4 and the proper acceleration is rank 1. Tensors never change their rank. They are different things, always.

 

[Dale]

If you want to be able to move along the timeline then you have to remove a spatial dimension.

Why would you think that? The flat spacetime metric in Rindler coordinates is:
$$ds^2 = -x^2 dt^2 + dx^2 + dy^2 + dz^2$$
Which clearly has 1 dimension of time and 3 dimensions of space. So why do you think you would have to remove a dimension in order to work in Rindler coordinates?

 

[Dale]

Work out the proper time using Rindler and Schwarzschild coordinates and you get exactly the same answer – it would take infinite proper time to reach an event horizon

I thought I may as well work this out so that you can see how the math goes. I will work it out for the Rindler coordinates only, but I encourage you to follow the same steps for the Schwarzschild coordinates. For convenience I will use units of years for time and light years for distance.

Again, the Rindler metric is:
$$\mathbf g = \left( \begin{array}{cccc} -x^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

The worldline of a free-falling clock which starts initially at rest at (t,x,y,z) = (0,1,0,0) is:
$$\mathbf r = (t,sech(t),0,0)$$

As expected
$$\lim_{t\to \infty } \, r^x = \lim_{t\to \infty } \, sech(t) = 0$$
meaning that it takes an infinite amount of coordinate time for the free falling clock to reach the Rindler horizon at x=0.

Now, to find the proper time we use formula 1.97 in Sean Carroll's lecture notes
$$\tau = \int \sqrt{-g_{\mu\nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}} \, dt = \int sech^{2}(t) \, dt = tanh(t)$$

Taking the limit of the proper time as the coordinate time goes to infinity we get:
$$\lim_{t\to \infty } \, \tau = \lim_{t\to \infty } \, tanh(t) = 1$$
meaning that it takes 1 year of proper time for the free falling clock to reach the Rindler horizon at x=0. The trip from x=1 to the event horizon, which takes an infinite amount of coordinate time, takes a finite amount of proper time.

Please do not repeat the false claim that the proper time for a free-faller to reach the horizon is infinite.

 

[PeterDonis]

The reason I insist on making this distinction is that tidal gravity can be observed with two neighboring objects that have no connection between them. For example, consider two rocks dropped from a height at the same instant of time (at that instant, they are mutually at rest), one slightly higher than the other. The rocks will separate--the lower one will fall faster than the higher one, even though they are both in free fall, and the distance between them will increase. But the lower rock is not pulling the higher one along. They are both in free fall and there is no force between them. Yet they still separate. *That* is tidal gravity. And since, as I noted above, such a situation, with two freely falling objects, initially at rest relative to each other, not staying at rest, simply cannot be modeled in SR. That's why we need GR to deal with gravity.

Btw, A-wal, I should add that this point has come up repeatedly in this thread, and every time it has, you have not disagreed with it--but after a few more posts go by, you have gone back to asserting that objects moving solely under the influence of tidal gravity feel a nonzero acceleration (proper acceleration), even though they are in free fall. At least once, when I explicitly asked you if, in the idealized case of "point-like" objects with no internal structure, the object would feel zero acceleration, you agreed. So presumably you would also agree that two such point-like objects in free fall, starting from mutual rest at slightly different heights, and remaining in free fall, would separate (i.e., they would not remain mutually at rest), even though they both continue to feel zero acceleration. And that, all by itself, is enough to show that SR cannot be valid in the presence of tidal gravity (because in SR, "point-like" objects with no internal structure, starting at mutual rest in free fall, are assumed to remain at mutual rest as long as they both remain in free fall). So I would like to put this point to bed one way or the other. Do you in fact agree with what I've just said about how two point-like freely falling objects would behave in the presence of tidal gravity? And if you do, can we stop talking about irrelevant complications like how individual parts of an extended object with internal structure would behave, since they do not affect the main point?

 

[A-wal]

But SR can't be used "as is" in the presence of tidal gravity. The normal way of saying that is to say that SR assumes that spacetime is flat, and in the presence of gravity spacetime is curved. But I just said you could avoid ever considering curvature. So instead I'll discuss the limitation of SR directly as it relates to tidal gravity.

Why can't SR be used "as is" in the presence of tidal gravity? Because SR assumes that, if two freely falling objects are at rest relative to each other at anyone instant of time, they will remain at rest relative to each other as long as they both remain freely falling. In the presence of tidal gravity, however, that assumption is false. In the presence of tidal gravity, you can have two objects, both freely falling, that start out at rest relative to each other but don't stay that way, even though they both remain freely falling. So SR is simply physically wrong in the presence of tidal gravity.

You appear to think that objects moving solely under tidal gravity will not be freely falling. I'll discuss that further below.

Which fits in exactly with what I've been saying. In the presence of tidal gravity two objects that start out at rest relative to each other but at different distances (one would have to be accelerating to start with) from the gravitational source will separate because the one closer to the gravitational source will be under the influence of a greater amount of tidal force and will therefore be accelerating harder than the more distant object, because tidal force is acceleration.

You appear to think that objects moving solely under the influence of tidal gravity can feel no acceleration?

Once again you're being careless about the word "acceleration", and also the word "energy", so I'm going to belabor those concepts some.

Suppose we drop a rock from a height, and consider the period of time during which the rock is freely falling. Assume air resistance is zero--we drop the rock in a very tall vacuum chamber, if you like. Assume also that the rock is small enough that tidal effects across it are negligible. Consider two questions: (1) What is the rock's acceleration? (2) What, if any, changes in energy are involved?

The standard GR answers to those questions are:

(1) The rock's *proper* acceleration is zero; it is freely falling. The rock's *coordinate* acceleration depends on the coordinates used; with respect to coordinates in which someone standing on the ground is at rest, the rock has a downward acceleration, but that acceleration is not felt by the rock.

(2) The rock's total energy is constant. Yes, the rock's kinetic energy increases as it falls; but its potential energy decreases by the same amount. So its total energy remains constant. (Edit: I should make clear that in the last sentence I am talking about kinetic and potential energy relative to coordinates in which the person standing on the ground is at rest. The point I was trying to make is that the individual "components" of the total energy can be frame-dependent--for example, the rock's kinetic energy does *not* change relative to an observer that is freely falling along with the rock. But the total energy, kinetic plus potential, is an invariant. I should also note that not all spacetimes even have an invariant energy in this sense; the spacetime surrounding a single gravitating body does because it has particular symmetry properties that not all spacetimes share.)

Now consider the case of a person standing on the ground watching the rock fall. Consider the same two questions. The standard GR answers for the person are:

(1) The person's proper acceleration is nonzero; the person is not freely falling, and feels the acceleration (more properly, the acceleration times his mass) as weight. The person's *coordinate* acceleration, relative to the ground, is zero, but that doesn't stop him from feeling weight.

(2) The person's energy is constant. He remains at the same height and doesn't move, so neither his kinetic nor his potential energy change.

So here we have two objects, one freely falling (zero proper acceleration) and one feeling weight (nonzero proper acceleration). But they both have constant energy. So how, again, does it require energy to accelerate something?

The free-faller is accelerated through tidal force. I know you said the tidal force is negligible but if there was no tidal force then they couldn’t be free-falling, so it can never really be negligible in this sense.

The person on the ground is accelerated by the ground itself because the ground is pushing them up in response to gravity pulling them down, which is why they feel weight.

Did you read in my last post where I talked about bodies with internal structure and subject to internal forces? That's what you're talking about here. The force experienced by each end of an extended object, in response to the other end "pulling it along" or "holding it back", is *not* tidal gravity. It's the internal forces between the atoms in the object. Those atoms, individually, are *not* in free fall; they have a nonzero proper acceleration (because they're being pulled on by other atoms), and they feel weight.

If the object as a whole is in free fall (i.e., if the only *external* force acting on it is gravity), then its center of mass will move just as if the internal forces inside the object didn't exist; i.e., the center of mass will be in free fall. So the center of mass of the object can be considered to be moving solely under the influence of gravity. But the object as a whole cannot be if you insist on considering its internal structure.

The reason I insist on making this distinction is that tidal gravity can be observed with two neighboring objects that have no connection between them. For example, consider two rocks dropped from a height at the same instant of time (at that instant, they are mutually at rest), one slightly higher than the other. The rocks will separate--the lower one will fall faster than the higher one, even though they are both in free fall, and the distance between them will increase. But the lower rock is not pulling the higher one along. They are both in free fall and there is no force between them. Yet they still separate. *That* is tidal gravity. And since, as I noted above, such a situation, with two freely falling objects, initially at rest relative to each other, not staying at rest, simply cannot be modeled in SR. That's why we need GR to deal with gravity.

Yes I read your post, and yes it's because of the internal forces, but that happens because of tidal force.

Why can't it be modeled in SR? If the amount of proper acceleration of two objects were to increase depending on their distance from a third object and they started off at different distances from that object then they would separate.

Only in the Newtonian approximation, when gravity is weak. Not in general. Yes, it's counterintuitive, but it's true.

Okay, more specifically, I don't see why the whole space-time isn't contained within the exterior of a black hole using Schwarzschild coordinates. It's right there!

You didn't answer this before. In the river model that I'm using would it be accurate to say that tidal force is the acceleration of the river?

Two points. First, there is no "hovering" observer at the horizon (because such an observer would have to move at the speed of light to "hover"), so there is no actual observer that ever sees a freely falling object moving inward at c. And inside the horizon, there are also no "hovering" observers (since not even moving at c will allow you to "hover"; you would have to move faster). So there is no observer anywhere that ever sees a freely falling object moving past him at c or faster.

Second, yes, the speed at which successive "hovering" observers will see a freely falling object fall past them does approach c as the horizon is approached. But it never reaches c, because there are no hovering observers at or inside the horizon. So again, there is no observer anywhere that ever actually sees a freely falling object moving past him at c or faster.

In other words, there's actually no problem. You would like to claim there is, but the laws of physics don't say what you think they say. They only say that two objects can't move faster than c relative to each other *locally*--when they're in the same local piece of spacetime, so they can directly observe each other's relative velocity. The laws do *not* say, and never did say, that an object in one region of spacetime can't move "faster than c" relative to another object in a different region of spacetime.

Again, this completely backs up what I've been saying. The speed that successive hovering observers would see freely falling objects pass them approaches c as the horizon is approached. But it never reaches c, so an object never reaches the horizon. How far away do you think two objects have to be to reach c relative to each other then?

Let's continue what I started above, and stop talking about "curved spacetime" since that term seems to produce confusion. Since curved spacetime, in standard GR, is equivalent to tidal gravity, let's just substitute the term "tidal gravity" for "curved spacetime" and ask your question above again. The question then is, why can't proper acceleration be thought of as tidal gravity? And the answer is, as I've said a number of times, that tidal gravity can be observed with objects that are in free fall and have no proper acceleration. So they can't possibly be the same thing.

I'm not confused by the concept of curved space-time. I actually really like that way of looking at it, as long as I keep in mind that describing the curvature between objects and describing the movement of objects through space-time as a world-line equate to the same thing. It's just a nice way of mapping proper acceleration.

Have you actually worked out the proper time using Rindler and Schwarzschild coordinates? If so, please post your math, if not, please stop making unfounded claims about what the math would work out to be.

The time dilation and length contraction that a distant observer measures for an in-faller approaching an event horizon in Schwarzschild coordinates would be the same as an inertial observer would measure for an accelerator approaching c. Rindler coordinates would obviously have to come up with the same answer.

I highly value math due to its being clear and unambiguous and you are completely misrepresenting the facts here. Your claim is not acceptable. You may freely claim what your opinion is, but if you attempt to give it a veneer of mathematical veracity or theoretical support then you should be prepared to back up your words with the claimed math.

I know you highly value maths and it is clear and unambiguous, but if you start to think that the maths determine reality rather than the other way round then you can end up going wrong somewhere and you'll have no way of knowing because the equations can't tell you. You can end up with an event horizon that allows you to break the light barrier.

Good analogy. You are "walking around with crap" in your head on the subject of GR because you are not reading GR in the original language, math. The translations from math to english are not very good.

I've noticed. One of the things I'm trying to do is make the translations clearer. Maths is not the original language. What actually happens is the original language and maths is just a tool to express the relationships in short-hand. It's precisely because I'm not walking round with that crap in my head that I'm able to see the mistake despite the fact that I've had no formal education in even the basics. I bet you even went to collage.

You are right. I should make a more specific claim and not generalize beyond what is warranted. The basic fundamental concepts of GR probably can be understood by some people without the equations, particularly people with a strong physics backgroud who can do the translation back to math in their heads. However, based on the past 350+ posts, you are not one of those people. Whether you are capable of understanding GR with math is not yet clear, particularly since you appear not to even be attempting it. I hope you do attempt it, because you appear to be interested in the subject, and learning completely new concepts is very enlightening and enjoyable.

You are wrong. The concepts of GR can be understood by practically anyone without using equations, particularly people without a strong physics background who don't do the translation to maths in their heads. However, based on the 350+ posts, you are not one of those people. Whether you are capable of understanding GR without the maths is not yet clear, particularly since you appear not to even be attempting it. I hope you do attempt it, because you appear to be interested in the subject, and this is a beautiful subject, one that deserves to not be cheapened by with numbers and symbols.

Let me be clear:
$${R^{\mu}}_{\nu\eta\kappa} \ne \nabla_{\tau} v^{\mu}$$
They are not equal under any circumstances whatsoever. The Riemann curvature is a rank 4 tensor and the covariant derivative of the tangent vector is a rank 1 tensor. You may solve the EFE including the stress-energy tensor of the accelerating object and you will still get that the curvature is rank 4 and the proper acceleration is rank 1. Tensors never change their rank. They are different things, always.

It's clear because squiggle, letter, arrow, squiggle, diagonal line, letter, letter, squiggle, squiggle, etc. Well I'm convinced.

Why would you think that? The flat spacetime metric in Rindler coordinates is:
$$ds^2 = -x^2 dt^2 + dx^2 + dy^2 + dz^2$$
Which clearly has 1 dimension of time and 3 dimensions of space. So why do you think you would have to remove a dimension in order to work in Rindler coordinates?

I never said I was talking about Rindler coordinates. That was you.

I thought I may as well work this out so that you can see how the math goes. I will work it out for the Rindler coordinates only, but I encourage you to follow the same steps for the Schwarzschild coordinates. For convenience I will use units of years for time and light years for distance.

Again, the Rindler metric is:
$$\mathbf g = \left( \begin{array}{cccc} -x^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

The worldline of a free-falling clock which starts initially at rest at (t,x,y,z) = (0,1,0,0) is:
$$\mathbf r = (t,sech(t),0,0)$$

As expected
$$\lim_{t\to \infty } \, r^x = \lim_{t\to \infty } \, sech(t) = 0$$
meaning that it takes an infinite amount of coordinate time for the free falling clock to reach the Rindler horizon at x=0.

Now, to find the proper time we use formula 1.97 in Sean Carroll's lecture notes
$$\tau = \int \sqrt{-g_{\mu\nu} \frac{dx^{\mu}}{dt} \frac{dx^{\nu}}{dt}} \, dt = \int sech^{2}(t) \, dt = tanh(t)$$

Taking the limit of the proper time as the coordinate time goes to infinity we get:
$$\lim_{t\to \infty } \, \tau = \lim_{t\to \infty } \, tanh(t) = 1$$
meaning that it takes 1 year of proper time for the free falling clock to reach the Rindler horizon at x=0. The trip from x=1 to the event horizon, which takes an infinite amount of coordinate time, takes a finite amount of proper time.

Please do not repeat the false claim that the proper time for a free-faller to reach the horizon is infinite.

But an object can't cross using Schwarzschild coordinates?

Btw, A-wal, I should add that this point has come up repeatedly in this thread, and every time it has, you have not disagreed with it--but after a few more posts go by, you have gone back to asserting that objects moving solely under the influence of tidal gravity feel a nonzero acceleration (proper acceleration), even though they are in free fall. At least once, when I explicitly asked you if, in the idealized case of "point-like" objects with no internal structure, the object would feel zero acceleration, you agreed. So presumably you would also agree that two such point-like objects in free fall, starting from mutual rest at slightly different heights, and remaining in free fall, would separate (i.e., they would not remain mutually at rest), even though they both continue to feel zero acceleration. And that, all by itself, is enough to show that SR cannot be valid in the presence of tidal gravity (because in SR, "point-like" objects with no internal structure, starting at mutual rest in free fall, are assumed to remain at mutual rest as long as they both remain in free fall). So I would like to put this point to bed one way or the other. Do you in fact agree with what I've just said about how two point-like freely falling objects would behave in the presence of tidal gravity? And if you do, can we stop talking about irrelevant complications like how individual parts of an extended object with internal structure would behave, since they do not affect the main point?

I do agree that two such point-like objects in free fall, starting from mutual rest at slightly different heights, and remaining in free fall, would separate. I don’t agree that, all by itself, is enough to show that SR cannot be valid in the presence of tidal gravity because they’re feeling different amounts of gravity (acceleration).

 

[Dale]

The time dilation and length contraction that a distant observer measures for an in-faller approaching an event horizon in Schwarzschild coordinates would be the same as an inertial observer would measure for an accelerator approaching c.

This is not correct. You should work it out, as you suggested.

if you start to think that the maths determine reality rather than the other way round

I have never said anything remotely similar to that, nor do I believe it. Math doesn't determine reality, and neither does English. They are both merely human tools used for describing nature. Math is simply the far better tool since it is unambiguous and enforces logic. Therefore mathematical descriptions are inherently clear and self-consistent, in contrast to English descriptions which usually have many layers of connotation and contradiction.

But an object can't cross using Schwarzschild coordinates?

Correct, as I showed above, the coordinate time is infinite, the proper time is finite (in this case 1 year).

English is pointless with you as we have seen here, and math seems pointless with you also since you are not willing to make even a small effort to work things out on your own. This discussion is plainly not going to progress in any possible mode.

If you ever become serious about learning GR please reach out again and I will be more than glad to help. In particular, if you want to learn how to work out the proper time vs. the coordinate time, or anything else we have discussed, I will be ready to assist.

 

[PeterDonis]

A-wal, I'm going to respond to your post in installments, because I see two issues that deserve separate treatment. The first is what ought to be a straightforward question of physical fact; the second is about the theoretical implications, but it's not even worth discussing those if we're not in agreement on the physical facts (and by "facts" I mean things that can be observed in our local piece of spacetime, not things like whether a black hole horizon is reachable). In this post I want to focus solely on the physical fact. I'm considering specifically the following excerpts from your post; I'll quote them all before commenting.

Which fits in exactly with what I've been saying. In the presence of tidal gravity two objects that start out at rest relative to each other but at different distances (one would have to be accelerating to start with) from the gravitational source will separate because the one closer to the gravitational source will be under the influence of a greater amount of tidal force and will therefore be accelerating harder than the more distant object, because tidal force is acceleration.

You appear to think that objects moving solely under the influence of tidal gravity can feel no acceleration?

I do agree that two such point-like objects in free fall, starting from mutual rest at slightly different heights, and remaining in free fall, would separate. I don’t agree that, all by itself, is enough to show that SR cannot be valid in the presence of tidal gravity because they’re feeling different amounts of gravity (acceleration).

The free-faller is accelerated through tidal force. I know you said the tidal force is negligible but if there was no tidal force then they couldn’t be free-falling, so it can never really be negligible in this sense.

As before, you continue to use the word "acceleration" without making it clear whether you mean actual proper acceleration or just coordinate acceleration. Do you honestly mean *proper* acceleration in any of the above? Because if so, we may have a very basic issue with physical facts.

Consider the two point-like objects in free fall. You say they are "feeling" different amounts of gravity. Do you mean they actually "feel" the gravity? In other words, do you mean that accelerometers attached to the objects would read something other than exactly zero? (More precisely, since you may be misunderstanding the scenario I posed: the two objects start out at mutual rest, and one way they could have gotten that way was by being artificially held at rest at slightly different heights--say they are attached to a long radially oriented boom on a rocket that is "hovering" at a high altitude--and while they are being held at rest that way, they will not be in free fall but will be accelerated, in the sense of having nonzero proper acceleration. That is because their motion is not solely due to gravity; they are being artificially held. But once the objects are released, from slightly different heights, they are moving solely under the influence of gravity, they are in free fall, and they have zero proper acceleration--at least, they are moving on worldlines that have zero proper acceleration; I'll discuss exactly what that means further below. It is during that time, when they are both freely falling, and are separating even though they are both in free fall, that I'm asking you whether you think accelerometers attached to the objects would read something other than exactly zero. Also, I say "exactly" because these are pointlike idealized objects, so they have no internal structure or forces; the only possible way their accelerometers could read something other than exactly zero is if gravity alone does it.)

If you say that accelerometers attached to idealized pointlike objects as I've specified above would in fact read something other than exactly zero, then I have to ask why you believe this. I am unaware of any experimental facts bearing directly on this issue; as far as I know, no one has directly done a real experiment similar to the scenario I described above, with accelerometers attached to the objects to see what they read. And even if they did, you could, I suppose, claim that the accelerometers simply weren't accurate enough to detect the small but non-zero accelerations caused by gravity alone.

However, physicists do have a powerful reason for believing that the accelerometers in the idealized case I described above would, in fact, read exactly zero--in other words, that bodies moving solely under the influence of gravity are, truly, in free fall, and truly feel *no* acceleration. The reason is simply that, mathematically, there is a simple criterion that corresponds to "zero proper acceleration" or "free fall"--it is that the worldline followed by the object has zero covariant derivative. (DaleSpam went into this a bit in one of his posts.) And, when worldlines of actual objects moving solely under the influence of gravity are observed and analyzed--for example, planets, moons of planets, asteroids, comets, satellites orbiting the Earth, spacecraft sent out into the solar system, etc.--all of them, without exception, meet that mathematical criterion. In other words, they are exactly the worldlines (those with zero covariant derivative) you would expect the objects to follow *if* they were truly in free fall, experiencing zero proper acceleration; but you would *not* expect them to follow those worldlines if they had *any* nonzero proper acceleration at all.

Therefore, by induction from many, many, many examples, I believe that the two idealized point-like objects in the scenario I described would, in fact, have exactly zero proper acceleration, and accelerometers attached to them would in fact read exactly zero. If you actually agree with that, then I'm stumped as to how you're working out the implications--but that's the theoretical issue I talked about, which I don't want to get into until we've got the factual issue cleared up.

If, however, you do *not* agree that the accelerometers attached to the two point-like objects would read exactly zero, then I can understand why you've been saying what you've been saying, but I still think you've got an issue you haven't deal with. After all, even if you think the accelerometers would read something other than exactly zero, the experimental facts, as I noted above, are abundantly clear that "freely falling" objects--objects moving solely under the influence of gravity--follow the *worldlines* that we predict them to follow based on the mathematical criterion of zero covariant derivative. So you would basically have to maintain that the physical interpretation of that mathematical criterion is not what everyone else thinks it is: that an object can follow a "zero proper acceleration" worldline (mathematically speaking), but still, physically, feel a non-zero acceleration.

The problem with that, however, is twofold. First, I have no idea how you would go about predicting what acceleration an object should feel. On standard GR, the answer is simple: just compute the covariant derivative along its worldline. This answer also agrees with experiment: we can use it to correctly calculate your weight, for example. But that rule, of course, tells you that a "freely falling" object, moving on a worldline with zero covariant derivative (like every orbiting satellite, like every planet, like the Moon, etc., etc.), feels exactly zero acceleration--and that's what you appear to be disputing. So what rule would you substitute, to replicate all the known experimental facts but still come up with the answer that the two idealized point-like objects in my scenario above would somehow feel a non-zero acceleration?

Second, if you agree that GR predicts the *worldlines* of objects correctly, then what acceleration they feel is actually irrelevant, because the mathematical criterion I gave to single out "freely falling" worldlines--zero covariant derivative--is still sufficient to show that SR is invalid. Just substitute "zero covariant derivative worldlines" for "freely falling worldlines" everywhere in what I posted before. SR still predicts that two worldlines with zero covariant derivative that are mutually at rest (i.e., parallel) at anyone instant of time, will remain mutually at rest forever. In the presence of gravity, that is false; the two idealized point-like objects are a counterexample, and you agree they will separate (i.e., that their worldlines, despite both having zero covariant derivative, will not remain parallel even though they start out that way). But that is getting into the theoretical issue again, which I don't want to pursue further until we've got the factual issue taken care of. (Though I have to comment that if you don't believe that objects moving on worldlines with zero covariant derivative feel zero acceleration, then I'm stumped as to how you are able to apply SR, since zero covariant derivative is just another way of saying "inertial frame" in SR, and the only physical way to pick out inertial frames is to pick out objects that feel zero acceleration.)

(One other note about the scenario. If the fact that I specified that the two objects were artificially held at rest before being dropped makes things messy, then I can revise the scenario to eliminate it as follows. Consider two idealized, point-like objects, which are both moving upward in the Earth's gravitational field in such a way that they both come to rest for an instant, at the same instant--"same instant" in the frame in which they are both mutually at rest for that instant--at slightly different heights. In this scenario the objects are moving solely under the influence of gravity for the whole time under consideration, without any external forces. The same conclusion follows as before: the worldlines are parallel at the instant when both objects are mutually at rest, but they are *not* parallel for their entire extent, as SR would predict they should be. So SR doesn't work in the presence of gravity.)

 

[PeterDonis]

One addition to my last post, to amplify on why physicists believe "zero covariant derivative" implies "zero proper acceleration". As I noted, in SR, "zero covariant derivative" worldlines are just the "inertial" worldlines--the worldlines that bodies follow when they are at rest in an inertial frame. And those bodies, physically, are the ones that feel exactly zero acceleration; that's the only way we can link up the theory of SR with actual physical facts.

But then, if in the presence of gravity, "zero covariant derivative" no longer equals "zero proper acceleration", then we would have a way to violate the equivalence principle. We know, experimentally, that objects moving solely under the influence of gravity move on zero covariant derivative worldlines. So if I'm floating in a box somewhere, with no view of the outside world, and there are no forces on me that I can observe (except possibly gravity, but by hypothesis I can't directly observe gravity without looking outside the box), I can tell by using an accelerometer whether the box is floating at rest deep in empty space (accelerometer reads zero) or is freely falling in a gravitational field (accelerometer reads nonzero). But the equivalence principle says that there is no local way to tell free fall in a gravitational field from floating at rest deep in empty space. So if we believe the equivalence principle, then "zero covariant derivative" has to equal "zero proper acceleration" or "accelerometer reads exactly zero".

 

[PeterDonis]

A-wal, this is a quick post in response to a couple of items in your post that are not really connected, at least not directly, to the main issue I talked about in my previous two posts. They're just other general clarifications of points you've raised.

In the river model that I'm using would it be accurate to say that tidal force is the acceleration of the river?

In the limit where gravity is weak (i.e., very far away from the hole), yes. Approaching and at the horizon and inside, no, at least not the way I think you mean "acceleration of the river" (i.e., acceleration relative to the "river bed"). That acceleration becomes infinite at the horizon, but as I've already pointed out, the tidal force at the horizon is finite (it goes like the inverse square of the mass), and the river model covers the horizon and the portion inside, so it is consistent with these predictions, meaning that the tidal force is not, in general, equal to the "acceleration of the river" in this sense.

The speed that successive hovering observers would see freely falling objects pass them approaches c as the horizon is approached. But it never reaches c, so an object never reaches the horizon.

How are you reaching that conclusion? The only way I can see that this would follow is if you assume that the freely falling object must, in principle, pass a hovering observer at every point on its worldline. That is just another version of the mistake you've been making, which I've pointed out before. In the standard GR model, that assumption is false; there are no hovering observers at or inside the horizon, so the freely falling object doesn't pass any for that portion of its worldline. If you have an argument, starting from premises we all accept, for why the assumption *has* to be true, then by all means post it; but you can't just assume it. You have to prove it.

 

[PeterDonis]

In the limit where gravity is weak (i.e., very far away from the hole), yes. Approaching and at the horizon and inside, no, at least not the way I think you mean "acceleration of the river" (i.e., acceleration relative to the "river bed"). That acceleration becomes infinite at the horizon

Oops, just realized that "acceleration relative to the river bed" in your version of the model is not the same as proper acceleration in GR. A "hovering" observer's "acceleration relative to the river bed" is zero, since the hovering observer is at rest in Schwarzschild coordinates. It doesn't change the main point I was making, but I phrased it wrong. Here's what I should have said:

In the limit where gravity is weak (i.e., very far away from the hole), I believe you can say that tidal force is equal to the *rate of change* of the "acceleration of the river relative to the river bed" (*not* the acceleration itself). I say "I believe" because I haven't actually worked out the math to confirm that, when you compute the "acceleration of the river relative to the river bed" far away from the hole, you do in fact get the correct Newtonian formula, - GM/r^2, for "acceleration due to gravity". If you do, then the (radial) tidal gravity, in the Newtonian approximation, is equal to the (radial) rate of change of that acceleration, i.e., 2GM/r^3.

If all that is the case, then the equality would continue to hold approaching, at, and inside the horizon, because, as I've already pointed out, the formula for tidal gravity remains the same as the Newtonian formula all the way into r = 0, even though the "acceleration due to gravity" does not (it acquires an extra sqrt(1 - 2GM/c^2r) term in the denominator, so the acceleration diverges as the horizon is approached). The formula for "acceleration of the river relative to the river bed" should also remain the same as the Newtonian formula all the way into r = 0 (since the formula for the velocity of the river remains the same, and the acceleration is just the radial rate of change of that velocity), so radial tidal gravity should continue to equal the radial rate of change of that acceleration. (This means, of course, that as the horizon is approached, reached, and passed, the "acceleration of the river relative to the river bed" is no longer equal to the correct relativistic "acceleration due to gravity", since that diverges as the horizon is approached.)

All this depends, however, on the formula for "acceleration of the river relative to the river bed" working out as I said it needs to above. I'll have to check that when I get a chance.

 

[Passionflower]

Technically true; you can simply avoid ever considering or utilizing the fact that tidal gravity is equivalent to spacetime curvature, and just work with tidal gravity directly without using any of the theoretical machinery of curvature. (It would be very difficult, but in principle you could do it.) But that won't change the physics. It will still be possible to reach and pass inside a black hole's horizon.

How would you model CTCs and wormholes?

 

[PeterDonis]

How would you model CTCs and wormholes?

Hmm, good point. I'm not sure how you could model spacetimes with non-trivial topology just by talking about tidal gravity. You could do it locally, but I don't see how you could capture the global properties that way.

 

[A-wal]

This is not correct. You should work it out, as you suggested.

This is the problem. A black hole occurs when the escape velocity needed to move away exceeds c. The trouble is that this can never happen because you would need to exceed c to be in that situation in the first place. You would need an infinite amount of energy to reach c in flat space-time because of time dilation/length contraction, which means c in fact represents infinite velocity. For an object to reach an event horizon the time dilation/length contraction associated with gravity means that the object would have to be moving at infinite velocity when it reaches the horizon. Therefore any maths or reasoning that allows an object to reach an event horizon is flawed in the same way that any form of maths or reasoning that allows an object to reach c in flat space-time is flawed.

A distant observer watches an object free-fall towards a black hole. Time dilation and length contraction don’t just make it appear that the object is slowing down as it approaches the event horizon. Time dilation and length contraction mean that the object falling towards the event horizon really is slowing down relative to the distant observer. It's exactly the same as it would be for an inertial observer watching an accelerator approaching c in flat space-time if its acceleration were to increase at a rate of an inverse square relative to the accelerators distance from another object. It will always reach that object before it reaches c relative to the distant observer, regardless of the distance to the object it's heading towards. The black hole will die before the in-faller reaches the horizon regardless of how long the black hole has left to live.

I worked it out ages ago. This is just a little experiment to see how long it takes and how much I have to spell it out before the penny drops for highly trained physicists.

I have never said anything remotely similar to that, nor do I believe it. Math doesn't determine reality, and neither does English. They are both merely human tools used for describing nature. Math is simply the far better tool since it is unambiguous and enforces logic. Therefore mathematical descriptions are inherently clear and self-consistent, in contrast to English descriptions which usually have many layers of connotation and contradiction.

Good point. Booth describe reality and each has its good and its bad points. Maths is completely clear and concise, but it still needs words to put it into context and tell you what the equations are describing. The biggest problem with maths is that it can work fine on paper but be completely impossible in reality.

English is pointless with you as we have seen here, and math seems pointless with you also since you are not willing to make even a small effort to work things out on your own. This discussion is plainly not going to progress in any possible mode.

English isn't pointless with me. Explain using words how an object could possibly reach an event horizon in time in a way that's self-consistent and that will be that. I don't think it can be done.

If you ever become serious about learning GR please reach out again and I will be more than glad to help. In particular, if you want to learn how to work out the proper time vs. the coordinate time, or anything else we have discussed, I will be ready to assist.

You think I can't be serious about learning GR if I don't want to do it using equations? You're trying to tempt me over to the dark side aren't you Mr Sith? It won't work. My heart is pure, ish. Seriously though, cheers for the offer but I don't see the point of learning the mathematics of something I don't even think is right.

A-wal, I'm going to respond to your post in installments, because I see two issues that deserve separate treatment. The first is what ought to be a straightforward question of physical fact; the second is about the theoretical implications, but it's not even worth discussing those if we're not in agreement on the physical facts (and by "facts" I mean things that can be observed in our local piece of spacetime, not things like whether a black hole horizon is reachable). In this post I want to focus solely on the physical fact. I'm considering specifically the following excerpts from your post; I'll quote them all before commenting.

I appreciate the time and effort you've taken to be as clear as possible.

As before, you continue to use the word "acceleration" without making it clear whether you mean actual proper acceleration or just coordinate acceleration. Do you honestly mean *proper* acceleration in any of the above? Because if so, we may have a very basic issue with physical facts.

Consider the two point-like objects in free fall. You say they are "feeling" different amounts of gravity. Do you mean they actually "feel" the gravity? In other words, do you mean that accelerometers attached to the objects would read something other than exactly zero? (More precisely, since you may be misunderstanding the scenario I posed: the two objects start out at mutual rest, and one way they could have gotten that way was by being artificially held at rest at slightly different heights--say they are attached to a long radially oriented boom on a rocket that is "hovering" at a high altitude--and while they are being held at rest that way, they will not be in free fall but will be accelerated, in the sense of having nonzero proper acceleration. That is because their motion is not solely due to gravity; they are being artificially held. But once the objects are released, from slightly different heights, they are moving solely under the influence of gravity, they are in free fall, and they have zero proper acceleration--at least, they are moving on worldlines that have zero proper acceleration; I'll discuss exactly what that means further below. It is during that time, when they are both freely falling, and are separating even though they are both in free fall, that I'm asking you whether you think accelerometers attached to the objects would read something other than exactly zero. Also, I say "exactly" because these are pointlike idealized objects, so they have no internal structure or forces; the only possible way their accelerometers could read something other than exactly zero is if gravity alone does it.)

Yes, I mean proper acceleration. I think tidal force and acceleration are the exact same thing under slightly different circumstances.

If you say that accelerometers attached to idealized pointlike objects as I've specified above would in fact read something other than exactly zero, then I have to ask why you believe this. I am unaware of any experimental facts bearing directly on this issue; as far as I know, no one has directly done a real experiment similar to the scenario I described above, with accelerometers attached to the objects to see what they read. And even if they did, you could, I suppose, claim that the accelerometers simply weren't accurate enough to detect the small but non-zero accelerations caused by gravity alone.

However, physicists do have a powerful reason for believing that the accelerometers in the idealized case I described above would, in fact, read exactly zero--in other words, that bodies moving solely under the influence of gravity are, truly, in free fall, and truly feel *no* acceleration. The reason is simply that, mathematically, there is a simple criterion that corresponds to "zero proper acceleration" or "free fall"--it is that the worldline followed by the object has zero covariant derivative. (DaleSpam went into this a bit in one of his posts.) And, when worldlines of actual objects moving solely under the influence of gravity are observed and analyzed--for example, planets, moons of planets, asteroids, comets, satellites orbiting the Earth, spacecraft sent out into the solar system, etc.--all of them, without exception, meet that mathematical criterion. In other words, they are exactly the worldlines (those with zero covariant derivative) you would expect the objects to follow *if* they were truly in free fall, experiencing zero proper acceleration; but you would *not* expect them to follow those worldlines if they had *any* nonzero proper acceleration at all.

Therefore, by induction from many, many, many examples, I believe that the two idealized point-like objects in the scenario I described would, in fact, have exactly zero proper acceleration, and accelerometers attached to them would in fact read exactly zero. If you actually agree with that, then I'm stumped as to how you're working out the implications--but that's the theoretical issue I talked about, which I don't want to get into until we've got the factual issue cleared up.

If, however, you do *not* agree that the accelerometers attached to the two point-like objects would read exactly zero, then I can understand why you've been saying what you've been saying, but I still think you've got an issue you haven't deal with. After all, even if you think the accelerometers would read something other than exactly zero, the experimental facts, as I noted above, are abundantly clear that "freely falling" objects--objects moving solely under the influence of gravity--follow the *worldlines* that we predict them to follow based on the mathematical criterion of zero covariant derivative. So you would basically have to maintain that the physical interpretation of that mathematical criterion is not what everyone else thinks it is: that an object can follow a "zero proper acceleration" worldline (mathematically speaking), but still, physically, feel a non-zero acceleration.

The problem with that, however, is twofold. First, I have no idea how you would go about predicting what acceleration an object should feel. On standard GR, the answer is simple: just compute the covariant derivative along its worldline. This answer also agrees with experiment: we can use it to correctly calculate your weight, for example. But that rule, of course, tells you that a "freely falling" object, moving on a worldline with zero covariant derivative (like every orbiting satellite, like every planet, like the Moon, etc., etc.), feels exactly zero acceleration--and that's what you appear to be disputing. So what rule would you substitute, to replicate all the known experimental facts but still come up with the answer that the two idealized point-like objects in my scenario above would somehow feel a non-zero acceleration?

I do believe that the accelerometers in your example would read a non-zero amount of proper acceleration because the tidal force they would be experiencing would be increasing, so they would be accelerating. You can have objects in free-fall that feel zero proper acceleration though, in an idealised setting. They would have to have a perfectly circular orbit so that the distance between them and the source of the gravity, and the amount of tidal force they experience remains constant and in fact cancels itself out if they're perfectly spherical objects.

Second, if you agree that GR predicts the *worldlines* of objects correctly, then what acceleration they feel is actually irrelevant, because the mathematical criterion I gave to single out "freely falling" worldlines--zero covariant derivative--is still sufficient to show that SR is invalid. Just substitute "zero covariant derivative worldlines" for "freely falling worldlines" everywhere in what I posted before. SR still predicts that two worldlines with zero covariant derivative that are mutually at rest (i.e., parallel) at anyone instant of time, will remain mutually at rest forever. In the presence of gravity, that is false; the two idealized point-like objects are a counterexample, and you agree they will separate (i.e., that their worldlines, despite both having zero covariant derivative, will not remain parallel even though they start out that way). But that is getting into the theoretical issue again, which I don't want to pursue further until we've got the factual issue taken care of. (Though I have to comment that if you don't believe that objects moving on worldlines with zero covariant derivative feel zero acceleration, then I'm stumped as to how you are able to apply SR, since zero covariant derivative is just another way of saying "inertial frame" in SR, and the only physical way to pick out inertial frames is to pick out objects that feel zero acceleration.)

SR predicts that two objects under the influence of different amounts of acceleration will separate. They can separate because one is closer than the other to a gravitational source, and so one is always undergoing a higher amount of acceleration. SR models it just fine.

(One other note about the scenario. If the fact that I specified that the two objects were artificially held at rest before being dropped makes things messy, then I can revise the scenario to eliminate it as follows. Consider two idealized, point-like objects, which are both moving upward in the Earth's gravitational field in such a way that they both come to rest for an instant, at the same instant--"same instant" in the frame in which they are both mutually at rest for that instant--at slightly different heights. In this scenario the objects are moving solely under the influence of gravity for the whole time under consideration, without any external forces. The same conclusion follows as before: the worldlines are parallel at the instant when both objects are mutually at rest, but they are *not* parallel for their entire extent, as SR would predict they should be. So SR doesn't work in the presence of gravity.)

One addition to my last post, to amplify on why physicists believe "zero covariant derivative" implies "zero proper acceleration". As I noted, in SR, "zero covariant derivative" worldlines are just the "inertial" worldlines--the worldlines that bodies follow when they are at rest in an inertial frame. And those bodies, physically, are the ones that feel exactly zero acceleration; that's the only way we can link up the theory of SR with actual physical facts

You can model acceleration with SR, so you can model gravity with it as well. It's just easier using GR because it looks at it as the space-time between objects being curved rather than the movement of the objects themselves. Ironically GR is literally just a generalisation of SR than modelling each objects acceleration individually.

But then, if in the presence of gravity, "zero covariant derivative" no longer equals "zero proper acceleration", then we would have a way to violate the equivalence principle. We know, experimentally, that objects moving solely under the influence of gravity move on zero covariant derivative worldlines. So if I'm floating in a box somewhere, with no view of the outside world, and there are no forces on me that I can observe (except possibly gravity, but by hypothesis I can't directly observe gravity without looking outside the box), I can tell by using an accelerometer whether the box is floating at rest deep in empty space (accelerometer reads zero) or is freely falling in a gravitational field (accelerometer reads nonzero). But the equivalence principle says that there is no local way to tell free fall in a gravitational field from floating at rest deep in empty space. So if we believe the equivalence principle, then "zero covariant derivative" has to equal "zero proper acceleration" or "accelerometer reads exactly zero".

You would be able to tell that you're accelerating if you're falling towards the source of the gravity, but not if you were in a circular orbit.

In the limit where gravity is weak (i.e., very far away from the hole), yes. Approaching and at the horizon and inside, no, at least not the way I think you mean "acceleration of the river" (i.e., acceleration relative to the "river bed"). That acceleration becomes infinite at the horizon, but as I've already pointed out, the tidal force at the horizon is finite (it goes like the inverse square of the mass), and the river model covers the horizon and the portion inside, so it is consistent with these predictions, meaning that the tidal force is not, in general, equal to the "acceleration of the river" in this sense.

I don't agree with this. I don't see how the tidal force could possibly anything other than infinite at the horizon because anything that made it that far would be moving at infinite speed relative to everything else. When you calculate the proper time and tidal force to be non-infinite you're overlooking the fact that the time dilation and length contraction are infinite at this range, which is why you get an event horizon in the first place. you can substitute any of those infinities with c because it's the same thing.

How are you reaching that conclusion? The only way I can see that this would follow is if you assume that the freely falling object must, in principle, pass a hovering observer at every point on its worldline. That is just another version of the mistake you've been making, which I've pointed out before. In the standard GR model, that assumption is false; there are no hovering observers at or inside the horizon, so the freely falling object doesn't pass any for that portion of its worldline.

No. A freely falling observer *can* in principle pass a hovering observer at every point on its worldline, which can't extend to the horizon. There doesn't need to be any observers at or inside the horizon since this is impossible anyway.

If you have an argument, starting from premises we all accept, for why the assumption *has* to be true, then by all means post it; but you can't just assume it. You have to prove it.

You want more? Do I have to do everything?