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

[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(<br /> \begin{array}{cccc}<br /> -x^2 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \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(<br /> \begin{array}{cccc}<br /> -x^2 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \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?

 

[Dale]

I worked it out ages ago.

Then please post your work.

The biggest problem with maths is that it can work fine on paper but be completely impossible in reality.

I agree and so does every scientist I know of, particularly physicists. That is why we value experiments so highly and why we continually strive to test and re-test every aspect of our equations to ever greater precision.

Thus far, the GR equations have passed all tests to date, so any alternative theory you would like to invent will need to reduce to the GR values for each experiment's conditions.

 

[PeterDonis]

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.

Ok, this clarifies where you're coming from. I wish we'd been able to get to this point a few hundred posts ago. That does lead to a couple of further questions, though.

First, do you have any way of quantitatively predicting what the accelerometers in my example should read, or do you just wave your hands and say, "Well, it's got to be smaller than whatever the most accurate accelerometers can read today"? I ask because, although nobody has done the precise experiment I described as far as I know, experimenters have certainly attached accelerometers to single objects in free fall (e.g., in orbit in the Space Shuttle), and those have read zero to within the precision of the instruments. So it seems to me that you would have to take the position that, whatever nonzero acceleration your model predicts, it must be smaller than the smallest acceleration that today's accelerometers can measure. In other words, is your model basically not empirically testable at our present level of technology?

(Btw, the orbits followed by satellites such as the Space Shuttle are never perfect circles, so by your reasoning there should be *some* nonzero acceleration that could in principle be read. I have no idea how your model would predict quantitatively what it would be, though.)

Second, even though you disagree with me (and everyone else who studies gravitation physics) about what the accelerometers in my example would read, do you agree that, even if GR predicts (incorrectly, by your reckoning) a zero acceleration reading on those accelerometers, it predicts correctly the worldlines that the objects in my example would follow? I don't see how you can disagree, since GR's predictions on this point have been verified countless times, but I'd like you to confirm that you agree. I'm going to assume you do for the rest of this post.

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.

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.

There's a problem here that you haven't considered. On your model, the "shape" of a worldline (its covariant derivative) is no longer equivalent to the acceleration measured by an accelerometer following that worldline. How, then, does your model predict the "shape" of an object's worldline? You can't just say "by its acceleration" any longer, because your premise is that they're not connected. But all the SR modeling you're referring to assumes they are; it assumes that the acceleration measured along a worldline is equal to its covariant derivative. For example, the SR prediction that objects under different amounts of acceleration will separate depends on that connection; if you remove that connection, the prediction is no longer valid, and your model can no longer rely on SR's modeling of acceleration. So what justifies your even using SR at all in your model? It seems to me that you need to start entirely from scratch, throw away SR (and GR), and develop your own account of how all this works--which, by the way, will still have to match all the experimental facts (like the fact I referred to in a previous post, that using the covariant derivative along a worldline to predict the acceleration measured is experimentally confirmed to work for accelerations we *can* measure, so it seems reasonable to predict that that method will still work for accelerations too small for us to measure, including the case of zero covariant derivative = zero acceleration).

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.

As I've already said many times, the GR calculations I've been referring to all already take into account all of the "time dilation" and "length contraction", and the answers I'm giving are all *after* all that has been taken into account. You don't understand how those calculations work, mathematically, so you're just going to have to take my word for it that I (and everyone else in gravitation physics--I've checked my work against multiple sources) am doing the math correctly. You may not agree with the result, but that's because you don't even agree on the basic physical fact that objects following freely falling worldlines experience exactly zero proper acceleration (accelerometers read zero), so from your point of view the math is starting from an incorrect premise, so it will give an incorrect result even if all the intermediate steps are done correctly.

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.

You want more? Do I have to do everything?

Not everything; just justify the assumption you just made in the above quote (again) without any justification. If you can prove it, fine. But you can't just assume it, because it is logically possible for it to be false, and assuming its truth is logically equivalent to assuming that the spacetime must end at the horizon. So you can't use this reasoning to prove that the spacetime ends at the horizon, because you're assuming what you're supposed to be proving. You have to find a starting assumption that is *not* logically equivalent to your conclusion.

 

[PeterDonis]

A-wal, one other question I just thought of. In an earlier post, you said:

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.

Actually this raises two questions (which may be connected, I'm not sure). First, what do you mean by "if there was no tidal force then they couldn't be free-falling"? I don't understand what that means, even in the light of your later post about what the accelerometers read.

Second, what does your model predict for the idealized case of a spacetime with absolutely no gravity present? What would an idealized point-like accelerometer, floating in empty space in such a spacetime, with no force acting on it, read, according to your model? Would it read exactly zero?

 

[PeterDonis]

experimenters have certainly attached accelerometers to single objects in free fall (e.g., in orbit in the Space Shuttle), and those have read zero to within the precision of the instruments.

Just to expand on this a little more, we have data about free-falling objects that are not on nearly circular orbits as well--for example, the Apollo spacecraft transiting to and from the Moon. The astronauts and everything else inside were weightless to within the precision of their measurements (I don't know for sure that there were actual accelerometers on board, but I would think there were, since NASA instrumented everything they could think of--certainly the astronauts reported *experiencing* weightlessness). It seems to me that your model would predict that those astronauts should have felt some non-zero acceleration, but again I have no idea how your model would predict quantitatively what it should have been.

 

[PeterDonis]

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.

And this brings up still another question. In SR, it's impossible for an object that feels zero acceleration to move in a circular orbit. It has to move in a straight line. Yet you say you can model gravity entirely with SR. How can that possibly work in the light of the above?