Black holes

[sol invictus]

No, it isn't a yes or no answer. Here's your question again:

"So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?"

The metric isn't something that actually exists, and whilst you employ it to describe spacetime, that doesn't actually exist either. What actually exists is space and light moving through it. Or not, as the case may be. And when it isn't moving, the observer sees nothing.

You're evading the question. Do observers see light moving at r=r0? At r<r0? Does time pass in those regions? Is there anything unusual or special about r=r0? Yes or no?

 

[Farsight]

How exactly do you define which way a beam of light points? That's not a trivial question. I suspect you haven't actually really thought about the answer.

It isn't a matter of definition, you just point it, and the beam shows you the direction. You're then free to change that direction in relation to for example the surface of the earth. If the earth is a long way from you you aim a laser at a point on the surface where the reflection comes straight back to you, then both the beam and its reflection are vertical with respect to the surface of the earth. Like, you aim for the middle.

Note that hilighted word: the coordinate speed of light varies with gravitational potential in Schwarzchild coordinates. It does not vary in Kruskal coordinates. You are favoring one set of coordinates over every other set.

I'm not. I'm looking at hard physical evidence that says the speed of light varies with gravitational potential regardless of any coordinate system. There is no coordinate system you can use to make the light traversing the lower parallel mirror get to the end before the light traversing the upper.

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I can get infinite time dilation for distant observers even without gravity. That really is no problem at all.

No you can't.

Every single thing you describe here applies to the event horizon created by acceleration.

No it doesn't. I can see you traversing the sky, but I cannot see your event horizon. It isn't something that has actually been created. Like I said, it isn't in the same league as a black hole.

You have made many fundamental mistakes. Including a number below, which I will describe to you... You are wrong on many, many counts. First off, you can't ever see a horizon, not even for a black hole.

But with a suitable telecope, I could see gravitational lensing around a black spot, and I could also see tight-orbiting stars from which I can estimate a mass. When I look at you zipping through space, what I see is you, I do not see your Rindler horizon.

Second, the Rindler horizon isn't a cone, it's a plane.

My mistake. Apologies.

Third, the stars you see behind your accelerating observer are the exact same stars that he sees (though he sees them redshifted).

I'm sitting here on earth watching the accelerating observer traverse the sky. The light from the stars in the background is not affected by his passage.

The light that you now see was emitted long ago, before they crossed the Rindler horizon. The Rindler horizon moves at the speed of light in your non-accelerating frame, and so you won't see light from stars on the other side of the Rindler horizon until you've crossed the Rindler horizon yourself.

I'm sitting here on earth. Perhaps you misread the scenario.

And that's exactly what will happen with a black hole. Drop a probe in and fall in after it, and you will NEVER lose sight of the probe.

Because I'll end up looking at it forever with my cold dead popsicle eyes, only I'm not really looking because all electromagnetic processes have ground to a halt.

In fact, if you pick the right trajectory, you won't even see it get red shifted.

No problem.

And lastly, you're even wrong about lensing: light WILL bend around a Rindler horizon, in the accelerating reference frame. The horizon doesn't have the same shape, so it won't the the same sort of lensing, but you'll get equivalent bending of the trajectory of light.

No you won't. Your rapid motion across the sky does not alter the motion of beams of light around you. Your observation of them changes, but they don't change.

You may have heard about Rindler horizons, but you clearly don't understand them.

I understand them enough.

 

[Farsight]

You did not answer sol invictus's question.

Here is the metric:

The actual question is

Or better yet: At what rate will a clock at r=0 be observed by an external observer to be ticking?

I thought he was referring to the black hole event horizon. I've just looked back at the posts, and it still sounds that way.

ETA: Hmmn, OK, maybe he wasn't. I'll look into it.

 

[sol invictus]

I thought he was referring to the black hole event horizon. I've just looked back at the posts, and it still sounds that way.

ETA: Hmmn, OK, maybe he wasn't. I'll look into it.

As my posts made quite clear, I was referring to the event horizon at r=r₀ in the metric I posted. I brought that up because it has all of the characteristics of black hole horizons that you refer to. I'm still waiting for you to answer some simple, yes/no questions about it.

 

[Ziggurat]

It isn't a matter of definition, you just point it, and the beam shows you the direction.

It's very much a matter of definition. You have defined it implicitly (though not very well), but you've still defined it. So let me make your definition explicit: a beam of light is pointed at an object if the world line of the emitted light intersects the world line of the object. Do you have an objection to this definition?

So what happens at the event horizon with this definition? Well, we find that at the event horizon, we cannot point "outwards" any more than we can point backwards in time. But the fact that you can't shine light backwards in time doesn't mean it's stopped.

I'm not. I'm looking at hard physical evidence that says the speed of light varies with gravitational potential regardless of any coordinate system.

The speed of light is zero at the event horizon in Schwarzchild coordinates. It's constant across the event horizon in Kruskal coordinates. That's what GR tells us. So either you're wrong, or GR is wrong. Those are the only two options. What will it be, Farsight?

There is no coordinate system you can use to make the light traversing the lower parallel mirror get to the end before the light traversing the upper.

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You're wrong. The events in question are space-like separated, so we are free to choose coordinates which will order the events in whatever manner we choose. Furthermore, it doesn't even mean what you think it means. You have once again chosen a privileged coordinate system, without even realizing it. The two ends of the mirror are stationary in the Schwarzchild coordinates, are they not? But they are NOT stationary in other coordinates. In fact, they aren't even moving at the same rate in other coordinates. Which means that the distance that light travels between the two mirrors is also not the same. This is exactly the issue which pops up in the classic special relativity problem of vertical mirrors inside a moving train: the clock on the train runs slower, but light doesn't slow down, because the path is longer.

No you can't.

Yes, I can. I already told you how.

No it doesn't. I can see you traversing the sky, but I cannot see your event horizon.

That's because it's moving at c. You can never see something approach if it's moving at c until it reaches you. And from beyond the event horizon, the event horizon will also always be invisible, since light CAN pass through the event horizon in that direction.

It isn't something that has actually been created. Like I said, it isn't in the same league as a black hole.

There are certainly differences. But everything you object to about time stopping at the event horizon also applies to the case of acceleration.

But with a suitable telecope, I could see gravitational lensing around a black spot, and I could also see tight-orbiting stars from which I can estimate a mass. When I look at you zipping through space, what I see is you, I do not see your Rindler horizon.

So what? The accelerating observer sees lensing. and even with a black hole, the lensing one observes changes with reference frame. One cannot cancel it globally because gravity is not uniform, but the essence of the event horizon problem remains the same.

My mistake. Apologies.

I'm not looking for an apology. I'm looking for you to learn something. Which you're apparently willing to do, provided it doesn't conflict with certain beliefs of yours. But when it conflicts with certain beliefs, then you refuse to learn.

I'm sitting here on earth watching the accelerating observer traverse the sky. The light from the stars in the background is not affected by his passage.

And if you and I fall into a black hole together, the light we exchange won't be affected either.

I'm sitting here on earth. Perhaps you misread the scenario.

I read your scenario perfectly. But you have not grasped what I said. The Rindler horizon moves at c in your earthbound reference frame. Before that horizon reaches you, it is impossible to see any light emitted from beyond that horizon, for the rather simple reason that light can't outrun the horizon.

Because I'll end up looking at it forever with my cold dead popsicle eyes, only I'm not really looking because all electromagnetic processes have ground to a halt.

That's not what GR says will happen.

No you won't. Your rapid motion across the sky does not alter the motion of beams of light around you. Your observation of them changes, but they don't change.

Yes, they do. When you accelerate, the trajectories of light become curved. This is an observation, but you seem to have forgotten what observation means in relativity. It doesn't mean simply what you see. It means what things really are in your reference frame.

I understand them enough.

No, you really don't, because you keep screwing up.

 

[Farsight]

I can't stop, but the wife's on the phone for a minute and I can't resist this:

Yes, they do. When you accelerate, the trajectories of light become curved. This is an observation, but you seem to have forgotten what observation means in relativity. It doesn't mean simply what you see. It means what things really are in your reference frame.

You're in space, I'm on earth, and I can see you. I can also see a beam of light orthogonal to you. I see you accelerate, and guess what? That beam of light doesn't change one jot. You only see it as curved because you're accelerating. And your reference frame doesn't exist, so don't try to tell me what things really are in it. Oooh look, there's Zig up in the sky with a rectangle around him.

SLAP!

Wake up Zig. Learn to distinguish reality from confection. You're moving that's all. You see things different when you move, because light moves too.

 

[Ziggurat]

I can't stop, but the wife's on the phone for a minute and I can't resist this:

You're in space, I'm on earth, and I can see you. I can also see a beam of light orthogonal to you. I see you accelerate, and guess what? That beam of light doesn't change one jot.

In your reference frame it doesn't. In mine it does. Welcome to relativity. That word is there for a reason.

You only see it as curved because you're accelerating.

And being "stationary" in a gravitational field is the same thing as accelerating as well. Gravitational lensing isn't any different than the lensing due to acceleration. That's the equivalence principle, the foundation of general relativity. If you want to argue that the equivalence principle is wrong, go ahead. But you can't do so while at the same time insisting that you're the only one here who correctly understands general relativity.

And your reference frame doesn't exist

Yes it does. It exists as much as your reference frame, without which you can't even claim that light ever goes straight.

so don't try to tell me what things really are in it.

Perhaps you're right, I shouldn't try to tell you what things really are, since you refuse to learn. Perhaps this really is all just a waste of my time.

Wake up Zig. Learn to distinguish reality from confection. You're moving that's all. You see things different when you move, because light moves too.

Again, they don't call it relativity for nothing.

 

[Reality Check]

One more time: Bob is a popsicle, and has been that way for a billion years. He isn't measuring anything.

And one more time - you are wrong.
Alisc is not a popsicle - she is an observer.
Bob is not a popsicle - he is an observer.
I will make this post:

Originally Posted by Reality Check
One more time:
Alice decides to use Schwarzschild coordinates and she measures that there is a singularity at the event horizon.
Bob decides to use Kruskal–Szekeres coordinates and he measures that nothing special happens at the event horizon.

even clearer:
Let there be an observer Alice who is using Schwarzschild coordinates.
Let there be an observer Bob who is using Kruskal–Szekeres coordinates
Neither observer is falling into the black hole. They are both observing the black hole from outside.
Alice measures that there is a singularity at the event horizon.
Bob measures that nothing special happens at the event horizon.

If Alice or Bob were falling into the black hole then GR states that they measure nothing special at the event horizon.

 

[Reality Check]

And your reference frame doesn't exist, ....

Farsight, are you really stating that you are the only observer in the universe with a reference frame ?

I have a reference farme.
I am in a gravitational field and I have chosen to use Kruskal–Szekeres coordinates. I observe that "the metric is perfectly well defined and non-singular at the event horizon".

 

[Octavo]

I'm quite sure this is off-topic and perhaps it is already well known, but I couldn't help mentioning stumbling across this when looking up Mr. Duffields credentials...

http://www.amazon.co.uk/RELATIVITY-Theory-Everything-John-Duffield/dp/0956097804

 

[tensordyne]

I am not sure how this fits in with the discussion thus far, but I could not help but disagree with the following statements.

The metric defines both the coordinates and the spacetime, and it lets you predict the results of all experiments.

The metric does not define coordinate system and space-time (or more appropriately, the geometry of a space-time), the metric is defined by which coordinate system and geometry of space-time one is considering. Coordinate systems (up to dimension and locally speaking) are independent of geometry. This is one of the main messages of GR, "your geometry and laws should be independent of coordinate system."

I find the claim that metrics let you predict the results of all experiments very unrealistic. There are lots of types of fields (scalar, vector, etc.) that might go into determining a metric as per Einstein's Field Equations. Going backwards from the metric to telling what the fields are, well, I do not see how this would be done.

One could determine the stress-energy tensor from the metric tensor, but there are many fields that might effect the form of the stress-energy tensor. When a field is substituted into the stress-energy tensor, I do not see any formal method in which one would then be able to go back to what the field is, just given the metric tensor.

The formalism of Quantum Mechanics does say that once one has the wave-function, one should in theory be able to predict any observable result. As far as I know, no similar claim is part of the current consensus interpretation of GR as per the metric performing a similar role in that context.

It is true that the metric touches upon just about every geometric entity one learns about in multivariate calculus.

 

[sol invictus]

The metric does not define coordinate system and space-time (or more appropriately, the geometry of a space-time)

Look at that metric. See those letters, like "t" and "r"? Those are called coordinates. As for spacetime geometry, one can compute everything about it from the metric (plus any boundary conditions or identifications, if any). It's true that there are infinitely many other metrics that describe the same spacetime in different coordinates, but any one is enough to find the spacetime geometry.

the metric is defined by which coordinate system and geometry of space-time one is considering. Coordinate systems (up to dimension and locally speaking) are independent of geometry. This is one of the main messages of GR, "your geometry and laws should be independent of coordinate system."

That's kind of true (although we're not "locally speaking", we're talking about the full spacetime), but it doesn't contradict anything I said - particularly if you don't rip it out of context.

I find the claim that metrics let you predict the results of all experiments very unrealistic.

Again, context. We were discussing experiments with light and clocks in spacetimes that are vacuum solutions to Einstein's equations.

 

[W.D.Clinger]

I'm quite sure this is off-topic and perhaps it is already well known, but I couldn't help mentioning stumbling across this when looking up Mr. Duffields credentials...

http://www.amazon.co.uk/RELATIVITY-Theory-Everything-John-Duffield/dp/0956097804

I think it's mostly off-topic, and would be entirely off-topic were it not for Farsight's habit of making claims backed solely by his own authority. I suppose those otherwise unsupported claims could be regarded as implicit references to his credentials.

In any case, his book has already been discussed. On 9 March, 2009, Farsight initiated a thread devoted to discussion of that book and his (John Duffield's) ideas.

In the first two posts of that thread, Farsight laid out a thesis that suggested John Duffield's knowledge of physics fell short of Maxwell's equations and the Lorentz force. By the eighteenth and (so far) final page of that thread, John Duffield's knowledge of physics and mathematics was being summarized by posts like this:

Just look at the images for electromagnetic spectrum. Pick any one at random. Look at the sinusoidal waveform. Look at the wave height. That's the amplitude. It's always the same regardless of frequency.

[qimg]http://www.antonine-education.co.uk/physics_gcse/Unit_1/Topic_5/em_spectrum.jpg[/qimg]

This thread just keeps getting better. It seems Farsight actually believes that all electromagnetic waves have the same amplitude because he's seen pictures explaining the idea of a spectrum of frequencies to school children. Just when you thought the ignorance couldn't get any more hilarious.

 

[tensordyne]

Look at that metric.

Using the imperative when making an argument is not good form. A detached voice is much more neutral in tone and much less hostile. "Consider the metric..." would be much more appropriate.

See those letters, like "t" and "r"? Those are called coordinates.

They are first and foremost variables. When one attaches the further meaning that each variable is part of a coordinate system by being a coordinate label, then one should say that t and r are not coordinates but that they are coordinate values or coordinate variables. Coordinates give a full specification for the location of a point. I do not see how "t" and "r" specify a point in 4D space-time for instance, therefore I disagree at this point that they are coordinates.

As for spacetime geometry, one can compute everything about it from the metric (plus any boundary conditions or identifications, if any). It's true that there are infinitely many other metrics that describe the same spacetime in different coordinates, but any one is enough to find the spacetime geometry.

For the record, here is the original quote:

The metric defines both the coordinates and the spacetime, and it lets you predict the results of all experiments.

With the metric one can in principle determine all geometry only related quantities from it, such as areas, volumes, test-particle paths, etc. There are lots of other things to predict experimentally that the metric alone would not give an answer for. Unless the context of "results of all experiments" is given to eliminate many experiments, the second part of the sentence above must be wrong.

A metric is a series functions that one can put into a square-shaped array. These functions are dependent upon a set of variables called coordinates. These coordinates can be thought of in various levels of abstraction related to how abstractly we are considering the functions.

One can either attach to each coordinate label a further meaning of how it is to be determined constructively speaking (take a ruler and a watch and then...), or not do so. If the set of coordinate labels is given this further meaning, it is a coordinate system. If one or more coordinate labels is not given a method for determination, such a system is an abstract coordinate system.

Since we are speaking about physics, unless one gives a constructive meaning to ones coordinates, there is no way to match up the coordinates with anything that might be measured because at least one of the "coordinates" will merely be a label.

There are also types of coordinate systems (not abstract coordinate systems as previously defined). A type of coordinate system is a common method of determining coordinates that does not take into consideration where the coordinate system will be used. Examples include polar, cylindrical, etc. etc.

If we take a particular coordinate system type and apply it about some point then there are many possibilities for what can be considered. Experimentally one could use the coordinate system to determine the metric, speed of some craft, etc. etc. Theoretically one could make some prediction that could subsequently be tested.

On a final note, abstract coordinates are good for stating laws of nature since they do not specify how coordinates are determined, they are free of dependence on any type of coordinate system. The first part of the sentence above ("The metric defines both the coordinates and the spacetime,") I still find troubling, even with the context as given below.

You can have a metric in terms of abstract or concrete coordinates. If one has the metric in terms of abstract coordinates, then there is no way of visualizing what an expression with a bunch of coordinate variables means observationally. If the metric is in terms of a concrete coordinate system (as it is in Schw. Sol. as per Schwarzschild Coordinates), then a specific metric is determined by both what coordinate system one is using and what the geometry is.

Case in point.

ds² = dx² + x² dy².

It is only convention that makes one think that x and y in the above case are Cartesian. To be safe though, until this is specified, the x and y are merely abstract labels. If I say that x and y are determined the same way as r and theta are in polar coordinates, then you would know what the geometry would be like. If on the other hand I were to say that x and y are Cartesian, then the geometry would be different than plane Euclidean.

The point is, the form of the metric is determined by the type of coordinate system used and the geometry present in a given patch of space where that coordinate system is employed. Conversely however, if one has the metric and what the coordinate system the metric is given in, then one can use that to help in understanding the geometry.

The metric helping to understanding the geometry is not the same as the metric determining the geometry. The geometry is independent of what coordinate system you put on it. In terms of functions the metric (in terms of the context given thus far) is dependent on what the geometry is and what coordinate system you use.

That's kind of true (although we're not "locally speaking", we're talking about the full spacetime), but it doesn't contradict anything I said - particularly if you don't rip it out of context.

Again, context. We were discussing experiments with light and clocks in spacetimes that are vacuum solutions to Einstein's equations.

Thank you for providing context in this case. I hope you will agree the statement I quoted originally by itself is most definitely wrong, as I agree that in the context just given, since there are no other fields to consider, the metric suffices to determine all of the experimental outcomes (as far as I know), within such a context.

A note on "locally", I included it without knowing the context to cover my bases. In general, coordinates are usually not global except in very simple cases (such as in the Schw. Sol.).

 

[DeiRenDopa]

I'm coming late to this thread, and to all threads in which Farsight's ideas are proposed and discussed. Apologies for going over ground that has, undoubtedly, been gone over before.

It's based on the hard scientific evidence that the speed of light varies with gravitational potential,

What is this "hard scientific evidence"?

I'd like to hear from Farsight, of course, but if others have already asked this of him, and can point to other posts in this forum where the topic is discussed ...

and a clear understanding that we define the second and the metre using the motion of light and so always deem the local speed of light to be 299,792,458 m/s,

I didn't think this is an accurate summary, so I went and consulted NIST. I found this:

The meter is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second.

Note that the effect of this definition is to fix the speed of light in vacuum at exactly 299 792 458 m·s^-1.

And this:

Considering that a very precise definition of the unit of time is indispensable for the International System, the 13th CGPM (1967) decided to replace the definition of the second by the following (affirmed by the CIPM in 1997 that this definition refers to a cesium atom in its ground state at a temperature of 0 K):

The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom.

So making an allowance for sloppy language ("using the motion of light" is not quite the same as "the length of the path travelled by light in vacuum", but close enough), Farsight is right about the meter.

But I can't see how he can possibly be right re the second (the unit of time), even allowing for sloppy language.

In any case, his stated conclusion (we "always deem the local speed of light to be 299,792,458 m/s") is correct (if not well expressed).

that clocks clock up regular cyclic motion rather than "the flow of time", and by reading the original Einstein to understand that that gravitational time dilation is the result of a reduced rate of motion caused in turn by a concentration of energy "conditioning" the surrounding space.

I have no idea what this is supposed to mean; does anyone (even Farsight)?

And what's the relevance of "the original Einstein"? I mean, he came up with GR, and published a paper or two on it. But he's not a god; his word is not inerrant.

From that I say that if gravitational time dilation goes infinite, the rate of motion goes to zero, and I test this by asking why a vertical light beam, which does not curve or fall back or decelerate, does not escape a black hole. Answer: because it's stopped.

Is it just me (that does not understand what Farsight is trying to say), or has he displayed a rather gross misunderstanding of relativity?

If nothing else, Farsight seems to be trying to have his GR cake and eat it.

"gravitational time dilation" is an effect you can derive from GR, and it is unambiguous. You can do experiments to test this GR prediction, and as far as I know, every such test has produced results consistent with GR (to the experimental uncertainties).

To understand what GR predicts concerning the observed behavior of light near black holes, you need to first understand GR (duh!). While the answers may be somewhat tricky to work out, and there will certainly be some subtleties, not least because the mathematical framework that GR is expressed in is not intuitive), they will nonetheless be unambiguous.

But Farsight seems to be introducing his own ideas - beyond GR - and mixing them in, without making any attempt to distinguish the two.

What am I missing?

"Spacetime manifolds are mathematical objects. We can prove their mathematical existence follows from the laws of logic and the axioms of mathematics, but we cannot use mathematics alone to prove a spacetime manifold accurately describes the physical universe. Whether something exists in a physical sense is a question for science, not mathematics".

What is this a quote from?

And in any case, what does "exists in a physical sense" mean, if not something to do with the results of (objective and independently verifiable) experiments and observations?

And if nothing else, then where are the experimental and observational results showing inconsistency with GR?

And please note that I've mentioned vacuum impedance before now.

What is Farsight referring to?

 

[sol invictus]

Case in point.

ds² = dx² + x² dy².

It is only convention that makes one think that x and y in the above case are Cartesian. To be safe though, until this is specified, the x and y are merely abstract labels. If I say that x and y are determined the same way as r and theta are in polar coordinates, then you would know what the geometry would be like. If on the other hand I were to say that x and y are Cartesian, then the geometry would be different than plane Euclidean.

Wrong. The only missing information is whether or not there are any identifications (like theta being identified modulo 2 pi) on x and y. Assuming there are no such identifications, that metric fully specifies the geometry, or at least a finite part of it. There is no need to state whether x and y are "Cartesian" or "polar".

I hope you will agree the statement I quoted originally by itself is most definitely wrong, as I agree that in the context just given, since there are no other fields to consider, the metric suffices to determine all of the experimental outcomes (as far as I know), within such a context.

You agree it was correct in context, but you think it was "most definitely wrong" when quoted by itself. Since you're the one that quoted it by itself without context, that makes any incorrectness entirely your error and your responsibility.

 

[Farsight]

I said I'd get back to you on Sol's expression and questions:

I've already given you an example. Instead of "knocking it down", you ignored it. Here it is again:

[latex]$ds^2 = -(r-r_0) dt^2 + dr^2/(r-r_0) + dy^2 + dz^2$[/latex]

The coordinate speed of light is zero at r=r0. Clocks (including light clocks) run slower and slower if they are held at fixed, smaller and smaller r. In fact, this metric has exactly the properties you keep referring to in the black hole.

Do you know what this metric describes, Farsight? It's not a black hole...

As I said in post #262, I didn't ignore it, and when he asked me this:

"So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?"

...over the weekend I thought he was talking about black holes. Anyway, starting from scratch with the standard expression for a spacetime interval in flat Minkowski spacetime:

[latex]$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$[/latex]

Here we have something related to Pythagoras' theorem, used in the Simple inference of time dilation due to relative velocity. We employ two twins each carrying parallel-mirror light clocks, called Zig and Zag. Zig goes off on an out-and-back trip whilst Zag stays at home with us. We observe the light in Zig's clock like this /\ and the light in Zag's clock like this ||. There's no literal time flowing in these clocks, merely light moving at a uniform rate through the space of the universe, tracing out straight worldlines through Minkowski spacetime. We know that Zig's total light-path-length is the same as Zag's. It is invariant, and is the underlying reality behind the invariant spacetime interval. Distant observers employing the CMB rest frame would assert that when Zig's macroscopic rate of motion through the universe is great his local rate of motion is less, though he of course is unaware of this.

Looking at sol's expression [latex]$ds^2 = -(r-r_0) dt^2 + dr^2/(r-r_0) + dy^2 + dz^2$[/latex], I don't recognise it. I thought it was a reference to the Rindler horizon. An inverse acceleration equates to a distance and the harder you accelerate the closer the Rindler horizon is to your rear, but I had a look on the internet but couldn't find the expression, so in response to this question...

"Do you know what this metric describes, Farsight? It's not a black hole..."

I have to say no, I don't know what it describes. I set rₒ to zero and said r²/r is the x term, a spatial distance, but that leaves me with a delta x rather than delta x², and an r multiplier on the delta -t². If go to the other limit and say r=rₒ I'm left with a zero delta -t² indicating travel at c and a division by zero giving me an undefined delta x akin to total length contraction, but I still can't give a positive answer. Hence when sol asks:

"So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?"

My answer is: I don't know.

 

[tensordyne]

Wrong. The only missing information is whether or not there are any identifications (like theta being identified modulo 2 pi) on x and y. Assuming there are no such identifications, that metric fully specifies the geometry, or at least a finite part of it. There is no need to state whether x and y are "Cartesian" or "polar".

I see that my careful exposition of terms and concepts was wasted. If I give you the metric in terms of some variables the geometry is determined in an abstract sense but not in a concrete way. Unless the geometry has a concrete meaning, I do not see how anyone can say they understand what such a geometry is really like.

The claim that one can understand what the geometry of a space is like when the metric is only given abstractly is false on its face.

You agree it was correct in context, but you think it was "most definitely wrong" when quoted by itself. Since you're the one that quoted it by itself without context, that makes any incorrectness entirely your error and your responsibility.

The first part of the quote is still wrong, the metric does not determine the coordinate system and the geometry. After the coma, assuming the context as given by you, it is correct (as far as I can tell as per Schwarzschild Solution context). Let's see, did I cover myself for this possibility:

I am not sure how this fits in with the discussion thus far, but I could not help but disagree with the following statements.

Yes, I did. I even thanked you for giving me it. Then I offered a method of understanding and conciliation. The response I got back was pretty immature, if I do say so myself.

If you can not understand why an abstract description does not allow for concrete understanding then that is not my problem. Maybe you are too stuck in abstraction to get my point. Either way, the metric does not determine the coordinate system used and the geometry, it only determines the coordinate labels being used (as well as other abstract relationships).

How are those coordinates measured? Unless one is told how coordinates are measured for specific cases in a given metric, I can guarantee you will not know what the geometry is really like, as I can introduce all sorts of screwy ways to measure an x, or a y, or a whatever. I guess that was not considered though by you perhaps?

Have a nice day.

 

[W.D.Clinger]

"So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?"

My answer is: I don't know.

That's the right answer.

It's also the right answer (from you) to the analogous question with regard to the Schwarzschild metric.

Some of us do know, and can give a definite answer (yes or no). We've done our homework.

Is it just me (that does not understand what Farsight is trying to say), or has he displayed a rather gross misunderstanding of relativity?

The latter.

"Spacetime manifolds are mathematical objects. We can prove their mathematical existence follows from the laws of logic and the axioms of mathematics, but we cannot use mathematics alone to prove a spacetime manifold accurately describes the physical universe. Whether something exists in a physical sense is a question for science, not mathematics".

What is this a quote from?

Farsight was quoting me, from my post #11 in the thread on "mathematics of black hole denialism".

And in any case, what does "exists in a physical sense" mean, if not something to do with the results of (objective and independently verifiable) experiments and observations?

The "mathematics of black hole denialism" thread is about the EU/Crothers claim that black holes (and the big bang) are incompatible with the mathematical theory of general relativity.

That's a purely mathematical claim, so I can refute it (and have) using mathematics alone, without relying on scientific experiments and observations. My purely mathematical refutation of that claim does not, by itself, prove that black holes exist in a physical sense. That's a question for science.

And if nothing else, then where are the experimental and observational results showing inconsistency with GR?

Farsight has no such results, but he's using a rhetorical tactic made famous by Michael Mozina: He's saying our mathematics may be right, but we don't understand the physics.

 

[Farsight]

In your reference frame it doesn't. In mine it does. Welcome to relativity. That word is there for a reason.

Not for that reason. Your reference frame does not actually exist. That light beam does not curve instantly the moment you accelerate towards it. You know that instantaneous action-at-distance does not occur. You know that the way you see the world is relative to your motion, not the way the world is.

And being "stationary" in a gravitational field is the same thing as accelerating as well.

It isn't the same thing. There's an equivalence, but the two situations are not identical.

Gravitational lensing isn't any different than the lensing due to acceleration.

Yes it is.

That's the equivalence principle, the foundation of general relativity. If you want to argue that the equivalence principle is wrong, go ahead. But you can't do so while at the same time insisting that you're the only one here who correctly understands general relativity.

It's the principle of equivalence, it draws a parallel, it doesn't say two situations are exactly identical. You don't need to do work to stand on the ground.

And your reference frame doesn't exist

Yes it does. It exists as much as your reference frame, without which you can't even claim that light ever goes straight.

It doesn't. Light exists. Space exists. You exist. But you are not surrounded by a real object called a reference frame. It's an artefact of your measurements, which you make using moving light, and those measurement change when you change your motion.

Perhaps you're right, I shouldn't try to tell you what things really are, since you refuse to learn. Perhaps this really is all just a waste of my time.

Ditto.

Again, they don't call it relativity for nothing.

It started out as the theory of invariance.