I am the first in the world

[Reality Check]

Because like sol said, coordinate systems are human conventions, they have no bearing on reality. I don't "obsess" with one of the infinite number of coordinate systems. ...usual rant snipped...

And you are quote mining because sol invictus actually said:

Originally Posted by sol invictus
That's exactly the point. Coordinate systems are human conventions, they have no bearing on reality. You can choose any coordinates you like - and it's only in the coordinate system you insist on always using that light slows down or time stops at the horizon. In an infinite number of other coordinate systems, all equally as valid as yours, nothing unusual happens at the horizon.

(enphasis added)

You are still using just this one coordinate system where there is an event horizon and ignoring the infinite number of coordinate systems where there is no event horizon, e.g. Kruskal-Szekeres coordinates.
That is an obsession.

 

[Reality Check]

I'm not.

You are wrong. You stated

Originally Posted by Farsight
In turn that tells me that the t=infinity is really a c=0.

Any one with a basic knowledge of mathematics knows that the "t=infinity" is wrong because infinity is not a number. You can never assign it to a variable.

Thus your "really a c=0" is also wrong.
Simple math and logic, Farsight.

You do get that c tends to zero correct. And like Zeno's arrow it never gets to zero. You thus are stating that you know that your "c=0" is wrong.

 

[ben m]

In turn that tells me that the t=infinity is really a c=0.

What clock is telling you t=infinity?

 

[ben m]

Because all the transformations in the world can't make a stopped clock tick.

Except that we've shown you this transformation repeatedly. The clock has not "actually stopped" (if that means anything at all); the fact that you think it's stopped is an accident of the clocks you chose to look at while asking "has the next tick happened yet".

 

[edd]

I think the only solution to this argument is if Farsight tests it for himself.

 

[W.D.Clinger]

Not so. There is no confusion on my part.

Bare assertion, contradicted by evidence.

I'm the one who's given the page from MTW where the Kruskal-Szekeres coordinates are is depicted, see post #135.

No, that page was given us by Misner, Thorne, and Wheeler circa 1973.

It may be new to you, but it isn't new to us. We did our homework years ago.

Since we also all know that vertical light emitted at the event horizon does not slow down and does not curve and does not escape a black hole, we can all work out that you haven't done your homework. And we can all see that you have evaded the issue and posed "exercises" instead.

Bare assertion, contradicted by evidence.

I worked through all of those exercises before I suggested them to you. If I hadn't done my homework, I wouldn't have been able to formulate those exercises.

You, however, have not done those exercises, nor have you done any similar homework. If you had done your homework, you wouldn't be making such elementary mistakes.

Please note that spacetime is static. There is no motion through spacetime, because it's an all-times "block universe" mathematical space. We draw worldlines in it, but objects do not move through it.

You must not understand what people mean when they say a spacetime is static.

Simplifying somewhat for the present audience, a spacetime manifold as described by some particular chart is said to be static if the metric's coordinate-dependent components are independent of the chart's timelike coordinate. It's a property of the chart as much as the spacetime.

In a Schwarzschild chart, for example, the spacetime submanifold on which the chart is defined is static (because all of the metric's coordinate-dependent coefficients are independent of t). In a Lemaître chart that's defined on that same submanifold, spacetime is not static (because the Lemaître metric depends on r, which depends on τ). The Schwarzschild and Lemaître metrics are equal wherever the Schwarzschild metric is defined (exercise 20), so whether you think a spacetime is static depends on your personal choice of coordinates.

No I'm not confused. I am crystal clear about this.

Bare assertion, contradicted by evidence.

Yawn. Spare me your intellectual arrogance and your you cannot hope to understand it erudition. Because I do.

Bare assertion, contradicted by evidence.

This is... banal. It's kid's stuff, Clinger.

It appears to have taken Einstein a couple of decades to understand that the Schwarzschild metric's singularity at r=2M is a mere coordinate singularity. He definitely understood this by 1935, as demonstrated by the Einstein-Rosen paper Vorpal quoted. That paper came two years after Lemaître's paper of 1933.

Now that Eddington, Lemaître, Einstein, Rosen, and many others have done the hard work of discovery, it's fair to say this stuff is elementary. As I have noted, anyone who's proficient in high school algebra and differential calculus can work through the exercises and come to understand what's wrong with your argument.

Somebody please spare me from this cargo-cult pseudoscience that traduces relativity.

I can't stop John Duffield from traducing relativity by promoting his cargo-cult pseudoscience. You could, but you haven't.

Clinger, you're parroting. Shut up, think it through. Then get back to me.

Bare assertion, unsupported by evidence.

If I'm parroting, without thinking it through, then you should be able to identify a source I'm repeating mindlessly.

In reality, I formulated the 23 exercises on my own, starting with little more than what's in the current Wikipedia article on Lemaître coordinates. A brief search (about 15 minutes) of my personal library and the web didn't yield much more in the way of information, so I worked through the calculations myself and drew the conclusions stated in the 23 exercises. That's an example of thinking it through.

Yawn. You're boring us to death. You haven't given a counter-argument, just exercises. They fool nobody. And pretence is what distinguishes quacks from sincere contributors.

Bare assertions, contradicted by evidence.

Not only have I given a counter-argument, I have summarized it in a list of 7 bullet items.

You, however, continue to give us many examples of what a counter-argument is not. Underlining the letter that distinguishes a Brit's spelling from an American's is not a counter-argument.

Note that ben has not replied, and has evaded my challenge to repeat another expression that I said I'd knock down like the rest of 'em. Instead he became abusive, calling me names. He's trying to blind you with mathematical smoke and mirrors. Don't be fooled by it.

It's no counter-argument at all. It's sophistry. If he had a counter-argument he'd be giving it in clear robust fashion that everybody could understand. Ain't gonna happen.

Bare assertions by the boatload, contradicted by evidence.

Guys: I smell desperation here, and sense capitulation. It's signalled by the evasion and abuse. I think one of you will crack and say I think Farsight's got a point soon. Then the floodgates will open, and then will come catharsis. You will have your OhMyGawd moment. It'll feel good, like the way you feel after a tear-jerker movie. And then I can rest easy knowing that you've got something out of this conversation. I know I have. Aw, it's my guilty pleasure. Yeah., I know I'm like a cage-fighter working out toddlers, but it's kinda fun all the same. LOL. Ciao for now!

Bare assertions by the boatload, contradicted by evidence.

 

[ben m]

Phooey. I said to ben your simple proof was wrong. When 2M/r = 1 at r = 2M you've got a division by zero in [latex]\[ \frac{dt}{d\tau} = \frac{e}{1-2M/r} \][/latex]. That's an undefined result, an infinity

I missed this until Clinger quoted it. Um, no I didn't. Why would I need to even compute dt'/dtau'---the ratio of two times according to some other observer---to find the trajectory of this observer? Why should "light I emit now will never get to observer X" interrupt me from obeying the laws of physics?

But I'm amused by the two sides to Farsight's argument.

a) Farside has one calculation in which something goes to infinity. Farsight: "There's an infinity there! I'm right"
b) Many other calculations do not have this infinity. But since (a) is true, that means there's an infinity there that they "skipped by" and are therefore wrong. Therefore (a) is true. "You don't have a counterargument! I'm right!"

 

[ctamblyn]

Specious straw-man garbage. You skip from a non-constant vacuum impedance to a scalar theory of gravity, and then declare it to be impossible. This is distasteful, CT.

I'm just spelling out to you what your ideas, as stated, imply. If you feel you have been misrepresented then perhaps you need to examine how you communicate your ideas.

You said (post #168 in this thread, and also during the old Relativity+ thread) that the effects of a black hole's gravity on an infalling object can be understood in terms of variations in the "vacuum impedance of space". You said it was "that simple". Those were your very words.

I repeat: If your theory can be understood in terms of a scalar quantity (e.g. vacuum impedance) that varies from point to point in space, it absolutely cannot be equivalent to GR.

So, is it really "that simple"?

(...snip...)
"According to this theory the metrical qualities of the continuum of space-time differ in the environment of different points of space-time, and are partly conditioned by the matter existing outside of the territory under consideration. This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that “empty space” in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gμν)..."

I never claimed that we live in a homogeneous or isotropic space. Since you took the trouble to post this, though, you should note that he refers to ten functions g_μν, these being the components of the metric tensor for a four-dimensional spacetime (in 3D we'd have just six functions). GR's field equations are fundamentally four dimensional.

Inhomogeneous space, CT. Not curved spacetime.

GR has inhomogenous, non-isotropic, curved 4D spacetime. Einstein just told you that, above, but you somehow missed it.

If your model doesn't, then it is not obviously equivalent to GR. If you think it is equivalent, you need to prove that rather than merely assert it. Stop hand-waving and using flawed car analogies (see below) and convince us with some proper analysis.

It is unlikely to be an easy task, but until you do it there is no rational reason for anyone to believe that you are merely reinterpreting GR as opposed to, say, making grandiloquent noises to impress the gullible.

Light curves when the space it moves through is not homogeneous. It veers like a car veers when it encounters gravel at the side of the road.

Your car analogy (why is it always cars?) says that you are still thinking in terms of something like a refractive index varying through space. It doesn't work, as we discussed before on the other thread.

 

[W.D.Clinger]

Some very minor corrections...

In my post #191, two days ago, I made several references to exercise 23. Some (but not all) of those should have been references to exercise 15 or 16. If you've done the exercises, you'll be able to figure it out. (If you haven't done the exercises, then you probably don't care.)

Phooey. I said to ben your simple proof was wrong. When 2M/r = 1 at r = 2M you've got a division by zero in [latex]\[ \frac{dt}{d\tau} = \frac{e}{1-2M/r} \][/latex]. That's an undefined result, an infinity

I missed this until Clinger quoted it. Um, no I didn't. Why would I need to even compute dt'/dtau'---the ratio of two times according to some other observer---to find the trajectory of this observer? Why should "light I emit now will never get to observer X" interrupt me from obeying the laws of physics?

In defense of Farsight, he was quoting Vorpal's post #64.

In defense of Vorpal, he was proving that "radial freefall is exactly Newtonian in Schwarzschild r and proper time τ. And so the proper time to reach the horizon (or the r = 0) is finite." If you ignore the parenthetical "(or the r = 0)", then r > 2M throughout the region that matters for Vorpal's proof, which establishes that a free-falling observer does indeed reach r=2M in finite proper time.

Farsight ignored that, of course, and jumped on the apparent division by zero that doesn't actually affect Vorpal's argument (apart from the parenthetical remark). Furthermore, Vorpal's argument can be extended all the way down to r=0 by any of several techniques that you, Vorpal, and I would regard as absolutely trivial, but would take a little time to explain to someone with Farsight's knowledge of calculus. That's why I answered Farsight as though he were talking about the division by zero in the Schwarzschild metric at the coordinate singularity r=2M.

That is, after all, the true source of the division by zero that Farsight our friendly neighborhood cage fighter insists he does not know how to avoid.

 

[Daylightstar]

...
And the wave can only propagate if space somehow "kicks back" via some kind of elastic response.

Evidently, this (and perhaps the word 'plastic' appearing somewhere on wiki) is what your 'elastic response' with respect to your 'space holding and/or supporting waves and fields' boils down to, a bare assertion.
No detailing because it would take too long, but a long (another) pointless message anyway.
What a waste of time.

Maybe it's better not to 'talk about it'.

 

[steenkh]

Farsight, the crux of the differences between you and the others seem to be that you believe that you are not using a coordinate system, whereas the others insist that you are. Perhaps you could show that your calculations of time stopping at the event horizon is not based on a coordinate system? That could clear up any misunderstandings. Thank you.

 

[Roboramma]

Farsight, the crux of the differences between you and the others seem to be that you believe that you are not using a coordinate system, whereas the others insist that you are. Perhaps you could show that your calculations of time stopping at the event horizon is not based on a coordinate system? That could clear up any misunderstandings. Thank you.

I think that farsight is saying that someone looking at an infalling observer will see (and measure time slowing. If he can look at an idealized light clock, he'll see the light taking more time (as measured on his clock) to get from one mirror to another. As it approaches the event horizon, that time becomes infinite.

He seems to be saying that this measurement is independant of coordinate systems, thus he's not using one.

I think he's half-right about that: it's true that this measurement is independent of coordinate systems: whatever coordinate system you use, you'll agree about what an observer far from the event horizon will measure with his clock.

But what farsight is missing (in my opinion) is that this doesn't tell us anything about the the infalling observer measures. And considering that that is what's under discussion, it's odd that he glosses over it and keeps harping about the measurements of someone else.

Of course, I'm no expert, but this is how I'm reading the thread, anyway. Hopefully I'll be corrected if I'm messing things up.

 

[Vorpal]

Simplifying somewhat for the present audience, a spacetime manifold as described by some particular chart is said to be static if the metric's coordinate-dependent components are independent of the chart's timelike coordinate. It's a property of the chart as much as the spacetime.

This is stationary, which is somewhat weaker than static. What makes it weaker is the possibility that you get metric components g_0i, i≠0, that are independent of the timelike coordinate x⁰ = t, but still nonvanishing. For an example of a chart on Schwarzschild spacetime in which the metric has a stationary but not static form, see the Gullstrand-Painleve coordinates (previous post).

In a Schwarzschild chart, for example, the spacetime submanifold on which the chart is defined is static ... so whether you think a spacetime is static depends on your personal choice of coordinates.

We should make the distinction between spacetime and metric/chart here. There's no problem in saying that a coordinate chart or metric is stationary or static if it has a particular form, and your comments apply in that sense. But in such a chart as you describe, ∂_t is a timelike Killing field, the existence of which can be taken to be as an intrinsic definition of a stationary spacetime, independently of any chart. And similarly, with some refinement, static spacetimes as well.

However, Farsight's statement is quite a bit more banal than this. He's merely saying that spacetime contains the entire history of any given particle, so the particle is not "really moving" through it. That's... it. It's not the kind of staticness that connects in any relevant way with the issues at hand.

Guys: I smell desperation here, and sense capitulation. It's signalled by the evasion and abuse. I think one of you will crack and say I think Farsight's got a point soon.

I have indeed capitulated, in the sense of complete resignation that you are too far gone into crank-land for any possibility of productive interaction. Many other posters here are not to that point, and that's their business.

 

[sol invictus]

You don't need a coordinate system to see that a clock clocks up the motion of light

The results of physical experiments are independent of coordinates, yes. But there are many physically different experiments one can do with light, clocks, and a black hole. GR predicts that some of those experiments will show clocks ticking slower near the horizon, and some will not. That's reflected in the fact that some coordinate systems have a coordinate singularity there, and some do not. (Those two statements are connected, because it's simple to describe experiments in coordinates in which most parts of the experiment are at rest.)

Let me add yet again that all of the above goes for perfectly flat spacetime. There is a horizon in flat spacetime (many actually, and your arguments apply just as well to the horizon of flat spacetime as they do to the horizons of black holes.

 

[steenkh]

I think he's half-right about that: it's true that this measurement is independent of coordinate systems: whatever coordinate system you use, you'll agree about what an observer far from the event horizon will measure with his clock.

My impression was the exact opposite: that it depends on the coordinate system what an observer from far from the event horizon will measure. And the point was that if you make such measurements you are using a coordinate system, even if you are not aware of it.

 

[steenkh]

The results of physical experiments are independent of coordinates, yes.

OK, I was wrong. Forget about my previous post!

 

[Farsight]

Farsight, the crux of the differences between you and the others seem to be that you believe that you are not using a coordinate system, whereas the others insist that you are. Perhaps you could show that your calculations of time stopping at the event horizon is not based on a coordinate system? That could clear up any misunderstandings. Thank you.

I can't steen, because the misunderstanding concerns time stopping. Only it isn't time that stops, it's light. The thing to note about a clock, is that what it's actually doing, is "clocking up" some kind of motion. A mechanical clock clocks up the motion of cogs and sprockets, a quartz wristwatch clocks up the vibrations of a crystal, and so on. So when a clock runs slower, the underlying motion is going slower. This is true for light clocks just like any other clock. So when a light clock stops it isn't time stopping, instead it's light. There is no mathematics that will reveal this, and there is no calculation that can substitute for looking at what a clock actually does. I've tried to get this across to sol, see posts 21 as well as posts 41 and 43. With this in mind look again at
[latex]\[ \frac{dt}{d\tau} = \frac{e}{1-2M/r} \][/latex]. The dt over dtau is a motion ratio, and when the divisor is zero the ratio is infinite. That means the motion associated with tau is zero. When we then say we're using light clocks, and try to define a coordinate system with light that doesn't move, we can't. The guys whitewash over this by saying that's OK, you've got forever for this light to move, the stopped light and eternity cancel each other out. Only they don't. Nothing happens, that's it.

 

[sol invictus]

OK, I was wrong. Forget about my previous post!

Think of changing from (say) meters to feet - that's a simple coordinate change. The numerical value of the length of something is of course different in feet versus meters. But obviously the physical object hasn't changed, and if you do a well-defined physical experiment - like seeing how many foot-long-ruler-lengths long an object is, by holding that ruler up to that object - your predictions for the results of that experiment won't depend on your nationality.

Still, Americans are likely to have foot-long rulers handy, and Europeans have more meter sticks lying around, so we do have to make sure we specify which experiment exactly it is (ruler or meter stick? what object?) that's being done, and bear in mind that describing meter-stick measurements may be simpler if we use meters.

Farsight insists on using one set of units that have the property that the tick marks on his rulers stretch out to infinity at a certain place (the horizon). As a result, he has trouble measuring, or even describing, anything that happens there, and he's fooled himself into thinking that this is somehow important.

 

[Farsight]

That just illustrates your inability to understand a point we have been making. There is no event horizon in Kruskal-Szekeres coordinates

Oh spare me your schoolboy ignorance, RC. Go and look at Kruskal-Szekeres coordinates on wiki. You'll read things like:

The location of the event horizon (r = 2GM) in these coordinates is given by
[latex]V = \pm U[/latex]

You know about Kruskal-Szekeres coordinates but are still ignoring that there is no event horizon in them .

Sheesh. You've linked to it, but you haven't read it.

All: you really should correct RC when he makes errors like this. When you don't, I'm afraid you damage your credibilty.

 

[Farsight]

What clock is telling you t=infinity?

No clock, the maths. The divisor is zero. Clocks clock up motion, dt over dtau is a ratio between one clock rate and another.

Except that we've shown you this transformation repeatedly.

And I've told you repeatedly you can't transform a stopped clock into a moving clock by adding a stopped observer who "sees the clock moving in his own frame of reference". It's a light clock. Light has stopped. Your observer can't see.

The clock has not "actually stopped" (if that means anything at all); the fact that you think it's stopped is an accident of the clocks you chose to look at while asking "has the next tick happened yet".

Our universe is 13.7 billion years old, and that next tick hasn't happened yet. Give it another 13.7 billion years and it still hasn't.