I am the first in the world

[ben m]

I've told you before, don't try hiding behind the maths, not with me. Look at this expression:

[latex]t_0 = t_f \sqrt{1 - \frac{2GM}{rc^2}} = t_f \sqrt{1 - \frac{r_0}{r}}[/latex]

Farsight, please write down the tension in the rope holding this observer in place. Got it? Good. Reexpress the above, not in terms of r but in terms of tension. Got it?

Farsight, please forget about the black hole; consider this same rope, under the same tension, towing this observer through flat space. Find the equivalent expression for this observer. (You'll want to compare his clock that of an inertial observe with respect to whom he is momentarily at rest.) Got it?

 

[Farsight]

The coordinate speed of light according to whom, Farsight? You can't throw three different observers into a problem---observers who, remember, will disagree on that speed---and declare that the speed "actually" has some special value.

According to distant observers. People who aren't falling into the black hole. Even if their measurements don't quite agree they can account for that from their knowledge of relativity - the coordinate speed of light is only constant in an inertial reference frame. In a non-inertial reference frame, such as the room you're in, it isn't constant. They can all see the black hole and the infalling observer, any differences in their measurements are miniscule and well-understood.

For any observer accelerating very strongly, your reference frame and coordinate system have "collapsed". That's true if you are accelerating in flat space (and picking up speed w/r/t reference objects). It's also accelerating near a black hole (and hovering just above the event horizon).

It isn't "collapsed", it's non-inertial.

For an inertial observer, your local space is approximately flat, your local light speed is c, Newton's Laws hold, etc.. That's true if you are drifting in (globally) flat space, it's true if you are falling inertially and just outside an event horizon..

Your local light speed is your measured speed of light. You always measure wave speed to be the same value because of the wave nature of matter. Remember pair production, and how you can create and electron (and a positron) from light? And how you can diffract electrons? And neutrons? And have you read what I've said about how the NIST caesium fountain clock works? In a nutshell, you calibrate your rods and clocks using the local motion of electromagnetic phenomena. Light in the wider sense. Then you use them to measure the local speed of light. That's why you always measure the same value. I'm not kidding you about this ben.

it's true if you are falling inertially and just inside the event horizon.

It isn't. Remember what sol said: all processes slow and stop on the horizon as viewed from outside. Now imagine that you're a light wave bouncing back and forth inside his parallel-mirror light clock. At the event horizon, you stop forever.

Your idea that "your coordinate system does something special at the horizon" is flatly wrong.

This isn't my idea, this is the original frozen star idea, in line with Schwarzschild and Einstein. Typical textbooks do that hop skip and a jump over the Schwarzschild event-horizon singularity and write it off as a coordinate artefact, missing the whole point that when light stops you don't have any coordinates any more. Again, see what I said to Vorpal. It's the GR equivalent of the SR case where you're travelling through flat space at c. You don't carry on observing as if nothing happened. You observe nothing.

It's not subtly wrong, it's straightforwardly wrong; and it's been derived explicitly in many places that you have access to; you have found no error in the derivations, you apparently just don't like the answer because your hobby-horse suggested a different one.

If it's "straightforwardly wrong", try explaining it. Or try addressing my straightforward argument. When you can't do either, stop digging, and yield.

 

[Vorpal]

I've told you before, don't try hiding behind the maths, not with me. Look at this expression:

[latex]t_0 = t_f \sqrt{1 - \frac{2GM}{rc^2}} = t_f \sqrt{1 - \frac{r_0}{r}}[/latex]

t₀ is the proper time between events A and B for a slow-ticking observer within the gravitational field,
t_f is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object...
r is the radial coordinate of the observer...
r₀ is the Schwarzschild radius...

It's a simple expression, and it tells you that when r is very very large, when you're a long way from the black hole, t₀ tends to t_f. It's the infalling observer's clock rate expressed as a fraction of the distant observer's clock rate.

Oh, Farsight. I'm afraid that's completely wrong. The infalling observer's clock rate expressed as a fraction of the distant stationary observer's clock rate is:
[latex]\[ \frac{d\tau}{dt} = \sqrt{\left(1-\frac{r_0}{r}\right)-\left(1-\frac{r_0}{r}\right)^{-1}\frac{dr^2}{dt^2} - r^2\left(\frac{d\theta^2}{dt^2}+\sin^2\theta\frac{d\phi^2}{dt^2}\right)} \][/latex]
Now, if I take a stationary observer, dr = dθ = dφ = 0, sure enough, that's [latex]$ d\tau = dt\sqrt{1-\frac{r_0}{r}}$[/latex]. But not an infalling observer.

What did you think the Schwarzschild metric was for, anyway? Window dressing?

There is actually something almost right about your post, but I'm afraid that if I tell you, you'll simply latch on to that and learn nothing, because your conceptual problems are quite deep. Let me instead ask you about how much you know of Einstein's work.
1) Are you aware of the geodesic equation and Christoffel symbols? They're featured quite prominently in Einstein's papers on GTR. I know--I read them.
2) Are you capable of calculating an orbit using them? Even a simple radial one, dθ = dφ = 0?

You know how you get all those cranks and crackpots who know b*ggar all physics and who insist that Einstein was wrong? Well, take a look at which side of the fence you're on.

You've not understood Einstein at all. You're just cherry-picking specific cases with no understanding of the bigger picture. Yes, the bigger picture Einstein himself explained.

 

[Farsight]

Farsight, please write down the tension in the rope holding this observer in place. Got it? Good. Reexpress the above, not in terms of r but in terms of tension. Got it? Farsight, please forget about the black hole; consider this same rope, under the same tension, towing this observer through flat space. Find the equivalent expression for this observer. (You'll want to compare his clock that of an inertial observe with respect to whom he is momentarily at rest.) Got it?

No. It isn't relevant. The coordinate speed of light varies with gravitational potential. The tension in the rope is merely an indicator of the difference in gravitational potential between the bottom and the top. Whether the observer at the event horizon has fallen, or has been gently lowered on a gedanken rope doesn't make any difference. Either way, he is at a place where the coordinate speed of light is zero. Please don't clutch at straws to cling to something that you've been taught. Think for yourself instead.

Gotta go. You know guys, I enjoy talking to you.

 

[ben m]

No. It isn't relevant. The coordinate speed of light varies with gravitational potential.

The coordinate speed of light is a weird, non-localized quantity that you cook up by comparing two particular observers. Yes, it does weird and divergent things. Many observer-observer comparisons do similarly weird things. You are pointing out one specific weirdness that occurs when you compare an observer in flat space to an observer hanging on a rope under diverging tension. Why should anyone care? "The coordinate speed of light varies with potential" is no more interesting a statement than "the tension on the rope varies with potential".

Whether the observer at the event horizon has fallen, or has been gently lowered on a gedanken rope doesn't make any different

Sure it does. A falling observer is in a different reference frame than the rope-hanging observer. You keep constructing "coordinate time" using the rope-hanging observer's clock, NOT the falling clock, so you do not have the luxury of saying "the rope makes not difference". It does make a difference---it imparts an acceleration, and acceleration affects clocks, and you're trying to prove everyone wrong by arguing about clocks. Under the circumstances, you do NOT have the luxury of ignoring differences between clocks.

 

[sol invictus]

No, I'm saying you don't understand this.

Me, and every other physicist since the 1960s? Ooookaaay.... <backs away slowly>

I've told you before, don't try hiding behind the maths, not with me. Look at this expression:

[latex]t_0 = t_f \sqrt{1 - \frac{2GM}{rc^2}} = t_f \sqrt{1 - \frac{r_0}{r}}[/latex]

t₀ is the proper time between events A and B for a slow-ticking observer within the gravitational field,
t_f is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object...
r is the radial coordinate of the observer...
r₀ is the Schwarzschild radius...

It's a simple expression, and it tells you that when r is very very large, when you're a long way from the black hole, t₀ tends to t_f. It's the infalling observer's clock rate expressed as a fraction of the distant observer's clock rate. I'm the distant observer, you're the other guy. As you fall, when you run the numbers for ever-smaller values of r expressed as a multiple of r₀, you can see the way t₀ trends.

When r is 4, t₀ = .866
When r is 3, t₀ = .816
When r is 2, t₀ = .707
When r is 1.5, t₀ = .577
When r is 1.1, t₀ = .301
When r is 1.01, t₀ = .099
When r is 1.00, t₀ = 0

At the event horizon, your clock rate, expressed as a fraction of mine, is zero. Like you said, all processes slow and stop on the horizon as viewed from outside. And your light clock is a parallel-mirror light clock, remember. With a clock rate of zero. I'll wait a million years. Has your clock ticked yet? No. I'll wait a billion years? Has your clock ticked yet? No.

Has the penny dropped yet?

Your math (or rather, your interpretation of it) is wrong. But never mind that, because there's a very simple argument that doesn't even involve doing any math. Look at the metric in post #17. It's exactly the same as the near-horizon metric of a black hole. But it's also a metric on flat spacetime, and I can put the horizon anywhere I want in flat spacetime. Therefore, any conclusion that the horizon must be special because of some feature in that metric is manifestly, unambiguously wrong.

You know how you get all those cranks and crackpots who know b*ggar all physics and who insist that Einstein was wrong? Well, take a look at which side of the fence you're on.

That's the classic "appeal to authority" logical fallacy. Just because Einstein said something - and so far, you haven't even presented a quote from him, and I don't take your word for it - doesn't make it true. Since Einstein was an expert on this topic it certainly makes it worth considering, and it means that establishing it as incorrect requires careful argument and the agreement of other experts. But that's been done, over and over and over again. This analysis is in every textbook on relativity. I've derived it myself. Vorpal has derived it. ben has derived it. In fact, the only one in this conversation that can't do the math and hasn't derived it is - you guessed it - you.

 

[ben m]

Has the penny dropped yet?

Here's the other thing, Farsight. You seem to think that you're presenting us stuff we've never through about before. You're expecting people to say "Holy cow---you're right! His clock hasn't budged! I never looked at it that way, that changes everything!

This is comically wrong.

This is where it becomes obvious that you've never taken a GR course or paged through a GR textbook. No, Farsight, none of your statements (the ones that are correct, anyway) are even the slightest bit surprising. You see, this is what learning GR consists of. You take a bunch of spacetimes, and a bunch of metrics, and you do dozens of exercises working through different scenarios. Your conclusion---minus the "OMG this changes everything" wrapper---is a familiar one, and it's one of dozens and dozens of exercises of this form that anyone who has taken a GR class has completed.

"Has the penny dropped yet?" No, Farsight. The penny has not dropped. Heck, the Canadian Tire Money has not dropped. There's nothing to drop; there's nothing confusing about any of this.

Here's an analogy to what you are doing.

"Look at this optics equation---if you put something right at the focal point of a lens, the magnification is INFINITY."
"Sure, so what?"
"Don't you see! That violates conservation of energy."
"No it doesn't."
"1/(f-d). Make f=d. Try it, I'll wait. Has the penny dropped yet?"

 

[alfa1]

And where is the information?

 

[Mashuna]

Here's the other thing, Farsight. You seem to think that you're presenting us stuff we've never through about before. You're expecting people to say "Holy cow---you're right! His clock hasn't budged! I never looked at it that way, that changes everything!

And that's why the arrow can never hit the tortoise! [/zeno]

 

[Vorpal]

"You know, what Mr. Einstein said is not so stupid..." -- Pauli

That's the classic "appeal to authority" logical fallacy. Just because Einstein said something - and so far, you haven't even presented a quote from him, and I don't take your word for it - doesn't make it true.

Logically, it could dismissed by those grounds, but there's indeed something pretty fishy and very questionable about all this attribution to Einstein in the first place, particularly in support of Farsight's rather strong misconceptions, such as

Your proper time is being measured on a stopped clock. Just because you're stopped too doesn't mean that nothing even slightly unusual happens. It means nothing happens.

and especially statements such as

At the event horizon it isn't nothing unusual happens, it's nothing happens. There are no more events.

It does not make that Einstein, who derived such results as this in collaboration with Rosen: Phys. Rev. 48, 73–77 (1935), would now suddenly take the point of view that spacetime just gets cut off at the horizon of black hole, and then suddenly advance a solution that amounts to gluing two Schwarzschild exteriors together at the horizon.

The Einstein-Rosen "bridge" paper also reveals Einstein's motivation for it: he may have disliked the Schwarzschild interior, his main problem with it was definitely the genuine curvature singularity at the center, not the horizon itself, which, after all, the "bridge" does not get rid of it at all, and rather puts a whole universe of events beyond it.

This also shows that whatever he might have thought about the Schwarzschild horizon at some point, he clearly didn't think it was a genuine physical singularity by at least 1935 (and I don't know if ever and to what extent). In fact, one can even quote Einstein pretty much explicitly telling Farsight that his argument is baloney, both in reasons and conclusion:

If, however, we replace the variable r by ρ defined by the equation

[latex]$\rho² = r - 2m$[/latex]​

we obtain

[latex]\[ds^2 = -4(2m+\rho^2)d\rho^2 - (2m+\rho^2)^2(d\theta^2+\sin^2\theta d\phi^2)+\frac{\rho^2}{2m+\rho^2}dt^2\][/latex]​

This solution behaves regularly for all values of ρ. The vanishing of the coefficient of dt² i.e. (g₄₄) for ρ = 0 results, it is true, in the consequence that the determinant g vanishes for this value; but, with the methods of writing the field equations actually adopted, this does not constitute a singularity.​

From Relativity Theory and Corpuscules essay included in many different books, such as this one.

 

[Jekyll's Guest]

Well I guess we know which side is Thal and which side is Kaled.

 

[alfa1]

The Big Bang

I answered to this question:
Physics major question: Big Bang: What Banged?

The Big Bang is the dominant (and highly supported) theory of the origin of the universe.
What Banged? In my view: a qubit of information (a photon) with a very high oscillation frequency.
The frequency of the qubit:
f = (Energy of the universe) / h = Muniverse × c^2 / h = 3.35 × 10^54 × (2.99792458 × 10^8)^2 / (6.62606896 × 10^(-34))
f = 4.544 × 10^104 Hz

Like I said before:
"In the beginning was the qubit" Adrian Ferent

 

[Reality Check]

What Banged? In my view: a qubit of information (a photon) with a very high oscillation frequency.

Your view is wrong since what Banged was spacetime.
There is also more than one photon in the universe!

The Big Bang was an explosion of spacetime. Ignorant cranks often think that it was an actual explosion of something within spacetime.

 

[alfa1]

The photon is an example of qubit
Are a lot of solutions for the Big Bang, this is related to the information…

 

[dafydd]

The Big Bang

I answered to this question:
Physics major question: Big Bang: What Banged?

The Big Bang is the dominant (and highly supported) theory of the origin of the universe.
What Banged? In my view: a qubit of information (a photon) with a very high oscillation frequency.
The frequency of the qubit:
f = (Energy of the universe) / h = Muniverse × c^2 / h = 3.35 × 10^54 × (2.99792458 × 10^8)^2 / (6.62606896 × 10^(-34))
f = 4.544 × 10^104 Hz

Like I said before:
"In the beginning was the qubit" Adrian Ferent

Who is this Adrian Ferret?

 

[steenkh]

The photon is an example of qubit
Are a lot of solutions for the Big Bang, this is related to the information…

So in your world, a single qubit is able to generate zillions of qubits of information?

 

[Farsight]

Oh, Farsight. I'm afraid that's completely wrong...

It isn't completely wrong. Yeah yeah, I should have said stationary observer, where his clock rate is a function of gravitational potential. But if you have that observer accelerating, his clock rate tends to zero even faster.

The infalling observer's clock rate expressed as a fraction of the distant stationary observer's clock rate is:
[latex]\[ \frac{d\tau}{dt} = \sqrt{\left(1-\frac{r_0}{r}\right)-\left(1-\frac{r_0}{r}\right)^{-1}\frac{dr^2}{dt^2} - r^2\left(\frac{d\theta^2}{dt^2}+\sin^2\theta\frac{d\phi^2}{dt^2}\right)} \][/latex]
Now, if I take a stationary observer, dr = dθ = dφ = 0, sure enough, that's [latex]$ d\tau = dt\sqrt{1-\frac{r_0}{r}}$[/latex]. But not an infalling observer.

So come on then, run the numbers like I did. Let's see that clock rate.

What did you think the Schwarzschild metric was for, anyway? Window dressing?

I'm the one who's saying it's more than just some coordinate artefact. People who swallow Kruskal-Szekeres coordinates treat the event-horizon singularity like window dressing to be disregarded.

There is actually something almost right about your post, but I'm afraid that if I tell you, you'll simply latch on to that and learn nothing, because your conceptual problems are quite deep.

Baloney. What I'm saying is right, you're dodging.

Let me instead ask you about how much you know of Einstein's work.
1) Are you aware of the geodesic equation and Christoffel symbols? They're featured quite prominently in Einstein's papers on GTR. I know--I read them.
2) Are you capable of calculating an orbit using them? Even a simple radial one, dθ = dφ = 0?
You've not understood Einstein at all. You're just cherry-picking specific cases with no understanding of the bigger picture. Yes, the bigger picture Einstein himself explained.

Aw here we go, hiding behind inscrutable mathematics. If I say yes you sidetrack the discussion demanding that I support my claim, if I say no you say well there you go then. It isn't good enough Vorpal. Everybody can see you're ducking the argument. Let me take a look at the other posts, then I'll nail you to the floor. Hurry up and start backing down, because it's going to be embarrassing.

 

[Farsight]

The coordinate speed of light is a weird, non-localized quantity that you cook up by comparing two particular observers. Yes, it does weird and divergent things. Many observer-observer comparisons do similarly weird things.

There's nothing "weird" about it all. The coordinate speed of light varies in a non-inertial reference frame, like the room you're in. Take a look at this report on a super-accurate optical clock. It's so precise that you can see two of these clocks losing synchronisation when they're separated by only a foot of vertical elevation. Now take a look at wiki re time dilation, and note the bit that says consider a simple clock consisting of two mirrors A and B, between which a light pulse is bouncing. When you simplify those optical clocks to parallel-mirror light clocks, when they lose synchronisation at different elevations this is what's happening:

|---------------|
|---------------|

That's the coordinate speed of light varying in a non-inertial reference frame. Only it isn't actually the coordinate speed of light, it's the speed of light. It varies with gravitational potential. If it didn't, those two optical clocks would stay synchronised. So please, spare us the "weird".

You are pointing out one specific weirdness that occurs when you compare an observer in flat space to an observer hanging on a rope under diverging tension. Why should anyone care? "The coordinate speed of light varies with potential" is no more interesting a statement than "the tension on the rope varies with potential".

See above. Don't dismiss this as uninteresting. This is physics ben. It's not what your good book tells you, but it's simple, and it's right.

Sure it does. A falling observer is in a different reference frame than the rope-hanging observer. You keep constructing "coordinate time" using the rope-hanging observer's clock, NOT the falling clock, so you do not have the luxury of saying "the rope makes not difference". It does make a difference---it imparts an acceleration, and acceleration affects clocks, and you're trying to prove everyone wrong by arguing about clocks. Under the circumstances, you do NOT have the luxury of ignoring differences between clocks.

I'm not ignoring the difference between clocks. I'm addressing it. Clocks clock up some kind of motion. All clocks do this. A light clock clocks up the motion of light. Nothing is weird, it's all very simple. When we see optical clocks at different elevations lose synchronisation, we say gravitational time dilation occurs. And if that goes infinite, the lower clock has stopped. And if light has stopped, it doesn't matter whether the light was falling into the black hole or bouncing back and forth between mirrors.

 

[Farsight]

Here's the other thing, Farsight. You seem to think that you're presenting us stuff we've never through about before. You're expecting people to say "Holy cow---you're right! His clock hasn't budged! I never looked at it that way, that changes everything! This is comically wrong.

You haven't thought it through. If you had, you wouldn't be trying the you haven't taken a GR course defence. You'd address the argument. Now come on, give it a shot.

This is where it becomes obvious that you've never taken a GR course or paged through a GR textbook. No, Farsight, none of your statements (the ones that are correct, anyway) are even the slightest bit surprising. You see, this is what learning GR consists of. You take a bunch of spacetimes, and a bunch of metrics, and you do dozens of exercises working through different scenarios. Your conclusion---minus the "OMG this changes everything" wrapper---is a familiar one, and it's one of dozens and dozens of exercises of this form that anyone who has taken a GR class has completed.

Pompous erudite flannel.

"Has the penny dropped yet?" No, Farsight. The penny has not dropped.

No, it hasn't, has it? It will.

Heck, the Canadian Tire Money has not dropped. There's nothing to drop; there's nothing confusing about any of this.

But you think the coordinate speed of light is weird.

"Here's an analogy to what you are doing. "Look at this optics equation---if you put something right at the focal point of a lens, the magnification is INFINITY."
"Sure, so what?"
"Don't you see! That violates conservation of energy."
"No it doesn't."
"1/(f-d). Make f=d. Try it, I'll wait. Has the penny dropped yet?"

Is that the best you can do? A febrile accusation instead of addressing the point of issue? Here's an analogy for what you're doing: Huff, puff, waffle, the man doesn't understand Latin.

 

[Farsight]

Me, and every other physicist since the 1960s? Ooookaaay.... <backs away slowly>

Not every physicist. Just those who thump your good book. Try emailing Steven Weinberg to see what he thinks.

Your math (or rather, your interpretation of it) is wrong. But never mind that, because there's a very simple argument that doesn't even involve doing any math. Look at the metric in post #17. It's exactly the same as the near-horizon metric of a black hole. But it's also a metric on flat spacetime, and I can put the horizon anywhere I want in flat spacetime. Therefore, any conclusion that the horizon must be special because of some feature in that metric is manifestly, unambiguously wrong.

No it isn't. I've said this and I'll say it again. Gravitational time dilation goes infinite at the event horizon, so your clock has stopped and you've stopped too. Light has stopped, so you can't measure anything, and there is no metric. Flat spacetime is irrelevant here because spacetime has gone. And you have in no way addressed the issue. The whole point of this discussion is the fallacy that a stopped clock carries on ticking. You said nothing unusual happens to observer. But everything has stopped, so nothing happens.

That's the classic "appeal to authority" logical fallacy.

Oh yes? Now remind me, what did you say above? "Me, and every other physicist since the 1960s? Ooookaaay.... <backs away slowly>". You've been hoisted by your own petard sol.

Just because Einstein said something - and so far, you haven't even presented a quote from him, and I don't take your word for it - doesn't make it true. Since Einstein was an expert on this topic it certainly makes it worth considering...

I haven't presented Einstein quotes because the scientific evidence is more important than Einstein's view. I can if you wish, but I seem to recall that you dismissed them as "out of context". Look at what I said to ben about the coordinate speed of light. It varies. You and I know this, and we know about the Shapiro delay. We also know that if you were travelling at the speed of light, you wouldn't see anything. We know that as far as you're concerned, it isn't nothing unusual happens, it's nothing happens. And we understand the principle of equivalence.

...and it means that establishing it as incorrect requires careful argument and the agreement of other experts. But that's been done, over and over and over again. This analysis is in every textbook on relativity.

Ah, the argument from authority again. I don't care what some expert declares to be right, I care about scientifc evidence, and logic, and sound argument. This is a discussion forum, not a scripture class.

I've derived it myself. Vorpal has derived it. ben has derived it. In fact, the only one in this conversation that can't do the math and hasn't derived it is - you guessed it - you.

Don't hide behind erudition sol. Instead just shoot a light beam between two stars. Slows down, doesn't it. And travels in a straight line. Now make the stars a bit more massive. Then repeat. You know where I'm going with this, so buck your ideas up, pull your finger out, and let's have a sincere discussion about a very interesting subject that takes us places.