Is 'charged black hole' an oxymoron?

[Q-reeus]

Excellent. If you use a formally correct derivation from correct axioms then you are guaranteed to get a correct result. Since the axioms are the EFE and ME, and since those have both been extensively validated (the EFE to the best experimental measurements possible and ME in the classical limit of QED), the axioms seem reasonably correct.

So we have a set of correct axioms, a formally correct derivation from those axioms, and therefore logically must have a correct result.

Not necessarily - I italicized *formally* on purpose. It's already been noted elsewhere recently how local things - conservation/divergence laws, can strictly hold locally yet globally fail in GR.

Q-reeus: "Usual interpretation of terms in final RN expression has Q contributing to SET as a function of r, and therefore to gravitating mass M = M(r) as a function of radius r. That is the 'one way' coupling part I referred to. Having trouble seeing where M is in turn effecting the value of Q - i.e. where do we have Q = Q(r) showing up?"

I don't know what you are talking about here. Neither M nor Q are functions of r, they are constant parameters which describe the entire spacetime. Where are you getting this idea that M is a function of r or that Q should be also?

There was discussion over the meaning of Q term as effective spatial SET contribution to M back in e.g. #20 and before/later. I see now from #345 Peter has pointed out that Q has an *effective* spatial variation even within RN model.

Q-reeus: "Thus imo it is indeed one-way coupling according to RN. (Aware there are other couplings to EM - e.g. 'light bending', but not relevant to static RN metric case)"
Did you miss the coupling I posted in 194?

You mean #191. Gets back to my first comments above.

That is why the logical derivations are so important. The logical derivation has been laid out for you, and I don't see any objections other than a kind of general personal distaste for formal derivations.

More than a distaste imo. Let's wait a bit for specific predictions for 'charge centred within a shell'. After that we will have more to go on.

 

[Dale]

Not necessarily - I italicized *formally* on purpose.

Do you agree or disagree with the following statement:

"If you use a formally correct derivation from correct axioms then you are guaranteed to get a correct result."

There was discussion over the meaning of Q term as effective spatial SET contribution to M back in e.g. #20 and before/later.

I looked at 20. That seemed to show that acceleration was a function of r, which I certainly agree to. It did not seem to show that M or Q are a function of r.

You mean #191.

Sorry, I meant 196. Please look back at that since you seem to have missed it. It shows the coupling back from the EFE to ME.

More than a distaste imo. Let's wait a bit for specific predictions for 'charge centred within a shell'. After that we will have more to go on.

Why? What is the purpose of the exercise? Is it not to determine self-consistency? We have a correct derivation from correct axioms, therefore it is self-consistent, your personal "more than a distaste" notwithstanding. The thought experiments are simply not relevant in determining self-consistency, for the reasons I pointed out above.

However, if the purpose is to learn some GR by doing some advanced "homework" problems, then I think that the scenarios are worth working through. However, I think that they are pretty advanced for me and most likely also too advanced for you. I am learning a lot by doing the parallel transport exercise.

 

[Q-reeus]

Do you agree or disagree with the following statement:
"If you use a formally correct derivation from correct axioms then you are guaranteed to get a correct result."

"Correct axioms" is imo the suspect link in that chain.

I looked at 20. That seemed to show that acceleration was a function of r, which I certainly agree to. It did not seem to show that M or Q are a function of r. Neither M nor Q are functions of r, so I don't see why their failure to be functions of r should be construed as any kind of a problem.

We are talking effective quantities. Nobody including myself is suggesting locally evident failure of anything. If coordinate acceleration a(r) is properly equated to effective gravitating mass M(r), need I say more?

Sorry, I meant 196. Please look back at that since you seem to have missed it. It shows the coupling back from the EFE to ME.

OK yes I did miss it. Advanced math notation already for me - but really you will surely agree that is a finished statement not a first-principles derivation.

Why? What is the purpose of the exercise? Is it not to determine self-consistency? We have a correct derivation from correct axioms, therefore it is self-consistent, your personal "more than a distaste" notwithstanding. The thought experiments are simply not relevant in determining self-consistency, for the reasons I pointed out above.

Relax DS. My first comment above stands for now. Look at it this way - PF is about educating, right? So if nothing else, if my Quoxotic crusade falls flat as you and hordes others expect, at least a detailed following through of said exercise will do it's little bit in fulfilling the educational PF goal here - yes?

 

[Dale]

"Correct axioms" is imo the suspect link in that chain.

So you agree with the statement, but think that the axioms for deriving the RN spacetime are not correct.

Can you please point out exactly which axiom is incorrect in your view?

We are talking effective quantities. If coordinate acceleration a(r) is properly equated to effective gravitating mass M(r), need I say more?

OK, I don't know the definition of "effective mass" or "effective charge". What is the equation for them to be "properly equated"?

OK yes I did miss it. Advanced math notation already for me - but really you will surely agree that is a finished statement not a first-principles derivation.

It is simply the source-free Maxwell's equations in curved spacetime. I consider that first principles. Do you object to Maxwell's equations as an axiom?

 

[Q-reeus]

So you agree with the statement, but think that the axioms for deriving the RN spacetime are not correct.
Can you please point out exactly which axiom is incorrect in your view?

You should have figured it by now - I work it 'backwards'. Setup a gedanken experiment. Check for consistent predictions. If lacking, trial various possible fixes. Key example here is my #248. Don't ask me to press that into a new axiom. I don't pretend to be a university educated boffin. Just try at being a logical thinker.

OK, I don't know the definition of "effective mass" or "effective charge". What is the equation for them to be "properly equated"?

My take: In #14, first 'a=' RN equation, imo whatever terms modify that expression from purely Schwarzschild solution involving just neutral M. Hence everything there with a Q expresses the effective mass contribution at r owing to distributed E field energy of Q up to that r (or the 'subtraction' from larger than r version). As for effective charge, perhaps you can give your own interpretation on what last expression in #345 is suggesting. My own idea is there in #248. Nothing rigorous, but it seems to be adding up better than elsewhere.

It is simply the source-free Maxwell's equations in curved spacetime. I consider that first principles. Do you object to Maxwell's equations as an axiom?

"In curved spacetime" - my little gedankens are worrying me about the marriage of ME's to such. Feeling a bit feisty at the moment btw?

 

[stevendaryl]

Well, the thing is that in GR the situation you describe in your first paragraph is made equivalent to the situation you describe in the second paragraph (at least locally but I'm taking the liberty of considering the book a local object) by virtue of the well known Equivalence principle, and that is the only way to consider gravity a fictitious force. See what I mean?

Not EVERYTHING is the same when comparing constant acceleration in flat spacetime with being at rest in a gravitational field. In particular, there is a big difference in that an observer in freefall in flat spacetime sees an unchanging metric, while an observer in freefall in the gravitational field of the Earth sees a time-varying metric.

Here's a way to sort of make sense of work and energy in curved spacetime.

Let's assume a special case to make things simple: In terms of Schwarzschild coordinates, we have a weak nongravitational force corresponding to a static electric potential $\Phi$. We'll assume that the force is so weak that its contribution to the metric is negligible, so we can use the Schwarzschild metric. Then the following quantity is conserved for a particle of mass m and charge q:

E = mc²√A/√(1-v²/(c²A²)) + q $\Phi$

where A = 1 - 2GM/(c²r) = the Schwarzschild factor. Notice that in the limit as GM → 0, A → 1 and we get the SR limit:

E = mc²/√(1-v²/c²) + q$\Phi$

When M is nonzero, the expression for E doesn't cleanly split up into "potential energy" and "kinetic energy"; the first term includes both velocity-dependent and position-dependent quantities. However, we can approximate E in the case where v/c << 1, and GM/(c²r) << 1. In that case, this expression simplifies to

E = mc² - GMm/r + 1/2 mv² + q $\Phi$

So in this limit, we can see that any work done by the electric field on the particle goes into either increasing the kinetic energy 1/2 mv² or in increasing the gravitational potential energy - GMm/r, just as in Newtonian physics. It's still the case that if r is constant, then no work is being done to the particle by the electric field.

In flat spacetime with an accelerating rocket, we can look at things from two different points of view: from the point of view of an inertial observer, the only energy is kinetic energy, and work must be done in order to keep the rocket accelerating.

From the point of view of an accelerating observer (Rindler coordinates), the rocket is stationary, and no work is being done on the rocket. So where does the energy used up by the rocket go? It goes into throwing exhaust gases backwards. Those exhaust gases are NOT stationary, and so work is done to get them moving backwards.

 

[stevendaryl]

I refer you to my 'intuitive' resolution in #248 for what that means re Q(r) 'in effect'.

Have you gone back as per directive in #340 and looked at the situations there? You have a different slant that actually makes sense? No equations per se there, but please think about it. And there is some supplementary considerations below.

You aren't starting with what General Relativity says about these things. You're making stuff up, and then reasoning from it in a sloppy, non-logical way. If you end up with a contradiction, you can't blame it on GR, because you're not using GR.

You object to using complicated mathematics, but mathematics is nothing more than being precise. By not using mathematics, you're not being precise, you're being sloppy. Sloppiness generates apparent inconsistencies where none exist.

 

[Q-reeus]

You aren't starting with what General Relativity says about these things. You're making stuff up, and then reasoning from it in a sloppy, non-logical way. If you end up with a contradiction, you can't blame it on GR, because you're not using GR.

You object to using complicated mathematics, but mathematics is nothing more than being precise. By not using mathematics, you're not being precise, you're being sloppy. Sloppiness generates apparent inconsistencies where none exist.

We've been here before haven't we? I won't debate that kind of polemic.

 

[stevendaryl]

We've been here before haven't we? I won't debate that kind of polemic.

It wasn't polemic--it was, I believe, an accurate description of what you are doing.

 

[stevendaryl]

It wasn't polemic--it was, I believe, an accurate description of what you are doing.

It's actually an explanation for why you are getting inconsistencies--they are your own creation, they are not from GR. That's harsh, but it's accurate.

 

[Dale]

You should have figured it by now - I work it 'backwards'. Setup a gedanken experiment. Check for consistent predictions.

This is a fundamentally impossible approach, for the reasons cited above. You cannot ever either prove or disprove self consistency in this manner.

"In curved spacetime" - my little gedankens are worrying me about the marriage of ME's to such.

OK, now we have something. You doubt that the expression in 196 correctly represents ME in curved spacetime. Do I understand your position correctly?

If you beleived that equation were the correct form of ME in curved spacetime then would you agree that the derivation is correct (ie correct axioms, formally correct derivation, therefore correct conclusions)? If not, then which other axioms are also suspect?

 

[TrickyDicky]

Not EVERYTHING is the same when comparing constant acceleration in flat spacetime with being at rest in a gravitational field. In particular, there is a big difference in that an observer in freefall in flat spacetime sees an unchanging metric, while an observer in freefall in the gravitational field of the Earth sees a time-varying metric.

Here you are objecting the equivalence principle when it says that the two situations are physically indistinguishable I guess.
Also your example uses a static spacetime too, no time-varying metric there either.

 

[stevendaryl]

Here you are objecting the equivalence principle when it says that the two situations are physically indistinguishable I guess.
Also your example uses a static spacetime too, no time-varying metric there either.

The equivalence principle doesn't say that curved spacetime is indistinguishable from flat spacetime. Of course, they aren't, because the curvature is an observable. The way that it comes into play in GR is just that "gravitational force" is actually a pseudo-force due to a particular choice of coordinates; it can be made to vanish locally by choosing locally inertial coordinates. That's really the only content of the equivalence principle.

As for my examples, yes in both Schwarzschild coordinates and in Rindler coordinates, the metric tensor is time-independent. And in both cases, there is no work done on an object at "rest" in those coordinates. So the idea that it takes work to keep an object in place is not correct.

 

[PeterDonis]

I got 2M in the limit.

Can you elaborate? K goes to 1 in the limit, so the first term in what I obtained is just M, and the other two cancel.

 

[PeterDonis]

My R referred to the shell radius specifically and it's that R that governs the depressed potential experienced by any charge inside said shell.

Ah, ok, I misread. Then I'll defer comment until I've had a chance to analyze the shell case.

 

[PeterDonis]

Interesting result there Peter.

I should note that on comparing this result for Q(r) with the equation I posted early in the thread for the proper acceleration on a test charge due to the charge of the hole, I'm not sure they're consistent. So I need to check things some more.

 

[Q-reeus]

Q-reeus: "You should have figured it by now - I work it 'backwards'. Setup a gedanken experiment. Check for consistent predictions."
This is a fundamentally impossible approach, for the reasons cited above. You cannot ever either prove or disprove self consistency in this manner.

There is a famous line that goes something like "A theory can be proved true a thousand times; but one counterexample and it's dead." I take it you reject that possibility outright - there can be no such thing as a counterexample. So you believe the one and only way to check on a theory is via experimental/observational evidence then?

Q-reeus: ""In curved spacetime" - my little gedankens are worrying me about the marriage of ME's to such."
OK, now we have something. You doubt that the expression in 196 correctly represents ME in curved spacetime. Do I understand your position correctly?

Yes, and not based on any direct analysis of that expression which as you know I can't even properly interpret. I base it on the problems imo raised as per everything previously presented in this thread - #1 and #109 summarise well enough. Do I need to repeat those yet again, or are you now as you should be, thoroughly familiar with exactly where I see inconsistencies appearing?

If you beleived that equation were the correct form of ME in curved spacetime then would you agree that the derivation is correct (ie correct axioms, formally correct derivation, therefore correct conclusions)?

Yes, with possible proviso it's unphysical boundary conditions that have been applied to RN case, and just how Poisson sets his psi = o to get F^tr =Q/r² in (5.22) on p177 of that previously linked article is perhaps under that category, or that best thought of as direct consequence of 'the coupling of EFE's to ME's', I'm not sure.

If not, then which other axioms are also suspect?

None I can think of. Now, with permission from the court I seek temporary leave persuant of legal council with an appointed barrister!:tongue:

 

[TrickyDicky]

The equivalence principle doesn't say that curved spacetime is indistinguishable from flat spacetime.

I never said the equivalence principle says that.

As for my examples, yes in both Schwarzschild coordinates and in Rindler coordinates, the metric tensor is time-independent.

So the "big difference in particular" that you claimed in your previous post between the gravitational field situation and the SR one is not a difference.

 

[stevendaryl]

So the "big difference in particular" that you claimed in your previous post between the gravitational field situation and the SR one is not a difference.

I was talking on the one hand about freefalling observers, and on the other hand about accelerated observers.

What I said was (and I quote)

"an observer in freefall in flat spacetime sees an unchanging metric, while an observer in freefall in the gravitational field of the Earth sees a time-varying metric"

That's true.

We have 4 cases:

1. Accelerated observer "at rest" in flat spacetime using Rindler coordinates.
2. Observer in "freefall" in flat spacetime using inertial coordinates.
3. Accelerated observer "at rest" in Schwarzschild spacetime using Schwarzschild coordinates.
4. Observer in "freefall" in Schwarzschild spacetime using locally inertial coordinates.

In all cases except the last, the metric is time-independent. In the last case, the metric is time-varying (as he gets closer to the center of the source of gravity, the curvature becomes stronger).

 

[TrickyDicky]

I was talking on the one hand about freefalling observers, and on the other hand about accelerated observers.

What I said was (and I quote)

"an observer in freefall in flat spacetime sees an unchanging metric, while an observer in freefall in the gravitational field of the Earth sees a time-varying metric"

That's true.

We have 4 cases:

1. Accelerated observer "at rest" in flat spacetime using Rindler coordinates.
2. Observer in "freefall" in flat spacetime using inertial coordinates.
3. Accelerated observer "at rest" in Schwarzschild spacetime using Schwarzschild coordinates.
4. Observer in "freefall" in Schwarzschild spacetime using locally inertial coordinates.

In all cases except the last, the metric is time-independent. In the last case, the metric is time-varying (as he gets closer to the center of the source of gravity, the curvature becomes stronger).

Basic facts of GR and differential geometry: Schwarzschild is an static spacetime, changes in coordinates cannot alter the physics nor the metric from time-independent to time-dependent. Do you agree with these facts or not?

 

[stevendaryl]

Basic facts of GR and differential geometry: Schwarzschild is an static spacetime, changes in coordinates cannot alter the physics nor the metric from time-independent to time-dependent. Do you agree with these facts or not?

I agree with the first, in the sense that if you express the physics in covariant form, then the physics is the same in all coordinate systems. But the description in terms of "gravitational potential energy" is NOT a covariant way of describing things. You can only use that description in special coordinates

Your second statement is completely wrong. Changes in coordinates can certainly change a time-independent metric into a time-varying one. If you are changing from one set of coordinates $X_{\mu}$ to another set of coordinates $X'_{\alpha}$, the metric tensor changes as follows:

$g'_{\alpha\beta} = \partial_{\alpha}X^{\mu} \partial_{\beta}X^{\nu} g_{\mu\nu}$

If the quantity $\partial_{\alpha}X^{\mu}$ is time-dependent, then $g'_{\alpha\beta}$ can be time-dependent, even if $g_{\mu\nu}$ is not.

For example, start with Rindler coordinates (in 2D spacetime, for simplicity) the coordinates are $X$ and $T$, and the metric components are:

$g_{TT} = X^{2}$
$g_{XX} = -1$

The metric components are time-independent. Now, switch to new coordinates $x$ and $t$ related to $X$ and $T$
$X = x + vt$
$T = t$

where $v$ is some constant. Then the metric in the new coordinates looks like this:

$g_{tt} = (x+vt)^2 - 1$
$g_{tx} = -1$
$g_{xx} = -1$

The metric component $g_{tt}$ is time-dependent.

 

[stevendaryl]

Basic facts of GR and differential geometry: Schwarzschild is an static spacetime, changes in coordinates cannot alter the physics nor the metric from time-independent to time-dependent. Do you agree with these facts or not?

I gave a detailed counterexample to your last claim, but really, you should see it immediately: whether or not something is "time-varying" depends on what you choose as your time coordinate. So being "time-independent" is not a coordinate-free notion.

 

[TrickyDicky]

Your second statement is completely wrong. Changes in coordinates can certainly change a time-independent metric into a time-varying one.

There is a recent thread devoted to clarify that changes of coordinates cannot change Killing vector fields.

 

[Dale]

So you believe the one and only way to check on a theory is via experimental/observational evidence then?

No. I also believe that it can be falsified by a rigorous formal proof.

Yes, and not based on any direct analysis of that expression which as you know I can't even properly interpret.

Well then I will endeavour to help you learn to interpret it.

For now, why don't you start with the relevant Wikipedia page for Maxwell's equations in tensor notation. Please let me know what, if anything, you have trouble with, and then I can point you to other appropriate resources:
http://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism

This page is just about the notation rather than any new physics. I.e. it is a tensor formulation for arbitrary coordinates in flat spacetime.

Yes, with possible proviso it's unphysical boundary conditions that have been applied to RN case

This I agree to without reservation. We have no direct experimental observation of a black hole let alone a charged black hole. The boundary conditions may very well be unphysical, that is an experimental matter which cannot be decided in a thread, but we shouldn't forget it either.

 

[stevendaryl]

There is a recent thread devoted to clarify that changes of coordinates cannot change Killing vector fields.

That doesn't have anything to do with what I said. What I claimed is that whether the metric components are time-varying (meaning: their derivative with respect to the time component is nonzero) is a coordinate-dependent fact. That's obviously true.

It's also true that if there is a timelike Killing vector field, then the metric is unchanged by translation along the Killing vector. But that doesn't mean that the metric is time-independent UNLESS the basis vector in the time direction happens to be the same as the Killing vector field; which is true of Schwarzschild coordinates and Rindler coordinates and inertial coordinates in flat spacetime.

 

[stevendaryl]

There is a recent thread devoted to clarify that changes of coordinates cannot change Killing vector fields.

This relates to the question of time-varying metric components in the following way:
If there is a time-like Killing vector field, then there exists a coordinate system in which the metric components are independent of time. You seem to be interpreting this as: If there is a time-like Killing vector field, then in EVERY coordinate system, the metric components are independent of time. That's clearly not true.

 

[TrickyDicky]

... That's clearly not true.

You might as well take a look at any definition of KV fields and specifically the fact they are coordinate-independent, you don't need to take my word for it.

 

[Dale]

Hi TrickyDicky, stevendaryl is correct. I think that you are confusing the coordinate independent concept of "static" and/or "stationary" with the coordinate dependent concept of "time varying".

 

[stevendaryl]

You might as well take a look at any definition of KV fields and specifically the fact they are coordinate-independent, you don't need to take my word for it.

For a number of exchanges, I have made a statement X, and you've said, No, Y is true. But Y doesn't mean that X is not true.

Statement X: Whether the components of the metric tensor is time-varying depends on which coordinate system you are using.

Statement Y: Whether there is a timelike Killing vector field does not depend on which coordinate system you are using.

Statement X is true AND statement Y is true. They don't contradict each other.

Now, to be fair, it's possible that some people use "static metric" to mean "there exists a timelike Killing vector field". But I explicitly said that I was talking about whether the components of the metric tensor are independent of the time coordinate. Those are two different things, and you act as if you don't understand the distinction. The first is a coordinate-independent notion, and the second is a coordinate-dependent notion.

 

[stevendaryl]

Hi TrickyDicky, stevendaryl is correct. I think that you are confusing the coordinate independent concept of "static" and/or "stationary" with the coordinate dependent concept of "time varying".

Thanks, Dale.

 

[TrickyDicky]

Hi TrickyDicky, stevendaryl is correct. I think that you are confusing the coordinate independent concept of "static" and/or "stationary" with the coordinate dependent concept of "time varying".

Nope, I never used the concept "time varying". Stevendaryl did.
He is saying a static spacetime can have metric components not time-independent and I was merely reminding him that the timelike KV of static spacetimes preserves the metric. Are you confused about this?

 

[TrickyDicky]

I was talking about whether the components of the metric tensor are independent of the time coordinate. Those are two different things, and you act as if you don't understand the distinction. The first is a coordinate-independent notion, and the second is a coordinate-dependent notion.

Both notions are the same notion.
A metric is called stationary if its components are time-independent. And all static spacetimes are stationary.

 

[TrickyDicky]

And just to be clearer, it is of course possible to use coordinates that don't reflect the time-independence of the spacetime but here we are trying to ellucidate the physics of the problem, not coordinate artifacts. I guess I took that for granted.