# Steven Crothers , against BB

Discussion in 'Pseudoscience' started by river, Nov 30, 2017.

1. ### Q-reeusValued Senior Member

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You will have likely better intel on that one. Fine by me whatever the case.

3. ### Xelasnave.1947Valued Senior Member

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No just guessing like I do with my science.
I do worry that something has happened to him if he is not around given how regularly he posts...
Alex

5. ### przyksquishyValued Senior Member

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I don't know where you're getting this from. The Reissner-Nordström metric is derived from the Einstein field equation and Maxwell's equations following the principle of general covariance. I.e., you assume that Maxwell's equations (in differential form) hold in any locally inertial reference frame, according to the equivalence principle. From there it is a routine (and solved) mathematical exercise to derive the form they take in any arbitrary noninertial coordinate system.

The main difference is that an electromagnetic field has energy/momentum density of its own, which has its own effect on the gravitational field, so the problem is to derive both a spacetime geometry and an electromagnetic field that together satisfy both the Einstein field equation and Maxwell's equations. This is what the Reissner-Nordström solution does.

7. ### Q-reeusValued Senior Member

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Yes the gravitational field is obviously modified by the presence of the E field energy density - but according to standard GR RN BH - NOT the other way round.
The actual field solution for a RN charged BH has the E field given simply as shown by (5.2.3) - top of p136 here:
http://www.physics.uoguelph.ca/poisson/research/agr.pdf
It's just the Coulombic field for flat spacetime! That Q is immune to gravity. Or so RN solution says. My thought experiment exposes the inconsistency.

As you evidently see no issue, be good enough to provide your own analysis of my spherical capacitor scenario. One free of paradox while within RN framework.
Or feel free to substitute with any similar gedanken experiment e.g. partial separation/contraction of an electric dipole, oriented wrt the BH any way you wish.

Do we at least agree it's not just legitimate but essential to test for theory self-consistency by such means?

8. ### przyksquishyValued Senior Member

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So? Like I told you, both the Reissner-Nordström metric and the associated electromagnetic field are derived as solutions to the Einstein field equation (general relativity) and Maxwell's equations (electromagnetism) in the static and spherically symmetric case. It's been nearly ten years since I studied general relativity, but the derivation of the Reissner-Nordström metric is one of the things I remember personally working through in detail and I do remember this is how it is done (or at least, this is one possible way of doing it). So the only way you can get a different result is if you -- somewhere -- introduce assumptions of your own that don't follow from one or both of these theories.

What about it? What you said about the scenario honestly made next to no sense to me.

9. ### Q-reeusValued Senior Member

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Really and truly? I notice you carefully avoided answering my last question in #84. Evidently then, having no trouble deriving RN solution, consideration of a very simple gedanken experiment testing it quite baffles you?! OK you prefer to rely on established solutions and see no merit in ever putting such to a simple self-consistency test. So be it.

10. ### przyksquishyValued Senior Member

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Yes. You do realise it is your responsibility to make sense in the first place, right?
I don't see much of a coherent argument in this.

What, this terrifying question:
<sarcasm>No, I think it's perfectly okay for a model to be self contradictory.</sarcasm>

Honestly, what answer were you expecting?

Really. I take an exact solution more seriously than handwaving, and you think this is irrational?

Last edited: Dec 5, 2017 at 2:27 AM
11. ### Xelasnave.1947Valued Senior Member

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Did anyone see Mr Crowthers review of the CBR map.
I was going to look again but on first view he seemed to have a point in respect to the map dealing with the milky way region.
I really didn't take much of what he said in such that I thought deeply on his reasoning but more that re the map I think it must be difficult if not impossible to get anything reliable from the region of the milky way.
Alex

12. ### Q-reeusValued Senior Member

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A lot of negative attitude statements. But here's how I will call you out. Provide your own ever-so-coherent worked-out scenario that refutes what I have claimed to show.
I even gave an obvious alternate - electric dipole. But really the spherical capacitor is a natural choice. Too hard for you? Or you just prefer to snipe away like above?
See by providing your own worked devastating counter-example the matter could be *objectively* decided. Or feel free to get down to some specific line-by-line critique of my #79.

I have a feeling you will do neither.

13. ### przyksquishyValued Senior Member

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Yes. Really and truly.

You are making arbitrary demands and trying to shift the burden of proof onto me for a claim you made. This is not "calling me out".

14. ### Q-reeusValued Senior Member

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Yes it is. A specific scenario revealing lack of self-consistency with NR BH solution was given in #79. Negative comments devoid of substance are no refutation. Asserting error is easy. Proving it via a clear and simple worked counter-example is another thing entirely. What's stopping you from just furnishing one?

15. ### przyksquishyValued Senior Member

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In my opinion, you sketched out a pseudo-calculation without bothering to explain what you were doing, what the point is, or justify how it is grounded in general relativity, in order to arrive at some "redshifted" E without any explanation of what that is, why it is relevant, or why anyone reading your post should care about it, never mind how that is supposed to "reveal a lack of self-consistency" with anything whatsoever.

Like I said, if you want people to take you seriously then it is your responsibility to make arguments that make a minimum amount of sense in the first place.

16. ### Q-reeusValued Senior Member

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Sincerity would be signaled by a genuinely objective critic asking for clarification on points not clear to such. The complete absence of that here means it's a waste of time to continue any further discourse. Maybe someone else here actually gets it. Then again SF is basically a hangout for idle chatter and PC rantings, so no high hopes.

17. ### NotEinsteinRegistered Senior Member

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This might just be my ignorance, but I don't see how some effective mean factor dealing with masses can be applied to the work done by electric fields?

18. ### Q-reeusValued Senior Member

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As you surely know well, a proper release of energy down in some gravitational potential, is redshifted i.e reduced/dilated when received at a higher potential.
I had assumed it would be understood that a sufficiently thin spherical capacitor can be considered to a good approximation as residing in a depressed metric that varies little over the depth of the capacitor - hence to save needless labour i.e. an exact integration over r, an effective average for √(g_00) applying at the mean radius of the capacitor was implied.
I suggest considering my next post, which adds some extra thoughts to #79.

19. ### Q-reeusValued Senior Member

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Continuing on from #79, a certain amount of further thinking is perhaps in order. Consider a single electric dipole, of proper moment qs, lying at a given static radius r and oriented transverse to r - defined wrt a spherically symmetric neutral mass M. The electric field owing to the dipole is weak and negligibly perturbs the Schwarzschild exterior metric owing to M. Transverse length in standard Schwarzschild metric is explicitly unaffected by gravitational potential.
A differential contraction of the dipole spacing ds will give off a proper i.e. local differential energy release dW as before, dilated by factor dW' = √(g_00)dW wrt a distant observer. But now it's completely clear the proper force of attraction F = Eq acting on each charge owing to the field of the other, is the quantity that has to be dilated by F' = √(g_00)F = E'q' to account for dW' = F'.ds' = E'q'ds', since ds = ds' explicitly.

That this coordinate determined reduction in F is 'real' can be made obvious. Just run a light string down to an anchored pulley, then to one charge of the dipole - anchored at the other charge. The force exerted 'out there' required to hold one charge from moving wrt the other (assuming no 'stick' holding them apart) is exactly the reduced F' defined earlier. This immediately shows the coordinate reduction in product Eq is quite real, and incompatible with RN result q and E are unaffected by depressed metric value √(g_00).

Which bizarre property of RN solution is acknowledged in section 5 here: http://arxiv.org/abs/gr-qc/0001010

Well RN solution thus implies that F' = F thus dW' = dW for either spherical capacitor or above dipole arrangements. Not hard to turn that into a cyclic energy creation scheme.

But that still leaves an ambiguity as to how dilation factor √(g_00) should be apportioned between E and q. One could argue E = E' is unchanged and each charge simply feels less coordinate determined force i.e. F' = Eq' = E{√(g_00)q}. Or vice versa, that charge is just q' = q and √(g_00) operates just on E such that F' = E'q = {√(g_00)E}q.

A third possibility is an even split: F' = E'q' = {√√(g_00)E}{√√(g_00)q}.
That last option has the advantage of seeming more compatible with a dilated E field energy density pov viz E'^2 = {√√(g_00)E}^2 = √(g_00)E^2.
On that basis, both charge Q and Field E for a charged mass are distantly reduced by the square root of √(g_00). Where √(g_00) applies at the location of source charge Q.

What cannot sensibly hold is the RN position that charge and it's field are unaffected by gravity. Because it implies zero gravitational redshift of energy. As per above.

Last edited: Dec 5, 2017 at 10:22 AM
20. ### NotEinsteinRegistered Senior Member

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From what I can tell, the charge (q) of a particle is indeed not changed by gravitational redshift, but I'm suspecting the E-field is. The E-field is not a four-vector, so it's not Lorentz invariant. Yes, this is hand-wavy, but I'm seeing multiple people saying that the energy stored in the E-field is affected by gravitational redshift. I suspect that the Reissner–Nordström doesn't keep the E-field the same under gravitational redshift either. I haven't read the whole thread; where is this claim coming from?

21. ### Q-reeusValued Senior Member

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Have you read section 5 re article linked to in #96? They make it clear RN metric has EM essentially immune to gravity. As they write: 'switch gravity on, and nothing changes re underlying EM field'. Which then leads to the nonsensical paradox shown earlier here. As for RN having Q or E changed - see #84 where eqn (5.2.3) and accompanying text in Eric Poisson's pdf 'An advanced course in GR' is given. Explicitly, Q and E are both unaffected by gravity according to RN. That EM 'occupies it's own spacetime metric independent of gravity' flies in the face of everything else a metric theory of gravity implies.

There is yet another way of seeing the absurdity. According to Birkhoff's theorem, an infalling spherical shell of 'dust' matter presents the same externally observed gravitational mass M at every instant - for a given observer at fixed r that is. Which is reasonable since the descent into a deeper and deeper potential and thus more and more depressed metric √(g_00), is exactly compensated for by the increase in KE of infall. To a locally hovering observer, shell mass density is greater than if it were at rest there by the inverse of √(g_00).
For the distant observer, the two effects cancel exactly. Which can only mean the count of 'gravitational field lines' reduces the further out one observes from.
Gauss's law for gravitational mass as 'charge' fails in GR. Gravitational 'field lines' occasionally terminate mid-air.

Now suppose that infalling dust is very weakly charged. Not enough to appreciably effect the dynamics of infall, but still there as source of an overall spherically symmetric E field. For the locally hovering observer, charged dust infalling past such has a locally measured charge-to-mass ratio reduced by factor √(g_00) relative to stationary dust there. Hence, one should expect that, given M is constant for a given external observer, Q as source of measured E effectively diminishes as infall progresses.
Not according to RN. It's ok for Gauss's law for gravity to fail, but mysteriously remains inviolate for electric charge. I don't think so.

22. ### NotEinsteinRegistered Senior Member

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Can you point me to the exact line? I can't find the search term "EM field" anywhere in that paper. I did find this in section 5: "Our conclusion is then that, as long as we opt for electric charge and magnetic ﬂux conservation, the Maxwell equations in gravity-free regions, i.e., in the Minkowski spacetime of special relativity, read dH = J and dF = 0; they remain the same irrespective of the switching on of gravity, be it in Einstein’s theory, in metric-aﬃne gravity (see [76]), or in any other geometrical theory of gravity."
But here they are not talking about the EM fields, just the Maxwell equations. I also don't see them mentioning the RN metric explicitly, but I guess that can be inferred from the sentence I quoted.

I don't see any explicit mention of the E-field in 5.2.3? Equation 5.2.3 gives the trace of the electromagnetic field tensor (if I understand their notation); is there a derivation from this to the E-field given elsewhere?

I'm pretty sure GR typically isn't done by counting field lines. Can you perhaps express this in more standard terminology?

Again, from what I can tell, charge doesn't get modified, only the E-field. In fact, that charge doesn't change is mentioned quite explicitly in the "An advanced course in GR" PDF: "where Q is a constant of integration, to be interpreted as the black-hole charge." (a couple of lines above eq. 5.2.1).

Q isn't supposed to change.

23. ### Q-reeusValued Senior Member

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I was to be honest paraphrasing from memory. Anyway if ME's are as claimed unaffected by gravity, in particular re RN then Gauss's law for electric charge is unaffected. Hence, charge is not a function of r regardless of gravitational potential, whereas mass is. Which answers your point later re eqn (5.2.3) - no it can't be interpreted as the trace since only Q not M is involved. That expression only makes sense as the Coulombic field E(r).

To cut to the chase regarding the rest there - 'gravitational field lines are not part of GR' etc. - see: https://en.wikipedia.org/wiki/Komar_mass
The appropriate GR analogue to Newtonian 'gravitational field lines' is set out explicitly there. Note the acknowledgement that on that basis, Keplerian gravitating mass is a reducing function of r.
And my point is - why should that not also hold for charge? You see no problem evidently. Despite the scenarios examined earlier. I do.
Especially in the extreme case of a 'charged BH' - one has an E field just hanging in space - causally cut off from it's source hidden behind an EH. Makes no real sense.