# Steven Crothers , against BB

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

1. ### river

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But physically what does this mean.

A vacuum solution the Einstein field equation ?

Perhaps you have , sure as far as the theory and science is concerned , no doubt .

But what in terms as a realistic theory , thats the question .

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He was right

5. ### arfa branecall me arfValued Senior Member

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If by "He" you mean Einstein, then no, he wasn't right.

That said, he did say that the inventor of a theory probably understands it the least. Good one Albert.

I think you can chalk that up to a truly critical thinker having the humility to accept they were wrong about something. Being unable to accept you're wrong is a sign of something else, methinks.

And as for Mr Crothers. I did think briefly about looking at some of his videos, I even started watching the one Xelasnave posted, but no equations (?!), just talk and lots of repetition. So instead I decided to just get on with my life.

Last edited: Dec 3, 2017

7. ### river

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Depends as always how one interprets red shifts from galaxies .

In the main , redshifts from galaxies means that they are moving away from us .

In another the redshifts are the indication of young galaxies being formed .

8. ### Xelasnave.1947Valued Senior Member

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I have been watching a few of the anti main stream utube presentations and strangely I am realising I know more than I thought I knew as I can work out where many of these folk have it wrong.

I like to see the positive side and think it is great many folk are so interested in science even if their interest manifests in them talking in ignorance.

I think an anti mainstream nutter is better to listen to than folk discussing football, or cricket ...at least they pursue something in my view worthwhile.
Alex

9. ### river

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I see

Science continues to grow

And that means looking at other perspectives on what is taken for granted , as science .

10. ### przyksquishyValued Senior Member

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It means that the Schwarzschild metric is derived as a solution to the Einstein field equation with the stress-energy tensor $T_{\mu \nu}$ set to zero everywhere (except at the singularity, where it is undefined). Setting $T_{\mu \nu} = 0$ in some region of space is how you express in general relativity that there is nothing there -- literal vacuum -- in that region. So the Schwarzschild metric describes the simplest possible situation, where you have a single black hole sitting in space surrounded by vacuum.

11. ### river

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So why set to zero ?

But not at the singularity ? And being undefined , means really it means nothing .

But there is always something there , there is no such thing as a pure vacuum , anywhere in space .

BH becomes nothing more than a mathematical imagnative concept .

Last edited: Dec 4, 2017
12. ### przyksquishyValued Senior Member

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Why not? If you're learning general relativity then you need to learn what kind of spacetime geometries are predicted by the theory. You learn this by learning to solve the Einstein field equation for different situations. Vacuum is simpler than nonvacuum so it makes sense to start learning about the vacuum solutions.

It's the same with anything else in physics. If you're a student starting to learn physics for example then one of the first things you learn to do is to solve simple projectile motion problems, e.g. a stone is thrown in a certain direction with a certain speed and you have to calculate where and when it will land. To keep the problem simple, you would normally be told to ignore air resistance. This is not because air doesn't exist but because taking it into account makes the problem more difficult to solve and often enough (e.g. if the stone is heavy and it isn't thrown too fast) it doesn't make a significant difference and it is not worth the effort.

No. Of course the universe is not empty except for just one black hole. So the Schwarzschild metric does not describe the whole universe. Sometimes this matters and sometimes it doesn't.

For example, you might have a large black hole in deep space. In that case you probably don't care about the relatively small gravitational influence of a small nearby spaceship, or a few stray cosmic rays, or of stars many lightyears away, so you could use the Schwarzschild metric as a good first approximation of the spacetime geometry in the vicinity of the black hole.

On the other hand, maybe you have a black hole tightly orbiting a large star, or another black hole, or you have an electrically charged black hole surrounded by a strong electromagnetic field. These are situations very different from what the Schwarzschild metric is derived for, so you can't use it and you just have to try to solve the Einstein field equation for those cases.

Last edited: Dec 4, 2017
13. ### river

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So there are no field equations , just A field equation .

Okay , now introduce nonvacuum spacetime geometries .

Which changes the dynamics of three dimensional objects

14. ### river

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Just one black-hole ....hmmmm
And how would a black-hole become electrically charged ?

15. ### river

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And That is what Steven argues against .

You should know that from Steven's presentations .

The BB and black-holes are contradictory .

Last edited: Dec 4, 2017
16. ### przyksquishyValued Senior Member

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This is a matter of how you want to count things. The Einstein field equation says that the Ricci tensor and the stress-energy tensor components are related by

$R_{\mu \nu} - \frac{1}{2} R \, g_{\mu \nu} = \kappa \, T_{\mu \nu} \,.$​

Each of the tensors has ten independent components (like how 3D vectors have $x$, $y$, and $z$ components). So you can count the Einstein field equation as ten differential equations for the ten different components or you can count them together collectively as one tensor equation.

You should be able to find examples of nonvacuum solutions to the Einstein field equation in any good reference on general relativity as well as the research literature.

However, things get complicated fast: the Einstein field equation is a very difficult equation to solve and it is not known how to solve it in general. Most of our understanding of general relativity is from situations that are simple enough that someone has been able to solve the equations or we can get away with making simplifying approximations. This is, by the way, the normal state of affairs in physics. It is not unique to general relativity.

But just to give one example: if you have a charged black hole surrounded by an electromagnetic field then this is not a vacuum situation (the electromagnetic field has nonzero energy associated with it, so it cannot be treated as vacuum). In that case the problem has been solved exactly and the solution, called the Riessner-Nordström metric, is one example of a nonvacuum solution to the Einstein field equation which is different from the Schwarzschild metric.

It might form from the collapse of charged matter, or charged matter might fall into it.

Then he is knocking down a strawman, since he is arguing against something nobody claimed in the first place.

Like I said, Crothers is complaining about something that actually has nothing specially to do with general relativity. Almost all applications of any physics, not just general relativity, involve some sort of idealisation or approximation. This is just a consequence of the real world being far too complicated to model in every detail, so we ignore the details that don't matter. The fact that Crothers thinks this can somehow be used as an argument specifically against black holes tells me one of two things about him: he either knows very little about science or he is counting on you knowing very little about science.

Last edited: Dec 4, 2017
17. ### river

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Indeed

White space ? Explain further .

18. ### river

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But nothing can escape the BH . How is charge escape velocity greater than the speed of light ?

19. ### przyksquishyValued Senior Member

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It isn't. Nothing is escaping it. It's just a black hole with a static electromagnetic field surrounding it.

20. ### river

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So why is static electromagnetic field , still not drawn into this BH ?

21. ### przyksquishyValued Senior Member

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Why are you assuming it should be "drawn in"? A static electromagnetic field, whether it is surrounding ordinary charged matter or a black hole, does not require any power to maintain. If it's there then it's just there. It is not like a rocket which would need to consume fuel to maintain a position outside a black hole.

22. ### Q-reeusValued Senior Member

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The basic premise allowing for a charged BH is that Gauss's law for electric charge is fundamental and holds globally not just locally, irrespective of any spacetime curvature present. That is, the net E-field flux-line count over any bounding enclosed surface is invariant for any gravitating charged body - including a BH.

I raised the matter of whether that made sense some years back at PhysicsForums. It was easy to show that it couldn't but not a single GR buff there bothered to respond to my critique. The easiest way to undermine the notion is to tackle the simpler case of a charged spherical capacitor. Overall neutral but having some mean field E acting everywhere radially between the inner and outer, oppositely charged conducting spherical surfaces A & B.

If the gap between A & B is say reduced by a differential increment dr, then locally, that represents a net release of energy dW = F.dr = qEdr. (It's understood that F really represents the net vector force if the spherical capacitor were unfolded to form a parallel-plate capacitor.)
But we are dealing with GR here as that capacitor is also a gravitating body residing in a gravitational potential that has 'redshifted' it's unassembled mass M by an effective mean factor M'/M = √(g_00). Where M' is the assembled i.e. gravitationally depressed mass, and where √(g_00) is just the usual frequency gravitational redshift factor. Hence, 'from afar', the net energy release should show as dW' = dW√(g_00) = q{√(g_00)E}.dr. Where use of curly brackets emphasizes that it is E that logically has to be 'redshifted'.
(Well ok as to whether one chooses q or E as ' fundamentally redshifted' can be down to a pov choice, but at any rate the far field and it's bounding flux has to less than commonly assumed.)

On the standard view E is mysteriously completely immune to otherwise universal action of √(g_00). Which situation would readily allow in principle a perpetuum mobile.

One could just as easily begin from a field energy density basis - the conclusion is the same. Gauss's law logically fails and consistency demands that effective charge globally is reduced by exactly the same amount as assembled mass M' reduces relative to unassembled mass M. One corollary is the initial assumption above of an overall uncharged capacitor fails - the charge residing on the outer shell, despite being equal and opposite to that on the inner shell locally, 'wins out' globally i.e. as determined from afar. As it resides in a higher gravitational potential thus less 'redshifted'.

Either you see the contradiction or you don't. This feature is generic to any reasonable metric theory of gravity not just GR, but as GR is the 800 lb Gorilla, almost all instances in the literature afaik place it in GR context. So, apart from other ways of poking holes in it, seems clear to me there can be no such beast as a 'charged BH'.

PS - hope above doesn't make one member in particular too indignant. Because you know - this was presented by a Tom, Dick, & Harry type not working at 'the coal face'.
PS2 - None of above should be seen as any real comfort for Crothers btw.

Last edited: Dec 4, 2017
23. ### Xelasnave.1947Valued Senior Member

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=working on dark matter (coal)?
Where is he?
Xmas coming up my guess is Fiji.
Alex