Solutions of the nonlinear Field Equations

[Chris Hillman]

Electrovacuum =/= Vacuum

Hi, notknowing,

I was mislead by the popular view that in a black hole a point of infinite density is obtained.

Well, if you know the math such statements are trying to describe, they have some value, but yes, this is just one illustration of why popular descriptions are almost always terribly misleading in some ways.

I recapitulate things to see if I understand it correctly. First you need a sufficient amount of mass to create a black hole in the first place ("feed" it). Then, through some magic, the matter is crunched together in such a way that it is transformed into pure gravitational energy such that "the curvature itself becomes the source of curvature" (the idea which I had originally to create a kind of "geon"). Of course, in a quantum theory of gravitation, things would probably turn out to be very different.

Well, I don't know why you say "magic" unless gravitation is "magical". No-one was claiming (or if they did, they probably should not have) that matter is "transformed to pure gravitational energy", unless that is how you want to summarize the discussion in the FAQ "How does gravity get out of a black hole?" and in my long extinct but archived post to sci.physics (both cited in a post to this forum made earlier today, 25 November--- happy birthday, gtr!).

I am not sure what you mean by your remark about quantum gravity. If you meant that quantum gravity should prevent the formation of an event horizon, that is probably not true. If you mean that quantum gravity should drastically alter physics as sectional curvatures approach the reciprocal of the Planck length squared, that is almost certainly true, although we won't know the details until a successful theory appears (and is understood at least in its most basic respects).

But what then about a charged black hole ? If the stress-energy tensor is also congruently zero everywhere in this case, such that one can not speek of mass in the usual sense, where is then the charge located ? It can probably not be considered to be "smeared out" in the region inside the horizon ?

The stress-energy tensor is NOT identically zero outside a charged black hole! That is because the electromagnetic field contributes a nonzero term to the stress-energy. This is why we speak of "electrovacuum" rather than "vacuum" in a region where an electromagnetic field but no matter or other non-gravitational fields are present.

In the case of a charged object such as a hollow spherical metallic shell with uniformly distributed charge (for example), the charge is located on the surface just as you would expect from Maxwell. This would be modeled by an electrovacuum exterior (a portion of the Reissner-Nordstrom electrovacuum) matched to a charged spherical shell (which we can model as "infinitely thin" for convenience) and to a vacuum interior (a portion of the Minkowksi vacuum, since there is no electric or gravitational field inside the shell, as in pre-relativistic physics).

Chris Hillman

 

[Chris Hillman]

Bianchi identities

Hi, swimmingtoday,

THERE IS NO "ALGEBRAIC" BIANCHI IDENTITIES. There is only a Bianchi Identities Ra [bcd;e]=0 and it's rsulting (Ruv- 1/2 guvR);v= 0

Careful, while this terminology is incompletely standardized, most authors DO distinguish between algebraic and differential Bianchi identities. These are of course completely different, but all happen to have been discussed by Luigi Bianchi, hence the names. See for example the fine discussion in MTW.

Someone, I think Robphy, said <<Weyl and Ricci are both zero, then Riemann must be zero>>, and of course that is true. The easiest way to remember this is to recall the Ricci decomposition I described in another post I made today in this thread.

If the Riemann tensor is such that R0101 = -(R0202 + R0303) then the Weyl tensor vanishes

That's not true as stated, of course (but maybe I have simply lost track of the context of the original remark?), since you only gave one algebraic condition, reducing the number of algebraically independent components from 20 to 19, which leaves plenty of choice for a nonzero Weyl tensor since the Ricci tensor can only account for 10 of these 19.

So it it easy to construct a situation that contradicts your "counting argument. This is yet another example of something I explained to you in great detail before , but you just are not able to get.

I think you'd need to clarify exactly what it is that you are claiming. Rob's counting argument is perfectly OK, but maybe we are both misunderstanding what you are trying to say.

<<Given the three nonzero components you specified, what are ALL of the other nonzero components of Riemann (implied by the algebraic symmetries of Riemann)?>>

They are all ZERO. I explained that to you ALSO before.

OH! Well, that makes a big difference! So your condition was actually suplemented by "all components which are algebraically independent of
$$R_{0101}, \, R_{0202}, \, R_{0303)$$ also vanish? Under some reasonable assumptions about your indexing convention and your choice of frame field, these would be associated with the electrogravitic tensor in the Bel decomposition taken with respect to $$\vec{X}=\vec{e}_0$$, the timelike unit vector in the frame, so you appear to be discussing a "nonrotating vacuum field", i.e. one in which the magnetogravitic tensor, taken with respect to a "spacetime symmetry adapted congruence" vanishes. Since the tidal tensor is traceless in a vacuum, the identity you wrote down holds automatically. The Schwarzschild vacuum solution, but not the Kerr vacuum solution, would answer to this description.

Well, I am still confused about what you were trying to get at, but I hope this post helps to clear the air!

Chris Hillman

 

[swimmingtoday]

<<OH! Well, that makes a big difference! So your condition was actually suplemented by "all components which are algebraically independent of
$$R_{0101}, \, R_{0202}, \, R_{0303)$$ also vanish? >>

Yes. I think I stated that.

<< The Schwarzschild vacuum solution, ... would answer to this description.>>

I stated that I thought my hypothetical situation was what was going in in the Schwarzschild vacuum solution

<<Well, I am still confused about what you were trying to get at>>

Someone was puzzled at how you could have a situation where the Ricci tensor vanished in a region but the Riemann tensor did not. I gave a simple hypothetical situation as a concrete example.

 

[Chris Hillman]

OK, I think we are all on the same page now--- whew!

Chris Hillman