Steven Crothers , against BB

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

  1. NotEinstein Registered Senior Member

    I don't think mass is either? Are you perhaps referring to relativistic mass; if so, don't, because that's not the mass meant in GR calculations.

    Then please explain to me what \(F^\text{tr}\) stands for?

    I don't see any mentioned of field lines in that article?

    I also don't see any mention of Keplerian gravitating mass in that article?

    (Look, my GR is very rusty; you're going to have to hold my hand a bit more, sorry.)

    I've never said that I see no problem; I don’t fully understand the problem you are pointing out yet and what GR has to say about it, so I have not yet formed an opinion about it.

    I agree that this seems weird, and at this moment I have no response to it; I'm still in the process of figuring out what GR has to say about it.
  2. Google AdSense Guest Advertisement

    to hide all adverts.
  3. przyk squishy Valued Senior Member

    Why? If you take electromagnetism at face value then all you need in order for there to be an electromagnetic field at some time \(t\) is that there was an electromagnetic field there before at time \(t - \mathrm{d}t\). From there, how the electromagnetic field configuration changes over time (in particular, whether it remains stable or whether it disperses or decays over time) is determined by Maxwell's equations.

    So if you want to know if you can have a black hole in GR with a stable electromagnetic field around it then the definitive way to find out is to determine if there is a space-time geometry describing a black hole and a static electromagnetic field that together satisfy both the Einstein field equation and Maxwell's equations. This is, as I pointed out some posts ago, exactly what the Reissner-Nordström solution is.
    Last edited: Dec 6, 2017 at 7:38 PM
  4. Google AdSense Guest Advertisement

    to hide all adverts.
  5. NotEinstein Registered Senior Member

    Right, I should have explained what I find weird about it. Take a black hole singularity and give it some charge. The EM-field will be carried by (virtual) photons, but these photons can't leave the event horizon. How then is it possible to propagate this EM-field outside of the black hole?
    Now, there has to be a way for the EM-field to extend outside of the event horizon, as the Reissner-Nordström solution indicates. One possible issue is that my description is mixing QFT with GR, which is a terrible thing to do. It's obviously wrong in some way, which can be seen if one makes the same argument with gravitons (I know, not proved to exist, but still); the gravity obviously extends outside of the event horizon. Is it that the empty space around the black hole starts emitting the EM-field, just like the gravitational fields around a black hole can start contributing to the mass. Another option is that infalling charged matter would never be seen to cross the event horizon, so the charge would always remain outside of it.
    What I find weird is that my intuitive picture/description above seems to natural (to me), but obviously has to be wrong if GR says differently. That's why I'm trying to figure out exactly what GR says about Q-reeus' scenario, because I clearly am missing some crucial argument here that I currently don't see, and I want to correct that.
  6. Google AdSense Guest Advertisement

    to hide all adverts.
  7. przyk squishy Valued Senior Member

    Well the first thing is: what does that mean? Charge conservation is baked into electromagnetic theory, so having some charge appear from nothing doesn't make sense even as a thought experiment. The situation contradicts Maxwell's equations right from the start.

    So I only see two situations that make sense: you have an eternal black hole (one that has just always been there and has always been surrounded by an electromagnetic field) or you have a black hole formed by charged matter collapsing or falling into it at some point, in which case you're just continuing to feel the electromagnetic field associated with the charge in the past before it fell behind the horizon.

    It's the same for the gravitational field itself: you can think of it as either feeling the gravitational field associated with whatever matter was there before it collapsed into a black hole, or you have an eternal black hole and a gravitational field around it that have both just always been there (or you are feeling the gravitational field associated with the associated white hole, if you like, if you're considering the extended Schwarzschild metric).

    I don't see any reason to bring quantum physics into this. The Reissner-Nordström solution is derived using only classical electromagnetism and GR. It doesn't involve any quantum theory.
  8. NotEinstein Registered Senior Member

    Let me rephrase: Assume there's a black hole singularity that has some non-zero charge.

    The latter scenario I can grasp intuitively, but with the former I hit the problem I described earlier.

    Same deal here: the former I can grasp intuitively, but with the latter I hit the problem I described earlier.

    Are you saying the QFT description of force propagation is unusable in the close proximity of black holes/event horizons?
  9. przyk squishy Valued Senior Member

    Well what does that mean physically? Operationally, we only know about "charge" by the electromagnetic fields around them, with the specific relation given by Gauss's law. (Imagine a charge without an electromagnetic field around it -- what does it even mean to call it a "charge" in that case?) If you claim that a black hole is "charged" then you're claiming there is a certain kind of electromagnetic field around it basically by definition. Or conversely, if you have a black hole with an electromagnetic field then the black hole is "charged" basically by definition. This does not mean there is any direct causal connection between what is happening inside the black hole and what is happening outside of it.
  10. river

    From post#74 by przyk

    White space ? Explain further .
    Last edited: Dec 7, 2017 at 1:49 AM
  11. przyk squishy Valued Senior Member

  12. river

  13. river

    Why not ?
  14. Q-reeus Valued Senior Member

    Well 'radial component of trace of EM component of stress-energy-momentum tensor' looks about right. So they leave out any vector notation there and somehow normalize the magnitude, but otherwise it's describing a Coulombic E field due to a charge Q. Since it's not Q(r) the implication is Gauss's law exactly holds.
    To check that is a correct interpretation, let's look elsewhere:–Nordström_metric
    Under 'Charged black holes', we find the only non-zero component of EM potential is just Q/r. Differentiating just recovers that disputed (5.2.3).
    The fact that a 'charged BH' actually exists immediately implies total imperiousness to source charge lying in an infinitely depressed metric! If one believes it.
    That and most of your other comments in #101 are dealt with in that referenced Wiki article on Komar mass. Did you not get the initial section there? Proper acceleration for a hovering observer serves as the equivalent of 'field line density' re Newtonian equivalent. Integrated over the bounding surface surrounding the central gravitating mass, one has an *almost* equivalent to Gauss's law for charge - but in this case 'gravitational charge' i.e. mass M(r). Yes M(r) because as shown there, the net flux is a decreasing function of r. Like I wrote way back, in GR 'field lines' (in a previous post you missed that I deliberately put that in quotes) occasionally terminate in mid-air.
    Keplerian mass is simply the M's that appear in celestial mechanics formulae e.g.
    And such M's - based as they are on Newtonian gravity where Gauss's law for gravity does strictly hold, are no longer constants independent of r in GR, as per that Komar mass article.

    Which is a bit of a problem in GR since, by definition of EFE's, the Ricci scalar R is zero everywhere in vacuo. Meaning, contrary to a false assertion made recently in another thread, gravity does NOT gravitate in GR. Not according to EFE's. Hence 'Gauss's law for gravity' *should* hold. In fact, as per that Wiki article on Komar mass, if anything gravity anti-gravitates in GR. A strange state of affairs.

    But the main point here is - how is it that electric charge is immune to any of this?
    You have made no comment on my considerations particularly in #98 on case of an infalling, weakly charged spherical dust shell. See no issue with constant M *and* constant Q during infall - as measured by an observer at fixed r and exterior to the infalling shell?
  15. NotEinstein Registered Senior Member

    If the singularity is what carries the charge, then there's some EM-field inside the event horizon. Are you saying it's possible for this EM-field not to extend outside of the event horizon?
  16. NotEinstein Registered Senior Member

    Actually, I just realized it's simply the time-radius component of the electromagnetic field tensor, and has nothing to do with the trace. (I promise I only just figured this out!) But you clearly didn't know this either. In fact, you have been using distinctly Newtonian notation instead of four-vectors. Is that the reason your scenario isn't expressed in standard GR terminology; because you aren't familiar with the mathematics of GR?


    Are you suggesting charge should disappear when a singularity forms?

    I did see that Komar mass is merely one way of looking at mass in GR; why are you using that one specifically?


    I'm not familiar with this other thread; what does "gravity does NOT gravitate" mean exactly?

    I have made no comments, because at this time, I have no comments to make. I'll have to brush up on my GR (which is what I'll be doing for the next days/weeks) before I can meaningfully talk about your proposed scenario.
  17. Q-reeus Valued Senior Member

    Oops, the trace of the Maxwell tensor is zero. So yes just the only non-zero component of field 4-vector.
    Haven't done a course in GR and scratchy re tensor maths but understand the basics. We are dealing with a static RN metric that lends itself to picking out 4-vector components separately. I like to make the conceptualization as physically intuitive as possible. I hate folks trying to baffle with BS which mathematically inclined often do. Thought experiments have a knack of cutting through that.
    I have been claiming all along that charge should be subject to 'dilation' according to it's level in a depressed metric. Singularities are moot here since they form only in so-named 'null future infinity' for the case of a 'realistic BH' formed via collapse of matter. The obvious barrier to a 'charged BH' making sense in that scenario is the formation of an EH, not a supposed 'singularity'. Time dilation is infinite at the EH - any 'machinery' such as assumed needed for 'virtual particle exchange' has become frozen solid wrt outside. Not that that is the appropriate argument. There is an oft cited article that claims to rationally explain the problem of 'virtual particles escaping from inside an EH':
    It's bunkum imo, but opinions vary.
    I concentrate on the imo absurd disparity between how 'mass charge' and electric charge are supposed to behave in situations nowhere near BH extremes. As per my given scenarios.
    ADM mass only concerns the mass observed at infinite radius. Bondi mass deals with non-static situations where say outflow of radiation is involved. Komar is the simplest case and the most appropriate here as it applies to a static arrangement.
    See here:
    Fine but just brushing up on GR won't change things. It all gets down to what is the appropriate conceptual basis for 'marrying' EM with GR. For the reasons I have outlined in earlier posts, in the standard treatment yielding RN solution, it's been done wrong. A supposed aloofness of EM field to spacetime curvature that everything else has to follow.
    Last edited: Dec 7, 2017 at 12:17 PM
  18. Q-reeus Valued Senior Member

    In #114, 'So yes just the only non-zero component of field 4-vector.' should have read 'Only non-zero gradient of EM 4-potential'.
  19. NotEinstein Registered Senior Member

    I've come to the same conclusion.

    If you haven't done the math, you can't make any claims about understanding the basics. Seriously, GR is so counter-intuitive at times, without the mathematical derivations you can't come to any valid conclusions. Thought experiment may work some of the time, but they often (and unexpectedly) fail. I'm not saying that's the case here, but I'm immediately suspicious when somebody makes claims about what GR says but can't actually show any maths to back up said claims.

    In that case, what did you mean when you said: "total imperiousness to source charge lying in an infinitely depressed metric"?

    But you are (per construction) in a situation where relativistic effects become significant. In other words, Newtonian physics can no longer be trusted to give the right answer to queries. I immediately become suspicious when somebody claim to have used Newtonian physics to disprove GR in a situation where GR has to be used for accurate predictions, without showing that under the specific circumstances using the Newtonian limit is valid.

    (No comments; I'll have to dive deeper into this first.)

    Thanks; I'll be sure to read that too!

    Actually, in order to know what GR says about the scenario, one needs to understand enough GR to be able to figure out what GR says about the scenario. I thought that would be self-evident?

    Yes, and that's a subject that's complicated enough that it needs a thorough understanding of GR in order to come to valid conclusions, in my opinion.

    (No comment at this time.)
  20. przyk squishy Valued Senior Member

    Why do you assume there is?
  21. przyk squishy Valued Senior Member

    Not so far as I know.

    This is determined by the principle of general covariance, which is just the mathematical statement of the equivalence principle and a basic and integral part of general relativity. Basically, you assume that Maxwell's equations of electrodynamics hold in any locally inertial (freefalling) reference frame. This is enough to determine how Maxwell's equations apply to general relativity in general.

    A basic introductory course or textbook on general relativity would normally explain how to do this systematically for any theory expressed in the Minkowski four-vector/tensor notation, not just electromagnetism. This is why references would probably not say much about how to apply electromagnetism to GR: if you're studying something like the Reissner-Nordstöm metric it would normally be taken for granted that you have already learned and understood this sort of thing.
    Last edited: Dec 7, 2017 at 11:32 PM
  22. Q-reeus Valued Senior Member

    The scenario's aka thought experiments covered earlier all refer to what GR predicts - or rather conflictingly predicts. If you think I'm using 'Newtonian physics' there then that is simply wrong. Go check back - #79, #96, #98. I did refer on occasion to Newtonian analogue re 'gravitational field lines' - quite appropriate in context.
    This covers your later repetition of that same type of comment in #116.
    'imperiousness' was my spell checker's flagging of 'imperviousness' - hastily accepted, assuming a typo was made initially. Bad decision. Evidently you didn't notice the context which that comment was set in. Of course I have consistently rejected the standard position as codified in RN metric. Any 'imperviousness' of Q/EM-field to depressed spacetime metric is physically unreasonable. As set out in #79, #96, #98.

    Other things that 'self-evidently' hold in flat spacetime necessarily fail in GR. For instance, a rod under tension from forces acting both ends can only be in static equilibrium if the forces acting are equal and opposite. Not so in GR - proper force lower down in a gravitational potential must exceed proper force higher up for equilibrium to hold (added effect of rod's own weight is a separate consideration). You might call that 'failure of stress flux law'.

    Similarly, in flat spacetime, a rotating shaft can't have different rotation rates one end to the other without experiencing ramping torsional strain. Not so in GR - proper rotational rate of lower end must exceed that at the higher end to avoid twist-up. You might call that 'failure of rotational flux law'.

    Similarly for flow of radiation - need I bring up well-known redshift of light for instance? Which requires a 'double redshift' when power of a radiation beam is considered. Or any other means of transporting energy/power from one gravitational potential to another.

    Static EM field is evidently a strange standout from that general feature - according to standard formulation anyway.

    I'm a great believer in checking the apparent watertight basis of any theory via 'test scenarios'. It's dead easy to find that for instance supposed divergence free nature of stress-energy-momentum tensor fails in certain easily demonstrated cases when stress is allowed as source term(s). Predicting a perpetuum mobile to be possible.
    I won't elaborate here but it is easy to show via thought experiments accounting for all relevant factors. Really. Something never uncovered if just trusting the standard formulation.
    Last edited: Dec 8, 2017 at 1:00 AM
  23. NotEinstein Registered Senior Member

    So a charge inside the event horizon can have its EM-field extend outside of it? But how is this then compatible with your statement that: "This does not mean there is any direct causal connection between what is happening inside the black hole and what is happening outside of it."? If a charged object inside the event horizon moves about, shouldn't we be able to see its EM-field change outside of the event horizon? (Yes, I know this isn't spherically symmetric or static anymore.)

    I am fully aware of this, but it's been so long ago that I'm too rusty (especially on the intricate and subtle details) that I'm having trouble translating what Q-reeus is describing into GR terms, figuring out what GR says, and then coming up with an answer that's more than just "GR says so". I'm trying to find a more intuitive answer than "the maths work out". (Which is obviously valid if true, but has very little convincing power, especially if others can't do the maths.)

Share This Page