(A friend at work kindly loaned me loaned me his copy of Kip S Thorne's "Black Holes & Time Warps". This may have been ill-advised... :)

BH&TW 1994 paperback p.410 Figure 11.5:

... all positive charge that enters the horizon from the outside Universe becomes attached to the horizon, and moves around on it, until it exits back into the outside Universe (in the form of of negative charge falling inward to neutralize the positive charge.

Whoa there buckaroos, QED this ain't. Protons are made of quarks, and quarks are individually charged at fractions of 1e. So, if I'm reading the above very straightforward statement the way it seem to be saying, every charge should be individually trapped by the horizon. So how does that nasty lumpy $-1e$ of a charge cancel out each of the fractional charges of the proton quarks, some of which (the d quarks) are negative BTW?

So while I'm mostly OK with protons and antiprotons doing this little dance and falling in, because in that case all you are really doing is shooting gammas into the interior of the horizon. Even for that case, wow, that one statement above would be skipping over a plethora of particle complexity for any real event. E.g. you really need to look at the quarks cancelling to get a complete picture, not just two particles with actual spatial size.

The Thorne book is nine years and intended to be a popular read, so am I safe to assume that someone has subsequently looked in gruesome detail at the troublesome issue how exactly you cancel the mismatched charges of ordinary matter in a way that allows out-of-balance all-matter (or all-antimatter) lumps and atoms to fall into the interior of an event horizon?

I note that the conservation principles in play for this question are the same ones that keep hydrogen atoms from turning into gammas. So while the interior of a black whole is a thing of beauty and pure energy and simplicity, messy matter first has to get there by getting, more or less, converted 100% into gamma radiation. Once you have converted matter into all gammas and shot those into the event horizon interior, you can spit back whatever matter-antimatter or radiation combo you want.

But if horizon charge attachment is real and fully self-consistent, I fully confess that I flatly do not see how you can ever get into that interior.

Also, I assume the membrane model has not been dramatically modified or abandoned since that time?

After all, the idea that horizons are conductive seems so relevant to how quasars are powered that I find it a bit hard to believe that the conductive-horizon idea could ever be abandoned, since in effect quasars qualify as decent experimental evidence for the idea. Also, for whatever it's worth, I found the arguments about misbehaving field lines leading to charge capture to be nicely self-consistent and persuasive.

So, more crisply:

If charge cancellation is required for particles to fall past an event horizon, how can ordinary matter with its odd mix of spatially separated fractional charges ever reach the 100% self-cancellation of charge needed for it to pass through the event horizon?

If appropriate, a reference to some widely accepted paper that adequately explains the nature of the whole-fractional charge cancellation mechanism would be fine.


1 Answer 1


The simplest answer is that the horizon doesn't hold the charged particles, but rather, the information about the charges of particles that have fallen into the horizon lives on the horizon itself.

In simpler terms, the electric monopole moment of all of the matter inside of the horizon can be inferred from the geometry and field in a neighborhood of the horizon.

So, the matter itself falls through the horizon, but you can consider the horizon to have a "charge", since gauss's law tells us that the field over a closed surface determines the charge inside.

  • $\begingroup$ How much time it will taken a particle to fall under the horizon? Infinite. No particle can ever fall inside. Period. -1 $\endgroup$
    – Anixx
    Commented Apr 26, 2014 at 5:52
  • $\begingroup$ @Anixx: nope. the horizon comes out to meet the particle. And even if you don't include this, its light redshifts to a wavelength of a parsec in a relatively short time, making it unobservable. $\endgroup$ Commented Apr 26, 2014 at 16:17

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