I Do black holes have magnetic fields?

[JimWhoKnew]

it's vacuum.

Even if you ignore the gravitational field of Schwarzschild BH, a Reissner-Nordstrom BH has EM field all the way to the singularity.

 

[PeterDonis]

The dying of the field

Which is just your handwaving. You have given no math showing "dying of the field".

the hair doesn't have to be completely radiated away. Some of it can be back-scattered into the hole.

No, no "hair" can be back scattered into the hole, because then it would be "hair" that the hole would have that it can't have by the no hair theorem.

How do electromagnetic fields actually work?

The way the math says they do.

 

[PeterDonis]

Even if you ignore the gravitational field of Schwarzschild BH

There is no such thing as a "gravitational field" in GR apart from the spacetime geometry. A Schwarzschild BH is pure spacetime geometry, i.e., vacuum.

a Reissner-Nordstrom BH has EM field all the way to the singularity.

Yes, for the charged case "electrovacuum" is the strictly correct term. But there is still no "location" that has any charge. We attribute a charge to the hole because it has a Coulomb field that is externally measurable.

 

[PeterDonis]

a good qualitative description

Has already been referred to in this thread, namely, Kip Thorne's description referenced in post #16 (which, as noted in that post, is based on published peer-reviewed literature).

 

[JimWhoKnew]

Which is just your handwaving. You have given no math showing "dying of the field".

This paper calculates the magnetic field due to a current loop in the external Schwarzschild spacetime. The ratio of the leading order magnetic dipole moment between the case in which the loop is positioned in $r_1$ and the case where it is in $r_2$ is approximated by$$\frac{^1 B_i(r)}{^2 B_i(r)}\approx\frac{I_1 r_1^2 \sqrt{1-\frac{2 m}{r_1}}}{I_2 r_2^2 \sqrt{1-\frac{2 m}{r_2}}} \quad .$$When $r_1\approx r_2 \approx 2 m$ , it is in accord with my hand waiving in #26.

However,
1) As discussed in the introduction, the approximation is limited, and may be far from valid in the present context.
2) The approximation is for a quasi-static case, while a collapsing star is a dynamic process. On the other hand, from the external observer's point of view, the late-time stage of the collapse takes a very long time. It might so happen that these calculations will turn out to be relevant after all.

No, no "hair" can be back scattered into the hole,

Box 32.2 in MTW discusses Price's theorem. It says:
"Price's theorem states that, as the nearly spherical star collapses to form a black hole, all things that can be radiated get radiated completely away - in part "off to infinity"; in part "down the hole"."

because then it would be "hair" that the hole would have that it can't have by the no hair theorem.

All I have to say in response is written in subsections A.6-A.9 of box 32.2 in MTW (too long to quote), and in the other subsections of that box.
"Result is destruction of all deformation IN EXTERNAL FIELD and IN HORIZON!"

 

[Ibix]

Box 32.2 in MTW discusses Price's theorem. It says:
"Price's theorem states that, as the nearly spherical star collapses to form a black hole, all things that can be radiated get radiated completely away - in part "off to infinity"; in part "down the hole"."

Yes, Black Holes and Time Warps also says that the EM radiation may go inwards into the hole. I think possibly just your wording "the hair doesn't have to be completely radiated away" (post #29) may be confusing, as it can be read as meaning that the black hole can still have hair at the end of the process. The hair all converts into radiation (or at least, something that can flow) and goes away, but some of the energy in the hair may go inwards into the hole and end up contributing to one of the non-hairy parameters.

 

[PeterDonis]

Box 32.2 in MTW

subsections A.6-A.9 of box 32.2 in MTW

None of these contradict what I said, or support your claim that "hair" such as a magnetic field can be radiated down the hole and become a property of a stationary hole. As I said, that would contradict the no hair theorem.

 

[JimWhoKnew]

your claim that ... and become a property of a stationary hole.

I claimed that?

 

[Ibix]

I claimed that?

I think you can read #29 that way, although I don't think you intended it so.

 

[PeterDonis]

I claimed that?

the hair doesn't have to be completely radiated away. Some of it can be back-scattered into the hole

The "hair" being back-scattered into the hole would mean it becomes a property of the hole. Perhaps what you intended is what @Ibix said in post #36, but if so, that was not at all clear.

 

[pines-demon]

The problem with stationary magnetic fields is that these are usually due to quantum mechanics (unless you have a dynamical system). So it is very possible that this question is not entirely possible to resolve without a valid theory for quantum gravity.

 

[JimWhoKnew]

I think you can read #29 that way, although I don't think you intended it so.

Yet, my understanding from the literature is that although hair can't fall into the hole to become a permanent property, it is not ruled out as a transient property.

Suppose an object falls radially (zero charge, zero angular momentum) into a perfect Schwarzschild BH. Between the crossing of the horizon and the crash at the singularity, would you consider it as hair?

 

[PeterDonis]

although hair can't fall into the hole to become a permanent property, it is not ruled out as a transient property.

Yes. If you read my previous posts where I talk about the no hair theorem, you will note that I say it is true for a stationary black hole. During the transient condition, the hole is not stationary.

Suppose an object falls radially (zero charge, zero angular momentum) into a perfect Schwarzschild BH.

It can't be a perfect Schwarzschild BH if an object is falling in, since the object affects the spacetime geometry.

Between the crossing of the horizon and the crash at the singularity, would you consider it as hair?

First, mass is already a property that a stationary black hole can have, so the "hair" question is irrelevant for it. (So are charge and angular momentum, so the same would apply if you postulated a charged object with nonzero angular momentum falling into a Kerr-Newman black hole.)

Leaving that aside, your question takes the wrong viewpoint. The right viewpoint is, what is the spacetime geometry? That is a 4-dimensional thing that already contains all the information about what fell in and what happened to it, and answers any "hair" questions one might have.

For the case of an object of non-negligible mass falling into a black hole, there is no known analytical solution for the spacetime geometry, so we either have to simulate it numerically or use approximations. The former has been done in the literature, but I don't have any references handy, so I'll stick to the latter. The obvious approximation is that we have a Schwarschild BH of mass $M$ in the far past, a transient region where an object of mass $m$ (or more precisely energy at infinity $m$) falls in, and a Schwarzschild BH of mass $M + m$ in the far future.

In this approximation, the hole has no "hair" in the stationary regions in the far past and far future, so any "hair" (note that, as above, mass itself is not considered "hair" since it is a property of a stationary black hole) carried by the object would have to go away in the transient region. As we've seen in the discussion in this thread, it would go away by being converted to radiation and either radiating away to infinity or going down the hole. Any radiation that goes down the hole adds further to the hole's mass, so as a hypothetical example, if the infalling object were carrying "hair" that converted to radiation with energy $E$, and of that, energy $R$ was radiated away to infinity, then the final mass of the hole in the far future stationary region would be $M + m + E - R$ instead of just $m$. In the transient region while the conversion of "hair" to radiation was still going on, the "hair" would be detectable, yes.

 

[PeterDonis]

Moderator's note: Thread level changed to "I".

 

[ohwilleke]

Mathematically, Kerr-Newman black holes which are charged and rotating, which would create a magnetic field, are theoretically possible.

But, observationally, we really don't know which of the four kinds of theoretically possible black holes are actually realized in the real world, how common each type is, and what processes tend to produce what kind of black hole. This is mostly because it is hard to tell.

Black holes, because they don't emit light, are tricky to detect at all, unless you have just the right circumstances (e.g. a black hole feeding on a nearby star in a binary system, or a merger of black holes leaving gravitational wave signal, or a supermassive black hole whose gravitational field impacts the motion of the stars in its vicinity).

Furthermore, the observation of whether it is charged or not is also confounded by the fact that black holes will frequently have charged particles orbiting them, often at very high speeds, near but outside the event horizon, which generates a magnetic field separate from the magnetic field of the black hole itself. So, for a solar system bound astronomer, distinguishing a magnetic field of the black hole itself from the magnetic field of charged particles surround it is challenging (i.e. virtually impossible).

Theory tells us what's possible, but when there are multiple possibilities, we can't necessarily to determine based upon pure logic alone what the actual reality looks like.

The conventional wisdom is that because observed astronomical objects do not possess an appreciable net electric charge (the magnetic fields of stars arise through other processes), that the Kerr–Newman metric is primarily of theoretical interest. But while this is a plausible conclusion from the available facts, we don't know for sure.

For example, it could be that actually, virtually all black holes are Kerr-Newman black holes, but that their electric charges, while non-zero, are so close to to zero (due to a typical net electric charge of perhaps one part per billion of negative charges in excess of positive charges in a pre-collapse star) that this electric charge and the resulting slight magnetic field can be ignored for all practical purposes.