I Do black holes have magnetic fields?

[Feynstein100]

I searched for this in the forum but apparently no one has asked this yet. So my question is, do black holes have magnetic fields and if so, does the magnetic field extend beyond the event horizon?

I have a thought experiment in mind. Consider a neutron star, which has a very powerful magnetic field that extends to some distance in space. Say we were to add some mass to the neutron star, enough to collapse it into a black hole. Its event horizon would necessarily have to be smaller than the size of the neutron star. But what would happen to its magnetic field? Would it suddenly disappear? That doesn't seem plausible to me.

And if the magnetic field were to remain unaffected, then it would still maintain its extents way beyond the event horizon. So what would actually happen?

This actually raises an interesting question. For charged bodies, the magnetic field is pretty straightforward. But for electrically neutral bodies that still have magnetic fields, for e.g. the earth/the sun, how would that work? I mean, from what I know, this magnetic field is only affected by two things: the rotation rate of the object and its internal temperature (read Mars lost its magnetic field because its interior cooled down). So if we were to compress the earth to let's say half its size, it would spin faster and its inside would get hotter. Would that mean its magnetic field would now be much stronger than before? And if that's true, would that also apply for our previous example of the neutron star that changed into a black hole?

 

[anuttarasammyak]

Charged BH in rotation generate magnetic field as well as electric field. Ref. https://en.m.wikipedia.org/wiki/Kerr–Newman_metric

 

[Feynstein100]

Charged BH in rotation generate magnetic field as well as electric field. Ref. https://en.m.wikipedia.org/wiki/Kerr–Newman_metric

No, I know that. Which is why I specifically mentioned the non-charged neutron stars and subsequent black holes. Would those have magnetic fields too?

 

[Vanadium 50]

So what exactly is your question? Its not the title. Does it really take 5 paragraphs to ask it?

 

[anuttarasammyak]

No-hair theorem or conjecture gives an negative view to your question of neutral BH with magnetic field. But it is not proved/disproved yet. Ref. https://en.m.wikipedia.org/wiki/No-hair_theorem

 

[PeterDonis]

Say we were to add some mass to the neutron star, enough to collapse it into a black hole. Its event horizon would necessarily have to be smaller than the size of the neutron star. But what would happen to its magnetic field? Would it suddenly disappear?

No. It would be radiated away as EM radiation. This is an example of the general result known as Price's Theorem, which says that when some object that has a bunch of properties collapses to a black hole, any of those properties that cannot be possessed by a stationary black hole (which means anything except mass, electric charge, and angular momentum) will be radiated away.

 

[Feynstein100]

No. It would be radiated away as EM radiation. This is an example of the general result known as Price's Theorem, which says that when some object that has a bunch of properties collapses to a black hole, any of those properties that cannot be possessed by a stationary black hole (which means anything except mass, electric charge, and angular momentum) will be radiated away.

That's the no hair theorem, right? And I have heard of that. Black holes cannot have any property except the 3 you mentioned. However, magnetic field lies in this kind of grey area since it's related to charge. I mean, a charged black hole would have a magnetic field by virtue of its charge. So that implies a black hole can have a magnetic field. I was just wondering if that extends to non-charged black holes as well. I guess that's a no then?
But what about the second part with the earth? Like if we could compress it to half its size (not turning it into a black hole), how would that affect its magnetic field?

 

[Feynstein100]

So what exactly is your question? Its not the title. Does it really take 5 paragraphs to ask it?

Yes, it is the title. Does an electrically neutral rotating black hole have a magnetic field?

 

[Feynstein100]

No-hair theorem or conjecture gives an negative view to your question of neutral BH with magnetic field. But it is not proved/disproved yet. Ref. https://en.m.wikipedia.org/wiki/No-hair_theorem

That doesn't really answer my question but thanks

 

[Ibix]

Does an electrically neutral rotating black hole have a magnetic field?

That's just a Kerr black hole, so no.

 

[Feynstein100]

That's just a Kerr black hole, so no.

Thank you. So in that case, the question with the neutron star. If it goes from having a magnetic field to not having a magnetic field, how does that work? Does the magnetic field get radiated away as PeterDonis mentioned? That sounds interesting. I've never heard that before. How does that actually work?

 

[phinds]

Yes, it is the title. Does an electrically neutral rotating black hole have a magnetic field?

No, it is NOT in the title. The title is "Do black holes have magnetic fields?" It does NOT say "electrically neutral rotating".

This is like asking if fruit is yellow and then insisting that what you REALLY asked was "are bananas yellow?"

 

[Vanadium 50]

No "Electrically neutral" is not in the title. We can see it.

If you think there is a magnetic field, where is the current loop?

 

[Feynstein100]

No, it is NOT in the title. The title is "Do black holes have magnetic fields?" It does NOT say "electrically neutral rotating".

This is like asking if fruit is yellow and then insisting that what you REALLY asked was "are bananas yellow?"

That's a great analogy because everyone knows that bananas are yellow and no one would ask that, since it is self-explanatory. And even if they did, it wouldn't be that hard to answer that most fruits aren't yellow, but some, including bananas, are.
[Mild insult deleted by the Mentors] My bad. I'll keep that in mind from now on.

 

[Feynstein100]

No "Electrically neutral" is not in the title. We can see it.

If you think there is a magnetic field, where is the current loop?

Assuming the black hole isn't a point and has some kind of internal structure, I'd estimate that the current loop resides internally, the same way it does in a neutron star, the sun or the earth.

 

[Ibix]

Does the magnetic field get radiated away as PeterDonis mentioned? That sounds interesting. I've never heard that before. How does that actually work?

The explanation in Thorne's popsci book Black Holes and Time Warps is that once the currents are inside the event horizon they can no longer anchor the magnetic field, and it is carried away as radiation. Thorne cites his student, Price, Nonspherical perturbations of relativistic gravitational collapse, Phys Rev D 5, 1972, p2419 and p2439 for the actual explanation, but that appears to be paywalled.

I would tend to suspect the process is loosely analogous to what happens to the magnetic field of a solenoid when you switch the current off. You get a short burst of EM generated by the no-longer-supported field as it collapses.

 

[Ibix]

current loop resides internally,

That would be the part of spacetime that cannot affect the exterior. So whether there are current loops there or not (my suspicion would be "not", except for transients) this cannot affect the magnetic field outside.

 

[Feynstein100]

Thorne cites his student, Price, Nonspherical perturbations of relativistic gravitational collapse, Phys Rev D 5, 1972, p2419 and p2439 for the actual explanation, but that appears to be paywalled.

That's what Scihub is for

once the currents are inside the event horizon they can no longer anchor the magnetic field

That sounds reasonable but I see a contradiction. Because if that were true then charged black holes shouldn't have a magnetic field either, assuming that the charge, and thus the current generated by it are inside the event horizon

 

[Ibix]

I would suggest you look up the maths then. I suspect you'll find that the behaviour of electromagnetic fields in non-static spacetime is not as trivial as you think.

I can't be sure without running through the maths myself, but I rather suspect that for a charged rotating black hole "has a magnetic field" will be a coordinate-dependent statement. But in systems where it manifests itself it should be possible to derive the appropriate Faraday tensor from the Kerr-Newman metric and generate the integral curves of the magnetic field, but that's quite a lot of work for a Sunday afternoon.

 

[weirdoguy]

and no one would ask that, since it is self-explanatory.

Ok, but your question in the title is not.

[Mild insult deleted by the Mentors] My bad. I'll keep that in mind from now on. My bad. I'll keep that in mind from now on.

What? You made a mistake asking incomplete question. That's on you, not on us.

 

[Vanadium 50]

Assuming the black hole isn't a point and has some kind of internal structure

And where exactly are you getting this from?

 

[PeterDonis]

That's the no hair theorem, right?

The no hair theorem is what says that the only properties a stationary black hole can have are mass, electric charge, and angular momentum. But that, by itself, doesn't tell you how an object with other properties that collapses to a black hole gets rid of those properties. Price's Theorem is what tells you that.

magnetic field lies in this kind of grey area since it's related to charge

No, this is not correct. Magnetic fields are not electric charge. The no hair theorem doesn't say a stationary black hole can have properties "related to" electric charge. It only says a stationary black hole can have electric charge, period.

what about the second part with the earth? Like if we could compress it to half its size (not turning it into a black hole), how would that affect its magnetic field?

That question belongs in a separate thread that wouldn't even be in the relativity forum since it has nothing really to do with relativity or with black holes.

 

[PeterDonis]

if that were true then charged black holes shouldn't have a magnetic field either

A Reissner-Nordstrom hole (i.e., a non-rotating charged hole) indeed does not have a magnetic field. It only has a static Coulomb electric field.

The rotating charged case would be a Kerr-Newman hole. You can look up the electromagnetic field tensor for that kind of hole and see whether it describes any magnetic field as well as an electric field.

assuming that the charge, and thus the current generated by it are inside the event horizon

That assumption is false. The charge of a charged hole does not have any location; it is a property of the global spacetime geometry, just like the mass. Black holes are not ordinary objects and your intuitions about ordinary objects do not work for them.

 

[PeterDonis]

Assuming the black hole isn't a point and has some kind of internal structure

This assumption is only half right. It is true that a black hole isn't a point; it's a global spacetime geometry. It is false that a black hole has internal structure; it's vacuum. There is no "stuff" inside it to have any structure. Again, black holes are not ordinary objects and your intuitions about ordinary objects do not work for them.

 

[Vanadium 50]

There is a difference between "your explanation conflicts with my calculation" and "your explanation conflicts with my guess." Thus far, I have seen no calculations from the OP.

 

[JimWhoKnew]

No, I know that. Which is why I specifically mentioned the non-charged neutron stars and subsequent black holes. Would those have magnetic fields too?

I think you can get some idea by hand-waiving: Suppose the collapsing star is neutral (charge-wise) with zero angular momentum, but there is some geophysical process that functions as a current loop and causes some magnetic field around the star. You can associate a frequency with the current loop, proportional to the angular velocity of the current, and to the magnetic field. As the collapse proceeds, before its surface passes the event horizon, the star shrinks and the "proper" frequency may grow larger. But this possible increase is expected to be bounded (angular velocity times Schwarzschild radius must not exceed the speed of light - this reasoning relies on unjustified rigidity, but, hey, it's only a hand-waiving). On the other hand, an external observer measuring the field at some constant distance, will see the "proper" frequency more and more redshifted, which will win over the former increase, since it goes all the way to zero. So this hand-waiving suggests that the external observer will see the magnetic field dying out (compatible with No-Hair).

 

[PeterDonis]

I think you can get some idea by hand-waiving

Your hand-waving ignores EM radiation, which, as has already been noted, is significant in this scenario.

You can associate a frequency with the current loop, proportional to the angular velocity of the current, and to the magnetic field.

This hand-waving also ignores how electromagnetic fields actually work. This is why we do physics with math, not hand-waving.

 

[Vanadium 50]

I'm not so sure that this thread needs more hand-waving and less specificity.

Some of the ideas here are wronger than wrong can be: a current loop in the interior of a black hole generating a field in the exterior is not that different from a current loop in the future creating a field in the past.

 

[JimWhoKnew]

Your hand-waving ignores EM radiation, which, as has already been noted, is significant in this scenario.

The dying of the field implies $~\frac{\partial\vec B}{\partial t}\neq 0~$, which implies $~\nabla\times\vec E\neq 0~$ . In addition, the energy density stored in the field goes somewhere else. Radiation?

(BTW, the hair doesn't have to be completely radiated away. Some of it can be back-scattered into the hole. I'm sure you know it)

Hand-waiving is no replacement for equations, but a good qualitative description (and I'm not saying mine is) can be more illuminating for a student than a forest of complex equations.

This hand-waving also ignores how electromagnetic fields actually work.

How do electromagnetic fields actually work?

 

[JimWhoKnew]

I'm not so sure that this thread needs more hand-waving and less specificity.

Since you've mentioned hand-waving, should I assume that you refer to #26 ?
If not, please ignore the rest of this post.

a current loop in the interior of a black hole generating a field in the exterior

Where in #26 did I talk of a current loop in the interior of a black hole generating a field in the exterior?

 

[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.