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!"