If a Kerr-Newman black hole is like a charged, spinning, heavy magnet, what kind of magnet is it like?

I was reading up on De Sitter spaces, which states that the gravitational effects from a black hole is indistinguishable from any other spherically symmetric mass distribution. This makes a lot of sense to me.

I'm now super curious, can we just formulate all the properties of a black hole beyond the limit at which GR is needed in a completely Maxwellian / Newtonian sense? The Wikipedia article on the Kerr-Newman metric seems to indicate so, but the equations are in terms of the GR metrics. This is not what I want, I want a simplification, a limit-case of that math.

Ted Bunn answered a part of my question in Detection of the Electric Charge of a Black Hole. Let me repeat the stupidly simple form of the gravitational and electric field for a black hole beyond the point at which GR is needed.

$$\vec{g} = \frac{GM}{r^2} \hat{r}$$

$$\vec{E} = \frac{Q}{4\pi\epsilon_0 r^2} \hat{r}$$

Correct me if I'm wrong, but these would be a meaningful and accurate approximation in many, in fact, most situations where we would plausibly interact with a black hole (if we were close enough that these are no longer representative, we'd be risking a date with eternity).

Question: Fill in the blank; what would the magnetic field $\vec{B}$ around a black hole be?

Here is why I find it non-trivial: Every magnet in "our" world has some significant waist to it. So here is a normal magnet.

What happens when this is a black hole? Would the approximation for $\vec{B}$ that I'm asking for have all the magnetic field lines pass through the singularity? Or would they all pass through the event horizon radius but not necessarily a single point?

I think most people who have understood this already, but ideally the answer would use the 3 fundamental metrics of a black hole. Mass $M$, charge $Q$, and angular momentum $L$. The prior equations for gravity and electric field already fit this criteria. So the answer I'm looking for should be doable in the following form.

$$\vec{B} = f \left( M, Q, L, \vec{r} \right)$$

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Keep in mind that rotating black holes won't have a single point singularity, but in fact will have a ring singularity, but that is a minor fact. –  Benjamin Horowitz Aug 29 '11 at 3:46
@Benjamin ooooh, you're right! That had not occurred to me. Well that's already on the way to answering the question! –  Alan Rominger Aug 29 '11 at 3:53
You should first ask yourself what you mean by a magnetic field. Maxwell theory is inherently Lorentz invariant, and it is well known that (even just in special relativity) changing the reference frame can change the measured electric/magnetic fields. en.wikipedia.org/wiki/… If you fixed your coordinates in Kerr-Schild form, doesn't the Kerr-Newman article you linked to already give you an expression of the magnetic field? –  Willie Wong Aug 29 '11 at 14:46
@Willie I believe that I can very explicitly define the reference frame in question here, which is the frame in which $\vec{p}=0$ for the BH and at a distance from the black hole at which $GM/r \ll c^2$. In that case, we can ditch the GR coordinates. So maybe we would be left with just the classical perfect magnetic dipole equation (which would be fine). But even IF this is the case, I don't know what the magnetic moment is since a self-gravitating loop singularity is impossible with Newtonian gravity. –  Alan Rominger Aug 29 '11 at 15:22
The answer is the same as for any other current loop with a magnetic dipole. There is nothing special about black holes. You just need the dipole moment to get the far field. The field lines end on the horizon for an external observer. –  Ron Maimon Sep 4 '11 at 2:10

It's a little tricky giving a formal answer to this, but here is a sketch:. The electromagnetic potential of the Kerr-Newman hole is given by:

$$A_{a}dx^{a}=-\frac{Qr}{r^{2}+a^{2}\cos^{2}\theta}\left(dt-a \sin^{2}\theta d\phi\right)$$

This field will acquire a magnetic field from the fact that $\frac{\partial A_{\phi}}{\partial r}$ and $\frac{\partial A_{\phi}}{\partial \theta}$ are both nonzero. The problem is that the magnetic field, when re-phrased in terms of vectors and not one-forms, will fall off as $\frac{1}{r^{3}}$. At that point, if we're keeping terms that fall off that quickly, then we need to have a discussion about that asymptotic form of the metric you imply above, because there are terms in the metric that we need to keep, arising from frame-dragging effects of the black hole. If you want, I can go into more detail.

EDIT:

OK, so once we have the vector potential, we can calculate the magnetic field according to the rule: $B^{i}=\frac{1}{\sqrt{\left|g\right|}}\epsilon^{ijk}\left(\frac{\partial A_{j}}{dx^{k}}-\frac{\partial A_{k}}{dx^{j}}\right)$, where $\epsilon^{r\theta\phi}=1$, $\epsilon^{ijk}=-\epsilon^{jik}=-\epsilon^{ikj}$ and $\epsilon^{ijk}=0$ if $i=j$, $i=k$ or $j=k$. So, now, we just plug in the above expression for the spatial components of $A_{a}$, the metric tensor, and turn the crank.

After doing this, and taking the limit that $r$ is larger than everything else, we find that

$${\vec B}=\frac{2Qa\cos\theta}{r^{3}}{\hat e}_{r} + \frac{Qa \sin \theta}{r^{3}}{\hat e}_{\theta}$$

Sensibly, this is zero if either $Q=0$ or $a=0$. And I will once again assert that there are $\frac{1}{r^{3}}$ corrections to the gravitational force that must be taken into account in your high $r$ limit if you are going to keep this magnetic field.

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I'm a little confused by the occurrence of $d\phi$, since I'd imagine that the solution will have rotational symmetry about the axis of rotation, GR or not. I think that solving for the one-form of magnetic scalar potential would be sufficient, since I think $\vec{B}$ follows directly from that unless I completely don't know what I'm talking about. But for a classical magnetic dipole, these things would fall off as $1/r^3$, although there are other terms. Right now I'm most interested to ask if there should be anything fundamentally different for a BH. –  Alan Rominger Aug 29 '11 at 14:57
fall off as $\frac{1}{r^{3}}$ though, so you would need to keep those if you are going to keep the magnetic field. –  Jerry Schirmer Aug 29 '11 at 15:27
@Zasso: the field is axially symmetric. But unlike the case of spherical symmetry, axially symmetric vector fields can have components tangent to the circle action. Just imagine the magnetic field generated by an infinitely long wire carrying current along the $z$ axis. For the multipole moments, there is a paper by Sotiriou et al in Class. Quan. Gravitiy in 2004 that does some of these computations. (For example in regards to the Gravitomagnetic effect that Jerry mentioned.) –  Willie Wong Aug 29 '11 at 15:28
can you please fix the answer to put the correct falloff from the comments in? It's like any other magnet. –  Ron Maimon Sep 3 '11 at 22:40
@Ron I know, and I thought someone would see this a an easy 50 reputation... I like how you said it "The answer is the same as for any other current loop with a magnetic dipole. There is nothing special about black holes." However, that statement is novel to me, and I want to select an answer that answers the question and is written by someone with a background in this (not me). –  Alan Rominger Sep 4 '11 at 20:08

I'm going to answer this question with the a priori assumption that if some amount of matter collapses with a given field, then those fields are maintained as it collapses into a black hole. I see nothing to refute this assumption, but the validity of it is beyond my knowledge. As I was pointing out before, this sufficiently far from the singularity such that GR specific concepts aren't needed. The cutoff for this is formalized by the following condition.

$$GM/r \ll c^2/2$$

Again, we have $Q$, $M$, and $L$ to describe the black hole. Obviously the black hole, and the singularity itself is some amount of mass rotating. I find this problematic due to difficulties with loop singularities in a completely Newtonian sense. Anyway, I need to make statements about the angular momentum.

$$L = M a V$$

This is to say, the angular momentum is due to the fact that the mass of the singularity is located at some position away from the axis of rotation, $a$ and spinning at speed $V$. With the above statement we can make some meaningful statement about the magnetic moment, $\vec{m}$. I'll assume the axis of rotation is the z-axis and denote that unit vector with $\hat{k}$. The following equation comes from Wikipedia. I will then combine this with the prior equation. Doing this algebra makes the assumption that the mass has the same distribution as the charge.

$$\vec{m} = \frac{1}{2} Q a V \hat{k}$$

$$\vec{m} = \frac{Q L}{2 M} \hat{k}$$

This does make intuitive sense to me. The more angular momentum, the stronger I expect the magnetic field from it to be. Also, since angular momentum has mass as a component, we basically have to divide that back out. Next, I would just use the equation for a perfect magnetic dipole. This is to say the waist is zero, basically assuming it's a singularity. I think this would be fine unless the rotational energy was a significant part of the energy contained in it. I'll just ignore the delta function for the infinite magnetic field at the origin, because this isn't valid there anyway.

$$\vec{B}(\vec{m}, \vec{r}) = \frac {\mu_0} {4\pi r^3} \left(3(\vec{m}\cdot\hat{r})\hat{r}-\vec{m}\right)$$

$$\vec{B}(M,Q,L,\vec{r}) = \frac {\mu_0 Q L} {8 \pi r^3 M} \left(3 \frac{r_z}{r} \hat{r}-\hat{k}\right)$$

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