# What happens if I put one pole of a magnet beyond the horizon of a black hole?

Let's assume I could lower a bar magnet slowly from some position outside of the Schwarzschild horizon. What happens if it crosses the horizon, so that one pole of the magnet is inside the black hole horizon and one side is still outside? Would the outside observer then see a magnetic monopole, because the inside pole becomes "disconnected" from the outside pole? Or would the outside observer basically see one pole sit on the horizon and the other pole/end of the magnet coming closer and closer to it until it vanishes completely behind the horizon?

• It is convenient to assign poles to a magnet but remember that what is observed is the sum of the magnetic moments of the constituent parts of atoms. – Farcher Sep 16 '17 at 22:12
• What do you think would happen if you simply take a saw and cut your magnet in half? You'd get two dipole magnets, not two monopoles. – safesphere Sep 16 '17 at 22:13
• @safesphere I'm not sure if an event horizon acts like a saw on a magnet. At least an observer sitting on the magnet shouldn't see any change while crossing the horizon. From outside an event horizon is probably a much sharper saw (thickness=planck length?) than anything else. So I think the answer is not so clear cut. – asmaier Sep 17 '17 at 7:58

Gravity isn't consistent across the length of the magnet. It results in 'tidal forces', because the further away you get from the centre of a mass, the weaker the gravitational attraction. For example, for the Earth-Moon system, whilst the force involved is not very large, the Earth is big enough for this differential between the force experienced on opposite sides of the Earth in the line of the Moon to matter, and this leads to the tides, hence the name tidal forces.

Now, a typical everyday bar magnet is much smaller than the Earth; however, the gravitational attraction due to a black hole is so much larger because the density of mass involved. This means that even the small length of a bar magnet would start to experience huge tidal forces as you got closer and closer to the Schwarzschild radius, as space-time is becoming more and more warped.

Note that at the Schwarzschild radius, the escape velocity becomes greater than the speed of light itself. This means nothing can overcome the attraction. Therefore, by this point or probably earlier, your magnet will have stretched and then snapped into two parts, then snapped into more (the name for this is "spaghettification" btw, great word there), until all of the bonds will break and a stream of particles enters the black hole's event horizon.

So as one of the commenters above pointed out, it's like when you saw a magnet in half here on Earth - you get two dipole magnets, not two monopoles.

• If the black hole and it's event horizon are huge compared to the bar magnet, there are no tidal forces which would destroy the magnet before reaching the horizon: en.wikipedia.org/wiki/… – asmaier Sep 17 '17 at 12:51

To add to the answer of ajd138, things are even more complicated. The point of no return around a black hole is not the event horizon, but the photon sphere that is 50% wider.

Anything between the photon sphere and event horizon must fall - only rocket engines can hold a spaceship on an orbit. However, from the standpoint of a free falling observer, the event horizon becomes smaller and smaller along the flight, so the observer, from his standpoint while falling, never crosses the event horizon until he hits the singularity.

On the other hand, if we look at this this from afar, we notice that the observer is moving slower and slower, as his time slows down while he approaches the event horizon from our viewpoint. Eventually we se heem frozen at the event horizon without crossing it ever.

As you can see, neither we nor the falling observer ever sees anything actually crossing the event horizon, such as a half of a magnet. However, even if hypothetically this were possible, cutting a magnet in half would only create two magnets.

• "There are no stable free fall orbits that exist within or cross the photon sphere. Any free fall orbit that crosses it from the outside spirals into the black hole. Any orbit that crosses it from the inside escapes to infinity. No unaccelerated orbit with a semi-major axis less than this distance is possible, but within the photon sphere, a constant acceleration will allow a spacecraft or probe to hover above the event horizon." The photon sphere is not a point of no return. – Johnathan Gross Sep 17 '17 at 5:15
• @JohnathanGross: Yes, you are right. Although the question implies hovering at the event horizon where the escape speed is near the speed of light. This does not affect my answer however, because, if you hover over and drop a magnet in, you would never see it crossing the horizon. From your stationary viewpoint above the horizon, time stops at the horizon and the magnet freezes in flight never crossing the horizon. – safesphere Sep 17 '17 at 5:31
• @JohnathanGross: Thanks for the edit. I accepted it and also fixed the "rocket engines"error you pointed out. I let the "point of no return" remain as a metaphor:) – safesphere Sep 17 '17 at 5:39