# B Rod in a black hole

#### black hole 123

if theres a really good rocket hovering just above a super massive black hole with very low tidal force, and it dips a rod just inside the event horizon, will the rod break? it seems a certainly but the tidal forces are very weak.

is it like dipping ur feet into piranha infested water? so when you lift the rod back up you see everything below the horizon gone missing?

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#### Ibix

will the rod break?
Yes. The acceleration needed to stop something falling into the hole approaches infinity as it approaches the horizon, so the rod must break.
u see everything below the horizon gone missing?
Rather like the letters y and o, yes. In practice a lot of stuff above the horizon will be gone as well because all materials have a finite tensile strength. But everything that crosses the horizon will necessarily not return. That's what crossing the event horizon means - it's not coming back.

#### PeterDonis

Mentor
the tidal forces are very weak.
What breaks the rod is not tidal forces. It's the acceleration of the rocket. The rocket is pulling at the rod very, very hard, and as the rod is lowered it gets more and more difficult for the points in the rod to keep up with the rocket. At some point, because the internal forces in the rod that are transmitting the pull of the rocket have a finite speed of propagation, the rod will break because the lower part of it simply can't keep up with the pull any more.

#### pervect

Staff Emeritus
if theres a really good rocket hovering just above a super massive black hole with very low tidal force, and it dips a rod just inside the event horizon, will the rod break? it seems a certainly but the tidal forces are very weak.

is it like dipping ur feet into piranha infested water? so when u lift the rod back up u see everything below the horizon gone missing?
The proper acceleration required for a point on the rigid rod to "hold station" in the limit of a very large black hole with "low tidal forces" will approach $c^2 / d$, where c is the speed of light, and d is the distance from the event horizon as measured by static observers.

To put some figures on this, one meter from the event horizon, the acceleration will be $\approx 9\,10^{16}$ m/s^2, or $\approx 9 \, 10^{15}$ , i.e. 9,000,000,000,000,000 earth gravities.

So as the distance d (as measured by static observers) approaches zero, the proper acceleration to hold station increases without bound. This means that no rocket at the bottom of the rod can accelerate hard enough to keep it from falling into the event horizon, and it also means that any rod that dangles down from the top of the rocket, no matter how strong, will break.

Trying to orbit the black hole won't work either, by the way - it makes things worse.

It may seem self contradictory to say that the tidal force is "small", if the proper acceleration is 9,000,000,000,000,000 gravities at 1 meter away from the horizon, but only 4,500,000,000,000,000 gravities 2 meters away from the horizon.

It's not really contradictory, though one could argue that the use of the term "tidal force" is a bit unfortunate. If we take the expression for proper acceleration, $c^2 d^{-1}$, and differentiate it, we get $-c^2 d^{-2}$

One might think that the tidal force "should be" the rate of change of proper acceleration with respect to distance away from the horizon, but this turns out not to be the case.

I will also point out that if we drop a spring through the event horizon of a black hole, in the specified limiting case of a very large black hole, the spring will not stretch. So there is something to the idea that the "tidal force" is zero in the case under consideration.

It's difficult to be more precise than this without going above the B level.

#### PeterDonis

Mentor
Trying to orbit the black hole won't work either, by the way - it makes things worse.
If by "orbit" you mean "free-fall orbit", that's not even possible that close to the hole's horizon.

#### pervect

Staff Emeritus
If by "orbit" you mean "free-fall orbit", that's not even possible that close to the hole's horizon.
Free fall orbits are not possible inside the photon sphere at 1.5x the Schwarzschild radius, as you say. Powered circular orbits which involving thrusting away from the black hole with rockets, are possible [add: inside the photon sphere], but such orbits demand more outward thrust than it takes to hover in place statically.

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#### PeterDonis

Mentor
Powered circular orbits , which involving thrusting away from the black hole with rockets, are possible, but such orbits demand more outward thrust than it takes to hover in place statically.
Yes, agreed.

#### black hole 123

thanks for the answer. i forgot its the pull of rocket not tidal force. pervect is that for a small black hole? a human can fall into a supermassive black hole and not be torn, and (most) humans are over 1 meter in length.

also about the inside of the black hole, can it be thought of as simply outside an external observer's world? because it doesnt make sense to ask "whats happening inside the black hole in our galaxy core right now at 6 pm".

#### Ibix

about the inside of the black hole, can it be thought of as simply outside an external observer's world?
Without a precise definition of "world" in this context, it's hard to say. You can always choose to enter a black hole, for example - so is it really a separate world?

The interior (or part of the interior, anyway) of a black hole is in your causal future - you can affect events there at least in principle. It is never in your causal past unless you fall in - you can never be affected by events inside the hole. That's probably the strongest statement you can make.

#### Pencilvester

is that for a small black hole? a human can fall into a supermassive black hole and not be torn, and (most) humans are over 1 meter in length.
An object in free-fall experiences no proper acceleration. What @pervect was talking about was objects experiencing extreme proper acceleration to “hover” just above the event horizon of any black hole. As he said, proper acceleration increases without bound for objects hovering closer and closer to the EH.

"Rod in a black hole"

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