Flywheel half-way through the event horizon of a black hole vs the equivalence principle

Question

Imagine I am inside an isolated rocket of arbitrarily small size, and I have a spinning flywheel right next to me. Now suppose my rocket passes through the event horizon / Schwarzschild radius of a simple Schwarzschild black hole.

By the equivalence principle, I should not notice myself and the rocket passing through the event horizon. However, since classically no object can escape the black hole once it passes the event horizon, it seems as though the flywheel should break as it passes through the event horizon, because for every piece going one way, the antipodal piece of it goes the opposite direction. Once the flywheel is half-way through the event horizon, the part of the flywheel inside the black hole cannot come out even though it must rotate, so it seems as though a part of the flywheel would split in half.

How does this square with the equivalence principle?

I am aware that the equivalence principle only applies locally in the limit of smaller and smaller regions. For example, tidal effects can allow you to distinguish regions with gravity and regions without gravity. However, I don't think that's enough to resolve my quandary. We can assume the black hole is sufficiently large so that no issues of tidal effects or spaghettifications occur. We can make the black hole as large as we like and the rocket as small as we like to remove second-order gravitational effects, and it seems like my paradox involving the flywheel crossing the Schwarzschild radius still exists. Am I wrong in this assertion?

Answer 1

since classically no object can escape the black hole once it passes the event horizon, it seems as though the flywheel should break as it passes through the event horizon, because for every piece going one way, the antipodal piece of it goes the opposite direction.

This analysis is incorrect. The event horizon is a lightlike surface. In a local inertial frame it moves outward at c. So while it is true that there is an antipodal piece going the other way it doesn’t matter. The antipodal piece is going slower than c in the local inertial frame. So the horizon is going faster and the antipodal piece cannot possibly cross back through the horizon. The flywheel continues spinning without interruption and without risk of crossing the horizon backwards.

Answer 2

This is not a direct answer, but an investigation of an analogous situation that helped me understand Dale's answer. I'm posting it here in case anyone else finds the discussion illustrative. (But you should still upvote Dave's response!)

This question is part of a more general class of phenomena: a spatially-extended system in which a conserved quantity travels in a loop. Examples include:

• a rotating flywheel (conserved quantity: mass),
• an electric circuit (conserved quantity: electric charge), and
• fluid pumped through a loop of pipe (conserved quantity: mass, vorticity, suspended particles — anything transported by the fluid).

It's very easy to analogize to the situation where the conserved quantity is electricity and flows counterclockwise. The nice conceptual feature of this analogous system is that it can give us a clear division into the black-hole-internal and -external portions.

To do this, consider an arbitrary division of the system into two sections. One section $L$ lies on the left; the other section $R$ on the right. We can assume that the division point occurs at position $x$ both on the top and bottom, where it crosses nothing more complicated than a wire. Importantly, however, we cannot assume that this division is time-invariant. The event horizon will pass through our system at lightspeed; $x$ must travel with it.

At any given point in time, we can then describe our system in terms of two quantities and four (signed) flows from right-to-left:

• $q_L$, the total charge on the left;
• $q_R$, the total charge on the right;
• $i_{T,F}$, the current along the top at $x$ (holding $x$ constant);
• $i_{T,B}$, the pseudo-current from momentarily fixing the charges and moving the boundary along the top;
• $i_{B,F}$, the current along the bottom at $x$; and
• $i_{B,B}$, the bottom pseudo-current.

Suppose temporarily that $x$ is constant. Then $i_{T,B}=i_{B,B}=0$. By the continuity equation, we have $$\frac{dQ_L}{dt}=i_{T,F}+i_{B,F}$$ But we cannot have unbounded charge accumulate on the left, so in the long-term equilibrium, we must have $$i_{T,F}=-i_{B,F}\tag{1}$$ Indeed, we may already assume that the system has reached this equilibrium. Since $i_{T,F}$ and $i_{B,F}$ are defined by holding $x$ constant, (1) must always hold. Since the current flows counterclockwise, each side of (1) is positive.

Now let the system fall into a black hole (at the left) and choose $x$ to always concide with the event horizon. By the equivalence principle, if our system is sufficiently small, this period should be "nothing special".

We can combine $i_{T,F}$ and $i_{T,B}$ to obtain the total charges falling into the black hole at top and bottom (respectively): \begin{align*} i_T&=i_{T,F}+i_{T,B} \\ i_B&=i_{B,F}+i_{B,B} \end{align*} But nothing can exit an event horizon, so we must have $i_T\geq0$ and $i_B\geq0$.

This, then, is the heart of the "paradox": our intuition is formed by situations in which $\frac{dx}{dt}$ is small, if not $0$. In that case, $$i_B\approx i_{B,F}<0$$ When we fall into the black hole, a large and positive $i_{B,B}$ must instead dominate.

But since $\frac{dx}{dt}=c$, a large, positive $i_{B,B}$ is not hard to arrange.