general relativity - Intuitively, why do attempts to delay hitting a black hole singularity cause you to reach it faster?

In general relativity, proper time is maximized along geodesics. Inside of a black hole, all future-oriented timelike trajectories end at the singularity. Putting these two facts together, we find that any deviation from geodesic free fall decreases the proper time before one hits the singularity, so as Carroll says, "you may as well sit back and enjoy the ride."

[Edit: As Dale points out, the Schwarzchild singularity does not consist of a single spacetime event, so this argument fails in general: one can in fact extend the proper time experienced by a free faller between the event horizon and the singularity to some extent by firing rockets inward. But this cannot occur appreciably in the limiting case where the free fall begins at rest just outside the horizon, which I'll assume to be the case.]

This is of course very much counter to nonrelativistic intuition. In Newtonian gravitation, if you fire your jetback inward, you slow your inward fall and buy yourself more time. Is there any physical intuition for why this isn't the case inside of a black hole (if you start free falling from rest at the horizon)?

Answer

Here's a partial answer, although it's still pretty formal. First define $$E := -\left( \frac{2GM}{r} - 1 \right) \frac{dt}{d\tau}, \qquad L := r^2 \frac{d\phi}{d\tau}$$ in the usual Schwarzschild coordinates. If you're free falling, then $E$ and $L$ are constant over your trajectory, but if you can fire your engines then they can change. We can expand out the normalization condition $U \cdot U = -1$ to $$\frac{1}{2} \left( \frac{dr}{d\tau} \right)^2 - \left(\frac{2GM}{r} - 1 \right) \left(1 + \frac{L^2}{r^2} \right) = \frac{1}{2} E^2,$$ which looks somewhat like the statement of conservation of energy (per unit mass) for a nonrelativistic particle with angular momentum $L$.

But there are two weird aspects to this equation:

1. 
   

   When you expand out the product of the two binomials on the LHS, you get a weird term $-2GM L^2/r^3$ that does not appear in the nonrelativistic case. Unlike the usual centrifugal angular momentum barrier, this is a centripetal angular momentum fictitious "force" that actually sucks the particle inward at small radii. This means that angular momentum is actually your enemy, not your friend, for avoiding the singularity - so you don't want to accelerate in a way that increases its magnitude.

   
   


2. 
   

   The effective total energy $\mathcal{E}$ isn't the physical mechanical energy $E$, but instead $\frac{1}{2} E^2$. In the standard nonrelativistic case, firing your engines to slow your infall decreases your total mechanical energy and helps you delay getting close to the center. (This may seem counterintuitive at first, because we associate highly negative energies with tightly bound orbits and positive energies with unbound orbits, so you might think you would want to increase your energy. But for the purpose of delaying getting close to the center, you actually want to brake and make your energy more negative, at the expense of trapping yourself deeper in the gravity well overall and spending more time near the center once you finally do get there.) But in the Schwarzschild case, $\mathcal{E} = \frac{1}{2} E^2$ means that your effective energy actually depends non-monotonically on your physical energy: if your physical energy $E$ is negative, then making it even more negative actually increases your effective energy $\mathcal{E}$. This means that minimizing your effective energy requires keeping your physical energy at $E = 0$, which indeed corresponds to the optimal geodesic which begins at rest infinitesimally outside the horizon. Any attempt to brake further will overshoot $E = 0$ and send $E$ negative, which will actually increase your effective energy and hurt you.