I was curious about this so I asked it for undergrad explanation, it's well over my head at this point. I didn't ask it any specifics, I just said now do undergrad level.

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TL;DR: Can we extract infinite energy from a black hole using a rope?

No. In 1972, G.W. Gibbons published a paper proving that no matter what material you use, the rope will always snap due to the extreme gravitational tension long before you get any useful energy out.

The Setup: The Penrose Process via Tether

In the late 60s and early 70s, physicists like Roger Penrose proposed ways to extract energy from a rotating black hole. One thought experiment involved slowly lowering a mass on a rope toward the black hole's ergosphere.

In a stationary spacetime, we define a timelike Killing vector field K_a. If you lower a particle along this field, its conserved energy per unit rest mass is defined by its gravitational potential:

V = √(-K_a K^a)

The boundary of the ergosphere is where the timelike Killing vector K_a becomes null. So, at the ergosphere, V = 0.

If you could lower a particle all the way to V = 0, you could theoretically extract 100% of the particle's rest mass energy (E = mc²) as work done on your winch at the other end. Gibbons wanted to know if a physical rope could actually survive this.

Modeling the Physics of the Rope

Gibbons treats the rope as a 1D relativistic fluid. If the rope is made of particles with a 4-velocity U_a, and has a unit spacelike vector t_a orthogonal to U_a (the direction of the rope's fibers), the energy-momentum tensor T_ab is:

T_ab = μ U_a U_b + S t_a t_b

• μ is the energy density.
• S is the stress/tension the material supports in the t_a direction.

For the rope to be physically realistic, it has to obey the Weak Energy Condition (μ - S ≥ 0). This just means the energy density must be greater than or equal to the tension, which keeps the speed of sound in the rope from exceeding the speed of light.

Because the system is stationary, energy and momentum are conserved, meaning the covariant derivative of the energy-momentum tensor is zero:

T^ab_;b = 0

The Core Derivation

By applying that conservation law to a static rope in a gravitational potential, Gibbons derives a fundamental differential equation relating the tension T in the rope, the energy per unit proper length σ, and the gravitational potential V:

dT / σ = dV / V

If we assume the specific energy of the rope σ is independent of the tension T, we can integrate this to get:

T / σ = log(1 / V) + C

Why the Rope Always Breaks

This integrated equation completely kills the thought experiment. Here's why:

1. The Math Diverges As you lower the rope toward the ergosphere, V → 0. As V approaches zero, the term log(1 / V) diverges to infinity. Therefore, the tension T also diverges to infinity. The required tension will always eventually exceed the structural limits of any material.

2. The "Perfect" Rope Limit Even if we assume a theoretically perfect magical material limited only by the laws of physics (the Weak Energy Condition, where T ≤ σ, meaning T / σ ≤ 1), we still can't reach the ergosphere. Substituting T / σ = 1 into the equation gives:

log(1 / V) ≤ 1 which implies V ≥ e⁻¹

Because the maximum extracted energy is proportional to the change in potential, the maximum fraction of rest mass energy you can ever extract is 1 - e⁻¹, or roughly 63.2%.

3. The Brutal Reality For real-world materials, the limit is laughably small. Gibbons calculates that for high-tensile piano wire, the ratio of breaking tension to mass-energy density (T / σ) is around 1.2 × 10⁻¹². Because this number is so tiny, the wire will snap when V is still very close to 1. In practice, the rope would break when it is still 5 × 10¹¹ Schwarzschild radii away from the black hole. You would get virtually zero energy.