# How much of a wave function must reside inside event horizon for it to be consumed by the black hole?

Related to this question in astronomy SE (https://astronomy.stackexchange.com/q/30611/10813) and in particular my attempt to answer it, I started to wonder which fraction of a waveform (for eg a photon) need to be inside event horizon in order to make its escape impossible? As far as I understand, there is infinitesimal, yet non-zero part of every photon inside event horizons even right now.

• As a GR layman, I would guess that any particle anywhere would have a non-zero probability of tunnelling into the attractive potential. – Pieter Apr 25 '19 at 11:44
• Good question, great answers, and insightful comments - all missing one critical point. In any valid frame of reference nothing ever crosses the horizon. So the wave function of any photon always remains entirely outside. A black hole is located in the infinite future of any external observer, so it cannot be observed consuming anything. – safesphere Apr 25 '19 at 17:22
• safesphere, if simultaneity is defined from the Schwarzschild $t$-coordinate, then it could be said "A black hole is located in the infinite future of any external observer". However under a definition of simultaneity using Gullstrand-Painleve coordinates (or various others) for example, this does not hold. I suggest the latter is more natural, e.g. because the Gullstrand-Painleve observers are freefalling whereas static observers (whose 4-velocity is orthogonal to Schwarzschild $t=\textrm{const}$ have unbounded acceleration as $r\rightarrow 2M$ – Colin MacLaurin May 1 '19 at 0:23

You are thinking of the photon as an extended classical wave, which is not correct within the standard model of particle physics. Photons are point like quantum mechanical particles, and quantum mechanics is all about probabilities.

A photon approaching a black hole will have a calculable probability to fall through and be captured, or join the photon sphere. This means that a distribution of photons with the same energy and boundary conditions has to be accumulated in order to check the calculation.

A single photon is described by a wavefunction $$Ψ$$ and the probability of finding a single photon at an (x,y,z) is given by $$Ψ^*Ψ$$, it is the probability that is a wave in space, not the photon. An ensemble of photons in quantum mechanical superposition builds up the classical electromagnetic wave in space and time in a mathematically consistent manner, using quantum field theory. ,

• how can a probability be a wave in space? Can you explain that further, please? And in this case, the photon is still a point like particle but the probability if there (where?) even is a photon is a wave? – undefined Apr 25 '19 at 13:11
• @undefined Wave is a mathematical entity that follows some rules. The probability evolves in time and space according to a wave equation, so it is a wave. If you want to get at least some understanding of quantum mechanics, you need to understand what a physicist means when they say "wave" first - it's a rather crucial part of the whole shebang. Things like the "rings" you see when you throw a stone in a pond are also waves (not quite, but that wouldn't fit a comment :), but don't get too stuck on that image - that's one specific case of the more general idea of a wave. – Luaan Apr 25 '19 at 14:48
• see the double slit experiment for single photon at a time. The accumulation shows wave interference, and the accumulation is a probability distribution. sps.ch/en/articles/progresses/… – anna v Apr 25 '19 at 15:17
• Quantum mechanics is all about amplitudes, not probabilities. – Roman Odaisky Apr 25 '19 at 20:41
• @RomanOdaisky The amplitude is the $Ψ$, yes, it is what is calculated,. It is not measurable directly, it is a complex number. It is the $Ψ^*Ψ$ that can be compared to measurements in space, and it is a probability distribution. – anna v Apr 26 '19 at 4:34

Introduction

First of all we need to establish what the question even means (I will restrict all formulas to non-rotating Schwarzschild black holes of mass $$M$$ for concreteness). There are no bound states of the electromagnetic field near a black hole, even quasi-stationary states of finite wavelength near the photon radius $$r = 3GM/c^2$$ will decay, a part falling into the black hole, a part to infinity.

So the question really is about scattering. It turns out that scattering of electromagnetic waves behaves differently based on wavelength. In SI units the wavelength is $$\lambda = c/(2\pi \omega)$$ and this is also the localization of the photon. For intuition we can picture the photon as a wavepacket localized in a sphere of radius $$\lambda$$. On the other hand, the typical length of the gravitational background is $$GM/c^2$$. The main parameter that controls the behavior of the scattering is thus the ratio of these lengths $$(GM/c^2)/(c/(2\pi \omega))$$. I will now switch to geometrized $$G=c=1$$ units and drop a $$2\pi$$ factor, and then the ratio of the wavelength to the size of the black hole is $$\sim M\omega$$.

Large wavelengths

For $$M\omega \to 0$$ the wavelength is much larger than the black hole, "a very small part of the photon is in the black hole". In this limit it can be computed that the total cross-section of an incoming monochromatic, asymptotically planar wave is $$\sigma = \left[4 \pi (2 M)^2 \right]\frac{4}{3}M^2 \omega^2 + \mathcal{O}(\omega^3)$$ It is difficult to interpret the leading-order term as some kind of "overlapping volume" of the wave and the black hole, since that would scale as $$M^3\omega^3$$. However, a sphere of radius $$R$$ will scatter particles with a cross-section $$4 \pi R^2$$. Taking a wave-packet with a centroid that would be classically absorbed by a rigid sphere of radius $$2M$$, the factor $$4 M^2 \omega^2/3$$ can be loosely understood as the probability it will actually be absorbed. Quite interestingly, the low-energy photon sees the black hole as a two-dimensional surface interacting with its wave-front rather than a three-dimensional volume.

Shorter wavelengths

However, this all changes with increasing frequency/decreasing wavelength. The intermediate regime is impossible to compute analytically, so I just put a plot from Crispino et al. (2007) here (I also recommend the paper for further references): In the plot, $$\sigma_\mathrm E^{(\omega = 0)}$$ is the low-frequency limit given above, $$\sigma_\mathrm E$$ is the total cross-section of monochromatic electromagnetic waves (computed numerically), $$\sigma_\mathrm S$$ is the same for massless scalar waves, and $$\sigma_\mathrm O$$ is the optical limit which you get from considering photons as point particles $$\sigma_\mathrm O = 27 \pi M^2$$ It is now easy to see that once the size of the wavepacket is comparable or smaller than the black hole, you can simply use the following approximate intuition: If the wavepacket centroid is within the photon sphere, it is going to fall inside it.

• this is within the classical electromagnetic interactions , fair enough, but one should not call them photons. Photons have to be treated quantum mechanically. and it propagates the false, imo, intuition that a photon can be partly in or out of the horizon. Individual particles are either in and absorbed, or out. It is the probability of this that can be calculated for individual photons. They do build up the electromagnetic wave, in QFT, but still they are individual measureable particles. The wave function of the photon is a wavefunction that builds a probability distribution, not space one – anna v Apr 25 '19 at 10:41
• See a wavefunction of a photon here cds.cern.ch/record/944002?ln=en . – anna v Apr 25 '19 at 10:45
• @annav The classical scattering cross-sections are exactly equal to their QFT counter-parts, since the theory is free quantum electrodynamics interacting semi-classically with the curved background. In other words, the total cross-section $\sigma(\omega)$ is a sum of the squared S-matrix elements corresponding to a photon of a sharp momentum and frequency $\omega$ getting absorbed by the black hole. The explanation in the answer just tries to provide a loose intuition for the results in a language similar to that of the OP. – Void Apr 25 '19 at 11:15
• @annav The classical Maxwell solution (the classical electromagnetic wave) is the wavefunction of a single photon in the sense of relativistic quantum mechanics. Saying that a photon is only a particle but not the wave is about as wrong as saying that it is only the wave but not the particle. I think that the history of quantum mechanics taught us this lesson very well. – Void Apr 25 '19 at 14:35
• @Void I am saying that a photon is a quantum entity, a particle when it interacts and a probability wavefor the accumulation of exact same interactions.. A singlephoton is not a wave in space, it has a wavelike probability to exist in an (x,y,z) in space, the individual photon energy is not spread out in space – anna v Apr 25 '19 at 15:15