The paper of Jahed Abedi, Hannah Dykaar, and Niayesh Afshordi, in turn, suggests that gravitational waves do indeed reflect back from the horizon:
Echoes from the Abyss: Tentative evidence for Planck-scale structure at black hole horizons
But is it really so that in classical general relativity, arbitrary waves should travel down the geometry of a black hole without reflecting back?
Let us study the Schwarzschild solution of the geometry.
Close to the horizon, the radial speed of light as measured in the global coordinates of the Schwarzschild solution goes as ~ d where d is the Schwarzschild global distance from the horizon and the tangential speed goes as ~ √d.
Let us study the Schwarzschild solution of the geometry.
Close to the horizon, the radial speed of light as measured in the global coordinates of the Schwarzschild solution goes as ~ d where d is the Schwarzschild global distance from the horizon and the tangential speed goes as ~ √d.
Definition 1. Let us have a ray of light whose wavelength far away in the Minkowski space is λ. Let the ray of light enter a static gravitational field. We can use the wavelength of the light as a measuring stick for distances. By optical coordinates we mean coordinates where spatial distances are measured in wavelengths λ, that is, we measure spatial distances by the optical path length of optics.
https://en.wikipedia.org/wiki/Optical_path_length
We can choose the wavelength λ of a radially ingoing wave such that, close to the horizon, each one wavelength step even closer will make the radial speed of light to halve, when measured in the global Schwarzschild coordinates. We call such a step a λ-step.
The λ-step will make the tangential speed of light to go to 1/√2 of the previous step. The wavelength λ far away from the black hole is of the order of the Schwarzschild radius.
Let us try to visualize the geometry of the Schwarzschild solution. We measure the radial and tangential distances in terms of the local length of the wavelength λ. By local we mean what is the wavelength of the incoming wave at that location, measured by a static observer. For tangential distances, we imagine that a mirror at a 45 degree angle is used to turn the radial ray of light into tangential.
Let z = 0 in the Schwarzschild solution. We try to visualize the x,y-plane. If we try to keep λ constant in the visualization, we get a kind of surface.
Our surface looks like a funnel with a wide upper end, the flat Minkowski space, and an infinitely deep exponentially widening pipe that descends to the horizon:
The pipe is infinitely deep before we reach the horizon. That is, the ingoing wave will have an infinite number of cycles λ before it reaches the horizon.
The width of the pipe grows exponentially when we move downward because the tangential λ decreases exponentially. We cannot really embed the pipe into a 3-dimensional Euclidean space because the circumference grows exponentially at each step λ closer to the horizon. The figure above is only to help imagination, not to be taken literally.
The funnel neck is at its narrowest at 3/2 Schwarzschild radii in the global Schwarzschild coordinates.
https://en.m.wikipedia.org/wiki/Fermat%27s_principle
Fermat's principle states that a ray of light can travel a path which is a stationary point of the optical length of the path, with respect to small variations in the path. A ray of light can have a circular orbit around the black hole at that radius.
Abedi et al. assume in their paper that part of outgoing waves are reflected back roughly at the narrowest neck. Abedi et al. consider the LIGO black holes which are rapidly spinning and have the Kerr solution while we have been looking at non-spinning Schwarzschild black holes.
Open problem 2. Where does the backreflection of outgoing waves actually occur? The geometry is not flat anywhere close to the horizon. The narrow neck in the figure is not special in that respect. In optical gravity, the optical density grows smoothly as we approach the horizon. Why would the narrow neck somehow produce more backreflection than other points around it? What does optics tell us? How does the spinning affect the backreflection?
Open problem 2 can probably be solved with a numerical simulation.
As an aside, note that if we measure the depth of the pipe with a ruler (= proper length) then the pipe is not infinitely deep. If an observer is at distance d from the horizon in global Schwarzschild coordinates, then the proper distance from the horizon is 2 * √(d * r_s), where r_s is the Schwarzschild radius. An ingoing wave will be blueshifted close to the horizon. That is why the pipe is infinitely deep if we measure in wavelengths λ.
https://en.wikipedia.org/wiki/Optical_path_length
We can choose the wavelength λ of a radially ingoing wave such that, close to the horizon, each one wavelength step even closer will make the radial speed of light to halve, when measured in the global Schwarzschild coordinates. We call such a step a λ-step.
The λ-step will make the tangential speed of light to go to 1/√2 of the previous step. The wavelength λ far away from the black hole is of the order of the Schwarzschild radius.
Let us try to visualize the geometry of the Schwarzschild solution. We measure the radial and tangential distances in terms of the local length of the wavelength λ. By local we mean what is the wavelength of the incoming wave at that location, measured by a static observer. For tangential distances, we imagine that a mirror at a 45 degree angle is used to turn the radial ray of light into tangential.
Let z = 0 in the Schwarzschild solution. We try to visualize the x,y-plane. If we try to keep λ constant in the visualization, we get a kind of surface.
Our surface looks like a funnel with a wide upper end, the flat Minkowski space, and an infinitely deep exponentially widening pipe that descends to the horizon:
____ ____
\ /
\ /
| |
/ \
/ \
/ \
. .
. .
. .
The pipe is infinitely deep before we reach the horizon. That is, the ingoing wave will have an infinite number of cycles λ before it reaches the horizon.
The width of the pipe grows exponentially when we move downward because the tangential λ decreases exponentially. We cannot really embed the pipe into a 3-dimensional Euclidean space because the circumference grows exponentially at each step λ closer to the horizon. The figure above is only to help imagination, not to be taken literally.
The funnel neck is at its narrowest at 3/2 Schwarzschild radii in the global Schwarzschild coordinates.
https://en.m.wikipedia.org/wiki/Fermat%27s_principle
Fermat's principle states that a ray of light can travel a path which is a stationary point of the optical length of the path, with respect to small variations in the path. A ray of light can have a circular orbit around the black hole at that radius.
Abedi et al. assume in their paper that part of outgoing waves are reflected back roughly at the narrowest neck. Abedi et al. consider the LIGO black holes which are rapidly spinning and have the Kerr solution while we have been looking at non-spinning Schwarzschild black holes.
Open problem 2. Where does the backreflection of outgoing waves actually occur? The geometry is not flat anywhere close to the horizon. The narrow neck in the figure is not special in that respect. In optical gravity, the optical density grows smoothly as we approach the horizon. Why would the narrow neck somehow produce more backreflection than other points around it? What does optics tell us? How does the spinning affect the backreflection?
Open problem 2 can probably be solved with a numerical simulation.
As an aside, note that if we measure the depth of the pipe with a ruler (= proper length) then the pipe is not infinitely deep. If an observer is at distance d from the horizon in global Schwarzschild coordinates, then the proper distance from the horizon is 2 * √(d * r_s), where r_s is the Schwarzschild radius. An ingoing wave will be blueshifted close to the horizon. That is why the pipe is infinitely deep if we measure in wavelengths λ.
If we have an incoming planar wave, how much of its energy can travel down the pipe? The Huygens principle tells us that we can calculate a new place for a wavefront by assuming that each point in the old wavefronts is a new source of oscillation.
Close to the horizon, the space is very much curved. If we try to draw a square where each side is λ and the upper line is horizontal, then the sides differ very much from the radial direction. The lower line would be √2 λ if we put the sides radially.
λ
_________
\ / A "square" close to the horizon
λ \_____/ λ
λ
According to the Huygens principle, there will always be waves reflected back if there is not a total destructive interference of the reflected waves. If a planar wave proceeds in the flat Minkowski space, the reflected waves are completely canceled out by destructive interference.
Intuitively, it is likely that the destructive interference cannot cancel out reflected waves in the very much curved geometry close to the horizon. A numerical simulation may confirm this.
The reflection would be very strong for waves of length λ, that is, waves whose length far away from the black hole is of the order of the Schwarzschild radius. The reflection might be up to 10 % (?).
For shorter waves the reflection is less, but since the distance to the horizon is infinite in terms of λ, even a small ratio of reflection will, after an infinite number of iterations, reflect everything back.
There is some uncertainty, how exactly we should simulate waves in a curved spacetime geometry. Our optical gravity hypothesis is one possibility.
Could it be that all energy of the downgoing waves will eventually be reflected back?
Open problem 3. If planar waves hit a black hole, will all wave energy be reflected back before the downgoing waves reach the horizon?
If the wavelength is of the order of the Schwarzschild radius, then Problem 3 can be solved with numerical methods. An analytic solution is unlikely.
For short waves, the Bogoliubov transformation might offer a way to estimate the reflection. We can use geometric optics to trace the wave a certain distance closer to the horizon. Then do the Fourier decomposition of the wave in a freely falling reference frame. The wave will appear as a chirp in such a frame. Its decomposition contains negative frequencies. The negative frequency waves can be interpreted as reflected waves that are traveling upwards.
If the wavelength is of the order of the Schwarzschild radius, then Problem 3 can be solved with numerical methods. An analytic solution is unlikely.
For short waves, the Bogoliubov transformation might offer a way to estimate the reflection. We can use geometric optics to trace the wave a certain distance closer to the horizon. Then do the Fourier decomposition of the wave in a freely falling reference frame. The wave will appear as a chirp in such a frame. Its decomposition contains negative frequencies. The negative frequency waves can be interpreted as reflected waves that are traveling upwards.
Open problem 4. Is there a depth where a sizeable portion of downgoing waves has been reflected back? Is the Planck length related to this depth in some way?
Open problem 5. If we have a static observer very close to the horizon, then any falling particle will have an almost infinite mass-energy as observed by the observer. How do these almost infinite masses affect the local geometry?
Critique of resonant cavity of Abedi et al.
Abedi et al. calculated that the echo repeat time should be of the order 0.1 seconds in the LIGO data. That is, if we assume reflection points at about 1.5 Schwarzschild radii and at 1 Planck length proper distance from the horizon.
In our model, the backreflection of waves may happen smoothly and not be concentrated at the narrow neck at 1.5 Schwarzschild radii.
Abedi et al calculated that in the LIGO data, the Schwarzschild distance from the horizon is of the order 10^-74 meters when the proper distance from the horizon is the Planck length 1.6 * 10^-35 meters. They expect the reflection to happen around that position.
Gravitational waves that the LIGO can observe have a long wavelength, that is, of the order of our λ above. Each one wavelength λ step towards the horizon halves the Schwarzschild distance from the horizon. If we start at a Schwarzschild distance of, say 10 kilometers from the horizon, it will require roughly 250 λ-steps to get to 10^-74 meters from the horizon in Schwarzschild coordinates.
If we assume a 10 % reflection at each λ-step, then the horizon has reflected most of the wave back already at 10 λ-steps, not 250, as Abedi et al. assume.
The author of this blog will next study the Kerr solution and the backreflection at 1.5 radii.
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