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This is not a duplicate. I do not ask about the gravitational field of the black hole, or why it extends farther out then the event horizon. I do understand that the gravitational field is described by a mathematical model called virtual gravitons. And that it has nothing to do with actual real gravitons. I do understand that nothing (no info, no real particles) can escape the event horizon. My question is about real gravitons (hypothetical).

I have read this question:

How does gravity escape a black hole?

Where John Rennie says:

The answer to your question is that nothing can travel faster than light, and light can't escape through the event horizon. Therefore gravitational waves can't escape either.

Now GWs are said to be made of real gravitons. They are said to be the quanta of gravity, just like photons are the quanta of electromagnetism, and the classical EM wave is built up by a herd of photons.

What is the difference between gravitational waves and gravitational distortions in spacetime?

where Bob Bee says:

On gravitons, probably a quantum theory of gravity will have to include gravitons, but till we figure out that theory we don't know for sure. All points to there probably being gravitons as the carriers of the gravitational waves, but we don't really know yet what that means.

What is the difference between gravitons and gravitational waves?

where annav says:

Gravitons are to gravitational waves the theoretical analogue of photons for electromagnetic waves. They are the proposed carriers of the gravitational interactions at the quantum level, and are expected to appear naturally in a future theory of quantized gravity. At the moment quantization of gravity can be accommodated in string theories, which are at the frontier of research for particle physics. The standard model involves only the three other forces , not the gravitational. A future standard model should have both the present standard model and quantization of gravity, a Theory Of Everything (TOE). So photons are the building blocks of light, and gravitons are (hopefully) the building blocks of gravitational waves.

Gravitons do interact with matter, but the interaction has a low crossection.

Unambiguous detection of individual gravitons, though not prohibited by any fundamental law, is impossible with any physically reasonable detector.[17] The reason is the extremely low cross section for the interaction of gravitons with matter. For example, a detector with the mass of Jupiter and 100% efficiency, placed in close orbit around a neutron star, would only be expected to observe one graviton every 10 years, even under the most favorable conditions.

https://en.wikipedia.org/wiki/Graviton

Real gravitons do have an extremely low cross section when interacting with matter.

Thus, they would easily just fly through a BH. The BH would seem transparent to a real graviton.

Question:

  1. if the real graviton has a extremely low cross section when interacting with matter, does that mean that the real graviton will just fly through the BH, like it (the BH) was transparent?
Árpád Szendrei
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  • Think about a similar question for photons and a transparent black hole. The photons fly in, but not back out. – mmesser314 Jul 04 '19 at 14:47
  • You don't need to say "real gravitons" to contrast with "virtual" ones. Just remember that most stories about virtual particles are just kind of made up to get some intuition about equations. In the actual formalism, only "real" particles are actually present at any time, so there's no need to specifically call them real. – knzhou Jul 04 '19 at 16:59
  • "the gravitational field is described by a mathematical model called virtual gravitons" - No, gravitons are hypothetical, because there is no valid theory that predicts them. "nothing (no info, no real particles) can escape the event horizon" - Only for true horizons, but they can't form in a finite time and don't exist. Stuff can escape real apparent horizons. A "BH would seem transparent to a real graviton" - A hypothetical graviton will never reach the horizon and for all practical purposes would appear absorbed 100%, unless it misses the BH (as @mmeent has correctly answered below). – safesphere Jul 06 '19 at 06:44
  • Based on your reference to virtual gravitons, you may be under the impression that the gravitational field of a black hole is sourced from inside the horizon (e.g. from the singularity), so that virtual gravitons can escape. This is a widespread misconception. A singularity doesn't cause gravity even inside and definitely not outside the horizon. The gravity of a black hole is sourced from outside the horizon. – safesphere Jul 06 '19 at 06:57
  • @safesphere "The gravity of a black hole is sourced from outside the horizon." I thought that the source of gravitational field is stress-energy. Most of the stress-energy of the BH comes from the matter it has inside the EH right? Is that a contradiction? – Árpád Szendrei Jul 06 '19 at 06:59
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    Gravity is curved spacetime. Imagine a stretched flat sheet of rubber. Pull it at a point and the sheet will curve. What curves it? At that very point, you do, but at any other point, rubber is pulled by rubber next to it, not by you. So stress-energy-momentum, where it is, causes gravity only there. Everywhere else, spacetime is curved by the curved spacetime next to it. This is why most black hole solutions are called "vacuum solutions", as they don't involve any stress-energy-momentum inside or outside. In a real BH, all matter is outside, as it takes an infinite time to cross the horizon. – safesphere Jul 06 '19 at 07:20
  • There is a lot of confusion about black holes, so many will disagree with the last statement above. – safesphere Jul 06 '19 at 07:37
  • @safesphere so basically since from an outside observer's view, everything freezes at the horizon, all the matter is there. So from the outside observer's view, the gravitational field of the horizon must be created by the matter outside the horizon. – Árpád Szendrei Jul 06 '19 at 09:05
  • Yes, sort of, with some caveats. Black holes produce weird effects. One could say that the energy of the falling things is converted to the energy of the gravitational field while these things become suspended out of existence by the infinite time dilation at the horizon. I believe their stress-energy becomes zero at the horizon for an outside observer. For example, an infalling photon is infinitely redshifted to a zero energy at the horizon. – safesphere Jul 06 '19 at 15:42

1 Answers1

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The scattering of gravitons of black holes is really no different than that of photons (or massless scalars for that matter). The scattering/absorbtion cross-sections are readily calculated using modern black hole perturbation theory. The only difference is the spin of the underlying Teukolsky wave equation.

The absorbtion cross-section depends on the wavelength of the graviton. If the wavelength of the graviton is much larger than the size of the black hole horizon, the absorbtion cross-section will be very small. However, rather than saying the black hole is transparent to the graviton, it is more accurate to think of this as the wave going around the black hole.

If the wavelength is much smaller than the size of the black hole horizon (but still larger than the Planck length), the absorbtion cross-section will be much bigger, and gravitons will have a high probability of being absorbed by the black hole, just as high as for photons. I.e. the effective cross-section of the black hole will be approximately the area of the black hole shadow (to the first approximation of the projection of the photon sphere).

(If the wavelength becomes comparable to the Planck length, the tacit approximation used above that non-linear effect can be ignored ceases to be valid and we would a full theory of quantum gravity to obtain the answer.)

Kasiéobì ùdumágà
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TimRias
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  • What would be the energy of a graviton (in kilograms) with the wavelength comparable to the Planck length? (+1) – safesphere Jul 06 '19 at 06:48
  • @safesphere That would be a Planck mass i.e. circa $2\cdot 10^{-8}$ kg. (Note that this assumes that the scale for quantum gravity is set by the Planck scale, in principle quantum gravity effects could set in earlier, e.g. if you have "large" extra dimensions.) – TimRias Jul 08 '19 at 07:19