Virtual particles and black holes

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SUMMARY

This discussion centers on the behavior of virtual particles, specifically virtual photons and gravitons, in relation to black holes and Hawking radiation. Participants clarify that virtual particle pairs are generated just outside the event horizon, with one particle potentially escaping while the other is trapped. The conversation also touches on the implications of charge, mass, and angular momentum in the context of the no-hair theorem, asserting that these properties remain detectable despite gravitational redshift over time. The semi-classical approximation is highlighted as a critical framework for understanding these phenomena.

PREREQUISITES
  • Understanding of Hawking radiation and its implications
  • Familiarity with the no-hair theorem in black hole physics
  • Knowledge of quantum field theory (QFT) principles
  • Basic concepts of general relativity (GR) and event horizons
NEXT STEPS
  • Study the semi-classical approximation in quantum field theory
  • Explore the implications of the no-hair theorem on black hole characteristics
  • Investigate the mechanisms of Hawking radiation in detail
  • Learn about gravitational redshift and its effects on particle detection
USEFUL FOR

Physicists, graduate students in theoretical physics, and anyone interested in the intersection of quantum mechanics and general relativity, particularly in the context of black hole research.

2Tesla
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Why are virtual photons and gravitons allowed to pass outwards through the event horizon of a black hole? Is it that it doesn't really make sense to assign a particular path to a virtual photon? This still isn't satisfying to me, as there is *no* path that allows a real photon to escape.

Further confusing me on this issue, part of the Hawking radiation process is for one of a virtual pair of particles to fall into the black hole and be trapped there. Why is that virtual particle trapped, while a virtual photon is free to escape?

And, if I can slip one more question into this already-crowded post: I've heard, as supposed proof for charge, mass and spin being exceptions to the no-hair rule, the fact that charge, energy and angular momentum wouldn't be conserved if these three characteristics were lost (radiated). This doesn't seem right to me, since I can put a charged object inside a shielded box and not violate conservation of charge. Is the real reason just that they can't be radiated?

(If people would prefer it, I can move these questions (or just the last one) over to the GR board.)
 
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This question concerns the so-called semi-classical approximation, which goes outside gtr to the realm of QFTs, so it can probably stay here.

Graduate students can see Jennie Traschen, An Introduction to Black Hole Evaporation, in Mathematical Methods of Physics, ed. by A. Bytsenko and F. Williams, World Scientific, 2000, for a very nice introduction.
 
Why are virtual photons and gravitons allowed to pass outwards through the event horizon of a black hole? Is it that it doesn't really make sense to assign a particular path to a virtual photon? This still isn't satisfying to me, as there is *no* path that allows a real photon to escape.
I believe what you are describing doesn't happen. The virtual pair is born just outside the event horizon. One particle falls in and the other doesn't.
 
mathman said:
I believe what you are describing doesn't happen. The virtual pair is born just outside the event horizon. One particle falls in and the other doesn't.

In the case of Hawking radiation, yes. But if the mass and charge of the black hole are able to exert gravitational and EM forces on outside particles, then virtual gravitons and photons (respectively) must be able to escape, as the force carriers.

The fact that one type of virtual particle is allowed to escape while the other is not, is exactly what is confusing me.

Thanks for the reference, Chris Hillman, I'll take a look.
 
how do you know
 
2Tesla said:
But if the mass and charge of the black hole are able to exert gravitational and EM forces on outside particles, then virtual gravitons and photons (respectively) must be able to escape, as the force carriers.
The force carrying particles don't need to cross the event horizon. The black hole would have been formed from collapsing matter, which would have had a given charge and mass and so emitted virtual gravitons and photons. As the collapse proceeds these will be more and more time dilated, and so the forces we see will be from virtual particles emitted by matter which hasn't yet crossed the event horizon.
2Tesla said:
This doesn't seem right to me, since I can put a charged object inside a shielded box and not violate conservation of charge. Is the real reason just that they can't be radiated?
I'm not sure that this is true, my understanding is that the charge would show up on the outside of the box.
 
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chronon said:
The force carrying particles don't need to cross the event horizon. The black hole would have been formed from collapsing matter, which would have had a given charge and mass and so emitted virtual gravitons and photons. As the collapse proceeds these will be more and more time dilated, and so the forces we see will be from virtual particles emitted by matter which hasn't yet crossed the event horizon.

Interesting! But it brings up another issue for me. Let's say that an isolated star collapses to a black hole, so that no other matter falls onto it afterwards. Then you say that any charge or mass we sense (let's talk about just charge, since it is simpler) would, from our far-away Minkowski perspective, come from the star matter as it falls ever-slower toward the event horizon, never quite reaching it. But, as you also say, these virtual photons will be more and more gravitationally redshifted with time. Eventually we will no longer be able to sense them, right? Does that mean that, in the limit of infinite time, an isolated black hole has no (detectable) charge (or even mass)? That can't be right, can it?
 
2Tesla said:
Then you say that any charge or mass we sense (let's talk about just charge, since it is simpler) would, from our far-away Minkowski perspective, come from the star matter as it falls ever-slower toward the event horizon, never quite reaching it. But, as you also say, these virtual photons will be more and more gravitationally redshifted with time. Eventually we will no longer be able to sense them, right? Does that mean that, in the limit of infinite time, an isolated black hole has no (detectable) charge (or even mass)? That can't be right, can it?
Well I realize that it seems counterintuitive, but the charge and the mass will indeed stay the same to an outside observer (neglecting Hawking radiation). Quite how this fits with it coming from ever more redshifted virtual particles I'm not sure, but I would guess that it would be possible, but tricky, to demonstrate how this comes out of the mathematics of the situation. It would be tricky because the normal way to think of gravity in this situation is as the curvature of spacetime rather than as a force or as mediated by gravitons.
 
mathman said:
I believe what you are describing doesn't happen. The virtual pair is born just outside the event horizon. One particle falls in and the other doesn't.

a black hole is an object having mass. so it shows a presence of some field consider it as a magnetic or a gravitational field. this field produces fluctuations in the form of particle antiparticle pair ON THE EVENTHORIZON . the antiparticle exists for a short period while the particle absorbs some energy and escapes out from the black holes event horizon. as this energy is in the packets called quanta it can radiate. this is called hawking radiation.my question is except hawking radiation black holes accreation disc also emits radiation. how can it be possible?
 
  • #10
doesnt matter because these things are highly theorotical.
 
  • #11
2Tesla said:
In the case of Hawking radiation, yes. But if the mass and charge of the black hole are able to exert gravitational and EM forces on outside particles, then virtual gravitons and photons (respectively) must be able to escape, as the force carriers.

The fact that one type of virtual particle is allowed to escape while the other is not, is exactly what is confusing me.

Thanks for the reference, Chris Hillman, I'll take a look.

its not actually a virtual particle but a particle antiparticle pair in which the antiparticle exists for a shorter period of time. and black hole doesn't exert any gravitational or em force outside. the only thing its field creates some fluctuations outside.
 

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