Black Holes and hair

[Aranfan]

I know that Black Holes have "No Hair", that the only things one can determine about them are mass, electric charge, and angular momentum. I also know that photons are the force carriers of the electromagnetic force.


Now, my question is how can the electromagnetic force get out of a black hole if the photons that carry the force can't?

[GrandMasterTerwynn]

I know that Black Holes have "No Hair", that the only things one can determine about them are mass, electric charge, and angular momentum. I also know that photons are the force carriers of the electromagnetic force.


Now, my question is how can the electromagnetic force get out of a black hole if the photons that carry the force can't?

Hawking radiation.

[Kanastrous]

What does the construction no hair mean, with regard to black holes?

[GrandMasterTerwynn]

What does the construction no hair mean, with regard to black holes?

The No-Hair Theorem postulates that a black hole can be completely characterized by its externally-observable characteristics. All other information about the black hole's component matter is presumed to have disappeared inside the black hole's event horizon.

[Aranfan]

I know that Black Holes have "No Hair", that the only things one can determine about them are mass, electric charge, and angular momentum. I also know that photons are the force carriers of the electromagnetic force.


Now, my question is how can the electromagnetic force get out of a black hole if the photons that carry the force can't?

Hawking radiation.

I fail to see how that explains anything.

[Kanastrous]

The force-carrying particles are the members of particle-antiparticle pairs that escape while their antiparticles are drawn into the singularity.

Is that right, Terwynn?

[GrandMasterTerwynn]

The force-carrying particles are the members of particle-antiparticle pairs that escape while their antiparticles are drawn into the singularity.

Is that right, Terwynn?

This is essentially correct. One half of the virtual particle pair gets promoted to a 'real' particle, and is boosted away from the black hole. This act transfers some of the black hole's mass-energy to the escaping particle (as mass-energy must be conserved.)

[Kanastrous]

What is the source of the energy providing the 'boost?' Dissimilar charges?

[Aranfan]

You still haven't explained how this lets the Black Hole have electric charge when photons can't escape, especially since electrons and positrons should make the jump from "virtual" to "real" at equal rates.

[GrandMasterTerwynn]

What is the source of the energy providing the 'boost?' Dissimilar charges?

It's a quantum effect. Usually, virtual particle-antiparticle pairs randomly appear from the vacuum on the account of vacuum energy. However, they almost instantaneously recombine and annihilate each other, meaning that the total sum of energy from this transaction is zero (as two particles come in, and two go back out.)

However, at the event horizon of a black hole, one of the particles falls into the black hole, 'disappearing' from the universe. However, the universe requires that all the books remain balanced (i.e., we can't have a net creation of energy from nothing.) Since one half of the virtual pair was pulled in by the black hole's gravitational gradient, the other half will carry the same gravitational energy as the first half that fell in, only with opposite magnitude. (So the black hole absorbs a particle of 1 unit of negative mass-energy, becoming lighter. To account for this, it emits a particle of 1 unit of positive mass-energy.)

Which is to say that the source of energy providing the boost is the black hole's own gravitational field.

You still haven't explained how this lets the Black Hole have electric charge when photons can't escape, especially since electrons and positrons should make the jump from "virtual" to "real" at equal rates.

Not necessarily. Hawking radiation mostly generates photons, and it does so on the usual black-body spectrum. Which is to say that the assertion that photons don't escape a black hole is false.

[Aranfan]

That doesn't explain how the Photons "know" which charge to carry. The photons should be generated in a completely random way. Or at least that was my understanding.

[starslayer]

Photons are not themselves charged. They simply mediate interactions between charged particles.

[Aranfan]

Yes, but how do the hawking photons "know" what to mediate? How do they "know" what charge the black hole has?

[Wyrm]

The reason a black hole can have an electric charge is the same reason why it can have a gravitational field, even though gravitational effects also only travel at the speed of light. The loss of an electrical charge measurable by a remote observer constitutes information about what's happening inside the black hole's event horizon, and you can't know what happens inside the event horizon.

For instance, if you drop a small, but measurable, test charge into the black hole, then if the charge stops being measurable, you know the charge is now within the event horizon. Except you can't know when something passes the event horizon, because the information about that event is trapped at the event horizon and can never escape. Therefore, you cannot observe this event, and therefore the charge cannot simply disappear.

In other words, this is yet another one of those "fossil fields" effects.

[Kuroneko]

Not necessarily. Hawking radiation mostly generates photons, and it does so on the usual black-body spectrum.

For large neutral black holes, yes, mainly because they're of such low temperature that they don't have much chance of generating much else. Charged black holes generate more charged particles asymmetrically, which actually tends to neutralize them.

Which is to say that the assertion that photons don't escape a black hole is false.

If it is understood to mean real photons, then the assertion is still true. Hawking radiation is a horizon effect; it's even present, although by a different name, even in ordinary special relativity (for the Rindler horizon in an accelerated frame of reference).

Yes, but how do the hawking photons "know" what to mediate? How do they "know" what charge the black hole has?

Mr. Terwynn was speaking a bit obliquely. Hawking radiation is not responsible for mediating the electromagnetic force from the black hole; rather, it is the same virtual nature of the particles that's responsible for them both. Virtual particles do not obey the speed of light limit. As for the classical answer to your question, see the post directly above this one: it's more accurate to say that black holes can have gravitational or electromagnetic fields because of their horizons rather than despite them. One can even get away with treating the black hole horizons as physical a two-dimensional membranes that are viscous, electrically charged and conductive, and of uniform temperature and finite, positive entropy. It's actually quite neat--from that viewpoint black holes can function in electrical circuits as a resistor or battery, or be rotors in electric motors.

[Aranfan]

Okay, that makes a lot more sense. But I thought that the gravitational pull of the black hole stopped stuff from escaping because of geometry, not because of some sort of speed limit.

How can the "virtual" photons get out of the black hole when "real" photons can't? I mean, they're both photons right?

[Kuroneko]

Okay, that makes a lot more sense. But I thought that the gravitational pull of the black hole stopped stuff from escaping because of geometry, not because of some sort of speed limit.

It is geometry--specifically, the light cones nearer the black hole "tilt" more and more toward the horizon; at the horizon, the edges of the light cones coincide with the horizon. That means that for a particle at or inside the horizon to escape the black hole, it must travel superluminally (i.e., outside the lightcone at its location).

How can the "virtual" photons get out of the black hole when "real" photons can't? I mean, they're both photons right?

By saying virtual photons don't obey the usual speed limit, I mean they can travel on spacelike paths, i.e., paths that go outside their local light-cones.

[Aranfan]

Okay, that makes a lot more sense. But I thought that the gravitational pull of the black hole stopped stuff from escaping because of geometry, not because of some sort of speed limit.

It is geometry--specifically, the light cones nearer the black hole "tilt" more and more toward the horizon; at the horizon, the edges of the light cones coincide with the horizon. That means that for a particle at or inside the horizon to escape the black hole, it must travel superluminally (i.e., outside the lightcone at its location).

How can the "virtual" photons get out of the black hole when "real" photons can't? I mean, they're both photons right?

By saying virtual photons don't obey the usual speed limit, I mean they can travel on spacelike paths, i.e., paths that go outside their local light-cones.

But doesn't that imply superluminal information transfer? Specifically about the Electromagnetic field?

[Kuroneko]

But doesn't that imply superluminal information transfer? Specifically about the Electromagnetic field?

No, it does not; at most it means that measurements of the field will not be independent. Virtual particles are never physically observed; it's not even physically meaningful to talk about a lone virtual particle. What happens in quantum mechanics is that its mathematics can be interpreted in the following (oversimplified) manner: the interaction of two charged particles is the result of a virtual photon going between them over all possible paths in spacetime. One can't send information in this manner simply because it's not even physically meaningful to have one virtual particle traveling some particular path; the interactions are actually the combined effect of virtual particles traveling all possible paths.

[Kuroneko]

I should probably clarify the connection between the light cones, virtual particles, and Hawking radiation, because at this point you may be wondering why Hawking radiation happens at all, given that virtual particles can travel superluminally and thus might be expected not to be at all inconvenienced by the horizon. The negative-energy virtual particle can be trapped inside the horizon not because of the speed limit per se, but because space and time reverse roles across the horizon [1] as the light cones "tip over". Because of this, the corresponding momentum component also inverts with energy [2], and because momentum is allowed to be negative, both particles can become non-virtual if the one outside has positive energy. The opposite isn't true--if the one that stays outside has negative energy, then it must stay virtual.

[1] The Schwarzschild metric is ds² = -(1-2M/r)dt² + dr²/(1-2M/r) + r²dΩ². Across the horizon r = 2M, the signs of the dt² and dr² terms reverse, so that r is now the timelike direction. This physically corresponds to the fact that for observers inside the horizon, moving in the r direction (and also inward) is mandatory, just as moving the t direction is mandatory outside the horizon.
[2] Energy is just the time-component of the momentum four-vector.

[erik_t]

That seems to imply that you can travel in either direction in time once inside the Schwarzschild radius. Is that a correct interpretation? That seems to be how the equation works out, since the coefficient of dt² would now be positive, allowing both positive and negative dt.

[Kuroneko]

That seems to imply that you can travel in either direction in time once inside the Schwarzschild radius. Is that a correct interpretation? That seems to be how the equation works out, since the coefficient of dt² would now be positive, allowing both positive and negative dt.

Not quite. It's true that in the Schwarzschild coordinate chart, inside the horizon one can go either in positive or negative direction of t. However, the t-direction doesn't represent time inside the horizon (in the sense that the corresponding vector isn't timelike). Internally, the it is the r-direction that's timelike instead.