Do Black holes emit radiation?

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  • #1
Jack
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In 'The universe in a nutshell' Hawking states that due to vacuum fluctuations black holes appear to emit radiation. This lead me to believe that the black holes does not actually emit radiation but they only appear to do so which I thought would be most likely since not even light can escape.

However he later goes on to say that due to this radiation black holes could get smaller.

Could someone please make this idea clearer to me whilst explaining it as simply as possible?
 

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  • #2
chroot
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A black hole is surrounded by a mathematical surface (not a "physical" surface) called the event horizon. Inside the event horizon, the escape velocity is greater than the speed of light -- which means that nothing can escape from within the event horizon. The event horizon is thus the "point of no return" for an object falling into the hole.

The vacuum is not really entirely empty -- it's a sea of particle pairs coming into and going out of existence. This same process, pair production, occurs near the event horizon of black holes.

Sometimes, one of the particles ends up crossing the event horizon and disappearing, while the other (on a different trajectory) manages to escape. The situation looks exactly as if the black hole had produced the escaped particle, and lost mass. Black holes "evaporate" by this method. Such escaped particles are called "Hawking radiation."

- Warren
 
  • #3
Jack
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But as you stated the situation only 'looks as if the black hole had produced the escaped particle, and lost mass' and so if it does not really then how do 'Black holes "evaporate" by this method'?
 
  • #4
chroot
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What I mean is this:

The situation a pair is created, one falls in, and the other flies away is physically indistinguishable from the situation the black hole emitted a particle.

- Warren
 
  • #5
LURCH
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I see a misscomunication here. One crucial ingrediant has been left out of the explanation.

Virtual particle pairs are formed as a result of the presence of energy in the vacuum. Near the event horizon of a black hole, there is tension on the vacuum caused by tidal forces in that region of space. This provides energy for the production of virtual particles. So the particles are energy from the black hole becoming matter (well, virtual matter). If both particles are made from energy that the black hole provided, and one of them escapes, the black hole loses half the energy it took to produce the pair. Since energy is mass (e=mc2), the loss of energy means a loss of mass.

This can also be explained in terms of negative energy, but most find the concept a little repellent.
 
  • #6
Jack
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Originally posted by chroot
The situation a pair is created, one falls in, and the other flies away is physically indistinguishable from the situation the black hole emitted a particle.

- Warren
physically indistinguishable

In what way is it distinguishable?
 
  • #7
ObsessiveMathsFreak
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Hey wait a minute.

If one flys off...
...And one falls in...

Then it isn't losing mass at all. It's actually gaining mass, dispite what you measure.
 
  • #8
chroot
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Originally posted by Jack
physically indistinguishable

In what way is it distinguishable?
Inside your own head.

- Warren
 
  • #9
chroot
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Originally posted by ObsessiveMathsFreak
Then it isn't losing mass at all. It's actually gaining mass, dispite what you measure.
Wrong.

The energy used to create the pair came from the black hole; half of that energy is lost. Energy and mass are the same thing. Since the black hole has lost energy, it has equivalently lost mass.

- Warren
 
  • #10
Jack
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Originally posted by chroot
Inside your own head.

- Warren

Eh? [?] [?] [?] [?]
 
  • #11
chroot
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Originally posted by Jack
Eh? [?] [?] [?] [?]
There are many situations in physics were two distinct mechanisms are physically indistinguishable. For example, say you have two particles -- like two electrons -- and you collide them. Say the one fired from the left is Electron #1. Say the one fired from the right is Electron #2.

When they collide, they "bounce" off each other, and fly off to some detectors, called also #1 and #2. But the electrons are indistinguishable particles -- your detectors record receiving the electrons, but there's no way to say whether #1 or #2 was detected in each one. So there are two mechanisms:

A) Electron #1 hits detector #1, and electron #2 hits detector #2
B) Electron #2 hits detector #1, and electron #1 hits detector #2

These two mechanisms are distinct inside your own head: you can imagine tagging an electron and watching it go through the entire collision, and keeping track of which is which. Physically, however, possibilities A and B are not distinguishable -- there's no way you can tell which one actually happened, because you can't tell one electron from another. Quantum mechanically, we'd say that the experiment would actually be a mixture (interference) of possibilities A and B.

- Warren
 

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