Username:

Password:

Altair AstroDIO DehumidifiersAtik CamerasModern AstronomyDavid HindsNe3 Filters

Author Topic: Black Hole Evaporation  (Read 2736 times)

0 Members and 1 Guest are viewing this topic.

Offline Kevbar

  • New Member
  • *
  • Posts: 45
Black Hole Evaporation
« on: 16:38:29, 03 November, 2012 »
Hullo Chaps/Chappesses,

I've been reading a bit about the evaporation of black holes via the process known as Hawking Radiation but can't quite get my head around one aspect.

I believe it works thusly:-

A matter particle and anti-matter particle combine to annihilate each other with a release of energy; conversely, pure energy can instantaneously develop two particles namely matter and anti-matter.

If the said two particles appear just outside of a black hole's event horizon, one particle could cross the boundary before re-combination occurs. The particle that remains outside the horizon could escape into the "aether". It would appear from an outsider (wrt the black hole) that the black hole has emitted a particle and therefore has lost mass/energy accordingly.

What I don't get (and its probably blatantly obvious though I fail to see) is that surely the emitted particle is negated by the effect of the particle that crossed the boundary regardless of their matter or anti-matter denomination.

+1 + -1 = 0
-1 + +1 = 0

 :confused:
It's so dark here, I can see everything.....

Offline andrew s

  • Full Member
  • ***
  • Posts: 373
Re: Black Hole Evaporation
« Reply #1 on: 19:11:37, 03 November, 2012 »
Although partices and anti particles annhihilate they both have "positive" mass and so one half of the energy used in creating the pair is carried off while one is captured back into the black hole. Andrew

Offline Kevbar

  • New Member
  • *
  • Posts: 45
Re: Black Hole Evaporation
« Reply #2 on: 19:32:31, 03 November, 2012 »
To quote Captain Mainwairing, "I was wondering when you'd spot that Wilson".

Obvious really, thanks Andrew for clearing that up.  :big_clap: Couldn't see it for looking!  :blindfold:
It's so dark here, I can see everything.....

Offline Kevbar

  • New Member
  • *
  • Posts: 45
Re: Black Hole Evaporation
« Reply #3 on: 13:10:51, 04 November, 2012 »
I've been thinking more on this subject and am still confused.

If the particles were created from energy outside the event horizon, then regardless of anything appearing to be emitted away from the black hole, the particle that is captured would add mass / energy to the black hole.

As I understand it, the previous explanation (Andrew's) would only work for particles created exactly at the event horizon interface and therefore still no net gain, or within it (which is impossible due to the inability of either particle to achieve the escape velocity).

I know I'm still missing something here, Andrew can you or anyone else elaborate more.
It's so dark here, I can see everything.....

Offline andrew s

  • Full Member
  • ***
  • Posts: 373
Re: Black Hole Evaporation
« Reply #4 on: 15:22:23, 04 November, 2012 »
Ok as there is no quantum theroy of gravity Hawking used some reasonable assumtions to get to his result. As far as I understand it the particles are created at the event horizon and one is just inside and one just outside allowing it to escape.


Regards Andrew

Offline Kevbar

  • New Member
  • *
  • Posts: 45
Re: Black Hole Evaporation
« Reply #5 on: 17:06:49, 04 November, 2012 »
hmmm, seems a bit of loose compromise.

I can't see any other end result other than no net mass gain no matter where the particles are created.

The only gain I see possible is if the particles are created at, as you say, the event horizon (EH). At the moment of creation, the EH lies neither within or without the Black Hole (ie it is a dimensionless boundary (like a light switch is either on or off, no in between), yet for the purposes of mass / energy measurement it is encompassed as part of the black hole such as the skin of a bubble , which has thickness.
It's so dark here, I can see everything.....

Offline Bazzaar

  • Poster God
  • *****
  • Posts: 979
Re: Black Hole Evaporation
« Reply #6 on: 19:35:07, 04 November, 2012 »
Doing a bit of Googling on this reveals that this explanation is a bit of a fudge.
It may be that it can only be explained in the language of maths and all attempts in plain english will fail.
It relies on the premise that the particle that fell into the black hole gained negative energy, whatever that is, as it fell down the gravity well and so, by conservation of energy, the escaping particle radiates the balance of real energy.
Hmm.
It also transpires that the amount of energy radiated is inversely proportional to the mass. Small black holes radiate more than large ones. So the black holes that are suspected to have significant effect on the universe, center of galaxies, collapsed massive stars, have virtually no chance of radiating anything. A black hole of thirty solar masses would emit one photon every 1061 years!

One for the theorists only methinks.

Barry
Tal 200k, SW 200P, iOptron iEQ45 pro, HEQ5 pro, QHY5III174m, QHY5L II mono, Samsung SBC 2001, SharpSky focuser, Zwo ADC,  Celestron 15x70 bins,

Offline Kevbar

  • New Member
  • *
  • Posts: 45
Re: Black Hole Evaporation
« Reply #7 on: 20:18:05, 04 November, 2012 »
The particle that gains negative energy as it falls in, is this irrelevant of its 'flavour'.

I still can't see any difference if the conservation of energy law still applies. Do you mean that although the black hole would 'gain' negative energy, the universe outside of the black hole in question would gain positive energy resulting in an overall net balance of zero when considering the system as a whole.
It's so dark here, I can see everything.....

Offline Bazzaar

  • Poster God
  • *****
  • Posts: 979
Re: Black Hole Evaporation
« Reply #8 on: 20:34:18, 04 November, 2012 »
The particle that gains negative energy as it falls in, is this irrelevant of its 'flavour'.

Yes I believe so.

The net result is energy is pumped out of the black hole. Energy and mass are interchangeable so the black hole looses mass.
Tal 200k, SW 200P, iOptron iEQ45 pro, HEQ5 pro, QHY5III174m, QHY5L II mono, Samsung SBC 2001, SharpSky focuser, Zwo ADC,  Celestron 15x70 bins,

Offline Ian Straton

  • Poster God
  • *****
  • Posts: 1131
    • Stoneage Observatory blog
Re: Black Hole Evaporation
« Reply #9 on: 23:27:50, 05 November, 2012 »
the main factor at work here (if I understand it correctly, which I probably don't!) is quantum tunnelling, that is the process by which sub atomic particles can appear to cross otherwise impermeable boundaries.  Effectively there is sufficient uncertainty in the position of sub atomic particles that when they are very near to a boundary their positional probabilities can include regions that are outside the boundary.  In the case of a black hole the formation of a matter/antimatter pair *within* the event horizon can due to the uncertainty principle have one of those particles "tunnel" through the event horizon thus moving mass from inside the EH to outside the EH whilst retaining the momentum it would have had within the EH thus giving it an escape velocity.  The larger the container (Black Hole) the less likely a tunnelling event is to happen because there are far more probably locations within the container than outside.

Offline Kevbar

  • New Member
  • *
  • Posts: 45
Re: Black Hole Evaporation
« Reply #10 on: 19:32:52, 12 November, 2012 »
Haven't heard of this 'tunnelling' before, but surely for a larger black hole, the effective surface area of the spheroidal event horizon would effectively be either the same on both sides or if the horizon has measurable thickness, greater on the outside. The outside will always be greater outside than inside no matter what size the BH.
It's so dark here, I can see everything.....

Offline Ian Straton

  • Poster God
  • *****
  • Posts: 1131
    • Stoneage Observatory blog
Re: Black Hole Evaporation
« Reply #11 on: 21:11:13, 12 November, 2012 »
Tunnelling is a direct consequence of the uncertainty principle: http://en.wikipedia.org/wiki/Quantum_tunnelling

whether a particle tunnels or not is an exercise in probabilities, the position of quantum particles are not certain, only probable, the probability of encountering a particular particle in a particular position can be readily calculated, where it gets odd is if you impose a barrier that a physical object would not be able to cross (for example an event horizon) that barrier doesn't prevent a quantum particle appearing outside the barrier in line with the calculated probabilities for finding the particle in any particular position.  The smaller the container the more likely a tunnelling event is because more of the probable positions are outside the container than are inside hence a larger black hole takes longer to evaporate.

In terms of what causes the BH to loose mass: energy within the black hole is converted into a pair of particles, one matter, one antimatter, just because that's what happens when energy is converted in matter, the really important point here is that the phrase "antimatter" refers only to the charge that the particle carries, both particles have + mass and + momentum. As long as the particles both remain inside the event horizon the total energy/mass of the BH doesn't change and the particles can freely decay back into energy and from the outside it looks like nothing has happened; If, however, one of those particles manages to tunnel through the EH it retains the same mass and momentum that it would have had inside the EH, now that it is outside the EH that momentum is enough for it to have escape velocity and so the mass of that particle is lost from the black hole that mass is equal to half the energy that was originally converted in to the particle pair, both particles can now follow whichever course of particle decay back down to photons without any need to annihilate with the other half of the pair.   Because mass is the sole variable in gravitational singularities the loss of energy/mass from the black hole reduces the radius of gravitational attraction and therefore the surface area of the event horizon, with each reduction in radius tunnelling becomes more likely and so given sufficient time and a lack of in-falling material the BH will eventually evaporate completely.

In a very simple case imagine that a BH contained just enough energy to create an electron/positron pair (both with + mass remember) E=mC^2 in this case E=2*(electron mass) the positron tunnels out of the EH leaving the electron behind, the radius of gravitational influence halves as there is now only half the original mass left within the event horizon.


Offline Kevbar

  • New Member
  • *
  • Posts: 45
Re: Black Hole Evaporation
« Reply #12 on: 22:07:48, 13 November, 2012 »
The smaller the container the more likely a tunnelling event is because more of the probable positions are outside the container than are inside hence a larger black hole takes longer to evaporate.



Couldn't this just be explained simply by the premise that a larger BH has more volume of mass to evaporate through a proportionally smaller surface area.

As to the rest of your explanation, this does in fact make sense to me now. As I posted earlier within this topic:-
As I understand it, the previous explanation (Andrew's) would only work for particles created exactly at the event horizon interface and therefore still no net gain, or within it (which is impossible due to the inability of either particle to achieve the escape velocity)., was almost correct other than that I was missing the option of particle creation effectively straddling the EH whereby one particle has a quiet night in, while the other has a quiet night out!
It's so dark here, I can see everything.....

Offline Ian Straton

  • Poster God
  • *****
  • Posts: 1131
    • Stoneage Observatory blog
Re: Black Hole Evaporation
« Reply #13 on: 11:19:32, 14 November, 2012 »

Couldn't this just be explained simply by the premise that a larger BH has more volume of mass to evaporate through a proportionally smaller surface area.

You could think of the effect it in those terms but it is not an accurate description of the mechanism.

Quote
I was missing the option of particle creation effectively straddling the EH whereby one particle has a quiet night in, while the other has a quiet night out!

Its not that the particle creation straddles the EH, both particles are created *within* the EH but then one  (indeed both could possibly tunnel its just much less likely than one of them doing it) is subject to the quantum tunneling effect which places it outside the EH and allows it to escape.

Offline markthais

  • New Member
  • *
  • Posts: 6
Re: Black Hole Evaporation
« Reply #14 on: 21:45:47, 17 November, 2012 »
Evaporation of a Black hole,
 Nature always uses the simplest solution to solve problems.  The definition of a black hole is a star that is compressed by gravity below it Schwarzschild radius. If you don’t believe that what enters the black hole leaves somewhere else in the universe.  Then if the black hole is not rotating as it gains mass its radius will increase beyond the Schwarzschild radius and will no longer a black hole.  If the black hole is rotating then as it gains mass it will slow down the angular momentum increase in radius and expand to a neutron star or something similar.
This is a nice simple hypothesis.
 It always fun to question the theories.
 It is assumed that anything near the black hole is drawn in. I would say that there is a plasma cloud at the event horizon that acts like an atmosphere if a particle does not enter at the right angle it is bound back into space. This would stabilize the black hole for a longer life.
This is my story and I’m sticking to it.
Mark

 

ukbuysellRemote Imaging from AustraliaSharpSkyblank APTUKAI on Facebook
Powered by SMF 2.0.15 | SMF © 2006, Simple Machines LLC
DarkBreak by DzinerStudio. Theme modified by The UKAI Team

Page created in 0.443 seconds with 37 queries.
TinyPortal © 2005-2012