Discussion:
Black holes can't evaporate through pair production
7
2018-02-07 22:29:31 UTC
Raw Message
Black holes can't evaporate through pair production
---------------------------------------------------

This idea that black holes can evaporate through
exploiting pair production is probably nonsense.

If a particle anti-particle pair were produced
and one fell into the black hole, then more
than likely another pair would be produced
and the opposite charge fell into the black hole
to statistically balance out.
So the black hole will gain mass?

To balance it, you might expect a particle
and its anti-particle to be emitted by
the black hole which would recombine
to emit gamma rays which is therefore 'evaporating'
the black hole. But it isn't evaporating - its gaining mass!

It is more than likely that pair production
and annihilation at the black hole boundary
is a vector and it points inward towards the centre
of the black hole or its reverse and the outwards
direction due to intense gravity,
and not some random direction such as tangential to the
surface of the black hole. This then guarantees
that pair production and annihilation happens in
an orderly way and doesn't result in neither black holes
growing in mass or evaporating.

Somewhere in there is a claim that matter can
escape a black hole if its done within a time
period as short as the pair production annihilation time
and distances.
Edward Prochak
2018-02-08 14:55:53 UTC
Raw Message
Post by 7
Black holes can't evaporate through pair production
---------------------------------------------------
This idea that black holes can evaporate through
exploiting pair production is probably nonsense.
If a particle anti-particle pair were produced
and one fell into the black hole, then more
than likely another pair would be produced
and the opposite charge fell into the black hole
to statistically balance out.
So the black hole will gain mass?
No.
Pair #1 electron escapes, taking -1 charge and 0.511MeV of mass.
Pair #2 positron escapes, taking +1 charge and 0.511MeV of mass.

Net charge total 0 (No gain or loss of charge)
Net mass loss of 1.022MeV (equivalent to 1.8E-30Kg)

Thing is, Mass is always positive.
Post by 7
To balance it, you might expect a particle
and its anti-particle to be emitted by
the black hole which would recombine
to emit gamma rays which is therefore 'evaporating'
the black hole. But it isn't evaporating - its gaining mass!
They annihilate forming photons with a 1.022MeV total mass.
Post by 7
It is more than likely that pair production
and annihilation at the black hole boundary
is a vector and it points inward towards the centre
of the black hole or its reverse and the outwards
direction due to intense gravity,
and not some random direction such as tangential to the
surface of the black hole. This then guarantees
that pair production and annihilation happens in
an orderly way and doesn't result in neither black holes
growing in mass or evaporating.
yes lots of pair production happens where nothing comes out
of the black hole. The probability of escape is very small,
but it is not zero.
Post by 7
Somewhere in there is a claim that matter can
escape a black hole if its done within a time
period as short as the pair production annihilation time
and distances.
The electron and positron half to be produced just outside
the event horizon. then the resulting photons can carry away
the energy (mass).

Note: the annihilation produces at least 2 photons.
It may be only one escapes. Then the energy (mass) loss
by the black hole will be less than 1.022MeV. But it will be > 0.

Enjoy.
ed
Volney
2018-02-08 13:10:22 UTC
Raw Message
Post by Edward Prochak
Post by 7
Black holes can't evaporate through pair production
---------------------------------------------------
This idea that black holes can evaporate through
exploiting pair production is probably nonsense.
If a particle anti-particle pair were produced
and one fell into the black hole, then more
than likely another pair would be produced
and the opposite charge fell into the black hole
to statistically balance out.
So the black hole will gain mass?
No.
Pair #1 electron escapes, taking -1 charge and 0.511MeV of mass.
Pair #2 positron escapes, taking +1 charge and 0.511MeV of mass.
Net charge total 0 (No gain or loss of charge)
Net mass loss of 1.022MeV (equivalent to 1.8E-30Kg)
Thing is, Mass is always positive.
Post by 7
To balance it, you might expect a particle
and its anti-particle to be emitted by
the black hole which would recombine
to emit gamma rays which is therefore 'evaporating'
the black hole. But it isn't evaporating - its gaining mass!
They annihilate forming photons with a 1.022MeV total mass.
Post by 7
It is more than likely that pair production
and annihilation at the black hole boundary
is a vector and it points inward towards the centre
of the black hole or its reverse and the outwards
direction due to intense gravity,
and not some random direction such as tangential to the
surface of the black hole. This then guarantees
that pair production and annihilation happens in
an orderly way and doesn't result in neither black holes
growing in mass or evaporating.
yes lots of pair production happens where nothing comes out
of the black hole. The probability of escape is very small,
but it is not zero.
Post by 7
Somewhere in there is a claim that matter can
escape a black hole if its done within a time
period as short as the pair production annihilation time
and distances.
The electron and positron half to be produced just outside
the event horizon. then the resulting photons can carry away
the energy (mass).
Note: the annihilation produces at least 2 photons.
It may be only one escapes. Then the energy (mass) loss
by the black hole will be less than 1.022MeV. But it will be > 0.
The description usually given for BH evaporation is that a virtual pair
forms just outside the event horizon, and one of the pair falls in. This
forces the other to become real and to have its normal positive mass
making the one that fell in have negative mass (how?). The negative mass
decreases the BH mass by the mass of the escaping particle.

I can understand the sums of the masses must be zero to conserve
mass-energy, but how can the mass of the infalling particle be forced to
be negative?
Edward Prochak
2018-02-08 18:13:12 UTC
Raw Message
[]
Post by Volney
Post by Edward Prochak
Note: the annihilation produces at least 2 photons.
It may be only one escapes. Then the energy (mass) loss
by the black hole will be less than 1.022MeV. But it will be > 0.
The description usually given for BH evaporation is that a virtual pair
forms just outside the event horizon, and one of the pair falls in. This
forces the other to become real and to have its normal positive mass
making the one that fell in have negative mass (how?). The negative mass
decreases the BH mass by the mass of the escaping particle.
I can understand the sums of the masses must be zero to conserve
mass-energy, but how can the mass of the infalling particle be forced to
be negative?
I think it is negative only in the accounting sense
where as you said that a virtual pair is formed.

Do they have to be virtual?
Consider a photon just inside the event horizon.
Through a it becomes a positron-electron pair.
(Standard QM stuff). The energy of the photon
is converted into the mass of the pair and any excess
becomes kinetic energy shared between the particles.
So the mass was there all along.

Kind of like dealing with positive holes in
solid state physics.

Ed
m***@gmail.com
2018-02-09 01:25:41 UTC
Raw Message
Post by Edward Prochak
[]
Post by Volney
Post by Edward Prochak
Note: the annihilation produces at least 2 photons.
It may be only one escapes. Then the energy (mass) loss
by the black hole will be less than 1.022MeV. But it will be > 0.
The description usually given for BH evaporation is that a virtual pair
forms just outside the event horizon, and one of the pair falls in. This
forces the other to become real and to have its normal positive mass
making the one that fell in have negative mass (how?). The negative mass
decreases the BH mass by the mass of the escaping particle.
I can understand the sums of the masses must be zero to conserve
mass-energy, but how can the mass of the infalling particle be forced to
be negative?
I think it is negative only in the accounting sense
where as you said that a virtual pair is formed.
Do they have to be virtual?
Consider a photon just inside the event horizon.
Through a it becomes a positron-electron pair.
(Standard QM stuff). The energy of the photon
is converted into the mass of the pair and any excess
becomes kinetic energy shared between the particles.
So the mass was there all along.
Kind of like dealing with positive holes in
solid state physics.
Ed
Why is negative only falling in?
Why is positive only escaping?
There should be an average
Roy? Do negatives exist?

Mitchell Raemsch
Edward Prochak
2018-02-09 15:16:15 UTC
Raw Message
On Thursday, February 8, 2018 at 8:25:46 PM UTC-5, ***@gmail.com wrote:
[]
Post by m***@gmail.com
Why is negative only falling in?
Why is positive only escaping?
There should be an average
Roy? Do negatives exist?
Mitchell Raemsch
First be clear: There is NO negative mass.
there is NO negative kinetic energy.

I already explained how the net loss of mass by
Hawking radiation can occur. It is a VERY SLOW process.
Don't expect to see black holes disappearing any time soon.

Regarding charge, there may be a bias.

Consider 2 cases:
1. Electron just outside the event horizon
2. Positron just outside the event horizon
Each has enough kinetic energy to escape the gravity of the black hole.

If the black hole has a net negative charge,

1. then the electron is also repelled by the charge
of the black hole. so it has a higher chance of escaping

2. then the positron is also attracted by the charge
of the black hole. so it has a lower chance of escaping

Over time the black hole loses its charge and becomes neutral.

HTH,
ed
reber G=emc^2
2018-02-09 18:08:31 UTC
Raw Message
Post by m***@gmail.com
Post by Edward Prochak
[]
Post by Volney
Post by Edward Prochak
Note: the annihilation produces at least 2 photons.
It may be only one escapes. Then the energy (mass) loss
by the black hole will be less than 1.022MeV. But it will be > 0.
The description usually given for BH evaporation is that a virtual pair
forms just outside the event horizon, and one of the pair falls in. This
forces the other to become real and to have its normal positive mass
making the one that fell in have negative mass (how?). The negative mass
decreases the BH mass by the mass of the escaping particle.
I can understand the sums of the masses must be zero to conserve
mass-energy, but how can the mass of the infalling particle be forced to
be negative?
I think it is negative only in the accounting sense
where as you said that a virtual pair is formed.
Do they have to be virtual?
Consider a photon just inside the event horizon.
Through a it becomes a positron-electron pair.
(Standard QM stuff). The energy of the photon
is converted into the mass of the pair and any excess
becomes kinetic energy shared between the particles.
So the mass was there all along.
Kind of like dealing with positive holes in
solid state physics.
Ed
Why is negative only falling in?
Why is positive only escaping?
There should be an average
Roy? Do negatives exist?
Mitchell Raemsch
HBs do not fade away.They implode Bert
hanson
2018-02-09 18:26:34 UTC