[Bored chemist]

"When particles escape, the black hole loses a small amount of its energy and therefore some of its mass (mass and energy are related by Einstein's equation E = mc2)."

Yes. That's the bit that says you are wrong.
Overall, the mass/ energy is conserved.
The BH loses a little mass an that is converted to the energy of a photon- which then escapes or the mass is converted to a particle which leaves.
But the sum of the energy/ mass remains the same.

Do you agree that hawking specifically claims for ejected particle with mass energy of E=mc^2?

It's complicated; the particle leaves with energy- it's moving.
And so the total mass is the rest mass plus the relativistic mass.
And that sum is what teh BH loses.

Do you consider it as a mass less photon or particle with real positive mass energy?

It's a photon, the same as the ones coming from your computer screen.

So, how can you claim for photon while Hawking is specifically discussing about a particle with positive mass energy?

A photon is a particle with (relativistic) mass.

[Kryptid]

Therefore, As the mass-energy in the ejected particle is E=mc^2, than the BH is losing exactly E=mc^2.
No more, no less.

It sounds like you are contradicting yourself. Here you seem to agree that black holes do not add any net mass/energy to the Universe over time (which is the correct viewpoint), but here...

However, as gravity force/energy is for free, than also this Et is for free!!!

...you seem to be claiming that black holes do increase the total mass/energy of Universe over time (which is the wrong viewpoint).

So which is it?

[Malamute Lover]

It seems that our spacetime does not allow negative energy.  If that is indeed the case, then the Hawking radiation process is in trouble.

Yes, I fully agree with you.
I will explain it later on.
However, concerning the kinetic and potential energy:

If I am reading this right, you are keeping all the kinetic energy on the way up the gravitational well but still adding potential energy. Not legal. One feeds the other at its own expense.

Well, why do you claim for: " kinetic energy on the way up"
I ONLY focus on the Ek and Et at the single moment of creation.
So, let's assume that the virtual particle had been converted to real particle at radius r.
Hence, before it starts its "way up" activity, do you confirm that its Ek + Ep was given by the gravity force energy as stated by Hawking?

You specifically have the particle ejected from near the event horizon where it is created into the lowest photon orbital level which is well above the event horizon by half the radius equivalent of the black hole. This is no longer the moment of creation. Kinetic energy is lost on the way up, while potential energy is increased.

Keep in mind that kinetic energy and potential energy are relative to the frame of reference. Throw a stone at someone and it hurts when it hits. Kinetic energy is expressed. But if you happen to be on a train going exactly as fast as the stone is thrown and you miss the target and the stone goes out the back door. What happens? It drops to the ground in a straight line as seen from an observer on the ground and the only energy expressed is from the fall. What happened to the kinetic energy? Likewise, potential energy is only meaningful if you specify how far the object will drop before it hits something. If it never hits something the potential is unrealized and there is no kinetic energy expressed.

A particle coming from the close neighborhood of an event horizon does not have a definable kinetic energy unless you say what it is going to hit. It would only have definable potential energy if it were to drop and hit something. It is legitimate to talk about changes in kinetic energy and potential energy as a particle goes up or down in a gravitational field. But you cannot assign a specific value unless you provide a frame of reference.

[Dave Lev (OP)]

You specifically have the particle ejected from near the event horizon where it is created...

Please stop at this moment.
You claim - "You specifically have the particle ejected from near the event horizon where it is created"
So, we have a new created particle near the event horizon.
Please, before you think about its wish to move upwards to the lowest photon orbital level:

into the lowest photon orbital level

This particle was boosted by the gravity energy in order to becoming real particle as stated by Hawking:
"being "boosted" by the black hole's gravitation into becoming real particles"
So, it gets its starting orbital velocity (V0 at T=0, moment of becoming real particle) by gravity force/energy.
Why is it so difficult to all of you to understand this clear explanation by Hawking?

Quote from: Dave Lev on Yesterday at 20:05:30
Therefore, As the mass-energy in the ejected particle is E=mc^2, than the BH is losing exactly E=mc^2.
No more, no less.

It sounds like you are contradicting yourself. Here you seem to agree that black holes do not add any net mass/energy to the Universe over time (which is the correct viewpoint), but here...

No, I don't
I claim that due to the conservation of energy, the mass-energy (E=mc^2) in the new created particle MUST come from the energy of the BH itself.
However - that is the only energy that is taken from the BH for the created new particle as based on Hawking, the velocity V0 (or boosted particle) and the radius of that new created particale at the moment of creation is clearly contributed by the gravity force/energy.
This is not based on my imagination. It is clearly based on Hawking explanation.
In the following message:
"When particles escape, the black hole loses a small amount of its energy and therefore some of its mass (mass and energy are related by Einstein's equation E = mc2)."
Hawking tells us that it is Particle (not photon). If he wanted to discuss about a photon, he could claim for photon.
This particle has a mass = m
Therefore, he specifically claims for the mass-energy in that new created particle mass (m)
Therefore, he also give us the formula for the mass-energy in that particle which is equal to Em=mc^2.
Hence, Hawking tells us the the energy that the BH is losing (let's call it E(BH energy lost) is equal to mc^2.
Therefore, there is no any request from the BH to "pay" extra energy for the velocity and location at the moment of creation.
On the contrary.
It is specifically claim that its velocity (or boosted) is due to gravity force/energy.
You have already confirmed that message by Hawking:

Quote from: Dave Lev on Yesterday at 15:59:05
Are you sure that the following statement by Hawking is incorrect:
1.  It "is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles"
2. "As the particle–antiparticle pair was produced by the black hole's gravitational energy"

They aren't incorrect.

However, some how you claim that the velocity due to that boosted activity (or actually its Kinetic energy at the moment of creation) which is based on V0 (Velocity at the moment of creation, T=0) and its potential energy Ep0 must also come from the BH energy itself
So, even as Hawking had stated that:
E (BH Energy Lost) = Em = mc^2
You claim that:
E (BH energy lost) = Em + Ek + Ep = Em + m c^2 / 2 + 3 M m G^2 /c^2

All of that fully contradicts the explanation by Hawking.
However, you insist to ignore the real meaning of "gravity energy" and the meaning of "being "boosted" by the black hole's gravitation into becoming real particles", while you clearly accept the explanation of hawking that:
It "is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles" and "As the particle–antiparticle pair was produced by the black hole's gravitational energy"

In the same token, as you also claim that:
E (BH energy lost) = Em + Ek + Ep = Em + m c^2 / 2 + 3 M m G^2 /c^2
While hawking tells us that
E (BH energy lost) = Em

Don't you see your clear contradiction with Hawking message/explanation?
So, what can I say?

[Kryptid]

Hawking tells us that it is Particle (not photon). If he wanted to discuss about a photon, he could claim for photon.

You know that photons are a kind of particle, don't you?

In the same token, as you also claim that:
E (BH energy lost) = Em + Ek + Ep = Em + m c^2 / 2 + 3 M m G^2 /c^2

This is the only way to satisfy conservation of energy. The total mass/energy in the Universe after the emission of the particle must be identical to the total mass/energy in the Universe before the emission of the particle. Anything else violates conservation of energy.

For the sake of argument, let's say the entire Universe had 10 GeV of mass/energy in it. If that number ever changed, then conservation of mass/energy would be violated. If it became 11 GeV, then conservation would be violated. If it became 9 GeV, conservation would be violated. Any number at all other than 10 GeV means violation.

While hawking tells us that
E (BH energy lost) = Em

Hawking never said that the emitted particles have more mass/energy than the black hole lost. If he did, then he would be claiming a violation of conservation of energy and would never have been taken seriously. But that's not what he claimed so that's not a problem.

[Bored chemist]

This particle was boosted by the gravity energy in order to becoming real particle as stated by Hawking:
"being "boosted" by the black hole's gravitation into becoming real particles"
So, it gets its starting orbital velocity (V0 at T=0, moment of becoming real particle) by gravity force/energy.
Why is it so difficult to all of you to understand this clear explanation by Hawking?

I see.
The problem is that you have misunderstood what he meant by "boosted".
In this case it doesn't mean a change in velocity (or energy) but a change in status.
It is boosted from being virtual to being real.

Why is it so difficult to all of you to understand this clear explanation by Hawking?

I don't know.
Perhaps you can tell us why you didn't understand it?
My guess would be that English isn't your first language and you missed that meaning.

This is not based on my imagination. It is clearly based on Hawking explanation.

It seems to be based on what you imagined Hawking meant.

[Malamute Lover]

You specifically have the particle ejected from near the event horizon where it is created...

Please stop at this moment.
You claim - "You specifically have the particle ejected from near the event horizon where it is created"
So, we have a new created particle near the event horizon.
Please, before you think about its wish to move upwards to the lowest photon orbital level:

Earlier you said:

The event horizon radius is:
https://steemit.com/space/@getonthetrain/can-light-orbit-a-black-hole
Rh = 2 M G / c^2
While the photon sphere radius is
Rf = 3/2 Rh =  3 M G / c^2
As the new created particle is ejected into the photon sphere, than its potential energy should be:
Ep = m G Rf = m G  3 M G / c^2 = 3 M m G^2 /c^2

You have the particle going from near the event horizon to the photon sphere. If the particle is not moving, you cannot speak of its kinetic energy. If it does not change its level in the gravitational gradient, you cannot speak of its potential energy.  This is what I was addressing.

Aside: While we are at it, it should be noted that only a Schwarzschild (non-rotating) black hole can have a photon sphere, in which a photon can have a circular orbit with any orientation. Real black holes would almost certainly be Kerr (rotating) black holes, since angular momentum of the components would never exactly cancel out and the spin would be amplified as the collapse took place. In a Kerr black hole, there are only 8 possible orbital paths of which only 2 are circular.

Not that any of that matters. Just being persnickety. What matters is that the particle moves upward away from the event horizon. More on that below.

into the lowest photon orbital level

This particle was boosted by the gravity energy in order to becoming real particle as stated by Hawking:
"being "boosted" by the black hole's gravitation into becoming real particles"
So, it gets its starting orbital velocity (V0 at T=0, moment of becoming real particle) by gravity force/energy.
Why is it so difficult to all of you to understand this clear explanation by Hawking?

Hawking does not use the term ‘boosted’. That term in the Hawking Radiation Wiki article has no supporting citation.  However it is clear that the ‘boosting’ refers to gaining real energy and becoming real and not to motion.

Hawking’s arguments are much denser and complex and not at all clear. These arguments involve spacetime curvatures in the neighborhood of the Planck length, such as could happen near the event horizon.

The virtual particle pairs that appear in the region of the event horizon can be split by the negative energy particle tunneling through the event horizon because it is within easy indeterminacy reach. The mechanism involves Killing vectors (named after someone named Killing) on the 4D spacetime manifold. The Killing vector is the effective direction of time at any given point on the manifold. At the event horizon spacetime has been turned around so much that Killing vector is spacelike instead of timelike. Going forward in time is the same as going down. There is no turning back.  The way back is in the past.

The upshot is that the negative energy entering the black hole via tunneling allows the positive virtual particle to become real. But this seems to amount to just bookkeeping, a way of accounting for the entropy that Bekenstein predicted. There is no real explanation of how the energy transfer takes place.

Some comments on the reply to Kryptid.

E = mc^2 relates the rest mass of a particle to its energy content. The particle could be motionless and E = mc^2 is still true. Photons has no rest mass but they still have energy in proportion to the frequency. BTW Hawking originally stated that the primary form of black hole radiation would be massless photons and neutrinos. (In the mid-1970s it was not yet realized that neutrinos have mass.)  Having an infinite Compton wavelength, like a photon, would make tunneling easier. 

However, any sort of particles might be generated by this process, including ones with mass. Hawking uses the term ‘emission to infinity’, implying that these particles climb out of the gravity well and escape to safer grounds. Why should they always be in motion and going upward at that? Photons travel at light speed and can always escape as long as they are outside the event horizon and heading the right way. But why should particles with mass necessarily escape? Even if they go the right direction, why should they have enough velocity to always get away? They were created very near the event horizon, and the escape velocity they would have to overcome would then be close to light speed. Anything with a lower speed and/or not exactly the right direction is going to fall toward the black hole. And why would the particle with mass move upward?

How then can ‘emission to infinity’ be justified? The Hawking process has the negative energy particle tunneling through the event horizon. Is this in any way comparable to falling into the black hole and gaining kinetic energy in exchange for potential energy? Remember that this is the negative energy particle, so its kinetic energy would be negative. Is there some mechanism whereby this energy could be balanced by the positive energy particle gaining positive kinetic energy and moving away from the event horizon? We are now in extreme speculation land. Hawking himself does not address this in any meaningful way.

[Malamute Lover]

Vaidya black hole.

Aside: While we are at it, it should be noted that only a Schwarzschild (non-rotating) black hole can have a photon sphere, in which a photon can have a circular orbit with any orientation. Real black holes would almost certainly be Kerr (rotating) black holes, since angular momentum of the components would never exactly cancel out and the spin would be amplified as the collapse took place. In a Kerr black hole, there are only 8 possible orbital paths of which only 2 are circular.

Not that any of that matters. Just being persnickety.

While we're being persnickety then, we're also talking about a Vaidya black hole. The Schwarzschild solution is a non-rotating static (without time evolution) solution, and a Kerr black hole is a rotating solution. The Vaidya solution is a dynamic solution (possibly non-rotating) that involves evaporation, and evaporation is the key feature on which Dave is hoping to pin his free-energy hopes. Not sure if Vaidya includes rotation or if there is another solution that includes both.

The formalism of the Schwarzschild solution introduces singularities (division by zero) at the event horizon. The Vaidya solution avoids this by using a different coordinate system so that the problem does not arise. There are other choices of coordinate systems than the rather exotic and hard to understand Eddington-Finkelstein version Vaidya used. But Thorne et al. used it in Gravitation, which is a very heavy book in more ways than one. It is also black. (1)

The interesting thing about the Vaidya metric is that the event horizon itself becomes describable and interesting properties emerge. There are rotating Kerr type solutions as well.

An absorbing Vaidya black hole is really no different from a conventional black hole. If I understand correctly, and I am not sure about that, a Vaidya blackhole in our universe with all sorts of stuff around to fall in (including the CMB), would typically absorb more than emit. In a large empty region of space, emission would presumably exceed absorption at least in lower mass entities.

The output of an emitting Vaidya black hole is called ‘null dust’ which is massless something or other. Photons or gravitons would work but EM radiation does not. Neither does anything with mass. What might be left after a Vaidya black hole evaporates? A naked singularity? A wormhole?

We need quantum gravity.


(1) It was reading a passage in Gravitation about the relativistic interval, the invariant spacetime distance between two events regardless of observer states, not having a physical interpretation that set me on the road to questioning some aspects of textbook physics.

[Dave Lev (OP)]

But why should particles with mass necessarily escape? Even if they go the right direction, why should they have enough velocity to always get away? They were created very near the event horizon, and the escape velocity they would have to overcome would then be close to light speed. Anything with a lower speed and/or not exactly the right direction is going to fall toward the black hole. And why would the particle with mass move upward?

Thanks for your great explanation.
So, in order for a new created particle (with real mass) to escape from near the event horizon, it must run for its life.
In other words, any new created particle pair must come with an orbital velocity which is almost as high as the speed of light.
If the orbital velocity will be too low, this pair will immediately fall down into the BH.
So, if the new created particles pair (positive and negative) are too lazy, and orbit at too low velocity, both of them must fall in and nothing could escape from the mighty BH gravity force.
Therefore, it is clear that at the moment of creation, the two positive and negative particles MUST be located near each other and Must also orbit at Ultra high velocity.
That activity must take place before one is move outwards and the other one moves inwards.

You have the particle going from near the event horizon to the photon sphere. If the particle is not moving, you cannot speak of its kinetic energy. If it does not change its level in the gravitational gradient, you cannot speak of its potential energy.  This is what I was addressing.

So, I claim that a new created particles pair without a velocity must fall in (both). Therefore, there is no meaning to discuss about new created particles without velocity.
Hence, at the same moment of creation, they must orbit at the same ultra velocity. Just after that one of them can move upwards, while the other one can move downwards.
Therefore, at the moment of creation they MUST have velocity even before one of them is moving outwards!!!
 

Quote from: Dave Lev on Yesterday at 04:11:02
In the same token, as you also claim that:
E (BH energy lost) = Em + Ek + Ep = Em + m c^2 / 2 + 3 M m G^2 /c^2

This is the only way to satisfy conservation of energy. The total mass/energy in the Universe after the emission of the particle must be identical to the total mass/energy in the Universe before the emission of the particle. Anything else violates conservation of energy.

So let me assume for one moment that your assumption is correct.
The negative particle has the following total energy:
E = Em + Ek + Ep = Em + m c^2 / 2 + 3 M m G^2 /c^2
However, how the BH is going to PAY for this extra Kinetic and potential energies?
Hawking only asked to pay by the negative mass which is Em = mc^2 as evaporation of mass from the BH.
So, are you sure that as the negative particle falls in with ultra high velocity, that velocity should increase the mass evaporation from the BH?
So let's assume that the negative particle is located at radius R1 and has a velocity of V1 during the moment of creation.
So, this particle must fall in. Let's assume that it hit the BH mass at R2 while its velocity is V2.
How the kinetic energy on the impact moment (due to V2) could be transformed into evaporation of more mass from the BH?

[Kryptid]

However, how the BH is going to PAY for this extra Kinetic and potential energies?

There is nothing to pay because the particle has a negative mass/energy equal in magnitude to the positive mass/energy of the particle that escapes. If the particle that escapes has a mass/energy of 10 MeV, then the infalling particle will have a mass/energy of -10 MeV. 10 plus -10 equals zero.

If the black hole has a mass/energy of, say, 1,000 MeV, then that negative energy particle will reduce the black hole's mass/energy to 990 MeV, because 1,000 MeV + (-10 MeV) = 990 MeV.

[Dave Lev (OP)]

There is nothing to pay because the particle has a negative mass/energy equal in magnitude to the positive mass/energy of the particle that escapes. If the particle that escapes has a mass/energy of 10 MeV, then the infalling particle will have a mass/energy of -10 MeV. 10 plus -10 equals zero.

Well, the total energy of the escape particle is:
Ee (Total Escape energy) = Em + Et = Em + Ek + Ep = 10 MeV
Now, just for the calculation, let's assume that:
Em = mc^2 =6 Mev
While
Et = 10Mev - Em = 10Mev - 6Mev = 4Mev

As you have explained due to the energy conservation, the antiparticle must have a negative identical energy

Therefore, based on this assumption, 
Ef (total falling particle energy) = - Ee (Total Escape energy) = - Em - Et = 10 MeV
Hence, the negative mass energy for the falling particle is,
Emf = -Em = - mc^2 = - 6mev
While, the energy due to the velocity of the falling particale is
Etf = -Et = -4Mev.

So, the negative mass of the falling energy must decrease the total mass of the BH by 6Mev
However, how the BH is going to lose any sort of mass due to the negative Etf energy (kinetic  + potential)?

If the black hole has a mass/energy of, say, 1,000 MeV, then that negative energy particle will reduce the black hole's mass/energy to 990 MeV, because 1,000 MeV + (-10 MeV) = 990 MeV.

Sorry, that calculation is incorrect:
I can accept the idea that the negative mass by itself should decrease the total mass of the BH.
Therefore,
1,000 MeV + (-6 MeV) = 994 MeV
However, how the total kinetic + potential energy of the falling particle which is -4Mev can reduce any sort of mass from the BH.
Can you show any formula that a velocity of falling negative particle could decrease a mass from the main object?
Hawking does not claim for it.
As I think about it, it seems to me that a falling in particle (or even antiparticle) must have a positive kinetic energy as it falls at high velocity.
Therefore,, the falling antiparticle should increase the black hole's temperature due to the collision with the BH at ultra high velocity.
Surprisingly, I have found that hawking fully agrees with me:
https://en.wikipedia.org/wiki/Hawking_radiation
"Unlike most objects, a black hole's temperature increases as it radiates away mass."
Actually, that message by Hawking shows that the system/BH gets extra energy due to the mass creation process. That by itself is a key contradiction with the conservation law.
In any case, even if you claim that this heat is negative (which is absolutely none realistic) at the most you can claim that this negative heat/energy decreases the black hole's temperature.
Therefore why are you so sure that the falling antiparticle must decrease extra mass from the BH just due to its falling in velocity?

[Kryptid]

If the total mass/energy (including its kinetic and potential energy) of the particle that falls in is added to the total mass/energy of the particle that escapes (including its kinetic and potential energy) and the result is any number other than zero, then conservation of mass/energy has been violated.

Sorry, that calculation is incorrect:

Then get a calculator, because 1,000 - 10 is, indeed, 990.

As I think about it, it seems to me that a falling in particle (or even antiparticle) must have a positive kinetic energy as it falls at high velocity.

The kinetic energy equation is E_k = 0.5mv², so if the mass (m) is negative, then the kinetic energy (E_k) must also be negative.

However, how the total kinetic + potential energy of the falling particle which is -4Mev can reduce any sort of mass from the BH.

It's called arithmetic.

[Bored chemist]

"Unlike most objects, a black hole's temperature increases as it radiates away mass."
Actually, that message by Hawking shows that the system/BH gets extra energy due to the mass creation process. That by itself is a key contradiction with the conservation law.

No, it doesn't.
The BH ends up hotter, but smaller.

[Malamute Lover]

But why should particles with mass necessarily escape? Even if they go the right direction, why should they have enough velocity to always get away? They were created very near the event horizon, and the escape velocity they would have to overcome would then be close to light speed. Anything with a lower speed and/or not exactly the right direction is going to fall toward the black hole. And why would the particle with mass move upward?

Thanks for your great explanation.
So, in order for a new created particle (with real mass) to escape from near the event horizon, it must run for its life.
In other words, any new created particle pair must come with an orbital velocity which is almost as high as the speed of light.
If the orbital velocity will be too low, this pair will immediately fall down into the BH.
So, if the new created particles pair (positive and negative) are too lazy, and orbit at too low velocity, both of them must fall in and nothing could escape from the mighty BH gravity force.
Therefore, it is clear that at the moment of creation, the two positive and negative particles MUST be located near each other and Must also orbit at Ultra high velocity.
That activity must take place before one is move outwards and the other one moves inwards.

Not orbital speed, but vertical speed. Orbits are not possible near the event horizon. It is too deep in the gravity well. Anyway, going round and round would not match Hawking’s ‘emission to infinity’.

But yes, the vertical speed would need to be almost light speed. A photon could do that, necessarily traveling at light speed. That would mean that the matching negative energy photon would go straight down. Since in Hawking’s model, the photon or whatever is tunneling through the event horizon, where the only possible direction is down, that is not an unreasonable thought. This would presumably apply to particles with mass as well, although I am not yet clear in my mind how the speed, i.e., kinetic energy, issue could be resolved.

Again, it is not just the bookkeeping aspect I am interested in, but the detailed mechanism.

You have the particle going from near the event horizon to the photon sphere. If the particle is not moving, you cannot speak of its kinetic energy. If it does not change its level in the gravitational gradient, you cannot speak of its potential energy.  This is what I was addressing.

So, I claim that a new created particles pair without a velocity must fall in (both). Therefore, there is no meaning to discuss about new created particles without velocity.
Hence, at the same moment of creation, they must orbit at the same ultra velocity. Just after that one of them can move upwards, while the other one can move downwards.
Therefore, at the moment of creation they MUST have velocity even before one of them is moving outwards!!![/quote]

Not necessarily have the correct velocity (speed and direction). The negative virtual particle is crossing the event horizon where down is the only direction. That would take care of the direction aspect. Since the gravitational gradient is so steep near the event horizon, even when one is near the event horizon, it is still a long way down in terms of kinetic energy gain. Perhaps a negative mass-energy particle could pick up enough negative kinetic energy falling down to account for enough positive kinetic energy in the positive mass energy particle to get it clear of the black hole neighborhood and so ‘emit to infinity’ as Hawking wants.

But again what is the mechanism involved that could move this energy around?

Just demonstrating that the energy balance sheets would come out right would take more math than I ever want to do again in my life. This is about as non-linear as GR gets. And there is quantum tunneling in the mix. I think I will just stick with armchair pipe dreaming.

[Dave Lev (OP)]

The kinetic energy equation is Ek = 0.5mv2, so if the mass (m) is negative, then the kinetic energy (Ek) must also be negative.

That is correct
However, how negative kinetic energy could evaporate any sort of mass from the BH?

The BH ends up hotter, but smaller.

Even if we assume that all the kinetic and potential energies of the negative particle are used to evaporate extra mass from the BH, how that BH ends up hotter? If it get hotter, than new energy is coming in. Don't you agree that this activity contradicts the conservation law.

Not orbital speed, but vertical speed. Orbits are not possible near the event horizon. It is too deep in the gravity well. Anyway, going round and round would not match Hawking’s ‘emission to infinity’.

Are you sure about it?
Vertical velocity is like a rocket moving upwards.

But yes, the vertical speed would need to be almost light speed. A photon could do that, necessarily traveling at light speed.

If a photon is boosted vertically upwards like a rocket from the event horizon at the speed of light c, it's energy would be so high that with all the good willing, it must be kicked out from the gravity impact of the BH.
Therefore, it won't be able to set even one orbital cycle around the BH.
However, we know that around a BH there is a photon sphere. In this sphere, photons orbit around the BH. In order to get there, they must orbit around the BH at the moment of their creation.
Therefore, this theory of vertical velocity is not applicable.
However, why do you highlight a photon?
We focus on a particle with real mass.
Do you see a possibility for vertical c velocity also with real mass particle?

The negative virtual particle is crossing the event horizon where down is the only direction. That would take care of the direction aspect. Since the gravitational gradient is so steep near the event horizon, even when one is near the event horizon, it is still a long way down in terms of kinetic energy gain. Perhaps a negative mass-energy particle could pick up enough negative kinetic energy falling down to account for enough positive kinetic energy in the positive mass energy particle to get it clear of the black hole neighborhood and so ‘emit to infinity’ as Hawking wants.

Why are you so sure that ONLY and ALWAYS the negative mass will fall in and the other one will be ejected outwards?

But again what is the mechanism involved that could move this energy around?

What kind of mechanism would prevent from the negative mass to be ejected outwards, while the positive mass to fall inwards.
Please remember, based on Hawking, first we get the two particles (negative and positive) and then the BH pays after the negative particle falls in.
So, could it be that the process works as follow:
The BH asks a permission from the conservation law to create two particles without any investment of extra energy.
The BH promise to the conservation law that after the creation it will pay back by losing some of its mass.
Therefore, Two particles are created without any need for external energy. One negative and one positive.
However, after the creation, without a clear mechanism to force the negative to all in, the positive activity could take place as the positive will fall in and the negative will be ejected outwards.
In this case, the BH will gain mass, while the Universe will lose mass.
Do you agree that this is not realistic?


With regards to Negative mass
Are you sure that this idea is real? Our scientists have never found any sort of negative mass in our universe. Therefore, could it be that the idea of negative mass is a pure fiction?
Hawking claims that in order to keep the conservation law, Positive + Negative particles must be created almost out of nothing and without any investment of real energy.
He claims that there is no violation of conservation law as those two particles carry exactly the same opposite energies.
However, the real payment for this activity will take place later on when the negative particle will hit the BH.
It sounds to me as a loan from the Universe.
You go to the bank, get 1000$ in cash (real particale) and also a bill of -1000$ (negative particale). You use the money and pay later on your bill.
Another example - You are going in the desert. There is no water. So, you set an agreement with the nature to get real water and antiwater. You drink the water and keep the antiwater. Later on when you get to the lake you will return the bill. Is it real?
In the same token, the BH takes a loan from the Universe.
It creates real particle (postitive) and get a bill in form of Antiparticle (negative).
The payment will be set after the collision between the negative particle with the BH.
Are you sure that the universe would accept this kind of agreement?

Sorry, our Universe is not a banking system. If something is created in our Universe, the payment in energy must come in advance and in cash.
Nothing would be created without payment in real energy in ADVANCE!
So, don't you agree that new particles/photons could be created just after a transformation of real energy (in advance) from the BH to that creation process (whatever  it is)?
Therefore, if you believe in BH mass evaporation process, than first the BH must evaporate some of its mass-energy and just then it can hope to get real particle/photon.

[Kryptid]

However, how negative kinetic energy could evaporate any sort of mass from the BH?

Because the negative mass necessarily gets absorbed by the black hole. Adding a negative number to a positive number makes the positive number smaller.

If it get hotter, than new energy is coming in.

No it isn't. All of that heat energy results in an equal amount of mass/energy being lost from the black hole. If the black hole radiates 10 MeV of heat energy, then it loses 10 MeV of mass.

Why are you so sure that ONLY and ALWAYS the negative mass will fall in and the other one will be ejected outwards?

The internal environment of the black hole is what allows that particle to have negative mass to begin with. Space-time is extremely distorted there.

Therefore, could it be that the idea of negative mass is a pure fiction?

If it is, then black holes can't produce Hawking radiation and thus cannot radiate particles. The existence of negative mass is required in order to keep the conservation of energy. The black hole can't radiate positive mass without losing an equal amount of mass itself.

[Malamute Lover]

Even if we assume that all the kinetic and potential energies of the negative particle are used to evaporate extra mass from the BH, how that BH ends up hotter? If it get hotter, than new energy is coming in. Don't you agree that this activity contradicts the conservation law.

The magnitude of the Hawking radiation effect is related to the gravitational gradient, the change in gravitational strength in the vertical direction. The small the black hole, the steeper the gravitational gradient. Small black holes radiate faster, making them hotter. As a result they shrink faster. Energy is conserved.

Not orbital speed, but vertical speed. Orbits are not possible near the event horizon. It is too deep in the gravity well. Anyway, going round and round would not match Hawking’s ‘emission to infinity’.

Are you sure about it?
Vertical velocity is like a rocket moving upwards.

Exactly. The particle is escaping from the region of the black hole, just like Hawking wants. ‘Emission to infinity’ is his phrase. Anyway, orbits are not possible near the event horizon. Even light, traveling at lightspeed, cannot maintain an orbit below half the radius above the horizon. The gravity well is too steep below that point.

But yes, the vertical speed would need to be almost light speed. A photon could do that, necessarily traveling at light speed.

If a photon is boosted vertically upwards like a rocket from the event horizon at the speed of light c, it's energy would be so high that with all the good willing, it must be kicked out from the gravity impact of the BH.
Therefore, it won't be able to set even one orbital cycle around the BH.
However, we know that around a BH there is a photon sphere. In this sphere, photons orbit around the BH. In order to get there, they must orbit around the BH at the moment of their creation.
Therefore, this theory of vertical velocity is not applicable.
However, why do you highlight a photon?
We focus on a particle with real mass.
Do you see a possibility for vertical c velocity also with real mass particle?

A particle with mass need not travel at lightspeed to escape, just very very close to it. The positive mass particle starts just above the event horizon, not on it. The escape velocity is almost light speed but not quite.

 And as before, orbits are not possible that low.

Hawking originally focused on massless photons because the already hairy math was easier that way and because massless photons would be the most common form of emission. He also mentioned neutrinos because at the time they were thought to be massless. The greater the mass of the real version of a virtual particle, the shorter its lifetime and range and therefore the less likely tunneling through the horizon would happen.

The negative virtual particle is crossing the event horizon where down is the only direction. That would take care of the direction aspect. Since the gravitational gradient is so steep near the event horizon, even when one is near the event horizon, it is still a long way down in terms of kinetic energy gain. Perhaps a negative mass-energy particle could pick up enough negative kinetic energy falling down to account for enough positive kinetic energy in the positive mass energy particle to get it clear of the black hole neighborhood and so ‘emit to infinity’ as Hawking wants.

Why are you so sure that ONLY and ALWAYS the negative mass will fall in and the other one will be ejected outwards?

I mentioned that earlier as a problem I had with the process. I do not see anything in the math that would mandate that, but it is very dense. One would think that a negative mass entity would go up in a strong gravitational field, making the positive mass partner be on the down side and more likely to be the one to tunnel through the event horizon. Virtual electron pairs do orient themselves in an electric field.

But again what is the mechanism involved that could move this energy around?

What kind of mechanism would prevent from the negative mass to be ejected outwards, while the positive mass to fall inwards.
Please remember, based on Hawking, first we get the two particles (negative and positive) and then the BH pays after the negative particle falls in.
So, could it be that the process works as follow:
The BH asks a permission from the conservation law to create two particles without any investment of extra energy.
The BH promise to the conservation law that after the creation it will pay back by losing some of its mass.
Therefore, Two particles are created without any need for external energy. One negative and one positive.
However, after the creation, without a clear mechanism to force the negative to all in, the positive activity could take place as the positive will fall in and the negative will be ejected outwards.
In this case, the BH will gain mass, while the Universe will lose mass.
Do you agree that this is not realistic?

There is no such thing as borrowing energy from the conservation law. It is not a bank. Virtual particles pairs can exist because they are balanced in every way even to the point of opposite mass-energy values. They add up to zero.

With regards to Negative mass
Are you sure that this idea is real? Our scientists have never found any sort of negative mass in our universe. Therefore, could it be that the idea of negative mass is a pure fiction?

Virtual particle pairs have one positive mass-energy partner and one negative mass-energy partner. Virtual particles are not ‘on the mass shell’, that is, they do not strictly obey energy-momentum laws. The total of the mass-energy is zero but it is composed of a plus and a minus not in the same exact place. If the total mass was not zero but was positive, and since virtual particle pairs are constantly appearing and disappearing, there would be a continuous positive surplus of mass-energy that would overwhelm the universe.

Negative mass-energy exists in virtual particle pairs.

Hawking claims that in order to keep the conservation law, Positive + Negative particles must be created almost out of nothing and without any investment of real energy.
He claims that there is no violation of conservation law as those two particles carry exactly the same opposite energies.
However, the real payment for this activity will take place later on when the negative particle will hit the BH.
It sounds to me as a loan from the Universe.
You go to the bank, get 1000$ in cash (real particale) and also a bill of -1000$ (negative particale). You use the money and pay later on your bill.
Another example - You are going in the desert. There is no water. So, you set an agreement with the nature to get real water and antiwater. You drink the water and keep the antiwater. Later on when you get to the lake you will return the bill. Is it real?
In the same token, the BH takes a loan from the Universe.
It creates real particle (postitive) and get a bill in form of Antiparticle (negative).
The payment will be set after the collision between the negative particle with the BH.
Are you sure that the universe would accept this kind of agreement?

Sorry, our Universe is not a banking system. If something is created in our Universe, the payment in energy must come in advance and in cash.
Nothing would be created without payment in real energy in ADVANCE!
So, don't you agree that new particles/photons could be created just after a transformation of real energy (in advance) from the BH to that creation process (whatever  it is)?
Therefore, if you believe in BH mass evaporation process, than first the BH must evaporate some of its mass-energy and just then it can hope to get real particle/photon.

I have also mentioned the time order issue earlier. I do not have an answer. If we ever get a viable quantum gravity theory, it might make sense. Especially if that involves a new concept of spacetime, which looks like it may very well be the case.

Again, my own solution to the whole ‘black holes eat entropy requiring black hole emission’ problem is that from the viewpoint of a distant observer, nothing ever reaches the event horizon because of the extreme time dilation. Everything that went in the direction of the black hole is still this side of it, entropy and all.

[Dave Lev (OP)]

1. Negative mass-energy particle

The internal environment of the black hole is what allows that particle to have negative mass to begin with. Space-time is extremely distorted there.

Virtual particles pairs can exist because they are balanced in every way even to the point of opposite mass-energy values. They add up to zero.

Outside the surface of the BH it is all about gravity (At least based on Hawking concept).
So, how the internal environment of the black hole can generate at the event of horizon (high above its surface) a negative mass particle?
Virtual particles can't be considered as real particles.
So, does real Negative mass-energy particle exist in our Universe?
The idea that you need it for your theory does not necessarily convert it to reality.
If you believe in negative mass, than you have to find it.
Have you ever found any sort of negative mass (or its impact as negative gravity) around a BH or elsewhere?

The existence of negative mass is required in order to keep the conservation of energy. The black hole can't radiate positive mass without losing an equal amount of mass itself.

If based on our modules it is there then why we have never ever found it?
Could it be that it is only exist in our wishful list?
How can you protect your theory without real observation of Negative mass particle?
As there is no Negative mass-energy particle in our universe, then don't you agree that this theory is none relevant?
Why are you so sure that a black hole can't radiate positive mass without losing an equal amount of mass itself?
Do you claim that there is only one theory for that radiation?
Sorry, if you need a negative mass for your theory, you must observe it.
If you can't observe it, then it's the time to change disc and look for better theory.

2. Perfect creation particles system
How do we know for sure that the total energy in the new created particles is absolutely zero?
If one particle is located even one micro of a Pico mm to the left with regards to the other one, than already its energy is different from the other particle.
In any real system, there is no perfect match. Even a very small difference of 10^1000 is still a difference.
So, how can you believe is such a perfect creation system?


3. Negative kinetic energy
You claim that the negative kinetic energy of the falling negative mass particle into the BH, can evaporate extra mass from the BH.
I can accept the idea that a negative mass particle can evaporate positive mass particle.
However, how its negative kinetic energy can evaporate any sort of mass?
Can you please prove it?
How could it be that a negative mass particle that is falling at velocity V can evaporate more mass than the same particle falling at a velocity V-v.?
Please be aware that the surface of the BH might not be so smooth.
Therefore, the same falling particle could collide with the surface of the BH at different H.
Hence, its velocity during the impact could be changed based on the location of the collision point.
Therefore, the same particle could collide with the surface of the BH at a different velocity (or kinetic energy).
This by itself sets a key violation in energy conservation - as you see it.

4. Gravity Energy
Hawking is specifically discuss on Gravity Energy. Not gravity force but gravity energy:
https://en.wikipedia.org/wiki/Hawking_radiation
"As the particle–antiparticle pair was produced by the black hole's gravitational energy"
If the gravity force is not energy than why he specifically highlights that the pair was produced by that gravitational energy?
So, why he is using the word energy instead of force for the key phase of pair production process?
Could it be that he knew that energy is needed to produce anything in our universe including new particle–antiparticle pair?
Therefore, why can't we understand that without investing energy, there will be no particle–antiparticle pair?
As he had no idea for the source of energy, he had selected the available gravity force as the source of the requested energy in that process.
If you claim that the gravity force has no energy, than how the universe could create this pair without investing energy?

5. Banking system Energy

Sorry, our Universe is not a banking system. If something is created in our Universe, the payment in energy must come in advance and in cash.
Nothing would be created without payment in real energy in ADVANCE!
So, don't you agree that new particles/photons could be created just after a transformation of real energy (in advance) from the BH to that creation process (whatever it is)?
Therefore, if you believe in BH mass evaporation process, than first the BH must evaporate some of its mass-energy and just then it can hope to get real particle/photon.

I have also mentioned the time order issue earlier. I do not have an answer. If we ever get a viable quantum gravity theory, it might make sense. Especially if that involves a new concept of spacetime, which looks like it may very well be the case.

Thanks
So, you also see a problem in this creation process.
The question is very simple:
When do we have to pay for in energy for the creation of the pair?
If it is just after the creation of the pair, than this is clearly a banking energy system.
Sorry, nothing in our universe would be created without a payment in energy in advanced.
I assume that even hawking knew it. Therefore, he called it Gravity Energy:
Therefore, in any production system, you must invest energy before you get something - even if you think that its energy is zero.
So, do you agree that without first investment with real energy - nothing could be created?

6. Energy waste/efficiency
In any system, there is no zero waste.
The sun converts Hydrogen to helium by fusion process. In that process we get some of the extra energy waste as heat.
Even in our car the energy efficiency is less than 100%.
Therefore, in any creation/transformation of energies some of the energy must be converted to heat and waste.
Therefore, in order to create a particle and antiparticle (even if we claim that their total energy is zero) the BH must invest more energy than zero.
However, you introduce a theory with 100% efficiency.

So, don't you agree that the idea that a BH creates real particles with total zero, with investment of zero energy, at zero waste of energy should be zero.
How can we accept/believe in such hypothetical idea?

[Kryptid]

How can you protect your theory without real observation of Negative mass particle?

It's in the math. I don't understand that level of math myself, but those who do seem to have no problem accepting it as plausible. If there was an obvious error, Hawking would have been called out on it and it wouldn't have become anywhere near as widely-accepted as it has been.

Please don't call it "my" theory. It belongs to Hawking. It just happens to be a widely-accepted mechanism for black hole evaporation in the physics community by those who do understand the math behind it. Could it be wrong? Yes. No one has observed Hawking radiation around a real black hole. They are too far away for us to do that.

As there is no Negative mass-energy particle in our universe

Now you are making a claim that it doesn't exist. So you now you have to demonstrate that it doesn't. "We have never seen any" is not enough of an argument to show that it doesn't exist.

The sun converts Hydrogen to helium by fusion process. In that process we get some of the extra energy waste as heat.
Even in our car the energy efficiency is less than 100%.
Therefore, in any creation/transformation of energies some of the energy must be converted to heat and waste.

When you add up all of the different forms of energy after the reactions (including heat), it equals the amount of energy before the reactions. That is required by the first law of thermodynamics.

Therefore, in any creation/transformation of energies some of the energy must be converted to heat and waste.
Therefore, in order to create a particle and antiparticle (even if we claim that their total energy is zero) the BH must invest more energy than zero.
However, you introduce a theory with 100% efficiency.

Matter-antimatter annihilation is 100% efficient too, so we know that such a thing can exist. Too bad you didn't think of that beforehand.

[Bored chemist]

So, how the internal environment of the black hole can generate at the event of horizon (high above its surface) a negative mass particle?

It doesn't need to.
Virtual particles pop in and out of existence anyway.