[Bored chemist]

This also does not break the laws of physics.
https://www.facebook.com/reel/324860803322438

[Dave Lev (OP)]

Ultimately, your model is going to depend on the energy coming from the accretion disk as well.

Dear kryptid
It seems that I was not clear enough.
In this thread I would like to understand the current modeling for the quasar energy
My model is absolutely irrelevant to this discussion. I do not wish to discuss about my personal modeling. If you insist to do so, then please reopen the locked thread and I would be more than happy to answer any relevant question.

I really want to thank you all for you excellent support.
After all the long discussion, please let me know if I understand correctly the current modeling theory for the quasar energy:
The basic Idea in this modeling is that the gravitational potential energy of the falling particles is transformed into kinetic energy while they fall in the direction of the quasar SMBH. This kinetic energy should give the particles which falls into the accretion disc most of their energy.
If some energy is missing in this process, then there are some options to add the missing the kinetic energy.
We will discuss it later on.

Let's first discover the basic energies:
The gravitational potential energy of a proton at two light-days from the MW black hole is: U = -3.71 x 10-16 joules.

The mass of the black hole is 886 million times that of the Sun. The Sun's mass is 1.9885 x 1026 kilograms. That makes the black hole's mass 1.722 x 1035 kilograms. This means that the gravitational potential energy of a proton at two light-days from the black hole would be:
U = (-GMm)/r
U = (-(-6.674 x 10-11)(1.722 x 1035)(1.673 x 10-27))/(5.18 x 1013)
U = (-(1.149 x 1025)(1.673 x 10-27))/(5.18 x 1013)
U = (-(1.9227 x 10-2)/(5.18 x 1013)
U = -3.71 x 10-16 joules

The estimated gravitational potential energy for a falling proton from a distance of 2-light days at the 3C 273 quasar is -8.37 x 10^-14 joules

The galaxy that contains 3C 273 has a mass of about 2 x 1011 solar masses. This is about 225.7 times the mass of the central black hole there....
So we just multiply the original numbers by 225.7: -3.71 x 10-16 joules x 225.7 = -8.37 x 10-14 joules....

The kinetic energy of a proton that orbits at almost the speed of light at the quasar accretion disc is approximately 9.087 x 10-10 joules.

Now how much kinetic energy does a proton traveling at 99% the speed of light have? Using this calculator: https://www.omnicalculator.com/physics/relativistic-ke The answer is approximately 9.087 x 10-10 joules.

Therefore, each falling proton from 2-light days with gravitational energy of -8.37 x 10-14 joules must increase its total energy by about 10,000 in order to achieve the orbital kinetic energy of 9.087 x 10-10 joules in the accretion disc?
9.087 x 10-10 / 8.37 x 10-14 = 1.08 * 10^4 = about 10,000

How the kinetic energy of falling proton could be increased by 10,000 without external energy source:

There are three options:
A. Transfer kinetic energy from one particle into the other particle:

The problem is that you are dealing with a system of many such particles. Let's consider two particles, as an example. If I take the kinetic energy out of one particle and put it into the other particle, then I can make the boosted particle travel faster than either of them were moving at the start without violating conservation of energy. All I've done is change the distribution of energy.
In the end, I get one particle going faster at the cost of making the other one go slower.

This also does not break the laws of physics.
https://www.facebook.com/reel/324860803322438

This is acceptable mechanism which could increase dramatically the kinetic energy of a falling proton.
Technically, we could claim that based on this mechanism, if 10,000 falling protons would transfer all their kinetic energy to one, then one proton can get the requested kinetic energy at the accretion disc.
Hence, out of 10,000 falling protons, 9,999 should be "eaten" by the SMBH and only one would be saved at the accretion disc.
However, the observation contradicts this assumption as only 1% of all the falling protons really drawn into the SMBH gravitational field and get consumed:

https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/
astronomers studying Sgr A* (the supermassive black hole at the centre of the Milky Way Galaxy) were surprised to notice that less than 1% of the gas and dust drawn into its gravitational field ever get consumed almost everything else gets ejected.

Therefore, the observation proves that out of 100 falling protons, only one would be drawn into the SMBH gravitational field and get consumed.
This by itself should kill the idea of Transferring kinetic energy.
Never the less, let's assume that out of 100 falling protons, 90 would be consumed by the SMBH and the other 10 would be survive.
Hence, out of 100 falling protons, 10 would increase its kinetic energy by 10. (Let's call them "Lucky Protons").
Therefore, each lucky proton can increase its kinetic energy by 10, from - 8.37 x 10-14 joules to - 8.37 x 10-13 joules, while all the other 90 would be consumed by the SMBH.

B. Fusion:
Fusion can contribute some degree of additional energy to the accretion disk:

I would also expect nuclear fusion to contribute some degree of additional energy to the accretion disk. It's much more than hot enough there for fusion to occur (even if fusion isn't the main source of power).

However, Fusion isn't the main source of power. therefore, let's assume that it can increase the kinetic energy by 100%.
Therefore, this process would increase the kinetic energy of the falling lucky proton from - 8.37 x 10-13 joules to - 1.677 x 10-12 joules.

C. Increasing the kinetic energy of the falling lucky protons by the spinning process of the SMBH that is called "Ergosphere".

Quote from: Dave Lev on 15/09/2023 16:12:41
However, how the SMBH's spin energy could be transformed into those falling particles?

It's something called the ergosphere. Anything falling into it must move with the black hole's rotation: https://en.wikipedia.org/wiki/Ergosphere

Hence, in order for the lucky proton to orbit at almost the speed of light at the accretion disc, it must increase its kinetic energy by about 500, from 1.677 x 10-12 joules to 9.087 x 10-10 joules.
9.087 x 10-10 joules / 1.677 x 10-12 = 5.42 x 10^2 = about 500

After all of that, please answer the following:
1. Drain energy away from the black hole's spin?

Let's say, for the sake of argument, that the black hole does indeed have its own magnetic field generated by its spin. Let's also say that it can interact with the magnetic field generated by the accretion disk, pull particles out of the disk, and launch them into beams. Okay, that's all fine and good. Now, when your magnetic black hole accelerates the material from the accretion disk into beams, it's going to have to transfer energy to the beams. That energy has to come from somewhere. That energy comes from the black hole's spin. This means that your model also predicts that the jets slowly drain energy away from the black hole's spin. That energy has to be replenished. The only thing around to replenish that energy is the accretion disk.

Sorry, I'm not fully sure that I understand correctly the above explanation.
Do you mean that if the lucky protons increase their kinetic energy by 500 times to the speed of light by using the SMBH' spinning motion, then this energy is for free, while if the magnetic fields is created by the same SMBH spin, then this magnetic fields drains energy away from the black hole's spin?
If so, why increasing the proton kinetic energy from the SMBH' spin doesn't drain energy away from the black hole's spin?

2. Do you agree to accept the following observation?

https://www.quantamagazine.org/physicists-identify-the-engine-powering-black-hole-energy-beams-20210520/
"Some 3 trillion trillion trillion joules of energy flow up the jet each second 500 trillion times more energy than the entire human population burns in a decade. How could something so tiny be so powerful?"

If you do, please think about the difficult process of increasing the proton potential starting gravitational energy of -8.37 x 10-14 joules to orbital kinetic energy of 9.087 x 10-10 joules (about 10,000 times).
One falling lucky proton in the accretion disc, adds only 9.087 x 10-10 joules to the disc.
Therefore, in order to add just one joule per second to the disc, one billion lucky protons must join the disc per second.
Based on the request for 3 trillion trillion trillion joules of energy flow up the jet each second, then technically 3 Billion trillion trillion trillion of just lucky particles are need to join the accretion disc per second.
As we already assume that for any lucky proton in the accretion disc, total of 10 particles should fall.
Then, is it realistic to hope that 3 tillion trillion trillion trillion of protons should fall from 2-light days per second in the direction of the quasar SMBH?

[Bored chemist]

If that is correct, then it is a clear indication that the matter in the accretion disc is affected by the quasar' magnetic fields and not vice versa.

If it's affected then that's because a force acts on it.
And Newton pointed out that any force has a reaction force.
So it's impossible for the matter to affect the field without the field affecting the matter, and vice versa.
It can't possibly be a "one or the other" situation.

[Kryptid]

However, the observation contradicts this assumption as only 1% of all the falling protons really drawn into the SMBH gravitational field and get consumed:

That statistic is from the Milky Way's black hole, not a quasar. Quasars have far more material in their accretion disk and consume far more of it.

Do you mean that if the lucky protons increase their kinetic energy by 500 times to the speed of light by using the SMBH' spinning motion, then this energy is for free, while if the magnetic fields is created by the same SMBH spin, then this magnetic fields drains energy away from the black hole's spin?

No, it isn't free in either scenario. Both of those processes cause the black hole to spin down over time if the rotational kinetic energy isn't replenished from an outside source.

If you do, please think about the difficult process of increasing the proton potential starting gravitational energy of -8.37 x 10-14 joules to orbital kinetic energy of 9.087 x 10-10 joules (about 10,000 times).
One falling lucky proton in the accretion disc, adds only 9.087 x 10-10 joules to the disc.
Therefore, in order to add just one joule per second to the disc, one billion lucky protons must join the disc per second.
Based on the request for 3 trillion trillion trillion joules of energy flow up the jet each second, then technically 3 Billion trillion trillion trillion of just lucky particles are need to join the accretion disc per second.
As we already assume that for any lucky proton in the accretion disc, total of 10 particles should fall.
Then, is it realistic to hope that 3 tillion trillion trillion trillion of protons should fall from 2-light days per second in the direction of the quasar SMBH?

I'm going to do some math.

The quasar 3C273 has a luminosity on the order of 2.5 x 10⁴⁰ watts: https://www.mssl.ucl.ac.uk/~gbr/Project%20Website/styled-5/index.html#:~:text=The%20fact%20that%20quasars%20are,is%203.9%20X%201026%20watts).

I'll explore possible energy sources now. If all of the mass being consumed by the black hole was converted completely into energy, then we can use E=mc² to calculate the needed mass flow: https://www.calculatorsoup.com/calculators/physics/emc2.php

The result is about 2.783 x 10²³ kilograms per second. That's about 0.05 Earth masses per second.

Another alternative would be that the energy all comes from gravitational potential energy. I'll calculate how much potential energy 1 kilogram has at the event horizon of the black hole (as a note, I think I miscalculated the mass of the black hole in past equations. I have hopefully corrected that here).

U = (-GMm)/r
U = (-(-6.674 x 10^-11)(1.722 x 10³⁹)(1))/(2.617 x 10¹²)
U = (-(1.149 x 10²⁹)(1))/(2.617 x 10¹²)
U = (-(1.149 x 10²⁹))/(2.617 x 10¹²)
U = -4.39 x 10¹⁶ joules

2.5 x 10⁴⁰ divided by 4.39 x 10¹⁶ equals 5.695 x 10²³ kilograms per second. That's about 0.07 Earth masses per second.

Now for nuclear fusion. Fusion of hydrogen produces about 0.7% of the energy that direct mass-energy conversion does. As such, this means about 143 times as much mass per second than that if fusion provided all the power (which would be 3.98 x 10²⁵ kilograms per second, about 6.675 Earth masses per second).

So the mass flow rate required to explain the power output of 3C273 is somewhere between 0.05 and 6.675 Earth masses per second, depending on the power source. Of course, inefficiencies could make the values higher than that.

[Dave Lev (OP)]

Quote from: Dave Lev on 23/09/2023 14:18:29
However, the observation contradicts this assumption as only 1% of all the falling protons really drawn into the SMBH gravitational field and get consumed:

That statistic is from the Milky Way's black hole, not a quasar. Quasars have far more material in their accretion disk and consume far more of it.

It is very clear that Quasars have far more material in their accretion disk
However, why are you so sure that they aren't picky eater?
In the following article they discuss about quasars that happen to have their jet streams of material pointed toward Earth that are called "blazing quasars."
https://www.wbur.org/npr/507594456/some-bizarre-black-holes-put-on-light-shows
"Isler specializes in the subset of quasars that happen to have their jet streams of material pointed toward Earth. These are called blazars, or "blazing quasars."
"They are billions of times the mass of our own sun," she says. "I like to call them 'hyperactive,' in the sense that they are just taking on a lot more than an average black hole."
"They're actually pretty picky eaters," says Jedidah Isler, an astrophysicist at Vanderbilt University. She spends most days chipping away at one of the universe's biggest mysteries: How do the huge, overactive black holes, known as quasars, work?"
Therefore, quasars are picky eaters!
I would like to remind you that also the MW' SMBH is called Picky eater:
https://www.urban-astronomer.com/news-and-updates/milky-ways-black-hole-a-picky-eater/
"astronomers studying Sgr A* (the supermassive black hole at the centre of the Milky Way Galaxy) were surprised to notice that less than 1% of the gas and dust drawn into its gravitational field ever get consumed ? almost everything else gets ejected. Who knew that a black hole could be such a picky eater!"
Picky eater means - less than 1% of the gas and dust drawn into its gravitational field ever get consumed.
Hence, although those Quasars have far more material in their accretion disk, they are still picky eater and therefore less than 1% of the gas and dust drawn into their gravitational field ever get consumed
In the article it is also stated that those astronomers don't really understand how this huge quasar really works"
"She spends most days chipping away at one of the universe's biggest mysteries: How do the huge, overactive black holes, known as quasars, work?"
"They are able to accelerate particles to 99.99 percent of the speed of light," Isler says. "How does that happen? So, I'm interested in where along that jet do we get this acceleration, and what is the physical mechanism that is responsible for the acceleration of particles that we see?"

Therefore, it is very clear that those astronomers don't know the physical mechanism that is responsible for the acceleration of particles o 99.99 percent of the speed of light that we see.
So, how can you claim that you know the quasar' physical mechanism and how it really works, while those astronomers (which specifically focus on quasars) don't know?

I'll explore possible energy sources now. If all of the mass being consumed by the black hole was converted completely into energy, then we can use E=mc2 to calculate the needed mass flow: https://www.calculatorsoup.com/calculators/physics/emc2.php
The result is about 2.783 x 1023 kilograms per second. That's about 0.05 Earth masses per second.

Please, try to take in your calculation the clear understanding that quasars are picky eater.
From any 100 falling protons in the direction of the Quasar SMBH, only one proton would be consumed, while all the other 99 would fall into the accretion disc and be considered as Lucky protons.
Therefore, somehow, this single proton that was consumed by the quasar SMBH, must explain the activity of all the 99 other lucky protons at the accretion disc.
Do you think that the energy in a single proton can achieve this goal?
Let's set the calculation based on the formula E = m c^2:
https://www.knowledgedoor.com/2/units_and_constants_handbook/proton-mass-energy-equivalent.html
1.503277616?10-10 joules
Are you sure that this single proton energy is good enough to accelerate all the other 99 Lucky protons to 99.99 percent of the speed of light at the accretion disc?

So the mass flow rate required to explain the power output of 3C273 is somewhere between 0.05 and 6.675 Earth masses per second, depending on the power source. Of course, inefficiencies could make the values higher than that.

Would you kindly reconsider your calculation?
If you still think that quasars aren't picky eater, then please prove it by real data/observation

[Kryptid]

If you still think that quasars aren't picky eater, then please prove it by real data/observation

There is another alternative: that these giant black holes start their lives with immense amounts of rotational kinetic energy and slowly lose it to the jets over billions of years. Once it's gone, then it's gone. I did find a source related to that: https://medium.com/amazing-science/rotating-black-holes-the-most-powerful-energy-generators-in-the-universe-832439add442#:~:text=This%20enormous%20amount%20of%20energy,%C3%97%2010%C2%B3%C2%B3%20erg%20per%20second.

Frustratingly, that source requires an account to be made in order to view it. But the Google search that led me there did have the relevant piece of information I was looking for. It says, and I quote:

This enormous amount of energy trapped in the black hole rotation implies a rotational energy of about 10⁶⁴ erg, comparable to the energy emitted by the brightest quasars over billion of years time-scales. Just for comparison, the Sun radiates about 3.8 ? 10³³ erg per second.

So it seems that the black hole may actually have enough energy in its spin to produce the jets for a very long time without needing to be "refueled".

[Dave Lev (OP)]

So it seems that the black hole may actually have enough energy in its spin to produce the jets for a very long time without needing to be "refueled".

Thanks for this important reply.
At last, you offer an idea of using the energy that is stored in the SMBH as the missing energy.
Therefore, I hope that you fully understand that the energy in the falling particles can't cover the total energy in the accretion disc + Jet stream and the missing energy is coming from the SMBH.

If the missing energy is coming from the energy that is stored in the SMBH at is first day (as some sort of one time charged battery) or if it is due to refueled' SMBH (as rechargeable battery) is less important at this phase.

Can we at last agree on the following highlight of our long discussion:
The energy in the falling particles can't cover the total energy in the accretion disc + Jet stream and therefore the SMBH must contribute the missing energy!

[paul cotter]

All the energy in the whole quasar system(bh, disc and jets) ultimately comes from what falls in.

[Bored chemist]

The energy in the falling particles can't cover the total energy in the accretion disc + Jet stream and therefore the SMBH must contribute the missing energy!

The energy of the smbh IS the energy of the particles that fell into it.

[Origin]

At last, you offer an idea of using the energy that is stored in the SMBH as the missing energy.
Therefore, I hope that you fully understand that the energy in the falling particles can't cover the total energy in the accretion disc + Jet stream and the missing energy is coming from the SMBH.

If the missing energy is coming from the energy that is stored in the SMBH at is first day (as some sort of one time charged battery) or if it is due to refueled' SMBH (as rechargeable battery) is less important at this phase.

Can we at last agree on the following highlight of our long discussion:
The energy in the falling particles can't cover the total energy in the accretion disc + Jet stream and therefore the SMBH must contribute the missing energy!

You seem to have convinced yourself that there is 'missing energy', I doubt anyone here agrees with that.
The exact mechanism of a jet is not known.  Jets are not that uncommon with a large mass that has an accretion disc, even star formation often have associated jets.  We can have fun speculating how the jets form, but if the physicist that study these jets haven't got the exact mechanism down, I certainly don't think we are going to figure it out.

[Kryptid]

Hence, although those Quasars have far more material in their accretion disk, they are still picky eater and therefore less than 1% of the gas and dust drawn into their gravitational field ever get consumed

That does not follow. Just because that's the number for the Milky Way's black hole doesn't mean it is automatically the same for quasars.

The energy in the falling particles can't cover the total energy in the accretion disc + Jet stream and therefore the SMBH must contribute the missing energy!

It could actually cover it just fine if the right ratio of material falling in to material being blown out into the jets is there. But let's say, for the sake of argument, that you are right and the energy all comes from the spin of the black hole. The spin is a finite source of energy. Once the spin stops, then the ergosphere and magnetic field produced by the black hole disappear as well. Without the ergosphere or magnetic field to flick particles away, material from the accretion disk is now free to fall right into the black hole. That infalling material then "recharges" the black hole by increasing its spin once more due to conservation of angular momentum. So in the end, the black hole gets its spin both from the material that formed it in the first place and material that eventually falls in later.

[Dave Lev (OP)]

Hence, although those Quasars have far more material in their accretion disk, they are still picky eater and therefore less than 1% of the gas and dust drawn into their gravitational field ever get consumed

That does not follow. Just because that's the number for the Milky Way's black hole doesn't mean it is automatically the same for quasars.

Dear Kryptid
The meaning of "picky eater" at the MW' SMBH is that - "less than 1% of the gas and dust drawn into their gravitational field ever get consumed"
For the sake of argument let's agree that you are right and the real meaning of picky eater is that 99% of the gas and dust drawn into the quasar' SMBH gravitational field and are consumed by the SMBH, while only 1% from all the gas and dust is actually falling in the accretion disc. (I hope that this kind of picky eater is ok for you).
Let's look again at the following data:

The estimated gravitational potential energy for a falling proton from a distance of 2-light days at the 3C 273 quasar is -8.37 x 10^-14 joules
The kinetic energy of a proton that orbits at almost the speed of light at the quasar accretion disc is approximately 9.087 x 10-10 joules.

Based on our maximal wish for picky eater meaning, and by the assumption that all the potential gravitational energy of 100 particles is transformed to only one of them, then its maximal kinetic energy could be:
-8.37 x 10^-14 joules * 100 = -8.37 x 10^-12 joules
Therefore, even in this most optimistic case, we still need to increase its kinetic energy by 100 times in order to get the observed kinetic energy of 9.087 x 10-10 joules at the accretion disc
9.087 x 10-10  / 8.37 x 10^-12 = 100

I hope that you agree that the SMBH is the only one that can increase that energy of the falling particles by 100 times.

It could actually cover it just fine if the right ratio of material falling in to material being blown out into the jets is there.

Why Don't you accept the clear message from the astronomies that are specialized in quasar that there is a problem with the current mainstream theory for the quasar activity?
Let's read it again:

In the article it is also stated that those astronomers don't really understand how this huge quasar really works"
https://www.wbur.org/npr/507594456/some-bizarre-black-holes-put-on-light-shows
"She spends most days chipping away at one of the universe's biggest mysteries: How do the huge, overactive black holes, known as quasars, work?"
"They are able to accelerate particles to 99.99 percent of the speed of light," Isler says. "How does that happen? So, I'm interested in where along that jet do we get this acceleration, and what is the physical mechanism that is responsible for the acceleration of particles that we see?"

Are you ready to accept the idea that they don't know the physical mechanism that is responsible for the acceleration of particles that we see?
Please, yes or no?
If it is still no, then at least, do you agree that based on the most optimistic assumptions for the meaning of the picky eater" quasar SMBH,  it must increase the energy of the falling gravitational potential particle energy by about 99%?
In other words, about 99% of lucky particle kinetic energies at the accretion disc must come from the spin of the quasar' SMBH & its magnetic energy.

But let's say, for the sake of argument, that you are right and the energy all comes from the spin of the black hole.

Well, I hope that you are ready to understand that based on the most optimistic assumptions for the meaning of that the picky eater" quasar SMBH,  at least 99% from the lucky particles kinetic energy at the accretion disc MUST come from the spin of the Black hole.

The spin is a finite source of energy. Once the spin stops, then the ergosphere and magnetic field produced by the black hole disappear as well. Without the ergosphere or magnetic field to flick particles away, material from the accretion disk is now free to fall right into the black hole.

Yes, I fully agree with this explanation.

That infalling material then "recharges" the black hole by increasing its spin once more due to conservation of angular momentum. So in the end, the black hole gets its spin both from the material that formed it in the first place and material that eventually falls in later.

This mechanism might be correct or incorrect. However, let's assume that it is correct.

So in the end, the black hole gets its spin both from the material that formed it in the first place and material that eventually falls in later.

How the black hole is "recharged" isn't relevant at this phase.

Once we agree that 99% from the kinetic particles energy at the accretion disc must come from the SMBH spinning motion we can find the real answer for the quasar' SMBH activity.

All the energy in the whole quasar system(bh, disc and jets) ultimately comes from what falls in.

The energy in the falling particles can't cover the total energy in the accretion disc + Jet stream and therefore the SMBH must contribute the missing energy!

The energy of the smbh IS the energy of the particles that fell into it.

We must distinguish between the gravitational potential energy of the falling particles to the SMBH spinning energy.
It is OK to believe that all of the SMBH spinning motion is due to the falling particles (Although it might be incorrect).

However, Do you agree by now that at the most optimistic scenario the lucky falling particles at the accretion disc can get maximal 1% of their energy from the falling particles due to the gravitational potential energy)?
Are you willing to accept the idea that 99% of all the energies in the accretion disc must come from the spinning
SMBH?
If So, why do we insist that the magnetic field is due to the kinetic energy of the falling lucky particles at the disc?
Why can't we understand that the real source of almost all the energies in the accretion disc + jet stream is the  mighty rechargeable spinning quasar' SMBH (Please feel free to charge it by any idea as you wish)?

[Kryptid]

I hope that you agree that the SMBH is the only one that can increase that energy of the falling particles by 100 times.

That's one possibility, but not the only one.

Why Don't you accept the clear message from the astronomies that are specialized in quasar that there is a problem with the current mainstream theory for the quasar activity?
Let's read it again:

The quote you provided does not say there is a problem with the modern understanding of how quasars work. What it says is that we don't know for sure how they work. It's entirely possible to have a plausible mechanism for how a phenomenon occurs without yet having obtained direct observational evidence for it. Until you get that evidence, what you have is technically a mystery. That is not the same as saying the proposed explanation has a problem.

Are you ready to accept the idea that they don't know the physical mechanism that is responsible for the acceleration of particles that we see?
Please, yes or no?

You're right, we don't know for sure. But we have some pretty good ideas.

If it is still no, then at least, do you agree that based on the most optimistic assumptions for the meaning of the picky eater" quasar SMBH,  it must increase the energy of the falling gravitational potential particle energy by about 99%?
In other words, about 99% of lucky particle kinetic energies at the accretion disc must come from the spin of the quasar' SMBH & its magnetic energy.

It could come from those things, but it doesn't have to.

However, Do you agree by now that at the most optimistic scenario the lucky falling particles at the accretion disc can get maximal 1% of their energy from the falling particles due to the gravitational potential energy)?

No, because, again, you are using numbers from the Milky Way's black hole (which is not a quasar).

Why can't we understand that the real source of almost all the energies in the accretion disc + jet stream is the  mighty rechargeable spinning quasar' SMBH (Please feel free to charge it by any idea as you wish)?

Because we don't have proof of that. We don't know that black holes even have magnetic fields.

[Dave Lev (OP)]

Quote from: Dave Lev on 27/09/2023 19:25:00
However, do you agree by now that at the most optimistic scenario the lucky falling particles at the accretion disc can get maximal 1% of their energy from the falling particles due to the gravitational potential energy)?

No, because, again, you are using numbers from the Milky Way's black hole (which is not a quasar).

I have used your calculation:

The mass of the black hole is 886 million times that of the Sun. The Sun's mass is 1.9885 x 1026 kilograms. That makes the black hole's mass 1.722 x 1035 kilograms. This means that the gravitational potential energy of a proton at two light-days from the black hole would be:

U = (-GMm)/r
U = (-(-6.674 x 10-11)(1.722 x 1035)(1.673 x 10-27))/(5.18 x 1013)
U = (-(1.149 x 1025)(1.673 x 10-27))/(5.18 x 1013)
U = (-(1.9227 x 10-2)/(5.18 x 1013)
U = -3.71 x 10-16 joules

Hence, the gravitational potential energy of a proton at two light-days from the MW' black hole  is -3.71 x 10-16 joules
Based on your following calculation about the 3C 273 quasar galaxy:

The galaxy that contains 3C 273 has a mass of about 2 x 1011 solar masses. This is about 225.7 times the mass of the central black hole there. So I can redo the calculations taking this into account. I am going to assume that all of that mass is concentrated at the center of the galaxy (it is, which means that my calculations will actually be an overestimate for how difficult it is for the proton to escape). So we just multiply the original numbers by 225.7: -3.71 x 10-16 joules x 225.7 = -8.37 x 10-14 joules, and -7.52 x 10-23 joules x 225.7 = -1.697 x 10-20 joules. That's a difference of 8.3699983 x 10-14 joules.

Hence, do you confirm that the gravitational potential energy of a proton at two light-days from the 3C 273 quasar black hole  is - -8.37 x 10-14 joules?
If no, please advise the correct value.

[Bored chemist]

U = (-GMm)/r
So the magnitude of that energy that gets smaller if we look at a larger radius.

If I drop a rock on my foot, it does more harm if I drop it from further up (i.e. with a larger radius)

So, apparently more damage results from less energy.

It's clear that U doesn't mean what Dave thinks it does.

In order to find out how hard something hits the ground you need to know where the ground is as well as from how far up you dropped it.

And as far as I can see (I may have missed it) Dave has not estimated the radius of the accretion disk (or whatever it is that get's "hit".

So he can't have worked out how much energy is available to do anything interesting like heat stuff or make magnetic fields.
Pity he didn't do the right calculation...
https://www.thenakedscientists.com/forum/index.php?topic=86495.msg712733#msg712733

Essentially, what you need is not a potential (which is undefined when something actually hits the BH and r= 0) but a potential difference.
.

[Kryptid]

I made a mistake with my original calculations. The Sun's mass is on the order of 10³⁰ kilograms, not 10²⁶ kilograms. So multiply the proton's energy by 10,000.

[Dave Lev (OP)]

I made a mistake with my original calculations. The Sun's mass is on the order of 10³⁰ kilograms, not 10²⁶ kilograms. So multiply the proton's energy by 10,000.

the gravitational potential energy of a proton at two light-days from the MW' black hole  is -3.71 x 10-16 joules

Thanks
Do you mean that the gravitational potential energy of a proton at two light-days from the MW' black hole should be:
U = -3.71 x 10-16 joules * 10,000 = -3.71 x 10-12 joules?
While the kinetic energy of a proton that orbits at almost the speed of light at the quasar accretion disc is approximately 9.087 x 10-10 joules (without any change)?

[Bored chemist]

Hence, do you confirm that the gravitational potential energy of a proton at two light-days from the 3C 273 quasar black hole  is - -8.37 x 10-14 joules?

Never mind the actual numerical value.
It's the energy required to remove a proton from 2 light days to infinity.
It has virtually nothing to do with the energy released when it falls in.

[Kryptid]

Do you mean that the gravitational potential energy of a proton at two light-days from the MW' black hole should be:
U = -3.71 x 10-16 joules * 10,000 = -3.71 x 10-12 joules?

Yes, but as Bored Chemist says, that's not the same as the energy you'll get from having the proton fall into the black hole.

While the kinetic energy of a proton that orbits at almost the speed of light at the quasar accretion disc is approximately 9.087 x 10-10 joules (without any change)?

That's the kinetic energy of a proton travelling at 99% the speed of light. Whether it is orbiting or not is irrelevant.

[Dave Lev (OP)]

Do you mean that the gravitational potential energy of a proton at two light-days from the MW' black hole should be:
U = -3.71 x 10-16 joules * 10,000 = -3.71 x 10-12 joules?

Yes, but as Bored Chemist says, that's not the same as the energy you'll get from having the proton fall into the black hole.

Thanks
Now it is clear:
the gravitational potential energy of a proton at two light-days from the MW' black hole should be:
U = -3.71 x 10-16 joules * 10,000 = -3.71 x 10-12 joules
Based on your following explanation, a proton at the 3C 273 quasar would have about 225.7 more potential energy

The galaxy that contains 3C 273 has a mass of about 2 x 1011 solar masses. This is about 225.7 times the mass of the central black hole there. So I can redo the calculations taking this into account. I am going to assume that all of that mass is concentrated at the center of the galaxy (it is, which means that my calculations will actually be an overestimate for how difficult it is for the proton to escape). So we just multiply the original numbers by 225.7: -3.71 x 10-16 joules x 225.7 = -8.37 x 10-14 joules, and -7.52 x 10-23 joules x 225.7 = -1.697 x 10-20 joules. That's a difference of 8.3699983 x 10-14 joules

Therefore, the gravitational potential energy of a proton at two light-days from the 3C 273  black hole should be:
U =  -3.71 x 10-12 joules * 225.7 = 8.37 10^10  joules.
That gravitational potential energy is almost identical to the kinetic energy of a proton that orbits at almost the speed of light at the quasar accretion disc (which is 9.087 x 10-10 joules).
 

While the kinetic energy of a proton that orbits at almost the speed of light at the quasar accretion disc is approximately 9.087 x 10-10 joules (without any change)?

That's the kinetic energy of a proton travelling at 99% the speed of light. Whether it is orbiting or not is irrelevant.

So, now the following process is very clear.
Proton at two light-days from the 3C 273  black hole with gravitational potential energy of -8.37 10^10  joules falls in the direction of the SMBH. As it falls inwards, its gravitational potential energy is transformed into kinetic energy. At the accretion disc, most of its potential energy had been transformed into kinetic energy, and therefore, it orbits there at almost the speed of light.
So far so good.

However, there are still two main problems:
1. Magnetic fields
We believe that the magnetic field is generated by motion of the protons at the accretion disc.
I hope that you agree that as the magnetic fields is created, it consumes energy from the motion of the protons.
Therefore, by definition it should slow down the velocity of the protons at the accretion disc.
However, we don't observe any slowdown in the velocity of the Protons at the accretion disc.
So, how a proton that orbits at almost a speed of light can generate magnetic field without losing its kinetic energy?

2. Potential / kinetic energy transformation & vice versa
We all agree that when the proton falls inwards into the quasar SMBH, its potential energy had been transformed to kinetic energy. therefore, by definition when the proton is ejected outwards, its kinetic energy should be transformed back to potential energy.
So, how could it be that the ejected protons keep their speed of light velocity as they get further away from the quasar SMBH?

Do you agree that a proton at 10-light days from the SMBH should have higher gravitational potential energy from a proton at two light-days from the SMBH?
If so, how could it be that the protons get more energy from the starting falling point, when they are ejected outwards?

Why the transformation of potential energy to kinetic energy is working perfectly when a proton is falling inwards, but a similar transformation of kinetic energy to potential energy isn't working when the proton is ejected outwards?

Why they don't just stop at the starting point of 2 light days?
Where the extra energy is coming from?