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Ultimately, your model is going to depend on the energy coming from the accretion disk as well.
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)/rU = (-(-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 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....
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.
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
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.
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).
QuoteQuote from: Dave Lev on 15/09/2023 16:12:41However, 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
Quote from: Dave Lev on 15/09/2023 16:12:41However, how the SMBH's spin energy could be transformed into those falling particles?
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.
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 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.
However, the observation contradicts this assumption as only 1% of all the falling protons really drawn into the SMBH gravitational field and get consumed:
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 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?
QuoteQuote from: Dave Lev on 23/09/2023 14:18:29However, 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.
Quote from: Dave Lev on 23/09/2023 14:18:29However, the observation contradicts this assumption as only 1% of all the falling protons really drawn into the SMBH gravitational field and get consumed:
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.phpThe result is about 2.783 x 1023 kilograms per second. That's about 0.05 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.
If you still think that quasars aren't picky eater, then please prove it by real data/observation
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".
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!
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!
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
Quote from: Dave Lev on 25/09/2023 04:57:01Hence, 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 consumedThat 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.
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.
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?"
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.
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.
All the energy in the whole quasar system(bh, disc and jets) ultimately comes from what falls in.
Quote from: Dave Lev on 25/09/2023 10:10:00The 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.
I hope that you agree that the SMBH is the only one that can increase that energy of the falling particles by 100 times.
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:
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.
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)?
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)?
QuoteQuote from: Dave Lev on 27/09/2023 19:25:00However, 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).
Quote from: Dave Lev on 27/09/2023 19:25:00However, 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)?
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.
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
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?
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)?
Quote from: Dave Lev on 02/10/2023 19:43:26Do 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.
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
Quote from: Dave Lev on 02/10/2023 19:43:26While 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.