Indeterminacy would apply equally to the mass of both particles, because the concomitant of indeterminacy is indistinguishability.
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Experience teaches that trying to forge any kind of link between this sort of mathematical concept and the “real world” leads nowhere.
the accretion disc temperature of the SMBH was 10 ^9 , and the temperature of a star's core is 10^6....I'm afraid not, because there is more to it than just temperature.
A star at 10^6 is fusing hydrogen into helium. At a higher temperature (an accretion disk) would be fusing the heavier elements
Quotefrom: alancalverd on 12/10/2018 09:49:23Using F = GmM/r2 you can calculate the force on a falling object of mass m in terms of M, the mass of the earth, and G, which we assume to be a universal constant.That doesn’t work. We’re trying to compute at least a rough G and M here. We don’t know either of them yet. We do know force F is 9.8 newtons for a 1KG mass. We can assume we know r. We therefore know the product of G and M, but not either separately.
QuoteAs F = ma, we can measure the acceleration of a falling object or the period of a pendulum to get a value for F/m
F=ma works (F 9.8 = 1 (mass) * 9.8 m/sec acceleration), but that doesn’t yield either mass of Earth M nor G, which are the two things we’re trying to determine here.
The pendulum thing is a function of acceleration (9., not of the mass of Earth. Put a pendulum in a rocket accelerating at that rate and it will have the same period as here on Earth. It tells you nothing about the mass of the Earth under you.
The illuminous region around the BH's gravitational field represents it's external boundaryThere is a region called the "photon sphere", which is outside the event horizon (at 50% greater radius). If you emitted a flash of light here, you would see a series of light echoes as photons from the flash repeatedly orbit the black hole.
The illuminous region around the BH's gravitational field represents ... is a reflection of the incoming light striking a mass.If there is no infalling matter (eg from a closely-orbiting companion star) then there is no mass between the photosphere and the event horizon (...if you ignore the hypothetical, microscopically thin, microwave-shifted skin of star matter that fell into the event horizon when the black hole first formed).
This reflection is similar to what occurs on a star, The cooler substrata below the star's surface creates a barrier that the hotter corona reflect off of. During a coronal () we see the cooler substrata mass.I am not sure what "coronal" event you are talking about - maybe a Coronal Mass Ejection, or the Corona that is visible during an eclipse by the Moon?
(A Black Hole) cannot be made of normal matter. That part is wrong. Normal matter of that mass cannot support its own weight, and breaks down even before the black hole forms.I think it is just a matter of grammatical tense here.
Is a BH empty? If it rotates, the answer needs to be no.We expect that most black holes will form with the angular momentum of their parent star, accretion disk or galaxy.
The black hole, is a singularity, it is comprised of unattached subatomic particlesA black hole singularity in general relativity is where all of the particles end up at a single point; all paths entering the event horizon end up at the singularity.
a chiralty gravitational effectGravitational waves are polarized.
the neutron lifetime puzzleI wondered why I kept seeing different figures for neutron lifetime.
create geometries very unlike such oxides at standard conditionsI agree that there will be crystal structures and molecules that would never be stable on Earth.
But the apparent speed of the CMBR is the same no matter which way we look at it. Velocity differs from speed in that it includes a direction. It looks to me that your reasoning would indicate that we are moving away from the source of the CMBR at roughly 368 km/s, but because the source is all around us, that presents a paradox: in which direction are we moving 368 km/s?Not quite
So what if we have the following formula?No
Δλ = 2πγ/k
Can we introduce gamma to modify the wavelength due to relative velocity? Can this be a valid way of viewing orbitals?
If gravitons exist (Gravity can be explained with a quantum theory), then gravitational waves would consist of gravitons.However, real gravitons propagate away to infinity, as oscillations on the gravitational field. This is what LIGO detected. (And real photons propagate away to infinity, as oscillations on the electromagnetic field. This is what telescopes detect.)
LIGO detected gravitational waves as predicted by EFE not gravitons. Gravitons are theorized to exist and have such low energy levels that we can not detect them with todays technology.
Gravitons can't, but since gravity would be mediated by virtual gravitons, this is no different from the fact that photons can't escape a BH, yet a BH can have a charge and an electromagnetic field mediated by virtual photons.
Bosons can not escape a BH, a graviton is a 2 spin boson how can it get out of a BH. A gamma ray cannot escape the event horizon of a BH and it has significantly more energy.
Gravitational waves definitely exist gravitons might not exist, and might not have enough energy to escape a BH.Gravitational waves can't escape a BH either (By this I mean that gravitational waves cannot get out from inside the event horizon anymore than electromagnetic waves can.)
This may be of interest.
This states that L = V + U.Correction. L = T - V, where T = Kinetic energy and V = Potential energy. If the potential is a velocity dependent one then the letter U is used. This is particularly important in electrodynamics.