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1

The metric of the Schwarzschild black hole is $$ ds^2=\left(1-\frac{2M}{r}\right)dt^2-\left(1-\frac{2M}{r}\right)^{-1}dr^2-r^2(d\theta^2+\sin^2\theta d\phi^2). $$ Therefore, the Lagrangian of a particle in this background is (with proper time $\tau$) $$ ...


0

The orbits in the video you saw, for all intents and purposes, are about the same as the orbits for objects like Halley's comet -- thank's to https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion">Kepler's second law, when the object is close in, then it moves much more quickly than when it's far out. So, why does this show a black hole? Well, ...


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Comment to the question (v2): One possible helpful comment is that $\varepsilon$ in Ref. 1 and Ref. 2 are different: In Ref. 1 with $(-,+,+,+)$ convention, the eq. (3.46) yields that $$\varepsilon~:=~n^{\alpha}n_{\alpha}~=~+1$$ is always $+1$ for a spacelike normal 4-vector to a timelike hypersurface (shell), which is the situation studied in Ref. 2. In ...


2

The simplest answer lies in a combination of Gauss's law and Birchoff's theorem. These say, alternately, that the electric field and gravitational field of a spherically symmetric charge distrubtion only depends on the charge and mass-energy enclosed in the spherical shell${}^{1}$. Therefore, if I have a spherically symmetric distribution of charged ...


2

A very brief, although somewhat incomplete, answer would be that charge is related to a local symmetry, therefore with a gauge field that acts in the whole spacetime, even if the charges are inside the event horizon you could use the electromagnetic field to probe it. Lepton and Baryon number, or other flavour related quantities are related to global ...


3

If you take two protons, they have a mass of about $3.34\times10^{-27}$ kg. The "Schwarzschild radius" of a two-proton black hole is given by $2GM/c^2 = 5 \times 10^{-54}$ m ! i.e. you have to get the protons closer together than this. Even ignoring that protons have a radius of $10^{-15}$ m and that there would be an enormous repulsive strong nuclear force ...


1

Theoretically, yes. Practically, no. In order to create a singularity you would have to compress mass so that it would have zero volume, and thus an indeterminate density. We simply do not have the capabilities to produce such results on our own. Edit: Density would approach infinity as the volume approaches zero.


0

Samir Mathur gives a very accessible (heuristic?) account of Hawking radiation and why it leads to the black hole losing mass. You can access the page here. It's a work in progress and is updated about once every couple of months. To summarize, the lifetime of particles is dictated by $\Delta E \Delta t < \frac{\hbar}{2}.$ Normally this means that the ...


1

I like Ben's answer, but here is my take on this. Degeneracy pressure is not due to a fundamental force; in fact in the simplest model, it occurs in ideal gases of non-interacting fermions. The simple quantum mechanics of particles in an infinite potential well (i.e. trapped in a volume) tells us that only certain quantised wavefunctions are possible. Each ...


2

The force we associate with the Pauli exclusion principle is not a fundamental force, connected with the four fundamental interaction, but rather an entropic force, a consequence of the limitations the principle puts on allowed wave functions. A more general statement of the principle states the the total wavefunction of a system of two (or more ) identical ...


2

All the known laws of physics, including the exclusion principle, are believed to be valid at all times during the collapse, up until the matter that you're talking is just about to hit the singularity. ("Just about to hit" may mean when the density reaches the Planck density, so that quantum gravity effects become important, or it may be a little earlier, ...


7

How does the Pauli Exclusion Principle actually create a force? The Pauli exclusion principle doesn't really say that two fermions can't be in the same place. It's both stronger and weaker than that. It says that they can't be in the same state, i.e., if they're standing waves, two of them can't have the same standing wave pattern. But for bulk matter, ...


0

I'll post the same answer here that I posted on Worldbuilding: Under the circumstances you describe, my immediate reaction is that it would not be possible. The issue here is that the cluster would be fairly unstable. The black holes would all be mutually attracted to each other, and would soon coalesce into one large black hole - taking the Imperial ...


2

Could it have been: $\large l_p = 1.61619926^{-35}m \small \quad \text{ (Planck Length)}$ At any rate, to create a black hole, you simply need enough energy density in a single area that its escape velocity (the speed at which the sums of $E_k$ and $E_p$ are $0$) is larger than the speed of light. As you should know, the photon has no mass. However, its ...


0

This question is unanswerable using current loop quantum gravity, which does not yet have a completely consistent way to couple gravity to matter.


2

Your first paragraph is not quite right. Gas pressure does not "stop" upon formation of an iron core, it is merely that the star cannot generate further heat from nuclear reactions and becomes unstable to collapse. i.e. The star does collapse! Perhaps what you mean is what halts the collapse (sometimes) before the star disappears inside its own event horizon ...


0

The Chandrasekhar mass is not the dividing line between those stellar remnants that will become black holes and those that will become something else. A compact, cold white dwarf (i.e. one supported by electron degeneracy pressure) may become unstable and collapse at close to the value of $M_{Ch}=1.44(\mu_e/2)^{-2}M_{\odot}$, where $\mu_{e}$ is the number ...


0

The big ticker is that no one really knows if evaporation of black holes will create stable planck-scale remnants, so the building blocks of holeum might not even exist. So until we've seen some black holes decay, and observed stable Planck-scale black holes formed, the hypothesis, though elegant, probably won't gain much traction.


0

To the extent that the black hole is not spherical (a disk), there is a line perpendicular to the disk, and passing through its center, that will experience the cancellation of the forces perpendicular to it (radial direction). So, as matter is converted into energy, some of it will be able to "escape" along said line.


0

Thank you both for your very considered replies. I've come up with a modified view based on both replies. It is natural to assume that: the train is falling freely. But if you think about it more closely, only the Centre of Gravity is in free fall (by definition). The CoG will also move toward the front (singularity end) as it increasingly ...


4

Yes. The electromagnetic field is nonzero, so the stress-energy tensor is equal to $\frac{1}{2}F_{ac}F^{c}{}_{b}- \frac{1}{8}g_{ab}F^{cd}F_{cd}$. And, of course, $8\pi G T_{ab} = R_{ab} -\frac{1}{2}Rg_{ab}$ EDIT: to clarify, in the riessner-Nordstrom spacetime, you have $g_{ab} = {\rm diag} (-\Delta, \frac{1}{\Delta}, {\rm sphere\; metric})$ and $F_{ab} = ...


0

I think you're misinterpreting what the no-hair theorems say. There's a no-hair theorem for stationary electrovac solutions. It applies to electrovac solutions, not to solutions containing any matter field you like; in fact, there are known counterexamples if you include certain types of matter fields. Also, it's a theorem about stationary solutions. A ...



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