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0

It is not possible to know. The reason they are called holes is that they are effectively holes in space/time so concepts like length, size, distance and time no longer have meaning in the conventional sense. Equally, while they may be the most compact objects known, there is no evidence to suggest they must be uniform. If you are talking about the event ...


8

The theory behind the trick is based on the Hellmann-Feynman (HF) theorem $$ \frac{dE_{\lambda}}{d\lambda}~=~\langle \psi_{\lambda} | \frac{d\hat{H}_{\lambda}}{d\lambda}| \psi_{\lambda} \rangle,\tag{A}$$ which works with a single derivative, but not with a square of a derivative, cf. OP's failed calculation (5) for the expectation value $\langle\frac{1}{r^2}...


4

That the trick works once is lucky coincidence, it is actually non-sense as presented. Consider, as an extreme counterexample, an operator $O$ independent from $e$, and a state $\lvert \psi\rangle$ whose eigenvalue w.r.t. $O$ is $e$. Then, by the same logic we have $$ 0 = \langle \psi \vert 0 \vert \psi \rangle = \langle \psi \vert \frac{\partial}{\partial ...


18

Yes, they are perfect spheres. But let's understand what is the sphere and why. The spherical surfaces are the horizons. They are surfaces in space that have perfect spherical geometry. Now, that only holds true for various conditions: 1) they are static black holes, and if they arose from matter/energy collapsing they are in their equilibrium states. ...


21

The shape of a black hole's event horizon depends on who is asking. Observers who are moving quickly towards a hole, for example, will see a different shape compared to those who are not. In the coordinates appropriate to very distant "inertial" observers, the event horizon of a nonspinning uncharged black hole in equilibrium is spherical. If the hole is ...


1

For question 1, it comes down to probability. I have two distinguishable particles, $a$ and $b$. The probability to find particle $a$ at $x_1$ is $$P_a (x_1)=\int dx_1 \Psi_a(x_1) \Psi_a^*(x_1),$$ and we have a similar expression for particle $b$ at $x_2$. The probability to find particle $a$ at $x_1$ and particle $b$ at $x_2$ is just the product of ...


1

If the state of two particles is the tensor product of the two single particle states, then the wave function of the two particles is the product of the two single particle wave functions. For indistinguishable particles it is an experimental fact that the final state must be either symmetric or antisymmetric with respect to the exchange of the two ...


0

Ok, I thnik I figured things out myself. I will follow the conventions of Wess & Bagger. If one wants to construct a free, $\mathcal N=1$ SUSY theory with a complex scalar and a Weyl Fermion, then the possible transformations are dictated by representation theory of the Lorentz group, dimensionalities, and the requirement that we have a free theory and ...


1

A part of the supersymmetry algebra is $$ \{Q_a,~{\bar Q}_{\dot b}\}~=~-2i\sigma^\mu_{a\dot b}\partial_\mu $$ which is a momentum operator $p_\mu~=~-i\partial_\mu$. The graded Lie algebra $g~=~h~+~k$ $$ [h,~h]~\subset~h,~[h,~k]~\subset~k,~\{k,~k\}~\subset~h, $$ where the last of these contains the above anti-commutator. This model has chiral symmetry. It ...


1

You should probably read up on the Stueckelberg action and the Affine Higgs mechanism it sends you to. Your boldface supposition "I'm not supposing that my matter has any global symmetry here, that I might be able to gauge" is unwarranted for the specific model you propose, $J_\mu=\partial_\mu\phi$. There is a global symmetry, $\phi \to \phi+\alpha$, whose ...


0

You're right that that Lagrangian isn't in general gauge invariant. In addition to making the $A^\mu$ terms gauge invariant, the $\mathcal{L}(J)$ term must also be gauge-invariant. And not just the equations of motion for $J_\mu$, either - the specific algebraic expression for $J_\mu$ in terms of fundamental fields must be such that if you literally plug ...


5

As an update on this old thread, the 2015 version of the Particle Data Group review on tests of conservation laws (the 2009 version of which was rightly pointed to by invisiblerhino) has an interesting update: The BABAR experiment has reported the first direct observation of $T$ violation in the $B$ system. The measured $T$-violating parameters in the ...


0

Before going into the details, let me describe pictorially how the Hamiltonian, the Symmetry group, and the Dynamical group look in a basis in which the Hamiltonian is diagonal. Hamiltonian $$ H = \begin{bmatrix} \begin{bmatrix} \lambda_1 \mathbf{1} \end{bmatrix} & & & \\ & \begin{bmatrix} \lambda_2 \mathbf{1} \end{...


1

The symmetry arises because of the boundary conditions, which are independent of $z$ and $r$. Let us place one conducting plate at $\theta=0$ with potential $\varphi=0$, and another at $\theta_0$ at potential $\varphi=V$. We now want to find $\varphi(r, \theta, z)$ in the gap. You are right that while $\varphi(z) = \varphi(-z)$ at arbitrary $(r,\theta,z)$, ...


0

Consider a planar crossection of the configuration. Let O be the origin at the intersection of the 2 planes. Suppose that there are electric field lines in the radial direction. This would imply that non zero work must be done to bring a charge q from infinity along the radial direction to O or in other words the potential at O (just above the plates ,not on ...


0

Dark matter is uncharged, it might be its own antiparticle. No relationship to matter and antimatter, where mainly its charge conjugation and parity. we don't know that dark matter has any antiparticle broken symmetry. And there is no known relationship between matter and dark matter, except they interact gravitationally and maybe through weak interactions....


0

No way, amount of dark matter is estimated to be 4 times the amount of normal matter. Even if we ignore all other arguments, it still does not restore the symmetry. Dark matter (if there is such a thing), is actually transparent matter (we can not see it) and cold matter (it does not absorb/emit any heat/radiation). That also means that you can not measure ...


4

No, we know enough of the "bulk properties" of antimatter to rule this out. Antimatter interacts with the electromagnetic field in exactly the same way as regular matter, just with the opposite charge. Therefore, antimatter should be detectable using most of the techniques we use to detect regular matter in astronomy. This works even if the antimatter is a ...


2

I think we can pretty safely say this is not the case. The main reason is that we have a pretty good idea of where dark matter is--to some degree, it can be reconstructed from the gravitational influence it has on surrounding matter. The dark matter appears to be distributed evenly throughout the galaxy. Thing is, a galaxy is pretty "dirty," as far as space ...


1

Like it was said before, there is no a priori reason why nature should treat everything symmetrically. Much to the contrary, we know several examples of P- and CP-violating processes. And in other cases we do not even know the reason why a process is "symmetric", when in principle it would be allowed to violate CP (see: the strong CP problem). I guess you ...


2

There's not reason to assume nature should treat everything symmetrically. There are many phenomena in nature that we actually know are asymmetric. For example the weak force violates parity symmetry (meaning the weak force has a preference for right or left handedness).



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