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21

Yes indeed to all your questions: mutually orbitting binaries do spin down, the system's orbital angular momentum thus decreases with time and the loss of energy and angular momentum is almost certainly owing to the emission of gravitational waves. Look up the Hulse-Taylor binary system: its spin-down has been carefully observed and measured since its ...


6

Am I missing something here? Photons and the double-slit experiments do the trick. Likewise, you can entangle the polarisation degrees of freedom of photons. This is inherently quantum and photons aren't charged. True, they are the excitations of electromagnetic fields, but they are not charged. Anyway, I guess the problem here is that except for the ...


5

As mentioned, this is not the first evidence for gravitational waves. The data from BICEP2 shows that there is a much higher amount of B-mode polarization than what is predicted by gravitational lensing alone. According to theory, this could only be due to higher amplitude tensor modes in the CMB than previously observed (or rather, lack of observed). These ...


4

Every particle that interacts with something is charged -- in some sense. We say that particles are charged under a certain interaction if they are affected by that interaction, and their charge measures how strongly they interact. But what you seem to be asking about is electric charge. Then one idea of an observable phenomenon could be quark confinement. ...


3

In answer to the edit, any transitions due to single-graviton exchange will involve energies that are just impossibly small. To convince yourself of this, remember that the energy levels of the hydrogen atom are given by: $$E = \frac{\mu k^{2}e^{4}}{2\hbar^{2}n^{2}} = \frac{13.6\,\,{\rm eV}}{n^{2}}$$ If you do the same for two solar mass neutron stars ...


3

Weak and Strong interactions do not involve charge. Strong interactions involve color charge, which is a different property than "normal" charge. The weak interaction mostly involves flavor change. As to your non-charge dependent Quantum Mechanical effects, we have tunnelling. See for example this link.


3

Recall that momentum $\pi^{ij}$ is a tensor density in 2-space, cf. e.g. Ref. 2. In other words, $$\frac{\pi^{ij}}{\sqrt{\det g}}$$ is a symmetric (2,0) tensor. Therefore the covariant derivative is $$(\nabla_{\ell}\pi)^{ij}~=~ \partial_{\ell} \pi^{ij} + \Gamma^i_{\ell k} \pi^{kj} + \pi^{ik}\Gamma^j_{\ell k} -\pi^{ij} \partial_{\ell}\ln\sqrt{\det g}.$$ ...



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