Recent content by Hans de Vries

Once you understand non-simultaneity it is really not that complicated. It is also very natural: Just what you would expect. You can also already see why non-simultaneity occurs without any reference to relativistic effects like Lorentz contraction and Time dilation. This because at low speeds...

This question is a simpler version of the always recurring question: "How to explain Magnetism as a relativistic side effect of the Electric Field" ---------------------------------------------------- The answer to the OP's question: - A test-charge at rest is only subject to an Electric...

. The equation for photons with mass is really the Proca equation ##\partial_\mu(\partial^\mu B^\nu - \partial^\nu B^\mu)+\left(\frac{mc}{\hbar}\right)^2 A^\nu=0## which is equivalent to a Klein Gordon equation for each of the four individual components of ##A^\nu## ##\left[\partial_\mu...

These are just arbitrary coupling constants. They originate from the self interaction energy V of the field ##\varphi## with itself in equation (2). It is a way of expressing the self interaction part of the arbitrary function V in a series of ##\varphi##.

If the spin-vector ##\vec{s}## is given by, ##\vec{s}~=~\left[\begin{array}{c}\xi^*\sigma_x\,\xi \\ \xi^*\sigma_y\,\xi \\ \xi^*\sigma_z\,\xi\end{array}\right]## then the rotated spin-vector ##\vec{s}'## is given by ##\vec{s}' ~=~ R_{ \hat{n} }\left[\begin{array}{c}\xi^*\sigma_x\,\xi \\...

Conservation of momentum is maintained at all times, always. One of the most basic laws of nature. Who would ever make such a ridiculous claim? - Momentum can also be absorbed by the silver mirrors. - Angular momentum can be absorbed by Wollaston prisms. So a loss of correlation at the...

Well yes, that is of course that's what I wanted to say: It's more than just momentum conservation. One would expect at least some "action-at-a-distance" correlation effect in need for an explanation. For instance a higher probability that both green detectors go off or both red detectors go...

It's not as simple as just energy conservation. Do you have an example of a correlation experiment using momentum as an entangled quantum state?

I would urge you to read carefully through the text as accurate and painstakingly as I have done (and anybody can do in the link at the bottom), before you come out in the aggressive way you did. Let's carefully go through the 3 statements (as I ascribed them to Carrol) one by one: 1) All...

I know Sean Carrol's arguments but quote: I don't see how Sean Carrol's following arguments help in anyway: 1) All universes are in superposition and superposition is normal in QM 2) In any universe all particles are entangled so they can't interact with particles in other universes. 3) And...

I should try not to get involved with quantum philosophical discussions, but okay, having done this sort of "hypothetical MWI" calculations in the past: - There are ~##10^{80}## elementary particles in the universe. - Interactions / State transitions take place at the femto second level in...

So, with the (correct) version of the post above we may write more general for ##P=\hat{p}_r##: ##\hat{p}_r~~=~~ i\hbar\left(\dfrac{\partial}{\partial r} +\dfrac{1}{r}\right)## For an arbitrary power ##\hat{p}_r^n## we can write ##\hat{p}^n_r~~=~~ (i\hbar)^n\left(\dfrac{\partial}{\partial r}...

Actually ##P## (or better ##\hat{p}_r##) here is given by the radial part of the Spherical Polar form of the Dirac equation. The following is from Paul Strange's book Look at (8.9) and concentrate on the essential radial part of ##\hat{p}##: $$\hat{p}_r...

According to Arfken & Weber (10.6) the operator ##p^2## is not self-adjoint in a Cartesian inner product but it is self-adjoint in a spherical radial inner product. See the ##r^2## factor in post #69.

We can not just use a specific eigenfunction ##\psi##, for instance the radial part of (##\ell=0##) in combination with the Self_Adjoint test because we need two independent wavefunctions. See post #73 The boundery-term in your calculation in #62 for ##\hat{p}^2## does not cancel in the case of...