Recent content by H_A_Landman

What goes in hard and dry and comes out soft and gooey? Bubble gum.

Let ##e## be a basis vector. Then ##\langle e|v\rangle## is the coefficient of ##|e\rangle## in the basis representation of ##v##. One can view ##|e\rangle\langle e|## as the projection operator onto ##e ##, that is ##|e\rangle\langle e|v\rangle## is the ##e## component of ##v## in the basis. If...

Everyone keeps saying that, but I don't see the math. De Broglie gives E=h𝜈. Schrödinger gives an angular frequency which also reduces to E=h𝜈, except the E includes the potential energy. So those seem nearly identical to me. What equations justify your statement? Well, first of all, I haven't...

OK, so your answer to my original question would be no, an electron does not have a definite phase frequency, because we can add an arbitrary constant to the energy? PeroK (in post #2) says that is not correct.

OK, let's make it simple. Take a hydrogen atom outside of a VDGG sphere, and another one inside the sphere, and charge the sphere to (say) +1 MV. Do you believe that the electrons in the two atoms have the same phase frequency, or different phase frequencies?

If you are intentionally trying to be as unhelpful as possible, you are succeeding.

Your answer is so terse that I can't even tell which one you claim is which, or how they supposedly differ. Care to elaborate?

I think you're saying that one is required to fix the gauge to the Coulomb/radiation gauge. But isn't that the same thing as saying that there is no gauge invariance for the potential? Consider an atom, or some simple electron experiment like a 2-slit interference or a charge-mass-ratio one...

The phase evolution is given by ##\psi(t) = e^{-iEt/\hbar}##. This is pretty much identical to ##E=h\nu##. How is that "very different"?

Stationary solutions to the Schrödinger equation factor into a spatial part, e.g. atomic and molecular orbitals, and a temporal part that gives the phase rotation frequency. It is often assumed that adding a constant to the potential leaves the physics unchanged. And clearly, any "spectroscopic"...

Nod. Curvature is defined in such a way that a cylinder has zero curvature and is not considered to be curved at all. However the forward implication ("A time dilation field with constant gradient ... has zero curvature.") is still correct, it's just the other direction that doesn't work.

Yeah, I had a brainfart there. Agree. A time dilation field with constant gradient ##\approx## a uniform gravtiational field, which has zero curvature. I should have said any change in the gradient by location implies curvature.

Constant time dilation doesn't. Time dilation that varies by location does; it's equivalent to a gravitational field, as was noted well before GR. (See e.g. Jun Ishiwara, “Zur Theorie der Gravitation.” Physikalische Zeitschrift 13: 1189–1193 (1912).) So the question of whether the interior is...

In the Newtonian limit, both ##1/r^2## forces from opposite parts of the shell would exactly cancel, so we'd only be left with the potentials. And so for ##Q=0## the interior metric would be flat Minkowski plus the gravitational time dilation, which would be constant inside the sphere. I think...

Maybe it would help to slice this a different way. Replace the "black hole" of mass ##M## and charge ##Q## with an insulating hollow sphere of radius ##R## and mass ##M## and charge ##Q##. Since ##M## and ##Q## are the same, presumably spacetime outside the sphere is still described by the same...