Recent content by anonymous299792458

Um, we've got two completely opposite answers here from pmb phy and vincenchan...

I wasn't talking about rotation. I was talking about the equivalence principle which states that a gravitational field is locally equivalent to an accelerating reference frame.

Yes, yes, that's what I would have thought too (and you may very well be right). But let's say we have a charge sitting on a table in the Earth's gravitational field. I don't think you'll measure any radiation, otherwise you'd have a source of "free" energy. Locally, this is equivalent to the...

If the acceleration of the charge is constant, it radiates. However, the radiation reaction force, which depends only on the third and higher derivatives of x with respect to t, is 0. How is this explained? Didn't Feynman say that, in fact, in the case of constand acceleration, there's NO...

If a charge is accelerating, but you are also accelerating WITH the charge, will you see any radiation? I.e. will the integral of the Poynting vector over a closed surface surrounding the charge be zero? It seems that it should be. Now let's say YOU are accelerating, but the charge is NOT...

1. Maxwell's equations do not hold in NON INTERTIAL reference frames, right?? 2. Let's say you have a charge which was briefly accelerated. You surround it by a closed surface CLOSE to the charge and integrate the Poynting vector over this surface and with respect to time to get the TOTAL...

So in an IDEALIZED case, the electrons wavefunction will be strictly in the n=2 state? Doesn't this violate the time-energy uncertainty relation? Assuming the transition (WE ARE STILL ASSUMING THE IDEALIZED CASE) from n=1 to n=2 happened 5 seconds ago, wouldn't the energy uncertainty be AT LEAST...

Well, I realize that an arbitrary wavefunction does not have a definite energy. But I thought that it could ALWAYS be represented as a superposition of energy eigenfunctions which DO have definite, DISCRETE energy values. Hence, it seems that the range of energies is discrete, not continuous as...

Lets take an electron which is "in orbit" around the nucleus. As far as I know, its (ANY conceivable) wavefunction can be represented as a superposition of energy eigenfunctions, which correspond to the discrete eigenvalues of the electron's enrergy. What I do not understand is where the natural...