Recent content by swansont

PhD in Atomic Physics

No. Visible spectrum photons' wavelengths are much too big to be able to image a nucleus or even an atom. Diffraction would wash out any image you might expect.

When you annihilate matter and antimatter, that's what you get. Two gammas, actually, because you must conserve momentum.

Uh, no. There are no stable isotopes of Uranium.

But do the motions of the quarks account for the neutron magnetic dipole moment, and if so, isn't that an answer to you question about the presence of circulating charges?

Except that "undiscovered" means you expect to find them, rather than having them be something excluded by Maxwell's equations.

The neutron isn't an elementary particle; it's made up of quarks. Wouldn't they be the source of the magnetic dipole moment?

No, it's still attracted, which is why there is a bound system. QM shows that in bound systems, energy levels are quantized. IOW, not all states are available to the electron - the ground state orbit is as low in energy as it can get.

There is no energy state available that has it sticking to the proton. This is one of the problems that led to QM, as has been noted. Only specific energy states can be occupied. The s-state electron's orbital actually has it in the nucleus some fraction of the time, but to actually combine...

Note that the experiment showed the light leaving before the peak of the pulse entered. It's anomalous dispersion - the pulse was reshaped. It's a mistake to think of the light pulses as infinitely narrow.

Surely you are aware that atomic clocks are shielded to prevent or limit any frequency changes due to the presence of or fluctuations in external electric or magnetic fields.

But moving clocks do indeed run according to relativity, as witness by GPS and other spaceborne clocks.

x1 and x2 are in a reference frame of an observer that is stationary with respect to the ground. If he wants to measure the relative speed between me and the car, he needs to know where I am at t1 -call that x3. Then the relevant distance is x1-x3. But I don't think that's the measurement you...

OK, thanks for clarifying. I wasn't sure you were actually measuring the car by collision - now I am. I think I understand your point. But x2 is where I am, so it's actually fixed in my coordinate system. If you want to look at it from another point of view, you have to properly transform...

It was actually Mrs. Schroedinger's cat.