Recent content by Dale

Yes. The result depends strongly on the initial velocity of the particle and the initial phase of the wave. The discussion above is about the specific conditions of the problem and does not generalize.

Good suggestion. Thread closed

And the output at that same point. As I already said above. But the actual problem is for ##t=0##, not ##t=0^+##. As I already said above. This is wrong; the limit does not exist. The function is discontinuous at ##t=0## with the limit from above being ##\lim_{t\rightarrow 0^+}v_0(t)=v_i/2## and...

As we have already discussed, the question is about ##t=0##, not ##t=0^+##. This is simply factually untrue. As the OP already discussed the problem comes from an EE forum, edaboard, where the answer to the problem is a matter of contention. The actual people on the actual EE forum did not...

As we have already discussed above, all of that is true. But the debate is specifically about the completely unrealistic idealized model. That is precisely why there is no solution. Also, the debate is about ##t=0##, not ##t=0^+##. As we have already discussed above. The Laplace transform also...

That changes the problem as already discussed repeatedly above. The actual problem has no solution

Neither causes the other.

Convection is easier with air because of that, IMO. But I guess “hold your hand above this flame” is probably not a good lab exercise.

Well, K&K’s book didn’t give you the answer, but they did prepare you with the skills that you needed to discover it, and they did point out that there was something interesting to discover there. So I wouldn’t exactly say that this is something that books don’t say. More like they chose to...

A couple of minutes found these: https://peer.asee.org/a-lab-experiment-involving-free-convection-heat-transfer-from-a-flat-horizontal-plate.pdf https://www1.eere.energy.gov/education/pdfs/basics_thermodynamicsguidestudent.pdf https://thehomeschoolscientist.com/convection-current-experiment/...

Sure, I have seen undergraduate labs on convection.

@Demiurge by way of brief explanation, the QM interpretations sub-forum is for discussing the published interpretations found in the professional scientific literature.

Yes. Note that each cycle accelerates and then decelerates the electron. The acceleration portion accelerates from zero to some positive velocity. The deceleration portion is symmetric. So it decelerates from the positive velocity back to zero. So the velocity never becomes negative. Without...

Look at ##\dot x## instead of ##x##. What is the range of ##\dot x##?

I don't see why a dipole moment anomaly would have any bearing on the monopole. Mathematically the various multipole terms are all orthogonal, i.e. they form an orthogonal basis.