What causes the divergence of acceleration in the parallel plate system?

[johne1618]

The problem, as stated, is one of classical EM, as is the paper you cited.

I just wanted to make sure the situation can be realized in a macroscopic system rather than one at the scale of subatomic particles.

 

[johne1618]

All of the bona-fide mathematical paradoxes in classical EM that I know about have to do with mathematical problems from treating classical point particles. They go away if you avoid classical point particles (which don't exist anyway). I suspect that is the case here also.

As far as I can see the calculation does not involve the action of a point particle's field on its self.

Only on the other particle which is always a finite distance away.

 

[Vanadium 50]

A couple of comments:

1) The fact is that we are guaranteed that there is no diverging acceleration by Poynting's theorem. Poynting's theorem follows directly from Maxwell's equations and the Lorentz force law. Therefore, if you are using some other method to calculate the force and you get a diverging acceleration then you know that your calculation method is flawed. (Thanks Dale!)

2) Writing textbook authors is not a good idea. Just because they wrote a textbook doesn't mean that they agreed to answer questions from any and all comers, and besides, this is something crackpots do.

3) The equation

$$a = \left(\frac{2c^2}{d}\right)\left[\left(\frac{e^2}{2mc^2d}\right)^{2/3}-1\right]^{1/2}$$

is very peculiar, as a is not proportional to 1/m. That should be a hint that something is wrong.

3) Looking at the paper of Cornish, it's clear that johne1618 neglected to mention something very important (naughty, naughty, John!): that the equation for acceleration has two solutions, and one of them is always a = 0. See Cornish's equation 16:

$$a \left(m + \frac{e_1 e_2 d^2}{2R^3 c^2} \right) = 0$$

The paradoxical acceleration comes from setting the second term to zero. Since the system has only one acceleration but two solutions, you need to pick one (just like you have to in Freshman kinematics when you have a quadratic) and the correct one is a =0.

 

[Dale]

I just wanted to make sure the situation can be realized in a macroscopic system rather than one at the scale of subatomic particles.

Obviously the situation cannot be realized in any system, microscopic or macroscopic. We don't see self-accelerating dipoles either macroscopically or microscopically.

This isn't a question of realization in macroscopic or microscopic systems, it is simply a mathematical exercise. As such, the domain of the mathematical exercise is classical EM.

Depending on time tomorrow I still may attempt it.

 

[Dale]

the equation for acceleration has two solutions, and one of them is always a = 0.

D'oh! Well, that is much more trivial than problems arising from using classical point particles.

 

[johne1618]

3) Looking at the paper of Cornish, it's clear that johne1618 neglected to mention something very important (naughty, naughty, John!): that the equation for acceleration has two solutions, and one of them is always a = 0. See Cornish's equation 16:

$$a \left(m + \frac{e_1 e_2 d^2}{2R^3 c^2} \right) = 0$$

The paradoxical acceleration comes from setting the second term to zero. Since the system has only one acceleration but two solutions, you need to pick one (just like you have to in Freshman kinematics when you have a quadratic) and the correct one is a =0.

But surely we want to know why EM is offering us a second solution that is clearly unphysical? Maybe there's something missing in the analysis (like not including the counterbalancing effects of advanced interactions for example).

 

[Vanadium 50]

We don't think the thread will lead to more relevant insights. Closed.

If you are interested in why equations of motion lead to spurious solutions, I suggest you start a new thread on that, based on mechanics. It's simpler, and the same thing happens.