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

[Nugatory]

Does that mean that light would be emitted with such momentum so that conservation of momentum would hold?
What about energy conservation? The system has more kinetic energy than was put in and then there is also the energy of the light.

Yes, both energy and momentum are conserved; the electromagnetic field can hold energy as well as momentum. The total increase in kinetic energy, plus the energy radiated away by light, plus the change in potential energy if the two charged particles change their distance, will add up to exactly the amount of work that was done to accelerate the first particle.

Really, this entire process is no more mysterious than what happens when you have two masses connected by a spring, you give one of the masses a shove and it takes a moment for the other one to start moving. At any moment you may not be able to make the energy and momentum of the two masses add properly, but that doesn't mean momentum or energy conservation is being violated; it means you forgot to add in the energy and momentum of the spring.

 

[chingel]

I'm saying that if the system accelerates itself and radiates away light, energy would not seem to be conserved and therefore it is questionable that it would accelerate itself.
However, as was previously posted, it seems to self-accelerate, about which an article was posted.

 

[Dale]

Hi all,

An acquaintence has pointed me towards "An electric dipole in self-accelerated transverse motion" by F. H. J. Cornish. Unfortunately you have to pay for the article or rent it for free for 5 mins.

http://ajp.aapt.org/resource/1/ajpias/v54/i2/p166_s1?isAuthorized=no&ver=pdfcov

A detailed relativistic calculation shows that the dipole should move with a constant proper acceleration given by:

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

where $e$ and $-e$ are the charges, each charge has mass $m$, $d$ is the distance between them and $d < e^2/2mc^2$ (cgs units used).

As the acceleration is constant the author states that the Lorentz-Dirac radiation reaction forces on each charge vanish and therefore do not contribute to the motion.

Very interesting. Too bad for the paywall.

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.

Unfortunately, I was planning on using the Lienard Wiechert potential in my solution, so my approach may wind up with the same problem.

 

[johne1618]

Very interesting. Too bad for the paywall.

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.

Unfortunately, I was planning on using the Lienard Wiechert potential in my solution, so my approach may wind up with the same problem.

I'm trying to understand if the problem is to do with quantum mechanics or classical EM.

If there is a problem then I think it's to do with high electric fields causing vacuum polarization.

If an electric field has a high enough energy density then electron-positron pairs will be generated which means that we must hand over the problem to quantum field theory.

Thus for vacuum polarization we have:

$$ \epsilon_0 E^2 = \frac{m_ec^2}{\lambda^3}$$
where $\lambda$ is the electron Compton wavelength given by $\lambda = h/m_ec$.

Thus

$$ E^2 = \frac{m_e^4 c^5}{\epsilon_0 h^3} $$
$$ E \approx 10^{15} V/m$$

In order to show the anomalous acceleration a system's electric potential energy has be roughly equal in magnitude to its rest mass energy.

Can I construct a parallel plate capacitor whose electrical energy is of the same order as its rest mass energy without using electric fields greater than $10^{15} V/m$?

If $\rho$ is the mass density of the plates, $w$ is the thickness of the plates, $A$ is the area of the plates and $d$ is the separation of the plates then a capacitor whose rest mass is equal to its electrical energy is given by:

$$\rho A w c^2 = \epsilon_0 E^2 A d$$

$$E^2 = \frac{\rho c^2}{\epsilon_0}\frac{w}{d}$$

Assuming $\rho=10^3 kg/m^3$ we have:

$$E = 10^{15} \times \sqrt{\frac{w}{d}}\ V/m$$

Thus as long as the ratio of plate thickness to separation is small enough we can avoid vacuum polarization electric field strengths inside the capacitor.

Thus I think the paradox lies with classical EM rather than with quantum field theory.

 

[Dale]

I'm trying to understand if the problem is to do with quantum mechanics or classical EM.

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

 

[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.