Capacitor paradox

**Privalov** · Jun21-09, 11:17 AM

Please see if you can resolve the following relativistic paradox for me.

Capacitor is being charged at rest and accelerated to relativistic velocity.

If we turn capacitor plates perpendicular to the direction of motion, width of dielectric contracts and amount of energy, stored in the capacitor, decreases. See left picture.

If we turn capacitor plates parallel to the direction of motion, length of capacitor contracts, identical charges become closer to each other and energy, stored in the capacitor, increases. See right picture.

If that’s correct, we can discharge capacitor at parallel state and recharge it at perpendicular state. We will get the perpetual motion machine.

**Naty1** · Jun21-09, 11:48 AM

I don't believe there is a paradox.

Relativistic length contraction affects any charge as well as the dielectric material....if one expands or contracts so does the other....it's space itself contracting or expanding....so I see no change in energy except for the kinetic energy of motion.

Besides, in the frame of the relativistic capacitor, all remains fixed...there is no contraction unless viewed from another reference frame.

**DaleSpam** · Jun21-09, 02:36 PM

You will find all of your missing energy in the magnetic field.

**Vanadium 50** · Jun21-09, 03:50 PM

Dale, I'm not sure that's correct.

Suppose in the rest frame, you have an area A and a separation d. If you are moving along the direction of the gap, you now have an area A and a separation $LaTeX Code: d/\\gamma$ for a total volume $LaTeX Code: Ad/\\gamma$ . Now, rotate it 90 degrees. Now you have an area $LaTeX Code: A/\\gamma$ and a separation of d, again for a total volume $LaTeX Code: Ad/\\gamma$ .

Since the energy stored in an electric field is

, since the field is the same and the volume is the same, energy is conserved. No magnetic field is required - at least not to solve this particular problem.

**Lok** · Jun21-09, 04:15 PM

As it is already said there is no paradox. relativity works in both ways so...

Imagine someone in a close to c spaceship passing close to you while you hold a capacitor and turn it which way u want. For the spaceship you will appear as to moving with a speed close to c. So the spaceship is the observer now. But you have no paradox in your reference frame as there is no energy difference.

**DaleSpam** · Jun21-09, 08:48 PM

Originally Posted by Vanadium 50

a total volume $LaTeX Code: Ad/\\gamma$ . ... since the volume is the same, energy is conserved.

Huh? $LaTeX Code: Ad/\\gamma \\ne Ad = V$ .

**diazona** · Jun21-09, 09:44 PM

Originally Posted by DaleSpam

Huh? $LaTeX Code: Ad/\\gamma \\ne Ad = V$ .

Actually $LaTeX Code: V = Ad/\\gamma$ , if I understood correctly.

Anyway, Vanadium 50 was really just saying $LaTeX Code: Ad/\\gamma = Ad/\\gamma$ . Can't argue with that...

**DaleSpam** · Jun21-09, 09:57 PM

Ahh, I was answering a different question! I understand Vanadium50's point now. He is absolutely correct. There is a magnetic field also in the moving frame, and like the electric field in the moving frame it has the same volume and same energy density in both the parallel and perpendicular orientations.

**Bob S** · Jun21-09, 10:43 PM

As long as the electric field in the capacitor is parallel to the direction of motion, the changes are limited to the relativistic contraction of the capacitor dimensions along the direction of motion. When the capacitor is rotated by 90 degrees, so that the electric field is perpendicular to the direction of motion, two additional things happen: See http://pdg.lbl.gov/2004/reviews/elecrelarpp.pdf
1) The component of E perpendicular to the motion, E_p, is increased by a factor γ, so that
E_p' = γE_p where the prime indicates the field observed in the system moving relative to capacitor.

In addition, the electric field perpendicular to the direction of motion creates a magnetic field (as suggested by daleSpam) B_p' given by
B_p' = -γ(1/c²)(v x E_p)
So the integral of the stored energy over the volume now becomes

W = (1/2) ∫(E_p' ² + B_p' ²) dV' = (1/2) ∫ (γ²E_p² + (γ²β²/c⁴)(c x E_p)²) dV' (where dV is a volume element)

[Edit] added μ₀ and ε₀
W = (1/2) ∫(ε₀E_p' ² + B_p' ²/μ₀) dV' = (1/2) ∫ (γ²ε₀E_p² + (γ²β²/c⁴)(c x E_p)²/μ₀) dV' (where dV is a volume element)
So
W = (1/2) ∫ (γ²ε₀E_p² + (γ²β²/c²)E_p²/μ₀) dV'

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
∂ C ∏ ∑

**Privalov** · Jun21-09, 10:55 PM

Originally Posted by DaleSpam

You will find all of your missing energy in the magnetic field.

I’m not missing energy; I’m getting infinite amount of it.

Originally Posted by Vanadium 50

Since the energy stored in an electric field is

Wikipedia seems to give slightly different formula:

http://en.wikipedia.org/wiki/Electri...electric_field

Anyhow, this is irrelevant minor correction.

Originally Posted by Vanadium 50

since the field is the same and the volume is the same, energy is conserved.

Volume is constant; however, electric field is not constant in this problem.

Electric field between (indefinitely large) capacitor plates is given by equation

$LaTeX Code: E = Q / A \\epsilon$

Where

is electric field;

is the charge (I believe it’s invariant and should be the same in all frames of reference);

is surface area of the plates;
$LaTeX Code: \\epsilon$ is permittivity.

http://www.ac.wwu.edu/~vawter/Physic...ParallCap.html

Surface area is subject to relativistic length contraction. Area depends on orientation of the capacitor. Therefore, stored energy can increase, when someone turns the capacitor, without any energy intake from external sources. I see a paradox here.

**meopemuk** · Jun22-09, 12:08 AM

Hi Privalov,

for the last 100 years your paradox has been known as the Trouton-Noble paradox. If you search literature, you can find many attempts to explain it from the point of view of classical electrodynamics. All these attempts look rather unconvincing.

**Phrak** · Jun22-09, 12:40 AM

Originally Posted by meopemuk

Hi Privalov,

for the last 100 years your paradox has been known as the Trouton-Noble paradox. If you search literature, you can find many attempts to explain it from the point of view of classical electrodynamics. All these attempts look rather unconvincing.

This would be shocking. Do you have a reference?

**Naty1** · Jun22-09, 08:43 AM

Wikipedia describes this:

http://en.wikipedia.org/wiki/Trouton-Noble_experiment

The Trouton–Noble experiment attempted to detect motion of the Earth through the luminiferous aether, and was conducted in 1901–1903 by Frederick Thomas Trouton (who also developed the Trouton's ratio) and H. R. Noble. It was based on a suggestion by George FitzGerald that a charged parallel-plate capacitor moving through the aether should orient itself perpendicular to the motion.

Sounds like a different effort?

**Naty1** · Jun22-09, 09:10 AM

Privalov posted:

Electric field between (indefinitely large) capacitor plates is given by equation

Where
is electric field;
is the charge (I believe it’s invariant and should be the same in all frames of reference);
is surface area of the plates;
is permittivity.

http://www.ac.wwu.edu/~vawter/Physic...ParallCap.html

Surface area is subject to relativistic length contraction. Area depends on orientation of the capacitor. Therefore, stored energy can increase, when someone turns the capacitor, without any energy intake from external sources. I see a paradox here.

[So why would this frame reference issue by any different than,say, length contraction...what you see depends on your frame??]

The above is basically what I was thinking in my post #2; but I now wonder if infinite plates are a suitable model since they will be infinite in any frame.

Here is what Peter Bergmann, a student of Einstein, says in his 1992 book THE RIDDLE OF GRAVITATION, page 45 which I believe relates. The overall description is quite complex:

..electric charge of a physical system is a scalar whose value is the same in all coordinate systems...the electric charge of an isolated physical system is conserved.... charge density (reflects) different volumes in different Lorentz frames...the charge density is not the same for all observers..

It goes on for another four or five long paragraphs regarding four dimensional vectors and I can't get the gist of it....I am not comfortable trying to summarize it...If anyone has access to the book and knows the math references, I believe it applies to this thread issue.

**Naty1** · Jun22-09, 11:28 AM

And while we are at it: isn't there an electric field in the frame of the capacitor, but no magnetic field, while the moving observer sees both E & M???

Also when charges are in relative motion, and I'm not sure if the observer in motion would observe such motion or not, there would be a current density, a flux. Is that observed or not?? And how might it relate to the OP questions??

**DaleSpam** · Jun22-09, 12:00 PM

There is clearly a magnetic field for the moving observer since the moving charge is, by definition, a current. Since energy is conserved in general for Maxwell's equations the details of the scenario are irrelevant. If you get that energy is not conserved then you have necessarily violated one or more of Maxwell's laws. I think that it is obvious that the OP's analysis violates Maxwell's equations since the energy in the magnetic field is not even considered, but I leave it up to the OP or other interested readers to derive the details.