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Capacitor paradox

  1. Jun 21, 2009 #1
    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.

    Attached Files:

  2. jcsd
  3. Jun 21, 2009 #2
    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.
  4. Jun 21, 2009 #3


    Staff: Mentor

    You will find all of your missing energy in the magnetic field.
  5. Jun 21, 2009 #4

    Vanadium 50

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    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 [itex]d/\gamma[/itex] for a total volume [itex]Ad/\gamma[/itex]. Now, rotate it 90 degrees. Now you have an area [itex]A/\gamma[/itex] and a separation of d, again for a total volume [itex]Ad/\gamma[/itex].

    Since the energy stored in an electric field is [itex]E^2V[/itex], 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.
  6. Jun 21, 2009 #5


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    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.
  7. Jun 21, 2009 #6


    Staff: Mentor

    Huh? [itex]Ad/\gamma \ne Ad = V[/itex].
  8. Jun 21, 2009 #7


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    Actually [itex]V = Ad/\gamma[/itex], if I understood correctly.

    Anyway, Vanadium 50 was really just saying [itex]Ad/\gamma = Ad/\gamma[/itex]. Can't argue with that...
  9. Jun 21, 2009 #8


    Staff: Mentor

    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.
    Last edited: Jun 21, 2009
  10. Jun 21, 2009 #9
    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, Ep, is increased by a factor γ, so that
    Ep' = γEp 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) Bp' given by
    Bp' = -γ(1/c2)(v x Ep)
    So the integral of the stored energy over the volume now becomes

    W = (1/2) (Ep' 2 + Bp' 2) dV' = (1/2) 2Ep2 + (γ2β2/c4)(c x Ep)2) dV' (where dV is a volume element)

    [Edit] added μ0 and ε0
    W = (1/2) (ε0Ep' 2 + Bp' 20) dV' = (1/2) 2ε0Ep2 + (γ2β2/c4)(c x Ep)20) dV' (where dV is a volume element)
    W = (1/2) 2ε0Ep2 + (γ2β2/c2)Ep20) 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 ∏ ∑
    Last edited: Jun 21, 2009
  11. Jun 21, 2009 #10
    I’m not missing energy; I’m getting infinite amount of it.

    Wikipedia seems to give slightly different formula:


    Anyhow, this is irrelevant minor correction.

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

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

    [itex]E = Q / A \epsilon[/itex]

    [itex]E[/itex] is electric field;
    [itex]Q[/itex] is the charge (I believe it’s invariant and should be the same in all frames of reference);
    [itex]A[/itex] is surface area of the plates;
    [itex]\epsilon[/itex] is permittivity.

    http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Capacitors/ParallCap.html [Broken]

    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.
    Last edited by a moderator: May 4, 2017
  12. Jun 21, 2009 #11
    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.
  13. Jun 21, 2009 #12
    This would be shocking. Do you have a reference?
  14. Jun 22, 2009 #13
    Wikipedia describes this:


    Sounds like a different effort?
    Last edited: Jun 22, 2009
  15. Jun 22, 2009 #14
    Privalov posted:

    [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:

    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.
    Last edited by a moderator: May 4, 2017
  16. Jun 22, 2009 #15
    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??
    Last edited: Jun 22, 2009
  17. Jun 22, 2009 #16


    Staff: Mentor

    There is clearly a magnetic field for the moving observer since the moving charge is, by definition, a current. Since http://farside.ph.utexas.edu/teaching/em/lectures/node89.html" [Broken] 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.
    Last edited by a moderator: May 4, 2017
  18. Jun 22, 2009 #17
  19. Jun 22, 2009 #18
    In my post above, I found that a) there is a magnetic field for the moving observer, and b) the stored energy may not be conserved. Specifically
    I showed
    W = (1/2) 2ε0Ep2 + (γ2β2/c2)Ep20) dV'
    which becomes, when substituting ε0μ0=1/c2

    W = (1/2)γ2ε0(1+β2)Ep2 dV'
    The real problem is whether there should be a + sign in (1+β2)? It arises from summing the stored energy from electric (E2) and magnetic (B2) fields. To make the stored energy independent of rotation, it should be a - sign.

    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
    Last edited by a moderator: May 4, 2017
  20. Jun 22, 2009 #19


    Staff: Mentor

    There must be a mistake then. Energy is provably conserved in general as I linked to above, so the specific situation does not matter. Also, the total energy is the sum of the energy in the electric and magnetic fields. That is what was proven to be conserved.
  21. Jun 22, 2009 #20
    According to your assumption, the energy of a moving capacitor depends on the orientation of its plates. This means that there should be a torque which tends to rotate the moving capacitor to a preferable orientation with respect to the movement direction, so that the energy is lowered. This is exactly the Trouton-Noble idea.

    Here are some references:

    F. T. Trouton and H. R. Noble, "The Mechanical Forces Acting on a Charged Electric Condenser Moving through Space", Phil. Trans. Roy. Soc. London A, 202 (1904), 165.

    G. Spavieri and G. T. Gillies, "Fundamental tests of electrodynamic theories: Conceptual investigations of the Trouton-Noble and hidden momentum effects",
    Nuovo Cim., 118B (2003), 205.

    J. Franklin, "The lack of rotation in the Trouton-Noble experiment", http://www.arxiv.org/abs/physics/0603110v3

    S. A. Teukolsky, "The explanation of the Trouton-Noble experiment revisited", Am. J. Phys., 64 (1996), 1104}

    O. D. Jefimenko, "The Trouton-Noble paradox", J. Phys. A: Math. Gen., 32 (1999), 3755.

    J. D. Jackson, "Torque or no torque? Simple charged particle motion observed in different inertial frames", Am. J. Phys., 72 (2004), 1484.
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