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An electromagnetic paradox?

  1. Sep 26, 2013 #1
    Consider the system below comprising a pair of oppositely charged parallel plates, connected by a rigid rod of length [itex]r[/itex], constrained to move in the x-direction.

    aVhRr.jpg

    The forces on the plates from the Coulomb electric fields and the magnetic forces are all in the vertical direction so that they do not affect the motion of the system.

    The only fields that do affect the system are the horizontal radiation fields due to the acceleration of the charged plates.

    Let us assume that we initially apply a force [itex]F_0[/itex] so that we have:

    [tex]F_0 = m a_0[/tex]

    As charge [itex]+q[/itex] has an acceleration [itex]a_0[/itex] then it will produce a horizontal electric field at the negative plate given by:

    [tex]E_+ = \frac{-qa_0}{4 \pi \epsilon_0 c^2 r}[/tex]

    This will lead to a horizontal force [itex]f_0[/itex] on the negative plate, pointing towards the right, given by:

    [tex]f_0 = \frac{q^2a_0}{4 \pi \epsilon_0 c^2 r}[/tex]

    Likewise as charge [itex]-q[/itex] has an acceleration [itex]a_0[/itex] then it will produce a horizontal electric field at the positive plate given by:

    [tex]E_- = \frac{qa_0}{4 \pi \epsilon_0 c^2 r}[/tex]

    Again this will lead to a horizontal force [itex]f_0[/itex] on the positive plate, pointing towards the right, given by:

    [tex]f_0 = \frac{q^2a_0}{4 \pi \epsilon_0 c^2 r}[/tex]

    Thus the total force on the system is now given by:

    [tex]F_1 = F_0 + 2f_0[/tex]

    Therefore the system has an increased acceleration given by:

    [tex]a_1 = \frac{F_1}{m}[/tex]
    [tex]a_1 = a_0 \left(1+\frac{q^2}{2\pi\epsilon_0c^2rm}\right)[/tex]

    The argument can be repeated indefinitely leading to the calculated acceleration diverging exponentially.

    What's gone wrong?

    There are no [itex]\dot{a}[/itex] terms in the calculation so I don't think the Abraham-Lorentz reaction force due to the production of electromagnetic radiation can save the day. A positive feedback involving forces and accelerations seems to preclude a consistent initial value for the acceleration.

    Here's my (rather radical) solution: perhaps the mass has to change to ensure that the acceleration stays constant.

    Perhaps we have:

    [tex]F_0 = m_0 a_0[/tex]
    [tex]F_1 = m_1 a_0[/tex]
    Therefore:

    [tex]\frac{m_1}{m_0} = \frac{F_1}{F_0}[/tex]
    [tex] \frac{m_1}{m_0} = \frac{1}{m_0a_0}\left(m_0a_0+\frac{q^2a_0}{2\pi\epsilon_0c^2r}\right)[/tex]
    [tex]m_1 = m_0 + \frac{q^2}{2\pi\epsilon_0c^2r}[/tex]
     
    Last edited: Sep 26, 2013
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  3. Sep 26, 2013 #2

    UltrafastPED

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    These are not static fields unless you are in the comoving frame.

    A relativistic treatment is required ... try it with a constant acceleration and see what you get.
     
  4. Sep 26, 2013 #3
    Can I just conduct the argument in a comoving frame then?

    I don't know how to do the argument relativistically with a constant acceleration.
     
  5. Sep 26, 2013 #4

    UltrafastPED

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    Last edited by a moderator: May 6, 2017
  6. Sep 26, 2013 #5

    Jano L.

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    Your calculation of the EM force neglects retarded character of EM field. If you calculate with exact expression for the retarded EM field, you will get many new terms that will influence the value of the EM force on the plates. In the first instant of period when the force is applied, the EM force has no acceleration component, so the acceleration is determined by F/m. The acceleration field kicks in only after time delay r/c.

    After this delay, the acceleration field acts on the plates, but then there is also the velocity dependent field, which has opposite direction. Perhaps it is this velocity field that prevents the acceleration to get too high.
     
  7. Sep 26, 2013 #6
    Thanks for the reply.

    I don't think the retardation time changes the effect. Also I think one could make the argument in a comoving frame where the velocities of the plates are zero.

    I think the best way to imagine the experiment is to simply drop the system in a gravitational field. The induced electric forces would cause the system to accelerate faster than gravity which of course is forbidden by the equivalence principle.
     
  8. Sep 26, 2013 #7

    UltrafastPED

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    Note that the comoving frame is stationary wrt the plates ... so you would both fall together.

    Then by the equivalence principle what would you expect to see?
     
  9. Sep 26, 2013 #8
    To be honest I'm not quite sure of the definition of comoving frame actually.

    Is it an inertial frame that has the same instantaneous velocity as the plates or is it a frame that has both the same velocity and the same acceleration as the plates?

    If it's the later then I expect the gravitational force + electromagnetic force on the plates to be balanced by an inertial force due to the plates' standard mass + an extra inertial force due to the plates' "electromagnetic mass".
     
    Last edited: Sep 26, 2013
  10. Sep 26, 2013 #9

    Jano L.

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    The retardation does not change the acceleration field much if the acceleration is constant, but it plays role in the other, velocity-dependent terms. You have to consider them thoroughly. When you do, you will most likely find that there is no such run away behaviour.



    Adding gravitational field means adding theory of gravity (you mention equivalence principle), which means things will get more unnecessarily complicated. It is better to consider your problem in an inertial frame, with external force F acting on the system.



    The other part of the problem in your consideration is, I think, the iterative way you use to find the acceleration. Consider this equation:

    $$
    ma = F + ka,
    $$

    which suggests this iterative equation:

    $$
    a_{n+1} = a_{n} + ca_{n}
    $$
    with ##a_0 = F/m##, ##c = k/m.##

    Using your iterative procedure, we would get

    $$
    a_1 = a_0 + c a_0 = a_0(1+c)
    $$
    $$
    a_2 = a_1 + ca_1 = a_0(1+c)^2
    $$
    ...

    and obviously sequence ##a_{n}## goes to infinity.

    However, the correct solution of the original equation is ##a = F/(m-k)##.

    Obviously the iterative way does not lead to correct solution.
     
  11. Sep 26, 2013 #10

    Dale

    Staff: Mentor

    Why would you think that?

    You don't have any sort of indefinite number of forces to repeat anything with. You have two forces, the mechanical force and the EM force. So your net force has two terms, one of which is a function of the acceleration. Plug in your net force on one side and ma on the other side and you have one equation in one unknown, a. Solve for a.

    That is it. No diverging exponential acceleration. This is no different from any other f=ma problem except that solving for a involves a little more algebra than simply dividing both sides by m.
     
    Last edited: Sep 26, 2013
  12. Sep 26, 2013 #11
    The argument can be repeated indefinitely leading to the calculated acceleration diverging exponentially.

    If the effect was instantaneous then as Jano says one would just solve the equation:
    [tex]m a = F + k a.[/tex]
    I think there is an iterative, positive feedback, effect here because of the time delay [itex]dt[/itex] between the plates accelerating and the subsequent electromagnetic force on the plates.

    [tex]F(t+dt) = F(t) + k a(t)[/tex]
    [tex]F(t) = m a(t)[/tex]
    [tex]F(t+dt) = m a(t+dt)[/tex]
    Thus we have
    [tex]a(t+dt) = a(t)(1+k/m)[/tex]
    [tex]\int da/a = \int k/m dt[/tex]
    [tex]a(t)=a_0e^{kt/m}[/tex]
     
  13. Sep 26, 2013 #12

    Dale

    Staff: Mentor

    The acceleration is uniform, is it not (it must be given the expression you used for the field)? Then the delay time is not important because nothing has changed during the propagation.

    Again, you don't have any sort of indefinite number of forces to repeat anything with. You have two forces, the mechanical force and the EM force. So your net force has two terms, one of which is a function of the acceleration. Plug in your net force on one side and ma on the other side and you have one equation in one unknown, a. Solve for a.

    That is it. No diverging exponential acceleration. This is no different from any other f=ma problem except that solving for a involves a little more algebra than simply dividing both sides by m.
     
    Last edited: Sep 26, 2013
  14. Sep 26, 2013 #13

    BruceW

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    this is an interesting 'problem'... The slight nudge will cause a slight change in the electric field so at a time ##r/c## later, each charge will feel (for a short time) an acceleration in the horizontal direction. and this acceleration causes another slight change in the electric field, which causes another acceleration at time ##r/c## later. So the charges will keep doing these little accelerations at every ##r/c## time interval.

    I think maybe the answer is simply that in the case when there is only one charge, when you give it a nudge, some of that energy just gets radiated outwards, and never comes back. But when you have the two charges in this set-up, some of that radiated energy will not escape to infinity, and it is this that causes the little 'hops' forward. I don't think the accelerations will diverge. I suspect the accelerations will slowly die down with time.

    Anyway, I'm not sure if any of this is right. It is just my first guess.
     
  15. Sep 26, 2013 #14

    Jano L.

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    Aha, so you do consider delays. That is better, but the above equation is not correct. You simply have to take the full expression for the electromagnetic force and then justify any particular simplification. The EM force is not just ka(t).
     
  16. Sep 26, 2013 #15

    Dale

    Staff: Mentor

    We are guaranteed that the accelerations will not diverge by Poynting's theorem.

    Under the steady acceleration case you have simply a f=ma where f is a function of a. One equation in one unkown. Obviously no diverging.

    In the impulsive acceleration case you will have to solve more complicated equations, but you are guaranteed that they will not diverge either.
     
  17. Sep 26, 2013 #16
    Having read your post about delays I now think they are an essential part of the argument. The delay between acceleration and electromagnetic force is what causes the positive feedback.

    I think my analysis is ok for non-relativistic velocities - I admit I don't know how to make a more accurate calculation.
     
  18. Sep 26, 2013 #17
    If you want to do it iteratively, you need to consider that with the new acceleration you only need to add the extra force coming from the field from the extra acceleration, but you added the whole force due to the field after every step, while you should have added just the extra force.
    So it is like [itex]a_{1}=a_{0}+a_{0}\times k/m[/itex], then [itex]a_{2}=a_{1}+(a_{1}-a_{0})\times k/m[/itex] and not [itex]a_{2}=a_{1}+a_{1}\times k/m[/itex], then [itex]a_{3}=a_{2}+(a_{2}-a_{1})\times k/m[/itex] etc and see where the values converge. Or the easier way is to do it the way as was pointed out before, [itex]a=a_{0}+a\times k/m[/itex] and solve for a.
    It does seem weird that the system seems to push itself. I think that if the charges give away radiation and energy, then this needs to be accounted for and would result in additional forces. However googling the radiation of an uniformly accelerating charge I get conflicting answers and it gets over my head into general relativity and stuff.
     
  19. Sep 27, 2013 #18
    Maybe I can just say that I start with an initial acceleration [itex]a_0[/itex] so that we have

    $$m a_n = k a_{n-1}$$
    $$a_n = (k/m)a_{n-1}$$
    $$a_n = (k/m)^n a_0$$

    Each time step takes a time [itex]r/c[/itex] so that we can write:
    $$a(t)=\left(\frac{k}{m}\right)^{ct/r}a_0$$
     
    Last edited: Sep 27, 2013
  20. Sep 27, 2013 #19

    BruceW

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    Just to make sure, 'k' is
    ##k=\frac{q^2}{2\pi\epsilon_0c^2r}##
    right? yeah, the iterative law ##a_n=(k/m)^na_0## looks right. But I am not sure about your last equation. I guess you are assuming the two charges are very close or something?

    Anyway, your equation ##a_n=(k/m)^na_0## is very insightful. If you think about it, to have a non-relativistic limit, we must have k<<m, so therefore the accelerations will get smaller and smaller. And if we have k to be the same order as m, then we have the relativistic case, and in that case, the equation
    ##E_+ = \frac{-qa_0}{4 \pi \epsilon_0 c^2 r}##
    Is not correct anyway, so the equation ##a_n=(k/m)^na_0## is also likely to be incorrect in the relativistic case.

    edit: to be clear, in the relativistic case, the relativistic mass of the charges will change, and other weird stuff will happen, so we would not be allowed to use the nice assumptions that we have been using. Although, maybe the equation
    ##E_+ = \frac{-qa_0}{4 \pi \epsilon_0 c^2 r}##
    will still work... It assumes that the change in velocity is small compared to the speed of light. So I guess this equation can still be true even when k>m
     
    Last edited: Sep 27, 2013
  21. Sep 27, 2013 #20
    If you are using iteration, then it should go something like this: the force [itex]m\times a_{0}[/itex] pushes the system, assume it to start accelerating with [itex]a_{0}[/itex]. But since the acceleration causes extra force which is proportional to the acceleration, you have an additional extra force [itex]k\times a_{0}[/itex]. Combining these forces you have a new acceleration [itex]a_{1}[/itex]. Now considering the system accelerating at [itex]a_{1}[/itex], you must now not add an entire new force from the electric field with magnitude [itex]k\times a_{1}[/itex], instead you have to add only the extra force that comes from the increased acceleration, so that it's magnitude is [itex]k\times (a_{1}-a_{0})[/itex]. You have to account for the increased force from the electric field, not add a whole new force, because otherwise you would count the same force many times.
    If you solve it, it is equivalent to solving this equation for a [itex]m\times a=m\times a_{0}+k\times a[/itex].

    However this whole analysis is problematic, because energy and momentum seem to not be conserved if it simply pushes itself. You have to account for the radiated energy and momentum as well, which should cause extra forces.
     
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