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Falling charged objects paradox?

  1. Feb 4, 2015 #1
    I have been thinking about a simple thought experiment in classical electromagnetism
    that seems to disobey conservation of energy.

    I'd be very interested to hear where people think I'm going wrong.

    Imagine that we have two oppositely charged objects with charges [itex]+q[/itex], [itex]-q[/itex] and
    masses [itex]m[/itex].

    Assume that they slid without friction on two vertical poles placed at a
    distance [itex]d[/itex] apart.

    First we lift them from the ground up to a height [itex]h[/itex] so that each one has a
    potential energy [itex]mgh[/itex].

    Now we let them drop simultaneously.

    The equation of motion of each object (in CGS units for clarity) is given by:

    [tex]m a = - mg + \frac{q^2}{d\ c^2} a + \frac{2}{3}\frac{q^2}{c^3} \dot{a}[/tex]

    The first term is the weight of the object.
    The second term is the force due to the electric field induced by the
    acceleration of the other charged object.
    The third term is the Abraham-Lorentz radiation reaction force.

    The above equation of motion has a constant acceleration solution given by:

    [tex]a = \frac{-g}{1 - q^2 / d\ m c^2}[/tex]

    The work done by each charged object as it falls a distance [itex]h[/itex] is given by:

    W = Force [itex]\times[/itex] distance

    [tex]W = \frac{-mg}{1 - q^2 / d\ m c^2} * -h[/tex]

    [tex]W = \frac{mgh}{1 - q^2 / d\ m c^2}[/tex]

    Thus the energy we get out is more than the energy that we put into the system in the first place.

    What's wrong with the calculation?
     
  2. jcsd
  3. Feb 4, 2015 #2
    May be the field Energy should be included in the analysis. I think there is a transitory period from a=0 to a=final value and I don't know if this has something to do with energy conservation. Sorry can not help you more
     
  4. Feb 4, 2015 #3
    It looks like you are mixing up directions. Acceleration has a direction and magnitude. Gravity accelerates things downward (by definition). Which way is the electric force pointed?
     
  5. Feb 4, 2015 #4
    I think the instant the objects are released they have acceleration [itex]\frac{-g}{1-q^2/d\ mc^2}[/itex]. This acceleration stays constant so no energy is radiated away into the field.
     
  6. Feb 4, 2015 #5
    The electric force is [itex](q^2/dc^2) a[/itex]. As the constants are positive then the electric force is in the same direction as the acceleration. The acceleration [itex]a[/itex] is negative (downwards) so the electric force is negative (like the gravitation force [itex]-mg[/itex]).
     
    Last edited: Feb 4, 2015
  7. Feb 5, 2015 #6
    Yes, I think you can impose this initial condition
    From the moment ##a\ne 0## you have radiation


    It is not obvious to me, from Lienard-Wiechert fields, that the electric force is in the same direction as acceleration. Besides you have a magnetic force
     
  8. Feb 5, 2015 #7
    From consideration of the Abraham-Lorentz reaction force, ##F_{rad}=(2q^2/3c^3)\ddot{v}##, I would say that charges only radiate electromagnetic energy into space when the power supplied to the charge by the Abraham-Lorentz force is negative i.e when:
    [tex]P_{rad}=F_{rad}v \propto \ddot{v}v < 0[/tex]
    This condition is met by an oscillating charge or a moving charge stopped abruptly by a target.

    This condition is not met in the present case of a falling charge where ##\ddot{v}## and ##v## have the same sign initially. In principle it is possible that ##P_{rad}>0## implying that the charges absorb converging electromagnetic radiation from space but this is not physically appropriate in this case . Thus the only option left is ##\ddot{v}=0## with no power radiated.

    The Lienard-Wiechert fields have two components: the static and radiation fields. The static electric and magnetic fields only produce horizontal forces. The radiation magnetic field force only acts in a horizontal direction too. This only leaves the electric radiation field. Because the charges are opposite each electric radiation field produces a force on the other charge that acts downwards.
     
    Last edited: Feb 5, 2015
  9. Feb 5, 2015 #8
    I was not considering the reaction force. I think this problem may be too complicated for me. Any way I think the Lienard-Wiechet electric field has a component in the vertical direction and the magnetic field lies in the horizontal plane. Let's wait and see if someone answer your question.
     
  10. Feb 5, 2015 #9
    By the way, as far as I understand it, Einstein's principle of equivalence states that an object falling in a gravitational field is equivalent to an object that is at rest in space but with the ground accelerating up to it. In that case no electric forces would be acting on the charged objects implying that they can only fall with the acceleration due to gravity, ##g##, in the equivalent Newtonian frame in which the ground is not accelerating. But in that case why are there no electric forces acting on the charged objects to accelerate them faster than ##g##?
     
    Last edited: Feb 5, 2015
  11. Feb 5, 2015 #10
    I do make the approximation that the objects do not move an appreciable distance in the time it takes the electromagnetic field to propagate from one charge to the other. This implies that any vertical component to the static fields is very small.
     
  12. Feb 5, 2015 #11
    Ok, now I understand
    All I know is that exist a long debate on this point, there is a whole book treating this problem "Uniformly Accelerating Charged Particles" by Stehen Lyle(Springer)
     
  13. Feb 5, 2015 #12
    Thanks for the reference!

    Stephen Lyle has a website:

    http://www.stephenlyle.org/
     
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