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Electrodynamics and Relativity Paradox?

  1. Apr 12, 2005 #1
    I have been thinking about this problem for a few days and it's been keeping me up; I must be missing something, I just don't know what. I would appreciate any one shedding some light on this:

    Assume an inertial frame of reference and all external fields are negligible.

    1. Say you have two equally charged particles traveling next to one another at a constant velocity. The velocity is tuned such that the repulsive force due to the electric field exactly balances the attractive force due to the magnetic field (created because the charges are moving). These particles should continue to travel parallel to one another.

    2. Now say you have the same situation as in (1) but the particles start with zero velocity. The particles will immediately fly apart due the repulsive force of the electric fields. There will be no magnetic field (to begin with) because the charges are not moving.

    3. Finally we can convert situation (1) into situation (2) by simply changing the frame of reference to that of the moving particles. For example surround the moving particles with a box that is traveling at the same velocity. To an observer inside the box the particles will appear to be stationary.

    What will happen to the particles in (1) when they are observed in (3)?
    A. Will they continue to appear to stand still due to the balanced forces, even though there should be no magnetic field generated by stationary charges?
    B. Will they suddenly fly apart because there is no magnetic field just because they now appear to be stationary to this observer?
    C. Something else?

    Given the choice, I would have to go with (A), but if that were true, that means that you should be able to measure “absolute velocity” by tuning the charge values until two identical particles at rest appeared to stop repelling each other. This can’t be right.

    According to Special Relativity (as I understand it), "The laws of physics are the same in any inertial frame, regardless of position or velocity". Since this is not the case in this situation, it appears to be a paradox, what am I missing? It’s probably some thing trivial isn’t it :-).

    Thanks for any help in understanding.
     
  2. jcsd
  3. Apr 12, 2005 #2

    jtbell

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    Have you tried calculating this velocity? :wink:

    In case you need it, you can find the equation for the magnetic field produced by a moving point charge on this page, a bit more than halfway down.
     
  4. Apr 12, 2005 #3

    Meir Achuz

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    1. The magnetic force between two particles cannot cancel the electric force. For the situation you describe, a relativistic calculation of the effect of the electric and magnetic fields of the particles gives
    [itex]\frac{d{\vec p}}{dt}=\frac{q^2{\vec r}}{\gamma r^3}.[/itex]
     
  5. Apr 12, 2005 #4

    pervect

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    This can't happen, because, as you point out, in the rest frame the particles will repel each other if they have the same charge.

    Perhaps you are thinking that the E-field remains constant for a moving particle. This is not the case - the correct formulas (which show that the transverse fields are boosted by a factor of gamma) are given

    on this page
     
  6. Apr 12, 2005 #5

    jtbell

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    In fact, both the relativistic E and B fields are increased by a factor of [itex]\gamma[/itex] over the non-relativistic formulas, if we confine our attention to the line that joins the two particles and is perpendicular to their direction of motion. The magnitudes are

    [tex]E = \gamma \frac {1}{4 \pi \epsilon_0} \frac {q} {d^2}[/tex]

    [tex]B = \gamma \frac {\mu_0} {4 \pi} \frac {q v} {d^2}[/tex]

    So if you calculate the electric and magnetic forces from those fields, and set them equal to each other, the [itex]\gamma[/itex]'s cancel. Therefore, if you start with the non-relativistic formulas, you fortuitously get the same result for [itex]v[/itex]. Interesting...
     
  7. Apr 12, 2005 #6

    axl

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    In the particle frame, the particles have no relative velocity, so there is no current. The particles will only be repeled by one another. You can evaluate the physics in the most convenient frame.
     
  8. Apr 12, 2005 #7

    Meir Achuz

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    "if you start with the non-relativistic formulas, you fortuitously get the same result for . Interesting..."
    The relativistic result for dp/dt is given in my previous post.
    It differs from the NR answer by a factor of gamma.
    In any event, for neither answer can the B effect cancel the E effect.
     
  9. Apr 12, 2005 #8

    krab

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    Your questions are good. If you calculate the velocity you need to exactly cancel the repulsive E force with the attractive B force, you'll find it is exactly the speed of light. I think if you think these things through (impossibility of massive objects reaching the speed of light, time dilation, etc.), all your paradoxes are resolved.
     
  10. Apr 12, 2005 #9
    Here are some applets that help visualize the problem, do they appear to be accurate?

    http://www.falstad.com/vector3de/
    http://www.falstad.com/vector3dm/

    Ok, lets ignore the force balancing and reduce the problem to one moving charged particle.

    First, I guess I don't understand how the electric field from a moving charge won't be constant. From the given Electromagnetic Field Tensor equations, it appears that [tex] \gamma [/tex] is constant for a constant velocity and so are the E and B components. Is the particle affected by it's own fields?

    In any case, am I to understand that the magnetic field is entirely a manifestation of relative velocity of a moving charge? i.e. In the particle reference frame there will be no magnetic field and at the same time to an outside observer there will be a magnetic field?

    So, if you had a hall effects sensor (that didn't have any effect on our particle) sitting in the box next to the particle, it would read a constant zero, while an outside sensor that was stationary relative to the external frame would see a pulse as the particle flew by?

    Now that I write it down, it actually makes more sense than what I was thinking before: That the magnetic field exists in it's own right the way the electric field does.

    It seems that the magnetic field is nothing more than an abstraction used to explain the behavior of the relative motion of the electric field.

    Is this correct or have I totally missed the point?
     
  11. Apr 12, 2005 #10
    SystemCrawler,

    "In any case, am I to understand that the magnetic field is entirely a manifestation of relative velocity of a moving charge?"

    You got it! And if you had figured that out 100 years and 1 month ago, you'd be famous instead of Einstein (well, he'd still be famous!).
     
  12. Apr 12, 2005 #11
    I agree that these classical eqns.will suffice since B fields develop even at low (non relativistic) velocities.

    However, what is necessary is the Lorentz relation. In the lab frame of the observer each charge moves in the E and B field of the other charge.
    The direction of the B field is perpendicular to v and to the radius.
    However, the resultant force from B is along the radius between the 2 charges in the positive direction (attractive), given by:

    [tex]F_B=qv X B[/tex], where [tex]X[/tex]is the cross product.

    The force exerted between the two charges by the E field is simply(repulsive):
    [tex]F_E=-qE[/tex]

    So the total perpendicular force F between the 2 charges is simply the Lorentz force:

    [tex]F=-qE + qv X B[/tex]

    Sustituting in for each value of E and B from the eqns. which were correctly given by jtbell, (without the [tex]\gamma[/tex] factor!):

    [tex]F = -q( \frac {1}{4 \pi \epsilon_0} \frac {q} {d^2}) + qv X (\frac {\mu_0} {4 \pi} \frac {q v} {d^2})[/tex]

    Setting the force equal to 0 and solving for velocity v :

    [tex]v = \frac {1}{\sqrt{\epsilon_o\mu_o}}= c[/tex]

    which is just the speed of light.
    IOW, the attractive magnetic force between the 2 charges equals the electric force at the velocity of light, (as stated by Krab).
    Interesting...
    Of course, this is reasonable since two photons in the same direction have no force between them-- at least not classically! :tongue:
    Creator :biggrin:

    --I don't suffer from insanity; I enjoy every minute of it.--
     
    Last edited: Apr 12, 2005
  13. May 30, 2007 #12
    This post has no answer to the problem !
    I'll try to put it in a easier way:
    Two system: A and B with a relaive velocity v
    In system A there's a charge.
    In system B there's a compass
    What happens whith the compass in B's system?
    From the A's viewpoint there's no magnetic field around...
    From the B's viewpoint the moving charge in A is creating a magnetic field, so the compass is oriented with it.

    How can we solve this paradox?
    Thanks for revive this post!
    enri.
     
  14. May 30, 2007 #13

    Claude Bile

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    The simple conclusion is that different observers in different reference frames will see different E and B fields due to that charge. This is essentially why we refer to E and B fields collectively as electromagnetic fields because they are really the same thing viewed under different reference frames.

    Claude.
     
  15. May 31, 2007 #14

    rbj

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    is that velocity not c? at least i know it is c when it is two infinite and parallel lines of charge (which i think makes for a better and simpler thought experiment) and i expect it to be the same in this case.
     
  16. May 31, 2007 #15
    OK, therefore, A (stationary) will see the charge is radiating, and B (co-moving with the accelerated charge) will see the carge isn't radiating.
    Then:
    A will see photons
    B will not see photons.
    Now, let me add something to the experiment, imagine a wall that can be perforated by a photon.
    A will see a hole and B won't see it???
     
    Last edited: May 31, 2007
  17. May 31, 2007 #16

    Claude Bile

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    A charge needs to be accelerated to radiate energy.

    Claude.
     
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