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Magnetism to test for movement?

  1. Aug 8, 2011 #1
    They say it is impossible to know whether you're moving at constant velocity or at rest unless you look outside.
    How about this:(no gravity please)
    I take two known charges and hold them at a distance one above the other.
    They repel.That is l I know.
    Someone else outside says,that the charges are moving at a velocity v and so he expects a magnetic force along with the electrostatic one.
    Now what really happens(in the both our frames).Do I feel the magnetic force?Why not?

    Suppose I do feel it,I would be able to tell that I'm moving without looking out.Right?
    Thank You
  2. jcsd
  3. Aug 8, 2011 #2


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    That's the whole point of relativity! The force due to a magnetic field on a charged particle is proportional to the speed of the particle. From that, it should be possible to determine an "absolute" speed. So experiments involving electro-magnetic properties were done to try to confirm that. Since light is waves in the electro-magnetic fields, experiments with light, such as the Michaelson-Morley experiment, were performed. They gave a "null result". Even though Maxwell's equations say we should be able to determine an "absolute speed", in fact, we can't. That was what led to the introduction of the theory of relativity.
  4. Aug 8, 2011 #3
    Now suppose we were to find the actual magnetic force.
    Then would it not fetch us an absolute velocity?

    What came out of them?
  5. Aug 8, 2011 #4


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  6. Aug 8, 2011 #5


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    You will never feel a magnetic force from a charge that's stationary with respect to you in your frame.

    To oversimplify slightly, a person in a different frame describes the same physical situation differently, they see both a magnetic and an electric force, but the sum of them does not change. So they describe it differently, but in the end it doesn't make any difference to what's measured.

    I've glossed over something here in the interest of simplicity - actually the total force doesn't change it's numerical value if and only if you use the four-force, and you probably think of force as the three force. This is similar to and a consequence of the fact that lengths and time intevals (rulers and clocks) change and intermix in relativity.

    This makes the calculation a bit harder, but you'll still get the same result, and you'll still only see an electric field in a frame where you and the charge are both at rest.

    If you do a transform to a frame where you and the charge are both moving, the numbers will change according to the specifications of relativity, but they'll change in such a manner that they describe the same actual physical events, just from a different perspective. THis is known as "covariance".
    Last edited: Aug 8, 2011
  7. Aug 12, 2011 #6
    ghwellsjr,I didn't completely understand that.

    Thank You for your explanation.
    But we have a formula of electrostatic force between two charges don't we?
    Are you saying that the constants in that change with the motion.(pi and epsilon?)
    I mean,doesn't the guy in motion have the authority to use the same equation of electrostatic force as the guy outside standing and watching?
    Please answer.
  8. Aug 12, 2011 #7
    No [itex] \mu_0 [/itex] and [itex] \epsilon_0 [/itex] do not change. But in other frames you might have a B field. But
    [itex] E \cdot B [/itex] will be the same in all frames.
    And yes stuff like gauss's law works for frames in motion.
    A better thing to analyze would be 2 line charges moving and have one positive and negative and compute the E field using Gauss's law and the B field with amperes law. And try it in different frames and have a test charge moving by the line charges and see how the forces compare in the 2 frames. And of course you have to use Einstien velocity additions when you talk about moving with respect to other stuff when you do this.
    And as people said above this was how special relativity was born and observers in different frames will get the same prediction but have different interpretations.
    Last edited: Aug 12, 2011
  9. Aug 12, 2011 #8


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    The electric field will be given by columb's law if and only if the charge is stationary in your frame of reference. If the charge is moving in your frame, the electric field will get squashed by the Lorentz contraction, looking something like this:


    http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_15.pdf for the full figure, the image above is a copy of part of the original, copied under the "fair use" provision and intended to inspire you to look at the original author's webpage.

    Note that there is in no sense any special frame of reference- when you move with the same velocity as the charge, i.e. you and the charge are co-moving, the field is symmetrical. When you move at some different velocity, the field of the charge is no longer symmetrical, it gets "squished" as above.

    And the numerical transformation laws are given a few slides earlier:

    the parallel component of the e-field, doesn't change, but the transverse component of the E-field gets multiplied by the relativistic factor gamma.

    Attached Files:

    Last edited by a moderator: Apr 26, 2017
  10. Aug 17, 2011 #9
    Oh Thank You pervect:)
    I am sorry for the late reply,I didn't have easy access to a computer.
  11. Aug 17, 2011 #10
    So in relativity, electromagnetic tensor is usually applied. After contracting indices with metric of moving reference frames, results of electromagnetic transformation is easily achieved. Is it correct to apply metric of different reference frames, since every inertial frame has the same transformation comparing to others?
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