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I Thought experiment: charged particles

  1. Apr 15, 2017 #1
    Suppose we have two charged particles on the laboratory and two observers A and B.

    ##A## is inside one of the charges (never mind how)
    ##B## is sitting at the laboratory

    In the lab reference frame we accelerate the particles. According to ##A## there will be only electric attraction or repulsion, depending on the sign of the charges. But according to ##B## there also will be a magnetic force due to the movement of the particles.

    Now suppose we stop the particles from moving (according to lab frame) and we measure their relative distance. Both observers ##A## and ##B## have to find out the same distance, but since ##F = ma##, in the frame that there were two forces (electric and magnetic forces) there was a greater acceleration and consequently the charges would be more distant from each other.

    How do we solve this problem?
     
  2. jcsd
  3. Apr 15, 2017 #2

    David Lewis

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    Even if acceleration is the same and starts at the same time for both particles, A and B will not agree on what at the same time means.
     
  4. Apr 15, 2017 #3

    Dale

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    A's frame is non inertial. None of the usual reasoning holds. I am not even sure that there is a way even in principle to separate electric and magnetic fields in such a frame.
     
  5. Apr 15, 2017 #4
    You are correct, after starting this thread I considered the situation more carefully. See below
    So are you saying that even that the particles don't move at all in ##A##'s frame, we still would have a magnetic field in that frame?

    Consider the following:

    In classical theory the time and (consequently) acceleration of the particles will be the same in both frames. So if ##F = ma## is still valid in this thought experiment, then the force of interaction between the particles is the same in both frames. But in ##B##'s frame there is a additional force, namely the magnetic force. The solution is that the electric field measured in ##B##'s frame differs from the one measured in ##A##'s frame. It's not surprising, because I guess since before Relativity coming up, physicists knew that a electrostatic field -the one that we can measure using Coulomb' theory- is different from the electric field of a moving charge.

    Now, let ##t_o##, ##E_o##, ##m_o##, ##a_o## be the time, electric field, mass and acceleration of a particle in ##A##'s frame and quantities without a subscript be the quantities in ##B##'s frame. Additionaly, let the particle charge be ##q## -it's the same in both frames-.

    We know that ##dt = \gamma dt_o## and so, ##a = a_o / \gamma##. If ##F = ma## is true, then ##E = E_o / \gamma##.
    Therefore for ##B## we have ##q( \gamma E_o + v \times B) = m a_o / \gamma##. Divinding by the force in ##A##'s frame and arranging terms, we get
    $$-v^2 \ \gamma \ \vec{E_o} = \vec{v} \times \vec{B}$$ where I've taken ##\vec{v}## to be the velocity of both particles relative to the lab.

    Is this correct?

    EDIT: Just realized that the electric field in ##B## would be ##E_o / \gamma##, and so the conclusion is that the magnetic field should be zero in the lab frame. So is relativity showing us that the magnetic field doesn't exist?
     
    Last edited: Apr 15, 2017
  6. Apr 15, 2017 #5

    Dale

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    No, I am saying that in non-inertial frames there isn't always something that you can even identify as a magnetic field.

    This is not true in non-inertial frames.

    It is not valid in non-inertial frames, that is what makes it non-inertial.

    None of this is necessarily correct in non-inertial frames. You actually have to work out each one of these expressions to see if they are valid in the specific non-inertial frame under consideration. You simply cannot use arguments based on inertial frames when working with non-inertial frames.
     
  7. Apr 15, 2017 #6
    Why the particle frame is non inertial? As its velocity is constant, I don't see why it's non inertial.
     
  8. Apr 15, 2017 #7

    Dale

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    Please clarify
     
  9. Apr 15, 2017 #8

    Nugatory

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    Because earlier you said "In the lab reference frame we accelerate the particles.", which is not consistent with the lab frame being inertial and the frame in which the particles are at rest also being inertial and the particle frame having a constant velocity. Only one of them can be inertial, and it will be the one in which objects at rest do not experience proper accelerations.
     
  10. Apr 15, 2017 #9
    Oh, ok. What I mean is in the lab frame we accelerate the particles up to a velocity ##v##, possibly with their velocities going from zero to ##v## in a very short time.
     
  11. Apr 15, 2017 #10

    Dale

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    So then the reference frame of the particles is non inertial. That is A's frame. The velocity is not constant.
     
  12. Apr 15, 2017 #11

    Nugatory

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    If you're looking at the problem that way, you can limit your attention to the period of time between the two accelerations. Solve the problem once using an inertial frame in which the lab is at rest, and again using an inertial frame in which the particles are at rest after they have accelerated (which is to say that before the first and after the second acceleration they are moving along with the lab). The amount of time that the particles are in motion and subject to magnetic forces will be different in the two frames, the magnitude of the electrical field will be different in the two frames, the magnitude of the magnetic field will be different in the two frames.... And all of these differences will come together in such a way that the invariants such as proper acceleration and proper time under acceleration are the same.
     
  13. Apr 15, 2017 #12
    This is what I did in post #4. Or that analysis of the problem isn't correct?
    The velocity is constant after some time
     
  14. Apr 15, 2017 #13

    Nugatory

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    Is it? There you assumed that the times and the coordinate accelerations would be the same in both frames.
     
  15. Apr 15, 2017 #14
    No, please read it again.
     
  16. Apr 15, 2017 #15

    David Lewis

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    Do you mean classical Newtonian theory of motion, or classical (non-quantum) Relativity?
     
  17. Apr 15, 2017 #16
    non - relativistic theory, ie. Newtonian theory
     
  18. Apr 15, 2017 #17

    Dale

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    Constant after some time isn't the same as constant. If you want to specify an inertial frame then you should say "moving uniformly" from the beginning.

    Plus you add this in, to further make the scenario non inertial.

    Why? Inertial frames don't have to agree on distance. So are you going back to a non inertial frame whose distances are defined the same as the lab frame?


    So now you are going back to inertial frames where distances are in fact different.

    Your scenario is hopelessly muddled now. I would recommend starting over in a new thread with a much more carefully arranged and described scenario.
     
    Last edited: Apr 15, 2017
  19. Apr 16, 2017 #18
    Perhaps it would be easier to describe and analyze if you used the principle of equivalence backwards:
    Two charges are resting in an upper floor of a tall building.
    Then the support is removed and the charges fall to the ground, where they stop.
    It makes it clear that the charges don't feel a magnetic field at any time.

    @davidge , you may want to write out a few intermediate steps of your derivation, I don't see how the "arranged" equation follows from what is said before.
    Also if you start another thread, don't name the lab frame B, or use the phrase "frame B" instead of just "B".
     
  20. Apr 16, 2017 #19

    Dale

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    @davidge it is perfectly acceptable to not attach a reference frame to the charge. You can simply state that there is a lab frame and a moving frame, both inertial, with relative velocity v. Then you can fully describe the acceleration and deceleration in one inertial frame, and transform into the other inertial frame.
     
  21. Apr 16, 2017 #20
    Thanks Dale and SlowThinker.
    I was looking the Feynman Lectures available on Caltech Website, where this topic is discussed. I will read it and start another thread if some doubt remain.
     
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