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Lorentz force in different frames of reference

  1. Jul 8, 2015 #1
    Hi, I have been studying lorentz force . The book says force experienced by a charge in magnetic field is
    lorentz-force-equation.gif
    But velocity is a relative concept . In one frame of reference(inertial frame) I might observe the charge moving with some velocity and in the 2nd frame(inertial too) I might observe the charge to be at rest . Does this mean that the force on the particle will be different in the 2 frames? .But since my frames are inertial I should observe the same force on the particle. How is this possible! please help.
     
  2. jcsd
  3. Jul 8, 2015 #2

    Orodruin

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    Just like the velocity, the fields are different in different frames.
     
  4. Jul 8, 2015 #3

    hilbert2

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  5. Jul 8, 2015 #4

    Orodruin

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    For some reason I doubt that this is at the appropriate level for the OP. Based on the question, I would be surprised if he/she knew what a tensor is.
     
  6. Jul 8, 2015 #5

    hilbert2

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    ^ That's possible, but the procedure is as simple as writing the transformation and field tensors in matrix form and doing a matrix multiplication.
     
  7. Jul 8, 2015 #6

    Orodruin

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    I know it is simple, if you know how to do it and how to interpret the result. If you have no idea what a tensor is, you would first have to learn that, learn tensor notation in SR, decipher the Wiki page, and then do the simple manipulations. If you skip any of the steps, you will just be left with a feeling of "what did I just do and why?"
     
  8. Jul 8, 2015 #7

    jtbell

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  9. Jul 8, 2015 #8
    An example in cartesian co-ordinates, E = ∫dE/dt dt = ∫(∂E/∂x * ∂x/∂t + ∂E/∂y * ∂y/∂t + ∂E/∂z * ∂z/∂t)dt, I will assume that the field is moving at a constant velocity v, ∂x/∂t = vx, ∂y/∂t = vy, ∂z/∂t = vz,
    E = ∫vx∂E/∂x dt + .....
    by this way you can transform a field from an INERTIAL frame to the other, Cheers !!,
     
    Last edited: Jul 8, 2015
  10. Jul 8, 2015 #9

    Orodruin

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    It is unclear how you are constructing that integral and what it actually means. It also seems to give you the wrong transformation properties. Transformation properties are not based on the change of variables in an integral, they are based on local coordinate transformations and how different frames relate to each other.
     
  11. Jul 8, 2015 #10
    Sorry if i have made ugly mistakes, All I have done is chain rule, I admit I have dropped the arrows of the vectors (Don't know how the write them) and this result isn't relativistic, but ∂x/∂t = vx isn't wrong (is it ?), for example if I have a an electric field E = xyz (just saying) that is moving a some velocity v in the x direction then E = v*∫yz dt= yz*vt + K., so sorry if I made other mistakes, but I think even tensors transform this way :/
     
  12. Jul 8, 2015 #11

    Orodruin

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    It is completely unclear why you are even doing an integral. Why are you differentiating the field with respect to time? It seemingly has absolutely nothing to do with a proper Lorentz transformation of the fields.
     
  13. Jul 8, 2015 #12
    The problem is in E = ∫dE/dt * dt ?, Ok thanks for the feedback, I think I will have to update by checking some Wikipedia pages :)
     
  14. Jul 8, 2015 #13
    The issue is in the phrase "since my frames are inertial I should observe the same force on the particle". Electromagnetic phenomena do not obey the principle of relativity in classical physics, and it was this kind of things that led to the development of special relativity.

    "It is known that Maxwell's electrodynamics—as usually understood at the present time—when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena". -http://fourmilab.ch/etexts/einstein/specrel/www/

    For a detailed discussion I think that this is not the right sub forum, but you can find the answer in §6 of Einstein's paper.
    In a nutshell, the lateral force will be different according to measurements in different frames, but because also time and distance are measured differently, still the same predictions will be made.
     
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