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Relativistically correct exression for Lorentz force

  1. May 6, 2008 #1
    I need to know the correct expression for the Lorentz force on a relativistic particle in an EM field, in cgs units, in terms of vector potential A and scalar potential phi, to prove that a given Lagrangian produces the correct equations of motion. I don't need it in covariant form with tensors and all of that because we haven't learned that yet. Thanks everyone.
  2. jcsd
  3. May 6, 2008 #2

    The the correct expression for the Lorentz force on a relativistic particle in an EM field is no different than the same expression for a non-relativistic particle, i.e. in MKS units

    f = dp/dt = q(E + vxB)

    This is not a 4-tensor equation though. If you're speaking about the tensor formulation then its taks on a different form. The tensor form is dP/d[itex]\tau[/itex] = qF*U where P = 4-momentum of particle, F = Faraday tensor and U = particle's 4-velocity.

  4. May 8, 2008 #3


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    In cgs gaussian units, just divide v by c, unless you have set c=1.
  5. May 22, 2008 #4
    Because of my poor editing, I cannot write it down here. In fact, I am not quite sure what you are asking. But I believe you can find your answer by looking at the differential expression of conservation of momentum of EM field which is in the form of divergence equation on the internet or other sources. The momentum density is something like 1/(4times pi times c) of cross product of E and B, the Lorentz force density is something like rho times E plus j/c cross B etc. If you have that equation in SI unit, you can transform it in Gaussian unit by replacing epsilon with 1/4pi, mu with 4pi / c^2, and B with B/c. If you find this is not helpful, I am sorry but I already did the best I can.
  6. May 22, 2008 #5

    Jonathan Scott

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    I'm interested to see that someone is being taught about this way of getting the Lorentz force law from a Lagrangian. It makes it possible to see how adding the charge times the four-potential to the four-momentum (as in the Pauli "minimal electromagnetic coupling" assumption) gives a Lagrangian which results in exactly the same law of motion for charged particles as the Lorentz force law. That puzzled me for a long time until I worked it out for myself, which isn't difficult but I didn't see it mentioned in the text books which I used.

    As mentioned in a previous response, the Lorentz force law is already relativistic. That means all you have to do is write out the Euler-Lagrange equations, recognize which partial derivatives of the four-potential correspond to the E and B fields and reorganize the result to show that it matches the Lorentz force law. I'm leaving the details as an exercise for the student.
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