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Charged particle in a B field, tensor notation

  1. Jul 21, 2015 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A charged particle of charge q with arbitrary velocity ##\vec v_0## enters a region with a constant ##\vec B_0## field.
    1)Write down the covariant equations of motion for the particle, without considering the radiation of the particle.
    2)Find ##x^\mu (\tau)##
    3)Find the Lorentz transformation such that the motion of the particle is restricted to a plane.
    4)Calculate the dipole term of the fields and the average radiated EM power by the particle.

    2. Relevant equations
    ##\tau=t\gamma##

    3. The attempt at a solution
    1)##\frac{dp^\alpha}{d\tau}=qU_\beta F^{\alpha \beta}## would be my reply.
    Where ##p^\alpha## is the four-momentum, ##U_\beta## is the four-velocity and ##F^{\alpha \beta}## is the electromagnetic tensor.
    2)I solved the equation found in 1) by writing the tensors under matrix form. I got that ##x^\alpha=(ct,x^1(t),x^2(t),0)## where ##x^1(t)=\frac{mv_{0x}}{qB_0}\sin \left ( \frac{qB_0 t}{m} \right ) + x^1_0## and ##x^2(t)=\frac{mv_{0x}}{qB_0}\cos \left ( \frac{qB_0 t}{m} \right ) + x^2_0##. Now I just realize that they asked in terms of ##\tau## and not t. Well I get ##x^1(\tau)=\frac{mv_{0x}}{qB_0\sqrt{\gamma}}\sin \left ( \frac{qB_0\sqrt{\gamma}\tau}{m} \right )##.
    3)I'm out of ideas on this one. I know I can express the tensor of the Lorentz transformation as a matrix (I guess this way: https://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_forms) but I'm not sure what they're asking for when they ask for the motion to be in a plane. Isn't the motion already in the x-y plane in my case? I picked ##\vec B_0## as ##B_0\hat z## originally.
     
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  3. Jul 22, 2015 #2

    Orodruin

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    Yea, your motion is in a plane, but it is not the most general solution to the equations of motion. In trying to solve 2, you solved the special case 3.
     
  4. Jul 22, 2015 #3

    fluidistic

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    I see. Is it because I chose the magnetic field to be oriented in the z direction? I would have thought that I'd need a boost in a particular direction to solve 3), not picking a convenient orientation.
     
  5. Jul 22, 2015 #4

    Orodruin

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    It is the combination of selecting the B-field in the z-direction and assuming that there is no velocity component in the z-direction.
     
  6. Jul 25, 2015 #5

    fluidistic

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    Well then I miserably failed, even at solving part 3 since there's no way I can pick the velocity to vanish in the z direction (unless I apply a boost maybe?). I'll try to solve the problem in a correct way. Which means, I believe, that I can pick the orientation for the B field to be in the z direction but I can't assume that there's no speed in the z direction.
     
    Last edited: Jul 25, 2015
  7. Jul 25, 2015 #6

    fluidistic

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    Ok I've redone part 2. I had made some mistakes previously I think.
    I get [tex]x^\mu (\tau) = (c\gamma \tau , \gamma ^2 v_{0x}\frac{B_0q}{m}\sin \left ( \frac{m\tau }{q\gamma B_0} \right ) +x_0 , -\frac{mv_{0x}}{qB_0}\cos \left ( \frac{m \tau}{q\gamma B_0} \right ) +y_0, z_0 )[/tex].
    The fact that I get different amplitudes in front of the sine and cosine really bothers me. This would make an elliptical trajectory in the x-y plane projection instead of circular path. Am I doing things wrong?
     
  8. Jul 25, 2015 #7

    Orodruin

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    The easiest way to check is to insert your result into the original differential equation ...
     
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