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I Vector to scalar potential, transformation of fields

  1. Sep 4, 2016 #1
    Hey guys. So, as i was going through Griffith's Electrodynamics, and i came across this problem:

    Screen_Shot_2016_09_04_at_18_36_08.png
    In the solutions:
    sol.png

    How to they actually get to that expression for V = (V(bar)+vAx(bar) )Ɣ? I understand everything after that, but this just made me very confused. How do they get this from the inverse Lorentz transformations?

    Thank you.
     

    Attached Files:

    Last edited: Sep 4, 2016
  2. jcsd
  3. Sep 4, 2016 #2
    The Lorentz transformation that goes from ##S## to ##\bar S## is given by the matrix:
    $$\Lambda(S\rightarrow \bar S)=\begin{pmatrix}\gamma& -\beta\gamma & 0 & 0\\ -\beta\gamma &\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\
    \end{pmatrix} $$
    The inverse is simply:

    $$\Lambda(\bar S\rightarrow S)=\begin{pmatrix}\gamma& \beta\gamma & 0 & 0\\ \beta\gamma &\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\
    \end{pmatrix} $$
    You can easily verify this by multiplying these two in any order and using ##\gamma=1/\sqrt{1-\beta^2}##. Basically it's the transformation with the velocity in the opposite direction, so ##\beta\rightarrow -\beta##, while ##\gamma## doesn't change since it's quadratic. If you take into account that
    $$
    \bar A^\mu=\begin{pmatrix}\bar V/c\\ \bar A_x \\ \bar A_y \\ \bar A_z\\
    \end{pmatrix} $$

    and ##A=\Lambda(\bar S \rightarrow S)\bar A##, you should be able to get the desired result.
     
  4. Sep 4, 2016 #3
    Thank you so much, it all makes sense now, even how to apply those!
     
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