Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lorentz transformations

  1. Apr 2, 2006 #1
    Show that [tex] \partial'_{\alpha} A'^\alpha (x') = \partial _{mu} A^{\mu}(x') [/tex]

    lets focus on the partial operator for now
    [tex] \partial'_{\alpha} = \frac{\partial}{\partial x'^{\alpha}} = \frac{\partial}{L_{\nu}^{\alpha} \partial x^{\nu}} [/tex]

    Now A represents the Scalar and vector fields of an EM field.

    [tex] A'^{\alpha}(x') = L_{\sigma}^{\alpha} A^{\sigma}(x') [/tex]
    is that fine?

    when i put them together
    [tex] \partial'_{\alpha} A'^\alpha (x') = \frac{\partial}{L_{\nu}^{\alpha} \partial x^{\nu}} L_{\sigma}^{\alpha} A^{\sigma}(x') [/tex]
    the argumetn is that both the L s represent the same dimensions thus the they are the same thing?

    But Since L is a matrix... i cant be int eh denominator... can it? Would it simply be represented as an inverse? The two Ls still turn in to idnetity matrix which is simply 1.

    your helpsi greatly appreciated!
  2. jcsd
  3. Apr 3, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    Well, the gradient on [itex] M_{4} [/itex] is a covector, so your first equation should read

    [tex] \frac{\partial}{\partial x' ^{\alpha}} =\Lambda_{\alpha}{}^{\mu} \frac{\partial}{\partial x^{\mu}} [/tex].

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook