Can Lorentz transformations be represented by matrices in EM fields?

[stunner5000pt]

Show that \partial'_{\alpha} A'^\alpha (x') = \partial _{mu} A^{\mu}(x')

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

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

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

when i put them together
\partial'_{\alpha} A'^\alpha (x') = \frac{\partial}{L_{\nu}^{\alpha} \partial x^{\nu}} L_{\sigma}^{\alpha} A^{\sigma}(x')
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 can't be int eh denominator... can it? Would it simply be represented as an inverse? The two Ls still turn into idnetity matrix which is simply 1.

your helpsi greatly appreciated!

 

[dextercioby]

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

\frac{\partial}{\partial x' ^{\alpha}} =\Lambda_{\alpha}{}^{\mu} \frac{\partial}{\partial x^{\mu}}.


Daniel.