# Lorentz transformations

1. Apr 2, 2006

### 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 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.

2. Apr 3, 2006

### 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.