Silly question about (Galilei) transformations

[haushofer]

Hi, I have a silly question concerning the chain rule. Imagine I have a time and space transformation as follows,

$$x^0 \rightarrow x^{'0} = x^0 + \xi^0, \ \ \ x^i \rightarrow x^{'i} = R^i_{\ j}(t)x^j + d^i (t) \ \ \ \ \ \ (1)$$
where xi^0 is constant, R is an element of SO(3) and d is a vector with arbitrary time dependence. Now I want to calculate how a potential term transforms under this group:

$$\frac{\partial \phi}{\partial x^{'i}} = \frac{\partial \phi}{\partial x^j}\frac{\partial x^j}{\partial x^{'i}} + \frac{\partial \phi}{\partial x^0}\frac{\partial x^0}{\partial x^{'i}}$$

The first term is OK, but I'm confused about the second,

$$\frac{\partial \phi}{\partial x^0}\frac{\partial x^0}{\partial x^{'i}} \,.$$

I know that for $x^0 = t$,

$$\frac{\partial x^{'i}}{\partial x^0} = \dot{R}^{i}_{\ j}(t)x^j + \dot{d}^i(t)$$

but should I invert this relation, or should I put $\frac{\partial x^0}{\partial x^{'i}}=0$? So, how does my potential transform under the group given in (1)?

 

[haushofer]

By the way, this is not the standard Galilei group, but it includes accelerations.

 

[haushofer]

Was it that silly? ;(

 

[haushofer]

It was indeed silly; I just had to calculate the Jacobian

$$J = \frac{\partial x^{'\mu}}{\partial x^{\nu}} \ \ \,,$$

and invert it. Then indeed one finds that the derivative transforms with the inverse of A, and the other term is zero:$$\partial'_{i} = [A^{-1}]^j{}_i \partial_j \ \ \ \ \ \,.$$