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,

<br /> x^0 \rightarrow x^{&#039;0} = x^0 + \xi^0, \ \ \ x^i \rightarrow x^{&#039;i} = R^i_{\ j}(t)x^j + d^i (t) \ \ \ \ \ \ (1)<br />
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:

<br /> \frac{\partial \phi}{\partial x^{&#039;i}} = \frac{\partial \phi}{\partial x^j}\frac{\partial x^j}{\partial x^{&#039;i}} + \frac{\partial \phi}{\partial x^0}\frac{\partial x^0}{\partial x^{&#039;i}}<br />

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

<br /> \frac{\partial \phi}{\partial x^0}\frac{\partial x^0}{\partial x^{&#039;i}} \,.<br />

I know that for x^0 = t,

<br /> \frac{\partial x^{&#039;i}}{\partial x^0} = \dot{R}^{i}_{\ j}(t)x^j + \dot{d}^i(t)<br />

but should I invert this relation, or should I put \frac{\partial x^0}{\partial x^{&#039;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

<br /> J = \frac{\partial x^{&#039;\mu}}{\partial x^{\nu}} \ \ \,,<br />

and invert it. Then indeed one finds that the derivative transforms with the inverse of A, and the other term is zero:<br /> \partial&#039;_{i} = [A^{-1}]^j{}_i \partial_j \ \ \ \ \ \,.<br />