haushofer
May26-11, 09:59 AM
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)?
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)?