saadhusayn
- 17
- 1
Riley Hobson and Bence define covariant and contravariant bases in the following fashion for a position vector $$\textbf{r}(u_1, u_2, u_3)$$:
$$\textbf{e}_i = \frac{\partial \textbf{r}}{\partial u^{i}} $$
And
$$ \textbf{e}^i = \nabla u^{i} $$
In the primed co-ordinate system the equations become
$$\textbf{e}^{'}_{i} = \frac{\partial \textbf{r}}{\partial u^{'i}} $$
And
$$ \textbf{e}^{'i} = \nabla u^{'i} $$
From the chain rule we have that
$$\frac{\partial u^j}{\partial x} = \frac{\partial u^j}{\partial u^{'i}} \frac{\partial u^{'i}}{\partial x}$$
The next step (which I do not understand) is this:
$$ \textbf{e}^j = \frac{\partial u^j}{\partial u^{'i}} \textbf{e}^{'i}$$
How does this last step follow from the previous one? Thank you.
$$\textbf{e}_i = \frac{\partial \textbf{r}}{\partial u^{i}} $$
And
$$ \textbf{e}^i = \nabla u^{i} $$
In the primed co-ordinate system the equations become
$$\textbf{e}^{'}_{i} = \frac{\partial \textbf{r}}{\partial u^{'i}} $$
And
$$ \textbf{e}^{'i} = \nabla u^{'i} $$
From the chain rule we have that
$$\frac{\partial u^j}{\partial x} = \frac{\partial u^j}{\partial u^{'i}} \frac{\partial u^{'i}}{\partial x}$$
The next step (which I do not understand) is this:
$$ \textbf{e}^j = \frac{\partial u^j}{\partial u^{'i}} \textbf{e}^{'i}$$
How does this last step follow from the previous one? Thank you.