If I have the variable
$g= r_g \cos(\theta_g)$
And then need to convert the operator:
$\frac{\partial^2}{\partial g^2}$
into terms of $r_g$ and $\theta_g$ only.
Is the following correct?
$\frac{\partial}{\partial g} = \frac{\partial \theta_g}{\partial g} \frac{\partial }{\partial \theta_g} + \frac{\partial r_g}{\partial g} \frac{\partial }{\partial r_g} $
$\frac{\partial^2}{\partial g^2} = \frac{\partial}{\partial g} \frac{\partial}{\partial g} =\bigg(\frac{\partial \theta_g}{\partial g} \frac{\partial }{\partial \theta_g} + \frac{\partial r_g}{\partial g} \frac{\partial }{\partial r_g} \bigg)^2 = \frac{\partial \theta_g}{\partial g} \frac{\partial }{\partial \theta_g} \frac{\partial \theta_g}{\partial g} \frac{\partial }{\partial \theta_g} + \frac{\partial r_g}{\partial g} \frac{\partial }{\partial r_g}\frac{\partial r_g}{\partial g} \frac{\partial }{\partial r_g} + 2\frac{\partial r_g}{\partial g} \frac{\partial }{\partial r_g} \frac{\partial }{\partial \theta_g} \frac{\partial \theta_g}{\partial g} = \bigg(\frac{\partial \theta_g}{\partial g} \bigg)^2 \frac{\partial^2 }{\partial \theta_g^2} + \bigg(\frac{\partial r_g}{\partial g}\bigg)^2 \frac{\partial^2}{\partial r_g^2} + 2 \frac{\partial r_g}{\partial g} \frac{\partial \theta_g}{\partial g} \frac{\partial}{\partial \theta_g} \frac{\partial }{\partial r_g}$
I am quite sure that it is, but applying this to a known function yields answers that I know are incorrect. I have checked >dozen times and obtain the same result each time, so I am full of doubt and have none to ask for this sanity check in person..... Please help?
$ \newcommand{\d}{\partial} \newcommand{\t}{\theta} \newcommand{\dg}{\partial_{g}} \newcommand{\dr}{\partial_{r}} \newcommand{\dt}{\partial_{\theta}} \newcommand{\dgg}{\partial^2_{gg}} \newcommand{\dg}{\partial_{g}} $ You are correct about the first derivative, but for the second one you need to use chain rule:
\begin{align} \partial_{g}\left(\partial_{g}\right) &= \partial_g \left(r_g\partial_{r_g} + \t_g\partial_{\theta_g} \right) = \partial_g \left(r_g\partial_{r_g} \right) + \partial_g \left(\t_g\partial_{\theta_g} \right) \\ &= r_{gg}\partial_{g} + r_g\partial_{g}\left(\partial_{r_g}\right) + \t_{gg}\partial_{\theta_g} + \t_g \partial_{g}\left(\partial_{\theta_g}\right) \\&= r_{gg}\partial_{g} + r_g\left(r_g\partial_{r_g} + \t_g\partial_{\theta_g}\right)\partial_{r_g} + \t_{gg}\partial_{\theta_g} + \t_g \left(r_g\partial_{r_g} + \t_g\partial_{\theta_g}\right)\partial_{\theta_g} %\\ = r_{gg}\partial_{g} + r_g^2\partial_{r_g}^2 + 2 r_g\t_g\partial_{r_g}\partial_{\theta_g} + \t_{gg}\partial_{\theta_g} + \t_g^2\partial_{\theta_g}^2 \end{align}
Expanding parenthesis and collecting alike terms yeilds $$ \partial_{g}\left(\partial_{g}\right) = r_g^2\partial_{r_g}^2 + 2 r_g\t_g\partial_{r_g}\partial_{\theta_g} + \t_g^2\partial_{\theta_g}^2 + r_{gg}\partial_{g} + \t_{gg}\partial_{\theta_g} $$