- #1
cc94
- 19
- 2
Homework Statement
Given
$$L = \left(\nabla\phi + \dot{\textbf{A}}\right)^2 ,$$
how do you calculate $$\frac{\partial}{\partial x}\left(\frac{\partial L}{\partial(\partial\phi / \partial x)}\right)?$$
Homework Equations
By summing over the x, y, and z derivatives, the answer is supposed to show that ##\nabla^2 \phi = 0##.
The Attempt at a Solution
My vector calculus isn't too good, so here's my guess ( ##\partial\phi / \partial x \equiv \phi_x##):
$$\begin{align}
\frac{\partial L}{\partial(\phi_x)} &= 2(\nabla\phi + \dot{\textbf{A}})\cdot\left[\frac{\partial}{\partial\phi_x} (\nabla\phi + \dot{\textbf{A}})\right] \\
&= 2\left(\langle\phi_x + A_x,\phi_y + A_y,\phi_z + A_z\rangle\right)\cdot\left[\frac{\partial}{\partial(\phi_x)}\langle\phi_x + A_x, \phi_y + A_y, \phi_z + A_z\rangle\right] \\
&= 2\left(\langle\phi_x + A_x,\phi_y + A_y,\phi_z + A_z\rangle\right)\cdot\langle 1,0,0\rangle\\
&= 2(\phi_x + \dot{A}_x)
\end{align}
$$
and so
$$\frac{\partial}{\partial x}\frac{\partial L}{\partial(\phi_x)} = 2(\phi_{xx} + \dot{A}_{xx})$$
and summing gives ##\nabla^2\phi + \nabla^2 \dot{A}##.
So did I calculate ##\frac{\partial L}{\partial(\phi_x)}## correctly? If so, why is there still an A term at the end?