# Euler-Lagrange Equation Proving Maxwell Equation [duplicate]

When quantizing the EM Field, we get the Lagrangian density,

$$L=\frac{1}{2}\left(\epsilon \vert E\vert ^2 - \frac{1}{\mu}\vert B\vert^2\right) = \frac{\epsilon}{2}\vert\nabla\phi + \dot{\textbf{A}}\vert^2 - \frac{1}{2\mu}\vert\nabla\times\textbf{A}\vert^2$$

My professor said that the first Maxwell equation, $\nabla \cdot E = 0$, is proved by the Euler-Lagrange equation for the above $L$ w.r.t. $\phi$. I.e.

$$\frac{\partial}{\partial t}\left(\frac{\partial L}{\partial \dot{\phi}}\right) + \sum\limits_{i=1}^{3}\frac{\partial}{\partial x_i}\left(\frac{\partial L}{\partial(\partial\phi / \partial x_i)}\right) - \frac{\partial L}{\partial \phi} = 0 \implies \nabla\cdot E = \nabla^2\phi = 0$$

I don't get exactly that result. I assume the first and last term are 0, since no phi or phi-dot appears in $L$. Using $\phi_x$ as the derivative, I get for the middle term (i=1), \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 + \dot{A}_x,\phi_y + \dot{A}_y,\phi_z + \dot{A}_z\rangle\right)\cdot\left[\frac{\partial}{\partial(\phi_x)}\langle\phi_x + \dot{A}_x, \phi_y + \dot{A}_y, \phi_z + \dot{A}_z\rangle\right] \\ &= 2\left(\langle\phi_x + \dot{A}_x,\phi_y + \dot{A}_y,\phi_z + \dot{A}_z\rangle\right)\cdot\langle 1,0,0\rangle\\ &= 2(\phi_x + \dot{A}_x) \end{align} Therefore,

$$\frac{\partial}{\partial x}\frac{\partial L}{\partial(\phi_x)} = 2(\phi_{xx} + \dot{A}_{xx})$$

And then summing i=1 to 3 gives

$$2(\nabla^2\phi + \nabla^2 \dot{A}) = 0$$

So in order to prove the Maxwell equation, I need to show that $\nabla^2\dot{A} = 0$. How do I proceed to do that?

## marked as duplicate by Qmechanic♦Apr 3 '18 at 11:04

• @HiddenBabel $,x$ as a subscript is a common notation for an $x$ derivative. – J.G. Apr 1 '18 at 22:03
Note that $$\sum\limits_{i=1}^{3}\frac{\partial}{\partial x_i}\left[\frac{\partial L}{\partial(\partial\phi / \partial x_i)}\right]=\mathrm{div}\left[\frac{\partial L}{\partial(\mathrm{grad}\phi)}\right]=\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\frac{\partial L}{\partial(\boldsymbol{\nabla}\phi)}\right] \tag{01}$$ and $$\frac{\partial L}{\partial(\boldsymbol{\nabla}\phi)}=\text{??? vector} \tag{02}$$