I started learning calculus of variations and found this question on my textbook:
The Lagrangian for a π-meson is given by $$L(x,t)=\frac{1}{2}(\phi'^2-|\triangledown\phi|^2-\mu^2\phi^2),$$ where x is a vector and $\mu$ is the menson mass and $\phi(x,t)$ is its wavefunction. Assuming Hamilton’s principle find the wave equation satisfied by $\phi$.
I think there are n Euler-Lagrange equations one for each component of the form:
$$\frac{\partial L}{\partial q_i}=\frac{d}{dt}(\frac{\partial L}{\partial q_i'}),\,\ i=1,2,..,n$$
How can I apply this to the problem?
Best Answer
The Euler-Lagrange equation for fields is given by \begin{align} \frac{\partial \mathcal{L}}{\partial \phi} = \sum^3_{\mu = 0}\partial_\mu\frac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)}. \end{align} Hence, in this case, we see that \begin{align} \frac{\partial \mathcal{L}}{\partial \phi}=-\mu^2\phi. \end{align} Likewise, we have that \begin{align} \frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)} = \dot\phi \ \ \text{ and } \ \ \frac{\partial \mathcal{L}}{\partial(\partial_i \phi)}= -\partial_i\phi \end{align} which means \begin{align} \partial_0\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)} = \ddot \phi \ \ \text{ and } \ \ \partial_i\frac{\partial \mathcal{L}}{\partial(\partial_i \phi)} = -\partial^2_i\phi. \end{align} Hence it follows \begin{align} -\mu^2\phi = \ddot\phi -\Delta\phi \ \ \implies \ \ \ddot\phi -\Delta\phi +\mu^2\phi = 0 \end{align} which is the Klein-Gordon equation.