Simple Lagrangian problem

Question

I am struggling with some very simple Lagrangian problems and I think this is because I am missing something very stupid.

I have to proof that particular Lagrangians lead to particular equations using Euler-Lagrange equation.

1. So, I have to proof that $\mathcal{L}^{free}_KG=\frac{1}{2}(\partial_\mu \phi)(\partial^\mu \phi)-\frac{1}{2}m^2\phi^2$ leads to Klein-Gordon equation.

So easily, $\frac{\partial \mathcal{L}}{\partial \phi} = -m\phi$.

$\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} = \frac{1}{2} [(\partial^\mu\phi)+(\partial_\mu \phi)\frac{\partial(\partial^\mu \phi) }{\partial (\partial_\mu \phi)}]= \frac{1}{2} [(\partial^\mu\phi)+(\partial_\mu \phi)\frac{\eta^{\mu \nu}\partial(\partial_\nu \phi) }{\partial (\partial_\mu \phi)}]$

Here I am stuck, because $\frac{\eta^{\mu \nu}\partial(\partial_\nu \phi) }{\partial (\partial_\mu \phi)}$ is not equal zero, if $\nu=\mu$ (at least this is how I understand it). So, I need to add $\delta_\nu^\mu$ or something like that and I have no idea how to get rid of it afterwards or how to evaluate $\eta^{\mu\mu}$.

2. The second problem is similar, but $\mathcal{L}_{EM}=-\frac{1}{4}(\partial^\mu A^\nu- \partial^\nu A^\mu)(\partial_\mu A_\nu-\partial_\nu A_\mu)-j^\mu A_\mu$

The first one is easy: $\frac{\partial \mathcal{L}}{\partial A_\mu} = -j^\mu$

but I don't know how to evaluate $\frac{\partial \mathcal{L}}{\partial (\partial_\mu A_\mu)}$. Do I have to open brackets in $\mathcal{L}_{EM}$ and treat $\frac{\partial(\partial_\nu A_\mu)}{\partial(\partial_\mu A_\mu)}$ as $\delta_\mu^nu$? $\partial^\nu A^\mu$ has to be treated as $\eta^{\mu \nu} \eta^{\nu \mu} \partial_\mu A_\nu = 4 \partial_\mu A_\nu$?

Answer 1

In the EL equation I use the following colored-labels

$$ \color{blue}{\frac{\partial}{\partial x^\mu}}\color{red}{\left(\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\right)} - \color{orange}{\frac{\partial \mathcal{L}}{\partial \phi}} = 0 $$

$\mathcal{L} = \mathcal{L}_{\rm KG}^{\rm free}$

For the Lagrangian

$$ \mathcal{L}_{\rm KG}^{\rm free} = \frac{1}{2}\partial_\nu\phi\partial^\mu\phi - \frac{1}{2}m^2\phi^2 $$

we can calculate

1. $\color{orange}{\rm orange}$:

$$ \color{orange}{\frac{\partial \mathcal{L}}{\partial \phi}} = -m^2\phi $$

2. $\color{red}{\rm red}$:

\begin{eqnarray} \color{red}{\frac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)}} &=& \frac{1}{2}\frac{\partial }{\partial (\partial_\mu\phi)}(\partial_\nu \phi \partial^\nu\phi) = \frac{1}{2}\eta^{\nu\alpha}\frac{\partial }{\partial (\partial_\mu\phi)}(\partial_\nu \phi\partial_\alpha \phi) \\ &=& \frac{1}{2}\eta^{\nu\alpha}(\delta^\mu_\nu \partial_\alpha\phi + \partial_\nu\phi\delta^\mu_\alpha) = \frac{1}{2}\eta^{\mu\alpha} \partial_\alpha\phi + \frac{1}{2}\eta^{\nu\mu}\partial_\nu\phi \\ &=& \frac{1}{2}\eta^{\mu\nu}\partial_\nu\phi +\frac{1}{2}\eta^{\mu\nu}\partial_\nu\phi \\ &=& \eta^{\mu\nu}\partial_\nu\phi = \partial^\mu\phi \end{eqnarray}

3. $\color{blue}{\rm blue}$:

\begin{eqnarray} \color{blue}{\frac{\partial}{\partial x^\mu}}\color{red}{\left(\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}\right)} &=& \partial_\mu\partial^\mu \phi = \eta^{\mu\nu}\partial_\mu\partial_\nu\phi \\ &=& \underbrace{\eta^{00}}_{1}\partial_0\partial_0\phi + \underbrace{\eta^{i0}}_{0}\partial_i\partial_0 \phi + \underbrace{\eta^{0i}}_{0}\partial_i\partial_j \phi + \underbrace{\eta^{ij}}_{-\delta^{ij}}\partial_i\partial_j \phi \\ &=& \ddot{\phi} - \nabla^2\phi \end{eqnarray}

The signs on this last expression may be different for you, depending on your choice of the metric. Putting everything together

$$ \ddot{\phi} - \nabla^2\phi + m^2\phi = 0 $$

$\mathcal{L} = \mathcal{L}_{\rm EM}$

In this case the Lagrangian is

$$ \mathcal{L}_{\rm EM} = -\frac{1}{4}(\partial^\alpha A^\beta - \partial^\beta A^\alpha)(\partial_\alpha A_\beta - \partial_\beta A_\alpha) - j^\alpha A_\alpha $$

and

1. $\color{orange}{\rm orange}$:

$$ \color{orange}{\frac{\partial \mathcal{L}}{\partial A_\nu}} = j^\alpha\delta^\nu_\alpha = j^\nu $$

2. $\color{red}{\rm red}$:

\begin{eqnarray} \color{red}{\frac{\partial \mathcal{L}}{\partial (\partial_\mu A_\nu)}} &=& -\frac{1}{4}\frac{\partial}{\partial (\partial_\mu A_\nu)}(\partial^\alpha A^\beta - \partial^\beta A^\alpha)(\partial_\alpha A_\beta - \partial_\beta A_\alpha) \\ &=&-\frac{1}{4}\frac{\partial}{\partial (\partial_\mu A_\nu)}(\eta^{\alpha\sigma}\eta^{\beta\rho}\partial_\sigma A_\rho - \eta^{\beta\rho}\eta^{\alpha\sigma}\partial_\rho A_\sigma)(\partial_\alpha A_\beta - \partial_\beta A_\alpha) \\ &=& -\frac{\eta^{\alpha\sigma}\eta^{\beta\rho}}{4}\frac{\partial}{\partial (\partial_\mu A_\nu)} (\partial_\sigma A_\rho - \partial_\rho A_\sigma) (\partial_\alpha A_\beta - \partial_\beta A_\alpha) \\ &=&-\frac{\eta^{\alpha\sigma}\eta^{\beta\rho}}{4}\frac{\partial}{\partial (\partial_\mu A_\nu)} (\partial_\sigma A_\rho \partial_\alpha A_\beta + \cdots) \\ &=&-\frac{\eta^{\alpha\sigma}\eta^{\beta\rho}}{4} (\delta^{\mu}_{\sigma}\delta^{\nu}_{\rho}\partial_\alpha A_\beta + \partial_\sigma A_\rho\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta} + \cdots) \\ &=&-\frac{1}{4}(\eta^{\alpha\mu}\eta^{\beta\nu}\partial_\alpha A_\beta + \eta^{\mu\sigma}\eta^{\nu\rho}\partial_\sigma A_\rho + \cdots ) \end{eqnarray}

I will leave the rest for you