# EL Equations for the modified electromagnetic field Lagrangian

1. Apr 16, 2012

### Irrational

Hi,

I'm trying to work through something and it should be quite simple but somehow i've gotten a bit confused.

I've worked through the Euler Lagrange equations for the lagrangian:

\begin{align*} \mathcal{L}_{0} &= -\frac{1}{4}(\partial_{\mu}A_{\nu})(\partial^{\mu}A^{\nu}) \\ &= \frac{1}{4}F_{\mu\nu}F^{\mu\nu} \end{align*}

getting:

$$\Box A_{\nu} - \partial^{\nu}\partial_{\mu}A^{\mu} = 0$$

I'm ok with this.

Then considering the modified lagrangian:

$$\mathcal{L}_{\xi} = \mathcal{L}_{0} + \frac{\lambda}{2}(\partial_{\sigma}A^{\sigma})^2$$

I'm trying to work out the EL equation components and as part of one of these calculations, I've to calculate:

\begin{align*} \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} \left[ \frac{\lambda}{2} (\partial_{\sigma}A^{\sigma})^2 \right] &= \frac{\lambda}{2} \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} \left[ ( \partial_{\sigma}A^{\sigma} ) ( \partial_{\rho}A^{\rho} ) \right] \\ &= \frac{\lambda}{2} \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} \left[ ( \partial_{\sigma}A_{\alpha} \eta^{\sigma \alpha} ) ( \partial_{\rho}A_{\beta} \eta^{\rho \beta} ) \right] \\ & = \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\rho \beta} \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} \left[ ( \partial_{\sigma}A_{\alpha} ) ( \partial_{\rho}A_{\beta} ) \right] \\ & = \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\rho \beta} \left[ ( \partial_{\sigma}A_{\alpha} ) \left( \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} ( \partial_{\rho}A_{\beta} ) \right) + \left( \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} ( \partial_{\sigma}A_{\alpha} ) \right) ( \partial_{\rho}A_{\beta} ) \right] \\ & = \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\rho \beta} \left[ ( \partial_{\sigma}A_{\alpha} ) \delta^{\mu}_{\rho} \delta^{\nu}_{\beta} + \delta^{\mu}_{\sigma} \delta^{\nu}_{\alpha} ( \partial_{\rho}A_{\beta} ) \right] \\ & = \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\rho \beta} ( \partial_{\sigma}A_{\alpha} ) \delta^{\mu}_{\rho} \delta^{\nu}_{\beta} + \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\rho \beta} \delta^{\mu}_{\sigma} \delta^{\nu}_{\alpha} ( \partial_{\rho}A_{\beta} ) \\ & = \frac{\lambda}{2} \eta^{\sigma \alpha} \eta^{\mu \nu} ( \partial_{\sigma}A_{\alpha} ) + \frac{\lambda}{2} \eta^{\mu \nu} \eta^{\rho \beta} ( \partial_{\rho}A_{\beta} ) \\ & = \frac{\lambda}{2} \eta^{\mu \nu} \left[ ( \partial_{\sigma}A^{\sigma} ) + ( \partial_{\rho}A^{\rho} ) \right] \\ & = \lambda \eta^{\mu \nu} ( \partial_{\sigma}A^{\sigma} ) \\ \end{align*}

Now I was hoping to get:

$$\lambda \partial^{\nu} A^{\mu}$$

as ultimately I need the EL equations to give me:

\begin{align*} \frac{\partial \mathcal{L}_{\xi}}{\partial A_{\nu}} - \partial_{\mu} \left( \frac{\partial \mathcal{L}_{\xi}}{\partial (\partial_{\mu} A_{\nu})} \right) &=\Box A^{\nu} - \partial^{\nu} ( \partial_{\mu} A^{\mu} ) - \lambda \partial^{\nu}(\partial_{\mu} A^{\mu}) \\ &= \Box A^{\nu} - ( 1 + \lambda ) \partial^{\nu} ( \partial_{\mu} A^{\mu} ) \\ &= 0 \end{align*}

Can anyone show me where i've gone wrong? I didn't stick this in the homework section as it's not homework. I'm just trying to work through the through missing steps from the text i'm reading.

Last edited: Apr 16, 2012
2. Apr 16, 2012

### Irrational

edit... now i'm finished stating the question.

Last edited: Apr 16, 2012
3. Apr 16, 2012

### Bill_K

Irrational, You've got exactly what you need. Well, up to a minus sign anyway. You've got λ ημν(∂σAσ). So plug this into the Euler-Lagrange equation:
-∂μ(λ ημν(∂σAσ)) = -λ ∂ν(∂σAσ) = -λ ∂ν(∂μAμ)

4. Apr 16, 2012

### Irrational

you have no idea how thick i feel right now. thanks for filling in the gap.