- #1

JD_PM

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- 158

- Homework Statement
- Given the Lagrangian density for the real vector field ##\phi^{\alpha} (x)##

$$ \mathcal{L} = - \frac 1 2 (\partial_{\alpha} \phi_{\beta}) (\partial^{\alpha} \phi^{\beta}) + \frac 1 2 (\partial_{\alpha} \phi^{\alpha}) (\partial_{\beta} \phi^{\beta}) + \frac{\mu^2}{2} \phi_{\alpha} \phi^{\alpha}$$

a) Show that the equation of motion is given by

$$\Big( \eta_{\alpha \beta} (\partial^{k} \partial_{k} + \mu^2) - \partial_{\alpha} \partial_{\beta} \Big) \phi^{\beta} = 0$$

b) Show that ##\partial_{\alpha} \phi^{\alpha} = 0##

*Source: QFT book by Franz Mandl and Graham Shaw; second edition

- Relevant Equations
- $$ \mathcal{L} = - \frac 1 2 (\partial_{\alpha} \phi_{\beta}) (\partial^{\alpha} \phi^{\beta}) + \frac 1 2 (\partial_{\alpha} \phi^{\alpha}) (\partial_{\beta} \phi^{\beta}) + \frac{\mu^2}{2} \phi_{\alpha} \phi^{\alpha}$$

a)

Alright here we have to use Euler-Lagrange equation

$$\partial_{\alpha} \Big( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} \Big) - \frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

Let's focus on the term ##\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )}##

I know that

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( - \frac 1 2 (\partial_{\alpha} \phi_{\beta}) (\partial^{\alpha} \phi^{\beta}) \Big) = - \partial^{\alpha} \phi^{\beta}$$

After reviewing my work, I think that my mistake has to be in computing the term

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( \frac 1 2 (\partial_{\alpha} \phi^{\alpha}) (\partial_{\beta} \phi^{\beta}) \Big)$$

What I have tried:

Note I have swapped ##\alpha \rightarrow \gamma## and ##\beta \rightarrow b## in the original Lagrangian

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( \frac 1 2 (\partial_{\gamma} \phi^{\gamma}) (\partial_{b} \phi^{b}) \Big) = \frac 1 2 \eta^{\gamma c} \partial_{b} \phi^{b} \frac{\partial(\partial_{\gamma} \phi_{c})}{\partial (\partial_{\mu} A_{\nu})} + \frac 1 2 \eta^{b d} \partial_{\gamma} \phi^{\gamma} \frac{\partial(\partial_{b} \phi_{d})}{\partial (\partial_{\mu} A_{\nu})} = \frac 1 2 \eta^{\gamma c} \partial_{b} \phi^{b} \delta_{\gamma}^{\mu} \delta_{c}^{\nu} + \frac 1 2 \eta^{b d} \partial_{\gamma} \phi^{\gamma} \delta_{b}^{\mu} \delta_{d}^{\nu} = \frac 1 2( \partial_{b} \phi^b + \partial_{\gamma} \phi^{\gamma}) \eta^{\mu \nu}$$

Mmm but I do not see where I got wrong...

Alright here we have to use Euler-Lagrange equation

$$\partial_{\alpha} \Big( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} A_{\nu})} \Big) - \frac{\partial \mathcal{L}}{\partial A_{\nu}} = 0$$

Let's focus on the term ##\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )}##

I know that

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( - \frac 1 2 (\partial_{\alpha} \phi_{\beta}) (\partial^{\alpha} \phi^{\beta}) \Big) = - \partial^{\alpha} \phi^{\beta}$$

After reviewing my work, I think that my mistake has to be in computing the term

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( \frac 1 2 (\partial_{\alpha} \phi^{\alpha}) (\partial_{\beta} \phi^{\beta}) \Big)$$

What I have tried:

Note I have swapped ##\alpha \rightarrow \gamma## and ##\beta \rightarrow b## in the original Lagrangian

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\alpha} \phi_{\beta} )} \Big( \frac 1 2 (\partial_{\gamma} \phi^{\gamma}) (\partial_{b} \phi^{b}) \Big) = \frac 1 2 \eta^{\gamma c} \partial_{b} \phi^{b} \frac{\partial(\partial_{\gamma} \phi_{c})}{\partial (\partial_{\mu} A_{\nu})} + \frac 1 2 \eta^{b d} \partial_{\gamma} \phi^{\gamma} \frac{\partial(\partial_{b} \phi_{d})}{\partial (\partial_{\mu} A_{\nu})} = \frac 1 2 \eta^{\gamma c} \partial_{b} \phi^{b} \delta_{\gamma}^{\mu} \delta_{c}^{\nu} + \frac 1 2 \eta^{b d} \partial_{\gamma} \phi^{\gamma} \delta_{b}^{\mu} \delta_{d}^{\nu} = \frac 1 2( \partial_{b} \phi^b + \partial_{\gamma} \phi^{\gamma}) \eta^{\mu \nu}$$

Mmm but I do not see where I got wrong...