Dual-tensors not in Lagrangian

[Pengwuino]

In Ryder's text, he defines the dual tensor as the anti-symmetric $$\tilde F^{\nu \mu} = \epsilon^{\nu \mu \alpha \beta} F_{\alpha \beta}$$. Later he plops down the complex scalar field Lagrangian as

$$L = (D_\mu \phi)(D^\mu \phi *) - m^2 \phi * \phi - \frac{1}{4} F^{\nu \mu}F_{\nu \mu}$$

where $$D_\mu$$ is the covariant derivative. So the thing i was wondering is why can't you have terms like $$\tilde F^{\nu \mu} \tilde F_{\nu \mu}$$? I did the work to figure out why you can't have terms like $$F^{\nu \mu} \tilde F_{\nu \mu}$$, but I just wanted to see what happens to the scalar term using two dual tensors.

 

[Jimmy Snyder]

I hope this answer doesn't make things worse. I think there is no difference between the two products.

$$\tilde F^{\nu \mu} \tilde F_{\nu \mu} = \epsilon^{\nu \mu \alpha \beta} F_{\alpha \beta} \epsilon_{\nu \mu \alpha \beta} F^{\alpha \beta}$$

You can freely contract the epsilons on mu and nu to get
$$-2(\delta_{\alpha}^{\alpha} \delta_{\beta}^{\beta} - \delta_{\alpha}^{\beta} \delta_{\beta}^{\alpha})$$
This formula is found in QFT by Mandl & Shaw first revised edition page 333, eqn (1.14c).
This is 0 when alpha is equal to beta, and -2 otherwise. I think there is an error somewhere in here because the factor should be 1, not 2. But the constant factor is irrelevant to the fact that there is no need to introduce the second product into the Lagrangian.

 

[hamster143]

IIRC,
$$\tilde F^{\nu \mu} \tilde F_{\nu \mu}$$

is equal to

$$F^{\nu \mu}F_{\nu \mu}$$
times some constant.

I hope this answer doesn't make things worse. I think there is no difference between the two products.
$$\tilde F^{\nu \mu} \tilde F_{\nu \mu} = \epsilon^{\nu \mu \alpha \beta} F_{\alpha \beta} \epsilon_{\nu \mu \alpha \beta} F^{\alpha \beta}$$

You can't do it exactly like that, indices can't repeat four times. The idea is correct though.

 

[Pengwuino]

I did the calculation out and got

$$\tilde F^{\nu \mu} \tilde F_{\nu \mu} = \epsilon^{\alpha \beta \nu \mu} \epsilon_{\gamma \delta \nu \mu}F_{\alpha \beta}F^{\gamma \delta}<br /> =-2(\delta^\alpha_\gamma \delta^\beta_\delta - \delta^\alpha_\delta \delta^\beta_\gamma)F_{\alpha \beta}F^{\gamma \delta}<br /> = -2F_{\gamma \delta}F^{\gamma \delta} + 2F_{\delta \gamma}F^{\gamma \delta}$$

But since $$F_{\delta \gamma} = F_{\gamma \delta}$$, everything vanishes.

Is this correct?

 

[ismaili]

I did the calculation out and got

$$\tilde F^{\nu \mu} \tilde F_{\nu \mu} = \epsilon^{\alpha \beta \nu \mu} \epsilon_{\gamma \delta \nu \mu}F_{\alpha \beta}F^{\gamma \delta}<br /> =-2(\delta^\alpha_\gamma \delta^\beta_\delta - \delta^\alpha_\delta \delta^\beta_\gamma)F_{\alpha \beta}F^{\gamma \delta}<br /> = -2F_{\gamma \delta}F^{\gamma \delta} + 2F_{\delta \gamma}F^{\gamma \delta}$$

But since $$F_{\delta \gamma} = F_{\gamma \delta}$$, everything vanishes.

Is this correct?

$$F_{\delta\gamma}$$ should be an anti-symmetric tensor.
And so $$F^2$$ is proportional to $$\tilde{F}^2$$

 

[Pengwuino]

oops, true story. There went that.