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I'm reading A. Zee's book, Einstein Gravity in a Nutshell. In problem 7 of chapter IV.2, it is said that the energy momentum tensor of the electromagnetic field \begin{align} T^{\mu\nu}=\eta_{\lambda\sigma}F^{\mu\lambda}F^{\nu\sigma}-\frac{1}{4}\eta^{\mu\nu}F_{\sigma\rho}F^{\sigma\rho} \end{align} can be written in the symmetric form using the dual electromagnetic tensor defined as $\tilde{F}_{\mu\nu}=-\frac{1}{2} \epsilon_{\mu\nu\lambda\sigma}F^{\lambda\sigma}$, \begin{align} T^{\mu\nu}=\frac{1}{2}\eta_{\lambda\sigma}(F^{\mu\lambda}F^{\nu\sigma}+\tilde{F}^{\mu\lambda}\tilde{F}^{\nu\sigma}). \end{align} It is mentioned in problem 6 that the key is the identity \begin{align} \eta_{\lambda\sigma}(F^{\mu\lambda}F^{\nu\sigma}-\tilde{F}^{\mu\lambda}\tilde{F}^{\nu\sigma})=\frac{1}{2}\eta^{\mu\nu}F^{\rho\tau}F_{\rho\tau} \end{align} However, I'm stuck when proving this identity. First problem is that I can't separate a $\eta^{\mu\nu}$ from the first term in left hand side. The second problem is the second term will have something like $\epsilon^{\mu\lambda\alpha\beta}\epsilon^{\nu}_{~\lambda\gamma\eta}$ after plugging the expression of dual tensor, but I don't know how to simplify it. Any hints about proof of this identity?

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  • $\begingroup$ What's your goal? If your goal is simply deriving a symmetric EM tensor for the EM field, then vary the EM action wrt the metric, e.g. as is done in Carroll's book in Sec 4.3. $\endgroup$ – WAH Apr 21 at 13:34
  • $\begingroup$ I'm just curious how the last identity is derived since I'm going through all the problems in Zee's book. It is one of the problem so I think it won't be too complicated. $\endgroup$ – Bruce Apr 21 at 13:40
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You can express the product of two $\epsilon$'s using the following: $$\epsilon^{\mu\lambda\alpha\beta} \epsilon_{\tau\lambda\gamma\eta}= \epsilon^{\lambda\mu\alpha\beta} \epsilon_{\lambda\tau\gamma\eta}=\epsilon^{\mu\alpha\beta} \epsilon_{\tau\gamma\eta}$$ $$=\delta^\mu_\tau(\delta^\alpha_\gamma \delta^\beta_\eta-\delta^\alpha_\eta \delta^\beta_\gamma)-\delta^\mu_\gamma(\delta^\alpha_\tau \delta^\beta_\eta-\delta^\alpha_\eta \delta^\beta_\tau)+\delta^\mu_\eta(\delta^\alpha_\tau \delta^\beta_\gamma-\delta^\alpha_\gamma \delta^\beta_\tau)$$

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  • $\begingroup$ Thanks. I think this can take care of the second term to produce what we want, but any suggestions about what to do with the first term? The free indices is on two F’s, and I think it’s crucial to move them out on to the metric. Still can’t figure out a way to do this. $\endgroup$ – Bruce Apr 22 at 13:39
  • $\begingroup$ Thanks, it turns out as long as the second term is simplified, it will give one term that will cancel the first term, and the reminding piece is what we want! $\endgroup$ – Bruce May 2 at 2:10
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The first step is to solve the second problem, which is to find the product $\tilde{F}_{\mu\nu}\tilde{F}^{\mu\nu}$. To this, you will need to calculate the product \begin{equation} \epsilon^{\alpha\beta\gamma\delta}\epsilon_{\mu\nu\rho\sigma}=\left( -1\right) ^{s}\delta_{\mu\nu\rho\sigma}^{\alpha\beta\gamma\delta} ,\tag{1}\label{eq1} \end{equation} where $s$ is the number of negative eigenvalue of the metric [1] and $\delta_{\mu\nu\rho\sigma}^{\alpha\beta\gamma\delta}$ is the generalized Kronecker's delta, which we can to expresses as \begin{equation} \delta_{\mu\nu\rho\sigma}^{\alpha\beta\gamma\delta}=\delta_{\sigma}^{\delta }\delta_{\mu\nu\rho}^{\alpha\beta\gamma}-\delta_{\sigma}^{\gamma}\delta _{\mu\nu\rho}^{\alpha\beta\delta}+\delta_{\sigma}^{\beta}\delta_{\mu\nu\rho }^{\alpha\gamma\delta}-\delta_{\sigma}^{\alpha}\delta_{\mu\nu\rho} ^{\beta\gamma\delta}.\tag{2}\label{eq2} \end{equation} You can find a general study of Kronecker's delta in [2].

Once knowing \eqref{eq1}, you will be able to calculate the product $\tilde{F}_{\mu\nu}\tilde{F}^{\mu\nu}$.

The first problem can be solved by means of a game of rising and falling indices, as long as you remember that \begin{equation} \eta_{\mu\rho}\eta^{\mu\rho}=\delta_{\mu}^{\mu}=4\rightarrow\dfrac{1}{4} \eta_{\mu\rho}\eta^{\mu\rho}=1.\tag{3}\label{eq3} \end{equation}

1 Sean Carrol, Spacetime and Geometry:An Introduction to General Relativity.

2 David Lovelock and Hanna Rund, Tensor, differential forms and variational principles, section 4.2, page 109.

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  • $\begingroup$ Thanks. I solve this problem following you suggestion. The two identities you show here is crucial! $\endgroup$ – Bruce May 2 at 2:11

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