How Can I Simplify an Equation Involving Levi-Civita Tensors and Indices?

[PhyAmateur]

I have an equation that says $$C_1\partial_{\mu}G^{\mu\nu}+C_2\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}\partial_{\mu}G_{\rho\sigma}=0$$ If I want to get rid of the ##\epsilon^{\mu\nu\rho\sigma}## in the second term, I know I must multiply the equation by some other ##\epsilon## with different set of indices, but I could use some help in knowing what those incides must be to avoid repeating dummy indices and at the same time being able to end up with an equation with a new epsilon present in the first term (the one with ##C_1##). My aim from all this process is to convert the first ##G^{\mu\nu}## to ##\star G^{\mu\nu}##, i.e., the Hodge dual of ##G^{\mu\nu}##. Any tip will do it, thanks guys!

 

[Orodruin]

I suggest starting by rewriting the ##G^{\mu\nu}## in terms of the components of the hodge dual, this should get you started.

 

[PhyAmateur]

Do you mean that ##\tilde{G}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}G_{\rho\sigma}##?

 

[Orodruin]

Yes, but if you want the left term to be expressed in the hodge dual, you need to invert that expression.

 

[PhyAmateur]

Yes, that is what I am having troubles in. To take the epsilon to the other side where the index placement is giving me a hard time.

 

[PhyAmateur]

I will give it a shot though I feel I am mistaken: Maybe, this would be more like: ##2\epsilon^{\mu\nu\rho\sigma}\tilde{G}_{\rho\sigma}=G^{\mu\nu}##? Is this by any chance correct? @Orodruin

 

[Orodruin]

How does the ##\varepsilon##-##\delta## relation look in four dimensions?

 

[Orodruin]

This part of the wikipedia entry on the Kronecker delta and its generalisation might help.

 

[PhyAmateur]

$$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho'\sigma'}=-2(\delta^{\rho}_{\rho'}\delta^{\sigma}_{\sigma'}-\delta^{\rho}_{\sigma'}\delta^{\sigma}_{\rho'})$$

I hope this is what you mean as I am new to those terminologies and to differential geometry in general. Please bear with me @Orodruin .