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Troubled with the indices

  1. Sep 20, 2015 #1
    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!
     
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  3. Sep 20, 2015 #2

    Orodruin

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    I suggest starting by rewriting the ##G^{\mu\nu}## in terms of the components of the hodge dual, this should get you started.
     
  4. Sep 20, 2015 #3
    Do you mean that ##\tilde{G}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}G_{\rho\sigma}##?
     
  5. Sep 20, 2015 #4

    Orodruin

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    Yes, but if you want the left term to be expressed in the hodge dual, you need to invert that expression.
     
  6. Sep 20, 2015 #5
    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.
     
  7. Sep 20, 2015 #6
    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
     
  8. Sep 20, 2015 #7

    Orodruin

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    How does the ##\varepsilon##-##\delta## relation look in four dimensions?
     
  9. Sep 20, 2015 #8

    Orodruin

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    This part of the wikipedia entry on the Kronecker delta and its generalisation might help.
     
  10. Sep 20, 2015 #9
    $$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho'\sigma'}=-2(\delta^{\rho}_{\rho'}\delta^{\sigma}_{\sigma'}-\delta^{\rho}_{\sigma'}\delta^{\sigma}_{\rho'})$$

    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 .
     
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