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How Do You Vary the Action of a Lagrangian for a Scalar Field?
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[QUOTE="romsofia, post: 5840634, member: 243887"] Hmm, can I do this then? Start with ## \int_R d(\delta \phi) \wedge \star d \phi + d \phi \wedge \star d(\delta \phi) ##. I split apart the integrals to get ##\int_R d(\delta \phi) \wedge \star d \phi + \int_R d \phi \wedge \star d(\delta \phi) ## on the 2nd integral, I claim that I can pull back, or hodge dual it, so.. ## \star \int_R d \phi \wedge \star d(\delta \phi) = \int_R \star d \phi \wedge \star \star d(\delta \phi) ## So, what is ## \star \star ##? We can determine if it's plus, or minus. We can tell by the end, it needs to be minus because we have ## d(\delta \phi) \wedge \star d \phi ## in the first integral, and a ## \star d \phi \wedge d(\delta \phi) ## so we will need to switch the places of the p-forms, which will carry along a minus. But, let's use the formula that ## \star \star = (-1)^{p(n-p)+s} = (-1)^{2(1)+1} = -1 ## S is the signature, p is the form, and (n-p) is the pull-back of the hodge dual. So, I think that this is right? It's been a while since I've done ## \star \star ##... Bringing it back to the full equation, we see that ##\int_R d(\delta \phi) \wedge \star d \phi - \int_R \star d \phi \wedge d(\delta \phi) = \int_R d(\delta \phi) \wedge \star d \phi -(-\int_R d(\delta \phi) \wedge \star d \phi) = \int_R d(\delta \phi) \wedge \star d \phi +d(\delta \phi) \wedge \star d \phi = \int_R 2d(\delta \phi) \wedge \star d \phi ## Which is what we want, hopefully this is better logic! [/QUOTE]
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How Do You Vary the Action of a Lagrangian for a Scalar Field?
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