Lagragian, scalar field

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Homework Statement


Vary the action of a Lagrangian for a scalar field. I kind of just need someone to read over this, i'm not sure if my steps are actually logical (especially the one before we do integration by parts).

Since this isn't actually homework, we can move it to the classical physics forum, but I felt it was more suited in this forum.

Homework Equations


## L = d\phi \wedge \star d\phi ##

The Attempt at a Solution


Okay, so let's do it.
Let ## \phi## be a scalar field. The lagrangian of the field is then ## L = d\phi \wedge \star d\phi ##. The action is then defined to be ## \int_R d\phi \wedge \star d\phi ##. Where R is the region we are integrating over. To vary the action, we must say that ## \phi_0 = \phi ##, so e can set ## \delta \phi = \frac{d \phi_{\lambda}}{d \lambda} |_{\lambda=0} ## so that ## \phi_{\lambda} =\phi_0 + \lambda \delta \phi ##

So, let's vary this action. ## \frac{ds}{d \lambda} |_{\lambda=0} = \int_R \frac{d( d\phi \wedge \star d\phi)}{d \lambda} = \int_R d(\delta \phi) \wedge \star d \phi + d \phi \wedge \star d(\delta \phi) ## To get the last step, we used a product rule and the fact that ## \delta \phi = \frac{d \phi_{\lambda}}{d \lambda} |_{\lambda=0} ##

This is a step I need help justifying, or maybe just need to be more confident in my logic. So, recall that for exterior differentiation, ## d^2 = 0 ## which we are going to use on that last term to make it go away. Since ## \delta \phi = \frac{d \phi_{\lambda}}{d \lambda} |_{\lambda=0} ##, the term ## d \phi \wedge \star d(\delta \phi) = 0 ## because ## \star d(\delta \phi) = 0 ## due to ## d^2 = 0 ##, so we can just throw that term out. Once again, I will be abusing the zero product rule on the first term, but in a different way. Since ## d(\delta \phi) \wedge \star d \phi = 0 ## due to the same reasoning above, we can just multiply it by two, since the 2 won't actually change anything, since 2*0=0. So our integral becomes ## \int_R 2 d(\delta \phi) \wedge \star d \phi ## The reason I'm unsure about this step is because of the 2. I don't know the reason for it, and it never is explained in the book. So... maybe it's a factored term? Not sure.

This looks like a place to use integration by parts, so I will. Recall the rule for integration by parts: ## \int_R d \alpha \wedge \beta = \int_R d(\alpha \wedge \beta) -(-1)^p \int_R \alpha \wedge d\beta ## Using ## d(\delta \phi) = \alpha ## and ##\star d \phi ## we see our integral become ##2 \int_R d(\delta \phi \wedge \star d \phi) - 2 \int_R \delta \phi \wedge d\star d \phi ## We drop the wedges in order to integrate, so the integrals become ##2 \int_R d(\delta \phi \star d \phi) - 2 \int_R \delta \phi d\star d \phi ##

We now invoke Stokes theorem on the first integral. Recall that Stokes is ## \int_R dF = \int_{\partial R} F ##. So our integral becomes ##2 \int_{\partial R} \delta \phi \star d \phi - 2 \int_R \delta \phi d\star d \phi ##

We now require these integrals to vanish. The second integral tells us that ## \phi d\star d \phi = 0 ## which is just saying that Laplace's equation must be satisfied. [Recall that ## d \star d = \nabla \cdot \nabla ## which is just the laplacian].

The first integral works on our boundary, which we assume is either ## \phi = 0 ## or ##\delta \phi = 0 ##. We don't HAVE to have this assumption, but I'll just end it here.

Is this good enough? I'm not sure.

Thanks for your time.
 
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Answers and Replies

  • #2
Orodruin
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You should not confuse ##\delta\phi## with the exterior derivative ##d\phi##, they are very different beasts. The first is the variation of the field with respect to some variation parameter while the second is the exterior derivative is a differentiation of an r-form (in this case r=0). It is not true that ##d\delta\phi = 0##. If it was true your variation would automatically be zero.

Your lower part looks fine to me.
 
  • #3
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You should not confuse ##\delta\phi## with the exterior derivative ##d\phi##, they are very different beasts. The first is the variation of the field with respect to some variation parameter while the second is the exterior derivative is a differentiation of an r-form (in this case r=0). It is not true that ##d\delta\phi = 0##. If it was true your variation would automatically be zero.

Your lower part looks fine to me.
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!
 
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