Real scalar field , Action, variation, deriving EoM

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Homework Help Overview

The discussion revolves around the derivation of equations of motion from a real scalar field's action and Lagrangian density. The original poster is attempting to understand the variation of the action with respect to the field and its derivatives, specifically focusing on the term involving the derivatives of the field.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster discusses the variation of the action and attempts to derive the necessary terms for integration by parts. They express confusion about how to manipulate the variation of the derivative terms.
  • Some participants question the correctness of the original poster's expressions and suggest that there may be missing factors in their calculations.
  • There is a mention of the need to clarify the meaning of the notation used, particularly regarding the raised and lowered indices in the derivatives.

Discussion Status

The discussion is ongoing, with participants providing feedback and corrections to the original poster's expressions. Some guidance has been offered regarding the handling of variations, but there is no explicit consensus on the resolution of the original poster's confusion.

Contextual Notes

The original poster acknowledges a potential typo in their calculations and expresses uncertainty about the notation used for derivatives. There is an indication that the discussion is constrained by the need for clarity in the mathematical expressions involved.

binbagsss
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## L(x) = L(\phi(x), \partial_{u} \phi (x) ) = -1/2 (m^{2} \phi ^{2}(x) + \partial_{u} \phi(x) \partial^{u} \phi (x))## , the Lagrange density.
## S= \int d^{4}(x) L (x) ##, the action.

## \phi -> \phi + \delta \phi ## (just shortened the notation and dropped the x dependence)

I have ## \delta S = \int d^{4} x ( -1/2( \delta (m^{2} \phi ^{2}) + \delta (\partial_{u} \phi \partial^{u} \phi )) =0 ## last equality by principle of least action .

I have ##\delta (m^{2} \phi ^{2}) = m^{2} \phi \delta \phi ## , which is fine.

I'm having problems with the next term:

##\delta (\partial_{u} \phi \partial^{u} \phi )) = \delta (\partial_{u} \phi ) \partial^{u} \phi + \partial_{u} \phi \delta ( \partial^{u} \phi ) ##

I need to show that is equal to ##\partial_{u} \phi \partial^{u} (\delta \phi) ## I am then fine with the rest of the derivation, which involves doing integration by parts on this term to 'change' ## \partial^{u} (\delta \phi) ## to ##\delta \phi## and then loosing this arbitrary variation in ##\phi## to get the equations of motion.

I am clueless how to get ## \delta (\partial_{u} \phi ) \partial^{u} \phi + \partial_{u} \phi \delta ( \partial^{u} \phi ) = \partial_{u} \phi \partial^{u} (\delta \phi) ##, any tips getting starting greatly appreciated.

Many thanks in advance.
 
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binbagsss said:
I have δ(m2ϕ2)=m2ϕδϕδ(m2ϕ2)=m2ϕδϕ\delta (m^{2} \phi ^{2}) = m^{2} \phi \delta \phi , which is fine
This is not fine, you are missing a factor of two. You are also missing the same factor of two in the other relation.

Try to use this to figure out how to solve your problem.
 
Orodruin said:
This is not fine, you are missing a factor of two. You are also missing the same factor of two in the other relation.

Try to use this to figure out how to solve your problem.

apologies that was a typo, well it cancels with the 1/2 anyway.
I don't see how it sheds any light on the issue I stated - the variation is acting separately on ##\partial_{u}## and ##\partial^{u}##.
I think that ## \delta ( \partial_{u} \phi ) = \delta \phi' ##, were ##\phi' = \partial x_{u} \phi ##, but I don't know what is ##partial^{u} \phi ##, actually...what is ## \partial^{u} ## ? , ##\partial_{u} x = \partial / \partial x^{u} ## and ##\partial x^{u} ## can be attained by raising a index in this? but I don't really know what it is, so I don't know what ##partial^{u} \phi ## is.

Many thanks
 
bump
 
it's okay got it. thanks for the help guys :thumbup:
 

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