Real scalar field , Action, variation, deriving EoM

In summary, the conversation discusses the Lagrange density and action, as well as the variation of the action and how it relates to the equations of motion. The key issue is finding a way to show that δ(∂ₓϕ)∂ᵤϕ+∂ᵤϕδ(∂ₓϕ)=∂ᵤϕ∂ₓ(δϕ). Through further analysis and understanding of the notation, the issue is eventually resolved.
  • #1
binbagsss
1,299
11
## 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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  • #2
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.
 
  • #3
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 seperately 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
 
  • #4
bump
 
  • #5
it's okay got it. thanks for the help guys :thumbup:
 

FAQ: Real scalar field , Action, variation, deriving EoM

1. What is a real scalar field?

A real scalar field is a mathematical concept used in physics to describe a physical quantity that varies over space and time, but has no direction. It is represented by a function that assigns a numerical value to each point in space.

2. What is the action in physics?

In physics, action refers to a mathematical quantity that describes the dynamics of a system. It is defined as the integral of the Lagrangian function over time and is used to derive the equations of motion for a physical system.

3. What is variation in the context of physics?

Variation, also known as infinitesimal variation, is a mathematical concept used in physics to describe small changes in a physical quantity. It is used in the calculus of variations to find the minimum or maximum of a functional.

4. How do you derive the equations of motion using the action?

To derive the equations of motion using the action, you must use the principle of least action, which states that the path a system takes between two points in time is the one that minimizes the action. This leads to a set of differential equations known as the Euler-Lagrange equations.

5. What are the implications of the Euler-Lagrange equations for physics?

The Euler-Lagrange equations have significant implications for physics as they provide a powerful tool for deriving the equations of motion for physical systems. They also allow for the formulation of the principle of least action, which is a fundamental principle in classical mechanics and is used in various other areas of physics.

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