Simplification of the Proca Lagrangian

AI Thread Summary
The discussion revolves around the derivation of a specific term in the Proca Lagrangian, particularly the origin of the factor -1/(4π) in the Euler-Lagrange equation. The user seeks clarification on how this term arises from the Lagrangian for a vector field and how to simplify the equation correctly. They explore the relationship between the Lagrangian and the field strength tensor, F_{μν}, and apply the product rule to differentiate the Lagrangian. The conversation includes attempts to derive the necessary terms and factors while expressing uncertainty about the tensor calculus involved. The user ultimately seeks validation for their updated solution and whether any steps are missing.
fabstr1
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Homework Statement
How do I simplify the Proca lagrangian for a spin-1 field
Relevant Equations
L = -(1/16*pi) * ( ∂^(μ)A^(ν) - ∂^(ν)A^(μ))(∂_(μ)A_(ν) - ∂_(ν)A_(μ)) + 1/(8*pi) * (mc/hbar)^2* A^ν A_ν
Hello,
I'm trying to figure out where the term (3) came from. This is from a textbook which doesn't explain how they do it.

∂_μ(∂L/(∂(∂_μA_ν)) = ∂L/∂A_ν (1)

L = -(1/16*pi) * ( ∂^(μ)A^(ν) - ∂^(ν)A^(μ))(∂_(μ)A_(ν) - ∂_(ν)A_(μ)) + 1/(8*pi) * (mc/hbar)^2* A^ν A_ν (2)

Here is Eq (1) the Euler-Lagrange equation and Eq (2) is the lagrangian for a vector field. In the textbook they just state the term

∂_μ(∂L/(∂(∂_μA_ν)) = -1/(4*pi)*(∂^(μ)A^(ν) - ∂^(ν)A^(μ)) (3)

Where does the term -1/(4*pi) come from, and how do I cancel out the rest of the term so that the equation becomes

∂_μ(∂^(μ)A^(ν) - ∂^(ν)A^(μ)) + (mc/hbar)^2* A^ν (4)

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Hints:

##\mathcal{L}= -\frac{1}{16\pi}F^{\mu\nu}F_{\mu\nu} ##

Now:
##F^{\mu\nu}= \eta^{\mu n}\eta^{\nu m}F_{nm}##

Now use this in above equation and:

##F_{nm}=\partial_n A_m- \partial_m A_n##
And
##F_{nm}= -F_{mn}##
 
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But what happends to the rest of the term in Eq (10.17), where is the -1/(4*pi) term coming from.

L = - (1/(16*pi)) * η^(μν)η^(νμ)*F_(μν)
 
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fabstr1 said:
But what happends to the rest of the term in Eq (10.17), where is the -1/(4*pi) term coming from.

L = - (1/(16*pi)) * η^(μν)η^(νμ)*F_(μν)
Ok. I will do one more step for you.

##\mathcal{L}= \frac{-1}{16\pi} \eta^{\mu n}\eta^{\nu m} F_{nm}F_{\mu\nu}##

Now, taking the differential and applying product rule to above term throws a factor of 2(Due to symmetry of product) so this becomes (Dropping the metric tensors for brevity):

##\partial\mathcal{L}= \frac{-1}{8\pi} F_{nm}\partial F_{\mu\nu}##

Now, use the defination of ##F_{nm}##
 
I haven't taken any tensor calculus before, so I'm not sure if I'm doing it right.

∂L/∂F_{ μν } = - 1/(16*pi) * η^(μν)η^(νμ)*F_(μν)* F_(μν) = - 1/(16*pi)*η^(μν)η^(νμ)* F_(μν)^2 = -1/(8*pi)*η^(μν)η^(νμ)* F_(μν) = - 1/(8*pi)*η^(μν)η^(νμ)*(∂^(μ)A^(ν) - ∂^(ν)A^(μ))
 
I have updated my solution shown below. Is it ok, or is there something that are missing ?

 
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