Simplification of the Proca Lagrangian

In summary, the term -1/(4*pi) comes from Eq (1), and is used to cancel out the rest of the term in Eq (10.17).
  • #1
fabstr1
4
1
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)

qft.jpg
 
  • Like
Likes JD_PM
Physics news on Phys.org
  • #2
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}##
 
  • Like
Likes JD_PM
  • #3
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)
 
Last edited:
  • #4
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}##
 
  • #5
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))
 
  • #6
I have updated my solution shown below. Is it ok, or is there something that are missing ?

 

FAQ: Simplification of the Proca Lagrangian

1. What is the Proca Lagrangian?

The Proca Lagrangian is a mathematical expression used in quantum field theory to describe the dynamics of a massive spin-1 particle, such as a photon or a W or Z boson.

2. Why is it important to simplify the Proca Lagrangian?

Simplification of the Proca Lagrangian is important because it allows for a more manageable and intuitive understanding of the behavior of spin-1 particles. It also makes calculations and predictions easier to perform.

3. How is the Proca Lagrangian simplified?

The Proca Lagrangian can be simplified by using gauge transformations to eliminate redundant terms and by making assumptions about the behavior of the particle, such as its mass and interactions.

4. What are the benefits of simplifying the Proca Lagrangian?

Simplifying the Proca Lagrangian allows for a deeper understanding of the underlying physics and can lead to more accurate predictions and calculations. It also makes the equations more elegant and easier to work with.

5. Are there any limitations to simplifying the Proca Lagrangian?

While simplification of the Proca Lagrangian can be beneficial, it may also result in loss of information or oversimplification of the system. It is important to carefully consider the assumptions and approximations made during the simplification process.

Back
Top