Proca Lagrangian (Math troubles with four vectors)

In summary: I'll write something up for you to help, let me compose it.In summary, Elwin explains that the last term in the Lagrangian, which represents the mass of the vector particle, does contribute something to the equation of motion. He writes that it is simpler to directly use Hamilton's principle to derive the eoms, and that using the west-coast metric, the mass of the vector particle is negative. He also mentions that a massive vector field is not necessarily an (at least abelian) gauge field.
  • #1
Elwin.Martin
207
0
I'm reading Griffith's Elementary particles and I'm stuck on the math for one of the examples, could anyone show me what I'm missing or point me in the right direction?

I attached a pdf (of the word doc I was using) that shows what I did so far since I'm really bad with LaTeX and it would've taken me an hour to write it for this post.

Thanks in advanced for any and all help/direction; I feel so useless for having to ask, but I need to get through this.

Elwin

**any review material on the math would be nice if you have recommendations!**
 

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  • #2
I'll write something up for you to help, let me compose it.

So we know the last part doesn't do anything, so let's not even put it in for now till we need it.
Lets multiply out the Lagrangian to see explicitly what's going on with the indices.
[tex]
\mathcal{L}=-\frac{1}{16\pi}(\partial^a A^b\partial_a A_b -\partial^a A^b \partial_b A_a -\partial^b A^a \partial_a A_b +\partial^b A^a \partial_b A_b)
[/tex]
Now anytime you sum over repeated indices, it doesn't matter what the letter used is, because you are summing over it. So change them so that you can simplify. If you look closer at this, there are really only two different expressions with the partial derivative and the vector potential: [itex]\partial^c A^d \partial_c A_d[/itex] and [itex]\partial^d A^c \partial_c A_d[/itex]. So you have two of these, that is,
[tex]
\mathcal{L}=-\frac{1}{8\pi}(\partial^c A^d \partial_c A_d-\partial^d A^c \partial_c A_d)
[/tex]
With this in mind, try to do what you did before with this new information in your mind.
 
Last edited:
  • #3
jfy4 said:
I'll write something up for you to help, let me compose it.

Thank you very much!
 
  • #4
Why shouldn't the last term contribute anything? Of course it does contribute something, namely the mass of the vector particle.
 
  • #5
Yes, of course. I meant for
[tex]
\frac{\partial\mathcal{L}}{\partial (\partial_a A_b)}
[/tex]
we can ignore it. It's definitely important, as the distinguishing factor between E&M.
 
  • #6
I see. BTW, it's definitely simpler to directly use Hamilton's principle to derive the eoms. I'll write the Proca action in modern Heaviside-Lorentz units, leading to

[tex]S[A_{\mu}]=\int \mathrm{d}^4 x \left (-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{1}{2} A_{\mu} A^{\mu} \right ).[/tex]

I use the west-coast metric. That's why I have a + sign in front of the mass term. Variation with respect to [itex]A_{\mu}[/itex] and using the anti-symmetry of the Faraday tensor, [itex]F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}[/itex] leads to

[tex]\delta S=\int \mathrm{d}^4 x \left ( -F^{\mu \nu} \partial_{\mu} \delta A_{\nu} + m^2 A_{\nu} \delta A^{\nu} \right )= \int \mathrm{d}^4 x \left (\partial_{\mu} F^{\mu \nu} + m^2 A^{\nu} \right ) \delta A_{\nu} \stackrel{!}{=}0.[/tex]

From this we have the equation of motion

[tex]\partial_{\mu} F^{\mu \nu} + m^2 A^{\nu}=0.[/tex]

First from this you get by taking the four-divergence of this equation, using [itex]\partial_{\mu} \partial_{\nu} F^{\mu \nu}=0[/itex]

[tex]m^2 \partial_{\nu} A^{\nu}=0.[/tex]

The Proca field with non-vanishing mass is thus necessarily transversal (contrary to the massless photon field).

From this we can rewrite the eom.
[tex]\partial_{\mu} F^{\mu \nu}=\partial_{\mu} (\partial^{\mu} A^{\nu}-\partial^{\nu} A^{\mu})=\Box A^{\nu}=-m^2 A^{\nu}.[/tex]

This shows that [itex]m[/itex] is indeed the mass of the vector particle.

One should also mention that a massive vector field is not necessarily an (at least abelian) gauge field as turns out to be the case for massless vector fields. However to build renormalizable interacting field theories one can make the Proca Lagrangian U(1)-gauge invariant by adding an additional scalar field (the Stueckelberg ghost) and then use the usual minimal-coupling approach to couple the massive vector field to other matter fields (e.g., scalar and/or Dirac fields) to get a gauge-invariant theory with massive gauge fields. It turns out that after gauge fixing the Stueckelberg ghost doesn't couple to the other fields as don't the Feynman-Faddeev-Popov ghosts in this abelian-gauge model.

This socalled Stueckelberg realization of massive vector fields doesn't work for non-abelian gauge theories, i.e., then the only way to make the non-abelian gauge fields massive without destroying local gauge invariance is via the Higgs mechanism, where you absorb the would-be Goldstone bosons into the gauge field but always keep at least one additional massive Higgs boson left in the physical particle spectrum.
 
  • #7
Thanks vanhees71,

That's nice to see :smile:
 

1. What is the Proca Lagrangian?

The Proca Lagrangian is a mathematical formulation used in theoretical physics to describe the dynamics of a vector field. It is named after Romanian physicist Alexandru Proca, who first introduced it in 1936.

2. How is the Proca Lagrangian used in physics?

The Proca Lagrangian is commonly used in theories involving massive vector particles, such as the electromagnetic field. It is also used in the theory of quantum electrodynamics to describe the interactions between charged particles and the electromagnetic field.

3. What are four-vectors in the context of the Proca Lagrangian?

Four-vectors are mathematical objects that have four components, including three spatial components and one time component. In the context of the Proca Lagrangian, four-vectors are used to describe the properties and behavior of vector fields in four-dimensional spacetime.

4. What are some issues or challenges with using the Proca Lagrangian?

One of the main challenges with using the Proca Lagrangian is that it only applies to vector fields with mass. This means that it cannot be used to describe massless particles, such as photons. Additionally, the Proca Lagrangian does not account for interactions between multiple vector fields, making it less applicable to complex physical systems.

5. How does the Proca Lagrangian relate to other Lagrangians?

The Proca Lagrangian is a special case of a more general type of Lagrangian known as a gauge theory. It is also related to other Lagrangians used in quantum field theory, such as the Yang-Mills Lagrangian. However, the Proca Lagrangian has its own unique properties and applications in theoretical physics.

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