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The most general Lorentz invariant, renormalizable Lagrangian for a spin 1 field $A_\mu$ reads

\begin{equation}\mathscr{L}_{\text{Proca}}= C_1 \partial^\mu A^\nu \partial_\mu A_\nu + C_2 \partial^\mu A^\nu \partial_\nu A_\mu + C_3 A^\mu A_\mu + C_4 \partial^\mu A_\mu.\end{equation}

However, in the textbooks this Lagrangian only shows up as a very special case

\begin{equation}\mathscr{L}_{\text{Proca}}= \frac{1}{2}(\partial^\mu A^\nu \partial_\mu A_\nu - \partial^\mu A^\nu \partial_\nu A_\mu ) + m^2 A^\mu A_\mu .\end{equation}

  • Why is the term linear in $A_\mu$ usually neglected, i.e. why is $C_4=0$ usually chosen? What would change for $C_4 \neq 0$?
  • What would go wrong or what would change if we wouldn't consider the very special case $C_2/C_1=-\frac{1}{2}$?
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    $\begingroup$ Some comments: 1) Note that the $C_4$ term doesn't enter into the equations of motion - it is a total derivative - so at the classical level it is irrelevant. 2) In the knowledge that we want this Lagrangian to describe a quantum field theory, we must worry about the number of degrees of freedom it contains. Choosing $C_1$ and $C_2$ appropriately, we can arrange for $\partial_\mu A^\mu = 0$ on-shell; this is important, since a spin-1 particle has at most three degrees of freedom. See Schwartz (QFT and the Standard Model) section 8.2 for more information. $\endgroup$
    – gj255
    Jul 26, 2017 at 13:20
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    $\begingroup$ @gj255 welp, that definitely looks like an answer to me. Consider posting it in the answer section at some point. $\endgroup$ Aug 3, 2017 at 11:33
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    $\begingroup$ @gj255 I also agree that your post is much more like an answer than like a comment. $\endgroup$
    – rob
    Aug 3, 2017 at 11:48

2 Answers 2

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The equations of motion (and thus the free particle states) will not be changed if we add a term to the Lagrangian which is in the form of a total divergence of a vector $\partial_\mu W^\mu$. Consider the $C_2$ term and rewrite it, up to divergences, as: $$\partial^\mu A^\nu \partial_\nu A_\mu = \partial^\mu(A^\nu \partial_\nu A_\mu) - A^\nu \partial^\mu \partial_\nu A_\mu = \partial^\mu(A^\nu \partial_\nu A_\mu) - \partial_\nu(A^\nu \partial^\mu A_\mu) - (\partial_\mu A^\mu)^2$$ We can now redefine the quantities in your Lagrangian as $$\epsilon = sign (C_1), \, \eta = sign(C_3), \xi = \frac{|C_1|}{C_1 + C_2}, m^2 = |\frac{C_4}{C_1}|, V_\mu = 2 \sqrt{|C_1|} A_\mu, F_{\mu\nu} = V_{\mu,\nu} - V_{\nu,\mu} $$ to obtain $$\mathcal{L}_\mathrm{Proca} = -\frac{1}{4}\epsilon F^{\mu\nu}F_{\mu\nu} + \frac{m^2}{2} \eta V^\mu V_\mu - \frac{1}{2 \xi} (\partial^\mu V_\mu)^2$$ As an exercise, you can compute the equations of motion of this Lagrangian, decompose the linear equations into a transversal solution $V^\mu_\mathrm{T}$ which has $\partial_\mu V^\mu = 0$, and longitudinal solutions $V^\mu_\mathrm{L}$ which have nonzero divergence, and you will obtain separate equations of motion in the form $$(\Box + m^2) V^\mu_\mathrm{T} = 0$$ $$(\Box + \eta \xi m^2) V^\mu_\mathrm{L} = 0$$ I.e., the transversal solutions are a Proca field of mass $m$. Further analysis shows you that $V^\mu_\mathrm{L}$ is in fact a spin-0 field of mass $\sqrt{\eta \xi} m$, $V^\mu_\mathrm{L} \sim \partial^\mu \phi$. This also explains why we write the Proca kinetic term as $\sim F^{\mu\nu} F_{\mu\nu}$, because it includes only transversal kinetics and $F^{\mu\nu}_\mathrm{L} \sim \partial_\mu \partial_\nu \phi - \partial_\nu \partial_\mu \phi = 0$.

The limit $\xi \to \infty$ makes this scalar mode infinitely heavy and thus inactive, but in some approaches to quantization of Proca field, $\xi$ is left finite, canonical quantization is executed, and only after that you take the $\xi \to \infty$ limit.

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  • $\begingroup$ Are you sure about $V_L^\mu \sim \partial^\mu \phi$? I ask because its kinetic term in the Lagrangian looks to be $4^{\mathrm{th}}$ order, which is considered to be unphysical because of a number of theorems related to the positivity of the Hamiltonian. $\endgroup$ Aug 26, 2017 at 13:13
  • $\begingroup$ Right now, I am not exactly sure how to write the Lagrangian above as a sum of the canonical Proca Lagrangian and a Lagrangian of the scalar field. However, the statement of which I am certain and which is easy to verify is: the general solution for $V^\mu$ can be written as $V^\mu = V^\mu_\mathrm{T} + \partial^\mu \phi$ where $V^\mu_\mathrm{T}$ would be the "usual" general transversal solution to the Proca-field equations and $\phi$ is the general solution to the scalar field equation with mass $\sqrt{\eta \xi}m$. $\endgroup$
    – Void
    Aug 26, 2017 at 13:47
  • $\begingroup$ The term with C2 is a total divergency but it does lead to a different equation of motion. With C1=1 and C2=C3=C4=0 you get the wave equation, with C1=-C2=1 and C3=C4=0 you get Maxwell's equations - which btw reduce to the wave equation if the Lorentz condition is applied. $\endgroup$
    – my2cts
    Apr 21, 2018 at 10:07
  • $\begingroup$ @my2cts No, it is not a total divergence, see the first line, it is $-(\partial_\mu A^\mu)^2$ plus terms that are a divergence. I agree with the rest of your statements. $\endgroup$
    – Void
    Apr 21, 2018 at 11:31
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I tend to believe that there is a mistake in the question. May you check ?

Since you have a factor 1/2 in front of the mass term, you are dealing with the complex case of field.

So in this case, how could you have a factor 1/2 in front of the kinematic term ?

Also, your sign seems in the wrong direction.

Have you really seen a book that contained your formula ? May you give the reference ?

Please find below my demonstration that there is a problem.

Do you see a problem in my derivation ? (I start from formula of wikipedia)

enter image description here

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