I feel a bit embarrassed that I can't figure this out on my own, but I am trouble deriving the equations of motion for the Lagrangian $$\mathcal{L}_{eff}=(\omega+eA_0)^2f(r)^2-(\partial_r f(r))^2-V(f)+\frac{1}{2}(\partial_r A_0(r))^2.$$ This effective Lagrangian comes from taking a Maxwell-scalar field theory and assuming a spherical ansatz for the scalar field and the $A_0$ component of the gauge field. If I treat $f(r)$ and $A_0(r)$ as the dynamical variables, then the Euler-Lagrange equations that yield the equations of motion should be \begin{align} 0&=\partial_\mu\frac{\partial \mathcal{L}}{\partial(\partial_\mu f)}-\frac{\partial \mathcal{L}}{\partial f}\\ 0&=\partial_\mu\frac{\partial \mathcal{L}}{\partial(\partial_\mu A_0)}-\frac{\partial \mathcal{L}}{\partial A_0}\\ \end{align} and since the only dependence in the Lagrangian is on the radial coordinate $r$, this should just reduce to \begin{align}0&=\frac{d}{dr}\left(\frac{\partial \mathcal{L}}{\partial f'}\right)-\frac{\partial \mathcal{L}}{\partial f}\\ 0&=\frac{d}{dr}\left(\frac{\partial \mathcal{L}}{\partial A_0'}\right)-\frac{\partial \mathcal{L}}{\partial A_0}. \end{align} The terms are $$\frac{\partial \mathcal{L}}{\partial f'}=-2f'$$ $$\frac{\partial \mathcal{L}}{\partial f}=(\omega+eA_0)^2\cdot 2f-\frac{dV(f)}{df}$$ and $$\frac{\partial \mathcal{L}}{\partial A_0'}=A_0'$$ $$\frac{\partial \mathcal{L}}{\partial A_0}=2ef(r)^2(\omega+eA_0).$$

Plugging these terms into the Euler-Lagrange equations just gives

\begin{align} 0&=f''(r)+f(r)(\omega+eA_0)^2-\frac{1}{2}\frac{dV(f)}{df}\\ 0&=A_0''(r)-2ef(r)^2(\omega+eA_0). \end{align}

The problem is that the correct equations of motion are apparently

\begin{align} 0&=f''(r)+\color{red}{2\frac{f'(r)}{r}}+f(r)(\omega+eA_0)^2-\frac{1}{2}\frac{dV(f)}{df}\\ 0&=A_0''(r)+\color{red}{2\frac{A_0'(r)}{r}}-2ef(r)^2(\omega+eA_0). \end{align} For example, this result is given as equations (7), (8) of of this paper (Gulamov 2015) and further corroborated by equations (4.1), (4.2) of this famous paper (Lee 1989).

I am obviously missing the $1/r$ terms, but I cannot see where this would come from. Can anyone provide see what I am missing in this simple derivation?

  • 2
    $\begingroup$ I am not accustomed to this notation, but isn't the $\partial_\mu \frac{\cdot}{\cdot}$ a divergence? Results of vector calculus operations in spherical coordinates are not trivial, maybe check the usual 3D versions? $\endgroup$
    – acarturk
    Commented Nov 3, 2019 at 18:55
  • 2
    $\begingroup$ The terms in red are additional terms due to the curvature in spherical coordinates, i.e $\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{df}{dr}\right)=\frac{d^2f}{dr^2}+\frac{2}{r}\frac{df}{dr}$, perhaps that gives you a lead $\endgroup$
    – nluigi
    Commented Nov 3, 2019 at 23:30

1 Answer 1



  1. Recall that the action $S=\int \! \mathbb{L}$ is an integral over the Lagrangian 4-form $\mathbb{L}~=~ d^4x~ {\cal L}$, where ${\cal L}$ is the Lagrangian density. It's a density in the sense that under a coordinate transformation $x\to x^{\prime}$, it transforms as ${\cal L}^{\prime}={\cal L}/J$ with the inverse Jacobian $J:=\det\frac{\partial x^{\prime}}{\partial x}$. It this way the action $S$ is a scalar/invariant under coordinate transformations.

  2. The inverse Jacobian from 3D rectangular to 3D spherical coordinates is $J^{-1}=\color{red}{r^2\sin\theta}$, so in spherical coordinates the Lagrangian density reads ${\cal L}^{\prime}=\color{red}{r^2\sin\theta} ~{\cal L}$.

  3. Now because of OP's spherical ansatz, the factor $\color{red}{\sin\theta}$ is not important, but the factor $\color{red}{r^2}$ is, and it leads to the sought-for correct EL equations.


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