What's the form of EL equation for KG field with gauge covariant derivative?

Question

In the situation when we have KG Lagrangian with normal derivative replaced by the covariant one (I'm using metric $\operatorname{diag}\{ +,-,-,- \}$ so $D_\mu = \partial_\mu + iq A_\mu,\ D^\mu = \partial^\mu + iq A^\mu$) $$ \mathcal{L}_{\text{KG}}' = \frac12 D_\mu \phi D^\mu \phi - \frac12 m^2\phi^2 = \frac12 \left( \partial_\mu \phi \partial^\mu \phi + iq A_\mu \phi \partial^\mu \phi + iq \partial_\mu \phi \cdot A^\mu \phi - q^2 A^2 \phi^2 \right) - \frac12 m^2 \phi^2, $$ I'm confused with which of the 2 forms (or some other one) below should be the proper one for EL equation (and why), say $$ D_\mu \frac{\partial \mathcal{L}_{\text{KG}}'}{\partial (D_\mu \phi)} = \frac{\partial \mathcal{L}_{\text{KG}}'}{\partial \phi} \quad \text{or} \quad \partial_\mu \frac{\partial \mathcal{L}_{\text{KG}}'}{\partial (\partial_\mu \phi)} = \frac{\partial \mathcal{L}_{\text{KG}}'}{\partial \phi}. $$ Both of them have the same result for RHS to be $$ \text{RHS} = iq A_\mu \partial^\mu \phi - q^2 A^2 \phi - m^2 \phi, $$ while results for LHS are different. To be specific, firstly I find it that $$ \frac{\partial \mathcal{L}_{\text{KG}}'}{\partial (D_\mu \phi)} = \frac{\partial}{\partial (D_\mu \phi)} \left( \frac12 D_\mu \phi D^\mu \phi - \frac12 m^2\phi^2 \right) = D^\mu \phi,\\ \frac{\partial \mathcal{L}_{\text{KG}}'}{\partial (\partial_\mu \phi)} = \frac{\partial}{\partial (\partial_\mu \phi)} \left( \frac12 \left( \partial_\mu \phi \partial^\mu \phi + iq A_\mu \phi \partial^\mu \phi + iq \partial_\mu \phi \cdot A^\mu \phi - q^2 A^2 \phi^2 \right) - \frac12 m^2 \phi^2 \right) = \partial^\mu \phi + iq A^\mu \phi = D^\mu \phi, $$ which has already seemed to be strange for me. Then by calculating the LHS of 2 possible forms above I obtain $$ \text{LHS}_{D_\mu} = D_\mu D^\mu \phi = \partial_\mu \partial^\mu \phi + iq \partial_\mu (A^\mu \phi) + iq A_\mu \partial^\mu \phi - q^2 A^2 \phi,\\ \text{LHS}_{\partial_\mu} = \partial_\mu D^\mu \phi = \partial_\mu \partial^\mu \phi + iq \partial_\mu (A^\mu\phi), $$ which can be combined with RHS to give $$ D_\mu\ \text{form}: (\partial^2 + m^2) \phi = -iq (\partial_\mu A^\mu + A^\mu \partial_\mu) \phi,\\ \partial_\mu\ \text{form}: (\partial^2 + m^2) \phi = -iq (\partial_\mu A^\mu) \phi - q^2 A^2 \phi. $$ However, the right answer seems to be (see Eq.(3.100) of Gauge Theories in Particle Physics (40th Anniversary Edition, Volume 1) by Ian J.R. Aitchison and Anthony J.G. Hey) $$ \boxed{(\partial^2 + m^2) \phi = -iq(\partial_\mu A^\mu + A^\mu \partial_\mu) \phi + q^2 A^2 \phi}. $$ So what's the right form of EL equation in this situation, and also what's my mistake in the calculation? Besides, what would be the case if we are dealing with (metric) covariant derivative $\nabla_\mu$ in GR, and maybe some more general one?

Answer 1

Remember EL equations are derived from the least-action principle:

$$ \frac{\delta S}{\delta\phi} = \frac{\delta \int{d^4x \mathcal{L}}}{\delta \phi}=0 $$

Expanding this out using your Lagrangian density gives, $$ -\partial_\mu\partial^\mu\phi + \frac{1}{2}\left(iqA^\mu\partial_\mu\phi - iq\partial_\mu(A^\mu\phi) - iq \partial^\mu(A_\mu\phi) + iqA^\mu\partial_\mu\phi \right) - q^2A^2\phi - m^2\phi = 0 $$

Simplifying:

$$ -\partial_\mu\partial^\mu\phi + iqA^\mu \partial_\mu\phi - iq \partial_\mu(\phi A^\mu) - q^2A^2\phi - m^2\phi = 0 $$

Further simplifying:

$$ -\partial_\mu\partial^\mu\phi -iq\phi\partial_\mu A^\mu - q^2A^2\phi - m^2\phi = 0 $$

So, it's the $\partial_\mu$ form of the E-L equation that is correct for the Lagrangian density you've written down.

With regards to the result from the textbook, I believe this result would be using a complex valued $\phi$ field, representing a charged particle.

The Lagrangian density should be

$$ \mathcal{L}=\frac{1}{2}D_\mu\phi\overline{D^\mu\phi}-\frac{1}{2}m^2\phi^2 =\frac{1}{2}\left( \partial_\mu\phi\partial^\mu\overline{\phi}+iq(\partial_\mu\phi)\overline{\phi}A^\mu -iq(\partial_\mu\overline{\phi})\phi A^\mu +q^2A^2\phi^2-m^2\phi^2 \right) $$

where we can take a variation of the action independently for the field $\phi$ and it's complex conjugate $\overline{\phi}$. Taking the variation of $S=\int{d^4x\mathcal{L}}$ by $\overline{\phi}$ and setting it to zero gives,

$$ \frac{1}{2}\left(-\partial_\mu\partial^\mu\phi+iqA^\mu\partial_\mu \phi + iq\partial_\mu(\phi A^\mu) + q^2A^2\phi -m^2\phi\right)=0 $$

which simplifies to

$$ (\partial_\mu\partial^\mu+m^2)\phi=iq(A^\mu\partial_\mu+\partial_\mu A^\mu)\phi +a^2A^2\phi $$

For a general coordinate system and/or spacetime metric, you can derive the EL equations by again using the least action principle $\delta S/\delta\phi=0$. In a general spacetime, one action that works is, $$ S=\int{d^4x \sqrt{-g}\left( D_\mu\overline{D^\mu}\phi -m^2\phi^2 \right)} $$

where $g = det(g_{ab})$. But this time when you apply the least action principle, one must be careful to remember that $g$ is a function of the coordinates.