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For a massless scalar field the equation of motion in a general curved Space time is $\frac{1}{\sqrt{g}}\partial_\mu(\sqrt{g}g^{\mu\nu}\partial_\nu\phi)=0$. Though, in the action, we can by hand include a term $\xi R\phi^2$ as the only possible local, dimensionally correct term which has the physical significance of coupling of the scalar field with the curvature of space time itself and it modifies the EOM as $\frac{1}{\sqrt{g}}\partial_\mu(\sqrt{g}g^{\mu\nu}\partial_\nu\phi)+\xi R\phi=0$.

Now for a massless spin-1 field, I can similarly first write down an EOM as $\frac{1}{\sqrt{g}}\partial_\mu(\sqrt{g}g^{\mu\nu}\partial_\nu A_\rho)=\Box A_\rho=0$ in Lorentz gauge. My idea was that now if I want, I can by hand put a coupling (with space time curvature) term which will modify the EOM likewise. Am I correct about that? For example Birrell-Davies book equation 3.184 says $F_{\mu\nu};\,^\nu+\zeta^{-1}(A^\nu_\,;\nu);\mu=0$ which goes with what I said (with gauge choice kept more general from what I said), but Eq. 3.185 says $A_{\mu;\nu}\,^\nu+R_\mu\,^\rho A_\rho-(1-\zeta^{-1})A_{\nu;}\,^\nu\,_\mu=0$.

Can we derive it from the previous equation? How? Sorry my references were not helpful. Synge's general relativity book might have the answer but I don't have it. So, I have now two questions. Whether it is just derivable from the previous equation 3.184 (in which case I will take 3.185 to be minimally coupled case EOM), or whether I need to introduce extra term (non minimal coupling term of gauge field with curvature just like scalar case) in the action to get 3.185. If so, then I guess it will be called the non-minimal contribution. I think its a non-minimal coupling term. And in this case I guess whether I introduce it or not is my choice (if I have an option to make this choice). Please verify this too.

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This is straightforward maths - mathematical identities - and maybe it doesn't really belong to Physics SE. You should first try to understand why your first equation of motion for the scalar should have been written intelligently as $\nabla^\mu\nabla_\mu \phi$ in terms of covariant derivatives. This will require you to learn how to raise and lower indices by the metric tensor (you avoided this notation, suggesting you don't know this basic point), the Christoffel symbol, covariant derivatives, and various identities how to translate between those things. –  LuboŇ° Motl Nov 24 '11 at 7:12
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For example, the covariant derivative $\nabla_\mu V^\mu$ of a vector may be written as a partial derivative $1/\sqrt{g} \partial_\mu (V^\mu \sqrt{g})$ - there should be $-g$ everywhere in the square roots, right? - which explains the first identity for the box of the scalar. A few more are needed to deal with the curvature etc. You will also have to learn why the Riemann curvature tensor is encoded in the "commutator" of two covariant derivatives as acting on a vector etc. Many identities but it's just maths. –  LuboŇ° Motl Nov 24 '11 at 7:14
    
Thanks a lot! I will definitely do the maths. So, I guess 3.185 (the last Eq. I wrote) is the general one, in the sense that there's no room for adding terms which couples curvature and A fields (like $\xi R\phi$ as in the scalar field). Because such terms are already produced from 3.184. So, is there any distinction between minimal and non-minimal coupling in case of $A_\mu$ fields (it seems it is in built non-minimal). Sorry for all my confusions. –  user1349 Nov 24 '11 at 16:51

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