Why the Proca Lagrangian has an inverted sign? I've recently finished reading Mark Thomson's "Modern Particle Physics". There is a question which is not answered in his book, and to which I couldn't find on the internet in the "introductory" parts of QFT.
My problem is with the form of the Proca Lagrangian. Indeed, for a massless field (Say in QED) the lagrangian is given in the book as:
$$L = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}.$$
I am placing myself in the vacuum here. As we can see, there is a minus sign in front of the "kinetic" term. This does not affect dynamics in this case, but it looked already unusual to me since usually the kinetic term is taken with a positive sign. If we generalise to a massive boson, we get:
$$L = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} +\frac{1}{2}m^2A^\mu A_\mu$$
with the $(+,-,-,-)$ Minkowski sign convention. Again, the overall sign is the opposite of the usual convention, otherwise nothing really strange here.
However, my problem arises when we consider the lagrangian of spin-half particle interacting with QED (but imagining the photon has a mass, I know it's non-physical but it's an easier example for me :D). Then, the total Lagrangian becomes, if I'm not mistaken:
$$ L =i\overline{\psi}\gamma^\mu D_\mu\psi-m\psi\overline{\psi}-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^2A^\mu A_\mu$$
And here, the seemingly arbitrary sign isn't anymore... What is the justification for this inverted kinetic and mass sign ? I guess that for massless bosons, the sign is still arbitrary since the $F_{\mu\nu}$ term decouples from the rest of the lagrangian, but I don't think it is with the mass term (haven't done the calculations though, so shame on my lazyness if that is the answer). I know my lagrangian is not U(1) gauge invariant, so my guess is maybe that strange sign comes from the spontaneous symmetry breaking from the Higgs mechanism ?
Edit: Some precisions about what I mean by "Inverted sign" :
If we take say a scalar field with a mass, it's lagrangian will be composed of a kinetic term $\frac{1}{2}\partial_\mu\phi\partial^\mu\phi$ and a mass term $\frac{1}{2}m^2\phi^2$. The lagrangian then is 
$$L = T-V = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2$$ 
with the $(+,-,-,-)$ Minkowski sign convention.
If I interpret $F_{\mu\nu}F^{\mu\nu}$ as a kinetic term, then we see that in the lagrangian I wrote up, the signs of the terms are inverted. I would like to know if this is a purely conventional choice (which I know it is, in the case of the isolated lagrangian, but not sure in the combined one), and if it's not, why it is written like that.
 A: The sign is not arbitrary. It is easiest to understand when you look at the propagator, i.e. inverse equation of motion. For a scalar field in momentum space
$$
\Delta_F(p) = \frac{i}{p^2-m^2}
$$
For a  vector field 
$$
\Delta^{\mu \nu}_F(p) = i\frac{-g^{\mu \nu}+\frac{p^\mu p^\nu}{m^2}}{p^2-m^2}.
$$
And now look at the physical components. For a scalar field, there is only one component, and the residue of the propagator is positive. This sign corresponds to the plus sign in front of the kinetic term in the Lagrangian. 
For a vector field, only physical components propagate with a positive sign. The $0$'th component is unphysical, and hence it has a wrong sign. Again this is a consequence of the sign in Lagrangian and the signs in the metric tensor. 
In other words, for both cases - scalar and vector, only the physical degrees of freedom have positive kinetic energy. 
The same argument applies to mass term. For a vector field 
$$
m^2 A^\mu A_\mu = m^2 A_0^2-m^2A_iA_i,
$$
so the physical components have the same sign of a mass term like the scalar field. The unphysical 0'th component has opposite sign.
Causality and unitarity depend on this sign. Kinetic energy and mass have to be non-negative for physical degrees of freedom. It is easy to see that this sign choice for the Lagrangian leads to a Hamiltonian that has non-negative eigenvalues; or more precisely, that is bounded from below. 
There are no physical polarization states associated with $A_0$, so this field is unphysical, it carries no energy or momentum. In the non-relativistic approximation (in certain gauge), it can correspond to instantaneous potential  (the so-called potential photons in NRQED) which violate causality because they generate non-local interactions. But this is just an approximation, expansion in small velocity compared to the speed of light.
For Proca field, we can use equation of motion 
$$
\partial_\mu A^\mu =0
$$
to show that only three out of four components of the field are independent. This means that using 4-component vector $A$ field to describe massive vector boson, we have one unphysical field and three physical fields. Note that this equation means that the longitudinal field is noninteracting.  
A: The answer sits in the fact that you want your real photons to have the correct sign. This can be realized by adding a gauge fixing term $\frac{1}{2}(\partial_\mu A^\nu)^2 = 0$. 
(The additional term is due to zero because I chose a gauge where it is!)
$$\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} - \frac{1}{2}(\partial_\mu A^\nu)^2 + \frac{1}{2}m^2A_\mu A^\mu$$
It is now easy to write out $F$ and perform one partial integration in order to end up with:
$$\mathcal{L} = -\frac{1}{2}\partial_\mu A_\nu \partial^\mu A^\nu + \frac{1}{2}m^2A_\nu A^\nu$$
Now we use our metric $$+ ---$$ to write out the summation over $\nu$ index:
$$\mathcal{L} = -\frac{1}{2}\partial_\mu A_0 \partial^\mu A^0 + \frac{1}{2}m^2A_0 A^0 + \frac{1}{2}\partial_\mu A_i \partial^\mu A^i - \frac{1}{2}m^2A_i A^i  $$
Which is to be understood as the Lagrangian of 4 decoupled scalar fields $A_\mu$. As you already noticed one of them has "flipped" sign and is therefore unphysical. We find another unphysical field that is tangent to the momentum of the photon. It turns out that these two unphysical fields will always cancel out in calculations !
So to recapitulate, there is indeed one field with it's sign "flipped" but it always cancels out against the longitudinal photon. We end up with 2 real photon polarizations as was to be expected :)
EDIT: notice that changing to a metric with (-+++) signature which is typically done in more theoretical papers will change a whole bunch of signs !
