Deviation of light rays in a scalar gravity theory (simple modification of Nordström theory) I'm considering a simple scalar theory of gravity in Minkowski spacetime, which isn't exactly the same as the old Nordström theory.  The scalar gravity field $\phi$ and the electromagnetic field $A_a$ Lagrangian is the following (I'm using units such that $c \equiv 1$ and metric signature $\eta = (1, -1, -1, -1)$):
$$\tag{1}
\mathcal{L} = \frac{1}{2 k} \, \eta^{ab} (\partial_a \, \phi)(\partial_b \, \phi) - \frac{1}{4} \, \mathcal{Q}(\phi) F_{ab} \, F^{ab} + \mathscr{L}_{\text{mat}},
$$
where $k = 4 \pi G$ is the gravitation coupling constant and $\mathcal{Q}(\phi)$ is an arbitrary function of $\phi(x)$.  From some arguments that aren't given here, we should use $\mathcal{Q}(\phi) = e^{\phi}$.  In the case of a weak field, we have $\mathcal{Q} \approx 1 + \phi$.  Neglecting the contribution of matter, an arbitrary variation $\delta\phi$ give the following equation for the dynamics of the scalar field:
$$\tag{2}
\square \, \phi = \frac{k}{\mathcal{Q}} \, \frac{d\mathcal{Q}}{d\phi} \, \mathscr{L}_{\text{EM}} = -\, \frac{k}{4} \, e^{\phi} F_{ab} \, F^{ab}.
$$
This is not the usual Nordström equation, which uses the trace of the energy-momentum tensor (which is 0 for the electromagnetic field).  Here, the electromagnetic field do generate some gravitation (unless $F_{ab} \, F^{ab} = 0$, as is the case for the usual plane wave solution).  So I'm expecting that a light ray may be subject to a deviation, in this scalar field theory, if the classical plane wave solution isn't a solution anymore, to the modified Maxwell equation! (see below).  An arbitrary variation $\delta A_a$ of the electromagnetic potential give the following equation:
$$\tag{3}
\partial_a (\mathcal{Q} \, F^{ab}) = 0 \qquad \Rightarrow \qquad \partial_a  F^{ab} = -\, F^{ab} \, \partial_a \, \phi.
$$
Hence, Maxwell's equation in vacuum is modified by the gravitational potential $\phi$ (there isn't any conflict with gauge invariance and charge conservation).
The usual Lagrangian of a massless particle predicts that there's no light rays bending (but it's not obvious that this Lagrangian should stay the same, with only a new $\mathcal{Q}(\phi)$ factor in it).  I'm getting something like
$$\tag{4}
\frac{d}{d\sigma} \Bigl(\mathcal{Q} \frac{d x^a}{d\sigma} \Bigr) = 0 \qquad \Rightarrow \qquad \mathcal{Q} \, k^a = \text{constant}.
$$
Apparently, this relation implies there should be some redshift, but no deflection.  It's not so obvious that (3) predicts the same (redshift, but no deflection), in the light rays approximation.
So here's the question:  How can we deduce that there shouldn't be any light ray deflection, from equations (2) and (3)?

EDIT: The ansatz $A_a(x) \approx \varepsilon_a \, e^{i k \cdot x}$ (with the usual constraints $\eta_{ab} \, k^a \, k^b = 0$ and $\eta_{ab} \, k^a \, \varepsilon^b = 0$) give the plane wave field
$$\tag{5}
F_{ab} = \partial_a \, A_b - \partial_b \, A_a \approx i (k_a \, \varepsilon_b - k_b \, \varepsilon_a) \, e^{i k \cdot x}.
$$
This plane wave isn't a solution to the modified Maxwell equation (3) because of the factor $\mathcal{Q}(x) = e^{\phi(x)}$.
 A: You are correct that strictly speaking, a general plane wave will not satisfy the equations of motion.  However, we can show that plane waves will approximately travel along null geodesics so long as their wavelengths are much smaller than the scale of variation of the scalar field $\phi$ — the so-called "geometric optics" approximation, which is what we're implicitly using when we say that "light rays travel along null geodesics."
To see this, consider an ansatz of the form
$$
A_a = \mathcal{A}_a  e^{i S}
$$
where $\mathcal{A}_a$ and $S$ are both functions of spacetime and the derivatives of $S$ are "large" compared to those of $\mathcal{A}_a$ and of $\phi$.  Plugging this equation into the equation of motion above, imposing the usual gauge condition $\partial_a A^a = 0$, and organizing the terms according to the number of derivatives of $S$ they contain, we get
$$
\left[ - (\partial_a S) (\partial^a S) + i \Box S \right] \mathcal{A}^b + \left[\text{terms containing only one derivative of $S$}\right] + \left[ \text{terms containing no derivatives of $S$} \right] = 0 \tag{1}
$$
Since the leading terms involve two derivatives of $S$, they must be "approximately" zero:
$$
(\partial_a S) (\partial^a S) \approx 0 \qquad \text{and} \qquad \Box S \approx 0.
$$
These two equations admit local solutions of the form $S = k_a x^a$, with $k_a k^a = 0$, and so solutions that locally look like plane waves are approximate solutions of the equations of motion.  Moreover, generally, we can define $k_a = \partial_a S$;  then we have
$$
0 = \partial_b \left[ (\partial_a S) (\partial^a S) \right] = 2 (\partial^a S) \partial_b \partial_a S = 2 (\partial^a S) \partial_a \partial_b S = 2 k^a \partial_a k_b
$$
and so $k_a$ satisfies the geodesic equation, i.e., light rays travel on null geodesics in this limit;  they are unaffected by the presence of the scalar field.
More precisely, the approximation above means "up to terms of order $\partial_a \mathcal{A}_b$ and $\partial_a \phi$".  All the terms in the second set of square brackets in (1) above will contain one derivative of either $\phi$ or $\mathcal{A}_a$, and the terms in the third set of square brackets will contain two such derivatives.  This implies that the leading order quantity in (1), $- (\partial_a S) (\partial^a S) + i \Box S $, will be of the same order as the derivatives of $\phi$ and $\mathcal{A}_a$.
If this seems suspicious, note that we have to make the same approximation when we try to use Maxwell's equations to prove that light rays follow null geodesics in curved spacetime—which is hopefully a non-controversial statement.  To prove this, we have to work in the limit that the curvature scale of the spacetime is much greater than the wavelength of the light (see, for example, §4.3 of Wald's General Relativity.)  Or, to put it another way, the derivatives of the metric and of the polarization & amplitude of the light must be small compared to the derivative of the phase — just like we needed to assume that the derivative of $\phi$ and $\mathcal{A}_a$ had to be "small" here.
