Although this is a standard derivation, you frequently don't see it in introductory electromagnetism courses, maybe because those courses shy away from the heavy use of vector calculus. Here's the usual approach. We'll find a wave equation from Maxwell's equations.

Start with

$\nabla \times \vec{E} = -\frac{\partial\vec{B}}{\partial t}$.

Take a partial derivative of both sides with respect to time. The curl operator has no partial with respect to time, so this becomes

$\nabla \times \frac{\partial\vec{E}}{\partial t} = -\frac{\partial^2\vec{B}}{\partial t^2}$.

There's another of Maxwell's equations that tells us about $\partial\vec{E}/\partial t$.

$\nabla \times \vec{B} = \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$

Solve this for $\partial\vec{E}/\partial t$ and plug into the previous expression to get

$\nabla \times \frac{(\nabla \times \vec{B})}{\mu_0\epsilon_0} = -\frac{\partial^2 \vec{B}}{\partial t^2}$

the curl of curl identity lets us rewrite this as

$\frac{1}{\mu_0 \epsilon_0}\left(\nabla(\nabla \cdot \vec{B}) - \nabla^2\vec{B}\right) = -\frac{\partial^2 \vec{B}}{\partial t^2}$

But the divergence of the magnetic field is zero, so kill that term, and rearrange to

$\frac{-1}{\mu_0 \epsilon_0}\nabla^2\vec{B} + \frac{\partial^2 \vec{B}}{\partial t^2} = 0$

This is the wave equation we're seeking. One solution is

$\vec{B} = B_0 e^{i (\vec{x}\cdot\vec{k} - \omega t) }$.

This represents a plane wave traveling in the direction of the vector $\vec{k}$ with frequency $\omega$ and phase velocity $v = \omega/|\vec{k}|$. In order to be a solution, this equation needs to have

$\frac{\omega^2}{k^2} = \frac{1}{\mu_0\epsilon_0}$.

Or, setting $v = 1/\sqrt{\mu_0\epsilon_0}$

$\frac{\omega}{k} = v$

This is called the dispersion relation. The speed that electromagnetic signals travel is given by the group velocity

$\frac{d\omega}{d k} = v$

So electromagnetic signals in a vacuum travel at speed $c = 1/\sqrt{\mu_0\epsilon_0}$.

**Edit**
You can follow the same steps to derive the wave equation for $\vec{E}$, but you will have to assume you're in free space, i.e. $\rho = 0$.

**Edit**
The curl of the curl identity was wrong, there's a negative number in there

To answer your title question, definitely not.
(Note: The title question has since been updated. The original was - "is it possible to deduce the constancy of the speed of light from Maxwell's equations?")

If you assume *Maxwell's equations hold for all inertial frames* as you are doing in your argument, then you are begging the question. You are making an assumption *further to* Maxwell's equations. That assumption cannot be deduced from Maxwell's equations themselves - you then need to ask how Maxwell's equations transform between different reference frames and that answer is outside the scope of Maxwell's theory.

In the 19th century, as hinted by dmckee's comment:

You can derive equations that give you the speed of sound in fluids and solids. These equation also hold for all observers in all reference frames (in the context of Galilean relativity). But such waves don't give rise to the surprises in SR. The 19th century physicists assumed they would deal with light in the same way that they had dealt with all the other wave phenomena they had met up to that time.

people simply assumed that Maxwell's equations held for an observer who was stationary with respect to the luminiferous aether, and that this aether would behave just as gas or the wood of a violin's sounding board for sound. Maxwell's equations would then change their form under the Galilean transformation. That was just "what waves did" to the mind of a physicist in 1862.

Physicists also assumed that Galileo's postulate, that only relative motion between observers were experimentally detectable, didn't hold for electromagnetism. Galileo, after all, knew nothing of electromagnetism.

And that is a perfectly plausible theory to postulate: from a logic standpoint, it is just as valid as Einstein's second postulate and the restoration of Galileo's relativity principle. These things cannot be settled by logic, but only by asking Nature for her take on these things. That is, they can only be settled by experiment - and we find that Maxwell's equations do indeed transform covariantly and that the transformation between relatively moving inertial frames is the Lorentz, not Galilee, transformation.

## Best Answer

Yes, GR predicts luminal propagation of gravitational waves, by now confirmed in LIGO. Linearized gravity is an adequate approximation to general relativity: the spacetime metric, $g_{ab}$, may be treated as deviating only slightly from a flat metric, $\eta_{ab}$, \begin{equation} g_{ab} = \eta_{ab} + h_{ab},\qquad ||h_{ab}|| \ll 1\;. \end{equation} The resulting Einstein tensor then is $$ G_{ab} = R_{ab} - \frac{1}{2}\eta_{ab} R \\ = \frac{1}{2}\left(\partial_c\partial_b {h^c}_a + \partial^c \partial_a h_{bc} - \Box h_{ab} - \partial_a\partial_b h -\eta_{ab}\partial_c\partial^d {h^c}_d + \eta_{ab} \Box h\right)\;. $$

In terms of the

trace-reversedperturbation $\bar h_{ab} = h_{ab} - \frac{1}{2}\eta_{ab} h$, this reduces to $$ G_{ab} = \frac{1}{2}\left(\partial_c\partial_b {{\bar h}^c}_{\ a} + \partial^c \partial_a \bar h_{bc} - \Box \bar h_{ab} -\eta_{ab} \partial_c\partial^d {{\bar h}^c}_{\ d}\right)\;.$$In Lorenz gauge, this can be further hacked down to $$ G_{ab} = -\frac{1}{2}\Box \bar h_{ab}\;, $$ a relativistic wave equation in vacuum, where

Gvanishes, $$ \left (\frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2\right ) \bar h_{ab}=0\;. $$_{ab}So this disturbance (gravitational wave) travels with the speed of light. Upon quantization, you can easily see that its quantum excitation, the graviton, is massless, and also travels with the speed of light, like all massless particles.

Note that the

scaleof gravity, Newton's constantG, or, equivalently, the Planck massonlyenter in the coupling of this wave or particle to matter (and energy) and not in its free propagation in space, in this weak field regime, so it decouples here.