$(c^{\mu}\partial_{\mu}-m)\phi(x)=0$ not Lorentz-invariant? I have a question about the following passage on pg. 89 of Zee's QFT in a nutshell:

At first sight, what Dirac wanted does not make sense. The equation is supposed to have the form "some linear combination of $\partial_\mu$ acting on some field $\psi$ is equal to some constant times the field." Denote the linear combination by $c^\mu\partial_\mu$. If the $c^\mu$'s are four ordinary numbers, then the four-vector $c^\mu$ defines some direction and the equation cannot be Lorentz invariant.

The simplest equation I can construct that he is referring to is
$$(c^{\mu}\partial_{\mu}-m)\phi(x)=0, $$
for $m>0$. Taking $\phi(x)\rightarrow\phi(\Lambda^{-1}x)$, and $c^{\mu}\rightarrow\Lambda^{\mu}_{\ \ \nu}c^{\nu}$. Then, the left term transforms like 
\begin{align*}
\Lambda^{\mu}_{\ \ \nu}c^{\nu}(\Lambda^{-1})^{\sigma}_{\ \ \mu}\partial_{\sigma}\phi(\Lambda^{-1}x) &= \delta^{\sigma}_{\ \ \nu}c^{\nu}\partial_{\sigma}\phi(\Lambda^{-1}x)\\
&=c^{\sigma}\partial_{\sigma}\phi(\Lambda^{-1}x).
\end{align*}
Then, we have that 
$$ (c^{\sigma}\partial_{\sigma}-m)\phi(\Lambda^{-1}x)=0.$$
But doesn't this show that the equation is Lorentz-invariant?
 A: It depends what you mean by "breaking Lorentz invariance".  As you say, it is certainly true that if you're given a vector $c^\mu$ at some point, you can construct Lorentz scalars from that vector (that is, objects that transform covariantly under Lorentz transformations).  The point Zee is getting at is this: physically, a theory being Lorentz invariant means that there is no notion of a preferred reference frame.  But if you specify some vector $c^\mu$ at a point, you automatically obtain a preferred reference frame constructed from it (for example, if $c^\mu$ is timelike, you can define a preferred reference frame as the one in which $c^t$ is the only nonzero component).  So while something like $c^\mu \partial_\mu$ may transform covariantly, any object you construct from $c^\mu$ cannot be Lorentz-invariant because the presence of $c^\mu$ chooses a preferred frame.
(By the way, the fundamental issue here has to do with needing to specify $c^\mu$ by hand; you can remedy the situation by, for instance, upgrading $c^\mu$ to be a dynamical vector field which is not fixed a priori but is instead determined as a solution to some equations of motion.)
A: In general, whenever you're in doubt, a more reliable way of doing transformations is, instead of replacing things by other things, defining a new coordinate $x'^\mu = \Lambda^\mu_\nu x^\nu$ and finding equalities. In this case we have that $\partial_\mu = (\Lambda^{-1})^\nu_\mu \partial'_\nu$, so let's take the equation
$$(c^\mu \partial_\mu - m) \phi(x) = 0$$
and use the two equalities I wrote in the above paragraph:
$$(c^\mu (\Lambda^{-1})^\nu_\mu \partial'_\nu - m)\phi(\Lambda^{-1}x') = 0$$
We can use that $c^\mu (\Lambda^{-1})^\nu_\mu = \Lambda^\mu_\nu c^\nu$, so that, if we define $\phi'(x') = \phi(\Lambda^{-1}x')$, we have
$$(\Lambda^\mu_\nu c^\nu \partial'_\mu - m)\phi'(x') = 0$$
So the transformed field $\phi'$ doesn't satisfy the original equation, because you need to transform the components of $c$.
A: You could argue that the $c_\mu \partial^\mu$ term must be a scalar. But then note that different observers would disagree about the constants $c^\mu$, as that ‘constant’ would obviously depend on the frame of reference. So the equation would look different to different observers, breaking the principle of relativity, viz (from wiki)

  
*
  
*First postulate (principle of relativity) 
  
  
  The laws of physics are the same in all inertial frames of reference.

You could get round this by promoting the constant $c^\mu$ to a dynamical vector field, which ultimately leads to a gauge theory.
You could say that they are constants that don’t transform as a four-vector. That would also break relativity, as it wouldn’t be Lorentz invariant.
