# $(c^{\mu}\partial_{\mu}-m)\phi(x)=0$ 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?

• Well, no. But doesn't Zee say that we care about it being invariant for SOME vector $c$? Couldn't we just choose a vector with all nonzero components? – InertialObserver Jun 16 '17 at 14:02
• – Alfred Centauri Jun 16 '17 at 14:32
• For this same problem, I find the argument in Peskin and Schroeder on the fact that we want to obtain the KG equation when we "square" the Dirac equation more convincing – user2723984 Aug 5 at 12:57

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.)

• I see what you're getting at, I believe. So you're making a distinction between "transforming covariantly" and being Lorentz-invariant. It seems that, in the literature, when they say "let's prove such and such is Lorentz-invaraint", all they do is check to make sure that transforms covariantly. They never mention anything about "checking for vectors" or anything like that. – InertialObserver Jun 16 '17 at 14:25
• Yes, that's essentially the distinction I'm making. While I do wish this distinction was clearer in the literature, I suspect the reason people use "transforms covariantly" as synonymous with "Lorentz-invariant" is that as a general rule, you almost never impose a field (vector or otherwise) by hand (there are exceptions to this that I won't get into, though). If no fields are specified by hand, then transforming covariantly does imply Lorentz invariance. – Sebastian Jun 16 '17 at 14:34
• I really appreciate your answer. So could you answer for me? If an equation contains an invariant tensor, then the equation can be Lorentz invariant? For example, in $2+1$ spacetime dimensions, $\epsilon^{uvp}\partial_u A_p+mA^v=0$ can be Lorentz invariant? Here $\epsilon^{uvp}$ is a levi-civita tensor with $\epsilon^{123}=1$. I know that this levi civita tensor is an invariant tensor in any coordinate systems. Thus does this make the equation I wrote Lorentz invariant? – Keith Mar 27 '18 at 16:27
• Yes, that equation is indeed Lorentz invariant. Another example is the Laplacian $g^{\mu\nu} \nabla_\mu \nabla_\nu$, where $g^{\mu\nu}$ is the (inverse) metric tensor. In fact, if $A_\rho$ is the vector potential for an electromagnetic field, then the object $\epsilon^{\mu\nu\rho} \partial_\nu A_\rho$ is the dual of the electromagnetic field tensor, since $\epsilon^{\mu\nu\rho} \partial_\nu A_\rho = \epsilon^{\mu\nu\rho} (\partial_\nu A_\rho - \partial_\rho A_\nu)/2 = \epsilon^{\mu\nu\rho} F_{\nu\rho}/2 = (\star F)^\mu/2$. – Sebastian Mar 28 '18 at 16:58

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 $c$.

• In reference to your first and second sentences, I have heard different. For instance, in Peskin, to show that an equation is Lorentz invariant we show that if an equation is satisfied by $\phi(x)$ then it is also satisfied by $\phi(\Lambda^{-1}x)$ under a Lorentz transformation. – InertialObserver Jun 16 '17 at 14:29
• Javier: Right, but if $c^\mu$ is a Lorentz vector (as it had better be if the equation has any chance of transforming covariantly), then its components do indeed transform as required for your field $\phi'$ to obey the original equation in the new coordinates. – Sebastian Jun 16 '17 at 14:38
• Oh, I see. I made a mistake. Since we are still evaluating this equation in the old coordinate system, i should not take $c\rightarrow \Lambda c$. – InertialObserver Jun 16 '17 at 15:34
• The equations look 'different' because that's precisely what you'd expect to happen with the transformation. (As a reference, compare with $(\partial_\mu \partial^\mu+m^2)\phi(x) = 0$.) The problem is what happens with the $c^\mu$, not the argument of $\phi$. This answer is pretty fundamentally flawed - the answer is correctly given by Sebastian. – Emilio Pisanty Jun 16 '17 at 16:44
• @EmilioPisanty that's a good point, but I think the rest of my answer is valid; Sebastian correctly explains why one would expect this equation to not be invariant, but doesn't go into the math. I show explicitly what happens when you change coordinates. – Javier Jun 16 '17 at 16:50

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)

1. 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.