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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?

Qmechanic
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InertialObserver
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3 Answers3

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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.)

Sebastian
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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 $c$.

Javier
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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.

innisfree
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