I'm trying to understand why you can't write down a first order equation of motion for a scalar field in special relativity.

Suppose $\phi(x)$ a scalar field, $v^{\mu}$ a 4-vector. According to my notes a quantity of form $v^{\mu}\partial_{\mu}\phi(x)$ will not be Lorentz invariant.

But explicitly doing the active transformation the quantity becomes

$$\Lambda^{\mu}_{\nu} v^{\nu}(\Lambda^{-1})^{\rho}_{\mu}\partial_{\rho}\phi(y) = v^{\nu}\partial_{\nu}\phi(y)$$

where $y=\Lambda^{-1}x$ and the partial differentiation is w.r.t. $y$. This seems to suggest that the quantity is a Lorentz scalar, so could be used to construct a Lorentz invariant first order equation of motion.

I'm clearly making a mistake here. But I don't see what I've done wrong. Am I wrong to think that $v$ transforms nontrivially under the active transformation? Maybe it shouldn't transform at all because it's just a vector, not a vector field?

Many thanks in advance for your help!

  • $\begingroup$ Your notes seem wrong. If $v^\mu$ is a 4-vector, then $v^\mu \partial_\mu = v \cdot \partial$ is indeed a Lorentz scalar. $\endgroup$
    – Vibert
    Commented Dec 23, 2012 at 22:43
  • $\begingroup$ @Vibert - then we could get a first order equation of motion for a scalar field though... Isn't that a problem? $\endgroup$ Commented Dec 23, 2012 at 22:52
  • $\begingroup$ You would only be able to get $v^\mu = x^\mu$ (what else?), and I don't think such a term could come from applying Euler-Lagrange to a translation-invariant Lagrangian. It's probably a good exercise to see if it's possible! $\endgroup$
    – Vibert
    Commented Dec 23, 2012 at 22:59
  • $\begingroup$ Why can't you choose $v^{\mu}$ to be another arbitrary four vector? $\endgroup$ Commented Dec 23, 2012 at 23:02
  • $\begingroup$ @Vibert $v^\mu=x^\mu$ is no less arbitrary than $v^\mu=(1,2,3,\text{cos}(x^2))$ since the origin is arbitrary anyway. $\endgroup$
    – sjasonw
    Commented Dec 23, 2012 at 23:07

1 Answer 1


The function $v^a \partial_a \phi$ is a scalar field. Nonetheless, an equation like this is ugly because $v^a$ points in some "preferred" direction.

Here's another point of view, and I think this gets at what you were saying about "no first order equations". Suppose that $v^a$ is not a vector but is instead just a collection of four fixed real numbers. Suppose we consider the equation $\sum _\mu v^\mu \partial_\mu \phi=0$ now. This equation is not Lorentz invariant anymore since the numbers $v^\mu$ don't change.

Another approach: think of $v^a$ as a new spacetime-dependent vector field. Then, $v^a \partial_a \phi=0$ is Lorentz invariant equation, but it involves two fields. This is nicer than choosing a preferred direction.

  • $\begingroup$ Aha this is the insight I was looking for. Think I was on the right lines in the question when I said it didn't transform because it wasn't a field. Cheers! $\endgroup$ Commented Dec 23, 2012 at 23:04

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