1
$\begingroup$

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!

$\endgroup$
8
  • $\begingroup$ 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
    Dec 23 '12 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? $\endgroup$ Dec 23 '12 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
    Dec 23 '12 at 22:59
  • $\begingroup$ Why can't you choose $v^{\mu}$ to be another arbitrary four vector? $\endgroup$ Dec 23 '12 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
    Dec 23 '12 at 23:07
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$ Dec 23 '12 at 23:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.