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In quantum field theory we have the commutators of fields must be zero outside the light cone.

$[\phi(x),\phi(y)]=0$ if $|x-y|^2<0$

How can one write this in polar coordinates or a general coordinate system of flat space-time say with flat metric $g^{\mu\nu}(x)$? One wonders if one must consider geodesics between the points $x$ and $y$?

Edit:

In the general case I'm thinking that one would have to write this locally like: (I don't think this is correct, probably need second derivatives:)

$[\phi(x), n^{\mu}\partial_\mu \phi(x)] = 0$ if $g_{\mu\nu}(x)n^{\mu}n^{\nu}<0$

For some general 4-vector $n$. Does this seem correct? Or is there a better way of writing this?

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  • $\begingroup$ I don't understand this question. Nothing about $\lvert x - y \rvert^2 < 0$ prescribes a particular coordinate system. Are you asking how to generalize this to spacetimes that are not $\mathbb{R}^{1,3}$? $\endgroup$ – ACuriousMind Mar 18 at 15:31
  • $\begingroup$ @ACuriousMind, Well for a general coordinate system with a flat metric $g(x)$ with very curved coordinate system of arbitrary complexity, it is not generally easy to find the distance $|x-y|^2$. Whereas it is fairly simple in cartesian coordinates. I guess the question comes down to describing $|x-y|$ in arbitrary coordinates. Which I guess is not that hard. $\endgroup$ – zooby Mar 18 at 15:33
  • $\begingroup$ @zooby The distance function $|X|^2$ is not a coordinate expression. In terms of coordinates it is the standard $g_{\mu\nu}X^\mu X^\nu$ $\endgroup$ – user195162 Mar 18 at 17:05
  • $\begingroup$ @Quantumness It's only $g_{\mu\nu}X^{\mu}X^{\nu}$ where $X=x-y$ if $g$ does not depend on $x$. $\endgroup$ – zooby Mar 18 at 19:25
  • $\begingroup$ @zooby sorry, I wrote down the infinitesimal version $ds^2=g_{\mu\nu}dX^\mu dX^\nu$. The finite version is a line integral over the path taken $S=\int_\Gamma \sqrt{g_{\mu\nu}dX^\mu dX^\nu}$, which in this case should be a geodesic connecting the two points in question. $\endgroup$ – user195162 Mar 19 at 18:20

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