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I'm coming at the following question from a mostly lay perspective (i.e. barely-undergrad physics), so please bear with the weirdness of it if possible.

I've generally been uncomfortable with the feeling that the -1 (i.e. $-+++$ or $+---$) is mathematically arbitrary because it fits the data, i.e. there's no mathematical reason there "should" be a sign flip (at least no obvious one to me). But I remembered a bit about path-ordering and was thinking that time is somehow different because there's a time-ordering operator which must be imposed on quantum evolution operators due to non-commutativity. Is there any evidence that this would have any link to the sign flip in relativity?

(of course any physical theory could be described as "mathematically arbitrary/just fits the data", but would be nice to have less axioms)

Edit (based on comment): The question is does the non-triviality of the time-ordering (which I would interpret as a kind of 'casual behavior') have any connection with the fact that timelike dimensions have the opposite sign of spacelike ones in Minkowski space?

Michael Seifert
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allidoiswin
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1 Answers1

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There are lots of answers already on the site for example here that discuss in varying levels of detail/nuance the ways in which the metric signature changes things. The common theme among all of them is that no physical predictions of the theory change when using an alternative metric signature, which ultimately means that any two such theories are physically indistinguishable (to the extent that one could call them "two" theories). I.e. there is no real-world experiment that could determine whether the universe works in the $(-,+,+,+)$ or the $(+,-,-,-)$ signature.

To your second point, the answer is no, the metric signature doesn't affect time ordering. The different metric signature does (if you like) change the direction in which the time axis points, but note that time ordering only cares about the relative positions of the time coordinates. A bit like if you point a stick in some direction and note that the end of the stick pointing away from you is further away than the end in your hand, then turn 180 degrees so the stick points in the opposite direction - the end in your hand is still closer than the end pointing away from you. The ordering has not changed under this transformation.

For what it's worth - though this likely goes beyond the level of technicality than you're asking - time-ordering isn't imposed because of non-commutativity, non-commutativity is the reason that time-ordering is non-trivial. Of course:

$$\prod_i^N\phi(x_i)=\phi(x_1)\phi(x_2)...\phi(x_N) \>\>\>\>\text{for }\>\>\>\> x_1>x_2>...>x_N$$

$$\text{if}\>\>\>\>[\phi(x_i),\phi(x_j)]=0\>\>\>\>\forall i,j$$

In words, if all operators commute, time ordering is trivial since you can quite literally just commute all operators into time order. Consequence of time-ordering being non-trivial include Wick's theorem.

Charlie
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