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For instance, a symmetric change from $(-,+,+,+)$ to $(+,-,-,-)$ in a metric, what are its physical implications? If there arent, why is that?

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The only situation I know of in which the sign convention has physical implications is when defining the Clifford algebra. The full Clifford algebras of $+{-}{-}-$ and $-{+}{+}+$ signature are not isomorphic. One consequence is that when, like Dirac, you try to write the "square root" of the Klein-Gordon equation, which is $\partial^2\phi = -||\hat t||^2m^2\phi$, you get a factor of $\sqrt{-1}$ if $||\hat t||^2 = +1$, and have to complexify the algebra, while if $||\hat t||^2 = -1$ you don't have to. This is perhaps evidence that the universe prefers $||\hat t||^2 = -1$. Of course, Dirac chose the "wrong" convention, probably because he wanted the Dirac equation to look like the Schrödinger equation that he was trying to replace.

The even Clifford algebras of $+{-}{-}-$ and $-{+}{+}+$ are isomorphic, so you'll probably never find a meaningful difference between the two signatures in the physics of integer spin.

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    $\begingroup$ This is nicely explained in this paper: arxiv.org/abs/math-ph/0012006 They even explain how a difference could be physically detectable. $\endgroup$ – NDewolf Aug 23 at 21:06
  • $\begingroup$ Thx for that link @NDewolf $\endgroup$ – planetmaker Aug 23 at 22:22
  • $\begingroup$ Sorry, what is $\hat{t}$? Which signature does $\|\hat{t}\|^2 = 1$ correspond to? $\endgroup$ – Brian Bi Aug 24 at 16:52
  • $\begingroup$ @BrianBi $\hat{t}$ is the unit vector in the time direction. This should also answer your second question. $\endgroup$ – NDewolf Aug 24 at 19:24
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In standard general relativity there is no physical implication. It is merely a sign convention. You arrive at the same predictions either way. My preference is $(-,+,+,+)$.

There may be some differences in other contexts.

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It depends on what OP means.

  1. One on hand, if OP literally means going from a spacetime with 3 spatial and 1 temporal directions to a spacetime with 1 spatial and 3 temporal directions, then it obviously has enormous physical consequences. Closed timelike loops for starters, cf. e.g. this Phys.SE post.

  2. On the other hand, if OP just means to change $\eta_{\mu\nu}\to -\eta_{\mu\nu}$ while at the same time change $ds^2=\eta_{\mu\nu} dx^{\mu}dx^{\nu}$ into $ds^2 =-\eta_{\mu\nu} dx^{\mu}dx^{\nu}$, then it is just a matter of conventions.

There is a separate issue with the definition of a Clifford algebra.

  1. On one hand, a signature reversal changes the real Clifford algebra, cf. e.g. this Phys.SE post.

  2. On the other hand, if OP just means to change $\eta_{\mu\nu}\to -\eta_{\mu\nu}$ while at the same time change the definition $\{\gamma_{\mu},\gamma_{\nu}\}_+=2\eta_{\mu\nu}{\bf 1}$ into $\{\gamma_{\mu},\gamma_{\nu}\}_+=-2\eta_{\mu\nu}{\bf 1}$, then it is again just a matter of conventions.

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  • $\begingroup$ Let me just recommend Greg Egan’s novel Dichronauts, set in a universe with a + + - - metric, and rigorously following the many physical implications. $\endgroup$ – Mike Scott Aug 24 at 13:52
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Answering: does the change in signature of the metric have a physical meaning.

Not for the example you cited. The example you cited is a choice of convention of which physics is independent. However this metric is very different from (+,+,+,+) or (-,-,+,+), as these do not differ by an overall sign. The former is known as Euclidean signature. Oftentimes physics in this signature is equivalent to physics in your (Lorentzian) signature but this equivalence is not guaranteed and is often unobvious and even enlightening. As just one example (out of many) the Schwarzschild black hole in Lorentzian signature is equivalent to polar coordinates (times a sphere) in Euclidean signature with the oddity that the origin is at $r=2GM$ (where $M$ is the mass of the black hole) instead of the usual $r=0$.

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  • $\begingroup$ I read here gregegan.net/ORTHOGONAL/04/EMExtra.html#CP that for the (++++) signature that, in the case of massive photons, that the direction of the electric field around an electric charge would vary with distance. Do you know if that's correct? $\endgroup$ – Anders Gustafson Aug 24 at 0:40

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