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A free particle moving from event $a$ to event $b$, which is timelike connected to $a$, on spacetime follows geodesics. In most cases it is a path that minimizes this integral

$$S=\int_a^b ds=\int_a^b \sqrt{g_{\mu\nu}dx^\mu dx^\nu}.\tag{1}$$

As a consequence of minimizing this integral, proper time of the particle is maximized

$$\tau_{ab}=\int_a^bd\tau=\int_a^b\sqrt{\frac{-ds^2}{c^2}}.\tag{2}$$

because of negative sign before $ds^2$. However, this result is based on the fact that the metric signature is $(-,+,+,+)$. I wonder what happens when the metric signature is $(+,-,-,-)$. Is proper time still maximized in this case?

I read this answer and it didn't address my doubt.

Qmechanic
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Jimmy Yang
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2 Answers2

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The physics cannot change!

The $g_{\mu\nu}$ has to encode spacetime curvature when there is curvature, and so it is convenient to still define the flat spacetime $\eta_{\mu\nu}$ which is the diagonal of the metric signature, i.e. constants of $\pm1$ or $0$ as appropriate for your spacetime.

Then the correct metric-signature-independent definition of proper time is $$c\,\mathrm d\tau=\sqrt{\eta_{tt}\,\mathrm ds^2}$$

Personally, I prefer to write $\eta^{tt}$ just to not have so many subscripts, but that is manifestly the same thing.

naturallyInconsistent
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TL;DR: The two signature conventions describe one and the same physical situation$^1$.

  1. In formulas, $$ \pm c^2(d\tau)^2~=~(ds)^2~=~g_{\mu\nu}dx^{\mu}dx^{\mu}, \tag{A}$$ if the Minkowski signature is $(\pm,\mp,\mp,\mp)$, respectively.

  2. So a finite lapse of proper time is $$ \tau~=~\frac{1}{c}\int \sqrt{\pm g_{\mu\nu}dx^{\mu}dx^{\mu}}.\tag{B}$$

  3. Note that the tangents of a worldline/curve should be everywhere non-spacelike in order to make the argument of the square root non-negative.

  4. The geodesics are given by stationary curves of the functional (B).

  5. That the proper time (B) is maximized in Minkowski space is proven in e.g. this & this Phys.SE posts.

  6. NB: There is no timelike curve that minimizes the proper time (B), cf. e.g. this Phys.SE post.

$^1$ We assume that OP is not considering a spacetime with 1 space and 3 time directions!

Qmechanic
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