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In his book, Sean Carroll, says p. 194 chapter 5:

To impose spherical symmetry, we begin b writing the metric of Minkowski space in polar coordinates $x^{\mu}=(t,r, \theta, \phi)$: $$ ds^2=-dt^2+dr^2+r^2d\Omega^2.$$ One requirement to preserve spherical symmetry is that we maintain the form of $d\Omega^2$. But we are otherwise free to multiply all of the terms by separate coefficients, as long as they are functions of the radial coordinate $r$: $$ds^2=-e^{2\alpha(r)}dt^2+e^{2\beta(r)}dr^2+e^{2\gamma(r)}r^2d\Omega^2.$$

What does expressing the functions as exponentials have to do with changing or preserving of the metric?

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The exponential function only returns positive numbers. Writing it this way ensures that the $dt$ term has a negative coefficient (because of the explicit minus sign) and the other terms have positive coefficients.

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  • $\begingroup$ Maybe also something about differentiability? $\endgroup$ – boyfarrell Oct 20 '15 at 15:48
  • $\begingroup$ The question was about the signature of the metric. That has nothing to do with differentiability. If you wanted to dig deeper into the why he did it this way, there might be some nice properties there too - but that's separate from what the OP asked. @boyfarrell $\endgroup$ – Brick Oct 20 '15 at 15:57
  • $\begingroup$ thank you Brick! I have a question though why is it that some times you rather see the metric without exponentials? Shouldn't one be concerned that all theories one is dealing with have preserved signature? $\endgroup$ – Beyond-formulas Oct 20 '15 at 16:37
  • $\begingroup$ @Beyond-formulas The metric should definitely be preserved. This way makes that manifestly true, but that doesn't mean that it isn't preserved when written differently. Some of it is a matter of style. Some relates to ease of calculation for what you're going to do next. (See, for example, comment by boyfarrel about differentiation.) The notation used here by Carrol, for example, makes the signature explicit but it pushes the dependence on $r$ into other functions that appear as arguments to the exponentials. Nothing wrong with that but it may (or may not) become cumbersome in calculations. $\endgroup$ – Brick Oct 20 '15 at 17:07

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