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Not much to add beyond the title. The Robertson-Walker metric solution to the field equations has the form

$$g_{\mu\nu}dx^\mu dx^\nu=-dt^2+a^2(t)\biggl(\frac{dr^2}{1-Kr^2}+r^2(d\theta^2+sin^2\theta \phi^2)\biggr)$$

in which the scaling factor $a(t)$ is squared. I cannot see any reason given for this so far, is this for dimensional reasons or is there a more important reason?

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    $\begingroup$ $a$ is at this point arbitrary, so it doesnt really matter. Redefine $\tilde a=a^2$ if it makes you any happy. The square is useful because it simplifies formulas later on (and also the factor is positive, so that the metric has the correct signature). $\endgroup$ Feb 28, 2020 at 3:10
  • $\begingroup$ Solved, thank you. $\endgroup$
    – Charlie
    Feb 28, 2020 at 3:12
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    $\begingroup$ If $b(t)\equiv a^2(t)$ would be in the metric, then space would scale with a factor of $\sqrt{b(t)}$, which is not as nice to handle (and look at). $\endgroup$
    – ersbygre1
    Feb 28, 2020 at 3:13

2 Answers 2

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Because $g_{\mu\nu} dx^\mu dx^\nu$ gives you the square of the distance. If the square of the distance increases by $a^2$, then the distance increases by $a$. That's why $a$ is called the scale factor.

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Updating to give the answer from @AccidentalFourierTransform

a is at this point arbitrary, so it doesnt really matter. Redefine a~=a2 if it makes you any happy. The square is useful because it simplifies formulas later on (and also the factor is positive, so that the metric has the correct signature).

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