# Why is the scaling factor in the Robertson-Walker metric squared?

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?

• $a$ 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). Feb 28, 2020 at 3:10
• Solved, thank you. Feb 28, 2020 at 3:12
• 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). Feb 28, 2020 at 3:13

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.

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).