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While answering this question about a hypothetical 3-sphere universe $S^3$ expanding with a constant acceleration $\phi$ from a zero initial speed

$$ r=\dfrac{\phi}{2}t^2$$

I started from a generic metric defined in the hyperspherical coordinates:

$$ ds^2 = - c^2 dt^2 + a(t)^2 r^2 d\mathbf{\Omega}^2 $$

Where r is the radius, $a(t)$ is a scale factor, and

$$ d\mathbf{\Omega}^2=d\psi^2 + \sin^2\psi\left(d\theta^2 + \sin^2\theta\, d\varphi^2\right) $$

By combining the formulas we obtain

$$ ds^2 = - c^2 dt^2 + (\dfrac{\phi}{2}t^2)^2 d\mathbf{\Omega}^2 $$

Is it possible to define the scale factor for this metric explicitly as follows?

$$ a=a(t) $$

Thank you for your insight.

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You cannot do it straight forward like this. The $R$ in the (FLRW) Robertsson metric is the comoving distance while in your case $r$ is probably the coordinate distance. These two are connected as $r= R \: a(t)$. In the Robertson metric, $t$ is also not coordinate time but comoving time. Even if you want to define something like $$ a \propto \dfrac{\phi}{2}t^2$$ then the metric is $$ds^2 = - c^2 dt^2 + (\dfrac{\phi}{2 r_0}t^2)^2 (dR^2 +R^2 d\mathbf{\Omega}^2)$$ where $r_0$ should be some value with the dimension of length. So $R$ can not be dropped anyhow as in your question.

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