# What is this metric's scale factor?

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?

$$a=a(t)$$

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.