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.


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.