A recent paper, arXiv:2403.01555, gives the equations for homogeneity and isotropy of an expanding 3-space as expressed in the following distance interval as $x^i = (t, \chi, \theta, \phi)$ and $x^i + dx^1 = (t +dt, \chi + d\chi, \theta + d\theta, \phi + d\phi)$: $$ds^2 = g_{ik} dx^i dx^k ≡ dt^2 - a^2(t)[d\chi^2 + f^2 (\chi)(d\theta^2 + sin^2 \theta d\phi^2)].$$ Where $t$ is the time variable, and the spatial coordinates $\chi$, $\theta$, and $\phi$ are connected with the standard Cartesian coordinates $(x^1, x^2, x^3)$ by the relations $$x^1 = a(t)f(\chi)sin\theta cos\phi, x^2 = a(t)f(\chi)sin\theta sin\phi, x^3 = a(t)f(\chi)cos\theta.$$ The quantity a(t) in the above distance interval equation has the dimension of length. It has the meaning of the radius of curvature of space. As to the function $f(/chi)$, it is defined as $f(\chi)$ = { $sin\chi, \kappa = 1$, $sin h\chi, \kappa = -1$, $\chi, \kappa = o$}, where $\kappa$ is the sign of curvature of the 3-space ($\kappa$ = 0 corresponds to the flat 3-space). What I am wondering about is if the distance interval parameters are changed in a very subtle way from $x^i = (t, \chi, \theta, \phi)$ and $x^i + dx^1 = (t +dt, \chi + d\chi, \theta + d\theta, \phi + d\phi)$ to $x^i = (t, \chi, \theta, \phi)$ and $x^i + dx^1 = (t +dt, \chi, \theta, \phi)$: does the distance interval then become $ds^2 = g_{ik}dx^idx^k ≡ dt^2 - 0$? The definition of the Cartesian coordinates for $\theta$, $\chi$, $\phi$ are confusing me. Is the distance interval solution I propose correct? This question is wondering if a change in time without a change in space, creates a change in space anyway, due to the expansion of the universe. The solution I propose above says, No. What is the correct answer?
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