\label{accretion_law} In literature the universe’s expansion is regarded as the expansion of a spherical surface \cite{milne1935}.
We see galaxies immersed in a 3-dim space. increasing volume, however, the cosmological principle imposes galaxies placed above a surface. If we add the time at every point of the universe, we obtain a 4-dim spherical surface.
If \(n\) is the increasing number of steps for each axis, the expansion will be obtained through the relation \(N = n^2\), where \(N\) are the points on the increasing spherical surface that rises with the growing number of points on the radius \(R\), following the law \(S = 4\pi R^2\).
Therefore a universe that would increases in space, according to the law \(N = n^2\), with \(N\) the number of particles and \(n\) the number of time-steps, would increase surface as happens to an expanding spherical surface.
However, in an isotropic universe any local observer \(O\) puts himself at the center of a sphere with radius \(R_i\) and places on its surface the distant galaxies.
An observer sees galaxies move away radially in a 3d space, but describes the isotropic expansion as dilatation of a 2d surface.
In a universe with increasing space, the radius \(R_i\) of any lattice-universe \(U_i\) would increase as \(R_i (n) = n \lambda_i\). It follows: \[S_n = 4 \pi n^2 \lambda^2\] \(S_n\) varies with increasing \(R_n\); then the index \(n\) will be a function of time, and so we can calculate the following derivatives: \[\begin{aligned}
\frac{\text{d} S_n}{\text{d} n} = & \frac{\text{d}}{\text{d} n} \left ( 4 \pi n^2 \lambda^2 \right ) = 8 \pi n \lambda^2 \label{surface01}\\
\frac{\text{d} S_n}{\text{d} t} = & \frac{\text{d}}{\text{d} t} \left ( 4 \pi R_n^2 \right ) = 8 \pi R_n v_n \label{surface02}
\end{aligned}\] Where \(v_n\) is the velocity of recession of a galaxy placed on the spherical surface of radius \(R_n\).
It’s clear that the number \(n\) of steps in the space expansion could be related to the time by the relation \[n (t) = \frac{t}{\tau}\] and so: \[\label{surface03}
\frac{\text{d} S_n}{\text{d} t} = \frac{\text{d}}{\text{d} t} \left ( 4 \pi n^2 \lambda^2 \right ) = 8 \pi n \lambda^2 \frac{1}{\tau}\] Combining (\ref{surface02}) with (\ref{surface03}) we obtain the velocity of recession \(v_n\): \[v_n = \frac{n \lambda^2}{R_n \tau} = \frac{n \frac{R_n^2}{n^2}}{R_n \tau} = \frac{R_n}{n \tau} = \frac{R_n}{\tau_n}\] Now, if we assume that \(H = \frac{1}{\tau_n}\), we obtain: \[v_n = H R_n\] finding again the Hubble’s law.