# Newtonian potential defines size of Universe?

As I understand it the Newtonian potential $\Phi$ at the position of an observer in space is approximately given by: $$\Phi \sim \frac{G M}{R},$$ where $R$ is the radius of a sphere centered at the observer and $M$ is the mass contained in the sphere.

As the dimensions of $\Phi$ is velocity squared could one argue that $\Phi$ must therefore be less than the velocity of light squared on the grounds that a value greater than $c^2$ would be unphysical?

If one could make such an argument then $R \sim GM/c^2$ would then be effectively the radius of our Universe.

PS This argument does not make the assumption that the recessional velocity $v = H r$ must be less that $c$. I'm trying to make a dimensional argument for the size of the Universe rather than simply define the Hubble radius $R_H = c / H$.

• 1. The Newtonian approximation does not hold for the universe as such. 2. Just because something has units of a velocity, it does not mean it can't exceed $c$. Oct 6, 2014 at 12:06

Only in the case of a static spacetime is the metric derivable from a scalar potential. Cosmological spacetimes aren't static, so they can't be derived from a potential.

There is no definite SIZE to the universe as such.

There is however a size to the OBSERVABLE universe. These are very different.

And indeed the observable universe is defined by Einstein information caveat where information cannot propagate faster than light.

Now, it is in this sense, that Newtonian mechanics fails us, as newtonian gravity (and grav. Pot.) are instantanous -> propagate instantly to infinity.

Thus the two terms $$\Phi \propto \frac{GM}{R}$$ and the speed of light are uncompatible.

However on CAN discuss the distance light has traveled from big bang, and this is basically given as $$\int \frac{c dt}{a(t)}$$ Where a is the cosmic scale factor