# Gravitation function (or space-time distorsion variation) below Schwarzschild radius

Let's consider a massive sphere (with an approximate constant density, not charged, not rotating, no anti-matter for example an Earth simplified model).

On the outside of this sphere, $g(r) = \alpha/{r^2}$, or I could say the space-time shrinks as $1/r^2$.

On the inside of this sphere, $g(r) = \beta.r$, or I could say the space-time expands as $r$.

And $g(0) = 0$ at the center, or I could say the space-time is plain flat.

Let's now consider a simplified non-rotating, non-charged, matter (and not anti-matter) black hole of Schwarzschild radius $r_s$, and let's imagine the geometry of this black hole is a sphere.

Is there any possible way to compute the variation of the gravitational field below this $r_s$ as $r \rightarrow g(r)$? ($g(r)$ being here a way to interpret the space-time expansion inside of a black hole, because as you progress inside a black hole the quantity of matter behind you increases and the quantity of matter forward you decreases).

I guess this is the solution of a differential equation where $g$, $d$ (density) and $r$ are related to themselves. But before attacking this level of math tools I am in search of an intuitive image of where the solution would be.

For symmetry reasons I imagine that at $r=0$, $g=0$ and there is no singularity. But I didn't find any proof of this or the contrary.

Where are the best publication on this computation of $g$ and models (simplifications) involved?

• What do you mean by $g$? The acceleration of gravity? This is going to be tough to define inside the horizon, since it's not possible to stand still. Not to mention there definitely is a singularity at $r=0$. – Javier Jun 18 '17 at 16:39
• [My Q & A are in the standard reading order.] I mean $g = || { \vec {g} } ||$. Yes. Nothing stands still within this neighbour of the universe, for example on earth, nonetheless, an acceleration of gravity (or movment) is possible to compute. I can't see the difficulty you see here. Could you explain it in simpler terms? – dan Jun 18 '17 at 22:03
• To amplify on what Javier said, you might want to read up on the equivalence principle. The equivalence principle tells us that g has no objectively defined value unless we pick a frame of reference. To measure g on the earth, we pick a frame of reference relative to the earth's surface, and in GR this is considered a noninertial (accelerating) frame. (The definition is different than the one in Newtonian mechanics.) In a good inertial (free-falling) frame, g=0. Inside the horizon, there is no frame that is at rest relative to the black hole. – Ben Crowell Jun 18 '17 at 23:09

Given a metric one can try to understand what the Newtonian potential is like by expanding the metric about flat space $\eta_{\mu\nu}$. If you can write $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} + O(h^2)$ where $h_{\mu\nu}$ is small in some sense then the gravitational potential is given by $h_{00} = - 2 \Phi$ (as you can derive from the Newtonian limit of Einstein's equations).
For the Schwarzschild metric inside the horizon, as $0\leq r \leq 2GM$ and the metric functions are like $1-2GM/r$, it will be very hard to define a sensible perturbation about flat space.