Does an observer experience stronger gravity on the way towards the singularity?

Question

I have read this question:

Clearly as you approach the singularity the tidal force on a body increases.

Are the fundamental forces unified in a black hole?

Based on the questions on this site, gravity inside a black hole is a phenomenon. It should actually increase as you (viewed as the infalling observer) approach the singularity.

Time dilation is said to be already infinite at the event horizon, so how could it become stronger inside?

As you approach the singularity:

1. tidal forces become stronger
2. time dilation is already infinite at the horizon
3. curvature could actually become stronger or weaker
4. the escape velocity is already the speed of light at the horizon, does it increase during the "trip" towards the singularity?

The confusion arises, because, after the observer falls into the black hole, and is on its way to the singularity (this voyage takes finite proper time), meanwhile, the gravitational field, and its effects can change. This is my question, whether these changes are experienced as caused by a stronger or weaker field.

That is, the question stated more precisely, do the effects of gravity get stronger while the observer moves towards the singularity?

Question:

1. Does an observer experience stronger gravity on the way towards the singularity?

Answer 1

The implication of your comparison with a neutron star is that you define "gravity" as the force required to keep an object at fixed radial coordinate.

By this definition, the "gravity" of a black hole is given by $$g =-\frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}\ , $$ which becomes infinite at the event horizon and is undefined below it, since nothing can remain at a fixed radial coordinate for $r<2GM/c^2$.

The notion of "gravity" is ill-defined below the horizon. However, the magnitude of the second derivative of the radial coordinate with respect to proper time ($d^2r/d\tau^2$) certainly does continue to increase with decreasing $r$ for a falling object and tends to infinity as $r\rightarrow 0$.

A falling (inertial) body does not "experience" any acceleration at all. However an object of finite size will experience non-inertial, tidal forces. These will increase smoothly as they cross the event horizon and will grow towards infinity as $r \rightarrow 0$ as the curvature of spacetime increases.

NB The shell theorem says that the gravitational force interior (anywhere) to a spherical shell of material is zero.

Answer 2

A black hole is a singularity in space; there is no going inside (defined as being surrounded by black hole matter). Rather, you could only get closer and closer to the singularity. When you get really close, no one knows what happens. But generally, the closer you get, the stronger the gravitational pull.

Answer 3

One of the great differences between classical and relativistic gravity is that in the former there is no side effect in supposing all matter concentrated at a point (well, except for the huge tidal forces). Elliptic orbits, for a small enough test mass to avoid being destroyed by them, can reach as close as we wish from the singularity.

In the Schwartschild solution there is the event horizon, and no matter the energy of the object, if it crosses the horizon it is trapped.

But the existence of the EH doesn't imply that there is something other than vacuum for $R < R_s$. The gravitational field increases continuously until the singularity.

Answer 4

Time dilation is said to be already infinite at the event horizon, so how could it become stronger inside?

Yes but that's seen from the outside. When you fall in you don't experience local time dilation. Only global time dilation, between you and the positions in front and in the back of you. So you only see tidal force.

What takes millions of years on the outside takes little time on the inside. If you fall in you are radiated away by Hawking radiation fast. On the outside it takes millions of years, if not more.