Event Horizon Issue

No.
Q_reeus has given you a metric specific counter, but since you are referring to Newtonian Force, it is very simple to deduce that force is inversely proprtional to mass of BH at EH. That is smaller force value on a test particle by a larger mass BH.

When the object is just at EH, the force on a test particle of unit mass shall be GM/R^2......but at EH R = 2 GM/C^2, so the force will be = c^4/4GM.

 

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double post.

 

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Force and velocity cannot be compared, but probably what you are saying is that since at EH irrespective of mass of BH, the escape velocity is c, so the force must be same at EH. But no, I have given you a caluclation in my earlier post where it is shown that force ~ 1/M.

But I must add on escape velocity. For a bound particle the total energy is negative, and the total energy is KE+PE, but since KE is always positive, the PE must be negative. Since you and me are bound to earth, we have a total negative energy. For us to escape away from earth's gravitational influence, some amount of KE must be imparted which will make our total energy zero. The speed required required for this KE is called as escape velocity. It is Sqrt(2GM/R) that is around 11.2 km/sec for earth and c at EH of BH. Force directly does not play any role here.

 

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It's a matter of how things are defined in GR. Great weight is placed on what are called 'invariants' - properties of spacetime that are independent of any coordinate chart being used. A feature of GR is that a simple fapp uniform gravitational field felt by us standing on earth's surface, disappears when in free-fall i.e when following a geodesic path in spacetime. But not perfectly, since there is slight non-uniformity in the gravitational field owing to a massive body like earth. It's that non-uniformity that remains when in free-fall that in GR-speak represents 'the gravitational field' i.e. the 'invariant' quantity, i.e. curvature even though as free-fall proceeds, it's value will change.
There is free license played even by experts who variously will call 'g-forces' gravity in one instance, and curvature i.e. tidal fields i.e. geodesic deviation, as gravity or the gravitational field, in another instance. So it's a matter of relying on the context very often to discern which use of gravity/gravitational field is being meant.
Anyway, that arXiv article by Baez & Bundy is likely to be of real help.

 

No. The proper acceleration needed to hover at constant r at BH EH is infinite. You will need to go down to a little over half-way through the following article to see how it's worked out:
http://mathpages.com/rr/s7-03/7-03.htm

 

You are right keeping in mind GR.

The biggest issue with metric theory is that they do not consider the physical aspect of the object. Thats why they allow the singularity to form. You know very well that for a large mass BH when at EH, the density can be as low as anything, so the conclusion (which is mathematically ok) looks incorrect, suggesting non applicability of maths here. I am of the opinion that as long as object density is not of the order of nucleus density when at EH, talking about very large curvature or infinities is bad.

 

Sticking to GR BH scenario, curvature as such can be arbitrarily small for an arbitrarily large BH, but that's about 2nd derivatives of the metric. What you are thinking of is the 'strength of gravity = g-forces' which is owing to a first derivative calculated for a stationary position wrt radius outside BH. That quantity is always infinite at supposed EH. That it is infinite is considered by some theorists to be a real issue indeed. But no point getting into that further here. Stick to GR picture.

 

In my WIP it is not infinite..but lets not discuss that as of now....