The key is to note that the coordinates you are using ( standard Schwarzschild coordinates) are bad coordinates at r= 2M , just as polar coordinates in 3 D Euclidean space are bad coordinates at r=0.

To deal with the region in question one needs to use good coordinates ( like cartesian coordinates en 3 D Euclidean space). There are various options .. perhaps the best are the Kruskal Seekers coordinates . Check those out. With those you will have no problem carrying out the construction of the Riemann Normal coordinates ( i.e once in which the metric looks Minkowski up to 2 order corrections) around any event and thus realize the Equiv Principle.. i.e. reduce GR to SR locally. Note however that the Riemann curvature involves "second derivatives" of the metric and thus it will not vanish even in those coordinates ( the curvature is a fully covariant tensor).

Daniel Sudarsky, thank you for your answer but in straight language, as we know all the terrestrial portions of GR is confirmed by the use of Schwarzschild line element, the results supplementation base for experimental confirmations, how could we ignore it and move for coordinate-transformations just to eliminate singularity problem? Further, my question is not to the straight point as you mentioned, I simply want to know, whether we could reduce the Schwarzschild solution directly to Minkowski for local frames of reference to validate SR. Like , for example, in vacuum we could write r=infinity which results direct conversion to Minkwoski.

Contrary to what is believed, Schwarzschild's solution doesn't represent an effective solution of Einstein's metric. In fact Schwarzschild metric is reduced to SR's linear metric only when M=0 and r=0.

It involves Schwarzschild's metric is compatible with Einstein's metric only for r=0.

This conclusion generates many doubts on the theoretical and real consistency of General Relativity besides on Schwarzschild's solution.

Daniele Sasso , thank you for your answer. I think you mean r=infinity not zero.

Dear Bishaf,

in my comment I have considered two conditions: r=0, M=0.

Under these conditions the case r=infinity loses meaning.

Daniele Sasso, I am not getting your point. To be linearized, the 2GM/r factor should be eliminated from the metric, which is possible if M=0 as you mentioned and if r=infinity. Please add your comment with a bit elaboration. The condition under r=0 is still the problem of singularity usually taken as natural singularity, how could it give linearized metric?

Dear Bishaf,

r is a variable, M is mass of celestial body. When M=0, in order to have a linear metric, the term 2MG/r is zero independently of any value of r, also for r=infinity that is a mathematical limit.

Anyway in my interpretation of physics, all spacetime metrics, including Einstein Schwarzschild Minkowski's metrics, are only kinematic mathematical models that don't represent thr physical reality.

Regards.

Schwarzschild's metric is

ds2=(1-2M/r)dt2 - (1-2M/r)-1dr2 - r2(d2(theta)-sin2(theta)d2(alfa))

in order to have Einstein's linear metric ds2=dt2 - dr2 it must be necessarily M=0 and r=0.

Regards.

Daniele Sasso, linearity occurs if and only if ds^2=dt^2 - dr^2 as you mentioned, where there will be no role of r and other angular terms as like your equation above but how you are obsessing the result is quite amazing...

I think I'm with Sasso, on this one, mostly... up to the point where he says " Anyway in my interpretation of physics, all spacetime metrics, including Einstein Schwarzschild Minkowski's metrics, are only kinematic mathematical models that don't represent the physical reality. "

... where I would only make the clarification "... are kinematic mathematical models that attempt to represent physical reality, but don't necessarily succeed."

In regards to the statement that General Relativity reduces to Special Relativity locally, the only interpretation I can come up with (that seems even remotely true) is to say that when you zoom in on a curve, (unless, perhaps it's generated from a fractal) you can always zoom in far enough, that at some point, that curve will appear as a straight line.

And then, when they say, such a thing is equivalent to SR, they are not talking about any usable part of SR, such as the Lorentz Transformation Equations... Rather, they're talking about the particular dogma of SR theory that says "SR does not handle accelerating frames". (These statements will come in pairs... If the author says "SR does not handle accelerating frames" he will then go on to say "GR reduces to SR locally".)

Essentially, any geodesic, taken for a short enough time, can be approximated as a straight line. And instead of saying that fairly obvious piece of information, somewhere along the line, someone decided it would be more fun to say "GR reduces locally to SR from the point of Schwarzschild line element".

That Daniele Sasso "obsesses" over the fact that the Schwarzschild Metric reduces to the Minkowski metric only when M=0 does not strike me at all as amazing. What strikes me as amazing is that his character is attacked as obsessive when he is noting a fairly simple fact, related to the mathematics.

It is possible that a few questions of physics were an obsession for me when I was a young researcher and I was amazed that postmodern mainstream physics could accept defined interpretations and Mathematical models for those questions. Now I am an aged researcher and my obsessions finished because I formulated different interpretations and Mathematical models for those questions that reconciled me with physics.

Jonathan Doolin, as you mentioned SR is not able to cover the acceleration but my question seeks more than that, please notice that the circular motion could be viewed from co-variance principle stating in terms of freefall (a condition of acceleration) where SR validates. see "Rohrlich, F.: The principle of equivalence. Ann. Phys. 22, 161-191 (1963)". But my quest is not addressed to this point now but I want to know whether that line element reduces to limit in SR, simply..

You should try to find a coordinate patch, in which the metric reduces to the fundamental one at some spacetime point covered by this patch. In other words, you want to construct a Riemann normal coordinate system. Please check Article Riemann Normal Coordinates

Thank you for your suggestion, Shaoqi. I would like to be more specific; please write me in your own words how does it exactly reduce as my question is addressed.