Dear Bernard,

maybe that is because the angular part of the metric is the same as in Minkowski, but the radial part of the metric is r-dependent.

If you want to calculate a radial distance in Schwarzschild metric you have to take into account also the term (1-2M/r)^(-1). You cannot simply take the difference of two circumferences as you would do in Minkowski, because that is a different metric and hence a different definition of distance.

Did I got you question right?

Greetings,

Giovanni

Dear Giovanni,

Yes, you got the question right. But, if I take the limits on the integral of the radial part from 0, the center of the field, out to the radius of the circle I will find that it is greater from the radius itself. How can this be?

The same information must be contained in the angular part of the metric as in the radial part. If space is uniform and isotropic, it should have constant curvature so I substitute the density for the mass in the term (1-2M/r) and come out with g=(1-k^2r^2), where k is a constant related to the constant negative curvature. If I consider the metric

ds=dr/g+r dphi/g^{1/2}

I will get if r=const, the result that the ratio of the circumference to radius is greater than 2pi r, while phi=constant, the integral of the first term will give me the difference of hyperbolic lengths, arctanh kr_2-archtanh kr_1. This is the only way that the two terms in the metric can be compatible for real k.

Saluti,

Bernard

Benissimo!

With regard to the first part of your answer: you mean that you are calculating the proper distance between two points by integrating "ds", keeping angular and temporal parts constant. That's ok for me.

But in the second part there are some points to be cleared:

you say "If space is uniform and isotropic, it should have constant curvature". But Schwarzschild, being certainly isotropic, is not uniform: it has a privileged center, i.e. the mass source in r=0. Moreover it doesn't have constant curvature, but an r-dependent curvature that even diverges in r=0.

The fact that the Schwarzschild metric is "curved" (whereas Minkowski is not) is the key point to the problem that bothers you: one has to give a different meaning to the radial coordinate.

In particular, there is a fundamental difference between radial coordinate "r" and proper radial distance (that is the integral of "ds"): the difference is that a physical ruler will measure the integral of "ds", not "r".

So (r2 - r1) does not describe a physical distance between two circumferences: the physical distance is greater, and this is because Schwarzschild metric has an r-dependent curvature.

Giovanni

Sorry, I forgot to link a page that could help!

http://physicspages.com/2013/04/05/schwarzschild-metric-radial-coordinate/

Giovanni, The inner solution of Schwarzschild has a constant mass density. Why should space change from one of nonconstant curvature to one of constant curvature just by going from the exterior to the interior solution?

Einstein's equations are solved with T_ij=0, more uniform and isotropic than that is hard to come by. Mass is only introduced as an arbitrary constant of integration, and then rationalized that Newtonian physics should apply at large distances and weak fields. But Newton's law vanishing at infinity applies to absolute space not to GR in which the g_ij have different values in different systems of coordinates at infinity. What makes me believe there is a mass, if there was none to begin with? What causes the nonconstant curvature? I would expect that deviations from Euclidean geometry would have constant curvature and the only two that are known are hyperbolic and spherical geometries. In the limit as the radius of curvature tends to infinity, you get back Euclidean geometry. Not so with Schwarzschild's solution to Einstein's equations.

Bernard Lavenda,

As Giovanni Acquaviva points out, the radial coordinate r in the Schwarzschild metric is not directly a distance; in fact it is defined so that the spherical shell around the origin that has area A is defined to be at radius r = (A/4 pi)^(1/2).

When you build Schwarzschild geometry, the outer region is vacuum, with no mass-energy at all, but in the inner region you choose a non-zero density, T_ij!=0, so you should expect a different solution with different curvature.

It's worth keeping in mind that when you build Schwarzschild geometry, you are not just picking a manifold, but also a particular coordinate system. You can of course change coordinates, but in the wrong coordinate system you lose all kinds of simplifications (like a diagonal metric, say). In fact, while the Schwarzschild solution goes singular at r=M (r=2M? don't remember), it turns out that if you switch to Kruskal-Szekeres coordinates this singularity goes away - but in these coordinates your metric doesn't look much like the Minkowski metric far from the hole. That said, flatness is a coordinate-independent property - your metric may look weird (take flat space in polar coordinates, for example) but the equations testing for flatness can tell you whether it's true in any system you like.

As for the outer region, there are in principle many possible solutions with T_ij=0, but the ones we're interested in are the ones that have flat spacetime far away from the origin. After all, if you're looking for something astrophysical, you want a spacetime where the curvature does go to zero far from the object. (Okay, well, you might have some cosmological curvature far away, but we neglect that because the problem is already too hard, and in any case, since that curvature is small our Schwarzschild solution should be approximately right.) Another way to think of it is that we're solving a partial differential equation on a manifold, so if we want a unique solution we need to apply some boundary conditions. If we're interested in the gravity around an isolated non-rotating star, well, flat space at infinity seems like a good choice, and the inner edge of the manifold will be a sphere. Exactly what boundary conditions we apply there are up to us; one answer would be to connect it to a spherically-symmetric solution with constant density inside.

Why does non-constant curvature arise? That's what the equations say. It comes from the requirements of spherical symmetry and flatness at infinity, and just pops out of the math. Constant curvature would make all points of space the same in some sense, but spherical symmetry picks out a particular location - r=0 - as special, so you should not expect translation invariance. And indeed, solving the equations finds a solution with non-constant curvature.

Anne Archibald

As late as 1936, Levi-Civita said "During a conversation I had with Einstein a few years ago I asked him.... if it is possible to give any concrete interpretation....to the elementary chronotopic interval ds^2....He replied to me that he did not know of any physical meaning...."

In the three static tests of Einstein's theory, the line element of the Schwarzschild metric is written in polar coordinates. So if r is not a radius, theta and phi are not angles. How then can the claim be made that the result for the perihelion procession is in "fullest agreement with the observed perihelion in the case of Mercury" or "the existence of light deflection in approximately the right amount" [quotes taken from Lanczos, "Variational Principles of Mechanics"].

Mass pops up from an arbitrary constant of integration. Supposedly it has been chosen so that "Einstein's law of gravitation goes over to Newton's when the field is weak and when it is static." [Dirac, "General Theory of Relativity"]. But Newton's theory deals with absolute space and "the 'Allgemeine Relativitatstheorie' is in fact entirely relative, and has no room for anything whatever that would be independent of the system of reference. The need for the introduction of distant masses arises from the wish to have the gravitational field zero at infinity in any system of reference." [de Sitter, "On the relativity of rotation in Einstein's theory"]

Then there is the fact that a mere change of coordinates can make singularities appear or disappear. This, alone, indicates the tenuous nature of the metrics involved. Finally, I can always go from M=const to one where the density is constant and this reduces the Schwarzschild metric to one of constant curvature, as he did when he went from his outer solution to his inner one. This does not coincide the correct uniform space metric ["A New Perspective on Relativity" Eqns (9.10.23) and (9.10.24)].

The Schwarzschild radius is the coordinate radius, so is the Schwarzschild circumference, ok? What is the real circumference at the local metrics? Infinite?

I don't understand what is a "local metric.". The optical metric corresponding to the Schwarzschild metric is the hyperbolic Beltramic metric, and the radius is not Euclidean. Ehrenfest explained the difference as a "Lorentz contraction" of rulers placed along the circumference in the direction of rotation. This was right, but for the wrong reason! The boundary of the disc where the metric  becomes singular is unattainable. There is nothing strange about that; it is the boundary of the hyperbolic world. 

With "local metrics" I mean measuring scales for individual observers in a gravity well, underlining the difference between distances measured by local observers  and coordinate distances (measured by distant observers).

If gravity is geometry then there is no distinction. Why do local and distant observers see different geometries?

For the same reason that observers see at relativistic speed each other as shorter, I think.

Why should gravity be a relativistic effect when it is related to the static geometry of the system?