Physics is not my primary scientific discipline.

However, about geodesics in general relativity you may find in the following book (you will understand that better than me) http://www.blau.itp.unibe.ch/newlecturesGR.pdf

Moreover, it is some geometrical invariances as I saw in this book. Anti-symmetric affine connection covers the theory of electromagnetism, but the symmetric affine connection covers the theory of gravitation (you know that better than me). Invariances obtained from curvature tensors you may find in this paper:

https://arxiv.org/pdf/1706.01784.pdf

I know that the physics of microspace and macrospace are covered by different laws. In the previous paper, there are two invariants obtained from the change of curvature tensor. For this reason, you maybe may apply these results in your research.

I hope that I helped you somehow.

Good Luck!

 What I am saying that Newton's second law generalizes to an equation for the four-force per unit mass in general relativity. The geodesic equation is usually taken to mean that the four-force vanishes. If it doesn't vanish in flat spacetime the Christoffel symbols vanish and supposedly Newton's second law is regained in the low speed limit where the proper time coincides with coordinate time. In Newton's second law the gravitational force is the external force. It should generalize for a four-force in general relativity since it reduces to it in flat spacetime at low speeds. Hence, Newton's law cannot "pop" out of the geodesic equation from the term involving the Christoffel symbols in the weak field limit. The metric involving the gravitational potential, i.e., the Schwarzschild metric is one of non-constant curvature which reduces to one of constant curvature when the mass is replaced by the density. The product of G and the density is the inverse of the square of the free-fall time and this is where free-fall applies, to a metric of constant curvature. In the FW metric, the geodesics are curves which cut the disc orthogonally or straight lines through the center of the disc.

I have written a paper recently that reasons from GENERAL Relativity a light speed limit to Escape Velocity in parallel to SPECIAL Relativity limit to simple vector velocity.   While it is mathematically reasoned from the General Relativity Time equation, there is also a logical argument for a limit to the escape velocity.  One of the very fundamental effects of Relativistic distortion is the slowdown of Bosons.  For General Gravitational Relativity that surely would include Gravitons.  An equation in General Relativity can be reasoned to make this argument directly.

Just consider the escape velocity equation 

            V_esc = (2GM/r)^.5

It can be re-phrased as 

            V_esc^2 = (2GM/r)

So a valid re-write of the General Relativity time distortion equation could be

            Time’ = Time/(1 – 2GM/rc^2)^.5

            Time’ = Time/(1 – (2GM/r)*1/c^2)^.5

            Time’ = Time/(1 – (V_esc^2/c^2)^.5

 The above argument is the same in structure as the Special Relativity time distortion equation.  It would slow Bosons with that energy passing to matter in the vicinity and allow it to eventually escape the Schwarzschild object it was contained in.  This argument is made in greater detail in my paper  A Relativistic Escape Velocity Maximum of Light Speed that appears in Journal of High Energy Physics, Gravitation, and Cosmology. Such a limit would resolve both the fundamental inconsistency in Big Bang Theory, but also answer the question as to why all the objects that have been identified as Schwarzschild Objects - from Quasars on down, are the brightest objects in their neighborhoods. A Relativistic Escape Velocity Maximum of Light Speed is linked below

http://www.scirp.org/journal/PaperInformation.aspx?PaperID=68063

If it is false, as you claim, say where instead of referring me to a bunch of articles!

You're first equation in your EL paper is not the metric used to derive your coefficient alpha. If you don't take into consideration the angular dependencies of the metric with undetermined coefficients you won't get the Schwarzschild radius condition when you integrate the R22=0 equation [c.f. eq (18.4) of Dirac's "General Theory of Relativity", or any other book on GR).

True, the R22=0 condition satisfies the R00=R11=0 condition, but so does an infinite number of other possibilities. It is the former condition that places a lower limit on the radial coordinate, namely 2m, where m is some arbitrary coefficient of integration. It is then another matter to identify m.

Moreover, in a 2D space, or spacetime which is what you are considering, only one component of the Riemann tensor is independent. All other non-zero components are related to this component. In other words, how can you use the full metric to derive the metric coefficients from Einstein's equations and throw out the precise terms that were used to derive the metric coefficient in your equation (1.1) (the R33=0 component would give the same condition)? For if you could, then why not start with your (undetermined) 2D spacetime metric and apply Einstein's condition of emptiness, R00=R11=0 ? You get an infinite number of other alpha's in your equation (1.1). 

Don't be surprised by the similarities with classical mechanics. They are superficial like associating the total energy with the the curvature index in the Friedmann solution to distinguish bounded from unbounded solutions.

You haven't answered my criticisms! 

Why do you need the rotational components of the Ricci tensor to derive your coefficient alpha and then discard all angular contributions to the metric? What is the physics in setting R22=R33=0? and why should an angular component of the Ricci tensor determine a lower bound on a radial coordinate?

There is absolutely no physics in what you are doing, just playing with mathematical expressions.

Then you don't need them to determine the metric. Try it!

Dear Bernard,

is it possible for you to check this out

https://www.researchgate.net/publication/312118218_Free_Fall_in_Gravitational_Theory

it relevant to your sentence "In contrast, Schwarzschild's solution is not compatible with free-fall,"

Best regards

Article Free Fall in Gravitational Theory

Adrian,

The radial Schwarzschild metric,

ds2= e2adt2-e2bdr2

where a and b are functions of r only. Einstein's condition, Rtt=Rrr=0 gives a+b=0 since both a and b are assumed to vanish as r tends to infinity. Full stop. No condition on r!

You need the vanishing of the Ricci components of the angular terms to get ea=1-2m/r.

The question you should ask is why should the angular components of the Ricci tensor determine a condition on the radial coordinate when the radial component does not?

Is your problem the English or the physics, or both?

Why not try my book, "A New Perspective on Relativity"?

To each his own!

Ciao Stefano,

Thanks for pointing out the paper; it is indeed interesting.

It illustrates the definite problem when phi is set equal to zero after the metric has been derived by setting all components of the Ricci tensor equal to zero. Notice that his geodesic equation (3) for angular momentum is in proper time, s, and not in coordinate time t. If he had considered the latter, he would not have obtained a conservation of angular momentum, but, rather, eqn (18) in chapter XII in Moller's book. Angular momentum is not conserved in coordinate time.

Later, when he considers radial motion by setting phi=0, he switches over to coordinate time, t,  in defining the velocity. If you consider s with phi=0 instead you  get dr/ds= (K-C2)1/2. Then differentiating with respect to s, you obtain Newton's law in proper time--exactly---without any approximation of a weak field. But, that doesn't correspond to any geodesic equation! (The Christoffel symbols are calculated in terms of coordinate time, t, and not proper time, s.)

This only reinforces my claim that the limit on the radius is due to the vanishing of the angular components, 33 or 22, of the Ricci tensor, and there is nothing physical in it. Yet, when you set phi=0--after you derive the metric--- and use coordinate time, t, you come out  with nonsense: the velocity, dr/dt, vanishing at the Schwarzschild radius together with its kinetic energy. There is no way that Newton's law in proper time, s, gives any hint to this unphysical behavior.

Saluti!

It behooves me to refer you to "your" literature, e.g. C. H. McGruder III, Phys. Rev. 25D (1982) 3191 and all the references therein.

Dear Bernard,

in the Paper of Engelhardt it comes out for sure one fact: the Einstein's mistake in trying to solve his EFE before Schwarzschild or the strong committment to make things come out as he needed (Gerber's formula).

Regarding the free fall it seems quite hard to most to understand that:

the approximation of test particle is something on which you cannot build up a theory. The geodesic motion is based on such approximation which holds always  for photons or entities with zero rest mass, but has clear limits on other cases.

Such model of geodesic free fall can be always falsified for adrons or leptons which get accelerated in any Celestial body Centered inertial frame and eventually alter the gravitational field of the heavy body, especially if they have a significative initial speed in that frame.

Any massive particle entering the picture distrupts the Schwartzschild symmetry if it is not included since the beginning.

Salve Stefano,

I've always worked in coordinate time like everyone else. If you work in proper time, Engelhardt's paper gives the conservation of angular momentum per unit mass (4) (without the imaginary i), and (7) is the "classical" conservation of energy for radial motion only. (It's better to switch signs in (7) so as to make s proper time and there is no i in (6).) Otherwise, you would have a K multiplying the square of the centrifugal term. But everyone works only to lowest order so the non-conservation terms do not appear.

This shows that without the angular part of the metric, there is no black hole at r=2m because there is no K. And if you were to solve the radial metric from the start, using Einstein's condition of flatness, there would also be no K. Physically, it makes no sense to attribute a limit on the radial coordinate to the centrifugal term since what counts is the radial infall of an (imaginary) "test" particle which actually doesn't exist because it does not create a gravitational field of its own.

By switching from coordinate to proper time, you can make a black hole disappear!  Only in the case of pure radial motion is Engelhardt's (7)  the exact conservation of classical energy, A; solving for dr/ds and taking the derivative with respect to s gives Newton's law exactly, again showing that there is no barrier at r=2m.

It is amazing how relativists rationalize the lack of conservation of angular momentum in coordinate time, as in eqn (18) of Ch XII in Moller's book. He writes: "the left-hand side of (18) cannot in general be interpreted as angular momentum, since the notion of a 'radius vector' has an unambiguous meaning only in a Euclidean space."

To make matters worse, Dirac (in "General Theory of Relativity", as everyone else does) writes Rab=0 (15.1) as the condition for "flat" space. "Flat space obviously satisfies (15.1)." But, we have just seen that the solution that Schwarzschild found is not a flat metric since the conservation of angular momentum and energy are violated.

Time does not enter into non-Euclidean metrics. As the relativists would say, the latter correspond to optical metrics. The debate over which time you use does not enter if you consider only null geodesics. That would imply that gravitational disturbances, seen through inhomogeneous indices of refraction, are propagated by electromagnetic waves and do not have a "life" of their own (i.e. gravitational waves).

Saluti

Salve Bernard,

"The debate over which time you use does not enter if you consider only null geodesics."

YES, exactly for this reason GR is a reliable theory in the case of geometrical optics like also Baryshev says.

"the left-hand side of (18) cannot in general be interpreted as angular momentum, since the notion of a 'radius vector' has an unambiguous meaning only in a Euclidean space"

Moller is one of the firsts to deal with GR, maybe somebody later rectified this. 

"the solution that Schwarzschild found is not a flat metric since the conservation of angular momentum and energy are violated."

it is not a flat metric but it satisfies the conservations, provided that the conditions in which an external body moves are stationary (ORBITING).

I have the impression that you think that GR is very wrong, it is a theory and it has limitations as every theory in representing reality.

It is also wrong to think that it is the ultimate theory or it is reliable for solving any gravitational problem.

"That would imply that gravitational disturbances, seen through inhomogeneous indices of refraction, are propagated by electromagnetic waves and do not have a "life" of their own (i.e. gravitational waves). "

I don't know, but I'm sure that EM waves have nothing to do with gravitational waves, except for their speed of propagation.

Yes, I think it is very wrong. I don't know of any two body solution in GR so test bodies have no physical features of their own. Moreover, when you read statements like Dirac's: "Flat space obviously satisfies (15.1). The geodesics are then straight lines and so particles move along straight lines." is completely void of meaning.

Friedmann used the half-plane and disc models of hyperbolic geometry as the most general homogeneous and isotropic models of space which are geometries of constant negative curvature. I would not agree with that statement about isotropy and homogeneity (e.g. the heated plane model of the half-plane certainly is certainly not homogeneous because the line at infinity represents absolute cold and a pseudosphere can hardly be called isotropic), but tagging on the Einstein equations to these models makes no sense. The geodesics are half circles that cut the line of infinity orthogonally (half-plane) or curves that cut the disc orthogonally (disc model). They are certainly not "straight" lines. They are, in fact, the trajectories of light rays. The key I think is the Beltrami metric which is

-the metric of a uniformly rotating disc,

- is conformal to the space part of the Schwarzschild metric, when density replaces mass (transforming it into constant negative curvature),

-and is  identical to it for null geodesics.

Time per se in the models of hyperbolic geometry doesn't exist, but the velocity space metrics coincide with Lobachevsky space as Fock emphasizes in his book.

I invite you to do the same instead of cannibalizing things with an insane and aggressive attitude and read carefully before posting papers which do not disconfirm what I said, because you continue to get a very meagre impression.

Bernard,

this paper is interesting

https://www.researchgate.net/publication/316451143_Diffraction_of_electromagnetic_waves_in_the_gravitational_field_of_the_Sun

Article Diffraction of electromagnetic waves in the gravitational fi...

To return to the original question...

“…Schwarzschild's solution is not compatible with free-fall, and does not yield geodesic paths...”

makes no sense at all to me! Geodesics exist in any Riemannian manifold and "free fall" in GR is geodesic motion. Free fall in a Schwartzschild spacetime is uncontroversial. Also, I'm utterly baffled by all the obscure talk of “non-constant curvature” creating difficulties. I just don't get it! It seems to be creating imaginary problems where none exist.

The derivation of Newton’s gravitational theory from the “weak field” approximation to GR is straighforward and well-known. A simple presentation is given here:

https://www.researchgate.net/publication/314232532_Poisson%27s_equation_from_GR

 (q.v.).

In empty space Newton’s theory can be expressed by the Laplace equation

∇2φ = 0

where φ is the gravitational potential φ = c2h00/2. The relevant solution for a central mass M is φ = GM/r, consistent with the weak Schwartzschild h00 = 2GM/c2r.

Note that the derivation of the Laplace equation from weak field GR involves a gauge condition, which is not satisfied by the weak field approximation to the Schwartzschild metric in standard Schwartzschild coordinates, because of the dependence of hrr on r. This is readily remedied by a gauge transformation to eliminate hrr. The only non-vanishing component of hij is then h00 = −2GM/c2r, the only non-zero Christoffel symbol is {r00} = ∂[GM/c2r]/∂r, and the geodesic equation is

d2r/d(ct)2 = ∂[GM/c2r]/∂r

which is the Newtonian equation for free fall

md2r/dt2 = −GMm/r2.

Where is the problem??

Method Poisson's equation from GR

Bernard> ...non-euclidean geometry is nothing more (or less) than the study of surfaces of constant curvature.

This is plain wrong. And I kind of doubt that Beltrami has ever claimed such a thing. Where?

Next time you eat a potato, look careful at its surface: Does it look like a constant-curvature surface? But it does look non-Euclidean, right?

Or, if you prefer, consider your boiled breakfast egg, and stretch a sewing thread between two points on its (non-Euclidean, non-constant curvature) surface. Congratulations, you have just constructed a geodesic for such a surface!

Of course. There, also,  do exist constant curvature solutions to Einstein's equations-the sphere and the one-sheeted hyperboloid are the typical examples.

It's, also, another standard exercise to show that Newton's equations can be written as the geodesic equations-and vice versa-thereby establishing the equivalence, which, in fact, can be found in the work of Hamilton and Jacobi. 

Newton's laws and the geodesic equations are classical. And the trajectory is a geodesic-as shown by Hamilton and Jacobi-for any potential. They established the quantitative correspondence between the  solutions of Newton's equations and light rays in media with a general refraction index. 

Non-euclidian geometry, incidentally, isn't identical with Lorentzian geometry. Lorentzian geometry is a special case of non-euclidian geometry, where the metric isn't positive--definite. 

Kare,

You're argument is not with me but with the mathematicians, in particular Liouville, Beltrami, Klein, Poincare', etc. As a historical survey see:

J. Stillwell, "Sources of Hyperbolic Geometry", especially his introduction to Beltrami's work on pp. 35ff. Beltrami's work dates back to 1868 where he showed that 2D non-euclidean geometry is nothing more or less than surfaces of constant curvature.

From the Encyclopedia of Mathematics:  A simply connected Riemannian space of constant sectional curvature C is isometric to the n-dimensional Lobachevskiian space when C0 to a n-dimensional sphere Sn in En+1 of radius 1/C1/2. The first and last are precisely non-euclidean geometries.

Also the Riemann metric, which he introduced in 1854, is one of constant curvature! (Wolf, "Spaces of Constant Curvature", Preface). The denominator defines for you the boundary of the disc in the hyperbolic plane. It is not compatible with Schwarzschild's outer solution; this does not belong to non-euclidean geometry! It is, however, compatible with his inner solution, which is one of constant negative curvature, GM/rc 2-> G rho (r/c)2, where 1/(G rho)1/2 is the Newtonian free-fall time. 

I know what geodesics are in hyperbolic geometry: they are rays that cut the boundary of infinity orthogonally, e.g. the Beltrami disc and the Poincare' half plane. When the denominator is the square root of the latter, the rays are cycloids but still intersect the line at infinity orthogonally.

For non-constant sectional curvature, the problem is much more complex. As far as I know, it is not known what function the sectional curvature must be in order for it to have a metric gij for which it is sectional curvature. (Encyclopedia of Mathematics)

Hence, my remarks refer to non-euclidean geometries which are surfaces of constant curvature (pseudo-and real-spheres) for they represent the most general homogeneous and isotropic spaces. Friedmann took that from Bianchi.

Bernard> my remarks refer to non-euclidean geometries which are surfaces of constant curvature

OK. It is very likely that the concept of "non-euclidean" (as I interpret it) has changed since the mid-1800's, when the focus was on the status of Euclid's fifth postulate. And when also the surface of a sphere was not consider to be a proper non-Euclidean^* space (since it violates Euclid's second postulate). From the point of view of general relativity I think the axioms of Euclid can just as well be forgotten, but for sufficiently large scale phenomena (perhaps be) replaced by the cosmological principles of homogeneity and isotropy.

Bernard>  it is not known what function the sectional curvature must be in order for it to have a metric gij for which it is sectional curvature

This is probably correct. Given a d-dimensional Riemannian space, with a metric gij, it may be impossible to specify this explicitly as a hyper-surface in higher-dimensional space. But I vaguely remember having heard about theorems that this can always be done, in principle, for a (2d+1)-dimensional imbedding space. And one may always go in the other direction, by starting with the imbedding and computing the metric from this (f.i., to describe the inhomogeneous surface of an egg). 

^*) So they only wanted to consider a-little-bit-non-Euclidean spaces :-). 

Eric,

 Regarding geodesic motion, you claim that Newton's law comes out in the weak field limit. By a theorem of Fermi, for any space with a symmetric connection it is possible to choose a coordinate system with respect to which the Christoffel symbols vanish at all points along the curve. Does that mean that the Newtonian force (your last equation) vanish in flat space? How can Newton's force vanish by a coordinate transformation?

There are significant differences between metrics of constant Riemannian curvature and those without. The former are governed by non-Euclidean geometries and the geodesics are shortest paths. For example,the geodesics the hyperbolic plane  are circular arcs that cut the boundary of the disc orthogonally or the line at infinity in the half-space model.  The circular functions have arguments r/c tN, where tN=1/(G rho)1/2, the Newtonian free fall time. To go from the Riemannian metric of constant curvature to the Schwarzschild metric one has to replace the mass density by the mass itself: G rho r 2->G M / r. Although the Schwarzschild metric has non-constant negative curvature,  it was derived from Einstein's condition of emptiness, Rij=0!

Yet, a homogeneous Riemann space is called Einsteinian when the Ricci tensor is proportional to the metric (the constant being the curvature). The description of such spaces is still an open question  (Encyclopedia of Mathematics). 

Therefore, unlike geodesics of non-euclidean geometries, I do not know what geodesics of homogeneous Riemannian spaces of non-constant curvature look like, or even if they exist at all.

Eric,

I looked at your paper: You say that when the accelerations can be neglected then the wave equation reduces to the Poisson equation. But I can always contract Einstein's equation with respect to a timelike unit vector and obtain that the scalar curvature of the spatial dimension is proportional to rho. This is independent of whether the accelerations can be neglected or not. 

The wave equation does not fix the speed of gravitational waves as coinciding with the speed of light. They can travel at any speed; that is, if they exist.

Gravitational waves don't satisfy an arbitrary wave equation-they satisfy a very particular one. In that equation the propagation speed is that of the speed of light in vacuum. And, contrary to popular belief, gravitational waves have been measured to have this property-for the first time more than 25 years ago. Were the period of the binary pulsars not varying as it did, this would have meant that gravitational waves don't travel at the speed of light. Same conclusion from the LIGO measurements.

This doesn't have anything to do with the subject of the question, once more. 

This paper: https://link.springer.com/content/pdf/10.1007/BF00764019.pdf might be useful. 

Dear Bernard ~

 The “Non-Euclidean Geometry” or “Hyperbolic Geometry” discovered by Bolyai and Lobatchevsky is a very special case (constant negative curvature) of Riemannian geometry. Constant curvature has very little relevance to General Relativity.  Riemannian geometry is essentially an extension to more than two dimensions of Gauss’s investigation of the intrinsic geometry of a curved surfaces. It is “Non-Euclidean”. The “non-Euclidean” geometry of General Relativity is a four-dimensional Riemannian geometry, not hyperbolic geometry!

“…unlike geodesics of non-euclidean geometries, I do not know what geodesics of homogeneous Riemannian spaces of non-constant curvature look like, or even if they exist at all.”

Then, clearly, you need to learn some elementary differential geometry!   

For general relativity what's relevant is pseudo-Riemannian geometry, since the metric doesn't have-Euclidian-signature (the metric is positive-definite), but Lorentzian signature (the metric has one negative eigenvalue). 

Bernard ~

“By a theorem of Fermi, for any space with a symmetric connection it is possible to choose a coordinate system with respect to which the Christoffel symbols vanish at all points along the curve. Does that mean that the Newtonian force (your last equation) vanish in flat space? How can Newton's force vanish by a coordinate transformation?"

According to Relativity flat spacetime correspond to the absence of a gravitational field. So, yes, of course the Newtonian force vanishes in flat spacetime!

Astronauts in “free fall” (in an orbiting space station for example) “float around”. For them, gravity is absent. We can choose a coordinate system for which the space station is “at rest” (eg. remaining at the origin of the coordinate system throughout its trajectory) and in which the Christoffel symbols vanish along its trajectory. The Newtonian “force” does not appear in that coordinate system.

(Recall the “Equivalence Principle” and Einstein’s “elevator” thought experiment.)

For these reasons, General Relativity does not regard gravity as a “force” in the Newtonian sense.

One shouldn't mix up space and spacetime. The Newtonian force vanishes in flat spacetime-not flat space. (In the Newtonian limit the scalar potential appears in the g00 component of the metric.)

In FLRW metrics the non-constant scale factor ensures that, while space is flat, spacetime isn't.

And while it is possible to eliminate curvature at any given point in spacetime, it's not possible to do so everywhere, unless spacetime is flat. 

In other words, it's possible to do so by a local transformation, always, but not by a global transformation, unless spacetime is flat. Cf. https://arxiv.org/abs/gr-qc/9712019

"In FLRW metrics the non-constant scale factor ensures that, while space is flat, spacetime isn't."

The FLRW metric has a constant k=0,1, or -1. Space is not flat in the last two cases! The scale factor is a function of time only! Actually, the scale factor and k are one and the same with k=-1, as Friedmann knew when he began with the Poincare' half-plane model as his model of a homogeneous and isotropic universe. The scale factor, which should be just that, just changes the magnitude of the constant negative curvature, k=-1. Making it a function of time is ridiculous since the x-axis is already infinitely far away in the x-y plane. In the heated plane analogy, the x-axis is infinitely cold (cf. Barankin, Am. Math. Monthly (1942) 4-14).

Shouldn't the question be "non zero curvature" instead of "non constant curvature".

Of course non constant curvature implies non zero curvature but excludes the global topological case of constant non zero curvature.

I'd keep it "nonconstant curvature". An example is a horocycle whose perpendicular geodesics all converge asymptotically to an ideal point. This occurs in the case of a bugle, a surface of revolution of negative constant curvature, not zero curvature.

Hi Bernard,

does your question address the curvature of space or the curvature of geodesics?

the curvature of space-time, if the latter has any meaning.

Hi Bernard,

what do you think about the physical impact of nonzero constant isotropic curvature? Would this produce a repressive force on moving matter?

Hi Wolfgang,

To answer that I would have to know the relation between force and curvature. I don't think there is any general relation between the two. To say that the Ricci tensor defines the geometry and the geometry acts on matter is too vague, and general, to be of any use.

Rather, the Gaussian curvature is well-defined as the ratio of the solid angle subtended by the normal projection of a small patch by the area of that patch. Gauss showed that the potential of a magnetic shell at any point is the product of the strength and the solid angle subtended by its edge at that point. Thus far, we have only an isolated magnetic dipole. Maxwell showed that the magnetic dipole m generated by a neighboring dipole m' is the rate of change of the potential of the isolated dipole with respect to the coordinate of m' of the dipole at m'. The resulting force is repulsive being a generalization of Ampere's law to higher multipoles. Transforming the space derivatives into time derivatives like Weber did in converting Ampere+Coulomb into his law of force raises the question of whether Newton's second law represents curvature. Instead of the second derivative of the radial coordinate r with respect to time, it would be the second derivative of 1/r with respect to time. The curvature is constant and positive since the law of cosines for a spherical triangle connects the accelerations and the velocities in the force law.

Dear Bernard,

you are definitively right, there is no general relation between force and curvature and a constant isotropic curvature cannot generate a force.

The crucial point is, that it might require energy to transport a mass across a curved space region with diverging geodesics, though no force acts on the mass at rest.

This means that the space has properties, which only concern movements or displacements. The geodesics at any point diverge in any direction, but only a moving probe is able to notice the divergency.

This leads to a tiny and slightly nonlinear gravitational impact, which brakes any movement of masses and even the dissemination of light.

Dear Wolfgang,

You are right, it is all about motion. GR is useless in the aspect because energy is not localized so it can't be used to transport mass.

In the case of constant curvature, the Beltrami metric describes a uniformly rotating disc. The metric is the Lobachevsky-Einstein metric (as Fock called it in his book on Space, Time and Gravitation.) It's all based on the velocity composition law of hyperbolic tangent, and I deal with it extensively in my book, A New Perspective on Relativity. The history goes back to Sommerfeld (1904), Varicak (1910) and Kaluza (1910). The restriction is to spaces of constant, negative curvature.

Dear Bernard,

the crucial point is that a metric by definition is always symmetric and therefore not direction dependent. If we only consider the metric property of a space, the impact of moving from A to B is always reverse to the impact of moving from B to A. This is valid in any kind of space and any kind of metric.

Therefore we need a another space property, a pseudo metric which is specifically direction dependent. As an example the quaternion multiplication in Minkowski space exactly provides such a property.

Description of the example:

[q=exp(t,ix,jy,zk), q(A,B)=q*(B,A); q(A,B) is the quaternion operator for the movement from A to B.

The quaternion multiplication property causes, that in the inverted multiplication, the time is also incremented, but the negative path direction is compensated by the adjoint operation.]

Dear Wolfgang,

If I know the curvature of the space why do I need a metric? Granted the metric coefficients can be used to derive the Gaussian or sectional curvature, but they are not necessary. And even if I have a metric, I still need second derivatives to determine curvature. As an example, the gravitational pseudo tensor of Einstein is not valid since it can be made to vanish by a coordinate transformation.

An often used analogy is a ball on a rubber membrane which causes indention in the rubber sheet. But this is misleading since it is intended that gravity causes the indention in the membrane. And if we know the cause of the curvature why do we need the membrane, or metric, at all?

Dear Bernard,

your question in that thread concerns geodesics. Geodesics embody the properties of the underlying geometry. Mentioning general relativity establishes a link to the space of our universe.

What you describe with the membrane and a ball are local disturbances. In the global context we can ask the following question:

Does our space have the following three properties or does a space exist with these properties?

- In any point and in any direction there begin closed geodesics, which finally intersect in their origin point. The geometric structure spanned by those geodesics is homogenous and isotropic.

- Geodesics, which begin in neighboured points in the same direction initially are divergent until they reach an antipode point.

- Physically only moving matter or energy can take notice of those geodesics. In a static sense those geodesics don't exist.

Dear Wolfgang,

What I asked in my Relativity book was whether through any given point is there only one line parallel to a given line? The velocity addition formula says no. That is your point 3.

Point 2 is whether Jacobi's equation is satisfied. This sets the curvature as -y"/y. But, I find an additional term, not included in GR, y'^2/y^2, which is Neumann's potential in Ampere's law. The same term also appears in Gauss's law for the curvature of a sphere.

Dear Bernard,

can we interpret your comment in the sense, that with your work you are able to prove the existence of a space with the three properties of geodesics:

"Closed", "divergent", "susceptible for moving matter or energy".

Dear Wolfgang,

I don't know what "closed", "divergent" and susceptible mean. Do they refer to elliptic, hyperbolic and flat?

Dear Bernard,

according to the three properties of geodesics mentioned before, closed means that a geodesic is like a circle, divergent means that the distance between originally parallel geodesics is increasing, only susceptible for moving matter means, that only moving matter can "feel" the properties.

Elliptic, hyperbolic, flat are geometrical space properties not necessarily applicable to geodesics in spacetime.

Dear Wolfgang,

Great circles on a sphere=elliptic, trajectories that satisfy the Jacobi equation=hyperbolic. We are saying the same thing in different jargon.