Dear Vladimir,

Thanks for your comments and mentioning your relevant work. I have been trying to understand Eisntein's theory (rather than showing anybody wrong). In this process, I found some conceptual problems in GR.

___Ram

Dear Manuel,

Thanks for your kind remarks. I shall read your paper.

___Ram

Dear Cristinel,

Thanks for your kind remarks. Riemann tensor can also be non-zero when Ricci tensor is vanishing. That will signify the net energy-momentum-angular momentum of the material and the gravitational fields at a point.

___Ram

Dear Satyavarapu,

In my essay, there is no problem of the violation of the conservation of energy or matter coming out from nowhere. If you read the article and the references therein, you will note that there are two time scales in the resulting cosmological theory. In terms of one of them, the universe becomes infinitely old without any singularity at any finite time in the past. So, the question of the origin of matter/universe becomes meaningless in this model.

___Ram

Dear Akinbo,

Thanks for your kind remarks. I shall read your paper.

___Ram

Dear Lawrence,

Thanks for your kind and knowledgeable comments. It would not be correct to say that all the solutions of Einstein equations are not meaningful. This will create doubt over the general validity of the theory. Then how can we be so sure that Schwazschild solution (for example) represents a meaningful solution. Just because, it seems consistent with experiments? May be, we have been unable to interpret other solutions correctly, which we claim unphysical. As an example, the Kasner solution (which is considered unphysical) in the new paradigm discovered, in the paper, represents a real big bang universe!

___Ram

Dear basudeba,

Thanks for your wonderful remarks. I shall read your paper.

___Ram

Dear Jonathan,

Thanks for your kind and inspiring remarks! I look forward to see your essay.

___Ram

Prof. R.G. Vishwakarma,

Thank you for your reply. You said "Riemann tensor can also be non-zero when Ricci tensor is vanishing.". When Ricci=0, Riemann=Weyl, so we are in complete agreement.

You mentioned the Kasner solution, and that the matter source (the singularity) exists only at the time t=-1/n, and yet it has effects at other times too. This in fact happens in the Schwarzschild solution as well. The singularity r=0 is not necessarily in the present of an observer affected by the black hole's gravity. This is obfuscated when using the Schwarzschild coordinates, but it is visible for example in the Kruskal-Szekeres coordinates. Another remark: I think something like this happens with the electromagnetic field. Imagine a pair electron/positron attract each other and annihilate. Yet, the electromagnetic field sourced by their charges exists an indefinite time after they were annihilated.

Best regards,

Cristi Stoica

Dear Cristinel,

Thanks for reminding me that the Weyl tensor (with 10 independent components) can be thought of as containing the information of the Riemann tensor (20 independent components) minus that of the Ricci tensor (10 independent components). So in the case of the vanishing Ricci, Riemann=Weyl.

I have tried to establish that space and `what fills space' are not two different entities. That is, by considering space means considering the accompanying fields as well. That the matter fields are present in the metric (without taking recourse to the energy-stress tensor), can also be proved by the conventional belief which considers singularity as the source of curvature in the absence of the energy-stress tensor. For this reason, I have considered Kasner, Kerr, Schwarzschild solutions. (For this purpose, I argue that if the source matter is present at the epoch of singularity, it must also be present at other times. That is, the source matter is already present through the metric field, in the Kasner solution.)

However, we must understand that the singularity is not the sole representative of curvature (in the absence of energy-stress tensor), since the conventional wisdom cannot explain the curvature of the Ozsvath-Schuckling solution (which is singularity-free). Hence the metric field (which has been shown to contain matter and gravitational energy in three cases) must be the real source of curvature.

Another reason, why the singularity is not an efficient representative of curvature, is the controversial character of the singularity (black hole) in the Schwarzschild solution. There are claims that the black hole mass must be zero [Narlikar & Padmanabhan, Foundations of Physics, Vol. 18, pp.659-668 (2008)].

The interpretation of the Schwarzschild solution, as a spacetime structure sufficiently away (of course that will be out of the event horizon) from some mass, is fine. In this case, the curvature present at those points can only be explained in terms of the gravitational energy (recalling that GR is a local theory). This is what I have emphasized in the essay.

The parallel you mention between the GR cases and the electromagnetic one, is interesting. Thanks. It may contain important information.

___Ram

Dear Jacek,

Thanks for your kind remarks and for your interest in my essay. I also thank you for mentioning Maluga's work on the geometrization of matter. I shall read them in time.

Best of luck on your ambitious endeavor. The theory based on R^{ik} = 0 might fit in it. It is scale invariant, describes not only the gravitational phenomena in the solar system, but the whole universe, as I have showed.

Best of luck on your essay.

___Ram

Ram,

If given the time and the wits to evaluate over 120 more entries, I have a month to try. My seemingly whimsical title, "It's good to be the king," is serious about our subject.

Jim

Dear Sreenath,

Thanks for your kind remarks. I don't understand how you can apply the evolving (time-dependent) Kasner solution inside a black hole represented by the static Schwarzschild (exterior) solution. The interior of a static spacetime is expected to be static. Anyway, the existing Schwarzschild interior solution, providing the standard representation of the interior of a static spherically symmetric non-rotating star, turns out to be unphysical, since the speed of sound (dp/d\rho) becomes infinite in the fluid with a constant density \rho.

While beauty should not be considered as a decisive factor for a physical theory, you seem to be unaware of the very Einstein's remarks of "low-grade wood" for the energy-stress tensor and the "fine marble" for the geometry (as he put it in a 1936 article in the Journal of the Franklin Institute). Thus shunning the "wood", ENHANCES THE BEAUTY of the "marble" in the extreme simplicity of the field equations R^{ik} = 0!

My pleasure - hope you like my essay too.

Antony

Dear Prof. R.G. Vishwakarma,

"we must understand that the singularity is not the sole representative of curvature (in the absence of energy-stress tensor)"

Yes, I understood this, because you've done such a good job in explaining everything well in the essay. I just wanted to make some remarks, which I hope may help. I will continue, although maybe you know the following.

In connection with explaining matter from spacetime (this time with nonvanishing Ricci), I find interesting the pioneering work of GY Rainich, who in 1924-1925 found the algebraic and geometric conditions curvature should satisfy, so that it corresponds to a source-free electromagnetic field. Wheeler and Misner built on this idea and that of Einstein-Rosen bridges in Classical physics as geometry, and coined the approach "charge without charge". Using a wormhole, the cohomology group of spacetime is changed, and we can have the appearance of an electric source, without having a charge.

Best regards,

Cristi Stoica

Prof. Ram,

Your essay is near and dear to my heart (as I am interested in how field theories are mathematically represented and the Cosmological Constant problems). I have some questions concerning your essay:

1. You state "in the absence of which the solution must have a singularity, serving as the source". What singularity are you referring to? I know of the singularity that arises in linearized gravity but that is based on

[math]G_{\mu\nu}\neq 0[/math]

. (I am not yet familiar with Osvath and Schucking so will have to review their material).

2. Your equation 4 uses an M which seems to be based on the Poisson equation definition of energy density within a volume. I see your disclaimers on LambdaCDM not conforming with the Poisson concept of energy density anyway but you seem to be equating the energy density of a gravitational field with the same energy density that mainstream physics is using for "vacuum" energy density. True or no?

3. How can a constant of integration represent the energy density of a gravitational field since the constant wouldn't change but the gravitational field would need to in order to do work on regular mass?

    Dear Jeff,

    Thanks for your wonderful remarks and interesting questions. My explanations to your queries are the following.

    1. I'm talking about the full GR and not any of its approximations, such as the linearized gravity. It is the conventional belief that the curved solutions of R^{ik}=0 (i.e., the solutions of Einstein's eq (1) in the absence of T^{ik}) must have a singularity. For example, the Schwarzschild and Kerr solutions have singularity at r=0, the Kasner solution (5) has singularity at t=-1/n. But this conventional wisdom does not seem correct as the Osvath-Schucking solution (a solution of R^{ik}=0) is curved but it is singularity-free.

    2. The value M in eq (4) comes by comparing geodesic equation in GR (considered in the case of a weak gravitational field) with the Newton's equation of motion. This gives the metric potential g_{00} in terms of the Newtonian gravitational potential. Remember that in the case of a weak field, eqs R^{ik}=0 reduce to the Laplace eq and not the Poisson eq. Though both these eqs (Laplace and Poisson) use the scalar gravitational potential, but in Poisson eq, it comes from a matter distribution, while in the other case, it results from a point mass [as in eq (4)].

    3. If you read carefully, you will find that the gravitational energy is represented IN TERMS OF the constant K, and NOT "by the constant" K. It is in fact K/r [(see the lines preceding eq (4 )].

    ___Ram