View Single Post
Nov11-05, 05:56 PM
sweetser's Avatar
P: 361

I think I am obligated to point out possible logical flaws in general relativity. After all, GR has done brilliantly for 90 years, passing many difficult tests. Many people work on the theory today. Its intellectual structure is elegant. However, I do see specific flaws that I will try to point out in this message.

The vector [itex]A^{\nu}[/itex] transforms like a tensor. The vector [itex]\partial_{\mu}[/itex] transforms like a tensor. The 4-derivative of a 4-vector, [itex]\partial_{\mu}A^{\nu}[/itex] does not transform like a tensor. Instead, the covariant derivative does:
[tex]\nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu }+\Gamma_{\sigma\mu}{}^{\nu}A^{\sigma}[/tex]
[tex]\Gamma_{\sigma\mu}{}^{\nu}A^{\sigma}= 1/2 g_{\beta\sigma}(\frac{\partial g_{\mu\beta}}{\partial q_{\nu}}+\frac{\partial g^{\nu}{}_{\beta}}{\partial q^{\mu}}-\frac{\partial g_{\mu}{}^{\nu}}{\partial q^{\nu}})[/tex]
is the Christoffel symbol of the second kind, a measure of how a metric [itex]g_{\mu\nu}[/itex] changes, as indicated by the three derivatives of the metric. The Christoffel symbol, complicated as it is, does not transform like a tensor. Instead, it must be teamed up with another non-tensor, [itex]\partial_{\mu}A^{\nu}[/itex] to transform like a tensor.

Up to this point, I am in complete agreement.

The next question is to ask: "What tensor can be formed out the the Christoffel symbol?" The correct answer provided in GR books is the rank 4 Riemann curvature tensor. No argument with that. I object to the question itself. Why not ask: "What tensor can I form with [itex]\partial_{\mu}A^{\nu}[/itex]?" There may be such a rank 4 tensor, but I am not aware of the question being raised.

If one works only with the Chrisoffel symbol and not the potential, then the potential and metric are effectively divorced. That is what I object to. The divorce is an accurate description of our current understanding of GR and EM. The Maxwell equations are a potential theory, and by extension, the standard model. EM requires a metric be provided as part of the background structure, a sure sign of a divorce. The standard model needs mass to be introduced via the Higgs mechanism. GR is exclusively about gravity. All efforts since 1930 have failed to unify GR with the rest of physics, particularly quantum mechanics. This is not a temporary separation. String theory in my opinion cannot bind the metric to the potential. In my GEM proposal, the changes in the potential and the changes in the metric are united at the most logical and elegant way, right in the asymmetric, reducible tensor [itex]\nabla_{\mu}A^{\nu}[/itex].

What is the Riemann curvature tensor? It is a measure of the amount of curvature at each point in spacetime. On essential looks at the differences between two paths. Here is the definition:
[tex]R^{\rho}{}_{\sigma \mu \nu}=\partial_{\mu} \Gamma^{\rho}{}_{\sigma \nu}-\partial_{\nu} \Gamma^{\rho}{}_{\sigma \mu}+\Gamma^{\rho}{}_{\mu \lambda}\Gamma^{\lambda}{}_{\nu \sigma}-\Gamma^{\rho}{}_{\nu \lambda}\Gamma^{\lambda}{}_{\mu \sigma}[/tex]
This tensor looks too complicated to me to ever understand in detail because each of those Christoffel symbols already has three metric derivatives inside it. None-the-less, the Reimann curvature tensor is the difference of two paths, which creates another problem in my opinion. Einstein's field equations conserve energy, a good thing. But at any point in spacetime, one can choose Riemann normal coordinates where the Christoffel symbol and all its derivatives are zero (but only one point in spacetime, since spacetime has to curve everywhere else). The energy density at that point will be zero. Thus energy density cannot be defined locally like it is for nearly all other field theories. People have gotten used to this difference in how energy is defined in GR, and do not consider it a flaw, just a property of the theory. I beg to differ because Nature i logically consistent. There should be no way to make a choice of coordinate frame such that the energy density is zero. In EM, one can choose difference coordinate frames, and the amounts of energy contributed separately be E and B fields cand shift, but not go to zero. In GEM, one can choose the Reimann normal coordinates, but the energy density would then by in the potential, and not zero. That to me is a good thing.

Since I am so close, I thought I'd sketch the rest of the way to Einstein's field equations for those reading this message and have not seen the path to those equations. Einstein figured Nature would want to use a simpler tensor to describe curvature. So he decided to use the Ricci curvature tensor, which is the Riemann curvature tensor with the first and third indices contracted with each other, [itex]R^{\rho}{}_{\sigma \rho \nu}=R_{\sigma \nu}[/itex]. A problem with the Ricci tensor is that its divergence is not zero, a problem for energy conservation. One needs to subtract the Ricci scalar to get to the zero, leading to Einstein's vacuum field equations:
[tex]R_{\sigma \nu}-1/2 g_{\sigma \nu}R=0[/tex]
Hilbert did it the proper way. He started with a super simple Lagrange density, [itex]\mathcal{L}_{GR}=R[/itex]. Varying the action with respect to the metric field [itex]g_{\mu \nu}[/itex], one gets the Einstein field equations.

One final clarification. I use the Chistoffel symbol in the gauge symmetric central to proposal, but at no time is the Reimann curvature tensor, or either of its contractions the Ricci tensor or Ricci scalar needed.