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Some questions about the Rainich-Misner-Wheeler theory

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As mentioned by @CristiStoica in the linked answer below, the Rainich-Misner-Wheeler (RMW) theory explains electromagnetism in a similiar way to the geometric description of gravity in general relativity. This seems rather surprising to me, as I remember having tried to do something similiar in the past, only to conclude that this is not possible because of the absence of an equivalence principle for electromagnetism. Therefore, I have a few questions:

  • How is the issue above explained, i.e. how does the RMW theory avoild the fact that there is no "equivalence principle" for electromagnetism, since the electric charge is not equivalent or proportional to the inertial mass of the object?
  • Is the RMW theory compatible with General Relativity? If so, I suppose that electromagnetic effects (I don't mean the gravitational effects of electromagnetism, of course) is not accounted for in the Riemann tensor in RMW theory. In that case,
    • What exactly (e.g. torsion, etc.) accounts for electromagnetism in RMW theory, and how can this quantity be tested?
asked Feb 1, 2015 in Theoretical Physics by dimension10 (1,950 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

The glib wrong answer to your first question is that Reinich Misner Wheeler theory is source free, so there is no equivalence principle for the motion of test charges, as there are no charges. This glib answer is wrong, because there are still electrovac black holes in the theory which serve as charges.

For the second question, What accounts for electromagnetism is that the vacuum equations aren't satisfied $R_{\mu\nu}\ne 0$, rather there is a nonzero stress tensor, which is the Einstein tensor found from the given solution. This stress tensor is equal to that produced by some F field obeying the free-space Maxwell equation.

Coming back to question 1--- not every geometric field can be reduced to 0 by a local coordinates, just the connection coefficients. Only the effect of the connection coefficients can be zeroed out. The curvature is covariant, and if it is nonzero in one frame, it is nonzero in all frames. A solution to the RMW equation gives you a metric compatible with a unique F tensor, determined up to coordinate covariance by the metric tensor.

How do we know that the F tensor is unique? Here is a physical argument, not a mathematical proof. You can always deform the given solution by adding the metric for a local charged black hole to the given solution of the theory. The result will have a deflecting black hole, and by studying the deflections of the infinitesimally perturbed solutions with a test-charge black-hole, you can uniquely determine the F tensor of the given solution, by looking at infinitesimally different solution. This argument can surely be turned into a proof, it would require the appropriate differentiation of the metric in the direction corresponding to adding an infinitesimal black hole solution, but for physics purposes this is sufficient to see why the answer must be unique (Edit: unique up to E-B duality rotation, with the corresponding electric-magnetic charge rotation for the test black hole--- this is obvious and mentioned in the other answer, but I didn't notice).

answered Feb 2, 2015 by Ron Maimon (7,535 points) [ revision history ]
edited Feb 2, 2015 by Ron Maimon
Thanks! However, do I understand correctly, that it is only possible to test the RMW theory in a vacuum, since the Einstein tensor is non-zero within a region with non-zero Energy-Momentum tensor in standard General Relativity anyway?
Yes, Rainich theory only makes sense as a classical "theory of everything", since if you have additional stuff, beyond a form field and gravity, the equation for the stress tensor is not constrained. It's just a weird way of writing electromagnetism enabled by the constraint-structure of GR, and it's really interesting (and I didn't know what it was when you asked the question, so thanks)

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