Are Christoffel symbols measurable?
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Feb16-12, 01:04 PM
I tend to agree with ApplePion. The usual comparison of gauge fields in electromagnetism and gravity is the electromagnetic potential A and the spacetime metric g.
If we do that, is the E field like the Christoffel symbol Г? The Г field can be considered a gravitational field in the sense of the equivalence principle in which acceleration = gravity, because both gravity and acceleration lead to non-zero Г (as Crowell himself has pointed out many times).
However, it's also bit different. If the E field is zero in an inertial frame, then we do say there is no electric force. OTOH, if Г is zero in a local inertial frame, there could still be tidal gravity, due to the derivatives of Г being non-zero.
The fact that local inertial frames in which Г is zero always exist is an implementation of the equivalence principle that says that gravity can always be cancelled away. The fact that although Г ("gravitational field") is zero, its derivatives ("tidal gravity") can still be non-zero shows that the "local" qualification is very important in the equivalence principle - a nonlocal experiment that looks at the derivatives of Г can still detect non-zero tidal gravity which cannot be cancelled away.
Also, usage differs. For example, it's also often said that the metric field is the "gravitational field". These are just terminology differences. Also, there are two meanings of "gauge field". Gravity is a "gauge field" only in one of the two sense of the word.
Also, the Aharonov-Bohm effect is a quantum effect, so the classical electromagnetic potential really only has effects when E and B are non-zero - there is no classical Aharonov-Bohm effect.
So what is the counterpart of Ricci curvature (or R^c (ab)) in electromagnetism if it is not E and B which is already taken up by the Christoffel symbols (and how can Cromwell be wrong in the table in page 173 so maybe I can tell him).