On 2/13/2014 11:02 AM, Jesse Mazer wrote:
Even though the curvature disappears in the first order terms, it remains in the higher order terms, whereas curvature is really zero in all terms for an accelerating observer in flat spacetime. So, the answer to your question is that acceleration does not in itself cause spacetime curvature, SR can handle acceleration just fine as discussed at http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html , but this isn't a violation of the equivalence principle since the mathematical formulation of the principle deals only with first-order terms.
As an example SR is used to calculate the size of proton bunches in the LHC even though they are at 9999% of the speed of light and subject to enormous acceleration at the turning magnets.
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