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Is it difficult to show that a Fermi-Walker "rotation" happens only in the plane formed by a particle four-acceleration and four-velocity?
Since it follows directly from the definition of Fermi-Walker transport, I would say no.Is it difficult to show that a Fermi-Walker "rotation" happens only in the plane formed by a particle four-acceleration and four-velocity?
I can't see directly from the definition. So I am trying to prove/show it. I guess we need to show that a vector in the plane formed by the four-acceleration and four-velocity, when rotated, still lies in the same plane. Correct?Since it follows directly from the definition of Fermi-Walker transport, I would say no.
The definition is ##D_F X = 0##, where ##D_F## is the Fermi derivative and ##X## is a vector field. (I'm leaving out indexes since just looking schematically at the definition is enough.) The Fermi derivative along a worldline with tangent vector ##U## and proper acceleration ##A = \nabla_U U## isI can't see directly from the definition.
No. You can Fermi-Walker transport any vector you like, even if it doesn't lie in the plane spanned by ##U## and ##A##. But the covariant derivative of that vector, if it's Fermi-Walker transported, will lie in the plane spanned by ##U## and ##A##; i.e., that's the plane it will "rotate" in.I guess we need to show that a vector in the plane formed by the four-acceleration and four-velocity, when rotated, still lies in the same plane. Correct?
I find it hard to show that an infinitesimal Lorentz boost in the ##u-a## plane gives as a result your equation 1.8.5.You can find a derivation of Fermi-Walker transport in terms of old-fashioned Ricci calculus here:
https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf