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kent davidge

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- #1

kent davidge

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- #2

PeterDonis

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Since it follows directly from the definition of Fermi-Walker transport, I would say no.

- #3

kent davidge

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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.

- #4

PeterDonis

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I can't see directly from the definition.

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## is

$$

D_F X = \nabla_U X - \left( X \cdot A \right) U + \left( X \cdot U \right) A

$$

If we set ##D_F X = 0##, we get

$$

\nabla_U X = \left( X \cdot A \right) U - \left( X \cdot U \right) A

$$

which just says that the covariant derivative of ##X## along ##U##, which is what you are calling "Fermi-Walker rotation", lies in the plane spanned by ##U## and ##A##.

- #5

PeterDonis

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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?

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

- #6

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https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

- #7

kent davidge

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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.

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

- #8

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This is discussed in the previous section (following Eq. (1.7.7)).

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