I Fermi-Walker transport and rotation

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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?
 
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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.
 
799
36
Since it follows directly from the definition of Fermi-Walker transport, I would say no.
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?
 
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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##.
 
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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 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.
 

vanhees71

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

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