# I Fermi-Walker transport and rotation

#### kent davidge

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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#### PeterDonis

Mentor
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.

• vanhees71

#### kent davidge

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?

#### PeterDonis

Mentor
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 $A$.

• kent davidge and vanhees71

#### PeterDonis

Mentor
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.

• kent davidge and vanhees71

#### vanhees71

Gold Member
• kent davidge