Parallel transport vs Fermi Transport

[davidge]

Since for a general contravariant vector, $\nabla_{\nu}V^{\mu}$ will not in general be zero, is it correct to say that all of them are transported by Fermi Transport? (With the only vector being parallel transported being the four velocity vector?)

 

[pervect]

Since for a general contravariant vector, $\nabla_{\nu}V^{\mu}$ will not in general be zero, is it correct to say that all of them are transported by Fermi Transport? (With the only vector being parallel transported being the four velocity vector?)

It's correct to say that parallel transport and Fermi-Walker transport both transport vectors along a curve. So both transport laws create a map between a vector defined at some point P to a vector at some other point Q, where P and Q are connected by some curve. Here the vector at point P is defined in the tangent space at point P, and the vector at point Q is defined in the tangent space at point Q.

It is also correct to say that they are in general different transport laws. When you say "all of them are transported by Fermi Transport", you apparently have some specific scenario in mind, but we don't know what it is. So we can't confirm or deny that in that in whatever specific application you're thinking of that Fermi-Walker transport is the correct transport law unless we understand more of what you're trying to do.

 

[davidge]

When I say all of them I'm really thinking of any kind of contravariant four vector that one can define. For instance, it could be the spin vector or even the four velocity vector.

 

[PeterDonis]

Since for a general contravariant vector, $\nabla_{\nu}V^{\mu}$ will not in general be zero, is it correct to say that all of them are transported by Fermi Transport? (With the only vector being parallel transported being the four velocity vector?)

You appear to be mixing up several things here.

First, when you talk about "transport" (parallel transport or Fermi-Walker transport), you are not talking about $\nabla_\nu V^\mu$ for some vector $V^\mu$. You are talking about a more specific derivative operator that involves $\nabla_\nu$: in the simplest case (see below), this operator is $u^\nu \nabla_\nu V^\mu$, i.e., the change in the vector $V^\mu$ along a specific curve whose tangent vector (I won't use the term "4-velocity" since there is nothing that requires the tangent vector to be timelike) is $u^\nu$.

The term "parallel transport" refers to the specific case where we have $u^\nu \nabla_\nu V^\mu = 0$. In the case where $V^\mu = u^\mu$, i.e., we are talking about the tangent vector to the curve itself, then parallel transport of the tangent vector means the curve is a geodesic. Note that this means that if the curve is not a geodesic, then the tangent vector is not parallel transported--contrary to the apparent assumption you made in your parenthetical comment in the quote above.

The term "Fermi-Walker transport" refers to the case where we have a frame, i.e., a set of four orthonormal vectors, one timelike and three spacelike, and we want to transport them all along the same curve, with the condition that the timelike vector of the frame is the tangent vector (which here is a 4-velocity because the curve must be timelike) to the curve. (The term "frame" is used because this is the best way mathematically to realize the intuitive concept of a "frame of reference".) The basic idea is that we want the only "change" in the vectors along the curve to be whatever change is induced by the path curvature of the curve itself, plus the requirement to keep all of the vectors orthonormal, with one timelike and three spacelike; i.e., we want the "minimum" change possible consistent with the vectors still forming a frame.

If the curve in question is a geodesic, then Fermi-Walker transport just reduces to parallel transport: all of the frame vectors satisfy the same condition that the tangent vector does, i.e., $u^\nu \nabla_\nu V^\mu = 0$ for all four frame vectors (of which the timelike one is just $u^\mu$ itself).

If the curve in question is not a geodesic, then the Fermi-Walker transport law becomes more complicated; it is

$$
u^\nu \nabla_\nu V^\mu - \left( V^\rho a_\rho \right) u^\mu + \left( V^\sigma u_\sigma \right) a^\mu = 0
$$

where $a^\mu$ is a shorthand for the proper acceleration vector $u^\nu \nabla_\nu u^\mu$ (which, remember, is nonzero because the curve is not a geodesic--in fact, you can see that if $a^\mu = 0$, the above law just reduces to the ordinary parallel transport condition). The factors in the parentheses are dot products, e.g., $V^\rho a_\rho = g_{\rho \sigma} V^\rho a^\sigma$.

 

[davidge]

Oh, thank you Peter. Now I understand it.

 

[PeterDonis]

Oh, thank you Peter. Now I understand it.

Good, glad I could help.

If you want an extra credit exercise, try this: write down the Fermi-Walker transport condition I gave for the case $V^\mu = u^\mu$, i.e., the Fermi-Walker transport law for the tangent vector itself, for the case where the curve is not a geodesic. Can you show that this condition reduces to the identity $u^\nu \nabla_\nu u^\mu = a^\mu$?