How Does Fermi Transport Describe a Gyroscope's Motion on a Rotating Disk?

[kontejnjer]

Homework Statement

Describe the motion of a gyroscope with center of mass fixed on a rotating disk in coordinates of an observer which is at rest on the disk in the absence of gravity.

Homework Equations

The hint given was to somehow use Fermi transport, so I'm guessing:
$\mathbb F_u X=\nabla_u X-\left<X,\nabla_u u\right>u+\left<X,u\right>\nabla_u u$

The Attempt at a Solution

I honestly have no clue. I suppose since $u$ is timelike and $X=S$ should be spacelike, the last term on the right side goes away. Can we also assume we Fermi transport $S$ along a curve (is this the curve the gyroscope traces out? My notes just say a "timelike curve" so it's not very clear) $\gamma(s)=\left(t(s),\cos\varphi(s),\sin\varphi(s)\right)$ with $\dot\gamma=u$, so $\mathbb F_u S=0$? I find this whole Fermi transport thing very confusing, so any tips or hints would be useful, I've been staring at this problem for days now and sadly made no significant progress.

 

[mark726]

Thank you for your post. The motion of a gyroscope with its center of mass fixed on a rotating disk can be described in the coordinates of an observer at rest on the disk in the absence of gravity using Fermi transport.

Fermi transport is a mathematical tool that allows us to transport vectors and tensors along a curve in a curved space. In this case, we can use Fermi transport to transport the gyroscope's orientation vector, which is perpendicular to the plane of the disk, along the curve traced out by the gyroscope.

To do this, we start by defining a timelike curve \gamma(s)=\left(t(s),\cos\varphi(s),\sin\varphi(s)\right) with \dot\gamma=u, where u is the observer's four-velocity and s is the proper time along the curve. We can then use the Fermi transport equation \mathbb F_u X=\nabla_u X-\left<X,\nabla_u u\right>u+\left<X,u\right>\nabla_u u to transport the orientation vector S along the curve.

Since the observer is at rest on the disk, their four-velocity u is tangent to the disk's surface. Therefore, the second term on the right-hand side of the Fermi transport equation goes to zero. Additionally, since the observer is at rest, their four-velocity is timelike and perpendicular to the disk's surface, so the last term on the right-hand side also goes to zero. This leaves us with \mathbb F_u S=\nabla_u S.

We can then use the definition of the covariant derivative to write this as \mathbb F_u S=\frac{dS}{ds}+\Gamma^i_{\mu\nu}u^\mu S^\nu, where \Gamma^i_{\mu\nu} are the Christoffel symbols. Since we are in a flat space with no gravity, the Christoffel symbols all go to zero, leaving us with \mathbb F_u S=\frac{dS}{ds}. This tells us that the gyroscope's orientation vector is being transported along the curve traced out by the gyroscope without any rotation or change in direction.

In summary, the motion of the gyroscope can be described as its orientation vector being Fermi transported along the curve traced out by the gyroscope, which is given by \