- #1

- 970

- 3

The equations don't seem to say it is.

The vector field would be [itex]\vec{V}=V(\theta)e_\theta [/itex]. The component of the vector only depends on theta, but at constant latitude theta is constant so the component of V doesn't change as you circle the earth. [tex]e_\theta=(cos \theta cos \phi, cos \theta sin \phi, -sin \theta) [/tex]

[tex]\partial_\phi e_\theta=cos \theta(-sin \phi, cos \phi, 0) [/tex]

So [itex]D\vec{V}=V(\theta)cos(\theta) (-sin \phi, cos \phi, 0) d\phi[/itex]. This is only zero at the equator. So is the vector field [itex]\vec{V}=V(\theta)e_\theta [/itex] not parallel transported at constant latitude?

I understand that you can integrate the above equation to get:

[itex]\vec{V}(\phi_2,\theta)-\vec{V}(\phi_1,\theta)=V(\theta)cos(\theta) [(cos \phi_2, sin \phi_2, cos \theta)-(cos \phi_1, sin \phi_1, cos \theta)][/itex].

Geometrically [itex]V(\theta)cos \theta[/itex] is the projection of your arm perpendicular to the z-axis (or parallel to the z=0 plane), and the difference [itex]\vec{V}(\phi_2,\theta)-\vec{V}(\phi_1,\theta)[/itex] can be attributed to the fact that the vector field rotates around in a circle about the z-axis.

But wasn't parallel transport [itex]D\vec{V}=0 [/itex]