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## Main Question or Discussion Point

If you walk at constant latitude with your arm always sticking towards the North pole, is that parallel transport of your arm?

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]

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]