As shown in the figure, there are two velocities on the sphere, one is the velocity along the meridian direction, the arc length is its size, the other is the velocity along the equatorial direction, and the arc length is its size. How to sum on a sphere? Can we use the parallelogram rule on the sphere?

What I'm talking about is the velocity on a two-dimensional sphere.

there are two velocities on the sphere

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    $\begingroup$ What is a two dimensional sphere? Or did you mean two dimensional surface of a sphere. $\endgroup$ – Bob D Apr 20 at 10:40
  • $\begingroup$ @BobD I mean, the sphere is two-dimensional, the velocity on this two-dimensional surface. $\endgroup$ – enbin zheng Apr 20 at 11:00
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    $\begingroup$ @BobD: It's standard to talk about "the 2-sphere" to mean the two-dimensional surface of a sphere. $\endgroup$ – Ben Crowell Apr 20 at 19:41
  • $\begingroup$ @Bob D : FWIW, in mathematics a 2-sphere $S^2=\partial B^3$ is the boundary of a 3-ball $B^3$. $\endgroup$ – Qmechanic Apr 21 at 18:27

You need to learn about the concept of tangent vectors and the tangent plane. If these two vectors are located at the same point, then we think of them as living in the plane tangent to the sphere at this point, and you add them as you would with any vectors in a plane.

If they're at different points on the sphere, then they live in different tangent spaces, and they can't be added. In order to add them, you would need to prepare them by moving one to the same place as the other, but this kind of parallel transport is path-dependent, so there's no unambiguous way to define how to do this.

  • $\begingroup$ The tangent plane you are talking about is not on the sphere. It has only one point on the sphere. $\endgroup$ – enbin zheng Apr 21 at 0:07
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    $\begingroup$ @enbinzheng every point on the sphere has a tangent plane. All the tangent planes are $\mathbb{R}^2$. $\endgroup$ – tfb Apr 21 at 18:25

For your problem, you have to calculate the geodetic equations for a sphere.

The sphere position vector with sphere radius equal one is:

$$\vec{R}= \left[ \begin {array}{c} \sin \left( \theta \right) \cos \left( \varphi \right) \\ \sin \left( \theta \right) \sin \left( \varphi \right) \\ \cos \left( \theta \right) \end {array} \right] \tag 1$$

from equation (1) you get the geodetic equations for $\theta(s)$ and $\varphi(s)$ where $s$ affine path parameter

$${\frac {d^{2}}{d{s}^{2}}}\theta \left( s \right) -\cos \left( \theta \left( s \right) \right) \sin \left( \theta \left( s \right) \right) \left( {\frac {d}{ds}}\varphi \left( s \right) \right) ^{2 }=0 \tag 2$$


$${\frac {d^{2}}{d{s}^{2}}}\varphi \left( s \right) +2\,{\frac {\cos \left( \theta \left( s \right) \right) \left( {\frac {d}{ds}} \varphi \left( s \right) \right) {\frac {d}{ds}}\theta \left( s \right) }{\sin \left( \theta \left( s \right) \right) }}=0 \tag 3$$

for a given initial condition,you can solve numerically equation (2) and (3) and get $\theta(s)\,,\varphi(s)$ and with equation (1) the path on the sphere.

enter image description here

the red line is the solution for the initial condition velocity $D(\theta)(0)=V/R\,,D(\varphi)(0)=0$, the blue line is the solution for the initial condition velocity $D(\theta)(0)=0\,,D(\varphi)(0)=U/R$.

the solution for both the initial conditions $D(\theta)(0)=V/R$ and $D(\varphi)(0)=U/R$ is the green line, which is a great circle on the sphere

  • $\begingroup$ The OP isn't asking how to combine displacements, they're asking how to combine tangent vectors. $\endgroup$ – Ben Crowell Apr 20 at 19:38
  • $\begingroup$ Can we use the parallelogram rule on the sphere? $\endgroup$ – enbin zheng Apr 21 at 11:59
  • $\begingroup$ No you can’t , you can use parallelogram rule only on a flat space or local on the tangent space $\endgroup$ – Eli Apr 21 at 13:01
  • $\begingroup$ @Eli Why can't we use the parallelogram rule? What will go wrong? $\endgroup$ – enbin zheng Apr 21 at 21:17
  • $\begingroup$ @BenCrowell Do you think his answer is wrong? $\endgroup$ – enbin zheng Apr 21 at 21:25

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