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So I was pondering about geodesic deviations and I'm confused about the following. Let's say I have $3$ geodesics $\gamma_1(t)$ , $\gamma_2(t)$ and $\gamma_3(t)$. I introduce a parameter $s$ such that $\gamma_{ij}(t,s)$ obeys:

$$ \gamma_{12}(t,0) = \gamma_1$$ $$ \gamma_{12}(t,1) = \gamma_2$$ $$ \gamma_{23}(t,0) = \gamma_2$$ $$ \gamma_{23}(t,1) = \gamma_3$$ $$ \gamma_{31}(t,0) = \gamma_3$$ $$ \gamma_{31}(t,1) = \gamma_1$$

Now, I know in the geodesic deviation $\zeta^\mu_{ij}$ between $\gamma_i$ and $\gamma_j$. In a flat spacetime intuitively I imagine the relation locally:

$$ \zeta^\mu_{12} + \zeta^\mu_{23} + \zeta^\mu_{31} = 0 $$

Similarly the relative velocities obey when summed around in a triangle obey:

How does this get modified in curved spacetime?

In other words, For each pair of $\gamma_i,\gamma_j$ I pick a congruence they're part of and get a deviation vector field $\zeta^\mu_{ij}$ on the inside of the triangle. Now I ask asking what happens when you sum these three vector fields in curved space time?

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    $\begingroup$ You may as well simplify your equations, e.g. the first becomes $\gamma_{12}(t,\,0)=\gamma_1$. $\endgroup$
    – J.G.
    Commented Oct 3, 2022 at 14:51
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    $\begingroup$ there is a bunch of mixed notation above and I cannot really read this. However, for this problem, my suggestion would be "do this on a globe with your three vertices being "the intersection of the prime meridian and the equator, (90 degrees west, 0 N), and the north pole", comparing the embedding in 3-d space to the geodesics along the globe. $\endgroup$
    – Zo the Relativist
    Commented Oct 3, 2022 at 15:39
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    $\begingroup$ or alternately, your comparison point could be "the triangle in "Euclidean lat-long space"", though that is a bit artificial for such a concrete problem. $\endgroup$
    – Zo the Relativist
    Commented Oct 3, 2022 at 15:41

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