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From wiki

To quantify geodesic deviation, one begins by setting up a family of closely spaced geodesics indexed by a continuous variable s and parametrized by an affine parameter $\tau$. That is, for each fixed $s$, the curve swept out by $\gamma _s(\tau)$ as τ varies is a geodesic. When considering the geodesic of a massive object, it is often convenient to choose $\tau$ to be the object's proper time. If $x^μ(s, τ)$ are the coordinates of the geodesic $\gamma_s(\tau)$, then the tangent vector of this geodesic is:

$$ T^\mu = \frac{\partial x^\mu(s,\tau)}{\partial \tau} $$

If $\tau$ is the proper time, then $T_\mu$ is the four-velocity of the object traveling along the geodesic. One can also define a deviation vector, which is the displacement of two objects traveling along two infinitesimally separated geodesics:

$$ X^\mu = \frac{\partial x^\mu(s,\tau)}{\partial s} $$

Now, naively

$$ v^\mu = T^\beta \nabla_\beta X^\mu$$

Should $v^\mu$ be the relative velocity? If so, how does that fit in with this definition of relative velocity? (when their geodesics intersect)

$$ v:=-\frac{1}{g(u^′,u)} u^′ - u \tag{1} $$

I feel I've managed to confuse myself.

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Indeed, you seem to have confused yourself :-)

I edited several of your formulas. Check them out.

The equation

$$ v^\mu = T^\beta \nabla_\beta X^\mu$$

measures the rate of separation between 2 neighboring geodesics, so it measures their relative 4-velocity. The quoted question and the paper mentioned therein take the matter to a deeper level, however... you have to realize that the "relative ordinary velocity" of an object passing close to you, regardless if it is moving on a geodesic or not, is simply the spacelike component of its 4-velocity (corrected by a $\gamma$ factor). This assumes that you are using a LIC (locally inertial coordinates system). You are gently guided to this formula in MTW Gravitation Ex. 2.5 page 65 and it coincides with your eq. 1.

So eq. 1 is simply the gamma corrected projection of the equation $$ v^\mu = T^\beta \nabla_\beta X^\mu$$

You can better see their relation, by observing that - since $X$ and $T$ vectors commute (they form a 2D coordinate system) - you have that $$ v^\mu = X^\beta \nabla_\beta T^\mu$$ which represent the difference in 4-velocity on neighboring geodesics, similar to eq. 1. You can read about this in MTW Box 11.4, for example.

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    $\begingroup$ Unlike in equation (1) where u and u' are specified, the quantity $v^{\mu}$ here represents relative velocity b/w which two neighboring tangent vectors? It seems that $v^{\mu}$ only provides the rate of change and not exactly the difference $\endgroup$
    – paul230_x
    Commented Oct 3, 2022 at 11:36
  • $\begingroup$ Indeed eq 1 shows the difference between 2 generic world lines, while $v^\mu$ shows the limit of (the difference between infinitesimally close world lines divided by $\Delta s$), so it is a ratio, but the idea is the same. Anyway, when I have more time I will improve the answer $\endgroup$
    – magma
    Commented Oct 3, 2022 at 12:23

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