# How to determine the three-velocity measured by a single observer?

In this question the issue of extracting the three-velocity of a test particle in some frame arose. The answer provided by @PaulT and in this paper (arXiv) seems to me to give rise to ambiguity when applied to measurements by a single observer.

Let $u^i$ be observer velocity, $t^i$ test particle velocity, and $T^a$ the observed three-velocity. Thus $a,b,\ldots \in \{1,2,3\}$ and $i,j,\ldots \in \{0,1,2,3\}$, and we take the frame to be a co-moving Lorentz frame so that $u^i = (1,0,0,0)$ and $g_{ij} = \eta_{ij}$ (where $\eta_{ij}$ is taken to be the Cartesian Minkowski metric). The answers provided then seem to me to both claim that $$T^a = \frac{t^a}{t^0},$$ which is all fine and well as long as the observer and test particle are coincident. This is normally the case if we consider a congruence of observers, but suppose we are interested in the measurements by a single observer. The single observer is defined by a single worldline, whence the vector field $u^i$ is only properly defined along that world line. We can define a co-moving frame by extending $u^i$, but then the measurement at some point not along the original world line will depend on the extension. Is the observed three-velocity only well defined when the observer and test particle are coincident?

The to me most intuitive way to solve this ambiguity would be to take $u^i$ at the point of the test particle to be defined by parallel transport of the original observer velocity along the null geodesic connecting the observer world line and the test particle world line. However, I am not sure if there already is a canonical way to deal with this situation nor if my approach would give the desired result.

The easiest way would be if we could extend $u^i$ in such a manner that it is parallel along all null geodesics that intersect the observer world line. I do not believe this can be done in general, since two such null geodesics may intersect at some point not on the worldline, and would produce the same vector at the intersection only if parallel transport along both paths were equivalent.

# Example: stationary observer in Schwarzschild metric

Consider the Schwarzschild solution in Eddington-Finkelstein coordinates: $$ds^2 = A^2(r)dv^2 - 2dvdr - r^2d\Omega,$$ where $d\Omega$ is the standard two-sphere metric, and $A(r) = \left(1 - \frac{r_s}{r}\right)^{1/2}$.

Let the observer be stationary at $r = r_0 > r_s$, with a velocity given by $u^v = A^{-1}(r_0) \equiv A_0^{-1}$ and $u^r = u^\theta = u^\varphi = 0$. As a co-moving Lorentz frame we take \begin{align*} e_0 &= A^{-1}\partial_v, \\ e_1 &= A^{-1}\partial_v + A\partial_r, \\ e_2 &= r^{-1}\partial_\theta, \\ e_3 &= \frac{1}{r\sin\theta}\partial_\varphi, \end{align*} with Ricci rotation coefficients \begin{align*} \gamma^{0}{}_{10} &= \gamma^{1}{}_{00} = A', & \gamma^1{}_{22} &= -\gamma^{2}{}_{12} = \gamma^1{}_{33} = -\gamma^3{}_{13} = -\frac{A}{r}, \\ \gamma^{2}{}_{33} &= -\gamma^{3}{}_{23} = -\frac{\cot\theta}{r}, \end{align*} where prime notation indicates differentiation with respect to $r$.

Suppose the observer releases a test particle at some point $v = v_0$. Since there is no angular motion the test particle will accelerate towards the gravitating mass, and its geodesic motion is given by \begin{align*} \dot{t^0} + A't^0t^1 &= 0,\\ \dot{t^1} + A'\left(t^0\right)^2 &= 0, \end{align*} where overdot notation indicates differentiation with respect to proper time of the test particle. By normalization we know that $\left(t^0\right)^2 = 1 + \left(t^1\right)^2$ and we also have $A' = \frac{\dot{A}}{At^1}$ whence \begin{align*} \frac{\dot{t^0}}{t^0} = -\frac{\dot{A}}{A} \end{align*} or \begin{align*} t^0 &= \frac{A_0}{A}, \\ t^1 &= -\sqrt{\left(\frac{A_0}{A}\right)^2 - 1}, \end{align*} where the integration constant was solved for by the initial conditions. Thus \begin{align*} T^1 = -\sqrt{1 - \left(\frac{A}{A_0}\right)^2}. \end{align*}

An outward going radial null geodesic is given by $n^i = (N(r),N(r),0,0)$, where \begin{align*} 0 &= N_{|1} + \gamma^0{}_{10}N \equiv N_{|1} + \gamma^{1}{}_{00}N \\ &= AN' + A'N \\ &= (AN)', \end{align*} so we may take $N = A^{-1}$ and we observe that the parallel transported vector is given by $\widehat{u}^i = (U(r), \sqrt{U^2 - 1},0,0)$ where \begin{align*} \frac{U'}{\sqrt{U^2 - 1}} = -\frac{A'}{A}, \end{align*} or \begin{align*} U &= \frac{A_0^2 + A^2}{2AA_0}, \\ \sqrt{U^2 - 1} &= \frac{A_0^2 - A^2}{2AA_0}. \end{align*} A co-moving Lorentz frame (along the geodesic) would thus be given by say \begin{align*} \widehat{e}_0 &= Ue_0 + \sqrt{U^2 - 1}e_1, \\ \widehat{e}_1 &= \sqrt{U^2 - 1}e_0 + Ue_1, \\ \widehat{e}_2 &= e_2, \\ \widehat{e}_3 &= e_3. \end{align*} In this frame \begin{align*} \widehat{t}^0 &= Ut^0 - \sqrt{U^2 - 1}t^1, \\ \widehat{t}^1 &= -\sqrt{U^2 - 1}t^0 + Ut^1, \end{align*} whence we find \begin{align*} \widehat{T}^1 &= \frac{A^2\left(A_0 - \sqrt{A_0^2 - A^2}\right) - A_0^2\left(A_0 + \sqrt{A_0^2 - A^2}\right)}{A^2\left(A_0 - \sqrt{A_0^2 - A^2}\right) + A_0^2\left(A_0 + \sqrt{A_0^2 - A^2}\right)} \end{align*}

In both cases the speed approaches the speed of light as the test particle approaches the event horizon, but until that point the two observed three-velocities do not match, as expected. I see no reason why the result $T^1$ should be correct for a single observer (i.e. if the observer and test particle are not coincident), but is the result $\widehat{T}^1$ any more correct?

## Statement of Questions

To clarify, these are the questions I have presented above:

1. Is the observed 3-velocity only well defined when the observer and test particle are coincident?

2. Is there otherwise a canonical way to deal with the ambiguity of the chosen reference frame when only a single observer is considered?

3. Does the method of parallel transport along null geodesics give a "correct" answer?