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\title{\href{https://www.physicsforums.com/threads/velocity-measurement-by-a-stationary-observer-in-gr.929633/}{How do we measure the velocity in curved space-time?}}
\author{Sergiy M. Ponomatenko}
%https://physics.stackexchange.com/questions/363452/how-do-we-measure-the-velocity-in-curved-space-time                                 
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\begin{document}
\maketitle
Let's consider a general metric:
\begin{equation}
	 ds^2 = g_{00}(dx^0)^2  + 2g_{0i}dx^0dx^i + g_{ij}dx^idx^j,
\end{equation}
where $i,j = 1 \ldots 3$ are spatial indices.


Tangential vector (4-velocity) to world-line: 
\begin{equation}
	v^{\nu} = \frac{dx^0}{ds}\vec{e}_0 + \frac{dx^i}{ds}\vec{e}_i
\end{equation} 
where $s$  is the affine parameter of an particle world-line (usually, proper time of particle).

Value of 3-velocity of a particle defined as 
\begin{equation}\label{3-velocity}
	v = \frac{d\ell}{d\tau} = \tanh\beta
\end{equation}
where $d\ell$~--- physical distance between two events, and $d\tau$~--- proper time of an observer, which is measures the velocity.

This definition follows from the\emph{ equivalence principle}.

Physical distance determine as:
\begin{equation}\label{distance}
	d\ell^2 = \left(-g_{ik} + \frac{g_{0i}g_{0j}}{g_{00}}\right)dx^idx^j.
\end{equation}

Let's consider 4-velocities of an observer $u^{\mu}$ and particle $v^{\nu}$:
\begin{align}
	u^{\mu} &=  \left\lbrace \frac{1}{\sqrt{g_{00}}},0,0,0\right\rbrace \\
	v^{\nu} &=  \left\lbrace \frac{dx^0}{ds},\frac{dx^i}{ds}\right\rbrace 
\end{align} 

  
Projection of $v^{\nu}$ by direction of $u^{\mu}$ is (look~\ref{std}):   
\begin{equation}
	\mathbf{g}(u^{\mu},v^{\nu}) = g_{00}u^0v^0 +   g_{0i}u^0v^i  = \sqrt{g_{00}}v^0 + \frac{g_{0i}}{\sqrt{g_{00}}} v^i = \sqrt{g_{00}}\left(v^0 + \frac{g_{0i}}{g_{00}} v^i \right)
\end{equation} 

From the another point of view, the \[\mathbf{g}(u^{\mu},v^{\nu}) =\cosh\beta= \frac{d\tau}{ds}  = \gamma = \frac{1}{\sqrt{1 - v^2}},  \] thus, the amount of observer's proper time elapsed from meeting particle at event $(x_0,x^i)$ to appearing it at distance $d\ell$, as measured by observer
\begin{equation}
	d\tau = \mathbf{g}(u^{\mu},v^{\nu}) ds = \sqrt{g_{00}}\left(dx^0 + \frac{g_{0i}}{g_{00}} dx^i \right). 
\end{equation} 

%---------------------------------------------------------
\begin{figure}[h!]\centering
    \begin{tikzpicture}[every node/.style={font=\normalsize},scale = 1.5]
	\pgfmathsetmacro{\angle}{80}
	\pgfmathsetmacro{\xmax}{4}
	\pgfmathsetmacro{\ymax}{8}
	\pgfmathsetmacro{\gtx}{cos(\angle)}
	 %---------------------- particle coordinates ----------------------------            
	\pgfmathsetmacro{\xparticle}{3}
	\pgfmathsetmacro{\yparticle}{4}
	\def\particle{(\xparticle,\yparticle)}
    %---------------------- functions definitions ----------------------------	
	\pgfmathsetmacro{\lightforward}{1/(\gtx + sqrt(\gtx^2 + 1))}
	\pgfmathsetmacro{\lightbackward}{1/(\gtx - sqrt(\gtx^2 + 1))}

	\pgfmathdeclarefunction{x}{2}{\pgfmathparse{#1+#2*\gtx}} % Real x coord of canvas
	\pgfmathdeclarefunction{y}{2}{\pgfmathparse{#2*sin(\angle)}} % Real y coord of canvas
	\def\point(#1,#2){({x(#1,#2)},{y(#1,#2)})}
	\pgfmathdeclarefunction{len}{2}{\pgfmathparse{sqrt(#1^2 + #2^2 + 2*#1*#2*\gtx)}}
	\pgfmathdeclarefunction{projection}{2}{\pgfmathparse{#2 + #1*\gtx}}
	
	% --------------------------- axis drawing ------------------------------
    \draw[-latex'] (0,0) coordinate (O) node [below left] {$x^0$} -- \point(0,\ymax) node[above] {$x^0$};
    \draw[-latex'] (O) -- \point(\xmax,0) node[right] {$x^i$};
    \draw[-latex'] (O) -- (-90+\angle:\xmax) node[right] {$l$};
    %---------------------------- grid --------------------------------------
    \foreach \i in {1,...,\xmax} {
        \draw[gray!50, ultra thin] \point(\i,0) -- \point(\i,\ymax);
    }
    
    \foreach \i in {1,...,\ymax} {
        \draw[gray!50, ultra thin] \point(0,\i) -- \point(\xmax,\i) ;
    }
    %----------------------------- ort --------------------------------------
        %\draw[-latex, blue] (0,0) -- (1,0) node[below] {$\vec{e}_i$};
        \draw[-latex, blue] (0,0) -- \point(0,1) node[left] {$u^{\mu}$};

	% ------------------Drawing v^{\nu} vector ------------------------------
    \draw [name path = velocity, blue, thick, -latex'](O) -- node [right] {$v^{\nu}$} \expandafter\point\particle coordinate (PE);
    \draw (\angle:0.55)  arc(\angle:{atan(y\particle/(x\particle))}:0.55) node[pos=0.7, above] {$\beta$};
    % --------------------------- main --------------------------------------

    \foreach \x [count=\i from 0] in {0,0.1,0.2,...,1}{
    \draw [densely dashed, red] \point(\x*\xparticle,\x*\yparticle) coordinate (V\i)  --  \point(0,{projection(\x*\xparticle,\x*\yparticle)}) coordinate (A\i) \ifnum\i=10 node [sloped, pos=0.5, above=-2pt] {\small simultaneity} \fi;
    \fill [red] (V\i) circle (0.05);
   	\ifnum\i=10
   	\draw [gray] \point(\x*\xparticle,\x*\yparticle) -- node [above, sloped] {\tiny light} \point(0,{\x*\yparticle-\x*\lightforward*\xparticle});
   	\draw [gray] \point(\x*\xparticle,\x*\yparticle) -- node [above, sloped] {\tiny light} \point(0,{\x*\yparticle-\x*\lightbackward*\xparticle});
   	\fi	
    }
   	
    %------------------ particle coordinates nodes --------------------------- 	
    \node [left] at \point(0,\yparticle) {$x^0 + dx^0$};
    \node [above] at \point(\xparticle,0) {$x^i + dx^i$};
    \ifnum\angle=90\relax\else\node [left] at (A10) {$\left( x^0 + dx^0\right)  + \frac{g_{0i}}{g_{00}}dx^0$};\fi
    
    %----------------- physical distance axis nodes -------------------------- 
    \foreach \x [count=\i from 0] in {0,0.1,0.2,...,1}{ 
    \coordinate (l\i) at ({-90+\angle}:{sqrt(1-\gtx^2)*\x*\xparticle});
    \draw [red, dashed] (V\i) -- (l\i);
      }
      
    \end{tikzpicture}
\caption{Space-time diagram}
\label{std}
\end{figure} 

\clearpage
\nocite{LL2E}
\printbibliography   
\end{document}
