Relativistic spacetime

It is a 4-dimensional topological manifold with a smooth atlas, $(\mathcal{M},\mathcal{O},\mathcal{A})$, carrying a torsion-free connection, $\mathrm{\nabla }$, which we require to be compatible with a Lorentzian metric $g$ (so in particular it is a Lorentzian manifold), and a so-called time orientation, $T$. So we need the sextuple $(\mathcal{M},\mathcal{O},\mathcal{A},\mathrm{\nabla },g,T)$.

Time Orientation

The absolute time function in Newtonian spacetime associates to each $p\in \mathcal{M}$ a time. That is, given any point you can just quote the time of that point unarguably. We used the absolute time function to define a future-directed vector field $X$ as $dt:X>0$. Pictorially, this is given by an arrow pointing to the "upper side" of a tangent plane to a constant $t$ surface.

\begin{document}
    \begin{tikzpicture}
        \draw[ultra thick] (0,0) .. controls (3,1) and (5,-1) .. (7,0.8);
        \draw[ultra thick] (0,1.5) .. controls (3,2.5) and (5,0.5) .. (7,2.3);
        \node at (7.5,0.8) {\Large{$t_1$}};
        \node at (7.5,2.3) {\Large{$t_2$}};
        \draw[ultra thick, blue, rotate around={10:(1,0.23)}] (-0.5,0.23) -- (2.5,0.23);
        \draw[->, ultra thick, red] (1,0.23) -- (1.5, 1.5);
        \node[circle, fill, inner sep=1.5pt, label={below:\Large{$p$}}] at (1,0.23) {};
        \node at (0.8,1) {\color{red}\Large{$X$}};
        \node at (-0.8, 0.2) {\color{blue}\Large{$dt$}};
    \end{tikzpicture}
\end{document}

We don't have an absolute time function for our relativistic spacetime $(\mathcal{M},\mathcal{O},\mathcal{A},\mathrm{\nabla },g)$, since the metric $g$ can only distinguish

• timelike vectors
• spacelike vectors
• lightlike vectors

\begin{document}
    \begin{tikzpicture}
  % Draw light cone
\draw[thick, dashed] (0,0) -- (3,3);  % Future light cone
\draw[thick, dashed] (0,0) -- (-3,3);
\draw[thick, dashed] (0,0) -- (3,-3); % Past light cone
\draw[thick, dashed] (0,0) -- (-3,-3);

% Draw the time axis
\draw[->] (0,-3.5) -- (0,3.5) node[anchor=south] {$t$};

% Draw the space axis
\draw[->] (-3.5,0) -- (3.5,0) node[anchor=west] {$x$};

% Timelike vector (within the light cone)
\draw[->, thick, blue] (0,0) -- (1,2) node[midway, above, sloped] {Timelike vector};

% Spacelike vector (outside the light cone)
\draw[->, thick, red] (0,0) -- (2,1) node[midway, above, sloped] {Spacelike vector};

% Lightlike vector (on the light cone)
\draw[->, thick, green] (0,0) -- (1.5,1.5) node[midway, above, sloped] {Lightlike vector};

% Labels for regions
\node at (1.5,2.5) {Future Light Cone};
\node at (1.5,-2.5) {Past Light Cone};
\node at (2.5,0.5) {Spacelike Region};  
    \end{tikzpicture}
\end{document}

So we need some other way to define what a future-directed vector field is. First, we need to restrict to Lorentzian manifolds which are oriented, $(\mathcal{M},\mathcal{O},{\mathcal{A}}^{\uparrow },g)$.
Definition (Time Orientation).
Let $(\mathcal{M},\mathcal{O},{\mathcal{A}}^{\uparrow },g)$ be an oriented Lorentzian manifold. Then a time orientation is given by a smooth vector field $T$ that

1. does not vanish anywhere, and
2. $g(T,T)>0$. For the signature $(-,+,+,+)$, the condition would be $g(T,T)<0$.}
   $◼$

It is the combination of the metric and the time orientation that allows us to define future/past/spatial directed vector fields in relativistic spacetime. The metric structure gives us a double cone structure in the tangent plane to each $p\in \mathcal{M}$. We want to identify one of these cones as the future and the other as the past. We know that a vector $X$ that satisfies $g(X,X){|}_p>0$ lies within either one of the two cones tangent to $p$. However, it doesn't tell us which cone it lies in, and so we don't know if it's future-directed or past-directed. We therefore need some method to select which cone is which. This is exactly what the time orientation does. Condition (i) tells us that it is defined everywhere, allowing us to define the future cone at each point, and condition (ii) tells us that $T$ lies within the cone (an obvious necessity). We then simply say, "whichever cone $T$ lies in, that is the future cone." The final, but very important, property is that $T$ is a smooth vector field. This means that the future cones at separate points are smoothly connected. That is, the selected cone doesn't suddenly "flip" as you move from point to point.

\begin{document}
    \begin{tikzpicture}[scale=1.3]
        \draw[draw=red, opacity=0.2, fill=red, fill opacity=0.2] (0,2) -- (-0.2,3.4) .. controls (2,4) and (3,1) .. (5,1.5) -- (4,0) .. controls (3,-0.5) and (2,3) .. (0,2);
        \node at (-0.5,3.5) {\color{red}\Large{$T_p$}};
        \node at (5.2,1.6) {\color{red}\Large{$T_q$}};
        \node at (2.2,2.1) {\color{red}\Large{$T$}};
        \draw [thick](-1,3) arc (180:0:1cm and 0.15cm);
        \draw[->,ultra thick,red] (0,2) -- (-0.2,3.4);
        \draw [thick](-1,3) arc (180:360:1cm and 0.15cm) -- (0,2) -- cycle;
        \draw [dashed,thick](-1,1) arc (180:360:1cm and 0.15cm) -- (0,2) -- cycle;
        \draw [dashed,thick](-1,1) arc (180:0:1cm and 0.15cm);
        \draw [thick,rotate around={-30:(4,0)}](3.5,1.5) arc (180:0:0.5cm and 0.1cm);
        \draw[->,ultra thick,red] (4,0) -- (5,1.5);
        \draw [thick,rotate around={-30:(4,0)}](3.5,1.5) arc (180:360:0.5cm and 0.1cm) -- (4,0) -- cycle;
        \draw [dashed,thick,rotate around={-30:(4,0)}](3.5,-1.5) arc (180:360:0.5cm and 0.1cm) -- (4,0) -- cycle;
        \draw [dashed,thick,rotate around={-30:(4,0)}](3.5,-1.5) arc (180:0:0.5cm and 0.1cm);
        \node[circle, fill, inner sep=1.5pt, label={left:\Large{$p \,\,$}}] at (0,2) {};
        \node[circle, fill, inner sep=1.5pt, label={right:\Large{$q$}}] at (4,0) {};
    \end{tikzpicture}
\end{document}

Note: We shall now simply refer to relativistic spacetime as just spacetime.

Stationary spacetime

Definition (Stationary Spacetime). A spacetime $(M,O,A,g,T)$ is called stationary if it admits a Killing vector field $K$ such that $g(K,K)<0$.

Claim A stationary spacetime is one where we can find a chart such that the components of the metric do not depend on time.
Proof. Recall a vector field is Killing if ${\mathcal{L}}_{K}g=0$. The exercise at the end of lecture 11 shows that in a chart this condition reads

Now imagine we pick a chart such that $T={\delta }_0^a{\partial }_a={\partial }_0$, then the second two terms vanish and we are simply left with

which is the statement that the metric components are time-independent in this chart. □