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Given any Lorentzian manifold containing three distinct time-like world lines $L$, $A$ and $B$ such that
$L$ and $A$ have exactly one common coincidence event, $\mathcal{E}_{AL}$,
$L$ and $B$ have exactly one common coincidence event, $\mathcal{E}_{BL}$, and
$A$ and $B$ have no coincidence event in common at all (therefore $\mathcal{E}_{AL} \ne \mathcal{E}_{BL}$),
is the following conclusion correct?:

There also exist four distinct time-like world lines $F$, $J$, $P$, and $U$ in the given Lorentzian manifold, which have the coincidence event $\mathcal{E}_{AL} \equiv \mathcal{E}_{ALFJPU}$ in common and
there exist another four distinct time-like world lines $G$, $K$, $Q$, and $V$ in the given Lorentzian manifold, such that

(1)
$F$ and $G$ have the coincidence event $\mathcal{E}_{FG}$ in common,
$J$ and $K$ have the coincidence event $\mathcal{E}_{JK}$ in common,
$P$ and $Q$ have the coincidence event $\mathcal{E}_{PQ}$ in common, and
$U$ and $V$ have the coincidence event $\mathcal{E}_{UV}$ in common,

(2)
there is no time-like world line at all having taken part in any two (or more) of the five coincidence events $\mathcal{E}_{FG}$, $\mathcal{E}_{JK}$, $\mathcal{E}_{PQ}$, $\mathcal{E}_{UV}$, or $\mathcal{E}_{BL}$
(therefore the coincidence events $\mathcal{E}_{FG}$, $\mathcal{E}_{JK}$, $\mathcal{E}_{PQ}$, $\mathcal{E}_{UV}$, and $\mathcal{E}_{BL}$ are all distinct; and the nine time-like world lines $F$, $G$, $J$, $K$, $P$, $Q$, $U$, $V$, and $L$ are all distinct from each other),

and such that

(3)
for any coincidence event $\mathcal{E}_{WX}$ in which any time-like world line $W$ took part which also took part in coincidence event $\mathcal{E}_{BL} \equiv \mathcal{E}_{BLW}$ there exists a time-like world line $Y$ which took part in coincidence event $\mathcal{E}_{WX} \equiv \mathcal{E}_{WXY}$ as well as in one of the five coincidence events $\mathcal{E}_{FG}$, $\mathcal{E}_{JK}$, $\mathcal{E}_{PQ}$, $\mathcal{E}_{UV}$, or $\mathcal{E}_{AL}$
?

user12262
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1 Answers1

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If no two points on your manifold $M$ can be joined by both a spacelike curve and a timelike curve, I believe I can construct curves satisfying (1) and (2). However, I suspect that (3) can never also be satisfied. [Edit: I misread (3), and now I think it may be possible to satisfy (3) with a variant of the construction I gave for (1) and (2). See comments for details.] Here's a rough sketch of my argument. I've made the answer community wiki so that anyone can add clarifications or corrections.

As far as I can tell, the curves $A$, $B$, $G$, $K$, $Q$, and $V$ serve only to mark points on the other curves, $L$, $F$, $J$, $P$, and $U$. Every point in $M$ has an infinite number of timelike curves running through it, so any point in $M$ can be marked by a timelike curve distinct from $L, \ldots, U$. Therefore, we don't have to worry about $A, \ldots, V$ at all; we can just keep track of the points they mark, which I'll call

  • $a = \mathcal{E}_{AL}$
  • $b = \mathcal{E}_{BL}$
  • $f = \mathcal{E}_{FG}$
  • $j = \mathcal{E}_{JK}$
  • $p = \mathcal{E}_{PQ}$
  • $u = \mathcal{E}_{UV}$.

Starting with the curve $L$ and its distinct marked points $a$ and $b$, let's construct $F$, $J$, $P$, and $U$. Let $L'$ be a compact segment of $L$ containing $a$ and $b$ (and extending out beyond $a$ and $b$, for good measure). You should be able to construct a spacelike vector field $\xi$ on some open neighborhood of $L'$ which vanishes at $a$ but does not vanish anywhere else on $L'$. Every point on $L'$ has a neighborhood in which the flow $\phi$ generated by $\xi$ is well-defined for more than zero time. Since $L'$ is compact, it follows that $L'$ has a neighborhood in which the flow is well-defined for more than zero time.

Under the flow $\phi$, the curve $L'$ moves smoothly through $M$, and its tangent bundle $TL'$ moves smoothly through $TM$. For every point $x$ on $L'$, the tangent space $T_xL'$ remains timelike for more than zero time during the flow. Thus, using comapctness, $L'$ remains timelike for more than zero time during the flow. Pick four times $t_F$, $t_J$, $t_P$, and $t_U$ at which $L'$ remains timelike, and call the images of $L'$ at these times $F$, $J$, $P$, and $U$.

Since $\xi$ vanishes at $a$, the curves $F, \ldots, U$ all pass through $a$. By tweaking $t_F, \ldots, t_U$, you should be able to ensure that $F, \ldots, U$ intersect $L$ only at $a$.

Let $f$ be the image of $b$ under the flow $\phi$ at time $t_F$. Define $j$, $p$, and $u$ similarly. Clearly $f$ lies on $F$, $j$ lies on $J$, and so forth. Notice that $b$, $f$, $j$, $p$, and $u$ lie on a spacelike curve: the path of $b$ under $\phi$. Therefore, by our initial assumption, no timelike curve can pass through two or more of these points.

We've now constructed curves with marked points satisfying (1) and (2).

Finally, here's why I suspect that (3) can never also be satisfied. [Edit: I misread (3), and now I think it may be possible to satisfy (3) with a variant of the construction I gave for (1) and (2). See comments for details.] Suppose we have some curves with marked points satisfying (1) and (2). I suspect that (2) prevents the light cones of $f$, $j$, $p$, and $u$ from covering all of $L'$. In fact, I think there has to be a neighborhood $\Omega$ of $b$ which does not intersect the light cones of $f, \ldots, u$. Draw any timelike curve $W$ through $b$ and pick a point $w = \mathcal{E}_{WX}$ in $W \cap \Omega$. Since $w$ lies outside the light cones of $f, \ldots, u$, it can't be connected to any of those points by a timelike curve $Y$.

Vectornaut
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