3
$\begingroup$

Consider a Pseudo-Riemannian Manifold with signature

$$ (\underbrace{+,\cdots,+}_p,\underbrace{-,\cdots,-}_q) $$

For any positive integers $p$ and $q$. Can this kind of manifold contain closed timelike curves (CTCs)? I know that if $p=1$, then we get a Lorentzian Manifold that can't contain CTCs, but I am interested in the cases where $p>1$.

$\endgroup$
4
  • 3
    $\begingroup$ Comment to the question (v1): If a pseudo-Riemannian manifold $(M,g)$ has at least two temporal dimensions, then it is trivially possible to fit a CTC within an arbitrary small open neighborhood. More on physics with multiple temporal dimensions: physics.stackexchange.com/q/43322/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Mar 3, 2014 at 22:51
  • 2
    $\begingroup$ What makes you think a Lorentzian manifold can't have CTC's? It seems you may be conflating two distinct concepts: the topology of the manifold $M$, and the geometry encoded in the metric $g$ on $M$. In particular, global AdS is an example of a Lorentzian spacetime with CTCs: en.wikipedia.org/wiki/Anti-de_Sitter_space#Global_coordinates $\endgroup$ Commented Mar 4, 2014 at 2:25
  • $\begingroup$ Isn't a Lorentzian Manifold simply a pseudo-Riemannian Manifold whose signature is (1,n-1)? If that's the case, how is it possible that with only a single time dimension we can have CTCs? $\endgroup$
    – Disousa
    Commented Mar 4, 2014 at 11:03
  • 1
    $\begingroup$ Because that single "time dimension" can be "curved" and "closed". Think of a cylinder obtained by identifying two different instants of time (for all points in space at those instants) in Minkowski spacetime with respect to a given Minkowskian coordinate system. $\endgroup$ Commented Mar 4, 2014 at 12:01

1 Answer 1

5
$\begingroup$

For $p=1$, CTC's do not exist in Minkowski spacetime. In other $1+3$ spacetimes, in principle they are admitted in the absence of further requirements (like globally hyperbolicity) on the causal structure of the spacetime. They must be present if the spacetime is compact, for instance.

For $p\geq 2$, the answer is obviously YES. Consider a manifold $M$ with metric $g$ with signature (p,q) and $p \geq 2$. In a $p+q$-dimensional neighbourhood $U$ of any point $s\in M$, using the exponential map at $s$ starting from a $p$ dimensional subspace generated by $p$ timelike vectors in $T_sM$, you can construct an embedded $p$-dimensional submanifold $N$ passing through $s$ and whose metric (induced by $g$) has signature $(p,0)$. This means that every vector tangent to a point in $N$, considered as a manifold on its own right, is timelike. In local coordinates on $N$ around $s\in N$, any circle surrounding $s$ is a closed timelike curve.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.