Spacelike curves definitions

Question

Well, I am looking for the definition, if there is any, of points separated by a spacelike curve in a Lorentzian or more generally in Semi-Riemannian space?

Answer 1

Your question is not very precise (not to mention the fact that the title question is somewhat different from the one in the body).

Here are several answers depending on how to read your question:

1. If you mean: "what is the definition of a spacelike curve?" The answer is simply: a space-like curve is a $C^1$ mapping $\gamma: (0,1) \to M$ such that $\dot{\gamma}$ is a spacelike vector (depending on convention, either $g(\dot{\gamma},\dot{\gamma}) > 0$ or $< 0$).

2. If you mean: "is there a definition that captures the fact that two points can be joined by a space-like curve?" then the answer is No. And for a very good reason: in Lorentzian geometry with more than 1 space-like dimensions (so with the exception of locally $\mathbb{R}^{1,1}$), any two points that are path-connected are connected by a space-like curve. Simply take a path connecting the two points. Now take a tubular neighborhood of the path. By replacing the path with a tightly wound helix, you can replace any time-like or causal segments by space-like curves.

   $\mathbb{R}^{1,1}$ is special. This is because in $\mathbb{R}^{1,1}$ the set of space-like vectors (not counting the zero vector) has two connected components. In higher spatial dimension $\mathbb{R}^{1,k}$ for $k > 1$, the set of space-like vectors is connected. And hence it is always possible to perturb a curve (in a way that is small in $C^0$ but large in $C^1$) to be space-like.

   Note that also the usual causal relations in Lorentzian geometry ($p \prec q$ standing for there exists a future directed time-like curve $\gamma$ such that $\gamma(0) = 0$ and $\gamma(1) = q$) cannot be used in general semi-Riemannian geometry for the same reason: if the set of time-like vectors (not counting the zero vector) form a connected set, as in the case when there are multiple temporal dimensions, you cannot define the notion of future and past. And the same argument above guarantees that you can always join any two (path-connected) points $p$ and $q$ by a $C^1$ time-like curve.

3. One more word about the case $\mathbb{R}^{1,1}$ or other $1+1$ dimensional Lorentzian manifolds. If you assume that your space-time is "spatially orientable" (meaning that there exists a non-vanishing space-like vector field $v$ on $M$), you can then say that "$p$ is to the left of $q$" when there exists $\gamma:[0,1] \to M$, $C^1$, with $\gamma(0) = p$ and $\gamma(1) = q$ such that $\dot{\gamma}$ is always space-like and such that $g(\dot{\gamma},v) > 0$. This is just taking the usual temporal causal relations and swapping the role of space and time (since in $1+1$ dimensions which direction you call space and which direct you call time are entirely symmetric). In this restricted case then yes, there is notion of two points being joined by a space-like curve.

4. The more usual notion of a "space-like" relation between two space-time events in Lorentzian geometry, however, is the following statement: "$p$ and $q$ are achronal if there does not exist a $C^1$, time-like curve joining $p$ and $q$." Again, in the case of general semi-Riemannian geometry, this definition is useless (see point 2.) Again, this uses the fact that in Lorentzian geometry the set of time-like vectors has two connected components.

If none of the above answers your question, please edit your original question and ping me below. I'll update the answer accordingly.

Answer 2

This varies depending on your convention - I use the convention that the metric in Lorentzian geometry has signature $(-,+,+,+)$, so timelike distances are negative.

A spacelike vector is a tangent vector $v$ such that $g(v,v) > 0$. A spacelike curve is a curve whose velocity vector is everywhere spacelike. Thus we can say two points are spacelike-separated (or separated by a spacelike curve) if they are the endpoints of a spacelike curve. Those who use the opposite sign convention would replace $>$ with $<$.