Is an achronal set contained in its own causal future?

+ 2 like - 1 dislike
1481 views

I use Wald's notation: $I^+$ is the chronological future and $J^+$ is the causal future.

My confusion arises from the following passage in Wald (1984):

Now, let $S$ be a closed, achronal set (possibly with edge). We define the future domain of dependence of $S$, denoted $D^+(S)$, by $$D^+(S)=\{p\in M|\, \text{Every past inextendible causal curve through p intersects S}\}$$ Note that we always have $S\subset D^+(S)\subset J^+(S)$.

I have to disagree with the last statement. We know that $S$ is achronal, i.e. $I^+(S)\cap S=\emptyset$. The relation $S\subset D^+(S)\subset J^+(S)$ implies $S\subset J^+(S)$, i.e. $J^+(S)\cap S\ne\emptyset.$ But I cannot see how a set can be both achronal and contained in its causal future. Hence the title of my question.

I think Wald meant to write $S\subset D^+(S)\subset \overline{J^+(S)}$.
This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7

edited May 2, 2015

+ 3 like - 0 dislike

I do not remember the precise definition in Wald's book, however $J^+(S)$, by definition, always includes $S$. By definition, it is made of $S$ and the points of the spacetime  which do not belong to $S$ and stay along the future directed (so that the trivial curve joining $p$ and $p$ itself is not such a curve) causal curves exiting from $S$.

By the way $J^+(S)$ is not closed in general, though it is true when the spacetime is globally hyperbolic as $J^+(S) = \overline{I^+(S)}$.

answered Feb 19, 2015 by (2,085 points)
+ 1 like - 0 dislike

As the causal future of $p$ is the set of points joined to $p$ by timelike or null curves, and the constant path $\gamma(t) = p$ joining $p$ to $p$ itself has vanishing tangent vector and hence is a null curve (though a rather silly one), $p \in J^+(p)$, and so, $S \subset J^+(S)$.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user ACuriousMind
answered Feb 15, 2015 by (910 points)
+ 0 like - 1 dislike

I'm fairly sure I got it.

The causal future $J^+(p)$ of a point $p$ is defined as the set of all points $q$ connected by a future pointing timelike or null curve to $p$. I think the secret lies in that this is a closed set in Minkowski spacetime. To see this, we see that the curves connecting the points in $J^+(p)$ are the timelike curves (negative length) plus the null curves (zero length). Included in the set of null curves is the trivial curve connecting $p$ to $p$, which has zero length. Thus $J^+(p)$ is closed. Since $J^+(S)$ is just the union of all $J^+(p)$, $p\in M$, it is also closed. This means it contains $S$, because $S\cap\partial J^+(S)\ne\emptyset$ and $J^+(S)$ is closed. I think this generalizes nicely to a general spacetime, because even though $J^+(S)$ need no longer be closed, $S\cap\partial J^+(S)\ne\emptyset$ should still hold.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7
answered Feb 15, 2015 by (50 points)
Careful, an infinite union of closed sets is not necessarily closed - just as an infinite intersection of open sets is not necessarily open, so you need another argument than just "$J^+(p)$ is closed" to conclude that $J^+(S)$ is closed.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user ACuriousMind
@ACuriousMind: Are there any simple counterexamples? I'm not sure how to refine my argument.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7
Take the intervals $[0,1 - \frac{1}{n}]$. They are closed, but the union over all $n$ is $[0,1)$, which is neither open nor closed. I am not a home with causal sets, so I cannot tell you which way this argument is supposed to go.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user ACuriousMind
@ACuriousMind: What if $S$ is compact? After all, $[0,1)$ is not compact IIRC.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7
@ACuriousMind: A more general question: What is the restriction needed on $\{U_i\}$ such that $\bigcup_iU_i$ is closed?

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7
Any compact set in $\mathbb{R}^n$ is bounded and closed, but this does not hold for all metric spaces, in particular, I think it indeed fails on Lorentzian manifolds. That said, I am confused what your actual question here is - you say, $S \subset J^+(S)$ doesn't make sense to you, and yet $J^+$ is the set of points joined by timelike or null curves, and the constant path joining $p$ to $p$ is a null curve, so $p \in J^+(p)$, and hence $S \subset J^+(S)$, right? Generally: Any finite union of closed sets is closed, but nothing else holds in arbitrary topologies.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user ACuriousMind
@ACuriousMind: Crap. After reading my question again I realize that I don't need $J^+(S)$ to be closed. Post your comment or whatever and I'll accept it as an answer. Thanks. Could you also please answer my more general question (comment above) in your answer? EDIT: Saw your edit.

This post imported from StackExchange Physics at 2015-02-15 12:13 (UTC), posted by SE-user 0celo7

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar\varnothing$sicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification