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This post concerns the causality of spacetime $\mathcal M$.

A future set $F$ is defined to be the chronological future of some set $S\in \mathcal M$, ie., $F=I^+[S]$. Similiarly, a past set $P=I^-[S']$ for some $S'\in\mathcal M$. The boundary of $F$ or $P$ is called an achronal boundary $B$. We can show that $\mathcal M=F\cup B\cup P$.

Due to the definition of $F$ and $P$, we find the following properties: \begin{gather} p\in F\Rightarrow I^+(p)\subset F \tag{1} \\ p\in F\Rightarrow I^-(p)\nsubseteqq P \tag{2} \end{gather}

You can write down similar expressions by interchanging $F$ and $P$, $I^+$ and $I^-$.

It is really easy to understand expression (1). Indeed, $I^+(p)$ should be completely contained in $F$.

What about $I^-(p)$? We know for sure that $I^-(p)$ is not a subset of $P$. Is it a subset of $F$? Intuitively, that is not true, but can we prove that?


I tried to prove that $I^-(p)\cap P\ne\emptyset$, but I cannot find any constradiations. Would you please give me some hints?

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  • $\begingroup$ The answer to both is "no". Alomost any example will do the job. $\endgroup$ – MBN Aug 8 '15 at 15:51
  • $\begingroup$ Why would these two sets F and P have the same boundary? Your spacetime could be Minkowski space and we could choose $S\subset M$ (not $S\in M$ like you wrote) to be the unit spatial ball at $t=5$ and $S'\subset M$ to be the unit spatial ball at $t=0$ and the boundaries are not equal. $\endgroup$ – Timaeus Aug 8 '15 at 20:42
  • $\begingroup$ @MBN Can you explicitly give me an example? Thank you! $\endgroup$ – Drake Marquis Aug 12 '15 at 16:02
  • $\begingroup$ @Timaeus It is always possible that P and F share the same boundary. In fact, there is a theorem stating that if there is a future set $F$, then there is a past set $P=I^-[T]$ where $T$ is the complement of $F$ and they share the same boundary. $\endgroup$ – Drake Marquis Aug 12 '15 at 16:06
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    $\begingroup$ @DrakeMarquis If you are just assuming random things randomly then how can anyone follow you? I thought you wanted to show the whole manifold is the union of those three things. Now it seems like you want to assume it, or maybe merely show that for any M it is possible to find an F and a P that make it work. No one thinks you are expressing yourself clearly enough. If you expect two sets to be arbitrary sets that satisfy a property, then say so. For instance if you need S and S' to be disjoint then tell us so. Otherwise we can make counter examples to (2) $\endgroup$ – Timaeus Aug 12 '15 at 16:22
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Assume you have a set $F\subset \mathcal M$ called a future set which is the chronological future of some set $S\subset \mathcal M$, ie., $F=I^+[S].$ Where $\mathcal M$ is your spacetime. Similarly, assume you have a past set $P=I^-[S']$ for some $S'\subset\mathcal M$.

So that we are clear, $F=I^+[S]$ means that $F$ is exactly those events $f$ where there is a future pointed timelike curve from a point $s\in S$ to $f.$

Furthermore assume the boundary of $F$ and the boundary of $P$ are equal. Call the boundary $B$ and call $B$ an achronal boundary. (Calling it achronal might be justified later on if we make enough additional assumptions, right now it is bogus.)

Assume further that $\mathcal M=F\cup B\cup P$.

What about $I^-(p)$? We know for sure that $I^-(p)$ is not a subset of $P$.

Except it could be unless you make additional assumptions.

Is it a subset of $F$?  Intuitively, that is not true, but can we prove that?

You say you can prove

\begin{gather} p\in F\Rightarrow I^+(p)\subset F \tag{1} \\ p\in F\Rightarrow I^-(p)\nsubseteqq P.\tag{2} \end{gather}

However, the second doesn't follow from your listed assumptions (let $\mathcal M$ equal Minkowski space, let $F$ equal the set of events with $t\geq 0$ and let $P$ equal the set of events with $t\leq 0$ and let $p$ be the origin). But we could make those assumptions too, or we could make ever more additional assumptions after assumption over and over again until we start being able to actually derive the things you claim to be able to prove, for instance we could assume that $F$ and $P$ are open, or that the unions in $\mathcal M=F\cup B\cup P$ were supposed to be disjoint unions $\mathcal M=F\uplus B\uplus P.$ Who knows what you mean to write? Your readers can't. When you wrote $S\in\mathcal M$ I assumed you meant $S\subset\mathcal M.$ But one can only correct so many things before one wonders if one corrected all of them correctly.

Moving on. You ask whether $I^-(p)$ is a subset of $F.$ Which is not necessarily true. Let $\mathcal M$ be Minkowski space and let $F$ be the set of events with $t>0$ (and $P$ is the set of events with $t<0$, which has to be the simplest possible example). Note that this time I made the sets disjoint, because you asked a question instead of telling me you could show it.

Then the past of any point in $F$ includes regions with negative time which are not in $F.$ So $I^-(p)\subset F$ does not have to hold even when $\mathcal M=F\uplus B\uplus P$.

So that's that question (the one in the original post if I correctly distinguished between assumptions and questions). And it is answered directly by the simplest possible example that I think meets your criteria.

I'm not even sure if you can come up with disjoint nonempty $F$ and $P$ that have $I^-(p)\subset F.$ So I'm not sure why you asked if you can't come up with a single example.

Particularly when the simplest example is a counter example.

I tried to prove that $I^-(p)\cap P\ne\emptyset$, but I cannot find any [contradictions]. Would you please give me some hints?

One hint is to find a single example before you make a conjecture. Another hint is to check the simplest possible examples to look for counter examples before you make a conjecture.

As for assuming your conjecture and then looking for a contradiction ... that whole methodology is bad. There aren't any contradictions to find because it is possible, the simplest possible example gave an example where $I^-(p)\cap P\ne\emptyset.$

The idea of assuming something and finding a contradiction only works in some situations. You need to assume the opposite of what you want. And you need to also have enough things already to force things to go the other way. And often it is a waste anyway to do proofs by contradictions since to get the contradiction you need a step where you first prove the actual thing you want and then have an additional step where you get a contradiction. Usually its for people that don't want to bother learning or using the contrapositive version of theorems.

Plus it is horrific for the student. When you make a false assumption and then spend a great deal of time working out the consequences you spend time exploring untruth. You develop incorrect intuition.

Stick to correct thinking and then your intuition will help you. Spend too much time thinking about untrue things and your intuition will remember these wrong things.

It is playing with fire. People encourage students to do it because "it can be easier" and that's the explanation for you. Laziness is the sole reason for the proof by contradiction technique. And the version you did isn't even that and doesn't work at all, assuming the thing you want and looking for a contradiction could at best show the negation of your conjecture it can never prove it (unless you work with a system known to be compete, which is not popular, though it is actually possible).

Assuming something and finding a contradiction only works if the thing you assume is wrong.

Now let's finally assume that $F\cap P\subset B \neq \emptyset$ in case we need it.

This last assumption is consistent with your canonical example where you start with an arbitrary $S\subset \mathcal M$, and let $F=I^+[S]$ be such that the complement of $F$ is nonempty and is your $P.$

You then ask in your comments whether

any point in $P$ [has] at least one future pointed timelike curve intersecting $F$.

Which is vague, you really really need to learn to speak precisely. At least if your goal is to understand why things must happen or can happen. If you mean is it the case whether every point in $P$ has a future pointing timelike curve intersecting $F$ or whether you mean if there is at least one point in $P$ with a future pointing timelike curve intersecting $F.$ These are very different statements.

Since you asked for hints. There is a clear answer in one direction for the simplest example. As always check Taub-Nut and the 2d version if simpler examples don't generate counter examples for the direction you are considering. Then you might also want to consider how your definitions deal with singularities.

For instance imagine your spacetime has a conformal diagram with a spacelike singularity in the $t=1$ hyperplane but only in the part with $x^2+y^2+z^2\leq 1.$ Then as long as your spacetime has regions in the $t=1$ hyperplane with $1<x^2+y^2+z^2<1+\epsilon$ then the future ($F: t>1$) is clearly connected to the past ($P: t<1$). But there are parts of the past that have all future pointing timelike curves within an event horizon because they are geodesically incomplete. In particular any point with $\sqrt{x^2+y^2+z^2}<t<1$ doesn't have future directed timelike curves that reach the future $F.$ But this a singular manifold, or at least a geodesically incomplete manifold.

But note this singular manifold is only a counter example of one way to read your ambiguously written comment that is unrelated to your original question.

Also note when I suggest Taub-Nut that was a general suggestion for any conjecture, I didn't check it myself for your example because your writing was too vague for me to know what you were asking for sure and I wasted my brain power trying to read your mind.

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  • $\begingroup$ Thank you! And sorry for asking such an ambiguous question. I feel shameful to tell you that when posting this question, I deserted from it... I learned a lot from your post. I should think about my way of learning new knowledge. Probably, I have to conquer a new problem directly instead of find contradictions so often. $\endgroup$ – Drake Marquis Aug 13 '15 at 1:51
  • $\begingroup$ @DrakeMarquis Everyone struggles sometimes and it is also frustrating to not succeed at communicating. But hard work helps, and learning good habits helps. And asking for help when you need it helps. And paying attention to what works and doesn't can also help. In this case precision matters. $\endgroup$ – Timaeus Aug 13 '15 at 6:22

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