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Philosopher JB Manchak discusses. This is over my head, so curious to hear what other philosophers of physics think. Do not comment unless you are a philosopher of science or physics (graduate students in these areas are welcome to comment as well) Use a full name and valid email address please (the email address will not appear).
One small remark and then a comment.
Remark: the editorial summary overstates the claim contained within the full text, that Heraclitus spacetimes are (likely to be) generic among the models of general relativity. The latter is a much weaker mathematical observation than the summary claim that “Manchak argues that the whole universe is like that” (which I, at least, read as a claim about how likely we ought to model our actual universe in the context of general relativity).
Comment: another subject not addressed in this essay, in context of that remark, however, is precisely the question of whether it is reasonable to suppose that our universe is best modeled by a Heraclitus spacetime. This touches on the more general matter of what makes a spacetime property (e.g. the Heraclitus property) a worthwhile representational choice for some aspect of some target system (e.g., the whole universe). Especially, one might be concerned with the justification for restricting to Heraclitus models of general relativity as a means of sidesteping concerns about cosmic underdetermination by all possible evidence. (This topic is alluded to in the essay.)
That a spacetime property is in some sense generic is sometimes considered a first check that the property itself isn’t the product of an overly idealized description of the given target system by a given model in the theory. But this seems to me to take for granted certain views on the importance of de-idealization, which are not universally agreed upon in the more general idealizations literature. Meanwhile, there’s something even more thought provoking about the specific case of a Heraclitus property: as hinted at in Manchak’s essay, literally all of the existing evidence for the descriptive adequacy of general relativity for gravitating modeling targets comes in the context of modeling target systems/phenomena in such a way as denies the Heraclitus property. To conclude that our universe is likely Heraclitus (should anyone want to! Again, this is not narrowly what Manchak writes) is therefore to draw on contested philosophical/metaphysical premises in support of “best modeling practices in mathematical relativity”, which are beyond methodology (i.e., beyond the “best evidential practices” of the sciences for generating scientific knowledge), if not squarely at odds with methodology (depending on one’s views about de-idealization).
Fun stuff to think about!
Thanks for your remark, Mike. As you say, I do not claim in the text that the actual universe is Heraclitus. Just to add a bit more context, my final submission to iai.tv had a completely different title/subtitle:
Heraclitus + Einstein = ?
Radical Asymmetry in General Relativity
It was quite a surprise to me to see the changes once the piece was published…
Using spectral color shading to represent the uniqueness of events in a two dimensional way is itself a contribution.
I’d say that in two ways this discussion reminds us that spacetime is a dynamical player in modern physics, not just the backdrop against which the dynamics plays out.
1) The Heraclitus property isn’t that surprising when applied to the *material contents* of spacetime (even putting general relativity aside). In that context it would say, roughly, that the distribution of matter is sufficiently variegated that no two small patches are *exactly* alike. That seems quite plausible (though on sufficiently small scales quantum theory would complicate stating it exactly) and as I recall it has a long pedigree – Leibniz asserts something similar when he discusses the principle of identity of indiscernibles in the Leibniz/Clarke correspondence, I believe.
Before general relativity, we thought of spacetime as a background, with a whole bunch of exact symmetries, so that small regions of spacetime are genuinely indistinguishable one from another. But of course if spacetime is dynamically interacting with matter, then if matter is sufficiently variegated, so will spacetime be.
2) This might be a bit more contentious, but I’d say that in physics, what matters are the symmetries of the laws, not of the objects. In pre-general-relativistic physics, the (spatiotemporal) symmetries of the laws basically coincide with the symmetries of spacetime, but that stops being true if spacetime itself is dynamical. In general relativity, a generic spacetime won’t have any symmetries at all, but the laws of general relativity do have a bunch of symmetries and these play an important role in understanding the theory, e.g. through the conservation laws they imply.
FWIW, I can’t follow the argument here. As David Wallace says, once you make the space-time structure dynamical—coupled to the matter distribution—then it is immediately plausible (although no doubt hard to prove) that generically no two open patches in a single global solution are completely isomorphic. (This would, I think, better be called “Leibnizian” (in the sense of the qualitative difference of all monads) than “Heraclitean”, but that’s not important.) It is also, of course, the case that generic solutions will (probably) be at thermal equilibrium, and that the actual universe isn’t, so the inference from “generic” to “probably ours” in not at all secure.
Anyway, that part seems plausible. But the part about being able to recover the global structure from the local is unclear to me. Of course, if no two open sets have the same structure, then from a collection of open sets that cover the manifold (an atlas of charts) the entire manifold is uniquely determined: any two overlapping charts will have a common open subset, and since all the open sets are unique it is determined how the charts overlap and hence what the manifold is. But in the other direction, I just can’t follow. Suppose two charts are identical, like identical jigsaw pieces. Then, of course, in a sense the jigsaw puzzle does not have a “unique” solution: either of the two numerically distinct isomorphic pieces can be put in the same place in a complete solution. But these only would count as “two different global solutions” if the *numerical identity* of the two pieces were relevant in what counts as the “correct solution”. But why should that be so? The normal thought would be that putting the pieces together, and then “switching the isomorphic pieces” count as *one and the same* global space-time, not two different global space-times. The two solutions, taken as wholes, are isomorphic and so represent the same situation. In which case the claim made in the article—if I follow it—is not correct.
Hi Tim. A couple of comments:
(1) I only claim one direction in the text: if a spacetime is Heraclitus, then a local to global recovery is possible. Regrading the other direction, I say “If the spacetime has non-trivial local symmetries, [a unique global recovery] may not be possible.”
(2) The other direction does, in fact, hold in the following sense: Suppose you give me a non-Heraclitus spacetime (M,g). I claim there is a collection of open sets in M that counts as an open cover for both (M,g) and some other (non-isometric) spacetime (M’,g’). Thus, a local to global recovery is impossible.
Proof: Suppose (M,g) is not Heraclitus. Then by definition, there will be open regions U and V in M and an isometry between them. Let {O_i} be a collection of open sets that cover of M which (i) includes both U and V as elements and (ii) is such that some portion of M is only covered by U — call this (topologically closed) portion K. (Such an open cover always exists.) Now I claim that there are distinct (non-isometric) spacetimes such that {O_i} is an open cover for both. Of course (M,g) is one such spacetime. The other is (M-K, g) which is topologically distinct from (and therefore non-isometric to) the spacetime (M,g). To see that {O_i} covers M-K, first note that {O_i} – {U} covers M-K just as in the spacetime (M,g). What to do with the “leftover” piece U? Glue it to the region V. The resulting structure is the spacetime (M-K, g). This procedure is sketched in the text where I construct a “hole” in the first river stepping example by removing a single piece and gluing it to a nearby isometric region. As I mention there: “complete overlap with other pieces is permitted”. By this I mean that the definition of open cover permits duplicate elements.
Stepping back, I agree with you that “switching the isomorphic pieces would count as *one and the same* global space-time”. I would just say that that is not my proof strategy. Instead, I glue one piece onto another to construct a “hole” that witnesses a topological difference between the two spacetimes. Does that help clarify things?
“Glue it to the region V”? I suppose you mean something by that but I cannot understand at all what you have in mind. V is already completely surrounded by other charts, so there is no obvious place to do the gluing without the resulting structure no longer even being a manifold.
I also don’t understand what you mean when you invoke M-K. You said that K is closed. So what makes you think that M-K can be covered by open sets from the original set of charts?
You write: ““Glue it to the region V”? I suppose you mean something by that but I cannot understand at all what you have in mind. V is already completely surrounded by other charts, so there is no obvious place to do the gluing without the resulting structure no longer even being a manifold.”
The gluing method is the standard one used in differential geometry. For a nice presentation, see O’Neil’s *Semi-Riemannian Geometry*, middle of page p. 7. (Note that he works with manifolds and smooth maps but the same idea is often used for spacetimes and isometries.) In our case, the gluing method will identify the region U with the region V in the spacetime (M-K,g). More formally, since there is an isometry f: U —> V, one can identify each point p in U with the point f(p) in V. The resulting structure is a spacetime.
You write: “I also don’t understand what you mean when you invoke M-K. You said that K is closed. So what makes you think that M-K can be covered by open sets from the original set of charts?”
By construction, the spacetime (M,g) can be covered — except for the closed region K — by the collection of open sets {O_i}-{U}. So the spacetime (M-K, g) can be completely covered by {O_i}-{U}. One then glues the open set U into (M-K,g) by the identification procedure mentioned above. So {O_i} covers (M-K,g).
I hope this helps! I am now off to the teach a group incarcerated students at Donovan State Prison. It is three hour seminar and a four hour drive there and back. This is all to say that if you reply I likely won’t be able to respond until tomorrow. My apologies!
Sorry, I’m still lost.
“In our case, the gluing method will identify the region U with the region V in the spacetime (M-K,g). More formally, since there is an isometry f: U —> V, one can identify each point p in U with the point f(p) in V. The resulting structure is a spacetime.”
If by “identify” you mean that they are one and the same point, then you have not done what you are supposed to do. In the original atlas, there are two isomorphic charts. Or, if you like, think of it as being given a multi-set. But in the constructed object, you don’t that two distinct regions—one for each chart—but just one. I would think that that violates the rules of the construction.
Thanks for your patience, Tim! Ok, I am trying to better understand your worry here. The gluing construction is common in general relativity and so my sense is that you are already familiar with it. (For example, it is used in the proof of a maximal Cauchy development for appropriate initial data wherein an infinite collection of developments are all glued together.) We can talk more about the details of the gluing construction if you like…
But given your last comment, perhaps the worry is that when U is glued to (M-K,g) to form a new spacetime (isometric to (M-K,g)), it can’t be the case that both U and V are members of an open cover since they have been glued together to form a single open set. Is that the worry? If so, I think I see it and I confess that I wasn’t very clear in my original statement about what it means for a collection of open sets in one spacetime to be an open cover for a different spacetime. My apologies for the confusion! One natural way of making this precise would be to say that a collection of open sets {O_i} in a spacetime (M,g) counts as an open cover for another spacetime (M’,g’) if there is an isometric embedding f_i: O_i —> M’ of each open set O_i such that the collection {f[O_i]} is an open cover for (M’,g’).
On this formulation, it is easily seen that that the collection of open sets {O_i} that I defined yesterday (with U and V as members) is both an open cover for (M,g) and also counts as an open cover for (M-K, g). Let f: U —> V be the isometry witnessing the non-Heraclitus property of (M,g). Let O_1=U, O_2=V. Except for i=1, let f_i: O_i —> M-K be the inclusion map. Let f_1=f_2 \circ f. Notice we are not doing any gluing here to get the result — it’s just not needed.
Stepping way back, I now think there is an even better way to formulate my claim (2) that there can be no local to global recovery for non-Heraclitus spacetimes. The new formulation has the benefit of being simple and clean and avoiding talk of open covers and gluing methods altogether. It is also directly analogous to the positive result for Heraclitus spacetimes. In order to state it, one just needs the definition of a local isometry between spacetimes: (M,g) and (M’, g’) are locally isometric if, for each point p in M, there is an open neighborhood O_p that is isometric to some open set O’ in M’ and, correspondingly, with the roles of (M,g) and (M’,g’) interchanged. Here is the new claim:
(2*) Suppose you give me a non-Heraclitus spacetime (M,g). I claim there is a spacetime (M’,g’) such that (M,g) and (M’,g’) are locally isometric but not isometric.
Hopefully it is clear that the spacetimes we have been working with — (M,g) and (M-K,g) — are locally isometric but not isometric. The proof mirrors the one just given that the collection of open sets {O_i} is both an open cover for (M,g) and also counts as an open cover for (M-K, g). It is natural to think of local properties of spacetime as those preserved under all local isometries and global properties as the non-local ones. Under this understanding, the local properties of a non-Heraclitus spacetime do not determine its global properties. So a local to global recovery is impossible in that sense. But the local properties of a Heraclitus spacetime do indeed determine its global properties. This is Corollary 1 of Manchak and Barrett (2023).
Thank you for helping me come to this cleaner alternate formulation!
OK, I think we are clearing things up! At least, I have a better sense of your claims, and as you say in the non-Heraclitean case you don’t make a categorical assertion.
I don’t think I ever thought deeply about the precise definition of an open cover, but I think that at least tacitly when one says that a set of charts forms an open cover of a manifold the claim is a little stronger than this as you intend it:
“One natural way of making this precise would be to say that a collection of open sets {O_i} in a spacetime (M,g) counts as an open cover for another spacetime (M’,g’) if there is an isometric embedding f_i: O_i —> M’ of each open set O_i such that the collection {f[O_i]} is an open cover for (M’,g’).”
I think that a set of charts that form an open cover is supposed to be minimal, in the sense that no proper subset is also an open cover. Or another way to put it is that in each chart there is at least one point that is not covered by another chart. By that definition, this construction is does not give an open cover of the second manifold. I haven’t checked in any text whether the condition of minimality is explicitly stated, but as I say I think it is at least assumed. In any case, I have always assumed it.
So I think this gets at the source of my puzzlement.
Yes, I think we are getting it sorted! Your distinction between open cover (as standardly defined) and a minimal open cover suggests an open question:
Question. Let (M,g) be a non-Heraclitus spacetime. Is there a collection of open sets {O_i} in M that counts as a *minimal* open cover for both (M,g) and some other (non-isometric) spacetime (M’,g’)?
I don’t see how to settle this one at the moment — something to think about! Stepping back, I think this discussion shows that there are several senses of local to global recovery that one might have in mind. Some certainly fail to hold for non-Heraclitus spacetimes while others may in fact hold. Again, thanks for your help with this. As I said right at the start of my first comment, this is not something I claim in the piece (or in any publication). This correspondence has spurred me to get clear on these issues and I appreciate it.
To my mind, the essay only covers the constructive direction (as, indeed, you indicate). But your counter-example “in the other direction” turns on what is merely an ambiguity in the essay. To make the full recovery claim explicit, we’d need a sharper characterization of what makes any two spacetimes either the same or different from each other—this is nothing troubling, it is what we usually talk about in terms of spacetime properties. But in present context, we don’t want to get distracted by all of the ways that two spacetimes may be differentiated in virtue of appeals to local states of affairs about one and the other, so we distinguish local versus global spacetime properties—again, nothing unusual. The “other direction” of the full recovery claim for you would then be an argument that, when considering only Heraclitus spacetimes, all properties (i.e., all ways of differentiating between pairs of spacetimes) are local properties, unlike the situation in general within general relativity. Naturally, this falls to how one defines the local/global properties distinction. But it is so, basically immediately, on one definition of local properties that has been used elsewhere. (Though it is not so, on another.)
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