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Einstein hole argument 
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#1
Aug808, 02:06 AM

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Could someone please help me understand the Einstein hole argument (as outlined by Norton, see below). In particular the step that says that the second solution within the hole is a valid solution to the generally covariant field equation. I think my understanding of general covariance is at fault here.
I'll summarise the argument as described by Norton: 1) g(x) is a solution in the hole in one coordinate system...ok 2) g'(x') is the same solution in another coordinate system...fine 3) g'(x), gained by using the function from 2 with the first coord system args, is a different gravitational field....fine 4) g'(x) is a solution of the field equations (what!?) How can he just say that g'(x) is a solution to the field equations? I can understand that the field equations are generally covariant and therefore take the same form in different coordinate systems. But I don't understand that a solution explicitly expressed in terms of one coordinate system can take the same form and be a solution in a different coord system. This is a rough paraphrase of my question... A generally covariant defintion of the circle is a curve equidistant from some point. 1)A solution in one coord system is x^2 + y^2 = 25 2)The same solution in another coord system is r=5 3)The equation x=5 is a different curve to 1) 4) The equation x=5 is a solution of the definition of a circle !? How can 4) be stated? 


#2
Aug808, 05:13 AM

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PF Gold
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What do you mean by "see below"? At least tell us what "Einstein's hole experiment" is!



#3
Aug808, 12:23 PM

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Excellent question. I have wondered about exactly that question for a while. Thanks for posting it. Hope someone will clarify. 


#4
Aug808, 01:24 PM

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Einstein hole argument
The curve x=5 is a circle, when the distance formula is given byYour definition of a circle is a predicate involving a set of points and a distance function. Your definition does remain unchanged when you apply a changeofcoordinate transformation  your problem is that you forgot to apply that changeofcoordinate transformation to all of the pieces. (of course, there are ugly issues involved in using polar coordinates in this way  but those difficulties are irrelevant to the question at hand) I believe the Einstein field equations are a criterion involving only the metric, the stressenergy tensor, and the cosmological constant. In the hole, we have that the criterion, the stressenergy tensor, and the cosmological constant are all invariant under coordinate changes. Therefore, if we have a metric satisfying the EFE, coordinatechanging it must also satisfy the EFE. To put it another way, I believe that in this context, general covariance means that physical laws depend only on the coordinate representation of a field, and not on the actual coordinate functions. The argument, as described, constructs two different metric tensors that have the same coordinate representation (under different coordinate charts). And since one was assumed to be a solution, the other must also be a solution. 


#5
Aug808, 11:28 PM

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#6
Aug908, 12:58 PM

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Hrm, let me try putting that another way. If you want to insist that the Euclidean distance function is invariant, the only coordinate changes you are permitted to do are elements of the Euclidean group  transformations built out of translations, rotations, and reflections. But if you want to allow arbitrary coordinate transformations, then you cannot consider the distance function as an invariant, and must instead consider it additional structure on your underlying manifold, and you must not have the expectation that the coordinate representation of the distance function remains the same in all coordinate charts. 


#7
Aug908, 06:39 PM

P: 16

nrqed, are you still perplexed like me? It's interesting to see the entry under Wikipedia for this and the discussion page. There is confusion in an attempt at explanation and one entry points out that the argument seems absurd nowadays, but I can't see how it wasn't absurd a 100 years ago either. I still think I'm missing something though as it's taken fairly seriously and once Einstein had come to reject the argument it wasn't a matter of finding a simple logic error but rather it taught him a "deep" physicophilosophical lesson about there being no spacetime without gravity. 


#8
Aug1008, 12:21 AM

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Yes, I am still perplexed and I am also missing something. But I am convinced it is something deep that is worth our efforts. I used to think that this was just invariance under coordinate reparametrizations but I am sure now that it's much deeper than that. Reading some parts of Rovelli's book on loop quantum gravity made me realize that there was something deep behind this. I think it's all about "diffeomorphism invariance" which is more than invariance underreparametrizations (even though some people use the two terms to mean the same thing which just adds to the confusion. I am unfortunately quite busy these days but as soon as I have a bit of time to ponder on this I will post some thoughts. Luckily, Hurkyl sounds like the right person to clear up things for us. regards 


#9
Aug1008, 04:16 PM

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But if you really were trying to do differential topology (and so expecting symmetry under diffeomorphisms is fair game), then you cheated in a different way. Your criterion really had two arguments to it: a distance function and a point set. But when you applied your diffeomorphism, you forgot to apply it to the distance function. When you apply the diffeomorphism to both parts, you get the observation I made above: x=5 is, in fact, a circle for the transformed distance function. 


#10
Aug1008, 07:23 PM

P: 16

OK PF Mentor, please do your mentoring! You seem to be hiding behind a lot of abstract mathematics here instead of thinking physics, you are redirecting the thread offtopic to abstract mathematics. In physics, distance is distance and is an observable, a circle is an observable. You cannot just reshape it in some abstract mathematical world, you cannot say in any way that x=5 as defined using an orthonormal x is a circle (and when I say circle I mean a circle!, a curve, a shape that everyone knows very well is a circle. When I say distance I mean physical distance, not something arbitrary). Think about the physics!! and don't get lost in topological morphisms and mathematical definitions, the hole argument as stated (by Norton and Einstein) is a lot simpler than that! Here's Norton... (After discussing g(x) as a solution and g'(x') as the same solution in different coords) "At this point Einstein effected a subtle manipulation that is the key to the hole argument. One could consider the symmetric matrix g(x) as a set of 10 functions of the variable x, and g'(x') as a set of 10 functions of the variable x' (x here is short hand for x subscripted with a dimensional index). One can now construct a new set of 10 functions g'(x). That is, take the 10 functions of the new matrix g' and consider them as functions of the old coordinates x....Einstein has presumed the field equations general covariant. Therefore, if they are solved by the g(x) then they must be solved by g'(x') and therefore also by the construction g'(x)". 


#11
Aug1008, 09:34 PM

P: 341

The page
http://www.seop.leeds.ac.uk/entries/...e_Passive.html (by none other than Norton) I found helpful. Norton agrees that the step from 34 isn't obvious, and only says that there are abstract considerations that show this in detail  it's a shame there's no reference to the general proof. But he illustrates the idea in terms of a toy example. Perhaps this is where the analogy between the two arguments you present breaks down: I take it that, in GR, we're given that certain laws involving the metric, stressenergy and maybe some other tensors, are covariant. It's then presumably just a matter of showing that these laws take the same form (as Norton shows for his toy law in the page above) in the two physically different situations described in your 2 and 3 (which you seem to agree are different gravitational fields?). In the toy example that Norton gives, although gradients can be expressed in a coordinate independent way, it's not the case that the tranformed line has the same gradient as the original. But that doesn't matter because all we are given is the covariance of certain (toy) equations. So though you may be right about what you say about circles, the argument isn't that, for all properties P expressible in a coordinate free way, the new object defined by the same equation in the transformed coordinate system has the same P's as the original object defined by the same equation in the original coordinate system. As I understand it, the reason why this argument is generally not accepted today (Einstein was putting forward this argument when his early attempts to find a covariant theory had broken down, explaining why a theory like general relativity couldn't be right!) is that 2 and 3 don't really represent different physical systems: the two fields that are constructed are diffeomorphic, and so, though mathematically distinct, aren't really physically distinct. 


#12
Aug1008, 11:49 PM

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You already seem to be aware of what is actually correct, so all that's left seems to be helping you understand what went wrong in your analogy with the circle, and my explanations were geared towards that. If you are really and truly thinking Euclidean geometry and restricting to orthonormal rectilinear coordinates, and all that jazz, then your entire problem is that you're only allowed to use Euclidean motions  transformations made out of translations, rotations, and reflections. The Cartesiantopolar transformation is not a Euclidean motion. Your argument is analogous to complaining that the EFE changed form under a nondifferentiable transformation! However, if you were thinking of distances as just being added structure to the plane (which you should usually be thinking if you have GR in mind), then your problem is that transformation invariance means that things remain the same if you transform everything  however, you transformed the circle without transforming the distance function, and thus things didn't match up. If you reconsidered your example, but only permitted Euclidean motions (e.g. in some other coordinate system related to the original by a Euclidean motion, your circle might become [itex](x'  4)^2 + (y' + 3)^2 = 25[/itex]), then everything would work out 'correctly'. Anyways, I want to take a stab in the dark, so I will rewrite the hole argument: Let g be the coordinate representation of a particular metric tensor, relative to the coordinates x. Let g' be the coordinate representation of the same metric tensor, relative to the coordinates x'. (So that g'(x') = g(x)) Let E be the coordinate representation of the Einstein field equations relative to the coordinates x. (So the statement E(h) is the assertion that h(x) satisfies the EFE) Let E' be the coordinate representation of the Einstein field equations relative to the coordinates x'. (So the statement E'(h) is the assertion that h(x') satisfies the EFE) Because g(x) = g'(x'), we must have E(g) = E'(g'). Because physical laws take the same form in all coordinate charts, we must have E = E'. Therefore, E(g) = E(g'). In other words, g(x) satisfies the EFE if and only if g'(x) satisfies the EFE. Incidentally, the above assumes we have already specified that the stressenergy is zero. Without that assumption, E needs to have an additional argument: the coordinate representation of the stressenergy tensor, so that E(g, T) would be the assertion that, together, g(x) and T(x) satisfy the EFE. The hole argument would fail, or at least take on a different form, because we can only conclude E(g,T)=E(g',T'). 


#13
Aug1008, 11:56 PM

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#14
Aug1008, 11:59 PM

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Yes I did. (Darned cutandpaste!) It's been corrected.



#15
Aug1108, 12:08 AM

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I'm surprised Einstein forgot that he didn't add the physics when he tried to show that GC was unphysical! 


#16
Aug1108, 12:11 AM

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Just to add more....
The whole point, as I understand it, is how it dramatically emphasizes the arbitrariness in representing the physical situation. In Newtonian mechanics, translation invariance leads to the realization that absolute position is 'physically' meaningless  you can't tell the difference between having everything in the universe shifted 1 meter in a particular direction and simply having the origin of your maps shifted 1 meter in the opposite direction. In special relativity, Lorentz invariance leads to the realization that things like absolute simultaneity are 'physically' meaningless. The hole argument is just the beginnings of doing the same thing for diffeomorphism invariance  it just has much furtherreaching implications because of the sheer generality of what can be done with a diffeomorphism. Because of that generality, it took a while before physicists were able to find anything 'physically' meaningful at all! 


#17
Aug1108, 12:18 AM

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You agree that g'(x') is a solution. So just rename x' > x, it's clear that this just a trivial relabelling and g'(x) is still a solution. The real problem is that, the story goes, g(x) and g'(x) describe exactly the same physics according to GR . To me, this is the tricky part. And to prove this, people have to invoke an active transformation ( as opposed to the passive transformation between step 1 and step 2 which is just a relabelling of the points in the manifold). By active transformation, I mean here that the points in the manifold are actually moved around (while keeping the grid fixed) so that a given spacetime point changes coordinates. But here the change of coordinate is active. In the passive case, the manifold itself is not deformed, only the grid of the coordinate system is changed. It is what happens when we do that active transformation that is still obscure to me. The way I see it is the following: we have a manifold with points corresponding to spacetime events. Now, we (or at least I) tend to think as there being already a notion of distance (spatial and temporal) between the spacetime points but the hole argument discredits that and says that it's the gravitational field itself that defines distances, without a metric there is no notion of distance and no notion of time at all. So that the manifold we start with has no notion of distance and time, we can deform it at will and it does not affect anything physical because the physics just enters the game once we have a metric introduced. This is the end result of the argument, as far as I understand it. But the proof is not clear in my mind. seeing that g'(x') and g(x) have the same physics is easy, conceptually. Seeing that g'(x') and g'(x) are both mathematical solutions is straightforward, I think. Seeing that g'(x) and g'(x') have the same physical implications is nontrivial, at least to me. Again, the key point is that the proof must involved active transformations. 


#18
Aug1108, 12:24 AM

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