# Einstein hole argument

1. Aug 8, 2008

### ecce.monkey

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. Aug 8, 2008

### HallsofIvy

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

3. Aug 8, 2008

### nrqed

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

4. Aug 8, 2008

### Hurkyl

Staff Emeritus
4) Can be stated as:

The curve x=5 is a circle, when the distance formula is given by
$$d(P, P') = x'^2 + x^2 - 2 x x' \cos(y - y')$$​

Your 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 change-of-coordinate transformation -- your problem is that you forgot to apply that change-of-coordinate 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 stress-energy tensor, and the cosmological constant. In the hole, we have that the criterion, the stress-energy tensor, and the cosmological constant are all invariant under coordinate changes. Therefore, if we have a metric satisfying the EFE, coordinate-changing 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.

Last edited: Aug 8, 2008
5. Aug 8, 2008

### ecce.monkey

This is ingenious. But is it allowed? Distance is an observable and is always defined the same way in the x,y coord system. Or in other words, no matter how one redefines the mathematics the fact remains when I "observe" the chart of x=5 it's not a circle.

I don't know how one assumes this though. It seems very forced to say that g'(x) is a solution, using a functional form that was meant for one coordinate chart and just forcing it on another. If it was g' without any coordinate arguments I could understand this, but then the rest of the Einstein argument falls down.

6. Aug 9, 2008

### Hurkyl

Staff Emeritus
No, it's not! The distance function is added structure. In a coordinate chart, the distance function on the Euclidean plane takes the form $d(P, P') = \sqrt{(x' - x)^2 + (y' - y)^2}$ if and only if (x, y) define an orthonormal affine coordinate chart. In other coordinate charts, the form of the distance function will vary. The distance function on the Euclidean plane is invariant under orthogonal coordinate transformations, but no other sorts of transformations. Incidentally, that's why rotations, reflections, and translations are so important for doing Euclidean geometry.

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.

Last edited: Aug 9, 2008
7. Aug 9, 2008

### ecce.monkey

That's exactly what I'm saying, you have misread my example. Step 3 uses the functional form g' but with the original Euclidean arguments (x,y) (though independent of y). This, to me, is what seems to be the original argument. We find g in one coord system, rewrite it as g' in another, and then unjustifiably use g' with the first coord system.

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" physico-philosophical lesson about there being no spacetime without gravity.

8. Aug 10, 2008

### nrqed

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. Aug 10, 2008

### Hurkyl

Staff Emeritus
If your trying to do your example in Euclidean geometry, then it's no fair using an arbitrary diffeomorphism -- Euclidean structure is only preserved by Euclidean motions. And you'll notice, for example, that if you translate your solution to the circle criterion, you get another solution to the circle criterion.

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. Aug 10, 2008

### ecce.monkey

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 off-topic to abstract mathematics. In physics, distance is distance and is an observable, a circle is an observable. You cannot just re-shape 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. Aug 10, 2008

### yossell

The page

http://www.seop.leeds.ac.uk/entries/spacetime-holearg/Active_Passive.html
(by none other than Norton)

Norton agrees that the step from 3-4 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, stress-energy 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. Aug 10, 2008

### Hurkyl

Staff Emeritus
What is the problem with my answers? I cannot effectively help if you simply hide behind vague accusations of 'abstraction' and not 'thinking physics'. :tongue: You can't just ignore the mathematics -- it's by far the best language we have for discussing such concepts.

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 Cartesian-to-polar transformation is not a Euclidean motion. Your argument is analogous to complaining that the EFE changed form under a non-differentiable 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 $(x' - 4)^2 + (y' + 3)^2 = 25$), 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 stress-energy is zero. Without that assumption, E needs to have an additional argument: the coordinate representation of the stress-energy 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').

Last edited: Aug 10, 2008
13. Aug 10, 2008

### nrqed

Did you mean E'(h) in this last statement?

14. Aug 10, 2008

### Hurkyl

Staff Emeritus
Yes I did. (Darned cut-and-paste!) It's been corrected.

15. Aug 11, 2008

### ecce.monkey

I think I agree with what you're saying. Their argument seems to me on the one hand we have a mathematical form without content (in the toy example case just the differential operators themselves and the constant 0), and on the other the math with the physics added. At least in the toy example, the purely mathematical form is the thing that is GC, not the physical field law and maybe that's what Einstein meant.

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

16. Aug 11, 2008

### Hurkyl

Staff Emeritus

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 further-reaching 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. Aug 11, 2008

### nrqed

To me, the problem is not in that g'(x) is a solution. That in itself is no problem I think.
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.

Last edited: Aug 11, 2008
18. Aug 11, 2008

### nrqed

Did you mean "I iwll rewrite the whole argument? :tongue2:
But this is where I get stuck. It would seem to me that one should also use h' since we have changed the coordinate system so it would seem to me that one should write E'(h') is the assertion that h'(x') satisfies the EFE. Of course, in that case the rest of the argument leads nowhere. SO I know this part is true but it is nontrivial.

19. Aug 11, 2008

### ecce.monkey

I'm really not sure what why you are talking about Euclidean motions. The transformations involved are inter-coordinate transformations, not intra. Forget about this, I don't think we're on the same wavelength here.

But you are trying to redefine distance within the Euclidean geometry. Do you really think step 4 is a solution, does it really look like a circle? Do you picture the big long vertical line as a circle? Remember in the hole argument as given by Norton, the step 4 is a transliteration of coordinate labels, the same function is just re-written using the old coordinate system. It's just a change of label. I think that's where our lack of synchronisation lies, you think step 4 is a transformation, when it's supposed to be a transliteration. Maybe I'm completely wrong about that, or maybe Norton is, maybe that's not what Einstein meant.

The problem here is that you have hidden a step away because of your short cut notation. Expanding it out reveals the same issue as always...

E(g(x))=E'(g'(x'))
E(g(x))=E(g'(x'))
You need an extra step (the transliteration) to get to...
E(g(x))=E(g'(x)).
which is not at all evident.

Last edited: Aug 11, 2008
20. Aug 11, 2008

### Hurkyl

Staff Emeritus
Because the Euclidean group is the appropriate symmetry group for doing Euclidean geometry... not the full diffeomorphism group, which you are trying to use.

(I don't know what precisely you mean by 'inter' versus 'intra')

This part is not relevant if you are intending for the distance function to be part of the ambient structure which must be preserved. This part is only relevant if you consider the distance function to be extra structure (which I had originally assumed, because we are talking about general relativity in which the metric tensor is merely 'extra structure', and you were considering nonEuclidean transformations, which almost by definition do not leave the Euclidean metric invariant)

That said.... Ignoring the technicalities involved with the fact the polar-to-Cartesian transformation is not one-to-one, the set of points R² with the distance function I proposed in post #4 IS the Euclidean plane. And if (x, y) are the standard coordinates, then x=5 IS a Euclidean circle.

Emphatically no! I was careful to say what I meant here.
g is a rank-2 tensor field on R^4.
x is a coordinate function on a neighborhood of your manifold.
g(x) is a rank-2 tensor field on your manifold.

By definition, for a tensor field A on your manifold, the coordinate representation of A relative to a coordinate function x is the unique tensor field f on R^n with the property that f(x(P)) is1 A(P). The notation f(x) is used to mean A1; the field given by the composition of the coordinate representation f with the coordinate function x.

So, when I said E(g), that's what I really meant. The coordinate representation of the criterion 'satisfies the EFE' is a condition on coordinate tensor fields on R^4... not a condition on tensor fields on your manifold. Given the definitions I used, E(g(x)) is nonsensical.

1: Technically speaking, I'm glossing over some canonical isomorphisms here -- in particular, I'm identifying tangent vectors on R^4 with vector-valued functions, and taking advantage of the fact the tangent bundle to the manifold on our neighborhood is canonically isomorphic to the standard tangent bundle on R^4. (And these extend to the entire tensor algebra) I suppose I can avoid taking this liberty if you prefer, but I don't believe it will make things more clear.

Last edited: Aug 11, 2008
21. Aug 11, 2008

### ecce.monkey

QED

22. Aug 11, 2008

### Hurkyl

Staff Emeritus
Neither an argument from personal incredulity nor an argumentum ad hominem affect the veracity of what I have posted.

23. Aug 11, 2008

### ecce.monkey

Yes I think you and I are in agreement on the sore spot nevertheless. To me it's not a trivial relabelling because it is specifically using the old coord system as the relabel, not just any label. This leads to a bending and breaking of the physics.

Well for me I'm much more satisfied that the hole argument is daft, especially after seeing Norton's toy example spelling the logic out. However this could be a fault of Norton, and I'll keep an open mind until I see a translation of Einstein's wording itself.

24. Aug 11, 2008

### Hurkyl

Staff Emeritus
h, here, is a dummy variable representing a tensor field on R^4. The statements
E'(h') is the assertion that h'(x') satisfies the EFE​
and
E'(h) is the assertion that h(x') satisfies the EFE​
say exactly the same thing.

25. Aug 12, 2008

### atyy

I think the hole argument is:

We start with a solution of the Einstein-Hilbert field equations which specifies the metric in two regions: one with matter, and one without (the hole); let the metric in the hole be G1. Now keep the coordinates in the region with matter unchanged, but change coordinates in the hole. Now solve the equations again, and you will get a different-looking metric in the hole; let the different-looking metric in the hole be G2. Now we haven't changed coordinates in the matter, so the metric there stays the same, so that means the distribution of matter there stays the same? Does that mean that one distribution of matter is consistent with two metrics G1 and G2 in the hole, and thus with two different motions of a test particle in the hole relative to the unchanged matter outside the hole?

The hole argument wrongly says "yes" to that question.

The right answer is that there is no notion of the matter distribution staying unchanged relative to the hole, nor of the motion of a test particle in the hole relative to the matter, until the metric in the hole AND the matter distribution is specified. Even in Newtonian physics, there is no meaning of relative unless we have a metric. Eg. the metric in cartesian coordinates says to measure distances with a rigid ruler (and not with, for example, chewing gum). If we start with a first spacetime with a metric (Riemannian manifold), and a second spacetime without a metric (a bare manifold), we can put a metric on the second spacetime that makes it physically equivalent to the first by using a "diffeomorphism" to map the metric from the first spacetime onto the second spacetime. I suppose this is why Smolin says that diffeomorphisms should be moded out: "But why should we mod out by diffeomorphisms? As Einstein intuited in his famous “hole argument”, and Dirac codified, one must mod out by diffeomorphisms if one is to have deterministic evolution from initial data (http://arxiv.org/abs/hep-th/0507235)" [Broken]

The diffeomorphism that carries the metric from the first spacetime to the second is equivalent to a coordinate change on the first spacetime. Take a look at the discussion on diffeomorphisms and coordinate transformations in <http://arxiv.org/find/all/1/all:+AND+carroll+lecture/0/1/0/all/0/1>. The appendix in General Relativity by Wald is very clear about this, to the point where you wonder why anyone was ever confused. I suppose if one did a coordinate change on the first manifold but interpreted it as a coordinate change on the second, then one would be confused.

If we start with two spacetimes both already with metrics, then they are physically equivalent only if they are "isometric", eg. Ellis and Hawking in the "The Large Scale Structure of Spacetime". They say "isometric" rather than "diffeomorphic" because a diffeomorphism doesn't by definition carry the metric along, although it can be made to.

Last edited by a moderator: May 3, 2017