Understand Einstein Hole Argument: Norton's Expln & General Covariance

[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?

 

[HallsofIvy]

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

 

[nrqed]

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?

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

 

[Hurkyl]

How can 4) be stated?

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.

 

[ecce.monkey]

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)

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 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.

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.

 

[Hurkyl]

This is ingenious. But is it allowed? Distance is an observable and is always defined the same way in the x,y coord system.

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.

 

[ecce.monkey]

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.

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.

 

[nrqed]

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.

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

 

[Hurkyl]

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.

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.

 

[ecce.monkey]

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.

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)".

 

[yossell]

The page

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

I found helpful.

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.

 

[Hurkyl]

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.

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').

 

[nrqed]

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)

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

 

[Hurkyl]

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

 

[ecce.monkey]

The page

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

I found helpful.

Yes very helpful thankyou.

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.

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!

 

[Hurkyl]

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 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!

 

[nrqed]

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!?)

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.

 

[nrqed]

Anyways, I want to take a stab in the dark, so I will rewrite the hole argument:

Did you mean "I iwll rewrite the whole argument? :tongue2:

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)

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.

 

[ecce.monkey]

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!

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.

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.

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.

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').

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=E' only leads to...
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.

 

[Hurkyl]

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.

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')

But you are trying to redefine distance within the Euclidean geometry.

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.

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=E' only leads to...
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.

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.

 

[ecce.monkey]

And if (x, y) are the standard coordinates, then x=5 IS a Euclidean circle.

Reductio ad absurdum.
QED

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.

Reductio Ad Charlatan

 

[Hurkyl]

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

 

[ecce.monkey]

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 .

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.

 

[Hurkyl]

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.

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.

 

[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)"

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.

 

[atyy]

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

Me too! The hole argument was wrong and Kretschmann was able to correct it at that time. And GR is not the only theory of gravity consistent with SR. There was Nordstrom's theory first, and Whitehead's theory later. Einstein was just fantasizing, not thinking logically - quite the opposite of SR, where his work is perfectly logical, following from experiments. In the end, the equivalence principle is moot, since the equations of motion fall out of the field equations, which are derived by reasoning not much better than dimensional analysis. The most important consequence of GR has been cosmology, which Einstein didn't foresee at all. I'm quite disappointed with how GR was achieved, compared to SR, though I suppose clarity also came to SR only after the discoveries of Lorentz and Poincare. But Einstein and Hilbert's (with Minkowski and Grossman's help) discovery is absolutely beautiful.

 

[Hurkyl]

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.

The mathematics of the hole argument are rock-solid: given a smooth manifold and a stress-energy tensor on that manifold containing a 'hole', there are many unequal metric tensors that satisfy the Einstein field equations. It's clear that the matter distribution does not uniquely determine the path a test particle takes through the hole. I really don't see how there can be any debate about that. (at least from anyone who follows the calculations)

Your objection doesn't really make sense. The two metrics we are comparing are defined on the hole, but that's sort of irrelevant anyways because we have an (absolute!) notion of the matter distribution being unchanged: the two structures we are considering have the same underlying manifold and the same stress-energy tensor.

 

[atyy]

The mathematics of the hole argument are rock-solid: given a smooth manifold and a stress-energy tensor on that manifold containing a 'hole', there are many unequal metric tensors that satisfy the Einstein field equations. It's clear that the matter distribution does not uniquely determine the path a test particle takes through the hole. I really don't see how there can be any debate about that. (at least from anyone who follows the calculations)

Your objection doesn't really make sense. The two metrics we are comparing are defined on the hole, but that's sort of irrelevant anyways because we have an (absolute!) notion of the matter distribution being unchanged: the two structures we are considering have the same underlying manifold and the same stress-energy tensor.

Yes, I agree with you (and my post was meant to even if it wasn't written clearly enough to seem that way).

 

[nrqed]

... It's clear that the matter distribution does not uniquely determine the path a test particle takes through the hole. I really don't see how there can be any debate about that. (at least from anyone who follows the calculations)

Right.
But that's highly nontrivial and should be disturbing (at least at first!).

I guess that's the difference between a mathematician's take on this and a physicisits's take.
The mathematician says :"the maths says that and that's the end of the story. No big deal."

The physicist says "the maths says that...What does it imply? At first it seems to imply that for the same matter distribution can lead to different motions of a test particle in the hole region...but that can't be! So is there something wrong with the equations?"

This is what was bothering Einstein. You make it sound as if Einstein must have been dumb for being bothered by this when you say that the whole thing is trivial. Einstein was not dumb at all and if the whole thing bothered him that much, there was a good reason for it!

 

[nrqed]

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.

I think that I finally convinced myself that I understood the solution.
You obviosuly agree that g(x) and g'(x') describe the same physics.
Do you agree that g(x) and g'(x) are both mathematical solutions to the (empty space) Einstein's equations? Then the next step is mathematically trivial but physically highly nontrivial. These two different metrics *have* to describe the same physics.

Now, obviously, if you draw a manifold and calculate, say, a geodesic using the two different metrics, the geodesics will be different! At this point we have two choices. Either give up the whole general covariance as a fundamental principle (which Einstein reluctantly did for a while if I understand correctly) or accept what this is telling us as physically correct. And what it is telling us is that there is no actual physical meaning to the actual points of the manifold (where there is no matter)! And I am not saying there is no meaning to the coordinates, I am saying there is no meaning to the actual points! This is very profound and is the issue people in loop quantum gravity for example keep mentionning and refer to as diffeomorphism invariance.

Now, in lqg, they say that upon quantization, they say that one must sum over diffeomorphic inequivalent metrics. But to me it feels like *all* metrics are then equivalent (as long as they are related by smooth transformations) so I am not sure what they mean by this.

Does that make sense?

By the way, I think you should have been a bit more respectful to Hurkyl :shy:
He/she has been very patient. It's normal that people use different languages sometimes and both parties must be patient and respectful in trying to understand each other's point of view.

Regards,

Patrick

 

[Hurkyl]

I guess that's the difference between a mathematician's take on this and a physicisits's take.
The mathematician says :"the maths says that and that's the end of the story. No big deal."

The physicist says "the maths says that...What does it imply? ...

I've been focusing on what 'the math says' because the validity of the mathematical argument was the question raised in the opening post! (And when I made the post you quoted, I thought atyy was denying its validity)

IAnd what it is telling us is that there is no actual physical meaning to the actual points of the manifold (where there is no matter)!

Well, we already new this, didn't we? We had the same conclusion from the translation invariance of Newtonian mechanics!

If I haven't made a mistake (and am remembering knot theory correctly), what the hole argument is telling us is this:

Suppose for simplicity that we have omniescient knowledge of the outside of the hole. We probe the hole by sending a test particle through it. The only physically meaningful thing we can say is where how it entered and exited the hole. There is no other physically meaningful information!

And, I believe it's also true that if we send in lots of point particles, the only information to be gained is where and how they entered and exitted, and if any of them crossed paths. (And the # of path crossings, the sequence of crossings a test particle experiences, and that sort of thing) But that's in a fuzzier area of my knowledge of knot theory.

(The above assumes that we are considering a hole consisting of one coordinate chart -- i.e. it's diffeomorphic to R^4. Otherwise, we might get a little bit of homotopical information too)

Now, in lqg...

In fact, while thinking about this yesterday, I realized that the whole thing I just described setup vaguely resembles a spin network, and I'm now wondering if this was relevant at all in the motivation of LQG.


Incidentally, if we can send test 'strings' through the hole (or enough test particles arranged so that we can convince ourselves they approximate a string), I believe there are a handful of more interesting things to say, but I know very little about that.

 

[ecce.monkey]

I think that I finally convinced myself that I understood the solution.
You obviosuly agree that g(x) and g'(x') describe the same physics.
Do you agree that g(x) and g'(x) are both mathematical solutions to the (empty space) Einstein's equations?

No I don't agree here. They may be solutions to the same empty mathematical form of an equation, but not to some concrete physical field equation. It's trivial that g'(x) is going to solve an equation of the same form as the EFE, but it's a bit hopeful that replacing x' by x using a simple transliteration is going to solve _the_ EFE. General Covariance (as I understand it) says an equation that is GC takes the same form in different coordinate systems. This is not the same as saying that GC equations of the same form are equivalent. You must take into account the content, in this case the physics, when measuring the equivalency. Especially if you are going to draw conclusions about the physics.

I'll illustrate my belief with Norton's toy example (see the above link posted by yossell) and my toy example...

In Norton's case he simply shows that an empty differential equation has two different solutions. He just finds two different curves whose second derivatives are zero, whoopy do. There's no other content in these equations. Or if you like he ends up finding solutions to two _different_ field equations, f''=0 and F''=0. He should be showing how to get two different curves for one equation.

In my case the only way to say x=5 is a circle is to change the physics. In that example the physics is the distance. Hurkyl shows you can call x=5 a circle by redefining distance. But I didn't stipulate a circle is a curve of points that have a constant any old thing. If you want to retain the physics, or the reality of it, if it's going to have some sense, then x=5 is not a circle.

So as I said before Einstein seems to have forgotten that he didn't add the physics when he tried to show that GC is unphysical.

Then the next step is mathematically trivial but physically highly nontrivial. These two different metrics *have* to describe the same physics.

Now, obviously, if you draw a manifold and calculate, say, a geodesic using the two different metrics, the geodesics will be different! At this point we have two choices. Either give up the whole general covariance as a fundamental principle (which Einstein reluctantly did for a while if I understand correctly) or accept what this is telling us as physically correct. And what it is telling us is that there is no actual physical meaning to the actual points of the manifold (where there is no matter)! And I am not saying there is no meaning to the coordinates, I am saying there is no meaning to the actual points! This is very profound and is the issue people in loop quantum gravity for example keep mentionning and refer to as diffeomorphism invariance.

Given there are two different geodesics, but the experience of a particle remains the same, can you elaborate on how this leads to a philosophical statement about meaning of the manifold points? To me of course it means there's just something fundamentally wrong with the logic that lead to the multiple geodesics.

By the way, I think you should have been a bit more respectful to Hurkyl :shy:
He/she has been very patient. It's normal that people use different languages sometimes and both parties must be patient and respectful in trying to understand each other's point of view.

I will attempt harder to contain myself, I tried but it came out. This is a public forum however, so there are bound to be all shades of characters, including (I only suspect mind you...) ones of a never-wrong, full-of-spin nature. I think we both pointed out the same flaw in Hurkyl's hole argument (essentially that he as overloading the meaning of his apostrophes) and as far as I can see the defence was to spin the language up quite a few notches.

 

[Hurkyl]

No I don't agree here. They may be solutions to the same empty mathematical form of an equation, but not to some concrete physical field equation.

I think you misunderstand what it means to be the same form.

Here's an example of a criterion being of the same form.

Consider two tangent vectors v and w to some point P of the Euclidean plane. Let $[v]_{x,y} = (v_x, v_y)$ denote the coordinate representation of v with respect to some orthonormal rectilinear coordinate system (with P not the origin), and let $[v]_{r, \theta} = (v_r, v_\theta)$ be its coordinate representation under the corresponding polar coordinates.

The condition that v and w are orthogonal is a condition that has the same form in both coordinate systems. In the (x, y) basis, the criterion is

$$v_x w_x + v_y w_y = 0$$​

and in the $(r, \theta)$ basis, the criterion is

$$v_r w_r + v_\theta w_\theta = 0$$​

In both cases, the relation has the same form

f(a, b) = 0​

where f is the function defined by

$$f(\mathbf{a}, \mathbf{b}) = a_1 b_1 + a_2 b_2$$​

For example, in the (x,y) basis, this relation is

$$f([v]_{x,y}, [w]_{x,y}) = 0$$​

which simplifies to

$$v_x w_x + v_y w_y = 0$$​

and in the $(r, \theta)$ basis, this relation is

$$f([v]_{r,\theta}, [w]_{r,\theta}) = 0$$​

Note that the only difference between these two relations is the basis with respect to which we represent the vectors as 2-tuples.

Correspondingly, we can observe things like how the coordinate vectors (1,0) and (0,1) represent orthogonal vectors both if we consider them as representing vectors in the (x,y) basis and as representing vectors in the $(r, \theta)$ basis.


An example of something not being of the same form is the distance formula of Euclidean geometry. In the (x, y) coordinate system, the distance between two points P and Q is:

$$d(P, Q) = \sqrt{ \left( x(P) - x(Q) \right)^2 + \left( y(P) - y(Q) \right)^2 }$$​

whereas in the $(r, \theta)$ coordinate system, the distance is

$$d(P, Q) = \sqrt{ r(P)^2 + r(Q)^2 - 2 r(P) r(Q) \cos\left( \theta(P) - \theta(Q) \right) }$$​

Again, let $[P]_{x,y} = (x(P), y(P))$ be the coordinate representation of the point P in the (x, y) coordinate system. Now, we can see that does not exist a function f(a, b) satisfying both:

$f([P]_{x,y}, [Q]_{x,y})$ is the distance between P and Q​

$f([P]_{r,\theta}, [Q]_{r,\theta})$ is the distance between P and Q​

On the other hand, if we demote the distance function to just another piece of added structure, then we can also ask for $[d]_{x,y}$ -- the coordinate representation of the distance function in (x,y) coordinates. (and similarly, in the $(r, \theta)$ coordinates) Then, the distance formula is given by the form

f(g, a, b) = g(a, b)​

and we see that

$f\left( [d]_{x, y}, [P]_{x,y}, [Q]_{x,y} \right) =$ distance between P and Q​

$f\left( [d]_{r, \theta}, [P]_{r,\theta}, [Q]_{r,\theta} \right) =$ distance between P and Q​

If it helps to read the above, the three arguments to f are, in order:
. a function that takes two 2-tuples as input
. a 2-tuple
. a 2-tuple


In case you're curious just what the heck the coordinate representation of d might be, it's a function that takes two 2-tuples as arguments and is given by:

$$[d]_{x, y}(\mathbf{a}, \mathbf{b}) = \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2}$$​

$$[d]_{r, \theta}(\mathbf{a}, \mathbf{b}) = \sqrt{ a_1^2 + b_1^2 - 2 a_1 b_1 \cos\left( a_2 - b_2 \right) }$$​

Hurkyl shows you can call x=5 a circle by redefining distance.

I had originally assumed you were taking this latter point of view, because:
(1) You're trying to understand an argument involving GR, where the metric tensor is merely a field which really and truly does have a different coordinate representation in different coordinate charts -- the analogous situation in your counterargument would be a distance function whose coordinate representation is also different in different charts
(2) You invoked a coordinate change that does not leave Euclidean geometry invariant

If you really and truly meant to consider the distance function of Euclidean geometry as having an absolute coordinate form, then you needed to restrict yourself to orthonormal coordinates -- your switch to polar coordinates was illegal, which is why you arrived at a problem in your counter argument.



P.S. I am avoiding tensor notation specifically to oppose the common abuse of thought that confuses a vector with its coordinate representation as a tuple of real numbers. While it is a very practical abuse of thought, it can also be a huge obstacle when you don't understand something. In that vein, I really shouldn't have used the notation $v_x$ and $v_y$ for the components of the coordinate representation of v w.r.t. (x,y), but alas, no convenient alternate notation sprung to mind.

 

[atyy]

So as I said before Einstein seems to have forgotten that he didn't add the physics when he tried to show that GC is unphysical.

Subtleties aside, it looks like everyone agrees on the hole argument (ie. metrics that are isometric represent the same physical spacetime), and the only real disagreement is whether Einstein made a clever mistake or a dumb mistake. If you think it's a clever mistake, then you say there's something to learn from it. If you think he made a dumb mistake, then there's nothing to learn from it, since the right answer was already obvious to you. It's funny to me that we say 'add the physics', when the solution is so dependent on Minkowski's geometrical formulation of special relativity, and Klein and Hilbert, both mathematicians around Einstein's time were more than aware of 'geometrical objects' that don't change with coordinates. Poincare wrote clearly and correctly that whether space is curved or not depends on whether we define light to follow straight lines. Lots of great physics, ie. connecting physical objects with mathematical objects, and being able to distinguish between mathematical convention and physical reality, came from mathematicians. But maybe it's ok to insult mathematicians - is it faint praise when Roger Penrose says that Ed Witten may be a better mathematician than Witten admits to being?

 

[ecce.monkey]

I think you misunderstand what it means to be the same form.

Here's an example of a criterion being of the same form.
...

No that is what I had in mind for form. However I am curious as to why you chose a distance formula as an example of something not retaining form...that would seem to be confusing the issue. I'm wondering if that's what you think I was saying, that the formula for distance was generally covariant? Well if so, no I was saying the definition of a circle as I gave it was (an attempt at being anyway) generally covariant.

I had originally assumed you were taking this latter point of view, because:
(1) You're trying to understand an argument involving GR, where the metric tensor is merely a field which really and truly does have a different coordinate representation in different coordinate charts -- the analogous situation in your counterargument would be a distance function whose coordinate representation is also different in different charts
(2) You invoked a coordinate change that does not leave Euclidean geometry invariant

If you really and truly meant to consider the distance function of Euclidean geometry as having an absolute coordinate form, then you needed to restrict yourself to orthonormal coordinates -- your switch to polar coordinates was illegal, which is why you arrived at a problem in your counter argument.

What do you mean distance having an absolute coordinate form? Again do you think I was saying distance was GC? Oh dear, re-reading your reply (#6) I think this is indeed what you thought I meant! Maybe, but then again you seemed to know I meant the circle definition previously. Maybe when I said we should use the same distance formula for step 3/4, when we are back in x,y, you thought we should use it all the time? I don't know, I just don't know where we misunderstand each other...