# A John Norton on the Hole Argument

1. Jun 8, 2016

### haushofer

Dear all,

I have a question on the hole argument as presented here by John Norton,

http://plato.stanford.edu/entries/spacetime-holearg/

in particular section 3. The hole argument, as I understand it, can be used to show that GR is a gauge theory with the metric as gauge field and general coordinate transformations (gct's) as gauge transfo's (interpreted actively). Basically, one performs a gct and the metric then transforms as

$$g_{\mu\nu} (x) \rightarrow g'_{\mu\nu} (x')$$

One can then evaluate the transformed metric $g'_{\mu\nu}$ in the old coordinate system $\{x\}$,

$$g'_{\mu\nu} (x)$$

which solves the Einstein equations if $g_{\mu\nu} (x)$ does. The difference between $g'_{\mu\nu} (x)$ and $g_{\mu\nu} (x)$ is a gauge transformation (which is a Lie derivative of the metric wrt the vector field used for the gct). My question is about the following statement of Norton:

The two different spreadings share one vital characteristic upon which the hole argument depends: the two spreadings agree completely on all invariant properties.

Here "spreadings" can be read as "$g'_{\mu\nu} (x)$ and $g_{\mu\nu} (x)$". And I guess "invariant properties" can be read as scalars, right? But scalars are invariant when you don't reset the new coordinate to the old one (the good old scalar transformation law). Scalars are not invariant under a Lie derivative. E.g., the two 'spreadings' have intervals

$$ds^2 = g_{\mu\nu}(x) dx^{\mu} dx^{\nu}$$

and

$$\tilde{ds}^2 = g_{\mu\nu}'(x) dx^{\mu} dx^{\nu}$$

respectively, which differ a gauge transformation (and hence are physically equal). Should I read "invariant" here then as "invariant up to a gauge transformation"? I understand that $ds^2$ equals
$$ds'^2 = g'_{\mu\nu}(x') dx'^{\mu} dx'^{\nu}$$

because it's a scalar, but that's not how we define the "spreadings"; we compare $g'_{\mu\nu} (x)$ and $g_{\mu\nu} (x)$ (and not $g'_{\mu\nu} (x')$ and $g_{\mu\nu} (x)$!).

Thanks for any insights!

Last edited: Jun 8, 2016
2. Jun 8, 2016

### haushofer

To be a little more specific: if an infinitesimal gauge transformation is induced by a vector field $\xi$, we have (the round brackets are symmetrization)

$$g'_{\mu\nu} (x) = g_{\mu\nu} (x) + 2 \nabla_{(\mu}\xi_{\nu)} \ ;$$

the second term is the Lie derivative of the metric wrt the vector field $\xi$. Then

$$\tilde{ds}^2 = g_{\mu\nu}'(x) dx^{\mu} dx^{\nu} = ds^2 + 2 \Bigl( \nabla_{\mu}\xi_{\nu}\Bigr) \, dx^{\mu} dx^{\nu}$$

So what exactly does Norton mean by "the spreadings agree on all the invariants"?

Last edited: Jun 8, 2016
3. Jun 9, 2016

### haushofer

*Bump*

4. Jun 10, 2016

### haushofer

Well, if somebody is still interested, I think Norton does not mean "numerically invariant", but invariant after the interpretation of the coordinates. So "invariant up to a gauge transformation". Although the arguments are the same in $g_{\mu\nu}(x)$ and $g'_{\mu\nu}(x)$ (namely, x), they don't represent the very same point. (That's the whole crux of the hole argument).

I think that could have been written more clearly.

5. Jun 10, 2016

### stevendaryl

Staff Emeritus
I don't have a lot to add, since I'm not sure I understand the hole argument, completely. At first blush, it seems completely obvious: Of course the facts about the universe everywhere except in the hole doesn't uniquely determine the metric inside the hole, because the metric is only defined up to a coordinate transformation. I think that Einstein is saying something more than that, but I'm not sure exactly what.

Okay, there is one subtle issue about general covariance, which is the distinction between "passive" and "active" coordinate changes. Covariance under passive coordinate changes is (in retrospect) obvious: If you choose a different coordinate system, then you still have the same physical situation, you just have to adjust your coordinate-dependent descriptions of that situation. Covariance under active coordinate changes is subtler. There, you're not keeping spacetime the same while adjusting the coordinates, you're actually considering a different spacetime, that is squashed or stretched or otherwise distorted compare to the original. Covariance under active transformations means that if you distort spacetime, and then adjust the metric (and all other fields) appropriately, then you can get back to a physical situation that is indistinguishable from the first. It's obvious (in retrospect) that theories can be written to be covariant under passive coordinate transformations, but I don't know whether it's obvious that theories should be covariant under active transformations (AKA diffeomorphisms). As I say, I don't completely understand it, but I think that Einstein's hole argument is showing how GR is covariant under active transformations, as well as passive.

I'm still not sure if the conclusion should be obvious, or not (which I guess means that it's not obvious).

6. Jun 10, 2016

### haushofer

I don't think so. In the argument one already assumes an active point of view. That passive and active are equivalent is shown in e.g. the appendix of Wald. The hole argument shows that that one cannot assign meaning to points on a manifold before considering the metric. Ontologically, spacetime is then the bare manifold PLUS the metric.

About your other question I have to think. As far as I can tell, one can also apply the hole argument to the Newton-Cartan formulation.

7. Jun 11, 2016

### stevendaryl

Staff Emeritus
This is getting beyond my expertise, but it seems to me that the notion of active diffeomorphisms completely resolves the hole "paradox". What Einstein was worried about was the nondeterminism: There seemed to be infinitely many solutions for $g$ inside the hole, while GR is supposed to be deterministic. But this is resolved if you note that all of the solutions are equivalent under diffeomorphisms.

That's an interesting point. I don't know if that's true or not, but it seems true.

What I thought was the case was that while all theories can be made covariant under arbitrary coordinate changes (Newton-Cartan shows how to do this for Newtonian gravity), GR was special in being invariant under active diffeomorphisms. But maybe actually every theory can be made invariant under both passive and active diffeomorphisms?

Maybe it's just a matter of coming up with the equivalent to Einstein's field equations (which Newton-Cartan does). If a theory has "fixed", non-dynamical scalar fields, vector fields or tensor fields, then that spoils the general covariance, but maybe it's always possible to get an equivalent theory in which those quantities are dynamic?

8. Jun 12, 2016

### haushofer

Yes. I'm not sure if you need the active interpretation at all, by the way.

Well, it depends on what you call dynamic. E.g., the Minkowski metric can be made 'dynamical' by replacing it by a general metric and imposing the EOM that the Riemann tensor vanishes; the only solution to that equation is the Minkowski-metric up to diffeomorphisms. That's a trivial thing to do.

9. Jun 12, 2016

### stevendaryl

Staff Emeritus
It's a little confusing to me, but I think that the active interpretation is crucial here.

A 4-D coordinate system is a smooth, invertible map between the manifold $\mathcal{M}$ and $R^4$. Now, suppose we fix the coordinate system. Then there should be no more freedom under passive coordinate changes. But GR still gives infinitely many solutions.

But I think the point of the hole argument is really that there is no such thing as "fixing" the coordinate system, because a point doesn't have any identity other than the values of fields and the metric in the neighborhood of the point.

The issue that I'm bringing up is the extent to which GR is somehow "more generally covariant" than other theories. One can always make a theory invariant under passive coordinate transformations, but I wasn't sure whether it's also true that any theory can be made invariant under active coordinate transformations. A distinction about GR that I've heard mentioned before is that GR has no non-dynamic scalar, vector or tensor fields (unlike SR, which has a non-dynamic metric tensor). But (1) I'm not sure whether it's trivial to make any theory into one with no non-dynamic fields, and (2) I'm not sure how that is related to general covariance.

10. Jun 12, 2016

### Staff: Mentor

Does it? Fixing the coordinate system should mean fixing the manifold $\mathcal{M}$ and the mapping from it to $R^4$. But doing that fixes a unique solution in GR, correct? For example, if we fix the manifold $\mathcal{M}$ to be the Schwarzschild geometry, and we fix the coordinates to be standard Schwarzschild coordinates, then we have fixed a unique solution, correct?

(Note that there is a possible counter-argument to this; see below.)

I'm not sure this means there is no such thing as fixing coordinates; it just means you need to be clear about what fixing coordinates consists of. Take the example of the Schwarzschild geometry again. Fixing the coordinates to be standard Schwarzschild coordinates means fixing what values of the coordinates correspond to what values of the geometric invariants, such as the areas of the 2-spheres, the invariant norm of the KVF that is labeled $\partial_t$ in standard Schwarzschild coordinates, the Kretzschmann curvature invariant, etc. Different coordinate charts will give different correspondences.

There is, as I said above, a possible counter-argument to this, namely that the Schwarzschild geometry has a high degree of symmetry--4 independent Killing vector fields at each point. So there will be infinite sets of points that all share the same values for all geometric invariants. So there are really an infinite number of ways of assigning Schwarzschild coordinates to this geometry, corresponding to where we pick the "origins" of each of the KVFs (for example, which hypersurface we call $t = 0$).

However, we can still imagine putting distinguishable test objects or test fields at different points in order to distinguish them--for example, have each integral curve of $\partial_t$ (at least outside the horizon) occupied by an observer carrying a clock, and using the clock readings to distinguish the different $t$ hypersurfaces. Any particular assignment of Schwarzschild coordinates to the Schwarzschild geometry, physically speaking, would have to presuppose something like this, in order to justify, for example, calling one particular hypersurface the $t = 0$ hypersurface rather than another. So I think that, when considering the hole argument, we have to imagine that we have dealt with any symmetries of the manifold in this fashion in order to make every individual point in principle distinguishable by some combination of geometric invariants and field values.

11. Jun 12, 2016

### haushofer

Well, this topic raises some questions which I have to think about, but maybe a clue is given by the question whether we could make GR less covariant, like making Newton-Cartan theory less covariant by fixing the general coordinate transformations to the Galilei transformations + (time dependent) accelerations. I think the failure to do so with GR makes GR "more covariant".

12. Jun 12, 2016

### stevendaryl

Staff Emeritus
Well, it's a little confusing, but it seems to me that there are two sources of freedom:
1. The choice of what coordinate system to use.
2. The choice of what metric to use.
These seem to be different things. Changing the coordinate system does not change the metric, it only changes the components of the metric.

The distinction might be clearer with a scalar field, rather than a tensor field. Then there is no change of components. Let's have an n-dimensional manifold, and a scalar field $\phi$ on that manifold. Suppose we have two points, $p_1$ and $p_2$ and we have two different field configurations:
1. $\phi(p_1) = A$, $\phi(p_2) = B$
2. $\tilde{\phi}(p_1) = B$, $\tilde{\phi}(p_2) = A$
To go from configuration 1 to configuration 2 is not a change of coordinates---I haven't even mentioned a coordinate system. In the above, I think of $\phi$ as a function from the manifold $\mathcal{M}$ to $R$. The change in going from 1 to 2, from $\phi$ to $\tilde{\phi}$ is an active transformation---you've actually changed the field $\phi$.

Now, we could also use coordinates: Let $X^\mu(p)$ be a function from $\mathcal{M}$ to $R^n$. Then we can describe the field configurations 1 and 2 in terms of components: Letting $X^\mu(p_1) = a^\mu$ and $X^\mu(p_2) = b^\mu$,
1. $\phi(a^\mu) = A$, $\phi(b^\mu) = B$
2. $\tilde{\phi}(a^\mu) = B$, $\tilde{\phi}(b^\mu) = A$
But these two field configurations can also be achieved by a coordinate transformation: Let $Y^\nu$ be a second coordinate system defined in terms of $X^\mu$ via: $Y^\nu = a^\mu + b^\mu - X^\mu$. Then going from 1 to 2 is just a coordinate change. It's still the case that the value of $\phi$ at point $p_1$ is $A$, it's just that $p_1$ has a different label: $b^\mu$ instead of $a^\mu$.

So conceptually, there is a difference between active and passive transformations. However, what I think might be true is that if you perform the appropriate active changes on EVERYTHING--fields, charge densities, metric, stress-energy tensor, etc., you get something that is equivalent to a passive transformation.

13. Jun 12, 2016

### Staff: Mentor

Agreed.

This can't be quite right as it stands, because $\phi$ doesn't map $R^n$ to $R$, it maps $\mathcal{M}$ to $R$. So what you actually are doing here, implicitly, is inverting the coordinate chart mapping, i.e., you are using a mapping $X^{-1}$ from $R^n$ to $\mathcal{M}$, and you are saying that we have two possible configurations:

$$1. \phi(X^{-1}(a^\mu)) = A, \phi(X^{-1}(b^\mu)) = B$$
$$2. \tilde{\phi}(X^{-1}(a^\mu)) = B, \tilde{\phi}(X^{-1}(b^\mu)) = A$$

Sure, because now you're using $Y^{-1}$ instead of $X^{-1}$ in 2. above; in other words, now you have

$$1. \phi(X^{-1}(a^\mu)) = A, \phi(X^{-1}(b^\mu)) = B$$
$$2. \tilde{\phi}(Y^{-1}(a^\mu)) = B, \tilde{\phi}(Y^{-1}(b^\mu)) = A$$

and you have $Y^{-1}(a^\mu) = X^{-1}(b^\mu)$ and vice versa. But you need that extra condition in order to call this a "passive transformation". In other words, on this view, a "passive transformation" is just an "active transformation" that happens to satisfy a particular constraint on the two coordinate charts.

Imposing the above extra condition on the two coordinate charts ensures all this; in fact, it amounts to specifying what "performing the appropriate active changes on everything" means.

14. Jun 12, 2016

### stevendaryl

Staff Emeritus
Yes, I meant that function $F$ from $R^n$ to $R$ such that $F(X^\mu(p)) = \phi(p)$. So I guess it's $(X^\mu)^{-1} \circ \phi$.

Yes.

No, switching from the use of $X^\mu$ to the use of $Y^\mu$ is purely a passive coordinate change. My point is that from the point of view of

$(X^\mu)^{-1} \circ \phi$

which is a function from $R^n$ to R, there is no difference between (1) the passive coordinate change from $X^\mu$ to $Y^\mu$, which leaves $\phi$ unchanged, and (2) the active change from $\phi$ to $\tilde{\phi}$, which changes $\phi$, but leaves the coordinates unchanged.

On the other hand, if there were other fields besides $\phi$, then you would have to perform active changes on all of them to get something that is equivalent to a passive change.

Last edited by a moderator: Jun 12, 2016
15. Jun 12, 2016

### Staff: Mentor

Only if $Y^{-1}(b^\mu) = X^{-1}(a^\mu)$. See below.

Let me restate this in more explicit functional composition notation.

You are defining a passive coordinate change, as above, to be a change from $X^\mu$ to $Y^\mu$ such that $Y^{-1}(b^\mu) = X^{-1}(a^\mu)$. So overall we are switching from a mapping $X^{-1} \circ \phi$ from $R^n$ to $R$, to a different mapping $Y^{-1} \circ \phi$ (note no tilde over $\phi$ here, so we are keeping the same mapping from $\mathcal{M}$ to $R$) from $R^n$ to $R$.

OTOH, we could keep the mapping $X^{-1}$ from $R^n$ to $\mathcal{M}$ the same, and switch to a different mapping $\tilde{\phi}$ from $\mathcal{M}$ to $R$. It seems evident that there will be some $\tilde{\phi}$ such that $X^{-1} \circ \tilde{\phi} = Y^{-1} \circ \phi$. If we compare this new mapping from $R^n$ to $R$ with the original mapping $X^\mu \circ \phi$, then we can view it two ways: as a "passive" change in coordinates from $X$ to $Y$, or as an "active" change in how we are looking at the manifold, from $\phi$ to $\tilde{\phi}$.

Let me restate this also in more explicit terms. We are now contemplating our "physical fields" as a map, not from $\mathcal{M}$ to $R$, but from $\mathcal{M}$ to some multi-dimensional space $S$, which contains enough information to specify all physical quantities of interest. (For example, a point of $S$ could contain enough information to specify the metric tensor, the electromagnetic potential, appropriate quantities for the other Standard Model fields, etc.) So an "active change" now corresponds to going from one map $\Phi$ from $\mathcal{M}$ to $S$, to a different map $\tilde{\Phi}$ from $\mathcal{M}$ to $S$. But everything else is exactly the same as above. Whether this change of maps is described as "performing active changes on multiple fields" seems to me to be a matter of terminology, not math or physics. Mathematically, all we are doing is changing the target space of the map from $R$ to $S$.

16. Jun 12, 2016

### stevendaryl

Staff Emeritus
I specifically constructed $Y^\mu$ to make that true: $Y^\mu = a^\mu + b^\mu - X^\mu$.

Exactly.

Yes. The point of switching from $R$ to $S$ is that an active change that only affects one field will result in a situation that is inconsistent with the laws of motion. If you change $\phi$ but you don't change the sources of $\phi$ and don't change the metric, then the resulting configuration will not satisfy the equations of motion.