There is a long history of debate in the philosophy of natural science concerning the epistemology of physical geometry. One venerable—if now unfashionable—position in this literature has held that the geometry of space and time is a matter of convention—that is, that geometrical facts are so radically underdetermined by possible empirical tests that we are free to postulate any geometry we like in our physical theories. Such a view, in various guises, has been defended by Poincaré (Reference Poincaré1905), Schlick (Reference Schlick1920), Carnap (Reference Carnap1922, Reference Carnap1966), Reichenbach (Reference Reichenbach1958), and Grünbaum (Reference Grünbaum1963, Reference Grünbaum1968), among others.Footnote ¹ All of these authors present the same basic argument. We may, by some process or other, come to believe that we have discovered some facts about the geometry of space and time. But alas, we could always, by postulating some heretofore unknown force or interaction, construct another physical theory, postulating different facts about the geometry of space and time, that is in principle empirically indistinguishable from the first.Footnote ²

Of course, at some abstract level of description, a thesis like this is irrefutable. But at that same level of abstractness, as has often been observed, it is also uninteresting. We can be conventionalists about geometry, perhaps, but in the same way that we could be conventionalists about anything. In this article we will take up a more focused question, inspired by the conventionality of geometry literature but closer to the ground floor of spacetime physics. The question is this. If one understands “force” in the standard way in the context of our best classical (i.e., nonquantum) theories of space and time, can one accommodate different choices of geometry by postulating some sort of “universal force field”? Surprisingly, the answer depends on the theory. In Newtonian gravitation, we will argue, there is a sense in which geometry is conventional, in precisely this way. But we will state and prove a no-go result to the effect that no analogous proposal can work in relativity theory. The upshot is that there is an interesting and perhaps tenable sense in which geometry is conventional in classical spacetimes, but in the relativistic setting the conventionalist’s position seems comparatively less appealing.Footnote ³

The strategy from here will be as follows. We will begin by discussing “forces” and “force fields” in Newtonian gravitation and relativity theory. We will then turn to an influential and unusually explicit version of the argument described above, due to Reichenbach (Reference Reichenbach1958).Footnote ⁴ Although the viability of Reichenbach’s recipe for constructing “universal force fields” is often taken for granted in the literature, we will present an example here that we take to show that the field Reichenbach defines cannot be interpreted as a force field in any standard sense.Footnote ⁵ We will then use the failure—for much simpler reasons—of an analogous proposal in the context of Newtonian gravitation to motivate a different approach to constructing universal force fields. As we will argue, this alternative approach works in the Newtonian context but does not work in relativity theory. We will conclude with some remarks on the significance of these results and a discussion of one option left open to the would-be conventionalist in relativity theory.

In what follows, the argument will turn on how one should understand terms such as “force” and “force field.” So we will now describe how we use these terms here.Footnote ⁶ By “force” we mean some physical quantity acting on a massive body (or, for present purposes, a massive point particle). In both general relativity and Newtonian gravitation, forces are represented by vectors at a point.Footnote ⁷ We assume that the total force acting on a particle at a point (computed by taking the vector sum of all of the individual forces acting at that point) must be proportional to the acceleration of the particle at that point, as in F = ma, which holds in both theories. We understand forces to give rise to acceleration, and so we expect the total force at a point to vanish just in case the acceleration vanishes. Since the acceleration of a curve at a point, as determined relative to some derivative operator, must satisfy certain properties, it follows that the vector representing total force must also satisfy certain properties. In particular, in relativity theory, the acceleration of a curve at a point is always orthogonal to the tangent vector of the curve at that point, and thus the total force on a particle at a point must always be orthogonal to the tangent vector of the particle’s worldline at that point.Footnote ⁸ Similarly, in Newtonian gravitation, the acceleration of a timelike curve must always be spacelike, and so the total force on a particle at a point must be spacelike as well.Footnote ⁹

A “force field,” meanwhile, is a field on spacetime that may give rise to forces on particles/bodies at a given point, where the force produced by a given force field may depend on factors such as the charge or velocity of a body.Footnote ¹⁰ We understand force fields to generate forces on bodies, and so there can be a force associated with a given force field at a point just in case the force field is nonvanishing at that point. (The converse need not hold: a force field may be nonvanishing at a point and yet give rise to forces for only some particles at that point.) A canonical example of a force field is the electromagnetic field in relativity theory. Fix a relativistic spacetime (M, g _ab). Then the electromagnetic field is represented by the Faraday tensor, which is an antisymmetric rank 2 tensor field F _ab on M. Given a particle of charge q, the force experienced by the particle at a point p of its worldline is given by , where is the unit tangent vector to the particle’s worldline at p. Note that since F _ab is antisymmetric, this force is always orthogonal to the worldline of the particle, because . In analogy with this case, we will focus attention on force fields represented by rank 2 (or lower) tensor fields.Footnote ¹¹

We can now turn to Reichenbach’s proposal.Footnote ¹² Suppose that the geometry of spacetime is given by a model of general relativity, (M, g _ab). Reichenbach claimed that one could equally well represent spacetime by any other (conformally equivalent) model,Footnote ¹³ , so long as one was willing to postulate a universal force field G _ab, defined by .Footnote ¹⁴ Various commentators have had the intuition that this universal force field is “funny”—that is, that it is not a “force field” in any standard sense.Footnote ¹⁵ And indeed, given the background on forces we have just presented, one can immediately identify some confusing features of Reichenbach’s proposal. For one, Reichenbach does not give a prescription for how the force field he defines gives rise to forces on particles or bodies. That is, he gives no relationship between the value of his field G _ab at a point and a vector quantity, except to say that the force field is “universal,” which we take to mean that the relationship between the force field and the force experienced by a particle at a point does not depend on features of the particle such as its charge or species. One might imagine that the relationship is assumed to be analogous to that between other force fields represented by a rank 2 tensor field, such as the electromagnetic field, and their associated forces at a point. But this does not work. Given Reichenbach’s definition, it is immediate that G _ab must be symmetric, and thus the vector can be orthogonal to at a point p for all timelike vectors at p—that is, for all vectors tangent to possible worldlines of massive particles through p—only if G _ab vanishes at p. These considerations should give one pause about the viability of the proposal. But they also make its full evaluation difficult, since it is not clear just how Reichenbach’s force is meant to work.

That said, there is a way to see that Reichenbach’s universal force field is problematic even without an account of how it relates to the force on a particle. Consider the following example. Let be Minkowski spacetime and let be the Levi-Civita derivative operator compatible with .Footnote ¹⁶ Choose a coordinate system t, x, y, z such that . Now consider a second spacetime , where for , and let be the Levi-Civita derivative operator compatible with . Then is a smooth timelike vector field on M with unit length relative to . Let be the maximal integral curve of through the point . The acceleration of this curve, relative to , is for all points on . Meanwhile, is a geodesic (up to reparameterization) of , the Levi-Civita derivative operator compatible with g _ab (see fig. 1). According to Reichenbach, it would seem to be a matter of convention whether (1) is the worldline of a free massive point particle in or (2) is the worldline of a massive point particle in , accelerating due to the universal force field . But now observe: along , the conformal factor is equal to 1—which means that along , and thus . And so, if one adopts option 2 above, one is committed to the view that the universal force field can accelerate particles even where G _ab vanishes.

Figure 1. Image of the maximal integral curve (depicted by the vertical line) passing through the point . According to , the acceleration of this curve is at every point (depicted by the arrows) even though the “force field” G _ab vanishes along the curve. Of course, can be reparameterized to be a geodesic according to the flat derivative operator .

This example shows that G _ab cannot be a force field in the standard sense (i.e., as described above), since a force field cannot vanish if the force it is meant to give rise to is nonvanishing (or, equivalently, the acceleration associated with that force is nonvanishing). It appears to follow that, whatever else may be the case about the conventionality of geometry in relativity theory, the universal force field Reichenbach defines is unacceptable.

The example is especially striking because, as we will presently argue, there is a natural sense in which classical spacetimes do support a kind of conventionalism about geometry, although the construction is quite different from what Reichenbach describes. To motivate our approach, we will begin by considering (an analog of) Reichenbach’s trade-off equation in classical spacetimes. Suppose the geometry of spacetime is given by a classical spacetime . Direct analogy with Reichenbach’s trade-off equation would have us consider classical metrics and and universal force fields F _a and G ^ab satisfying and . We might want to assume that G ^ab must be symmetric, since is assumed to be a classical spatial metric. And as in the relativistic case, we might insist that these new metrics preserve causal structure—which here would mean that the compatibility condition must be met, and that simultaneity relations between points must be preserved by the transformation, which means that and . Together, these imply that .

Given these trade-off equations, a version of Reichenbach’s proposal might go as follows: the metrics (t _a, h ^ab) are merely conventional since we could always use instead, so long as we also postulate universal forces F _a and G ^ab. One could perhaps investigate this proposal to see how changes in the classical metrics affect the associated families of compatible derivative operators, or even just to understand what the degrees of freedom are.Footnote ¹⁷ But there is an immediate sense in which this proposal is ill formed. The issue is that the metrical structure of a classical spacetime does not have a close relationship to the acceleration of curves or to the motion of bodies. Acceleration is determined relative to a choice of derivative operator, and in general there are infinitely many derivative operators compatible with any pair of classical metrics. All of these give rise to different standards of acceleration. And so it is not clear that the fields F _a and G ^ab bear any relation to the acceleration of a body. As in the relativistic example given above, this counts against interpreting them as force fields at all.

These considerations suggest that Reichenbach’s force field does not do any better in Newtonian gravitation than it does in general relativity. But it also points in the direction of a different route to conventionalism about classical spacetime geometry. The proposal above failed because acceleration is determined relative to a choice of derivative operator, not classical metrics. Could it be that the choice of derivative operator in a classical spacetime is a matter of convention, so long as the choice is appropriately accommodated by some sort of universal force field? We claim that the answer is yes.

This proposition means that one is free to choose any derivative operator one likes (compatible with the fixed classical metrics) and, by postulating a universal force field, one can recover all of the allowed trajectories of either a model of standard Newtonian gravitation or a model of geometrized Newtonian gravitation. Thus, since the derivative operator determines both the collection of geodesics—that is, nonaccelerating curves—and the curvature of spacetime, there is a sense in which both acceleration and curvature are conventional in classical spacetimes. Most important, the field G _ab makes good geometrical sense as a force field. Like the Faraday tensor, the field defined in proposition 1 is an antisymmetric, rank 2 tensor field; moreover, this field is related to the acceleration of a body in precisely the same way that the Faraday tensor is (except that all particles have the same “charge”), which means that the force generated by the field G _ab on a particle at some point is always spacelike at that point. Thus G _ab as defined in proposition 1 is not a “funny” force field at all.Footnote ¹⁹

It is interesting to note that from this perspective, geometrized Newtonian gravitation and standard Newtonian gravitation are just special cases of a much more general phenomenon. Specifically, one can always choose the derivative operator associated with a classical spacetime in such a way that the curvature satisfies the geometrized Poisson equation and the allowed trajectories of bodies are geodesics (yielding geometrized Newtonian gravitation), or one can choose the derivative operator so that the curvature vanishes—and when one makes this second choice, if other background geometrical constraints are met, the force field takes on the particularly simple form , for some scalar field that satisfies Poisson’s equation (yielding standard Newtonian gravitation). These are nontrivial facts, but they arguably indicate that some choices of derivative operator are more convenient to work with than others (because the associated G _ab fields take simple forms), and not that these choices are canonical.Footnote ²⁰

Now let us return to relativity theory. We have seen that in classical spacetimes, there is a trade-off between choice of derivative operator and a not-so-funny universal force field that does yield a kind of conventionality of geometry. Does a similar result hold in relativity? The analogous proposal would go as follows. Fix a relativistic spacetime (M, g _ab), and let be the Levi-Civita derivative operator associated with g _ab. Now consider another torsion-free derivative operator .Footnote ²¹ We know that cannot be compatible with g _ab, but we can insist that causal structure is preserved, and so we can require that there is some metric such that is compatible with .Footnote ²² The question we want to ask is this. Is there some rank 2 tensor field G _ab such that, given a curve , is a geodesic (up to reparameterization) relative to just in case its acceleration relative to is given by , where is the tangent field to with unit length relative to ? The answer is no, as can be seen from the following proposition.

Proof.— Since g _ab and are conformally equivalent, their associated derivative operators are related by , where . Moreover, given any smooth timelike curve , if is the tangent field to with unit length relative to g _ab, then is the tangent field to with unit length relative to . A brief calculation reveals that if is a geodesic relative to , then the acceleration of relative to is given by . Now suppose that a tensor field G _ab as described in the proposition existed. It would have to satisfy for every unit (relative to g _ab) vector field tangent to a geodesic (relative to ). Note in particular that G _ab must be well defined as a tensor at each point, and so this relation must hold for all unit timelike vectors at any point p, since any vector at a point can be extended to be the tangent field of a geodesic passing through that point. Pick a point p where is nonvanishing (which must exist, since we assume is nonconstant), and consider an arbitrary pair of distinct, cooriented unit (relative to g _ab) timelike vectors at that point, and . Note that there always exists some number such that is also a unit timelike vector. Then it follows that

But since G _ab is a linear map, we also have

These two expressions must be equal, which, with some rearrangement of terms, implies that

But this expression yields a contradiction, since the left-hand side is a vector with fixed orientation, independent of the choice of and , whereas the orientation of the right-hand side will vary with and , which were arbitrary. Thus G _ab cannot be a tensor at p. QED

So it would seem that we do not have the same freedom to choose between derivative operators in general relativity that we have in classical spacetimes—at least not if we want the universal force field to be represented by a rank 2 tensor field.

One might think there is a certain tension between proposition 1 and proposition 2. To put the point starkly, proposition 2 could be immediately generalized to semi-Riemannian manifolds with metrics of any signature. It shows that, in the most general setting, the relationship between two derivative operators compatible with conformally equivalent metrics can never be captured by a rank 2 tensor. And yet, proposition 1 appears to show that in the case of classical spacetimes, two derivative operators compatible with the same metrics (which are trivially conformally equivalent) can be captured by an antisymmetric rank 2 tensor. It is this freedom that allows us to accommodate different choices of derivative operator by postulating a universal force field with relatively natural properties. But why does this not yield a contradiction—that is, why is proposition 1 not a counterexample to proposition 2 (suitably generalized)?

The answer highlights an essential difference between relativistic and classical spacetime geometry. Although proposition 2 could be generalized to nondegenerate metrics of any signature, it cannot be generalized to degenerate metrics of the sort encountered in classical spacetime theory. Indeed, this is precisely the content of proposition 1. The important difference is that in relativity theory, the fundamental theorem of Riemannian geometry holds: given a metric, there is a unique torsion-free derivative operator compatible with that metric. Thus if one wants to adopt a different choice of derivative operator, one must also use a different spacetime metric. And varying the spacetime metric puts new constraints on what derivative operators may be chosen. In the case of a degenerate metrical structure, as in classical spacetimes, none of this applies. A given pair of classical metrics may be compatible with a continuum of derivative operators. A different way of putting this point is that insofar as the metric in relativity theory is determined by certain canonical (idealized) experimental tests involving, say, the trajectories of test particles and light rays, then the derivative operator and curvature of spacetime are also so determined. But in classical spacetimes, even if one could stipulate the metric structure through empirical tests, the derivative operator and curvature of spacetime would still be undetermined.Footnote ²⁴

We take the results here to settle the question posed at the beginning of the article. But as we emphasized there, the considerations we have raised do not refute conventionalism. For instance, one might argue that the senses of “force” and “force field” that we described above, which play an important role in our discussion, are too limiting, and that there is some generalized notion of force field that could save conventionalism. An especially promising option would be to argue that a force field need not be represented by a rank 2 tensor field.Footnote ²⁵ And indeed, given a relativistic spacetime (M, g _ab), a conformally equivalent metric , and their respective derivative operators, and , there is always some tensor field such that we can get a “funny force field” trade-off. Specifically, a curve will be a geodesic relative to just in case its acceleration relative to is , where is the unit (relative to ) vector field tangent to , and .Footnote ²⁶ That the field exists should be no surprise—it merely reflects the fact that the action of one derivative operator can always be expressed in terms of any other derivative operator and a rank three tensor.Footnote ²⁷ This field presents a more compelling force field than the one Reichenbach defines, for instance, since will always be proportional (in a generalized sense) to the acceleration of a body, just as one should expect. In particular, it will vanish precisely when the acceleration of the body does, which as we have seen is not the case for Reichenbach’s force field.

Ultimately, though, the attractiveness of a conventionalist thesis turns on how much one needs to postulate in order to accommodate alternative conventions. In some sense, one can be a conventionalist about anything, if one is willing to postulate enough—an evil demon, say. The considerations we have raised here should be understood in this light. From the perspective of the broader literature on the conventionality of geometry, what we have done here is to clarify the relative costs associated with conventionalism in two theories. We have shown that in the Newtonian context, one does not need to postulate very much to support a kind of conventionalism about spacetime geometry: one can accommodate any torsion-free derivative operator compatible with the classical metrics so long as one is willing to postulate a force field that acts in many ways like familiar force fields, such as the electromagnetic field. Of course, one may still resist conventionalism about classical spacetime geometry by arguing that even this is too much. But whatever else is the case, it seems that the costs of accepting conventionalism about geometry in relativity theory are higher still. As we have shown, Reichenbach’s proposal requires a very strange sense of “force/force field”; meanwhile, if one wants to maintain the standard notion of “force field,” then the universal force field one needs to postulate cannot be represented by a rank 2 tensor field. So one must posit something comparatively exotic to accommodate alternative geometries in relativity theory—which, it seems to us, makes this view less appealing.