# The Physical Basis of Our Assumptions about Physical Space

Let $\mathcal{S}$ represent the set of all points in physical space. Using measuring rods and assuming our use of them does not depend on time, we can establish a one-to-one correspondence between $\mathcal{S}$ and $\mathcal{R}^3$, the latter considered merely as the vector space of all real three-tuples. This procedure involves several assumptions, perhaps the most noteworthy being the continuum hypothesis: that between each pair of points on a straight line in space (the definition of which assumes some facts about rods) lies another point and that each real number can be mapped to a point on the line. We choose an arbitrary point $\mathcal{O}$ in $\mathcal{S}$ as the origin and define $\phi_{\mathcal{O}}:\mathcal{S}\rightarrow\mathcal{R}^3$ by writing $\phi_{\mathcal{O}}(P)=\vec{r}_P$ for each $P$ in $\mathcal{S}$ and requiring that $\phi_{\mathcal{O}}(\mathcal{O})=\vec{0}$.

We define the displacement from $P$ to $Q$ in $\mathcal{S}$ to be the ordered pair of points $(P, Q)$. The "natural" progression is to define the corresponding quantity in $\mathcal{R}^3$ to be the vector $\vec{r}_P-\vec{r}_Q$; it appears, however, that there is no compelling reason for this step so we more generally define a function $\oplus:\mathcal{S}\times\mathcal{R}^3\rightarrow\mathcal{S}$ writing $\oplus(P,\vec{v})=P\oplus\vec{v}$. We assume this displacement function'' has the following properties:

a. For each choice of $P$ and $Q$ in $\mathcal{S}$ there exists a unique vector $\vec{v}$ in $\mathcal{R}^3$ such that $P\oplus\vec{v}=Q$. b. For any choice of $P$ and all choices of $\vec{u},\vec{v}$ we have $P\oplus(\vec{u}+\vec{v})=(P\oplus\vec{u})\oplus\vec{v}$.

If these properties are satisfied then $(\mathcal{S},\oplus)$ becomes an affine space.

I have several questions at this point. Is there indeed a compelling physical reason to define the displacement function in the usual way'' by $P\oplus\vec{v}=\vec{r}_P+\vec{v}$? If not, are there physical reasons leading to some particular definition of $\oplus$? Indeed, are there physical reasons for assuming the affine axioms at all?

Once this question has been resolved the issue of length, or measure of displacement, arises. If we chuck the affine axioms and assume space is well modeled by a Riemannian manifold, size of displacement is modeled by a second order covariant tensor. But this is a big leap, not only in topology but in the length concept as well. It seems more fundamental to assume that $\mathcal{S}$ is a metric space, the metric being a function $\rho :\mathcal{S}\times\mathcal{S}\rightarrow\mathcal{R}$ having the following properties:

a. $\rho(Q,P)=\rho(P,Q)$ for all $P$, $Q$

b. $\rho(P,Q)=0$ if and only if $P=Q$

c. $\rho(P,Q)\leq\rho(P,R)+\rho(R,Q)$ for all $P$, $Q$, and $R$.

It is a known though somewhat obscure fact that if $\rho$ is translation invariant, $\rho(P\oplus\vec{v},Q\oplus\vec{v})=\rho(P,Q)$, and scalable, $\rho(P,P\oplus\lambda\vec{v})=\mid\lambda\mid\rho(P,P\oplus\vec{v})$, then $\rho(\mathcal{O},P)=\mid\vec{r}_P\mid$ has all the properties of a norm. Furthermore, if this norm satisfies the parallelogram law, $\mid\vec{r}_P+\vec{r}_Q\mid^2+\mid\vec{r}_P-\vec{r}_Q\mid^2=2\mid\vec{r}_P\mid+2\mid\vec{r}_Q\mid$, then $\langle P, Q\rangle=\frac{1}{4}\left[\mid\vec{r}_P+\vec{r}_Q\mid^2-\mid\vec{r}_P-\vec{r}_Q\mid^2\right]$constitutes a well defined inner product.

Thus, we can start with more fundamental assumptions about a metric defined on the points of physical space and, by imposing additional restrictions, reach the inner product, aka the metric tensor.

Now I realize that a metric tensor on a Riemannian space is, in general, a function of the point at which it is evaluated. But I am trying to correlate the physical assumptions and/or operational definitions one makes about physical measurements to the mathematical tools which result. We are here discussing only classical Newtonian (Galillean?) space and undoubtedly the more powerful machinery of Riemannian space and metric tensors are only required when we allow spatial displacement/length measurements to become intertwined with time; however, I am trying to understand the physico-mathematical underpinnings. Can anyone help? Please do not respond with a welter of equations and associated tacit assumptions. I am looking for simple, down-to-earth explanations. I can supply the mathematical detail. Thanks.

This wasn't supposed to be an answer, but just a comment... it's just too long for that, so please bear with me. I think one can agree without too much omission in precision that the mathematical construction for Newtonian space doesn't result in anything interesting and that we have been doing physics very successfully for something like 300 years without it, which should be enough factual proof that it's not urgently needed. Parallel transport is both trivial and naive in Newtonian space/time and so one can glance over all of this without loss of functionality. This, of course, is not true any longer in spacetime with curvature.

So why metric theories? Because they work. You just saw the announcement of the observation of gravitational waves. The physically interesting part of that is not the existence of the waves themselves, those are weak field phenomena, but in the match between the observed data and the numerical gravity model for the strong-field merger of black holes. That was an untested regime that is now open for precision physics. If you put these results together with past tests of general relativity, then the ultimate outcome (if I had to bet on it) will probably be that general relativity "just works", which means that a (the most important) metric theory works. That is enough support to justify the choice.

From a wider physical perspective, though, I think the bigger question lies in the opposite direction. Why (in a mathematical sense) is the "general case" (as of today) defined by a metric space and not a more general construction? As a trivial extension one can introduce torsion and arrive at a non-trivial theory with interesting possibilities, Einstein-Cartan, which you may already know. I would not consider that as an obvious endpoint, still. One could ask the question which structures lie (in some way naturally) beyond metric spaces (with or without torsion)?

I don't have the answer to that, but it seems to me that if we want to understand spacetime as an emergent structure, then we have to let go of distance measurement in the conventional manner, i.e. the metric tensor is insufficient and there may not even be a "simple" local correspondence between elements of the tangent spaces and (infinitesimal) transport. In the "classical limit" the new structure will have to lead back to a metric theory, of course.

In the most simple case one could, for instance, consider parallel transport as defined trough expectation values rather than a differential geometry construction. I have a feeling this may not lead to a physical theory because, if I am not mistaken, loop quantum gravity, string theory and other approaches are arriving more or less independently at quantized area and volume elements, instead, so parallel transport may not even be the correct measure for the "local" properties of quantized or emergent spacetime. What should one replace parallel transport with as the fundamental mechanism that stitches local "spacelets" together into a globally defined dynamic space?

The point is that the interesting question of physics, at the moment, doesn't lie in properties of metric spaces (and their reduction to an affine Galilean space) anymore, but it lies in what's beyond.

• CuriousOne: Thanks for the comments. I do understand your stance. For a practicing physicist yours is without a doubt the most effective approach. I don't recall the exact quote, but Thorne et al in "Gravitation" say something like "That day is past when one makes a simple postulate and goes from there. Nowadays, one posits a more general theory and just dives in!" I know that is a very loose quote, but it captures tge idea, I think. But I am a retired educator who would like to see more motivation in our basic texts. – user59591 Feb 14 '16 at 13:54
• @Heaviside: That's a nice quote. From an experimental physicists perspective, of course, the motivation has to be experimental. I think this would involve more teaching of how one can test trivial looking postulates (like that spacetime is flat). It would also involve proper teaching of what physicist actually mean by properties (i.e. approximations that are "good enough") for the present situation together with a basic measure for when they may not be. That's the physical path to manifolds and back, I believe. The question is, on what level can you teach that? – CuriousOne Feb 14 '16 at 19:34

This procedure involves several assumptions, perhaps the most noteworthy being the continuum hypothesis:

Please do not use that terminology. The continuum hypothesis is a specific famous mathematical statement that is about a completely different topic.

We choose an arbitrary point $\mathcal{O}$ in $\mathcal{S}$ as the origin and define $\phi_{\mathcal{O}}:\mathcal{S}\rightarrow\mathcal{R}^3$ by writing $\phi_{\mathcal{O}}(P)=\vec{r}_P$ for each $P$ in $\mathcal{S}$ and requiring that $\phi_{\mathcal{O}}(\mathcal{O})=\vec{0}$.

This is completely vague. You are merely defining notation, for every point $P$ you are telling us that you call the vector $\phi_{\mathcal{O}}(P)$ by the name $\vec{r}_P.$ And in general you could start with a physical space or with a vector space (a model of physical space) and all you are doing is saying you have a bijection. That's essential, but not much.

We define the displacement from $P$ to $Q$ in $\mathcal{S}$ to be the ordered pair of points $(P, Q)$.

Again, this is a mathematician's defining. You are defining terminology and notation. You are choosing to denote something by $(P,Q)$ and choosing to call it a displacement. But nothing is happening. At least before when you had a function you said it was a bijection. This time you are just giving names, nothing more.

The "natural" progression is to define the corresponding quantity in $\mathcal{R}^3$ to be the vector $\vec{r}_P-\vec{r}_Q$; it appears, however, that there is no compelling reason for this step

The is no compelling reason for anything at this point: all you have done is assign names to vectors based on where your bijection sends them. And then made up some names for some notations. In particular you have never made any connection to anything physical that can be done or measured. So there is no compelling reason to do anything yet. We don't even know that the bijection you made is useful even at a topological level.

And that is fairly basic to Physics. When you measure some object with a ruler in the lab you are stating that the end point of the object lies between two markings on the ruler. This level of betweenness is essential. It is not necessarily captured by your bijection and so your bijection isn't motivated to have betweenness of vectors be related to betweenness of points.

so we more generally define a function $\oplus:\mathcal{S}\times\mathcal{R}^3\rightarrow\mathcal{S}$ writing $\oplus(P,\vec{v})=P\oplus\vec{v}$.

We assume this displacement function'' has the following properties:

a. For each choice of $P$ and $Q$ in $\mathcal{S}$ there exists a unique vector $\vec{v}$ in $\mathcal{R}^3$ such that $P\oplus\vec{v}=Q$. b. For any choice of $P$ and all choices of $\vec{u},\vec{v}$ we have $P\oplus(\vec{u}+\vec{v})=(P\oplus\vec{u})\oplus\vec{v}$.

If these properties are satisfied then $(\mathcal{S},\oplus)$ becomes an affine space.

But again, this affine structure could be based on the $\vec r_P$ which might not even respect the betweenness of physical space, so these pseudo displacements aren't necessarily related to measurements in any way. Imagine you randomly assigned a vector to every point in physical space, but did it with a bijection. It wouldn't be useful for anything. And the differences would satisfy your assumption above, but again, aren't related to any measurements.

You should notice that you have defined some notations and exhibited some bijection, but never connected anything to any measurements. So its not physics.

I have several questions at this point. Is there indeed a compelling physical reason to define the displacement function in the usual way'' by $P\oplus\vec{v}=\vec{r}_P+\vec{v}$?

If you made your bijection between points and vectors so that nearby points were sent to nearby vectors then there are physical reasons. Because then you get small displacements between nearby points. But otherwise there is no physical reason to do anything. You haven't connected any of your bijections to measurements.

If not, are there physical reasons leading to some particular definition of $\oplus$? Indeed, are there physical reasons for assuming the affine axioms at all?

There is no physical reason for assuming the affine axioms, and in general we don't even expect them to be true. If the universe is spatially like a Pac-Man game. Then there are many displacements between two points. Displacements are useful for local displacements, i.e. velocities.

Once this question has been resolved the issue of length, or measure of displacement, arises.

You have it backwards. You should have started with measurements to make sure your bijection made mathematical relations that are useful for the physical measurements.

It seems more fundamental to assume that $\mathcal{S}$ is a metric space, the metric being a function $\rho :\mathcal{S}\times\mathcal{S}\rightarrow\mathcal{R}$ having the following properties:

a. $\rho(Q,P)=\rho(P,Q)$ for all $P$, $Q$

b. $\rho(P,Q)=0$ if and only if $P=Q$

c. $\rho(P,Q)\leq\rho(P,R)+\rho(R,Q)$ for all $P$, $Q$, and $R$.

Again, a global function is fairly irrelevant to local measurements, which is what physics is based on.

It is a known though somewhat obscure fact that if $\rho$ is translation invariant, $\rho(P\oplus\vec{v},Q\oplus\vec{v})=\rho(P,Q)$, and scalable, $\rho(P,P\oplus\lambda\vec{v})=\mid\lambda\mid\rho(P,P\oplus\vec{v})$, then $\rho(\mathcal{O},P)=\mid\vec{r}_P\mid$ has all the properties of a norm.

A norm isn't a useful thing unless it relates to measurements. And there is no reason to think space is not curved and that it satisfies these axioms. You should have a topology and you should be able to have curves and talk about betweenness along a curve and even have some parameterization along a curve that can be connected to clocks and rulers (since we use clocks and rulers to make measurements). But we shouldn't start out by assuming the clocks and rulers have to do something in particular other than agree with what actual clocks and rulers do. And real ones don't follow the rules you are giving.

Thus, we can start with more fundamental assumptions about a metric defined on the points of physical space and, by imposing additional restrictions, reach the inner product, aka the metric tensor.

Keep in mind that what a physicist calls a metric and what a mathematician calls a metric are two different things.

I am trying to correlate the physical assumptions and/or operational definitions one makes about physical measurements to the mathematical tools which result.

The point is that physical curves can have their properties be measured by physical clocks and physical rulers and have a physical betweenness along the curve. And you want to make a mathematical structure that has a betweenness that corresponds and a length (of curves) that corresponds to the physical correspondence.

So it isn't about what is fundamental or what comes first. The idea is that physical space has points and curves and betweenness along the curve and distances along the curves and that these have operational ways to measure them. Then you want to make a mathematical structure that has corresponding points. Corresponding curves, and corresponding distances along those curves. So you need the mathematical structures to allow you to make models that have various testable properties and then you use the models to test specific theories.

• Well, I gather that your opinion is that my question is incompetent, irrelevant, and immaterial---as the late Erle Stanley Gardner would have Perry Mason say! And perhaps you are right. My intention was to provoke a discussion about what physical measurements imply what about the mathematics. I yearn for the modern equivalent of turn of the century works on relativity which had an observer going forth with measuring rods and clocks and using them to establish a mapping from physical space (or, later, spacetime) to ordered tuples of real numbers. Sorry if I failed to communicate that. – user59591 Feb 15 '16 at 4:42