# Why do we need coordinate-free descriptions?

I was reading a book on differential geometry in which it said that a problem early physicists such as Einstein faced was coordinates and they realized that physics does not obey man's coordinate systems.

And why not? When I am walking from school to my house, I am walking on a 2D plane the set of $\mathbb{R} \times \mathbb{R}$ reals. The path of a plane on the sky can be characterized in 3D parameters. A point on a ball rotates in spherical coordinates. A current flows through an inductor via cylindrical coordinates.

Why do we need coordinate-free description in the first place? What things that exist can be better described if we didn't have a coordinate system to describe it?

• Since when is the earth you walk on a 2D plane? sure, you can approximate it as such for small distances, but it starts to deteriorate rather fast. Your last example, the current flow in an inductor might be a good example of where it makes sense to talk about the current flow without the specific coordinates of the inductor, depending on your use case, i.e. an EE might not care if it's a toroid or a coil, and simply treat it in terms of it's inductance. – user2813274 Jun 12 '15 at 11:43
• @FemaleTank: even when I am a mathematician, I have never spotted a real number or and ordered pair of real numbers nowhere near my house. And I wouldn't even know how to get on top of $\mathbb R\times\mathbb R$ to walk on it. – Martin Argerami Jun 12 '15 at 16:27
• The advantage of having coordinate-free description is, well, that they are coordinate-free (as coordinates depend on the observers). – gented Dec 2 '16 at 14:25

That's a very good question. While it may seem "natural" that the world is ordered like a vector space (it is the order that we are accustomed to!), it's indeed a completely unnatural requirement for physics that is supposed to be built on local laws only. Why should there be a perfect long range order of space, at all? Why would space extend from here to the end of the visible universe (which is now some 40 billion light years away) as a close to trivial mathematical structure without any identifiable cause for that structure? Wherever we have similar structures, like crystals, there are causative forces that are both local (interaction between atoms) and global (thermodynamics of the ordered phase which has a lower entropy than the possible disordered phases), which are responsible for that long range order. We don't have that causation argument for space (or time), yet.

If one can't find an obvious cause (and so far we haven't), then the assumption that space "has to be ordered like it is" is not natural and all the theory that we build on that assumption is built on a kludge that stems from ignorance.

"Why do we need coordinate free in the first place?"... well, it's not clear that we do. Just because we have been using them, and with quite some success, doesn't mean that they were necessary. It only means that they were convenient for the description of the macroscopic world. That convenience does, unfortunately, stop once we are dealing with quantum theory. Integrating over all possible momentum states in QFT is an incredibly expensive and messy operation that leads to a number of trivial and not so trivial divergences that we have to fight all the time. There are a few hints from nature and theory that it may actually be a fools errand to look at nature in this highly ordered way and that trying to order microscopically causes more problems than it solves. You can listen to Nima Arkani Hamed here giving a very eloquent elaboration of the technical (not just philosophical) problems with our obsession with space-time coordinates: https://www.youtube.com/watch?v=sU0YaAVtjzE. The talk is much better in the beginning when he lays out the problems with coordinate based reasoning and then it descends into the unsolved problem of how to overcome it. If anything, this talk is a wonderful insight into the creative chaos of modern physics theory.

As a final remark I would warn you about the human mind's tendency to adopt things that it has heard from others as "perfectly normal and invented here". Somebody told you about $\mathbb R$ and you have adopted it as if it was the most natural thing in the world that an uncountable infinity of non-existing objects called "numbers" should exist and that they should magically map onto real world objects, which are quite countable and never infinite. Never do that! Not in physics and not in politics.

• 40 billion light years? what about 13.7 billons? how can we see far away than the universe size itself.. im confused. – rnrneverdies Jun 19 '15 at 11:03
• It is confusing. The universe has been expanding for 13.8 billion years and the oldest lights we can see is therefor that old, however, space is expanding everywhere, so the surface of last scattering that created this light and is the edge of the "visible universe" for us is now moving away from us faster than at the speed of light, so the metric distance in an expanding universe can be larger (much, much larger) than the age and the speed of light would suggest. See e.g. en.wikipedia.org/wiki/…. It takes a while to get it sorted out! – CuriousOne Jun 19 '15 at 17:26
• Regarding the last point, yes, it should be emphasized that $\mathbb{R}$ is just a model for space. It's probably not the best and final one, but rather simply one that is fairly easy and convenient to use for most uses. Models should never be taken more seriously than they are supposed to be. – The_Sympathizer Aug 1 '16 at 2:39

Why do we need coordinate free in the first place?

Let me tell you about a related experience I have had teaching students. When I ask them to define the scalar product then the vast majority will write down something along the lines of $$\vec{v}\cdot \vec{w} = v_x w_x + v_y w_y + v_z w_z \qquad (1)$$ i.e. a coordinate-based description. However, every once in a while a student will slip up and write something like this $$\vec{v}\cdot \vec{w} = |(v_x w_x , v_y w_y , v_z w_z)| \qquad (2)$$ (It's surprisingly common; I teach an introductory maths methods class and I see it at least once a semester...)

Of course, you can teach them to do (1) instead of (2), but explaining why (1) is 'right' and (2) is 'wrong' in the coordinate description is quite challenging. Why shouldn't we combine vectors like (2)?

What I do in this case is go into the coordinate-free description. There it is obvious that $\vec{v}\cdot\vec{w} = |v||w|\cos\theta$ is invariant under rotations, whereas you can easily check (2) is not. Since physical laws are rotationally invariant any mathematical expression of them can only use operations which preserve the rotational symmetry. Therefore (1) is fine whereas (2) is no good.

So this shows at least one advantage of coordinate-free descriptions: they make the symmetries of the equations immediately clear.

• I'm not sure I'd accept the answer for 3-vectors when the answer for all vectors is more succinct: $\vec v \cdot \vec w = \sum_i v_i w_i$ – user121330 Jun 12 '15 at 18:20

Let me first tell you that what you read is very vague. In GR the laws of physics are assumed to be independent of the observer. An observer is represented by the a reference frame the observer uses to measure physical phenomena.

There are a set of coordinate transformations that relates observables for different observers. Say for example the speed of a car as observed by a stationary observer and an observer moving at a constant velocity with respect to the stationary observer are related by the Galilean transformations(This is just an example don't read too much into this).

When we say physics is coordinate free we mean the equations of physics retain their form when we transform from one reference frame to another.

To put it in less formal manner, if there are two observers with different space time coo ordinates and they try to derive the laws of physics through a series of experiments in their respective reference frames, they should get the same laws of physics or the equations they find for the laws of physics should be the same. Say for electrodynamics both should get Maxwell's equations.

Representing the coordinate independence in an obvious form is called covariant representation( which you can check out if interested).

As you are reading differential equations the closest thing I can think of for you to understand this is to look for the coordinate free representation of the geodisc equation and try to make an arbitrary coordinate transformation and observe how the form is same.

As to why we need coordinate free description, its just because the coordinate free description correctly produces all experimental results and a more complicated coordinate dependent description is not required.

This is a great question, with three great answers, and here I am, a bit late to the party. There are two crucial things which the above answers don't seem to address, so I am going to try to give a really simple explanation at those levels.

# Locality / Manifolds

I'm going to come close to giving you one technical definition of a manifold, following Penrose's book Spinors and Space-Time, but I will try to keep this light and common-sense. Let's start by forgetting what else we know about space for a second and focus on one thing: spacetime is a set of "points". Sure, it's not a "discrete" set (it has an uncountably infinite number of points in it), but spacetime is fundamentally defined by the fact that it contains these other objects, which are "points" in spacetime, and we can't really define those points further. They're just some sort of atomic thing that we can talk about, and spacetime is a set $\mathcal S$ of those points.

## Arrows as information

The simplest form of information is to create the allowable scalar fields $\mathcal A \subseteq (\mathcal S \rightarrow \mathbb R)$, which means "$\mathcal A$ is some subset of (or equal to) the functions which take points in $\mathcal S$ to numbers in $\mathbb R$." We could call these "scalar fields" but maybe an even better thought is to steal some category theory and call them "arrows". By restricting the arrows we choose in the set $\mathcal A$ we can actually tune how fussy/persnickety we're being about our descriptions of $\mathcal S$, since they're our only tool to describe $\mathcal S$. It is nonetheless good to make some really basic assumptions. One broad assumption is about closure: suppose $f$ is a smooth function $\mathbb R^n \rightarrow \mathbb R$, then it makes sense that if $a_1, \dots a_n$ are all arrows (scalar fields in $\mathcal A$), then the scalar field $p \rightarrow f(a_1(p), \dots, a_n(p))$ should also be in $\mathcal A$. I'm going to use these smooth-functions again, so let's write $p \rightarrow f(a_1(p), \dots, a_n(p))$ as $f[a_1, \dots, a_n]$ with square-brackets, to remind ourselves that we're "lifting" the definition of $f$ from $\mathbb R^n \rightarrow \mathbb R$ to $\mathcal A^n \rightarrow \mathcal A$ in the "most obvious" way we can.

That closure axiom gives us useful operations like $+$ and $\cdot$ on arrows, "lifting" those operations from reals to scalar fields in $\mathcal A$. In fact the closure of $\mathcal A$ gives us something even deeper: the "kernel closed-set topology". Define a set $s \in \mathcal S$ as "closed" if there exists an arrow $a \in \mathcal A$ such that $\operatorname{ker} a = s$ (in other words, $s$ is the set of points that $a$ takes to the number $0$). Define a set as open if its set-complement is closed. This set of definitions is called a topology (if it meets certain axioms, which the closure axiom guarantees) and it lets you define what it means for maps to be "continuous" (which, it turns out, all arrows are). It does this with some other definitions: for example a "neighborhood" of a point is an open set which contains that point.

So if $\mathcal S$ is a 2-sphere (i.e. a surface of a 3-ball), the obvious thing is to use 3D coordinates and assign values $x$, $y$, and $z$ to the sphere, allowing any smooth functions $\mathbb R^3 \rightarrow \mathbb R$ of those three numbers to be our "arrows".

## Coordinates as arrows

And now we can finally talk about what it means to have coordinates by adding another axiom on $\mathcal A$: for every point $s \in S$ then it has a neighborhood $N_s$ and some arrows $c^s_1, ... c^s_d$ such that if two points $p, p' \in N_s$ are different ($p \ne p'$), then some coordinate is different ($c^s_i(p) \ne c^s_i(p')$ for some $i$). So the arrows can be used, locally, to tell points apart.

Notice that we can do this with 2 coordinates on the 2-sphere -- we can do it locally, but not globally. ($\theta$ is not an arrow; it has a discontinuity.) So if a point is in the Northern or Southern hemisphere we can use $x, y$ as coordinates. If it's on the equator we either need to use $x, z$ or $y, z$ as coordinates. There are continuous functions which map $x \le 0$ to $0$ and so forth, so that's all we need: the neighborhoods are hemispheres.

This gives a very technical but 100% valid reason for "why do we need coordinate-free descriptions": once you know what coordinates are, the fact that they are local to certain points in spacetime means that any coordinate-specific definition is only valid in a neighborhood of your current position. You start giving a 2D description of your world in terms of North/South and East/West, but your description fails when you start going more than 10,000 km East or West. Once you define your function well enough, it becomes an arrow and therefore "coordinate-free".

Similarly you often see people say that "it'd take forever to reach a black hole" or "time stands still at the event horizon" or similar things. We've actually known for a long time now that that's not true, and it's an error of the same sort. If you're standing still and you're far-away from the black hole, you have certain coordinates which are "natural" for describing it. It turns out that your coordinates can't pass through the event horizon of the black hole. But that doesn't mean that matter can't. In fact, we discovered that there are "co-moving coordinates", a set of coordinates used by someone who is falling naturally into the black hole, which work just fine for getting across the event horizon of a black hole. It's just that when you fall past the event horizon at your local time $t$, you stop being able to send light of yourself to the person who is far away from the black hole: so the last image that they see of you is at your time $t$, and because the light from that time needs to escape more-and-more gravity the closer you get to the horizon, images from that time take longer and longer to get there. From their perspective, then, of course it sort of "looks like" it takes you forever to reach the black hole and like everything at the event horizon "stops changing" and all that. But that's because their coordinates can't pass through it.

## Vector fields make it even more obvious

A different example: you can certainly start to define vector fields now on our $(\mathcal S, \mathcal A)$ topological-space, as derivations: functions from arrows to arrows which satisfy a "chain rule" in the sense that, if $f$ is a smooth function $\mathbb R^n \rightarrow \mathbb R$ (remember those?) where $f_(i)$ is the first derivative of $f$ with respect to its $i^{\text{th}}$ parameter, then $$\mathbf v(f[a_1,\dots,a_n]) = \sum_{i=1}^n f_{(i)}[a_1, \dots, a_n] \cdot \mathbf v(a_i).$$ Notice again that this means that $\mathbf v(a + b) = \mathbf v(a) + \mathbf v(b)$ and $\mathbf v(a \cdot b)$ = $a \cdot \mathbf v(b) + \mathbf v(a) \cdot b$. We get a lot from these smooth functions $\mathbb R^n \rightarrow \mathbb R$! (If you've never seen a vector field as a derivation before, in Euclidean space with your usual orthonormal coordinates, inner product $\langle\cdot,\cdot\rangle$ and vector fields, the derivation corresponding to field $\vec v$ is the function $\mathbf v(f) = \langle \vec v, \nabla f \rangle$.

Now, locally, you can define nonzero-everywhere vector fields on your coordinates, fitting those rules: but on the 2-sphere, there are no vector fields which are nonzero everywhere. This has the charming name of the "hairy ball theorem" and if you interpret the vector fields on the 2-sphere as wind profiles, it says that there are essentially always at least two "cyclones" (measured by counting eyes-of-storms where the wind speed is 0) on the sphere.

You wouldn't have expected this with your coordinate-based descriptions of the wind, would you?

# Generalized coordinates

It's actually not just manifolds and general relativity where you benefit from coordinate independence: much earlier in their school time we start teaching students about variational calculus and generalized coordinates. The usual way to motivate these is the Brachistochrone problem:

Consider a purely-gravity-fed friction-free roller-coaster starting at $(-L, 0)$ and ending at $(L, 0)$. Find the track $y(x)$ that the roller coaster has to take, so that the trip from start to finish takes the least amount of time.

We can see two extremes. Think of a track that goes straight down some very height $h$, followed by a sharp turn that converts it all into forward momentum at $(-L, -h)$, followed by a sharp turn that converts it all into upward momentum at $(L, -h)$. From the energy, we know that the speed at the bottom of this track is $v^2 / 2 = g h$ or $v = \sqrt{2 g h}$. The free-falls speed linearly up from $0$ to $h$, so they average half this speed, $\sqrt{g h / 2}$. So the total time for this sort of track is$$T = \frac{2 h}{\sqrt{g h / 2}} + \frac{2 L} {\sqrt{2 g h}} = \frac {4 h + 2 L} {\sqrt{2 g h}}.$$

The two extremes here are $h \rightarrow 0$ where $T \rightarrow \infty$ (no free-fall, but also not enough forward speed) and $h \rightarrow \infty$ where $T \rightarrow \infty$ also (moving infinitely fast, you cross the $2L$ distance in no time -- but you take forever to fall far enough to get there. With some calculus you can even find a minimum $T$ between those, assuming that form of the path -- but there are other, more-curvy paths to investigate between these straight lines.

The problem is that those "sharp turns" in the above example are the only constraint forces which are "nice". A constraint force has to operate in tandem with a particle's momentum, pointing that momentum into the allowed-directions. And if you don't yet know the track then you don't yet know the direction of the constraint force -- much less the magnitude! So the most important force to this problem (in fact, the only force other than gravity) is actually pretty hard to figure out even if we assume that you have a form for $y(x)$. It's hard to do this problem with classical, coordinate-based methods and forces and such.

Something similar happens for double-pendulums: the constraint force that keeps the two parts of the pendulum at the same distance is varying its direction constantly; how are you going to handle it? Well, wouldn't it be nice if you had a coordinate-independent understanding of the physics so that you could choose coordinates which enforce the constraints -- in this case $\theta_1, \theta_2$, and then do physics with those? After all, if you choose the right coordinates then you don't have to model the constraint force. And then you can get some differential equations for the double pendulum, prove that it's chaotic, and build two of them to show chaos to your classroom.

• You may want to use \mapsto instead of \rightarrow in definitions of functions like $p \mapsto f(a_1(p), \dots, a_n(p))$ instead of $p \rightarrow f(a_1(p), \dots, a_n(p))$, otherwise it can be confused with declaration of $f:\mathbb{R}^n\to\mathbb{R}$. Also, instead of \rightarrow in the function declarations, you can use \to, which seems more logical (and might be spaced more correctly, although I don't see too much difference). – Ruslan Jun 13 '15 at 19:45

Why do we need coordinate-free description in the first place?

In physics, we have coordinate-free description in the first place.

When I am walking from school to my house

The description of this consist of

• you and the school (building) departing from each other,
• you and your house meeting each other,

and, filling in a few more details,

• your feet meeting and passing particular constituents of "the pavement"; between your indication of departing from school and your indication of arriving at your house.

(Even more details might be filled in, for instance concerning what else you had observed in coincidence with any particular one of these "steps" you took;
and what the school and your house and the various distinct identifiable constituents of the pavement observed of you and of each other.)

And that's it, as far as the physical, geometric description is concerned.
There is nothing meaningful added by sprinkling the elements of the description with any tuples of real number values as coordinates.

I am walking on a 2D plane

Indeed, the identifiable constituents of the pavement (which your feet had touched, or even beyond that) may for instance determine that they were (at least to some approximation) at rest to each other, and plane to each other.

Again: if this has been measured (derived from what the relevant participants had observed of each other) then this result is not altered by assigning coordinates in any way whatsoever.

Now, of course coordinates may be assigned in order to (more or less) represent geometric results. For instance:

• you may prefer to assign to (the steps of) your walk coordinate values $t$ such that they increase monotonously with respect to the order of your steps (which you know best); or

• the identifiable constituents of the pavement may be assigned pairs of real numbers as coordinates (rather than only single numbers, or triples, etc.) because (in order to represent) the result that they were plane to each other.

Even more:
Given the distances between all constituents of the pavement,

$$d : \mathcal P \times \mathcal P \rightarrow \mathbb R,$$

any one-to-one coordinate assignment

$$c : \mathcal P \rightarrow \mathbb R^2$$

induces a particular distance representation function

$$\mathfrak g : \mathbb R^2 \times \mathbb R^2 \rightarrow \mathbb R,$$

such that for any two (not necessarily distinct) constituents $A$ and $B$

$$\mathfrak g[~c[~A~], c[~B~]~] = d[~A, B~].$$

Then you may for instance prefer such coordinate assignments for which $\mathfrak g$ is a smooth function of both of its arguments.

Deciding whether and in which sense result values may be represented by certain coordinate values (with their implicit properties concerning topology and algebra) is always subsequent to obtaining such values, and especially to deciding and commiting to how to measure them in the first place.

## protected by Qmechanic♦Jun 12 '15 at 17:32

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).