1
$\begingroup$

A coordinate system allows one to define points in the space as tuples where at least one of these is defined as a distance function from origo (e.g. $(\color{orange}{r},\varphi)$). At least that is my understanding, given the definition on Wikipedia:

In geometry, a coordinate system is a system which uses one or more numbers, or coordinates, to uniquely determine the position of the points or other geometric elements on a manifold such as Euclidean space.

However, tuple points also allows us to define $\Delta$-vectors; the difference between two vectors. E.g. given two points $(3,2)$ and $(-1,1)$ in euclidean coordinates, we have the distance vector $\pm(4,2)$. One could argue that this represents the point $(4,2)$ (or $-(4,2)$), but it seems unnatural to apply this logic; we have only defined the points $(3,2)$ and $(1,-1)$ in our space, we don't even know if the point $(1,-1)$ exists.

To give a practical example of $\Delta$-vectors, consider the difference between the melting point of water and wolfram in Celcius $\Delta = 0℃-3422℃ = -3422℃_\Delta$. Using Kelvin instead yields $273.15 K - 3697.15K = -3422K_\Delta$. Both are outside the domain of the temperature scales, so it's meaningless (at least physically) to assign a position to these points. And though the two coordinates systems have different origos, the distance is independent of this (because they use the same scale).

It would also be convenient to say that the $\Delta$-vectors are equal, i.e. $-3422K_\Delta = -3422℃_\Delta$, even though they origin from different coordinate systems. A change of basis $(X K + 273.15K_\Delta)\frac{℃}{K} = Y ℃$ would also incorrectly shift the position of a $\Delta$-vector, so these seem to be "disconnected" from origo. So how can $Delta$-vectors be formalized and interpreted?

$\endgroup$
  • $\begingroup$ I don't quite follow: are you asking for intuition behind sums of vectors? Formally, "Delta" vectors would just be ordinary elements of the underlying vector space (which is closed under vector addition and contains an additive inverse for every vector). Because of this, an origin is not of significance to a vector space, as every element could just be translated to adjust to the new $O$ and the vector space will remain unchanged. $\endgroup$ – SystematicDisintegration Jun 14 '18 at 10:41
1
$\begingroup$

Because this has almost nothing to do with vector algebra, and everything to do with applied vector algebra, many mathematicians will be puzzled about this question.

The approach that I have found best solves this dilemma in applied vector algebra is:

There are two types of vectors: position vectors, and direction vectors.
A position vector is "fixed", and identifies a location.
A direction vector is "relative", indicates a direction and magnitude; for example the direction of an instantaneous change, and its magnitude, or for example the distance and direction from one location to another.

There is even a mathematical basis for this.

For example, let's assume we have a triangle in a 2D Cartesian coordinate system, with vertices at $\vec{p}_1 = \left[\begin{matrix}x_1\\y_1\end{matrix}\right]$, $\vec{p}_2 = \left[\begin{matrix}x_2\\y_2\end{matrix}\right]$, and $\vec{p}_3 = \left[\begin{matrix}x_3\\y_3\end{matrix}\right]$. We have a point $(u_p , v_p)$ and a direction $(u_\Delta , v_\Delta)$ in barycentric coordinates (with respect to this triangle), and we want to convert them to Cartesian coordinates, $(x_p , y_p)$ and $(x_\Delta , y_\Delta)$, respectively.

In 2D barycentric coordinates, one axis is $\vec{u} = \vec{p}_2 - \vec{p}_1$, and the other is $\vec{v} = \vec{p}_3 - \vec{p}_1$, and the locations of the vertices are $(0,0)$ for $\vec{p}_1$, $(1,0)$ for $\vec{p}_2$, and $(0,1)$ for $\vec{p}_3$. The barycenter of a triangle is always at barycentric coordinates $\left(\frac{1}{3},\frac{1}{3}\right)$; the triangle itself covers barycentric coordinates $0 \le u, v, u+v \le 1$.

The formula to convert any point from barycentric coordinates $(u_p, v_p)$ to Cartesian coordinates $(x_p, y_p)$ is $$\left\lbrace\begin{aligned} x_p &= x_1 + u_p (x_2 - x_1) + v_p (x_3 - x_1) = (1 - u_p - v_p) x_1 + u_p x_2 + v_p x_3 \\ y_p &= y_1 + u_p (y_2 - y_1) + v_p (y_3 - y_1) = (1 - u_p - v_p) y_1 + u_p y_2 + v_p y_3 \end{aligned}\right.$$ However, the formula for converting a direction from barycentric coordinates $(u_\Delta , v_\Delta)$ to Cartesian coordinates $(x_\Delta , y_\Delta)$ is $$\left\lbrace\begin{aligned} x_\Delta &= u_\Delta (x_2 - x_1) + v_\Delta (x_3 - x_1) = (- u_\Delta - v_\Delta) x_1 + u_\Delta x_2 + v_\Delta x_3 \\ y_\Delta &= u_\Delta (y_2 - y_1) + v_\Delta (y_3 - y_1) = (- u_\Delta - v_\Delta) y_1 + u_\Delta y_2 + v_\Delta y_3 \end{aligned}\right.$$ because these vectors are relative, and not anchored to any specific point. (This sense is where the terms "fixed" and "relative" come from.)

Mathematically all vectors behave the exact same way, it is only their interpretation that changes. That is, the math does not care; it is only us humans using math to describe and model phenomena, who care. That is also why the same notation is used for all vectors. And also why this only matters when applying vector algebra to describe something, or solve a problem.

In particular, when we consider the domains where an interpretation is valid (say, temperature, or (absolute) pressure), we specifically refer to the domain of the position or fixed vectors, because the interpretation of the direction/relative vectors is different. For example, it is perfectly physically valid to report that from time $t_1$ to time $t_2$, the temperature change was $-420 K$, because it is not a physical point or sample that is reported, but the change; the $-$ indicates the direction of the change, and $420 K$ the magnitude of the change. The domain for temperature changes (or relative temperatures) is therefore the entire real axis, not just the positive real axis.

$\endgroup$
  • $\begingroup$ I realized this is related to Affine spaces & Euclidean vectors. I might add an answer myself, but I thought I should mention it. $\endgroup$ – Frank Vel Jun 16 '18 at 14:41
  • $\begingroup$ @FrankVel: Do write an answer yourself, please! As far as I can tell, the definition of affine spaces is basically a mathematical way to separate points and vectors, my applied vector algebra description being an analog or simplification of it. However, I'm not a mathematician, and have never needed to delve into abstractions like affine spaces; it certainly would be, in my humble opinion, be useful/helpful to see an answer from a pure mathematical, non-applied math, viewpoint. $\endgroup$ – Nominal Animal Jun 16 '18 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.