Cartesian coordinates

Cartesian coordinates are a system of coordinates defined by an orthonormal Basis where any given point in space is determined by a set of real numbers, called coordinates, that represent the distance between the point and a basis vector.

Motion

Planar motion (2D)

Motion of a particle in two dimensional Cartesian coordinates can be represented like

r(t)=x(t)i^+y(t)j^\mathbf{r}(t)=x(t)\hat{i}+y(t)\hat{j}

with i^\hat{i} and j^\hat{j} being the basis vectors (1,0)(1,0) and (0,1)(0,1). The velocity is

v(t)=r˙=x˙i^+y˙j^\mathbf{v}(t)=\dot{\mathbf{r}}=\dot{x}\hat{i}+\dot{y}\hat{j}

and acceleration

a(t)=v˙=r¨=x¨i^+y¨j^\mathbf{a}(t)=\dot{v}=\ddot{r}=\ddot{x}\hat{i}+\ddot{y}\hat{j}

Given a trajectory curve s(t)s(t), a tangent unit vector an be defined for each point as

T(t)=v∣v∣=(cos⁡(ϕ(t)), sin⁡(ϕ(t)))\mathbf{T}(t)=\frac{\mathbf{v}}{|\mathbf{v}|}=(\cos(\phi(t)), \ \sin(\phi(t)))

and also a normal vector by differentiating each component

N(t)=(−sin⁡(ϕ(t)), cos⁡(ϕ(t)))\mathbf{N}(t)=(-\sin(\phi(t)),\ \cos(\phi(t)))

Since differentiating the components of a unit vector is equivalent to rotating it 90° counterclockwise, the normal vector is π/2\pi/2 from the tangent vector, as expected1. The tangent and normal vectors form a Cartesian basis that moves with the particle, with its origin determined by the origin vector r(t)\mathbf{r}(t). These three vectors form a Moving frame

{r(t),T(t),N(t)}\{\mathbf{r}(t), \mathbf{T}(t),\mathbf{N}(t)\}

center

Velocity and acceleration can be expressed in the moving frame as

v=∣v∣T=s˙T,a=v˙=s¨T+s˙2κN\mathbf{v}=|\mathbf{v}|\mathbf{T}=\dot{s} \mathbf{T}, \quad \mathbf{a}=\dot{\mathbf{v}}=\ddot{s}\mathbf{T}+\dot{s}^{2} \kappa \mathbf{N}

where κ\kappa is the curvature of the curve at arc length ss. The first component in the acceleration is called tangential acceleration and the second centripetal acceleration.

These are valid:

dTds=dϕds(−sin⁡ϕ,cos⁡ϕ)=κN(s),dNds=dϕds(−cos⁡ϕ,−sin⁡ϕ)=−κT\frac{d\mathbf{T}}{ds}=\frac{d\phi}{ds}(-\sin\phi,\cos\phi)=\kappa \mathbf{N}(s), \quad \frac{d\mathbf{N}}{ds}=\frac{d\phi}{ds}(-\cos\phi,-\sin\phi)=-\kappa \mathbf{T}

which can be expressed in matrix notation as

(dTdsdNds)=(0κ−κ0)(TN)\begin{pmatrix} \frac{d\mathbf{T}}{ds} \\ \frac{d\mathbf{N}}{ds}\end{pmatrix}=\begin{pmatrix}0 & \kappa \\ -\kappa & 0\end{pmatrix}\begin{pmatrix}\mathbf{T} \\ \mathbf{N}\end{pmatrix}

where the curvature matrix is antisymmetric.

Spatial motion (3D)

In three dimension, the position vector becomes

r(t)=x(t)i^+y(t)j^+z(t)k^\mathbf{r}(t)=x(t)\hat{i}+y(t)\hat{j}+z(t)\hat{k}

with i^=(1,0,0)\hat{i}=(1,0,0), j^=(0,1,0)\hat{j}=(0,1,0) and k^=(0,0,1)\hat{k}=(0,0,1) as the basis vectors. The velocity then is

v(t)=r˙=x˙i^+y˙j^+zk^\mathbf{v}(t)=\dot{\mathbf{r}}=\dot{x}\hat{i}+\dot{y}\hat{j}+z\hat{k}

and acceleration

a(t)=v˙=r¨=x¨i^+y¨j^+z¨k^\mathbf{a}(t)=\dot{v}=\ddot{r}=\ddot{x}\hat{i}+\ddot{y}\hat{j}+\ddot{z}\hat{k}

Given a trajectory curve s(t)s(t), the tangent vector can be defined as

T(t)=v∣v∣\mathbf{T}(t)=\frac{\mathbf{v}}{|\mathbf{v}|}

In three dimensions, there are infinitely possible normal vectors to T\mathbf{T}, which are all the unit vector on the plane perpendicular to T\mathbf{T}. Conventionally, the normal vector is chosen by seeing that dT/dsd\mathbf{T}/ds is perpendicular to T\mathbf{T} (by virtue of being a unit vector) and the normal vector is just, given the curvature κ(s)\kappa(s),2

dTds=κ(s)N(s)\frac{d\mathbf{T}}{ds}=\kappa(s)\mathbf{N}(s)

The curvature can also be computed as

κ=σ∣v×a∣∣v∣3\kappa=\sigma \frac{|\mathbf{v}\times\mathbf{a}|}{|\mathbf{v}|^{3}}

where σ=±1\sigma=\pm1 is a sign parameter designed to keep the normal vector continuous. Thus, we can write

N=σ(v×a)×v∣v×a∣∣v∣\mathbf{N}=\frac{\sigma(\mathbf{v}\times\mathbf{a})\times\mathbf{v}}{|\mathbf{v}\times\mathbf{a}||\mathbf{v}|}

The third and final basis vector can be taken as the cross product of tangent and normal like

B=T×N\mathbf{B}=\mathbf{T}\times\mathbf{N}

called the binormal vector. These three, plus the position vector, make a moving frame:

{r(t);T(t),N(t),B(t)}\{\mathbf{r}(t);\mathbf{T}(t),\mathbf{N}(t),\mathbf{B}(t)\}

As B\mathbf{B} is perpendicular to both T\mathbf{T} and N\mathbf{N}, we can write

0=dBds⋅T+B⋅dTds=dBds⋅T+κB⋅N=dBds⋅T0=\frac{d\mathbf{B}}{ds}\cdot\mathbf{T}+\mathbf{B}\cdot \frac{d\mathbf{T}}{ds}=\frac{d\mathbf{B}}{ds}\cdot\mathbf{T}+\kappa \mathbf{B}\cdot \mathbf{N}=\frac{d\mathbf{B}}{ds}\cdot\mathbf{T}

Differentiating by ss and diving by 2 we obtain

0=B⋅dBds0=\mathbf{B}\cdot \frac{d\mathbf{B}}{ds}

which, when combined with the last equation, tells us that the derivative of B\mathbf{B} is perpendicular to both B\mathbf{B} and T\mathbf{T}. It must therefore be parallel to N\mathbf{N}, so

dBds=−τN\frac{d\mathbf{B}}{ds}=-\tau\mathbf{N}

where τ\tau is the torsion of the curve. The minus sign is conventional. We can also compute

dNds=B×dTds+dBds×T=κB×N−τN×T=−κT+τB\frac{d\mathbf{N}}{ds}=\mathbf{B}\times \frac{d\mathbf{T}}{ds}+ \frac{d\mathbf{B}}{ds}\times \mathbf{T}=\kappa\mathbf{B}\times\mathbf{N}-\tau\mathbf{N}\times\mathbf{T}=-\kappa\mathbf{T}+\tau\mathbf{B}

The three derivatives can be put into a matrix:

(T˙N˙B˙)=(0κ0−κ0τ0−τ0)(T(s)N(s)B(s))\begin{pmatrix}\dot{\mathbf{T}} \\ \dot{\mathbf{N}} \\ \dot{\mathbf{B}}\end{pmatrix}=\begin{pmatrix}0 & \kappa & 0 \\ -\kappa & 0 & \tau \\ 0 & -\tau & 0\end{pmatrix}\begin{pmatrix}\mathbf{T}(s) \\ \mathbf{N}(s) \\ \mathbf{B}(s)\end{pmatrix}

which is also antisymmetric. These are the Frenet-Serret equations in matrix form.

Footnotes

  1. A rotation by −π/2-\pi/2 is completely equivalent and would end up with a clockwise-rotation system. A π/2\pi/2 rotation gives a counterclockwise-rotating system. It's convention to use the counterclockwise form. ↩

  2. Like in the 2D case, there are actually two possibilities, N\mathbf{N} and −N-\mathbf{N}. If the curve is planar, we can just pick the counterclockwise direction again. If it's nonplanar, this reasoning cannot apply and the choice should be made to keep N\mathbf{N} continuous (if possible). ↩