# Triad T, N, B and path parametrization

• I

## Main Question or Discussion Point

Hello,

In 3D, the trajectory, which is a curve, represents all the points that an object occupies during its motion. Given a certain basis (Cartesian, cylindrical, spherical, etc.), the instantaneous position of the moving object, relative to the origin, along its trajectory can always be identified by three functions that are the components of the position vector $P(t)$.
For example, in Cartesian coordinates, the objects' position is given instant by instant by the three parametric functions $$P(t) = x(t) \hat{\textbf x} +y(t) \hat{\textbf y} + z(t) \hat{\textbf z}$$ or $$P(t)=r(t) \hat{\textbf r}+\theta(t) \hat{\theta}+ z(t) \hat{\textbf z}$$ in cylindrical coordinates, etc.

But the scalar parameter can be anything, correct? The same curve/trajectory can be parametrized with the parameter being time $t$, distance travelled $s$ or any other constructed parameter. Is that correct? Are there other specific parameters that are commonly used besides time and distance travelled $s$?

When intrinsic coordinates $\textbf (T, N, B)$ are used, it is common to use the distance travelled as the scalar parameter $s$ instead of the time parameter $t$. Why? The vector $T$ is the tangential unit vector tangent to the trajectory at a specific point, $N$ is the unit vector perpendicular to the trajectory and $B$ is normal to both $T$ and $N$.

How is the position of an object along its trajectory represented using the unit basis vectors $\textbf (T, N, B)$ and their components $(a,b,c)$? For example, how would $$P(s)= a(s) \textbf T +b(s) \textbf N +c(s) \textbf B$$ ? I am not sure how the vector $P(t)$ expressed this way would identify the object in space along its trajectory...

jambaugh
Gold Member
[...]
But the scalar parameter can be anything, correct? The same curve/trajectory can be parametrized with the parameter being time $t$, distance travelled $s$ or any other constructed parameter. Is that correct? Are there other specific parameters that are commonly used besides time and distance travelled $s$?
Very often one of the coordinates are used for the parameter. You can for example imagine the graph y=f(x) to be a 2-dim parametric curve:
y= f(t), x=t, but just call t x.

When intrinsic coordinates $\textbf (T, N, B)$ are used, it is common to use the distance travelled as the scalar parameter $s$ instead of the time parameter $t$. Why?
When this is done the formulas become simpler since the first vector derivative is automatically the unit vector T plus as a constant length vector its derivative will be normal to it meaning the second derivative is already perpendicular to it and thus in the direction of N.

How is the position of an object along its trajectory represented using the unit basis vectors $\textbf (T, N, B)$ and their components $(a,b,c)$? For example, how would $$P(s)= a(s) \textbf T +b(s) \textbf N +c(s) \textbf B$$ ? I am not sure how the vector $P(t)$ expressed this way would identify the object in space along its trajectory...
This typically is not done since the basis is also dependent on the parameter. Typically either the position vector is referred to by name directly or given in a standard basis. Note that the next step in generalization is to move from flat spaces to curved manifolds and the points are no longer identified with position vectors. You still can describe vector valued derivatives and the local T N B basis is a natural one for this purpose.

• fog37
jambaugh
Gold Member
I will also add that in the generalization to manifolds, one must construct the tangent and cotangent vector spaces in which these vectors will reside. If you hear reference to the "tangent bundle" that's the fiber bundle of tangent spaces for every point on the curve manifold. Imagine a curved surface and the tangent planes at each point on the surface.

• fog37
Hello jambaugh,

I am aware that parametrization is a different topic than coordinate systems but I am still confused on the difference between the difference between a global, fixed, Cartesian coordinate system and moving, local, coordinate systems. I guess moving body-fixed coord. system are also called intrinsic...

First of all, I think it is important to remember that the concepts of frame of reference and coordinate system are different: a frame of reference is a framework from which an observer studies the physical phenomenon. A frame of reference always needs and uses a particular coordinate system or another to attribute numerical values to the points of space. The time variable $t$ is measured by a clock (assuming the time is the same for all spatial points).

Going back to moving coordinate systems like the triad $T,N,B$, what advantage does a moving, body-fixed coordinate system provide in comparison to a fixed, global coord system? Is the $T,B,N$ triad essentially a local Cartesian coordinate system glued to the moving particle?
• Does a moving coordinate system, like the triad $T,N,B$, automatically implies a moving frame of reference relative to the fixed, global coord. system which implies a reference frame that is stationary relative to the moving body?
• In 2D, is the polar coordinate system an example of a moving, local coordinate system since its radial and polar unit vectors change and are different for different spatial points the same way the vectors $T,B,N$ change for different spatial points along the particle's trajectory?
Thank you for any enlightenment.

jambaugh
Gold Member
The T,N,B triad doesn't in and of itself define a moving coordinate system but defines a spatial frame. Imagine the moving particle as moving along a fixed track placed in space. At each point on the track we can define the unit T vector pointing along the direction of the track, the unit N vector pointing perpendicular to the track in the direction toward which it curves and the B unit vector pointing in the third direction as defined by the right hand rule TxN=B. There is a distinctly defined T,N,B triad for each point along the track (excepting where it is straight for long enough that there not distinct direction of curvature). No time is passing here yet and we are defining these purely in the timeless geometry of the curve. But as you then let the particle zoom along the track you will note that its velocity when it passes each point on the track will be parallel to that point's T vector, and its non-tangential component of acceleration will be in the N direction. Let distinct particles move along the path with different varying speeds and you get different velocities and accelerations but still at each point on the curve the same triad as a function of position.

I think this should resolve for you the main questions you asked about these three vectors. They are purely geometric objects for a given spatial curve.

As to the relationship between observers and coordinate systems that's a whole chapter of a book and I just deleted a long exposition on it when I realized how long it was getting. Suffice it to say that for flat space or flat space-time we can define a linear observer or inertial observer respectively by giving a linear coordinate system, namely by defining a set of coordinates which are the coefficients of the basis representation of a displacement vector. The basis defines the observer's frame and with it his coordinate system.

For spatially non-linear observer frames and in space-time non-inertial observer frames we consider a curvilinear coordinate system for flat space/space-time and these are the only type of coordinate systems possible in curved space or space-time. It is by comparing how the same set of observations compare in different observers frames that we formulate our understanding of local spatial or space-time geometry and dynamics.

That's a book in and of itself but let me finish on how we define local linear coordinate frames relative to general curvilinear coordinates when we have a flat geometry. Start with some standard orthonormal basis [ $\mathbf{i},\mathbf{j}, \mathbf{k}$ ] and corresponding rectilinear coordinate system $(x,y,z)$. Define your gradient of a scalar field, $\nabla f(x,y,z) = f_x\mathbf{i}+f_y\mathbf{j}+f_z\mathbf{k}$. Notice that the coordinates themselves can be viewed as scalar fields e.g. $f(x,y,z) = x$ and as such their gradients defines the standard basis:
$$\nabla x = \mathbf{i}, \quad \nabla y = \mathbf{j}, \quad \nabla z = \mathbf{k}$$
For any change of coordinates we have the gradient chain rule:
$$\nabla f(u,v,w) = f_u\nabla u + f_v \nabla v + f_w\nabla w$$
Except at coordinate singularities the gradients of our coordinates at a given point define a basis at that point and with them we get a local linear coordinate system (local to that point treated as our new origin). This is called the coordinate basis, and we can use it to define vector fields in the coordinate basis. This is in point of fact the modern definition of coordinate differentials.

$$d\mathbf{r} = dx\mathbf{i}+dy\mathbf{j}+dz\mathbf{k} = du\nabla u + dv\nabla v + dw\nabla w$$

They are not "infinitesimals" they are auxiliary variables to express the position of a second point, relative to the first in this local linear coordinate system centered at the first and tangent to our general coordinates at the first point. Naturally the basis is position dependent. As an example, for polar coordinates in the plane we have:
$$\nabla r = \frac{x}{r}\mathbf{i} +\frac{y}{r}\mathbf{j} =\frac{1}{r} \mathbf{r}=\hat{\mathbf{r}},\quad \nabla \theta = \frac{-y}{r^2} \mathbf{i} + \frac{x}{r^2}\mathbf{j} = \frac{1}{r}\hat{\theta}$$
Note that the coordinate basis for polar coordinates is orthogonal but not ortho-normal as the gradient of $\theta$ is not a unit vector. When we normalize it we no longer are using the coordinate basis and thus expressing the gradient gets a little more complicated.

But if you want to express the gradient, divergence and curl in polar or spherical coordinates then start with the corresponding coordinate basis and work out the cross products and dot products between components. You will also need to work out the curl and divergence of the standard basis vectors as well since they too are vector fields given they depend on position. That's a fun and very useful exercise.

(Note that by choosing the coordinate basis all the curls will be zero since they are curls of gradients! You'll then want to use the appropriate vector calculus product rule to work out the curls of the normalized versions.)

• fog37
Hello jambaugh. Thanks again. I was reflecting on the Frenet triad...still a work on progress for me :)

Let me try to connect things: I agree that a curve $C$ in space is a geometric entity with its local tangent and normal vectors at every point on the curve (from the derivatives of the tangent and normal vectors we get the curvature and torsion of the curve).

Then we introduce kinematics and motion with a particle traveling along that specified curve $C$. The particle's instantaneous position along the curve can be parametrized as $r(t) = [x(t, y(t), z(t)]$ using the time $t$ parametrization. Given $r(t)$, we can get $v(t)$ and $a(t)$ at every instant of time $t$.
The particle can travel along the same curve with a different instantaneous speed and acceleration (it depends on the parametric equations $x(t), y(t),z(t)$. Same curve $C$ but different motions.

Another approach stems from the fact that a geometric curve $C$ has infinite parametrizations. One other possible parametrization is the arc-length parametrization using the scalar $s$. So the particle's position can be expressed as $r(s)$, i.e. as function of the traveled distance along the curve. There is a one-one relation linking the parameters $t$ and $s$ since the particle travels a unique distance $s$ along the curve at time $t$ and vice versa: $t=f(s)$ or $s= f^-{1} (t)$. Is that correct?

Back to the triad $T,N,B$ now. For simplicity, let's discuss planar motion. The acceleration of the particle along the trajectory $C$ could be written as $$a(t) = v (t) T(t) + k(t) [v(t)]^2 N(t)$$ where $T$ and $N$ are unit vectors which change direction along the curve and can be parametrized using the arc length $s$...

What about expressing the acceleration as $$a(s) = v (s) T(s) + k(s) [v(s)]^2 N(s)$$?

Does $$a(t) = v (t) T(s) + k(t) [v(t)]^2 N(s)$$ which mixes the two parametrizations, make any sense?

Thank you!

Homework Helper
Gold Member
• fog37

I notice that at #5, "An item of interest: I saw the first of these equations in a calculus textbook by Purcell where they write velocity →v=(ds/dt)^Tv→=(ds/dt)T^ and ^T=cosϕ^i+sinϕ^jT^=cos⁡ϕi^+sin⁡ϕj^. "

You mention that the velocity vector is written as ds/dt (speed) times the unit tangent vector. Which parametrization being used there? The time $t$ or the arc length $s$ parametrization?

I understand that the word/concept "speed" means distance over time. For example, if the position vector is given as a function of arc length $s$, i.e. $r(s)$, then the derivative w.r.t to $s$ is the vector $T= \frac {dr(s)}{ds}$ which is not really a speed (since it is not a time rate of change). Nonetheless, the vector $T$ is a unit vector that is locally tangent to the curve because the vector difference $ds = r(s+\Delta s) - r(s)$ is tangent to the curve...

Homework Helper
Gold Member
You mention that the velocity vector is written as ds/dt (speed) times the unit tangent vector. Which parametrization being used there? The time ttt or the arc length sss parametrization?
With this formalism, the parametrization can be viewed as one of $s$ or $t$. In the case of the Insights article https://www.physicsforums.com/insights/frenet-equations-2-d-result-cornu-spiral/ with the Cornu spiral, $s=v_o t$, with $v_o$ constant. $\\$ One additional item: Velocity vector $\vec{v}=\frac{d \vec{r}}{dt}=(\frac{d \vec{r}}{ds})(\frac{ds}{dt})=(\frac{ds}{dt}) \hat{T}$.

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jambaugh
Gold Member
[...]
What about expressing the acceleration as $$a(s) = v (s) T(s) + k(s) [v(s)]^2 N(s)$$?

Does $$a(t) = v (t) T(s) + k(t) [v(t)]^2 N(s)$$ which mixes the two parametrizations, make any sense?
Your conceptualization is correct so far as I understood you. As to your expressions of acceleration... hmmm..

For $t$ parameterization as "time" (giving meaning to "velocity" and "acceleration")...
$$\vec{v}(t) = \dot{\vec{r}}(t) = v(t)\mathbf{T}(t)$$
with $v=\lVert \dot{\vec{r}}\rVert$.
$$a(t) = \ddot{\vec{r}}(t) = \dot{\vec{v}}(t) = \dot{v}(t)\mathbf{T} + v(t)\dot{\mathbf{T}}$$
To express in terms of curvature with dot still being the (arbitrary) "time" parameter deriv. and $v(t) = \frac{ds}{dt}$:
$$\frac{\dot{\mathbf{T}}}{v(t)} \frac{d}{ds}\mathbf{T} = \kappa(s)\mathbf{N}$$
Which gives us:
$$\vec{a}(t) = \dot{v}(t)\mathbf{T} + v^2(t)\kappa(t)\mathbf{N}$$
So you've lost a t-derivative on your tangent component but otherwise you're formula is correct.

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jambaugh
Gold Member
[...] which mixes the two parametrizations, make any sense?
Mathematically it's somewhat bad form to switch independent variables using the same function names... or more to the point to use the same name for variable and function. Such as: $x= x(t) = x(s)$ where $s=s(t)$ or something similar. The critical issue is in the distinction of the various derivatives so playing the physicist you thumb your nose at the mathematicians who insist distinct functions have distinct names BUT you be very careful to distinguish the derivatives. Using Leibniz notation emphasizes variables over functions and you're fine. But using Newtons notation, it is best to use distinct "primes" for distinct independent variables. So use $\dot{x}=dx/ds$ and $x'=dx/dt$... in short be very careful what is the independent varable when you are taking derivatives and you are ok. (Particularly when there is a single independent variable in any one context!)

It does get much trickier in multivariable calculus so it is best to be more the mathematician and distinguish function names and variable names when you invoke partial derivatives and gradients and such unless one is never ever changing the set of independent variables.

[edit: appendix] You might note this issue being very sticky when doing thermodynamics of gasses when P,V, S,T an maybe even N or rho are variables some two or three of which are independent in a given context. Specifically look at the different versions of "Free Energy" in this context.

I think this should read $\frac{\dot{\hat{T}}}{v(t)}=\frac{d \hat{T}}{ds} = \kappa(s) \hat{N}$ .