# What is the purpose of Arc-Length Parameterization?

• I
My teacher just briefly introduced arc length parameterization and went on to frenet serret frames, without any explanation or motivation. What is the purpose of arc length parameterization? What role does it play in TNB? What is the purpose of TNB frames anyways?

jedishrfu
Mentor
From my recollection, the parametrization was for use in describing trajectory motion where you could imagine an airplane flying a path matching the curve and the TNB (tangent, normal and binormal) described a convenient frame of reference for the pilot of the plane.

You can get a better description on Wikipedia:

https://en.m.wikipedia.org/wiki/Frenet–Serret_formulas

Is it convenient mathematically speaking? Why is TNB given with respect to arc length and not time?

Homework Helper
Gold Member
Is it convenient mathematically speaking? Why is TNB given with respect to arc length and not time?
The parametrization w.r.t. arc length will tell the shape of the curve. In two dimensions ## \kappa=d \phi / ds=1/r ## where ## r ## is the instantaneous radius of curvature. ## ds/dt ## gives the speed that the path is being traversed. Even in two dimensions the TNB formulation is quite useful where it just uses T and N. I first saw the TNB formalism in 2-D in a calculus book by Purcell. It always helps when a useful application is found that uses the formalism. I can give you a problem you might find of interest in two dimensions that I came up with that I solved using the TNB. Start with object mass m that has velocity ## v_0 ## (in the x-direction) and experiences a force perpendicular to its path that increases linearly with time, so that acceleration ## \vec{a}=(bt) \, \hat{N} ## for some constant b. ## \ ## ## \ ## ## \ ## 1) Does the speed of the object change? and ## \ ## 2) Determine its path. Starting with ## \vec{v}=(ds/dt) \, \hat{T} ##, ## \ ## where ## \hat{T}=\cos(\phi) \hat{i}+\sin(\phi) \hat{j} ##, and writing out the expression for ## \vec{a}=d \vec{v} /dt ##, the TNB formalism allowed for a simple solution. In particular, ## d \hat{T}/dt=(d \hat{T}/ds)(ds/dt)=(d \hat{T}/d \phi)( d \phi /ds)(ds/dt) =(\hat{N}) \kappa (ds/dt) ## is a very useful result that I first saw in the Purcell calculus book. Thereby ## \vec{a}=(d^2 s/dt^2) \hat{T}+(ds/dt)^2 \kappa \hat{N} ##. A couple months ago, someone posted about the Frenet equations: https://www.physicsforums.com/threa...t-equations-using-the-vector-gradient.876724/

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mathwonk
Homework Helper
to add to the previous answer, in motion along a curve, acceleration comes from two sources, change in speed along the curve, and change in direction of the curve, i.e. curvature. If we parametrize by arc length, i.e. move at constant unit speed, we remove the contribution from change of speed and thus isolate out the contribution from the shape of the curve, thus allowing us to measure its curvature.

jedishrfu
Mentor
Integrating over the arc length parameter gives distance traveled. Also arc length parameter can always be replaced by a function involving time.

OK, so calculating TNB gives us a coordinate system wrt to the cur, and we can get at the TNB using only the equation of the curve. Basically, the curve at any instant in time is resolved into 3 components (TNB), and the derivatives of these components gives us the information about how the curve is changing with arc length or time? Wouldn't defining the Frenet apparatus with respect to time be easier because some arc length integrals are hard to solve?

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