# If acceleration is decomposed into the T and N directions, why can an object leave the plane?

I'm reading Thomas' Calculus, and had a question similar to this question,

why-is-there-no-b-component-of-acceleration-in-my-multivariable-calculus-class

I understand the math part, but cannot quite figure the physics out.

$$\mathbf{a} = a_T\mathbf{T} + a_N\mathbf{N}$$

Say I have an object following the curve:

• $$a_T$$: change the norm of speed
• $$a_N$$: change the direction of speed

But then it seems the result would keep the object in the $$\mathbf{T}, \mathbf{N}$$ plane since it only has force in these two directions, and the speed is in $$\mathbf{T}$$ direction.

So I am confused now, what force caused the object to go out of the plane? I kind of feel my problem is mixing the wrong physics thoughts here.

• Locally, every curve is planar and so the acceleration at any point can be expressed in those planar coordinates for that point.This frame itself does change from point to point, and measuring this change tells you about an object "leaving the plane". Torsion will tell you how much your plane is twisting and so encodes that "leaving the plane" phenomena. Indeed, torsion is basically how much the binormal vector is changing with respect to arclength. May 29 '20 at 21:54

The key fact here is that the $$\bf{T}$$ and $$\bf{N}$$ vectors are defined at an instant, and change over time according to these equations—which are simply scaled forms of the Frenet-Serret equations:

$$\frac{d\bf{T}}{dt} = \kappa v\mathbf{N}$$

$$\frac{d\bf{N}}{dt} = -\kappa v\mathbf{T} + \tau v \mathbf{B}$$

$$\frac{d\bf{B}}{dt} = -\tau v\mathbf{N}$$

At any given instant, the particle will only travel on the plane defined by the $$\mathbf{T}$$ and $$\mathbf{N}$$ vectors; but as those vectors change over time, the particle travels outside of the plane defined by the vectors as defined at that moment in time of the past.

Physically, what happens is that the effects of "non-planar" movement pop in as a result of the derivative of the acceleration, or jerk. You can see this by taking the derivative of the acceleration with time:

$$j = \dot{a_T}\mathbf{T} + a_T\frac{d\mathbf{T}}{dt} + \dot{a_N}\mathbf{N} + a_N\frac{d\mathbf{N}}{dt}$$

$$j = \dot{a_T}\mathbf{T} + a_T\kappa v\mathbf{N} + \dot{a_N}\mathbf{N} - a_N\kappa v\mathbf{T} + a_N\tau v \mathbf{B}$$

As you can see, there's a component of the jerk pointing towards the binormal direction, which means that the acceleration won't always point towards the plane generated by the instantaneous $$\bf{T}$$ and $$\bf{N}$$ vectors.