In our class, we're learning that you can split up the acceleration, $\mathbf{a}$, of a particle into two convenient components, like so:

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

Where $a_T$ is the "tangential component" of acceleration, $a_N$ is the "normal component", and $\mathbf{T}$ and $\mathbf{N}$ are the unit tangent and unit normal vectors to the curve $\mathbf{r}(t)$, respectively.

But we also learned earlier about a third kind of vector, $\mathbf{B}$ - the "binormal vector" - which is orthogonal to both $\mathbf{N}$ and $\mathbf{T}$.

Why isn't the formula thus

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

Note: I know that the binormal vector $\mathbf{B}$ is not generally defined as a unit vector for the purposes of Multivariable Calculus classes. But in this instance, just assume $\mathbf{B}$ represents the unit binormal vector, and that $a_B$ represents the "binormal component" of acceleration.

I have a sneaking suspicion that the jounce, $\mathbf{j} = \mathbf{r}^{(3)}(t)$, of the particle moving along $\mathbf{r}(t)$ is, in fact, defined by

$$\mathbf{j} = j_T\mathbf{T} + j_N\mathbf{N} + j_B\mathbf{B}.$$

...since, well,

$$\mathbf{v} = \Vert\mathbf{v}\Vert\mathbf{T} = v_T\mathbf{T}$$


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

It just seems like each new order of derivative taken of $\mathbf{r}(t)$ adds to the equation a new, orthogonal component of motion. If that's the case, why??

  • 2
    $\begingroup$ Aren’t $\mathbf a$, $\mathbf T$ and $\mathbf N$ by definition coplanar? $\endgroup$
    – amd
    May 16 '18 at 1:43
  • 1
    $\begingroup$ I removed the "algebra-precalculus" tag and replaced it with "differential-geometry"; this post is mos' def' "algebra-precalculus". Nice question, though, endorsed!!! Cheers! $\endgroup$ May 16 '18 at 2:52

First, a note to your note: the way the binormal vector $\mathbf{B}$ is defined, it is automatically a unit vector (whenever it's well-defined); but for some historical reason (which I don't know) the word "unit" isn't used in its name. More specifically, $\mathbf{B}=\mathbf{T}\times\mathbf{N}$ and $\|\mathbf{B}\|=\|\mathbf{T}\|\|\mathbf{N}\|\cos(\pi/2)=1\cdot1\cdot1=1$.

Regarding your first question: in fact, we can say that $\mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}+a_B\mathbf{B}$, because the three vectors $\mathbf{T}$, $\mathbf{N}$, and $\mathbf{B}$ (when they are well-defined) form an orthonormal basis. But it turns out that $a_B=0$ always, hence the standard expression $\mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}$.

Why is $a_B=0$? I presume you've read calculations leading to this formula, so instead I'll try to describe my personal way of understanding this intuitively. The two vectors $\mathbf{v}=\mathbf{r}'(t)$ and $\mathbf{a}=\mathbf{r}''(t)$ typically span a plane (unless they are collinear or one is zero), called the osculating plane. So to describe any vector in that plane we only need two basis vectors. And $\mathbf{T}$ and $\mathbf{N}$ do just that — they form an orthonormal basis for this plane. We will need a third basis vector $\mathbf{B}$ for any vectors sticking out of this plane, such as $\mathbf{r}'''(t)$, for example, which may or may not be in the same plane.


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