7
$\begingroup$

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,

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

and

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

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??

$\endgroup$
2
  • 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.