I'm reading my notes on geometry of curves and trying to understand the intuition behind the definition of torsion. Specifically, I don't understand why $\|b\|=1$ implies that $b\perp{}b'$ and hence implieds that $b'\parallel{}n$. I understand the proceeding line though.
Any help in ways of thinking about this would be much appreciated.
On
For any parametrized, differentiable unit vector $X(s)$ we have
$X(s) \cdot X(s) = \Vert X(s) \Vert^2 = 1, \tag 1$
and if we differentiate this equation with respect to $s$ we obtain, via the product rule for derivatives,
$2X^\prime(s) \cdot X(s) = X^\prime(s) \cdot X(s) + X(s) \cdot X^\prime(s) = 0, \tag 2$
which forces
$X^\prime(s) \cdot X(s) = 0, \tag 3$
that is,
$X^\prime(s) \bot X(s). \tag 4$
Clearly this argument applies to $\mathbf b$, whence
$\mathbf b^\prime \bot \mathbf b. \tag 5$
Since, as our OP kam has shown,
$\mathbf b^\prime \bot \mathbf t, \tag 6$
the fact that we are working in $\Bbb R^3$ forces
$\mathbf b^\prime \parallel \mathbf n; \tag 7$
in greater detail, if we write
$\mathbf b^\prime = p \mathbf t + q \mathbf n + r \mathbf b, \tag 8$
generally possible since $\mathbf t$, $\mathbf n$, $\mathbf b$ form an orthonormal frame in $\Bbb R^3$, then it is clear from (5), (6) that
$p = r = 0, \tag 9$
leaving us with
$\mathbf b^\prime = q \mathbf n, \tag{10}$
and now a simple change of notation
$q = -\tau \tag{11}$
allows us to write
$\mathbf b^\prime = -\tau \mathbf n. \tag{12}$
The perhaps more conventional derivation of this equation stems from expressing $\mathbf n^\prime$ in terms of $\mathbf t$, $\mathbf n$, and $\mathbf b$; indeed, we define $\mathbf n$ via
$\mathbf t^\prime = \kappa \mathbf n, \; \Vert \mathbf n \Vert = 1 \tag{13}$
when
$\kappa = \Vert \mathbf t^\prime \Vert > 0; \tag{14}$
since
$\Vert \mathbf t \Vert = 1, \tag{15}$
as in (1)-(4) we have
$\kappa \mathbf n \cdot \mathbf t = \mathbf t^\prime \cdot \mathbf t = 0, \tag{16}$
whence in light of (14),
$\mathbf n \cdot \mathbf t = 0, \tag{17}$
from which we have, again in light of the product rule,
$\mathbf n^\prime \cdot \mathbf t + \mathbf n \cdot \mathbf t^\prime = 0, \tag{18}$
or
$\mathbf n^\prime \cdot \mathbf t = -\mathbf n \cdot \mathbf t^\prime = -\mathbf n \cdot \kappa \mathbf n = -\kappa; \tag{19}$
from (13), again as in (1)-(4),
$\mathbf n^\prime \cdot \mathbf n = 0. \tag{20}$
We have thus obained the components of $\mathbf n^\prime$ along $\mathbf t$ and $\mathbf n$; but since we are working in $\Bbb R^3$ we may need a third component to fully express $\mathbf n^\prime$; we acccomodate this prospect by defining
$\mathbf b = \mathbf t \times \mathbf n, \tag{21}$
so that
$\mathbf b \cdot \mathbf n = \mathbf b \cdot \mathbf t = 0, \tag{22}$
and in light of (17),
$\Vert \mathbf b \Vert = 1; \tag{23}$
thus $\mathbf t$, $\mathbf n$, and $\mathbf b$ form an orthonormal triad in $\Bbb R^3$ and so we may complete the description of $\mathbf n^\prime$ by taking
$\mathbf n^\prime \cdot \mathbf b = \tau, \tag{24}$
which defines the torsion of our (tacitly) given curve. Then
$\mathbf n^\prime = -\kappa \mathbf t + \tau \mathbf b, \tag{25}$
and from (22) we find
$\mathbf b^\prime \cdot \mathbf n + \mathbf b \cdot \mathbf n^\prime = 0, \tag{26}$
which via (24) yields
$\mathbf b^\prime \cdot \mathbf n = -\mathbf b \cdot \mathbf n^\prime = -\tau. \tag{27}$
Since by (1)-(6)
$\mathbf b' \cdot \mathbf b = \mathbf b^\prime \cdot \mathbf t = 0, \tag{28}$
we conclude from (27) that
$\mathbf b^\prime = -\tau \mathbf n, \tag{29}$
which in turn gives us (7).
Note Added in Edit, Friday 20 December 2019 10:06 AM PST: The arguments (1)-(4) that $X^\prime(s) \bot X(s)$, (17)-(19) that $\mathbf n^\prime \cdot \mathbf t = -\mathbf n \cdot \mathbf t^\prime$ and (24), (26)-(27) that $\mathbf b^\prime \cdot \mathbf n = -\mathbf b \cdot \mathbf n^\prime$ may in fact be succinctly subsumed under the observation that, for any two parametrized, differentiable vectors $X(s)$ and $Y(s)$ such that
$X(s) \cdot Y(s) = c, \; \text{a constant}, \tag{30}$
we have
$X^\prime(s) \cdot Y(s) = -X(s) \cdot Y^\prime(s), \tag{31}$
which follows from (30) upon differentiation using the product rule; then taking
$Y(s) = X(s) \tag{32}$
yields
$X^\prime(s) \cdot X(s) = 0, \tag{33}$
i.e.,
$X^\prime(s) \bot X(s). \tag{34}$
End of Note.
If $\|b\|=1,$ then $b\cdot b=1.$ If you differentiate this, you get $2b\cdot b'=0,$ so $b\perp b'.$ Since we also know that $b\perp t,b\perp n$, this tells us that $b'$ must be parallel to either $t$ or $n$ (we're in $\mathbb{R}^3$, so non-zero elements can only be perpendicular to two directions). Since the first part of the proof showed that $b'\perp t$, we must have that $b'$ is parallel to $n$.