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Given a curve $\gamma (t) \in \mathbb{R}^3$, when working out the Frenet frame my lecture notes define the unit normal $$\tau : = \frac{\gamma ' (t)}{\| \gamma'(t)\|}$$ and then the principle unit normal as $$n : = \frac{\dot\tau}{\|\dot\tau\|}.$$

Surely it would give the same answer for the normal if I just calculated the unit normal by $$ n = \frac{\gamma''(t)}{\|\gamma''(t)\|}?$$

Because differentiating $\tau$ is far more complicated, is this right?

Edit: So from the comments it would seem as if they aren't the same, and when trying to work it out as in @William's comment, I came out with a huge monster I couldn't simplify... but why (if so) aren't they the same? Surely it's just two different ways of working out the same formula; the formula for the unit normal at any particular point with respect to $t$. And there can't be two different unit normals..

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Why don't you try? Differentiate $\tau$ and compare $\tau'/\|\tau'\|$ with $\gamma''/\|\gamma''\|$. –  William Jan 13 '13 at 19:06
How do you differentiate $\tau$? –  user53076 Jan 13 '13 at 19:09
I guess that should have been your first question. Notice that $\|v(t)\|$, where $v$ is differentiable in $t$, is also a differentiable function. Assuming the norm is defined in the standard way, you get $\|v(t)\| = \sqrt{v_1^2(t) + \cdots + v_k^2(t)}$, where $v(t) = (v_1(t), \dots, v_k(t))$ is a vector in $\mathbb{R}^k$. On the other hand each $v_i$ is a differentiable function from $\mathbb{R}$ to $\mathbb{R}$. –  William Jan 13 '13 at 19:13
well it doesn't look like they're equating.. –  user53076 Jan 13 '13 at 19:26
unless I'm not seeing an obvious simplification –  user53076 Jan 13 '13 at 19:26
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5 Answers

up vote 2 down vote accepted

I won't produce more formulas, but I can tell you why they aren't the same:

Given a parametric representation $t\mapsto \gamma(t)$ of a curve $\gamma$ in ${\mathbb R}^3$ the velocity vector $\gamma'(t)={d\gamma(t)\over dt}$ encodes information about the speed $\|\gamma'(t)\|$ of the moving point $\gamma(t)$ as well as information about the tangential direction of the curve $\gamma$ at the point $P=\gamma(t)$. The normalized vector $$\tau:=\gamma'(t)/\|\gamma'(t)\|=\dot\gamma$$ is the unit tangent vector to $\gamma$ at $P$.

The $\dot{}$ has the following meaning: Among all parametric representations ("timetables") of one and the same geometric curve $\gamma$ there is a distinguished one: the representation with respect to arc length $s$ along $\gamma$. This special representation is distinguished by the property $\|\dot\gamma\|=\left\|{d\gamma(s)\over ds}\right\|\equiv1$, and differentiation with respect to this special parameter is denoted by a $\cdot\ $.

The acceleration vector $\gamma''(t)$ encodes the infinitesimal change of the velocity vector $\gamma'(t)$. This change has a tangential component which is related to a change of speed, and a normal component which is related to a change of direction, and therewith to curvature. When you normalize this vector to $n^?:=\gamma''(t)/\|\gamma''(t)\|$ you get a certain unit vector incorporating information about speed change and direction change. Its physical or geometrical meaning is unclear.

On the other hand the vector $\tau:=\gamma'(t)/\|\gamma'(t)\|=\dot\gamma$ contains only information about the tangential direction of $\gamma$ at the point $P=\gamma(t)$, but no information about the speed with which the moving point passes the point $P$ . The vectors $\tau'={d\tau\over dt}$ and $\dot \tau={d\tau\over ds}$ therefore only contain information about the change of direction along $\gamma$, whereby $\tau'$ measures this change in relation to time and $\dot\tau$ in relation to arc length. The relation between $\dot\tau$ and $\tau'$ is given by $$\dot\tau ={d\tau\over ds}={d\tau\over dt}\bigg/{ds\over dt}={\tau'\over\|\gamma'\|}\ .$$ In any case $\dot\tau$ is geometrically more relevant than $\tau'$; in fact $\|\dot\tau\|$ is the curvature of $\gamma$ at the point $P$.

The vector $n:={\dot\tau\over\|\dot\tau\|}$ is a unit vector pointing into the direction into which $\tau$ changes to. It is automatically orthogonal to $\tau$ since we took care to keep $\tau$ of unit length at all times.

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Thank you for your answer, it has made it much clearer. One thing: you've used both $\tau'$ and $\dot \tau$, what is the difference? I thought they were both just $\tau$ differentiated with respect to $t$? –  user53076 Jan 16 '13 at 17:19
@user53076: See my edit. –  Christian Blatter Jan 16 '13 at 20:54
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Let's show that $\frac{\gamma''(t)}{\|\gamma''(t)\| }$ and $n$ are generally not the same vector.

By definition we know that $$\gamma'(t) =\|\gamma'(t) \|\tau. \quad (1)$$ If we diferentiate $(1)$ we get $$\gamma''(t) =\|\gamma'(t) \|'\tau +\|\gamma'(t) \|\tau'. \quad(2)$$ But by definition $\tau'=\|\tau'\|n$, so substituting in $(2)$ we get $$\gamma''(t) =\|\gamma'(t) \|'\tau +\|\gamma'(t) \|\|\tau'\|n. \quad(3)$$ Taking the norm of $\gamma''(t)$ we get $$\|\gamma''(t)\| =\sqrt{(\|\gamma'(t) \|')^2+(\|\gamma'(t) \|\|\tau'\|)^2}. \quad(4)$$ Using equations $(3)$ and $(4)$ we get $$\frac{\gamma''(t)}{\|\gamma''(t)\| } =\frac{\|\gamma'(t) \|'\tau +\|\gamma'(t) \|\|\tau'\|n}{\sqrt{(\|\gamma'(t) \|')^2+(\|\gamma'(t) \|\|\tau'\|)^2}}. \quad(5)$$ If $\|\gamma'(t) \|'=0$ we can conclude from $(5)$ that $$\frac{\gamma''(t)}{\|\gamma''(t)\| } = n$$ otherwise $\frac{\gamma''(t)}{\|\gamma''(t)\| }$ and $n$ are not the same vector, since $\frac{\gamma''(t)}{\|\gamma''(t)\| }$ will have two perpendicular components $\frac{\|\gamma'(t) \|'}{\sqrt{(\|\gamma'(t) \|')^2+(\|\gamma'(t) \|\|\tau'\|)^2}}\tau$ and $\frac{\|\gamma'(t) \|\|\tau'\|}{\sqrt{(\|\gamma'(t) \|')^2+(\|\gamma'(t) \|\|\tau'\|)^2}}n$.

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When you have a unit-speed parametrization of the curve, $\gamma'$ is the same as $\tau$ and you can use: $$ n = \frac{\gamma''(t)}{\|\gamma''(t)\|} $$

However, this is not true in general. To see why, compare the two values for $\gamma(t) = (t, \cosh t)$ which isn't a unit-speed parametrization.

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The simplest way to understand why this happens may be the basic principle that the derivative of a vector function of unit length is always orthogonal to the function; that is, if $|v(t)| =1$ for all $t$, $v(t)\cdot \dfrac{dv(t)}{dt} = 0$; you can prove this by differentiating the equation $|v(t)|^2=1$ and using the product rule.

By contrast, when a vector function isn't of unit (or more generally, constant) length, its temporal derivative has a component in the (pointwise) direction of the function, corresponding to the change in length. This is why, to get a frame, it's important to normalize the tangent vector to the curve before differentiating it to get the normal; otherwise the normal won't actually be orthogonal to the tangent vector at points where the 'speed' of the curve changes.

More broadly, this leads to the concept of parametrization invariance which explains why we want the normalized versions in the first place: since we're concerned with properties of the curve, it shouldn't matter how fast we go along it. That is, if our original curve is $\vec{\gamma}(t)$ (with tangent $\vec{\tau}(t)$ and normal $\vec{\nu}(t)$) and we define a new curve $\vec{\delta}(t)$ by $\vec{\delta}(t) = \vec{\gamma}(g(t))$ for some function $g(t)$, then we want the tangent to $\vec{\delta}$ to be given by $\vec{\tau}(g(t))$ and the normal to be given by $\vec{\nu}(g(t))$. Unfortunately, if we try to take the normal by simply taking the second derivative and normalizing, the 'cross terms' in $g$ coming from the chain rule start to get in the way: $\vec{\delta}'(t)$ = $g'(t) \vec{\gamma}'(g(t))$, so $\vec{\delta}''(t)$ = $g''(t)\vec{\gamma}'(g(t))+\left(g'(t)\right)^2\vec{\gamma}''(g(t))$. This is no longer a simple multiple of $\vec{\gamma}''$, so we can't get rid of the $g$-dependence just by normalizing it.

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For further beyond good answers above see the following books which are classic references on Differential geometry of curves in $\mathbb{R}^3$: Modern Differential Geometry of Curves And Surfaces With Mathematica and A Comprehensive Introduction to Differential Geometry by Michael Spivak.

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