# Why isn't torsion the magnitude of the derivative of the binormal vector?

I am learning about the $$TNB$$, curvature and torsion and I found this weird derivation about torsion:

$$\tau=-B'(s)\cdot \hat{N}(s).$$

And the explaination I got basically is that since $$|\hat{B}(s)|=1,$$ then $$B'(s)\perp \hat{B}(s).$$

Also using some algebraic manipulation you get that $$\hat{T}(s)\perp B'(s).$$

Therefore $$B'(s)$$ must be parallel to the normal vector $$\hat{N}(s)$$ since it's perpendicular to both $$B$$ and $$T$$.

So you could say that $$B'(s)$$ is some constant $$\tau$$ multiplied by $$\hat{N}$$, and that's how you get to the first formula.

So the question is, why is it done this way? Isn't it posible to say "torsion is the speed of change of the binormal vector", write it down as $$\tau=|B'(t)|$$? Am I missing something? Is there a historic or practical reason why it's defined this way?

It just seemed very counterintuitive.

The norm $$\|B'(t)\|$$ tells us only something about the size of the change of $$B(s)$$, it doesn't tell you in which directions the binormal changes. For this reason it is natural to have a look at the components of $$B'(s)$$ w.r.t. the Frenet frame: $$B'(s) = \bigl( B'(s)\cdot T(s) \bigr) T(s) + \bigl( B'(s)\cdot N(s)\bigr) N(s) + \bigl(B'(s) \cdot B(s)\bigr) B(s).$$ Since we have shown that the 1st and 3rd component are zero, we have $$B'(s) = \bigl( B'(s)\cdot N(s)\bigr) N(s). \tag{*}$$ Since $$B'(s)$$ only has one component, your characterization $$\tau(s) = \|B'(s)\|$$ agrees with the definition, at least up to the sign. Also, $$\tau = \|B'(s)\|$$ doesn't specify a sign for $$\tau$$: a norm is just always non-negative, but $$B'(s)\cdot N(s)$$ can be negative.
Wikipedia and my old differential geometry notes claim that the sign is chosen historically. However, the sign intuitively makes sense. The torsion is a measure of how hard the curve is moving out of the $$TN$$-plane (osculating plane) at a point. If $$\tau > 0$$, it goes in the direction of $$B$$; if $$\tau < 0$$, it goes in the direction of $$-B$$.
• DoCarmo's book seems to be at odds with most of the rest of us with regard to the sign choice. The best thing to do is think about what torsion is for a helix and make sense of the sign there: Typically, a right-handed screw has $\tau>0$ and a left-handed one has $\tau<0$. :) I suppose that he likes the symmetry of the equations $$\kappa = T'\cdot N \quad\text{and}\quad \tau = B'\cdot N.$$ Apr 26 '20 at 20:00
• @TedShifrin that's really interesting. In my course $k$ is defined as $k=|T'(s)|$ so it made sense that $\tau=|B'(s)|$. Your commend shed some light into the whole thing too Apr 27 '20 at 18:58
• So curvature is nonnegative by definition if we want $N$ to point in the direction in which $T$ is turning. Once we have $T$ and $N$ we define $B$ so as to have a right-handed system. Thus, whether $B$ twists toward or away from $N$ makes there be a sign on $\tau$. Apr 27 '20 at 19:07
• That makes sense, I was just realizing that. so the whole $k=T'.N$ is just to have symmetry in the equations. thanks! Apr 27 '20 at 19:09