The curvature and torsion of the tangent indicatrix

Let $\alpha$ be a unit speed curve. Its tangent indicatrix $\sigma$ is defined by $\sigma(t)=T(t)$. Find torsion and curvature of $\sigma$ with respect to the torsion and curvature of $\alpha$.

There is an answer on my book. But I don't understand (*) part

$$\kappa_\sigma=\frac{||\alpha'\times \alpha''||}{||\alpha'||^3}$$

$$\tau_\sigma=\frac{\left<\alpha'\times\alpha'',\alpha'''\right>}{||\alpha'\times\alpha''||^2}$$

$$\sigma'(t)=T'(t) \Rightarrow \sigma'=T'=\kappa N$$

....

we find $\sigma'\times\sigma''=\kappa^3B+\kappa^2\tau T$

$$||\sigma'||=\sqrt{\left<\sigma',\sigma'\right>}=\sqrt{\left<\kappa N,\kappa N\right>}={\kappa^2\left<N,N\right>}=|\kappa|=||\alpha''||$$

$$||\sigma'\times\sigma''||=\sqrt{(\kappa^2\tau)^2+(\kappa^3)^2}=\kappa^2\sqrt{\tau^2+\kappa^2} \tag{*}$$

$$\kappa_\sigma=\frac{\sqrt{\tau^2+\kappa^2}}{\kappa}$$

...

I couldn't understand the equation (*). Can you explain it please? (expression "tangents indicator" may not true. I translated into English and $\kappa$ is the curvature and $\tau$ , $T$ tangent vector, and $B=T\times N$

• $B$ and $T$ are orthogonal unit vectors. You know that $\|a+b\|^2 = \|a\|^2 + \|b\|^2$ if $a$ and $b$ are orthogonal, right? Apply it to $\|\sigma'\times\sigma''\| = \|\kappa^3B+\kappa^2\tau T\|$. – Rahul Jan 4 '14 at 14:39
• In English this is called the tangent indicatrix. Your $x$s need to be written as '\times' :) – Ted Shifrin Jan 4 '14 at 14:39
• @RahulNarain thanks. appreciated. – lyme Jan 4 '14 at 15:15
• @TedShifrin I see. Could you suggest me a book or an article about this subject? – lyme Jan 4 '14 at 15:15
• You just have to use the formulas for curvature/torsion of non-arclength parametrized curves. The tangent indicatrix appears in curve theory, but only as a tool. What book are you studying from? – Ted Shifrin Jan 4 '14 at 16:04