Bertrand Curves

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Bertrand curves and its properties

A pair of space curves $C$ and $C_1$ and are called Bertrand curves if principal normal to $C$ is also principal normal to $C_1$ , i.e, $\overrightarrow{n_1}=\overrightarrow{n}$

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Principal Noemals are Same

0,0

–o+←↓↑→

az = 1.00

el = 0.30

\[ec{n|=n₁\] = 2.00

P

P₁

Principal Noemals are Same

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Properties of Bertrand curves

Let $C$ and $C_1$ are Bertrand curves then

1. Distance between corresponding points of Bertrand curves is constant
2. Tangent at corresponding points of Bertrand curves inclined at constant angle
3. Curvature and torsion of either curves are connected by linear relation
4. Torsion of Bertrand curves have same sign and their product is constant

Let $C$ and $C_1$ are Bertrand curves then mathematics of above statements are

1. If $\lambda$ is distance between corresponding points of Bertrand curves, then ${\lambda }^{\prime }=0$
2. if $\overrightarrow{t}$ is tangent to $C$ and $\overrightarrow{t_1}$ is tangent to $C_1$, $\frac{d}{ds}(\overrightarrow{t}.\overrightarrow{t_1})=0$
3. if $\kappa$ and $\tau$ are curvature and torsion of $C$ then
   $a\kappa +b\tau +c=0$ ; $\kappa$ and $\tau$ are connected by linear relation
   if ${\kappa }_1$ and ${\tau }_1$ are curvature and torsion of $C_1$ then
   $a_1{\kappa }_1+b_1{\tau }_1+c_1=0$ ; ${\kappa }_1$ and ${\tau }_1$ are connected by linear relation
4. if $\tau$ and ${\tau }_1$ are torsion of $C$ and $C_1$ then $\tau$ and ${\tau }_1$ have same sign, means $\tau .{\tau }_1=positive$ and $\tau .{\tau }_1=constant$

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Proof

1. Distance between corresponding points of Bertrand curves is constant
   
   0,0

   –o+←↓↑→

   O

   P

   $$P_1$$

   $$\overrightarrow{r}$$

   $$\overrightarrow{r_1}$$

   $$\overrightarrow{n}=\overrightarrow{n_1}$$
   , 7+2*Math.sin(t)];}; function (t) { return [ 7+5*Math.cos(t), 7+5*Math.sin(t)];}; board.create('curve',[r, 0, Math.PI],{strokeColor:'red'}); board.create('functiongraph',[r1, 0, Math.PI],{strokeColor:'red'}); Let $C$ and $C_1$ are Bertrand curves then
   $\overrightarrow{n_1}=\overrightarrow{n}$(i)
   Let $P$ be a point on $C$ and $P_1$ be its corresponding on $C_1$ with
   $\overrightarrow{O{P}_1}=\overrightarrow{r_1}$
   $\overrightarrow{OP}=\overrightarrow{r}$
   Then,
   $\overrightarrow{P{P}_1}$ is principal normal to C
   or $\overrightarrow{P{P}_1}=\lambda \overrightarrow{n}$
   where $\lambda$ is distance between corresponding points $P$ and $P_1$
   or $\overrightarrow{O{P}_1}-\overrightarrow{OP}=\lambda \overrightarrow{n}$
   or $\overrightarrow{r_1}-\overrightarrow{r}=\lambda \overrightarrow{n}$
   or $\overrightarrow{r_1}=\overrightarrow{r}+\lambda \overrightarrow{n}$ (ii)
   Differentiating (ii) w. r. to. s , we get
   $\overrightarrow{t_1}\frac{d{s}_1}{ds}=\overrightarrow{t}+{\lambda }^{\prime }\overrightarrow{n}+\lambda (\tau \overrightarrow{b}-\kappa \overrightarrow{t})$
   or $\overrightarrow{t_1}\frac{d{s}_1}{ds}=\overrightarrow{t}+{\lambda }^{\prime }\overrightarrow{n}+\lambda \tau \overrightarrow{b}-\lambda \kappa \overrightarrow{t}$
   or $\overrightarrow{t_1}\frac{d{s}_1}{ds}=(1-\lambda \kappa )\overrightarrow{t}+{\lambda }^{\prime }\overrightarrow{n}+\lambda \tau \overrightarrow{b}$
   Taking dot product by $\overrightarrow{n_1}=\overrightarrow{n}$, we get
   $(\overrightarrow{t_1}\frac{d{s}_1}{ds})\overrightarrow{n_1}=[(1-\lambda \kappa )\overrightarrow{t}+{\lambda }^{\prime }\overrightarrow{n}+\lambda \tau \overrightarrow{b}]\overrightarrow{n}$
   or $0=[0+{\lambda }^{\prime }1+0]$ $\overrightarrow{t_1}.\overrightarrow{n_1}=0,\overrightarrow{t}.\overrightarrow{n}=0,\overrightarrow{b}.\overrightarrow{n}=0$
   or $0={\lambda }^{\prime }$
   It shows that, distance between corresponding points of Bertrand curves is constant
   The first property established.

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2. Tangent at corresponding points of Bertrand curves inclined at constant angle
   Let $\overrightarrow{t}$ is tangent to $C$ and $\overrightarrow{t_1}$ is tangent to $C_1$,
   and also let, $\alpha$ is angle between $\overrightarrow{t}$ and $\overrightarrow{t_1}$, then
   $cos\alpha =\frac{\overrightarrow{t}.\overrightarrow{t_1}}{|\overrightarrow{t}|.|\overrightarrow{t_1}|}$
   or $cos\alpha =\overrightarrow{t}.\overrightarrow{t_1}$
   Differentiating w. r. to s, we get
   $\frac{d}{ds}(cos\alpha )=\frac{d}{ds}(\overrightarrow{t}.\overrightarrow{t_1}$
   or $\frac{d}{ds}(cos\alpha )=(\frac{d}{ds}\overrightarrow{t}).\overrightarrow{t_1}+(\frac{d}{ds}\overrightarrow{t_1})\overrightarrow{t}$
   or $\frac{d}{ds}(cos\alpha )=(\kappa \overrightarrow{n}).\overrightarrow{t_1}+({\kappa }_1\overrightarrow{n_1}\frac{d{s}_1)}{ds})\overrightarrow{t}$
   In Bertrand curves, $\overrightarrow{n_1}=\overrightarrow{n}$ , so
   Replacing $\overrightarrow{n}$ by $\overrightarrow{n_1}$ and also replacing $\overrightarrow{n_1}$ by $\overrightarrow{n}$ in the right, we get
   $\frac{d}{ds}(cos\alpha )=(\kappa \overrightarrow{n_1}).\overrightarrow{t_1}+({\kappa }_1\overrightarrow{n}\frac{d{s}_1)}{ds})\overrightarrow{t}$
   or $\frac{d}{ds}(cos\alpha )=0+0$
   or $\frac{d}{ds}(cos\alpha )=0$
   It shows that,
   derivative of $cos\alpha$ is zero
   so
   $cos\alpha$ is constant
   or $\alpha$ is constant
   Hence, yangent at corresponding points of Bertrand curves inclined at constant angle
   The second property established.

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3. Curvature and torsion of either curves are connected by linear relation
   Let $C$ and $C_1$ are Bertrand curves then using (A), we write
   $\overrightarrow{t_1}\frac{d{s}_1}{ds}=\overrightarrow{t}+{\lambda }^{\prime }\overrightarrow{n}+\lambda \tau \overrightarrow{b}-\lambda \kappa \overrightarrow{t}$
   $\overrightarrow{t_1}\frac{d{s}_1}{ds}=(1-\lambda \kappa )\overrightarrow{t}+0\times \overrightarrow{n}+\lambda \tau \overrightarrow{b}$
   or $\overrightarrow{t_1}\frac{d{s}_1}{ds}=(1-\lambda \kappa )\overrightarrow{t}+\lambda \tau \overrightarrow{b}$ (iii)
   Also, we know that,
   0,0

   –o+←↓↑→

   az = 1.00

   el = 0.30

   rotate = 2.00

   t_1

   b_1

   t

   b

   n=n_1
   • principal normals to $C$ and $C_1$ are same
   • tangents to $C$ and $C_1$ are inclined at constant angle $\alpha$
   • binormals to $C$ and $C_1$ also inclined at same constant angle $\alpha$

   Hence,
   $\overrightarrow{t_1}=\cos \alpha \overrightarrow{t}+\cos (90-\alpha )\overrightarrow{b}$
   or $\overrightarrow{t_1}=\cos \alpha \overrightarrow{t}+\sin \alpha \overrightarrow{b}$(iv)
   Comparing (iii) and (iv), we get
   $\frac{d{s}_1}{ds}=\frac{1-\lambda \kappa }{\cos \alpha }=\frac{\lambda \tau }{\sin \alpha }$
   We have to show relation between $\kappa$ and $\tau$, so
   Taking, second and third part, we get
   $\frac{1-\lambda \kappa }{cos\alpha }=\frac{\lambda \tau }{\sin \alpha }$
   or $(1-\lambda \kappa )sin\alpha =\lambda \tau \cos \alpha$
   or $(-\lambda \sin \alpha )\kappa +(-\lambda \cos \alpha )\tau +sin\alpha =0$
   This shows that $\kappa$ and $\tau$ are connected by linear relation in the form
   $a\kappa +b\tau +c=0$ where $a,b,c$ are constants given by $a=(-\lambda sin\alpha ),b=(-\lambda \cos \alpha ),c=\sin \alpha$
   Next,
   Since relation between $C$ and $C_1$ are reverse in terms of $\alpha$ and $\lambda$ we have
   $\lambda$ will be $-\lambda$ and $\alpha$ will be $-\alpha$ for $C_1$
   Thus, while writting expression for ${\kappa }_1$ and ${\tau }_1$ , we get
   $(\lambda \sin (-\alpha )){\kappa }_1+(\lambda \cos (-\alpha )){\tau }_1+\sin (-\alpha )=0$
   or $(-\lambda \sin \alpha ){\kappa }_1+(\lambda \cos \alpha ){\tau }_1+(-\sin \alpha )=0$
   This shows that ${\kappa }_1$ and ${\tau }_1$ are connected by linear relation in the form
   $a_1{\kappa }_1+b_1{\tau }_1+c_1=0$
   where $a_1,b_1,c_1$ are constants given by
   $a_1=(-\lambda sin\alpha ),b_1=(\lambda \cos \alpha ),c_1=(-\sin \alpha )$
   The third property established.

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4. Torsion of Bertrand curves have same sign and their product is constant
   Let $C$ and $C_1$ are Bertrand curves then using (B), we get
   $\frac{d{s}_1}{ds}=\frac{1-\lambda \kappa }{cos\alpha }=\frac{\lambda \tau }{sin\alpha }$
   We have to show relation between $\tau$ and ${\tau }_1$, so
   Taking, first and third part, we get
   $\frac{d{s}_1}{ds}=\frac{\lambda \tau }{sin\alpha }$ (v)
   Since relation between $C$ and $C_1$ are reverse in terms of $\alpha$ and $\lambda$ . Thus, while writing expression for ${\tau }_1$ , we get
   $\frac{ds}{d{s}_1}=\frac{-\lambda {\tau }_1}{sin(-\alpha )}$ (vi)
   Multiplying (v) and (vi), we get
   $\frac{d{s}_1}{ds}\frac{ds}{d{s}_1}=\frac{\lambda \tau }{sin\alpha }\frac{-\lambda \tau }{sin(-\alpha )}$
   or$1=\frac{\lambda \tau }{sin\alpha }\frac{\lambda {\tau }_1}{sin\alpha }$
   or$1=\frac{{\lambda }^2\tau {\tau }_1}{si{n}^2\alpha }$
   or$\tau {\tau }_1=(\frac{sin\alpha }{\lambda }{)}^2$
   Torsion of Bertrand curves have same sign and their product is constant
   The forth property established.