Question

Show that the geodesics on a circular cylinder (with elements parallel to the z axis) are helics $\mathit{a}\mathit{z}\mathbf{+}\mathit{b}\mathit{\theta }\mathbf{=}\mathit{c}$, where a,b,c are constants depending on the given endpoints.(Hint: Use cylindrical coordinates) Note that the equation $\mathit{a}\mathit{z}\mathbf{+}\mathit{b}\mathit{\theta }\mathbf{=}\mathit{c}$includes the circles $\mathit{z}\mathbf{=}$const.(for $\mathit{b}\mathbf{=}\mathbf{0}$), straight lines $\mathit{\theta }\mathbf{=}$const.(for $\mathit{a}\mathbf{=}\mathbf{0}$), and the special heclices $\mathit{a}\mathit{z}\mathbf{+}\mathit{b}\mathit{\theta }\mathbf{=}\mathbf{0}$.

Short answer

It is proved that geodesics on a circular cylinder is $az+b\theta =c$, which represents helics.

Step-by-step solution

Given Information.

It is given to use cylindrical coordinates to prove that geodesics on a circular cylinder is$az+b\theta =c$, which represents helics

Definition of Euler equation

In the calculus of variations andclassical mechanics, the Euler equations is a system of second-orderordinary differential equations whose solutions arestationary points of the givenaction functional.

Use Euler equation

Geodesics on a circular cylinder is to be found out using cylinderical coordinates. So distance integral is to be minimized.

$$\begin{gathered} \int ds=\int \sqrt{d{z}^2+R^{2}d{\theta }^2} \\ =\int \sqrt{1+R^2\theta {\text{'}}^2}dr \\ \theta \text{'}=\frac{d\theta }{dz} \end{gathered}$$

Let $F=\sqrt{1+R^2\theta {\text{'}}^2}$

Euler equation for coordinates$\left(z,\theta \right)$is$\frac{d}{dz}\left(\frac{\partial F}{\partial \theta \text{'}}\right)-\frac{\partial F}{\partial \theta }=0$.

Calculate the required derivatives.

$$\begin{gathered} \frac{\partial F}{\partial \theta \text{'}}=\frac{R^2\theta \text{'}}{\sqrt{1+R^2\theta {\text{'}}^2}} \\ \frac{\partial F}{\partial \theta }=0 \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{d}{dz}\left[\frac{R^2\theta \text{'}}{\sqrt{1+R^2\theta {\text{'}}^2}}\right]=0 \\ \frac{R^2\theta \text{'}}{\sqrt{1+R^2\theta {\text{'}}^2}}=C \\ \theta {\text{'}}^2=\frac{C^2}{R^2\left(R^2-C^2\right)} \\ \theta \text{'}=\frac{C}{R\sqrt{R^2-C^2}} \end{gathered}$$

Where$C$is constant.

Integrate $\theta \text{'}=\frac{C}{R\sqrt{R^2-C^2}}$to get the desired result

$$\begin{gathered} \theta =\int \frac{C}{R\sqrt{R^2-C^2}}dz \\ =\frac{C}{R\sqrt{R^2-C^2}}z+B \end{gathered}$$

Where is the integration constant. Since $C,R$and $B$are constants. Constants can be redefined as

$$\begin{gathered} C\equiv -a \\ R\sqrt{R^2-C^2}\equiv b \\ B\equiv \frac{c}{b} \end{gathered}$$

Thus,

role="math" localid="1665034834432" $$\begin{gathered} \theta =-\frac{a}{b}z+\frac{c}{b} \\ az+b\theta =c \end{gathered}$$

Therefore, It is proved that geodesics on a circular cylinder is $az+b\theta =c$, which represents helices.