91Ó°ÊÓ

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

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.

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

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

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

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

Calculate the required derivatives.

$$\begin{gathered} \frac{∂F}{∂\mathrm{θ}\text{'}}=\frac{R^2\mathrm{θ}\text{'}}{\sqrt{1+R^2\mathrm{θ}{\text{'}}^2}} \\ \frac{∂F}{∂\mathrm{θ}}=0 \end{gathered}$$

Therefore,

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

Where$C$is constant.

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

$$\begin{gathered} \mathrm{θ}=∫\frac{C}{R\sqrt{R^2-C^2}}dz \\ =\frac{C}{R\sqrt{R^2-C^2}}z+B \end{gathered}$$

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

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

Thus,

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

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