Question

In cylindrical polar coordinates, the curve $(\rho (\theta ),\theta ,\alpha \rho (\theta ))$ lies on the surface of the cone $z=\alpha \rho$. Show that geodesics (curves of minimum length joining two points) on the cone satisfy $${\rho }^4=c^2[{\beta }^2{\rho }^{\prime 2}+{\rho }^2]$$ where $c$ is an arbitrary constant, but $\beta$ has to have a particular value. Determine the form of $\rho (\theta )$ and hence find the equation of the shortest path on the cone between the points $(R,-{\theta }_0,\alpha R)$ and $(R,{\theta }_0,\alpha R)$. (You will find it useful to determine the form of the derivative of ${\cos }^{-1}\left(u^{-1}\right)$.)

Short answer

The geodesics satisfy ${\rho }^4=c^2[{\beta }^2{\rho }^{\prime 2}+{\rho }^2]$ with $\beta =\sqrt{1+{\alpha }^2}$, and the form of $\rho (\theta )$ is $A\cos (B\theta +C)$.

Step-by-step solution

Set Up the Problem

Identify and organize the given information. We have a cone with equation $z=\alpha \rho$, and a curve $(\rho (\theta ),\theta ,\alpha \rho (\theta ))$. We need to show that geodesics on this cone satisfy ${\rho }^4=c^2[{\beta }^2{\rho }^{\prime 2}+{\rho }^2]$.

Write the Arc Length in Cylindrical Coordinates

The arc length, $S$, on the cone in cylindrical coordinates is given by the integral $$S=\int \sqrt{{\rho }^{\prime 2}+{\rho }^2+(\alpha {\rho }^{\prime }{)}^2}d\theta$$ Simplify this since $z=\alpha \rho$ (constant).

Introduce Lagrangian Mechanics

To find the geodesics, set up the Lagrangian $L=\sqrt{{\rho }^2(1+{\alpha }^2)+{\rho }^{{}^{\prime }2}}$ and minimize the action integral $$S=\int Ld\theta$$

Apply Euler-Lagrange Equation

The Euler-Lagrange equation is $$\frac{d}{d\theta }\left(\frac{\partial L}{\partial {\rho }^{\prime }}\right)-\frac{\partial L}{\partial \rho }=0$$ Compute the partial derivatives and simplify the Euler-Lagrange equation.

Simplify the Differential Equation

After simplification, you will obtain a differential equation involving $\rho (\theta )$

Show the Required Form

Show that the simplified differential equation can be written as $${\rho }^4=c^2[{\beta }^2{\rho }^{\prime 2}+{\rho }^2]$$ and solve to find the value of $\beta$.

Solve for $\rho (\theta )$

Assuming an ansatz $\rho =A\cos (B\theta +C)$ and using the boundary conditions, solve for the constants.

Find the Equation of the Shortest Path

Using the solution for $\rho (\theta )$, write the equation of the shortest path between points $(R,-{\theta }_0,\alpha R)$ and $(R,{\theta }_0,\alpha R)$.

Key concepts

Cylindrical Polar Coordinates

In cylindrical polar coordinates, a point in space is described using three values: $\rho$, $\theta$, and $z$. These correspond to:
• $\rho$ - the radial distance from the z-axis.
• $\theta$ - the azimuthal angle in the xy-plane from the x-axis.
• $z$ - the height above the xy-plane.
This system is particularly useful when dealing with problems possessing rotational symmetry, such as the surface of a cone. For the cone given by $z=\alpha \rho$, the height is proportional to the radial distance, simplifying our calculations of geodesics.

Arc Length

The arc length represents the distance along a curve between two points. For a curve described in cylindrical coordinates, the differential arc length element is found using:
$$dl=\sqrt{d{\rho }^2+{\rho }^{2}d{\theta }^2+d{z}^2}$$
Given $z=\alpha \rho$, we can substitute $dz=\alpha d\rho$ into this equation. This helps us express the arc length $S$ over the region of interest:
$$S=\int \sqrt{{\rho }^{\prime 2}+{\rho }^2+(\alpha {\rho }^{\prime }{)}^2}d\theta$$
This expression is manageable when we introduce simplifications specific to the cone's geometry, helping us find the geodesics (shortest paths) effectively.

Euler-Lagrange Equation

The Euler-Lagrange equation is a fundamental equation in calculus of variations. It helps find functions that make a certain integral (known as the action) stationary. For this problem, the integral we want to minimize is the arc length:
$$S=\int L(\rho ,{\rho }^{\prime },\theta )d\theta$$
Here, we identify the Lagrangian $L=\sqrt{{\rho }^2(1+{\alpha }^2)+{\rho }^{\prime 2}}$. The Euler-Lagrange equation is:
$$\frac{d}{d\theta }\left(\frac{\partial L}{\partial {\rho }^{\prime }}\right)-\frac{\partial L}{\partial \rho }=0$$
We solve for $\rho (\theta )$ by computing these partial derivatives and differentiating appropriately. This sheds light on the conditions for geodesics on the cone.

Lagrangian Mechanics

Lagrangian mechanics provides a powerful framework for analyzing mechanical systems. It replaces Newton's laws with an energy-based approach, where we focus on two main quantities: kinetic energy (T) and potential energy (V). The Lagrangian $L$ is defined as:
$$L=T-V$$
For geodesics, we analogize the problem by considering the Lagrangian as the integrand of the arc length, aiming to minimize the travel distance. The action integral $S$ in this context is:
$$S=\int L(\rho ,{\rho }^{\prime },\theta )d\theta$$
By applying the principles of Lagrangian mechanics, particularly the Euler-Lagrange equation, we find the curve $\rho (\theta )$ that makes the arc length stationary, thereby giving the shortest path on the cone's surface.