Question

In spherical coordinates, find the \(\theta\) Lagrange equation for a particle moving in the potential field \(V=V(r, \theta, \phi) .\) What is the \(\theta\) component of the acceleration? Hint: The \(\theta\) Lagrange equation is the \(\theta\) component of \(m \mathbf{a}=\mathbf{F}=-\nabla V ;\) for components of \(\nabla V,\) see Chapter \(6,\) end of Section \(6,\) or Chapter \(10,\) Section 9.

Short answer

\[m r^2 \ddot{\theta} + 2m r \dot{r} \dot{\theta} - m r^2 \sin(\theta) \cos(\theta) \dot{\phi}^2 + \frac{\partial V}{\partial \theta} = 0\]

Step-by-step solution

Write the Lagrangian

First, write down the Lagrangian for a particle in spherical coordinates. The Lagrangian is given by:\[ L = T - V \]where \(T\) is the kinetic energy and \(V\) is the potential energy.In spherical coordinates, the kinetic energy \(T\) is:\[ T = \frac{1}{2}m\left(\frac{dr}{dt}\right)^2 + \frac{1}{2}m r^2 \left(\frac{d\theta}{dt}\right)^2 + \frac{1}{2}m r^2 \sin^2(\theta) \left(\frac{d\phi}{dt} \right)^2 \]

Compute the Lagrange's equations

Compute Lagrange's equation for \(\theta\) by using: \[\frac{d}{dt}\left( \frac{\partial L}{\partial\dot{\theta}} \right) - \frac{\partial L}{\partial\theta} = 0\]where \(\dot{\theta} = \frac{d\theta}{dt}\).

Derive \(\frac{\partial L}{\partial\dot{\theta}}\)

We calculate \(\frac{\partial L}{\partial \dot{\theta}}\) by differentiating the Lagrangian with respect to \(\dot{\theta}\):\[ \frac{\partial L}{\partial \dot{\theta}} = m r^2 \dot{\theta} \]

Compute \(\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{\theta}} \right)\)

We take the time derivative of \(\frac{\partial L}{\partial \dot{\theta}}\):\[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{\theta}} \right) = \frac{d}{dt} (m r^2 \dot{\theta}) = m (2r \dot{r} \dot{\theta} + r^2 \ddot{\theta}) \]

Derive \(\frac{\partial L}{\partial \theta}\)

We calculate \(\frac{\partial L}{\partial \theta}\) by differentiating the Lagrangian with respect to \(\theta\):\[ \frac{\partial L}{\partial \theta} = \frac{\partial}{\partial \theta} \left( \frac{1}{2} m r^2 \sin^2(\theta) \dot{\phi}^2 \right) = m r^2 \sin(\theta) \cos(\theta) \dot{\phi}^2 - \frac{\partial V}{\partial \theta} \]

Combine results and simplify

Combine the results from previous steps to obtain the \(\theta\)-component of the Lagrangian equation:\[ \boxed{m r^2 \ddot{\theta} + 2m r \dot{r} \dot{\theta} - m r^2 \sin(\theta) \cos(\theta) \dot{\phi}^2 + \frac{\partial V}{\partial \theta} = 0} \]We simplify to find the equation of motion.

Key concepts

Lagrangian Mechanics

Lagrangian mechanics is a reformulation of classical mechanics introduced by Joseph-Louis Lagrange. This approach provides powerful methods for analyzing physical systems, especially those with constraints. The Lagrangian, denoted as \(L\), is defined as the difference between kinetic energy (\(T\)) and potential energy (\(V\)). Therefore, the Lagrangian is given by:

\[ L = T - V \]

The equations of motion are derived from the Lagrangian using the Euler-Lagrange equation:

\[ \frac{d}{dt} \frac{\text{∂}L}{\text{∂}\text{∂}\theta} - \frac{\text{∂}L}{\text{∂}\theta} = 0 \]

This method is particularly useful in complex systems where Newton's laws are difficult to apply directly.

Spherical Coordinates

Spherical coordinates are a three-dimensional coordinate system where points are represented by three numbers: the radial distance (\(r\)), the polar angle (\(\theta\)), and the azimuthal angle (\(\phi\)). This system is ideal for problems with spherical symmetry.

In spherical coordinates:

• \(r\) measures the distance from the origin.
• \(\theta\) (colatitude) measures the angle from the positive z-axis.
• \(\phi\) (longitude) measures the angle in the x-y plane from the positive x-axis.

These coordinates simplify the description of systems with radial potentials and allow for easier computation of kinetic and potential energies in such systems.

Kinetic Energy

Kinetic energy (\(T\)) is the energy that a particle has due to its motion. In classical mechanics, it is given by the formula:

\[ T = \frac{1}{2} m v^2 \]

where \(m\) is the mass of the particle and \(v\) is its velocity. In spherical coordinates, the kinetic energy is a combination of the radial, polar, and azimuthal components of motion:

\[ T = \frac{1}{2}m\bigg(\bigg(\frac{dr}{dt}\bigg)^2 + r^2 \bigg(\frac{d\theta}{dt}\bigg)^2 + r^2 \text{sin}^2(\theta) \bigg(\frac{d\text{ϕ}}{dt}\bigg)^2\bigg) \]

This expression captures the energy due to motion along the radial distance, the polar angle, and the azimuthal angle.

Potential Energy

Potential energy (\(V\)) is the energy stored due to the position or configuration of a particle within a force field, like gravity. In the given problem, the potential energy is a function of spherical coordinates \(r\), \(\theta\), and \(\phi\):

\[ V = V(r, \theta, \phi) \]

Potential energy plays a crucial role in the Lagrangian, which is defined as the difference between kinetic and potential energy. The gradients of the potential energy contribute to the forces acting on the particle, as derived from Lagrange's equations.

Partial Derivatives

Partial derivatives are essential in Lagrangian mechanics for dealing with functions of several variables. They represent the rate of change of a function with respect to one variable while keeping the others constant. In the Euler-Lagrange equation, we use partial derivatives to describe how the Lagrangian changes over time:

\[ \frac{\text{∂}L}{\text{∂}\theta}, \frac{\text{∂}L}{\text{∂}\text{∂}\theta} \]

Additionally, partial derivatives are used to find the gradient of the potential energy:

\[ \frac{\text{∂}V}{\text{∂}r}, \frac{\text{∂}V}{\text{∂}\theta}, \frac{\text{∂}V}{\text{∂}\text{ϕ}} \]

Understanding partial derivatives allows us to derive the necessary equations of motion by tracking how changes in one coordinate affect the overall system.