Question

(a) For a system described in terms of coordinates \(q_{1}\) and \(t\), show that if \(t\) does not appear explicitly in the expressions for \(x, y\) and \(z\left(x=x\left(q_{i}, t\right)\right.\), etc. \()\) then the kinetic energy \(T\) is a homogeneous quadratic function of the \(\dot{q}_{1}\) (it may also involve the \(q_{i}\) ). Deduce that \(\sum_{i} \dot{q}_{i}\left(\partial T / \partial \dot{q}_{t}\right)=2 T\). (b) Assuming that the forces acting on the system are derivable from a potential. \(V\), show, by expressing \(d T / d t\) in terms of \(q_{1}\) and \(\dot{q}_{1}\), that \(d(T+V) / d t=0\).

Short answer

The kinetic energy \( T \) is a homogeneous quadratic function of \( \dot{q_i} \) and \( \sum_{i} \dot{q_i} \left( \frac{\partial T}{\partial \dot{q_i}} \right) = 2T \). Considering forces derivable from potential energy, \( \frac{d(T+V)}{dt} = 0 \).

Step-by-step solution

Express the Kinetic Energy

The kinetic energy of a system can be expressed as \[ T = \frac{1}{2} m (\frac{dx}{dt}^2 + \frac{dy}{dt}^2 + \frac{dz}{dt}^2). \]Since we are given that the coordinates \(x, y, z\) are functions of \(q_i\) and \(t\), but \(t\) does not explicitly appear, the velocities \( \frac{dx}{dt}, \frac{dy}{dt} \), and \( \frac{dz}{dt} \) can be expressed as functions of \( \frac{dq_i}{dt} = \dot{q_i} \).

Kinetic Energy as a Homogeneous Quadratic Function

Since the velocities \( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \) can be expressed as linear functions of \( \dot{q_i} \), the kinetic energy \(T\) becomes a homogeneous quadratic function of \( \dot{q_i} \). Symbolically, we write:\[ T = \frac{1}{2} \textbf{A}_{ij}(q_k) \dot{q_i} \dot{q_j}, \]where \( \textbf{A}_{ij}(q_k) \) is a symmetric matrix depending on the coordinates \( q_k \).

Deduce the Summation Involving Partial Derivatives of Kinetic Energy

To show that \[ \sum_{i} \dot{q_i} \left( \frac{\partial T}{\partial \dot{q_i}} \right) = 2 T, \]we use the fact that \(T\) is a homogeneous quadratic function of \(\dot{q_i} \). By Euler's theorem for homogeneous functions:\[ T(\alpha \dot{q_i}) = \alpha^2 T(\dot{q_i}). \]Taking the derivative with respect to \(\alpha\) and setting \(\alpha = 1\), we get:\[ \sum_{i} \dot{q_i} \left( \frac{\partial T}{\partial \dot{q_i}} \right) = 2 T. \]

Express Potential Energy of the System

Assume that the potential energy of the system \(V\) is a function of the coordinates \(q_i\) only: \[ V = V(q_i). \]Since \(t\) does not explicitely appear in \(x, y, z\), the potential energy does not depend on \(\dot{q_i} \).

Time Derivative of Kinetic Energy

To express \( \frac{dT}{dt} \) in terms of \( q_i \) and \( \dot{q_i} \), apply the chain rule: \[ \frac{dT}{dt} = \sum_{i} \left( \frac{\partial T}{\partial q_i} \dot{q_i} + \frac{\partial T}{\partial \dot{q_i}} \ddot{q_i} \right). \]

Time Derivative of Total Energy

To find \( \frac{d(T+V)}{dt} \), consider:\[ \frac{d(T+V)}{dt} = \frac{dT}{dt} + \frac{dV}{dt}. \]For the potential \(V\), \[ \frac{dV}{dt} = \sum_{i} \frac{\partial V}{\partial q_i} \dot{q_i}. \]

Use Lagrange's Equations

Using Lagrange's equations, \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}} \right) = \frac{\partial L}{\partial q_i} \) for the Lagrangian \( L = T - V \), we equate:\[ \frac{\partial T}{\partial q_i} + \frac{\partial V}{\partial q_i} = \sum_{i} \left( \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q_i}} \right) + \frac{d}{dt} \left( \frac{\partial V}{\partial \dot{q_i}} \right) \right). \]Since \( \frac{dV}{dt} \) involves \(\dot{q_i}\) only and not \(\dot{q_i} \), \[ \frac{d(T+V)}{dt} = 0. \]

Key concepts

Kinetic Energy

In physics, kinetic energy is the energy an object possesses due to its motion. To calculate the kinetic energy, you can use the formula: \[ T = \frac{1}{2} m ( \frac{dx}{dt}^2 + \frac{dy}{dt}^2 + \frac{dz}{dt}^2 ) \]. In Lagrangian mechanics, coordinates often rely on generalized coordinates denoted as \( q_i \). If time \( t \) doesn't appear explicitly in the expressions for the coordinates, then the kinetic energy can be expressed purely in terms of the derivatives of these coordinates with respect to time. This makes the kinetic energy a homogeneous quadratic function of the velocities \( \dot{q_i} \). The quadratic nature indicates the kinetic energy depends on the product of the velocity terms.

Potential Energy

Potential energy represents the stored energy an object has due to its position in a force field, such as gravity. In this exercise, potential energy \( V \) is a function of the coordinates \( q_i \) and does not depend on their velocities \( \dot{q_i} \). Mathematically, this is expressed as \[ V = V(q_i) \]. When forces derive from a potential energy function, they simplify the analysis within Lagrangian mechanics. The relation \( \frac{d(T+V)}{dt} = 0 \) emerges, where the total mechanical energy of the system (kinetic + potential) remains conserved over time.

Euler's Theorem

Euler's theorem for homogeneous functions is a key concept in this exercise. It states that for any homogeneous function of degree \( n \), the following identity holds: \[ f( \alpha x ) = \alpha^n f(x) \]. In the context of kinetic energy, which is a homogeneous quadratic function (degree 2), Euler's theorem helps us deduce that: \[ T( \alpha \dot{q_i} ) = \alpha^2 T( \dot{q_i} ) \]. By differentiating with respect to \( \alpha \) and setting \( \alpha = 1 \), we get the important result: \[ \sum_i \dot{q_i} \left( \frac{\partial T}{\partial \dot{q_i}} \right) = 2T \]. This equation reinforces the relationship between kinetic energy and its partial derivatives.

Lagrange's Equations

Lagrange's equations form the foundation of Lagrangian mechanics, offering a powerful method to derive the equations of motion for a system. For a Lagrangian \( L = T - V \), where \( T \) is kinetic energy and \( V \) is potential energy, Lagrange's equations are given by: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}} \right) - \frac{\partial L}{\partial q_i} = 0 \]. This results in the equations that describe the dynamics of a system. The terms \( \frac{\partial L}{\partial q_i} \) and \( \frac{\partial L}{\partial \dot{q_i}} \) turn into forces and momenta. Using these equations in the exercise, we find: \[ \frac{d(T+V)}{dt} = 0 \], emphasizing that the total energy of a system remains constant over time when only conservative forces (derivable from a potential) are acting.