Question

A system has Lagrangian \(L=\frac{1}{2} T_{a b} v_{a} v_{b}\) where the \(T_{a b} \mathrm{~s}\) are functions of the \(q_{a} \mathrm{~s}\) alone. Show that $$ \frac{\mathrm{d} L}{\mathrm{~d} t}=0 $$ Show that if \(f\) is any function of one variable, then \(L^{\prime}=f(L)\) generates the same dynamics.

Short answer

In conclusion, we have shown that the time derivative of the given Lagrangian \(L\) is zero, by calculating its time derivative and demonstrating that all terms cancel out to zero. Furthermore, we have proven that a transformed Lagrangian \(L'=f(L)\) generates the same dynamics as the original Lagrangian \(L\), by showing that their respective Euler-Lagrange equations are equivalent.

Step-by-step solution

Write down the expression for L and take its time derivative

Given the Lagrangian \(L = \frac{1}{2} T_{ab} v_a v_b\), where \(T_{ab}\) is a function of the generalized coordinates \(q_a\) alone, and \(v_a\) and \(v_b\) represent the generalized velocities. Let's take the time derivative of the Lagrangian \(L\): $$ \frac{\mathrm{d} L}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{2} T_{ab} v_a v_b\right) $$

Calculate the time derivative of L

Now, we need to compute the time derivative of \(L\), using the product rule for differentiation and remembering that \(T_{ab}\) depends only on the generalized coordinates \(q\): $$ \frac{\mathrm{d} L}{\mathrm{d}t} = \frac{1}{2} \frac{\mathrm{d} (T_{ab} v_a v_b)}{\mathrm{d}t} = \frac{1}{2}\left( \frac{\partial T_{ab}}{\partial q_c} \frac{\mathrm{d} q_c}{\mathrm{d}t} v_a v_b + T_{ab} \frac{\mathrm{d} v_a}{\mathrm{d}t} v_b + T_{ab} v_a \frac{\mathrm{d} v_b}{\mathrm{d}t}\right) $$ Notice that the summation over repeated indices is implied. Since \(T_{ab}\) depends only on \(q_a\), the term \(\frac{\partial T_{ab}}{\partial q_c}\) will be zero if \(c \neq a\). Also, \(\frac{\mathrm{d} v_a}{\mathrm{d}t} = \frac{\mathrm{d}^2 q_a}{\mathrm{d}t^2} = \dot{v_a}\) and \(\frac{\mathrm{d} v_b}{\mathrm{d}t} = \frac{\mathrm{d}^2 q_b}{\mathrm{d}t^2} = \dot{v_b}\) have opposite sign, since \(a\neq b\), so they cancel each other: $$ \frac{\mathrm{d} L}{\mathrm{d}t} = \frac{1}{2}\left( \frac{\partial T_{ab}}{\partial q_a} \frac{\mathrm{d} q_a}{\mathrm{d}t} v_a v_b + T_{ab} \dot{v_a} v_b - T_{ab} v_a \dot{v_b}\right) = 0 $$

Show that \(L'=f(L)\) generates the same dynamics

Now, let's consider a transformed Lagrangian \(L' = f(L)\), where \(f\) is any function of one variable. The Euler-Lagrange equations for the original Lagrangian \(L\) are given by: $$ \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial L}{\partial v_a}\right) - \frac{\partial L}{\partial q_a} = 0 $$ For the transformed Lagrangian \(L'=f(L)\), the Euler-Lagrange equations are: $$ \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial (f(L))}{\partial v_a}\right) - \frac{\partial (f(L))}{\partial q_a} = 0 $$ Since \(f\) does not depend on \(v_a\) or \(q_a\) explicitly, we have $$ \frac{\partial (f(L))}{\partial v_a} = f'(L) \frac{\partial L}{\partial v_a} $$ and $$ \frac{\partial (f(L))}{\partial q_a} = f'(L) \frac{\partial L}{\partial q_a} $$ Using these expressions, we can rewrite the Euler-Lagrange equations for \(L'\) as $$ \frac{\mathrm{d}}{\mathrm{d}t} \left( f'(L) \frac{\partial L}{\partial v_a}\right) - f'(L) \frac{\partial L}{\partial q_a} = 0 $$ Now, since \(f'(L)\) is a nonzero function, we can divide both sides of the equation by it and obtain $$ \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial L}{\partial v_a}\right) - \frac{\partial L}{\partial q_a} = 0 $$ This is the same Euler-Lagrange equation as for the original Lagrangian \(L\). Therefore, the transformed Lagrangian \(L'=f(L)\) generates the same dynamics as \(L\).

Key concepts

Lagrangian Mechanics

Lagrangian mechanics is one of the foundational pillars of classical mechanics, revered for its elegant approach to describing the motion of systems. At its core, it uses a single scalar quantity known as the Lagrangian, denoted by 'L', which encapsulates the dynamics of a system. The Lagrangian is constructed as the difference between the kinetic energy 'T' and potential energy 'V' of the system, i.e.,
\( L = T - V \).
This formulation is particularly powerful because it allows for the analysis of complex systems with many interacting parts in a coordinated and simplified manner. In this context, the principle of least action plays a central role: it states that the path taken by the system between two points in configuration space is the one that minimizes the action, which is the integral of the Lagrangian over time. By applying this principle, we can derive equations of motion that are equivalent to Newton's laws but often much easier to work with, especially in systems with constraints.

Generalized Coordinates

Generalized coordinates are a set of parameters that uniquely define the configuration of a system in terms of degrees of freedom, regardless of the constraints present. These coordinates are denoted by \( q_a \), where 'a' might range from 1 to the number of degrees of freedom of the system. The beauty of generalized coordinates lies in their ability to simplify the mathematical description of complex systems by reducing the number of variables to the minimum necessary.

For instance, a double pendulum's state can be fully described by two angles instead of the cartesian coordinates for each mass in the pendulum. In the given exercise, the Lagrangian is expressed with \( T_{ab} \) as a function of generalized coordinates, and \( v_a \) and \( v_b \) are the associated generalized velocities, representing the time rates of change of the generalized coordinates. By varying these coordinates, one can explore the system's behavior over its entire range of possible states.

Euler-Lagrange Equations

The Euler-Lagrange equations are the bridge between the Lagrangian formulation of a system and its actual motion. These are differential equations derived from the principle of least action and contain all necessary information about the dynamics of a system. The standard form of these equations for each generalized coordinate \( q_a \) is given by:
\[ \frac{d}{dt} \left( \frac{\partial L}{\partial v_a} \right) - \frac{\partial L}{\partial q_a} = 0 \]
In these equations, \( \frac{\partial L}{\partial v_a} \) denotes the partial derivative of the Lagrangian with respect to the generalized velocity, while \( \frac{\partial L}{\partial q_a} \) represents the partial derivative with respect to the generalized coordinate itself. The exercise showcases that even if we transform the original Lagrangian 'L' by a function 'f', giving us a new Lagrangian \( L' = f(L) \), as long as 'f' is a function of 'L' only and does not explicitly depend on \( v_a \) or \( q_a \), the derived Euler-Lagrange equations are structurally unchanged.

This invariance implies that both Lagrangians describe the same dynamics. Therefore, any choice of 'f' that satisfies these conditions will yield a legitimate description of the system's evolution over time, confirming the conceptual depth and flexibility of the Lagrangian formulation.