Question

The dynamics of a system with \(n\) degrees of freedom are governed by a Lagrangian \(L(q, v, t)\). Show that if \(f(q, t)\) is any function on \(C T\), then $$ L^{\prime}=L+\frac{\partial f}{\partial q_{a}} v_{a}+\frac{\partial f}{\partial t} $$ generates the same dynamics.

Short answer

Question: Show that if we have a Lagrangian \(L(q, v, t)\) for a system with \(n\) degrees of freedom, and \(f(q, t)\) is any function on \(C T\), then the modified Lagrangian \(L^{\prime}=L+\frac{\partial f}{\partial q_{a}} v_{a}+\frac{\partial f}{\partial t}\) will generate the same dynamics. Answer: By applying the Euler-Lagrange equation to both the original and modified Lagrangians and showing that they result in the same equations of motion, we can conclude that the modified Lagrangian \(L^{\prime}\) generates the same dynamics as the original Lagrangian \(L\).

Step-by-step solution

Write down the Euler-Lagrange equation for the original Lagrangian

To find the equations of motion for the original Lagrangian, we first apply the Euler-Lagrange equation: $$ \frac{d}{dt}\frac{\partial L}{\partial v_a} - \frac{\partial L}{\partial q_a} = 0 $$ This equation represents the dynamics of the system for each degree of freedom \(a\).

Write down the modified Lagrangian

Now we have the modified Lagrangian \(L^{\prime}\) defined as follows: $$ L^{\prime}=L+\frac{\partial f}{\partial q_{a}} v_{a}+\frac{\partial f}{\partial t} $$

Write down the Euler-Lagrange equation for the modified Lagrangian

First, we will calculate \(\frac{\partial L^{\prime}}{\partial q_a}\), \(\frac{\partial L^{\prime}}{\partial v_a}\), and \(\frac{d}{dt}\frac{\partial L^{\prime}}{\partial v_a}\) and then, we will apply the Euler-Lagrange equation for the modified Lagrangian \(L^{\prime}\). $$ \frac{d}{dt}\frac{\partial L^{\prime}}{\partial v_a} - \frac{\partial L^{\prime}}{\partial q_a} = 0 $$

Calculate necessary derivatives of \(L^{\prime}\)

First, we find the derivatives of \(L^{\prime}\) with respect to \(q_a\) and \(v_a\). $$ \frac{\partial L^{\prime}}{\partial q_a} = \frac{\partial L}{\partial q_a} + \frac{\partial^2 f}{\partial q_a^2}v_a + \frac{\partial^2 f}{\partial q_a \partial t} $$ $$ \frac{\partial L^{\prime}}{\partial v_a} = \frac{\partial L}{\partial v_a} + \frac{\partial^2 f}{\partial q_a^2} $$ Now, we find the time derivative of \(\frac{\partial L^{\prime}}{\partial v_a}\). $$ \frac{d}{dt}\frac{\partial L^{\prime}}{\partial v_a} = \frac{d}{dt}(\frac{\partial L}{\partial v_a} + \frac{\partial^2 f}{\partial q_a^2}) $$ $$ \frac{d}{dt}\frac{\partial L^{\prime}}{\partial v_a} = \frac{d}{dt}\frac{\partial L}{\partial v_a} + \frac{d}{dt}(\frac{\partial^2 f}{\partial q_a^2}) $$

Apply the Euler-Lagrange equation for the modified Lagrangian

Now, plug the derivatives we found into the Euler-Lagrange equation for \(L^{\prime}\). $$ \frac{d}{dt}\frac{\partial L^{\prime}}{\partial v_a} - \frac{\partial L^{\prime}}{\partial q_a} = \frac{d}{dt}\frac{\partial L}{\partial v_a} + \frac{d}{dt}(\frac{\partial^2 f}{\partial q_a^2}) - \frac{\partial L}{\partial q_a} - \frac{\partial^2 f}{\partial q_a^2}v_a - \frac{\partial^2 f}{\partial q_a \partial t} = 0 $$ Notice that the terms \(\frac{d}{dt}\frac{\partial L}{\partial v_a} - \frac{\partial L}{\partial q_a}\) represent the Euler-Lagrange equation for the original Lagrangian, which is equal to zero.

Simplify and show that the dynamics are the same

Since the Euler-Lagrange equation for the original Lagrangian is equal to zero, we have: $$ \frac{d}{dt}(\frac{\partial^2 f}{\partial q_a^2}) - \frac{\partial^2 f}{\partial q_a^2}v_a - \frac{\partial^2 f}{\partial q_a \partial t} = 0 $$ This equation does not depend on the original Lagrangian \(L(q, v, t)\), which implies that the dynamics generated by the modified Lagrangian \(L^{\prime}\) are the same as the dynamics generated by the original Lagrangian \(L\).