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 $CT$, 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 $CT$, 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 $L$.