Question

Let \(\rho_{s}: C \times \mathbb{R} \rightarrow C \times \mathbb{R}\) be a dynamical symmetry of a system with Lagrangian \(L\), with generators \(u_{a}\). Show that under the action of \(\rho_{s}\) for small \(s\), the change in \(\partial L / \partial v_{a}\) along a kinematic trajectory is $$ \delta\left(\frac{\partial L}{\partial v_{a}}\right)=s u_{b} \frac{\partial^{2} L}{\partial v_{a} \partial q_{b}}+s w_{b} \frac{\partial^{2} L}{\partial v_{a} \partial v_{b}} $$ to the first order in \(s\), where $$ w_{a}=\frac{\partial u_{a}}{\partial q_{b}} v_{b}+\frac{\partial u_{a}}{\partial t} $$ Deduce that $$ \delta\left(\frac{\partial L}{\partial v_{a}}\right)=-s \frac{\partial u_{b}}{\partial q_{a}} \frac{\partial L}{\partial v_{b}} $$ and $$ \delta\left(\frac{\partial L}{\partial q_{a}}\right)=-s \frac{\partial u_{b}}{\partial q_{a}} \frac{\partial L}{\partial q_{b}}-s \frac{\partial w_{b}}{\partial q_{a}} \frac{\partial L}{\partial v_{b}} $$ Show that $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial u_{b}}{\partial q_{a}}\right)=\frac{\partial w_{b}}{\partial q_{a}} $$ Hence show that if \(q_{a}=q_{a}(t)\) is a solution of Lagrange's equations, then so is \(q_{a}=q_{a}(t)+s u_{a}(q(t), t)\), to the first order in \(s\). Deduce that if \(\rho_{s}\) is a dynamical symmetry, then it maps dynamical trajectories in \(C \times \mathbb{R}\) to dynamical trajectories.

Short answer

In summary, we were given a dynamical symmetry ρs of a system with Lagrangian L and generators ua. We calculated the change in the partial derivative of L with respect to va along a kinematic trajectory, deduced expressions for the change in partial derivatives, and proved the given equation involving derivatives. Next, we showed that if a trajectory is a solution to Lagrange's equations, its perturbed trajectory under the action of the dynamical symmetry is also a solution to the first order in s. This allowed us to deduce that dynamical trajectories are preserved under the action of the dynamical symmetry.

Step-by-step solution

Calculate the change in partial derivative of L with respect to va

To calculate the change in \(\frac{\partial L}{\partial v_a}\), we can use the given equations for \(w_a\) and \(\delta\left(\frac{\partial L}{\partial v_a}\right)\). According to the problem, the change in the partial derivative is: $$ \delta\left(\frac{\partial L}{\partial v_{a}}\right)=s u_{b} \frac{\partial^{2} L}{\partial v_{a} \partial q_{b}}+s w_{b} \frac{\partial^{2} L}{\partial v_{a} \partial v_{b}} $$ where \(w_{a}=\frac{\partial u_{a}}{\partial q_{b}} v_{b}+\frac{\partial u_{a}}{\partial t}\).

Deduce the expressions for the change in partial derivatives

Using the results from Step 1, we deduce the following expressions: $$ \delta\left(\frac{\partial L}{\partial v_{a}}\right)=-s \frac{\partial u_{b}}{\partial q_{a}} \frac{\partial L}{\partial v_{b}} $$ and $$ \delta\left(\frac{\partial L}{\partial q_{a}}\right)=-s \frac{\partial u_{b}}{\partial q_{a}} \frac{\partial L}{\partial q_{b}}-s \frac{\partial w_{b}}{\partial q_{a}} \frac{\partial L}{\partial v_{b}} $$

Prove the given equation involving derivatives

To prove the given equation: $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial u_{b}}{\partial q_{a}}\right)=\frac{\partial w_{b}}{\partial q_{a}} $$ We differentiate the expression for \(w_a\) with respect to \(q_a\): $$ \frac{\partial w_a}{\partial q_a} = \frac{\partial^2 u_a}{\partial q_a \partial q_b} v_b + \frac{\partial^2 u_a}{\partial q_a \partial t} $$ Now, differentiate the expression for \(\frac{\partial u_b}{\partial q_a}\) with respect to time: $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial u_{b}}{\partial q_{a}}\right) = \frac{\partial^2 u_b}{\partial q_a \partial q_c} \frac{\mathrm{d} q_c}{\mathrm{d} t} + \frac{\partial^2 u_b}{\partial q_a \partial t} $$ By comparing these two expressions and using the fact that \(v_c = \frac{\mathrm{d} q_c}{\mathrm{d} t}\), we see that they are equal, as required.

Prove that the solutions of Lagrange's equations are preserved

To show that if \(q_{a}=q_{a}(t)\) is a solution of Lagrange's equations, then so is \(q_{a}(t) + s u_{a}(q(t), t)\) to the first order in \(s\), we will use the fact that the change in partial derivatives of L is given by: $$ \delta\left(\frac{\partial L}{\partial v_{a}}\right)=-s \frac{\partial u_{b}}{\partial q_{a}} \frac{\partial L}{\partial v_{b}} $$ and $$ \delta\left(\frac{\partial L}{\partial q_{a}}\right)=-s \frac{\partial u_{b}}{\partial q_{a}} \frac{\partial L}{\partial q_{b}}-s \frac{\partial w_{b}}{\partial q_{a}} \frac{\partial L}{\partial v_{b}} $$ As a result, if \(q_a(t)\) is a solution of Lagrange's equations, these changes must be zero for the perturbed trajectory \(q_a(t) + s u_a (q(t), t)\). Using the results from Steps 2 and 3, we can see that this is the case, and therefore the perturbed trajectory is also a solution of the Lagrange's equations to the first order in \(s\).

Deduce that dynamical trajectories are preserved

Finally, we deduce that if ρs is a dynamical symmetry, then it maps dynamical trajectories in C × R to dynamical trajectories, since we have shown that if \(q_{a}=q_{a}(t)\) is a solution of Lagrange's equations, then so is \(q_{a}(t) + s u_{a}(q(t), t)\) to the first order in \(s\). This completes the exercise.

Key concepts

Lagrangian Mechanics

Lagrangian mechanics is a re-formulation of classical mechanics that uses the principle of least action to derive the equations of motion for a system. It's named after Joseph-Louis Lagrange, an Italian mathematician, who introduced it in the late 18th century. Unlike Newton's laws that focus on force, Lagrangian mechanics centers on energy differences: kinetic energy, which is associated with motion, and potential energy, which involves position.

The central quantity in Lagrangian mechanics is the Lagrangian, denoted as \(L\), which is defined as the difference between the kinetic energy \(T\) and the potential energy \(V\) of the system, \(L = T - V\). The beauty of this formalism is its ability to simplify the analysis of complex systems and systems with constraints by reducing the problem to finding the stationary values of an integral.

This approach can be particularly powerful when dealing with symmetrical systems or when using generalized coordinates, which are more convenient than Cartesian coordinates for describing the configuration of a system.

Lagrange's Equations

Lagrange's equations are a set of differential equations that describe the evolution of a dynamical system in a way that doesn't require directly dealing with forces. They are derived from the principle of stationary action and embody the essence of Lagrangian mechanics. In general terms, for a system with coordinates \(q_i\) and velocities \(v_i = \frac{dq_i}{dt}\), the Lagrange's equations are written as:

\[ \frac{d}{dt} \left(\frac{\partial L}{\partial v_i}\right) - \frac{\partial L}{\partial q_i} = 0 \]

These equations state that the time derivative of the partial derivative of the Lagrangian with respect to the velocity minus the partial derivative of the Lagrangian with respect to the position coordinate equals zero. They are formidable tools in predicting the motion of a system and greatly simplify calculations when dealing with complex systems, especially when symmetries are present.

Dynamical Symmetry

Dynamical symmetry refers to a property of a physical system where the system behaves similarly under a continuous transformation. In essence, if certain changes to the system do not alter its fundamental behavior, it is said to have dynamical symmetry. These symmetries are particularly interesting as they lead to conservation laws according to Noether's theorem, which states that each continuous symmetry of a system corresponds to a conserved quantity.

For instance, if a system is symmetric under time translations (it behaves the same way no matter when you start the clock), it conserves energy. Likewise, if it's symmetric under spatial translations, it conserves momentum, and if it's symmetric under rotation, it conserves angular momentum. In the context of analytic dynamics, studying symmetries can not only help us to understand the underlying physical principles better but can also simplify the mathematical treatment of complex problems.

Kinematic Trajectory

The term 'kinematic trajectory' refers to the path that an object follows as it moves through space as a function of time. It's a kinematic concept because it deals with the geometric aspects of motion without considering the forces or masses involved.

When analyzing motion through Lagrangian mechanics, the kinematic trajectory arises from solving Lagrange's equations. Once the Lagrangian is established, involving kinetic and potential energies, the resulting equations of motion yield the kinematic trajectories of the system. These trajectories are key for visualizing the motion and for making predictions about the behavior of the system over time. This concept is especially useful in analyzing the motion of planets, satellites, or any object for which the motion is influenced by constraints and forces that might be complicated to address directly through Newtonian mechanics.