Question

Given the Lagrangian L(q,˙q,t)=˙q2q−V(q)+˙qt+qL(q,q˙,t)=q˙2q−V(q)+q˙t+q depending on the generalized coordinateq(t)q(t), its total time derivative˙q(t)q˙(t)and timett. Is the Hamiltonian energyHHequal to the total energy EE for this system?

Is˙q2q+V(q)q˙2q+V(q)a conserved quantity?

Answer 1

The Hamiltonian energy is not necessarily equal to the total energy for this system and the equation is not a conserved quantity.

Step-by-step explanation:

The Lagrangian of a system is a function that describes the dynamics of the system in terms of generalized coordinates, their time derivatives, and time. In this case, the Lagrangian is defined as L(q,˙q,t)=˙q²q−(q)+˙qt+q.

The Hamiltonian of a system is defined as the sum of the generalized momentum multiplied by the corresponding generalized velocity minus the Lagrangian.

For this system, the Hamiltonian H is given by H = p˙q−L(q,˙q,t), where p is the generalized momentum conjugate to q. The total energy of a system is the sum of its kinetic and potential energy.

In this case, the total energy E is given by E = 1/2 m˙q² + V(q), where m is the mass and V(q) is the potential energy. So, to answer the first question, the Hamiltonian energy H is not necessarily equal to the total energy E for this system.

The equation ˙q²q+V(q) is not a conserved quantity as it does not remain constant throughout the motion.