The curve is the graph of the function DF. The left-hand side depends only on the point labeled by w and v and the right-hand side depends only on the point labeled by w 0 and v 0 , so these must be constant, independent of the variable endpoints. For the active arguments the derivation goes through as before. The first argument to F and G is just along for the ride—it is a passive argument. The partial derivatives with respect to the passive arguments are related in a remarkably simple way.

The last relation is not as trivial as it looks, because x enters the equations connecting w and v. With this symmetrical form, we see that the Legendre transform is its own inverse. For each of the following functions, find the function that is related to the given function by the Legendre transform on the indicated active argument.

## Lagrangian and Hamiltonian mechanics. Solutions to exercises - PDF Free Download

Show that the Legendre transform relations hold for your solution, including the relations among passive arguments, if any. We can use the Legendre transformation with the Lagrangian playing the role of F and with the generalized velocity slot playing the role of the active argument.

The Hamiltonian plays the role of G with the momentum slot active. The coordinate and time slots are passive arguments. This relation is purely algebraic and is valid for any path. The passive equation 3. This equation is valid only for realizable paths, because we used the Lagrange equations to derive it.

We have found that if the Lagrangian has no explicit time dependence, then energy is conserved. So if the Hamiltonian has no explicit time dependence then it is a conserved quantity. Using Hamilton's equations, show directly that the Hamiltonian is a conserved quantity if it has no explicit time dependence. We cannot implement the Legendre transform in general because it involves finding the functional inverse of an arbitrary function. However, many physical systems can be described by Lagrangians that are quadratic forms in the generalized velocities.

For such functions the generalized momenta are linear functions of the generalized velocities, and thus explicitly invertible. More generally, we can compute a Legendre transformation for polynomial functions where the leading term is a quadratic form:. Because the first term is a quadratic form only the symmetric part of M contributes to the result, so we can assume M is symmetric. So if M is invertible we can solve for v in terms of w. Thus we may define a function V such that.

## Lagrangian and Hamiltonian Mechanics Solutions to the Exercises

We implement the Legendre transform for quadratic functions by the procedure The procedure Legendre-transform takes a procedure of one argument and returns the procedure that is associated with it by the Legendre transform. We can use the Legendre-transform procedure to compute a Hamiltonian from a Lagrangian:.

Notice that the one-argument Legendre-transform procedure is sufficient. The passive variables are given no special attention, they are just passed around. For example, the Hamiltonian for the motion of the point mass with the potential energy V x , y may be computed from the Lagrangian:. And the Hamiltonian is, as we saw in equation 3. A uniform cylinder of mass M , radius R , and height h is mounted so as to rotate freely on a vertical axis. What are the degrees of freedom of this system? Pick and describe a convenient set of generalized coordinates for this problem.

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Write a Lagrangian to describe the dynamical behavior. It may help to know that the moment of inertia of a cylinder around its axis is 1 2 M R 2. You may find it easier to do the algebra if various constants are combined and represented as single symbols. Make a Hamiltonian for the system. Write Hamilton's equations for the system. Are there any conserved quantities?

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Consider a point particle of mass m constrained to move in a bowl and acted upon by a uniform gravitational acceleration g. Make a Hamiltonian for this system. Can you make any immediate deductions about this system? The previous two derivations of Hamilton's equations made use of the Lagrange equations. Hamilton's equations can also be derived directly from the action principle.

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The action is stationary with respect to variations of a realizable path that preserve the configuration at the endpoints for Lagrangians that are functions of time, coordinates, and velocities. The Legendre transformation construction gives. Rearranging terms, the variation of the action is. As a consequence of equation 3. We are left with. For the variation of the action to be zero for arbitrary variations, except for the endpoint conditions, we must have. Figure 3. The wires represent identifications of the quantities on the terminals that they connect. For example, there is a box that represents the Lagrangian function.

Other terminals of the Lagrangian carry the values of the partial derivatives of the Lagrangian function. The upper part of the diagram summarizes the relationship of the Hamiltonian to the Lagrangian. This is the active part of the Legendre transform.

The passive variables are related by the corresponding partial derivatives being negations of each other. In the lower part of the diagram the equations of motion are indicated by the presence of the integrators, relating the dynamical quantities to their time derivatives. One can use this diagram to help understand the underlying unity of the Lagrangian and Hamiltonian formulations of mechanics.

We see that the two formulations are consistent. One does not have to abandon any part of the Lagrangian formulation to use the Hamiltonian formulation: there are deductions that can be made using both simultaneously. Here we introduce the Poisson bracket, in terms of which Hamilton's equations have an elegant and symmetric expression.

Consider a function F of time, coordinates, and momenta. If the phase-space path is a realizable path for a system with Hamiltonian H , then Dq and Dp can be reexpressed using Hamilton's equations:.

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Note that the Poisson bracket of two functions on the phase-state space is also a function on the phase-state space. According to equation 3. We have. The Poisson bracket of any function with itself is zero, so we recover the conservation of energy for a system that has no explicit time dependence:. Let F , G , and H be functions of time, position, and momentum, and let c be independent of position and momentum.