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A roller coaster of mass \(m\) moves along a frictionless track that lies in the \(x y\) plane \((x\) horizontal and \(y\) vertically up). The height of the track above the ground is given by \(y=h(x) .\) (a) Using \(x\) as your generalized coordinate, write down the Lagrangian, the generalized momentum \(p,\) and the Hamiltonian \(\mathcal{H}=p \dot{x}-\mathcal{L}\) (as a function of \(x\) and \(p\) ). (b) Find Hamilton's equations and show that they agree with what you would get from the Newtonian approach. [Hint: You know from Section 4.7 that Newton's second law takes the form \(F_{\text {tang }}=m \ddot{s},\) where \(s\) is the distance measured along the track. Rewrite this as an equation for \(\ddot{x}\) and show that you get the same result from Hamilton's equations.]

Short Answer

Expert verified
The Hamiltonian is \( \mathcal{H} = \frac{p^2}{2m(1 + (\frac{dh}{dx})^2)} + mgh(x) \), and Hamilton's equations match Newton's second law.

Step by step solution

01

Write Down Kinetic and Potential Energy

For a roller coaster moving on a frictionless track in the xy-plane, the kinetic energy (T) is given by: \[ T = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) \] Since the track height is \(y = h(x)\), \( \dot{y} = \frac{dh}{dx} \dot{x} \). Therefore, \[ T = \frac{1}{2} m \dot{x}^2 \left( 1 + \left( \frac{dh}{dx} \right)^2 \right) \] The potential energy (U) is given by: \[ U = mgh(x) \]
02

Construct Lagrangian

The Lagrangian \( \mathcal{L} \) is defined as the difference between kinetic and potential energy: \[ \mathcal{L} = T - U = \frac{1}{2} m \dot{x}^2 \left( 1 + \left( \frac{dh}{dx} \right)^2 \right) - mgh(x) \]
03

Determine Generalized Momentum

The generalized momentum \(p\) is the derivative of the Lagrangian with respect to \(\dot{x}\): \[ p = \frac{\partial \mathcal{L}}{\partial \dot{x}} = m \dot{x} \left( 1 + \left( \frac{dh}{dx} \right)^2 \right) \]
04

Write Hamiltonian

Hamiltonian \(\mathcal{H}\) is defined by \(\mathcal{H} = p \dot{x} - \mathcal{L}\). Start by solving for \(\dot{x}\): \[ \dot{x} = \frac{p}{m \left( 1 + \left( \frac{dh}{dx} \right)^2 \right)} \] Then, substitute \(\dot{x}\) into \(\mathcal{H}\): \[ \mathcal{H} = \frac{p^2}{2m (1 + \left( \frac{dh}{dx} \right)^2)} + mgh(x) \]
05

Derive Hamilton's Equations

The Hamilton's equations are given by: \[ \dot{x} = \frac{\partial \mathcal{H}}{\partial p}, \quad \dot{p} = -\frac{\partial \mathcal{H}}{\partial x} \] Calculate these partial derivatives: \[ \dot{x} = \frac{p}{m(1 + (\frac{dh}{dx})^2)} \] \[ \dot{p} = -\left( \frac{d}{dx} \left( \frac{p^2}{2m (1+(\frac{dh}{dx})^2)} \right) + mg\frac{dh}{dx} \right) \]
06

Relate to Newton's Second Law

According to Newton's second law, \[ F_{\text{tang.}} = m \frac{d^2 s}{dt^2} \] Express \(\ddot{s}\) in terms of \(\ddot{x}\): \[ m \ddot{x} (1 + (\frac{dh}{dx})^2) = p \cdot \frac{d}{dt}\left( \frac{\dot{x}}{\dot{s}} \right) \] This demonstrates consistency with results from Hamilton's equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hamiltonian Mechanics
Hamiltonian mechanics is an alternative to the classical Lagrangian mechanics. It provides a powerful framework for analyzing dynamic systems. In Hamiltonian mechanics, the Hamiltonian function \( \mathcal{H} \) plays a central role. The Hamiltonian often represents the total energy of the system, encompassing both kinetic and potential energy.

Unlike the Lagrangian, which depends on generalized coordinates and their velocities, the Hamiltonian is expressed in terms of generalized coordinates and momenta. This places it in the phase space rather than configuration space, as in Lagrangian mechanics.

A key feature of Hamiltonian mechanics is its formulation of conservative motion using symplectic geometry. This means that the Hamiltonian's equations of motion directly connect to the conservation of physical quantities, such as energy.

This framework provides insights into quantum mechanics and statistical mechanics because of its abstraction and generality. Understanding the Hamiltonian allows us to describe systems ranging from simple pendulums to more complex particle fields.
Generalized Momentum
Generalized momentum, denoted \( p \), is a crucial component in both Lagrangian and Hamiltonian mechanics. It is defined for each generalized coordinate \(q\) as the derivative of the Lagrangian \( \mathcal{L} \) with respect to its corresponding generalized velocity \( \dot{q} \). The formula is:
  • \[ p = \frac{\partial \mathcal{L}}{\partial \dot{q}} \]

In the context of the roller coaster problem, the generalized coordinate is \( x \), the horizontal position. Thus, the calculation of generalized momentum becomes:
  • \[ p = m \dot{x} \left( 1 + \left( \frac{dh}{dx} \right)^2 \right) \]

This expression arises because the kinetic energy involves the motion not only along \( x \) but also in \( y \) direction, as determined by the track's shape. Generalized momentum encompasses more than just typical linear momentum. It accounts for the potential contribution from constraints and forces depending on the system's configuration.

Generalized momentum is essential for deriving Hamilton's equations, and it is useful for transitioning from Lagrangian to Hamiltonian mechanics.
Hamilton's Equations
Hamilton's equations provide a set of first-order differential equations used to describe the evolution of a system over time. They link the Hamiltonian with the changes in generalized coordinates and momenta. The equations are:
  • \[ \dot{x} = \frac{\partial \mathcal{H}}{\partial p} \]
  • \[ \dot{p} = -\frac{\partial \mathcal{H}}{\partial x} \]

The first equation determines how the position changes with time, while the second equation determines how the momentum changes. They replace Newton’s laws of motion in this framework.

For the roller coaster example, solving the Hamiltonian with these equations results in expressions for \( \dot{x} \) and \( \dot{p} \) that are consistent with classical mechanics. It ties back to Newton’s second law, demonstrating that despite the different mathematical structures, they lead to the same physical dynamics.

Hamilton’s equations highlight the power of using energy functions to describe mechanical systems, offering elegant solutions especially when dealing with complex potentials or non-Cartesian coordinates.

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Most popular questions from this chapter

Here is a simple example of a canonical transformation that illustrates how the Hamiltonian formalism lets one mix up the \(q^{\prime}\) s and the \(p\) 's. Consider a system with one degree of freedom and Hamiltonian \(\mathcal{H}=\mathcal{H}(q, p)\). The equations of motion are, of course, the usual Hamiltonian equations \(\dot{q}=\partial \mathcal{H} / \partial p\) and \(\dot{p}=-\partial \mathcal{H} / \partial q .\) Now consider new coordinates in phase space defined as \(Q=p\) and \(P=-q .\) Show that the equations of motion for the new coordinates \(Q\) and \(P\) are \(\dot{Q}=\partial \mathcal{H} / \partial P\) and \(\dot{P}=-\partial \mathscr{H} / \partial Q ;\) that is, the Hamiltonian formalism applies equally to the new choice of coordinates where we have exchanged the roles of position and momentum.

Find the Lagrangian, the generalized momenta, and the Hamiltonian for a free particle (no forces at all) moving in three dimensions. (Use \(x, y, z\) as your generalized coordinates.) Find and solve Hamilton's equations.

Evaluate the three-dimensional divergence \(\nabla\). \(\mathbf{v}\) for each of the following vectors: \((\mathbf{a}) \mathbf{v}=k \hat{\mathbf{x}}\) (b) \(\mathbf{v}=k x \hat{\mathbf{x}},(\mathbf{c}) \mathbf{v}=k y \hat{\mathbf{x}} .\) We know that \(\mathbf{\nabla} \cdot \mathbf{v}\) represents the net outward flow associated with \(\mathbf{v} .\) In those cases where you found \(\nabla \cdot \mathbf{v}=0,\) make a simple sketch to illustrate that the outward flow is zero; in those cases where you found \(\nabla \cdot \mathbf{v} \neq 0,\) make a sketch to show why and whether the outflow is positive or negative.

Evaluate the three-dimensional divergence \(\nabla \cdot\) v for each of the following vectors: (a) \(\mathbf{v}=k \mathbf{r}\) (b) \(\mathbf{v}=k(z, x, y),(\mathbf{c}) \mathbf{v}=k(z, y, x),(\mathbf{d}) \mathbf{v}=k(x, y,-2 z),\) where \(\mathbf{r}=(x, y, z)\) is the usual position vector and \(k\) is a constant.

The general proof of the divergence theorem $$\int_{S} \mathbf{n} \cdot \mathbf{v} d A=\int_{V} \nabla \cdot \mathbf{v} d V$$ is fairly complicated and not especially illuminating. However, there are a few special cases where it is reasonably simple and quite instructive. Here is one: Consider a rectangular region bounded by the six planes \(x=X\) and \(X+A, y=Y\) and \(Y+B,\) and \(z=Z\) and \(Z+C,\) with total volume \(V=A B C .\) The surface \(S\) of this region is made up of six rectangles that we can call \(S_{1}\) (in the plane \(x=X\) ), \(S_{2}\) (in the plane \(x=X+A\) ), and so on. The surface integral on the left of (13.63) is then the sum of six integrals, one over each of the rectangles \(S_{1}, S_{2},\) and so forth. (a) Consider the first two of these integrals and show that $$ \int_{S_{1}} \mathbf{n} \cdot \mathbf{v} d A+\int_{S_{2}} \mathbf{n} \cdot \mathbf{v} d A=\int_{Y}^{Y+B} d y \int_{Z}^{Z+C} d z\left[v_{x}(X+A, y, z)-v_{x}(X, y, z)\right] $$ (b) Show that the integrand on the right can be rewritten as an integral of \(\partial v_{x} / \partial x\) over \(x\) running from \(x=X\) to \(x=X+A .(\text { c) Substitute the result of part }(b)\) into part (a), and write down the corresponding results for the two remaining pairs of faces. Add these results to prove the divergence theorem (13.63).

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