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Find the Lagrangian, the generalized momentum, and the Hamiltonian for a free particle (no forces at all) confined to move along the \(x\) axis. (Use \(x\) as your generalized coordinate.) Find and solve Hamilton's equations.

Short Answer

Expert verified
The Lagrangian is \(L = \frac{1}{2} m \dot{x}^2\); generalized momentum \(p = m \dot{x}\); Hamiltonian \(H = \frac{p^2}{2m}\); \(\dot{x} = \frac{p}{m}\), \(\dot{p} = 0\).

Step by step solution

01

Define the Lagrangian

A free particle moving along the x-axis has kinetic energy but no potential energy. The Lagrangian, which is the difference between kinetic and potential energy, is given by:\[ L = T - V = \frac{1}{2} m \dot{x}^2 - 0 = \frac{1}{2} m \dot{x}^2, \]where \( m \) is the mass and \( \dot{x} \) is the velocity of the particle.
02

Determine Generalized Momentum

The generalized momentum \( p \) is derived from the Lagrangian and is given by the derivative of the Lagrangian with respect to the velocity \( \dot{x} \):\[ p = \frac{\partial L}{\partial \dot{x}} = m \dot{x}. \]
03

Derive the Hamiltonian

The Hamiltonian \( H \) is obtained by transforming the Lagrangian and is defined as:\[ H = \sum_{i} p_{i} \dot{q}_{i} - L. \]For our system, it translates into:\[ H = p \dot{x} - \frac{1}{2} m \dot{x}^2. \]Substituting \( p = m \dot{x} \) in, we have:\[ \dot{x} = \frac{p}{m} \]and substituting into the expression for Hamiltonian:\[ H = p \left(\frac{p}{m}\right) - \frac{1}{2} m \left(\frac{p}{m}\right)^2. \]Simplifying gives:\[ H = \frac{p^2}{m} - \frac{p^2}{2m} = \frac{p^2}{2m}. \]
04

Solve Hamilton's Equations

Hamilton's equations of motion are:\[ \dot{q} = \frac{\partial H}{\partial p} \quad \text{and} \quad \dot{p} = -\frac{\partial H}{\partial q}. \]For the free particle:\[ \dot{x} = \frac{\partial H}{\partial p} = \frac{p}{m}, \]\[ \dot{p} = -\frac{\partial H}{\partial x} = 0, \]The first equation reaffirms that \( \dot{x} = \frac{p}{m} \), which is consistent with our earlier findings. The second equation indicates that the momentum \( p \) is constant over time, as expected for a free particle.

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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 a reformulation of classical mechanics that offers a powerful and abstract framework for solving mechanical problems. While Lagrangian mechanics focuses on the difference between the kinetic and potential energy, Hamiltonian mechanics refines this approach by using energy conservation and symplectic geometry to make predictions about a system's future state.

The essence of Hamiltonian mechanics lies in its function, the Hamiltonian, denoted by \( H \). This function is typically expressed in terms of general coordinates and generalized momenta, and it often represents the total energy of the system (kinetic plus potential energy). For a free particle, the potential energy is zero, so the Hamiltonian consists entirely of kinetic energy, giving a simpler picture of the system's motion.

Working in the Hamiltonian framework is advantageous for complex systems because it transforms differential equations of motion from potentially second-order (in the Lagrangian form) to a set of first-order equations (Hamilton's equations). This simplification can make calculations more tractable and provides better insights into conservation principles.
generalized momentum
Generalized momentum is a crucial concept in both Lagrangian and Hamiltonian mechanics, bridging the gap between the two approaches. Unlike the conventional momentum we encounter in Newtonian physics, generalized momentum \( p \) may be associated with the derivative of the Lagrangian with respect to a generalized velocity \( \dot{q} \).

For our free particle moving along the \( x \)-axis, the generalized momentum is defined as:
  • \( p = \frac{\partial L}{\partial \dot{x}} = m \dot{x} \)
This equation shows that generalized momentum takes the intuitive form of mass times velocity, matching our classical understanding of linear momentum.

The importance of generalized momentum is highlighted when transitioning to Hamiltonian mechanics. Here, \( p \) becomes one of the primary variables in the Hamiltonian function, capturing the kinetic aspects of a mechanical system and facilitating the conversion of mechanical descriptions from configuration space (using velocities and positions) to phase space (using positions and momenta).

Understanding generalized momentum aids in unraveling complexities encountered in advanced topics like quantum mechanics and provides a solid foundation for analyzing dynamic systems.
Hamilton's equations
Hamilton's equations provide a set of first-order differential equations vital for predicting the behavior of dynamic systems. They describe how the system evolves over time in Hamiltonian mechanics, offering a clear path from initial conditions to the system's future states.

These equations are given by:
\[ \dot{q} = \frac{\partial H}{\partial p} \quad \text{and} \quad \dot{p} = -\frac{\partial H}{\partial q} \]For our scenario of a free particle:
  • The equation \( \dot{x} = \frac{\partial H}{\partial p} = \frac{p}{m} \) confirms that position evolves according to the particle's momentum and mass.
  • \( \dot{p} = -\frac{\partial H}{\partial x} = 0 \) reveals that the momentum is constant, aligning with expectations for a system without external forces.
These results emphasize the elegance and precision of Hamilton's equations in encapsulating motion, especially for systems with conserved quantities.

Hamilton's equations underscore the core principle of deterministic evolution in physics, where fixing the current state suffices to predict future states. Such robustness is pivotal in fields like celestial mechanics, where Hamiltonian mechanics offers insights into orbits and stability analysis.

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Most popular questions from this chapter

13.13 \star\star Consider a particle of mass \(m\) constrained to move on a frictionless cylinder of radius \(R\), given by the equation \(\rho=R\) in cylindrical polar coordinates \((\rho, \phi, z) .\) The mass is subject to just one external force, \(\mathbf{F}=-k r \hat{\mathbf{r}},\) where \(k\) is a positive constant, \(r\) is its distance from the origin, and \(\hat{\mathbf{r}}\) is the unit vector pointing away from the origin, as usual. Using \(z\) and \(\phi\) as generalized coordinates, find the Hamiltonian \(\mathcal{H}\). Write down and solve Hamilton's equations and describe the motion.

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).

A beam of particles is moving along an accelerator pipe in the \(z\) direction. The particles are uniformly distributed in a cylindrical volume of length \(L_{\mathrm{o}}\) (in the \(z\) direction) and radius \(R_{\mathrm{o}} .\) The particles have momenta uniformly distributed with \(p_{z}\) in an interval \(p_{\mathrm{o}} \pm \Delta p_{z}\) and the transverse momentum \(p_{\perp}\) inside a circle of radius \(\Delta p_{\perp} .\) To increase the particles' spatial density, the beam is focused by electric and magnetic fields, so that the radius shrinks to a smaller value \(R\). What does Liouville's theorem tell you about the spread in the transverse momentum \(p_{\perp}\) and the subsequent behavior of the radius \(R ?\) (Assume that the focusing does not affect either \(L_{\mathrm{o}}\) or \(\Delta p_{z}\).)

A bead of mass \(m\) is threaded on a frictionless wire that is bent into a helix with cylindrical polar coordinates \((\rho, \phi, z)\) satisfying \(z=c \phi\) and \(\rho=R,\) with \(c\) and \(R\) constants. The \(z\) axis points vertically up and gravity vertically down. Using \(\phi\) as your generalized coordinate, write down the kinetic and potential energies, and hence the Hamiltonian \(\mathcal{H}\) as a function of \(\phi\) and its conjugate momentum \(p\) Write down Hamilton's equations and solve for \(\ddot{\phi}\) and hence \(\ddot{z}\). Explain your result in terms of Newtonian mechanics and discuss the special case that \(R=0\).

The simple form \(\mathcal{H}=T+U\) is true only if your generalized coordinates are "natural" (relation betweeen generalized and underlying Cartesian coordinates is independent of time). If the generalized coordinates are not "natural," you must use the definition \(\mathcal{H}=\sum p_{i} \dot{q}_{i}-\mathcal{L} .\) To illustrate this point, consider the following: Two children are playing catch inside a railroad car that is moving with varying speed \(V\) along a straight horizontal track. For generalized coordinates you can use the position \((x, y, z)\) of the ball relative to a point fixed in the car, but in setting up the Hamiltonian you must use coordinates in an inertial frame \(-\) a frame fixed to the ground. Find the Hamiltonian for the ball and show that it is not equal to \(T+U\) (neither as measured in the car, nor as measured in the ground-based frame).

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