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Consider a mass \(m\) moving in two dimensions, subject to a single force \(\mathbf{F}\) that is independent of r and \(t\). (a) Find the potential energy \(U(\mathbf{r})\) and the Hamiltonian \(\mathcal{H}\). (b) Show that if you use rectangular coordinates \(x, y\) with the \(x\) axis in the direction of \(\mathbf{F},\) then \(y\) is ignorable. (c) Show that if you use rectangular coordinates \(x, y\) with neither axis in the direction of \(\mathbf{F},\) then neither coordinate is ignorable. (Moral: Choose generalized coordinates carefully!)

Short Answer

Expert verified
(a) \( U(\mathbf{r}) = Fx + C \); \( \mathcal{H} = \frac{1}{2}m(v_x^2 + v_y^2) + Fx + C \). (b) \( y \) is ignorable. (c) Neither \( x \) nor \( y \) is ignorable.

Step by step solution

01

Identify the Force and Potential Energy

The potential energy, \( U(\mathbf{r}) \), is related to the force \( \mathbf{F} \) by the equation \( \mathbf{F} = -abla U(\mathbf{r}) \). Since \( \mathbf{F} \) is constant (independent of \( \mathbf{r} \) and \( t \)), it must take the form \( \mathbf{F} = -F\hat{x} \) in some direction (e.g., along the \( x \)-axis). The potential energy function that gives rise to this constant force is linear in the corresponding coordinate, hence \( U(\mathbf{r}) = Fx + C \).
02

Find the Hamiltonian

The Hamiltonian \( \mathcal{H} \) of a system is given by the sum of the kinetic and potential energies. The kinetic energy for a mass \( m \) moving in two dimensions can be expressed as \( T = \frac{1}{2}m(v_x^2 + v_y^2) \). Therefore, the Hamiltonian is: \[ \mathcal{H} = T + U = \frac{1}{2}m(v_x^2 + v_y^2) + Fx + C. \]
03

Show y is Ignorable (Aligned with Force)

If the \( x \)-axis is aligned with the force \( \mathbf{F} \), then the potential energy \( U \) depends only on \( x \), not on \( y \). Thus, the Hamiltonian \( \mathcal{H} = \frac{1}{2}m(v_x^2 + v_y^2) + Fx + C \) does not explicitly involve \( y \), making \( y \) an ignorable coordinate.
04

Show No Ignorable Coordinates (Not Aligned with Force)

If neither the \( x \)-axis nor the \( y \)-axis is aligned with \( \mathbf{F} \), the force could be expressed as \( F_x \) and \( F_y \) (i.e., \( \mathbf{F} = -F_x\hat{i} - F_y\hat{j} \)). Then, the potential energy \( U = F_x x + F_y y \) depends on both \( x \) and \( y \). Hence, the Hamiltonian \( \mathcal{H} = \frac{1}{2}m(v_x^2 + v_y^2) + F_x x + F_y y + C \) involves both coordinates, making neither \( x \) nor \( y \) ignorable.

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Key Concepts

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

Potential Energy
Potential energy is an essential concept in studying Hamiltonian mechanics. It is a type of energy that is stored within a system due to its position or configuration. When a mass moves under the influence of a force that does not depend on time or the specific path taken, potential energy becomes especially important. For a force \( \mathbf{F} \), which is defined mathematically as \( \mathbf{F} = -abla U(\mathbf{r}) \), the change in potential energy is related to the work done by the force.

This relationship implies that the force \( \mathbf{F} \) is derived from a scalar potential function \( U(\mathbf{r}) \). If the force is constant as implied in our scenario and aligns along a single direction, the potential function \( U \) becomes linear with respect to that direction. In our problem, since the force is constant along the x-axis, the potential energy is expressed as \( U(x) = Fx + C \), where \( C \) is a constant. Here, only the x-coordinate contributes to the potential energy, simplifying many calculations in physics.

Understanding how forces relate to potential energy functions helps in predicting the movement and behavior of objects within a system. It is crucial in solving many physical problems, especially in conservative force fields where energy is conserved.
Force in Two Dimensions
When analyzing forces acting on a body in two dimensions, each force can be broken down into components along the x and y axes. This helps in simplifying the calculations and applying the principles of mechanics effectively.\(\mathbf{F} = -F \hat{i}\), assumes the simplest case where the force acts in one direction only, namely the x-direction.

If you do not align your coordinates along the force's direction, the components become \( F_x \) and \( F_y \). Thus, the system becomes slightly more complicated because the potential energy needs to be recalculated across both directions \(x\) and \(y\). It is expressed as \( U(x, y) = F_x x + F_y y \).

In an environment where neither axis aligns with the force direction, every point in space is influenced by components along both axes. This means both coordinates \(x\) and \(y\) are relevant for determining the dynamic behavior of the system. Such a setup is an effective way to understand how forces interact with bodies in more complex settings. Breaking down forces into two dimensions forms the basis for analyzing a variety of real-world problems.
Ignorable Coordinates
In Hamiltonian mechanics, an ignorable coordinate is a coordinate that does not explicitly appear in the calculation of potential energy or Hamiltonian of a system. This simplification occurs because changes along this coordinate do not affect the energy of the system. Hence, the equations of motion become easier as no dynamics are reflected along this dimension.

Consider the example from the exercise where \( y \) is ignorable. When the force is constant and aligned along the x-axis, the Hamiltonian, \( \mathcal{H} = \frac{1}{2}m(v_x^2 + v_y^2) + Fx + C \), does not involve the \( y \) coordinate. This is because the potential energy \( U \) only depends on \( x \), allowing \( y \) to be omitted from any changes in potential.
  • Benefits of Ignorable Coordinates: Simplifies analysis and reduces computational complexity.
  • When Ignorable Coordinates Arise: Occurs often when systems exhibit symmetries, such as uniform circular motion.
If neither coordinate axis aligns with the force, both must be considered, as effects from the force contribute to changes in both coordinates. Thus, understanding when and how coordinates can be ignored is crucial in simplifying and solving mechanical problems effectively.

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Most popular questions from this chapter

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).

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.

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.]

Consider a particle of mass \(m\) moving in two dimensions, subject to a force \(\mathbf{F}=-k x \hat{\mathbf{x}}+K \hat{\mathbf{y}}\) where \(k\) and \(K\) are positive constants. Write down the Hamiltonian and Hamilton's equations, using \(x\) and \(y\) as generalized coordinates. Solve the latter and describe the motion.

In the Lagrangian formalism, a coordinate \(q_{i}\) is ignorable if \(\partial \mathcal{L} / \partial q_{i}=0 ;\) that is, if \(\mathcal{L}\) is independent of \(q_{i}\). This guarantees that the momentum \(p_{i}\) is constant. In the Hamiltonian approach, we say that \(q_{i}\) is ignorable if \(\mathcal{H}\) is independent of \(q_{i}\), and this too guarantees \(p_{i}\) is constant. These two conditions must be the same, since the result " \(p_{i}=\) const" is the same either way. Prove directly that this is so, as follows: (a) For a system with one degree of freedom, prove that \(\partial \mathcal{H} / \partial q=-\partial \mathcal{L} / \partial q\) starting from the expression (13.14) for the Hamiltonian. This establishes that \(\partial \mathcal{H} / \partial q=0\) if and only if \(\partial \mathcal{L} / \partial q=0 .\) (b) For a system with \(n\) degrees of freedom, prove that \(\partial \mathcal{H} / \partial q_{i}=-\partial \mathcal{L} / \partial q_{i}\) starting from the expression (13.24).

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