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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).

Short Answer

Expert verified
The conditions \( \partial \mathcal{H} / \partial q_{i} = 0 \) and \( \partial \mathcal{L} / \partial q_{i} = 0 \) are equivalent.

Step by step solution

01

Recall the Hamiltonian Definition

The Hamiltonian \( \mathcal{H} \) is defined in terms of the Lagrangian \( \mathcal{L} \) by the equation \( \mathcal{H} = \sum_{i} p_{i} \dot{q}_{i} - \mathcal{L} \), where \( p_{i} = \frac{\partial \mathcal{L}}{\partial \dot{q}_{i}} \). This expression is crucial for deriving the relationship between the partial derivatives of \( \mathcal{H} \) and \( \mathcal{L} \).
02

Single Degree of Freedom (a)

For a system with one degree of freedom, the Hamiltonian can be expressed as \( \mathcal{H} = p \dot{q} - \mathcal{L} \). Taking the partial derivative of \( \mathcal{H} \) with respect to \( q \), we get:\[\frac{\partial \mathcal{H}}{\partial q} = \frac{\partial}{\partial q}(p \dot{q}) - \frac{\partial \mathcal{L}}{\partial q}.\]Since \( p \) is defined as \( \frac{\partial \mathcal{L}}{\partial \dot{q}} \), and assuming \( \dot{q} \) is independent of \( q \), the first term is zero, leading to:\[\frac{\partial \mathcal{H}}{\partial q} = -\frac{\partial \mathcal{L}}{\partial q}.\]
03

Generalize to n Degrees of Freedom (b)

For n degrees of freedom, the expression for Hamiltonian becomes \( \mathcal{H} = \sum_{i=1}^{n} p_{i} \dot{q}_{i} - \mathcal{L} \). The partial derivative with respect to one coordinate \( q_{i} \) is:\[\frac{\partial \mathcal{H}}{\partial q_{i}} = \sum_{j=1}^{n} \frac{\partial}{\partial q_{i}}(p_{j} \dot{q}_{j}) - \frac{\partial \mathcal{L}}{\partial q_{i}}.\]Since \( p_{j} = \frac{\partial \mathcal{L}}{\partial \dot{q}_{j}} \) does not depend on \( q_{i} \) when \( \mathcal{L} \) is independent of it, each term in the sum involving \( j eq i \) vanishes, and we are left with:\[\frac{\partial \mathcal{H}}{\partial q_{i}} = -\frac{\partial \mathcal{L}}{\partial q_{i}}.\]
04

Conclusion

We have shown through direct differentiation that \( \frac{\partial \mathcal{H}}{\partial q_{i}} = -\frac{\partial \mathcal{L}}{\partial q_{i}} \) holds for both single and multiple degrees of freedom situations. Consequently, \( \frac{\partial \mathcal{H}}{\partial q_{i}} = 0 \) implies \( \frac{\partial \mathcal{L}}{\partial q_{i}} = 0 \), and vice versa, for all coordinates. This proves the equivalence conditions for ignorable coordinates in the Lagrangian and Hamiltonian formalisms.

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Key Concepts

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

Lagrangian Formalism
The Lagrangian formalism is a fundamental concept in classical mechanics, introduced by Joseph Louis Lagrange. It elegantly describes the dynamics of a system by using the Lagrangian function, denoted by \( \mathcal{L} \). This function is the difference between the kinetic and potential energies of the system: \( \mathcal{L} = T - V \), where \( T \) is the kinetic energy and \( V \) is the potential energy.

One of the major advantages of the Lagrangian formalism is its ability to handle complex systems, especially those with constraints. A coordinate \( q_i \) is considered ignorable if the Lagrangian \( \mathcal{L} \) does not depend on it. When a coordinate is ignorable, it implies that the partial derivative \( \frac{\partial \mathcal{L}}{\partial q_i} = 0 \). Consequently, the canonical momentum \( p_i \) associated with this coordinate remains constant.

Key characteristics of the Lagrangian formalism include:
  • Focuses on energy differences (\( T - V \)) rather than just energetic states.
  • Incorporates constraints naturally using generalized coordinates.
  • A powerful tool in deriving equations of motion through the Euler-Lagrange equations.
Hamiltonian Formalism
The Hamiltonian formalism is another pivotal framework in classical and modern physics. It translates the energy dynamics of a system into a different yet insightful perspective using the Hamiltonian function, \( \mathcal{H} \). The Hamiltonian is defined by the Legendre transformation of the Lagrangian, and in many cases, it represents the total energy of the system: \( \mathcal{H} = \sum_{i} p_{i} \dot{q}_{i} - \mathcal{L} \).

In this formalism, a coordinate \( q_i \) is ignorable if the Hamiltonian \( \mathcal{H} \) does not depend on it, which leads to the conclusion that the momentum \( p_i \) is conserved or constant over time. Hamiltonian formalism is particularly useful for systems where energy conservation and symmetries are important.

Notable features of the Hamiltonian formalism involve:
  • Focuses on phase space, comprising coordinates and momenta.
  • Provides a holistic view of both classical and quantum mechanics.
  • Emphasizes the conservation laws and symmetries via Poisson brackets and Hamilton's equations.
Degrees of Freedom
Degrees of Freedom in a physical system refer to the number of independent variables that uniquely define the state of the system. Each degree of freedom corresponds to a coordinate \( q_i \) that can vary independently. Understanding the degrees of freedom is crucial when analyzing the dynamics of systems in classical mechanics.

For systems with multiple degrees of freedom \((n)\), each can contribute to kinetic and potential energy in the Lagrangian or Hamiltonian approaches. By taking into account all degrees of freedom, you can fully describe the system’s mechanics.

Key insights into degrees of freedom include:
  • Indicate the number of ways a system can move or be configured.
  • In Lagrangian systems, they correspond to the number of generalized coordinates involved.
  • In Hamiltonian mechanics, degrees of freedom map to coordinates and canonical momenta.
Partial Derivatives
Partial Derivatives are invaluable in the study of complex systems, especially those encountered in classical mechanics, where multiple variables are at play. A partial derivative denotes the rate at which a function changes as one of the variables changes, keeping all other variables constant.

In the context of Lagrangian and Hamiltonian mechanics, partial derivatives are used to determine dependencies between variables. For instance, if the partial derivative of the Lagrangian with respect to a coordinate \( \frac{\partial \mathcal{L}}{\partial q_i} = 0 \), it indicates that the system's behavior does not rely on this specific coordinate, marking it as ignorable.

Essential aspects of partial derivatives include:
  • Partial derivatives help uncover relationships between different variables in multi-dimensional functions.
  • Used extensively to derive equations of motion in both Lagrangian and Hamiltonian frameworks.
  • Crucial for determining conservation laws, such as conserved momenta, through simplifications in functional dependencies.

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Most popular questions from this chapter

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.

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.]

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) Evaluate \(\left.\nabla \cdot \mathbf{v} \text { for } \mathbf{v}=k \hat{\mathbf{r}} / r^{2} \text { using rectangular coordinates. (Note that } \hat{\mathbf{r}} / r^{2}=\mathbf{r} / r^{3} .\right)\) (b) Inside the back cover, you will find expressions for the various vector operators (divergence, gradient, etc.) in polar coordinates. Use the expression for the divergence in spherical polar coordinates to confirm your answer to part (a). (Take \(r \neq 0\).)

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.

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