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Let \(F=F\left(q_{1}, \cdots, q_{n}\right)\) be any function of the generalized coordinates \(\left(q_{1}, \cdots, q_{n}\right)\) of a system with Lagrangian \(\mathcal{L}\left(q_{1}, \cdots, q_{n}, \dot{q}_{1}, \cdots, \dot{q}_{n}, t\right) .\) Prove that the two Lagrangians \(\mathcal{L}\) and \(\mathcal{L}^{\prime}=\mathcal{L}+d F / d t\) give exactly the same equations of motion.

Short Answer

Expert verified
Both Lagrangians have the same equations of motion because the total derivative \(\frac{dF}{dt}\) doesn't affect them.

Step by step solution

01

Understand the problem

We're given a Lagrangian \(\mathcal{L}\) and a second Lagrangian \(\mathcal{L}' = \mathcal{L} + \frac{dF}{dt}\), where \(F\) is a function of generalized coordinates. The task is to show that these two Lagrangians produce the same equations of motion.
02

Write the Euler-Lagrange equations

The Euler-Lagrange equations for a system are given by: \[\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right) - \frac{\partial \mathcal{L}}{\partial q_i} = 0\] for each coordinate \(q_i\). We need to apply this to both \(\mathcal{L}\) and \(\mathcal{L}'\).
03

Differentiate the new Lagrangian

For the new Lagrangian \(\mathcal{L}' = \mathcal{L} + \frac{dF}{dt}\), we find: \[\frac{d}{dt} \left( \frac{\partial \mathcal{L}'}{\partial \dot{q}_i} \right) = \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} + \frac{\partial}{\partial \dot{q}_i} \left( \frac{dF}{dt} \right) \right)\] Notice that the term \(\frac{\partial}{\partial \dot{q}_i} \left( \frac{dF}{dt} \right)\) involves differentiating a total derivative, which only depends on \(q_i\) and not on \(\dot{q}_i\), thus it does not contribute.
04

Simplify total derivative effect

For \(\frac{dF}{dt}\), since it is a total time derivative, differentiate with respect to \(\dot{q}_i\) leads to zero because \(\frac{dF}{dt} = \sum_i (\frac{\partial F}{\partial q_i} \dot{q}_i) + \frac{\partial F}{\partial t}\). Time derivative of \(F\) affects only \(q_i\) but doesn't affect \(\dot{q}_i\), thus \(\frac{\partial}{\partial \dot{q}_i} \left( \frac{dF}{dt} \right) = 0\).
05

Ensure both constraints are satisfied

After computing \(\frac{d}{dt} \left( \frac{\partial \mathcal{L}'}{\partial \dot{q}_i} \right) = \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right)\) and \(\frac{\partial \mathcal{L}'}{\partial q_i} = \frac{\partial \mathcal{L}}{\partial q_i}\), so the Euler-Lagrange equation for \(\mathcal{L}'\) will be: \[\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right) - \frac{\partial \mathcal{L}}{\partial q_i} = 0\]which is the same as before for \(\mathcal{L}\).
06

Conclusion

The total time derivative \(\frac{dF}{dt}\) does not affect the Euler-Lagrange equations. Both Lagrangians \(\mathcal{L}\) and \(\mathcal{L}'\) result in the same equations of motion for the system.

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Key Concepts

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

Generalized Coordinates
In Lagrangian mechanics, the concept of generalized coordinates is central to describing motion in dynamic systems. Instead of using traditional coordinates like Cartesian coordinates, generalized coordinates take the complexity of a system and simplify it by representing all the possible ways a system can move. These coordinates, denoted usually by \( q_1, q_2, \ldots, q_n \), are a set of variables that uniquely describe the configuration of the system.

Generalized coordinates can be any parameters that define a system's position, such as angles for rotational systems or distances for translational ones. The advantage of using them is that they provide a tailored approach:
  • They reduce the number of variables needed, focusing only on the ones necessary for describing the specific motion.
  • They can be chosen to simplify the equations of motion, making solving them more straightforward.
For example, when analyzing the motion of a double pendulum, using angles as generalized coordinates can make the process more intuitive than using Cartesian coordinates. By tailoring variables specifically to the system in question, Lagrangian mechanics becomes a powerful tool.
Euler-Lagrange Equations
The Euler-Lagrange equations are pivotal in deriving the equations of motion for a system using Lagrangian mechanics. Derived from a principle of stationary action, these equations help find paths which make the action integral stationary, leading to the natural evolution of the system.

Mathematically, the Euler-Lagrange equation for each generalized coordinate \( q_i \) is given by:\[\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right) - \frac{\partial \mathcal{L}}{\partial q_i} = 0.\]Here, \( \mathcal{L} \) represents the Lagrangian, a function that encapsulates the dynamics of the system. This equation essentially states:
  • The change in the derivative of the Lagrangian concerning the generalized velocities and time equals the derivative of the Lagrangian concerning the coordinates.
  • It involves both temporal and spatial derivatives that describe how the system's state evolves over time.
  • By solving these equations, one can find the trajectory or path of the system under given conditions.
The beauty of these equations lies in their generality, working for a variety of systems without needing explicit force calculations as in Newtonian mechanics.
Total Time Derivative
A total time derivative in dynamics involves understanding how a function changes when all its variables are functions of time. In this context, it is essentially a rate of change of a function as seen from a moving reference point.

In Lagrangian mechanics, the total time derivative of a function, such as \( F(q_1, q_2, \ldots, q_n, t) \), is expressed as:\[\frac{dF}{dt} = \sum_{i} \left( \frac{\partial F}{\partial q_i} \dot{q}_i + \frac{\partial F}{\partial \dot{q}_i} \ddot{q}_i \right) + \frac{\partial F}{\partial t}.\]Here’s why it’s noteworthy:
  • It incorporates the rates of all variables — both generalized coordinates and time — capturing how a function’s value evolves.
  • This derivative plays a crucial role in mechanics because it can modify the form of the Lagrangian without changing the equations of motion.
  • Specifically, if a Lagrangian \( \mathcal{L}' = \mathcal{L} + \frac{dF}{dt} \) only differs by a total derivative \( \frac{dF}{dt} \) of a function of coordinates, it results in the same equations of motion as \( \mathcal{L} \). This universality is critical as it allows for transformations of the problem that simplify the description without altering the physical content.
Thus, the total time derivative is not just a mathematical artifact but a gateway to understanding fundamental characteristics of motion unaffected by specific reference choices.

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Most popular questions from this chapter

Prove that the potential energy of a central force \(\mathbf{F}=-k r^{n} \hat{\mathbf{r}}(\text { with } n \neq-1)\) is \(U=k r^{n+1} /(n+1)\). In particular, if \(n=1,\) then \(\mathbf{F}=-k \mathbf{r}\) and \(U=\frac{1}{2} k r^{2}\).

A small cart (mass \(m\) ) is mounted on rails inside a large cart. The two are attached by a spring (force constant \(k\) ) in such a way that the small cart is in equilibrium at the midpoint of the large. The distance of the small cart from its equilibrium is denoted \(x\) and that of the large one from a fixed point on the ground is \(X,\) as shown in Figure \(7.13 .\) The large cart is now forced to oscillate such that \(X=A \cos \omega t,\) with both \(A\) and \(\omega\) fixed. Set up the Lagrangian for the motion of the small cart and show that the Lagrange equation has the form $$\ddot{x}+\omega_{0}^{2} x=B \cos \omega t$$ where \(\omega_{\mathrm{o}}\) is the natural frequency \(\omega_{\mathrm{o}}=\sqrt{k / m}\) and \(B\) is a constant. This is the form assumed in Section 5.5, Equation (5.57), for driven oscillations (except that we are here ignoring damping). Thus the system described here would be one way to realize the motion discussed there. (We could fill the large cart with molasses to provide some damping.)

Using the usual angle \(\phi\) as generalized coordinate, write down the Lagrangian for a simple pendulum of length \(l\) suspended from the ceiling of an elevator that is accelerating upward with constant acceleration \(a\). (Be careful when writing \(T\); it is probably safest to write the bob's velocity in component form.) Find the Lagrange equation of motion and show that it is the same as that for a normal, nonaccelerating pendulum, except that \(g\) has been replaced by \(g+a\). In particular, the angular frequency of small oscillations is \(\sqrt{(g+a) / l}\).

Write down the Lagrangian for a projectile (subject to no air resistance) in terms of its Cartesian coordinates \((x, y, z),\) with \(z\) measured vertically upward. Find the three Lagrange equations and show that they are exactly what you would expect for the equations of motion.

The "spherical pendulum" is just a simple pendulum that is free to move in any sideways direction. (By contrast a "simple pendulum"- unqualified - is confined to a single vertical plane.) The bob of a spherical pendulum moves on a sphere, centered on the point of support with radius \(r=R\) the length of the pendulum. A convenient choice of coordinates is spherical polars, \(r, \theta, \phi,\) with the origin at the point of support and the polar axis pointing straight down. The two variables \(\theta\) and \(\phi\) make a good choice of generalized coordinates. (a) Find the Lagrangian and the two Lagrange equations. (b) Explain what the \(\phi\) equation tells us about the \(z\) component of angular momentum \(\ell_{z^{*}}\) (c) For the special case that \(\phi=\) const, describe what the \(\theta\) equation tells us. (d) Use the \(\phi\) equation to replace \(\dot{\phi}\) by \(\ell_{z}\) in the \(\theta\) equation and discuss the existence of an angle \(\theta_{\mathrm{o}}\) at which \(\theta\) can remain constant. Why is this motion called a conical pendulum? (e) Show that if \(\theta=\theta_{0}+\epsilon,\) with \(\epsilon\) small, then \(\theta\) oscillates about \(\theta_{\mathrm{o}}\) in harmonic motion. Describe the motion of the pendulum's bob.

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