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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

(a) Write down the Lagrangian for a particle moving in three dimensions under the influence of a conservative central force with potential energy \(U(r),\) using spherical polar coordinates \((r, \theta, \phi)\). (b) Write down the three Lagrange equations and explain their significance in terms of radial acceleration, angular momentum, and so forth. (The \(\theta\) equation is the tricky one, since you will find it implies that the \(\phi\) component of \(\ell\) varies with time, which seems to contradict conservation of angular momentum. Remember, however, that \(\ell_{\phi}\) is the component of \(\ell\) in a variable direction.) (c) Suppose that initially the motion is in the equatorial plane (that is, \(\theta_{0}=\pi / 2\) and \(\dot{\theta}_{0}=0\) ). Describe the subsequent motion. (d) Suppose instead that the initial motion is along a line of longitude (that is, \(\dot{\phi}_{0}=0\) ). Describe the subsequent motion.

[Computer] Consider a massless wheel of radius \(R\) mounted on a frictionless horizontal axis. A point mass \(M\) is glued to the edge, and a massless string is wrapped several times around the perimeter and hangs vertically down with a mass \(m\) suspended from its bottom end. (See Figure 4.28.) Initially I am holding the wheel with \(M\) vertically below the axle. At \(t=0,\) I release the wheel, and \(m\) starts to fall vertically down. (a) Write down the Lagrangian \(\mathcal{L}=T-U\) as a function of the angle \(\phi\) through which the wheel has turned. Find the equation of motion and show that, provided \(m

Noether's theorem asserts a connection between invariance principles and conservation laws. In Section 7.8 we saw that translational invariance of the Lagrangian implies conservation of total linear momentum. Here you will prove that rotational invariance of \(\mathcal{L}\) implies conservation of total angular momentum. Suppose that the Lagrangian of an \(N\) -particle system is unchanged by rotations about a certain symmetry axis. (a) Without loss of generality, take this axis to be the \(z\) axis, and show that the Lagrangian is unchanged when all of the particles are simultaneously moved from \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}\right)\) to \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}+\epsilon\right)\) (same \(\epsilon\) for all particles). Hence show that $$\sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}}=0.$$ (b) Use Lagrange's equations to show that this implies that the total angular momentum \(L_{z}\) about the symmetry axis is constant. In particular, if the Lagrangian is invariant under rotations about all axes, then all components of \(\mathbf{L}\) are conserved.

Find the components of \(\nabla f(r, \phi)\) in two-dimensional polar coordinates. [Hint: Remember that the change in the scalar \(f \text { as a result of an infinitesimal displacement } d \mathbf{r} \text { is } d f=\nabla f \cdot d \mathbf{r}.]\)

A mass \(m\) is suspended from a massless string, the other end of which is wrapped several times around a horizontal cylinder of radius \(R\) and moment of inertia \(I\), which is free to rotate about a fixed horizontal axle. Using a suitable coordinate, set up the Lagrangian and the Lagrange equation of motion, and find the acceleration of the mass \(m\). [The kinetic energy of the rotating cylinder is \(\frac{1}{2} I \omega^{2} .\)]

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