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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

Consider two particles moving unconstrained in three dimensions, with potential energy \(U\left(\mathbf{r}_{1}, \mathbf{r}_{2}\right) .\) (a) Write down the six equations of motion obtained by applying Newton's second law to each particle. (b) Write down the Lagrangian \(\mathcal{L}\left(\mathbf{r}_{1}, \mathbf{r}_{2}, \dot{\mathbf{r}}_{1}, \dot{\mathbf{r}}_{2}\right)=T-U\) and show that the six Lagrange equations are the same as the six Newtonian equations of part (a). This establishes the validity of Lagrange's equations in rectangular coordinates, which in turn establishes Hamilton's principle. since the latter is independent of coordinates, this proves Lagrange's equations in any coordinate system.

Write down the Lagrangian for a cylinder (mass \(m\), radius \(R\), and moment of inertia \(I\) ) that rolls without slipping straight down an inclined plane which is at an angle \(\alpha\) from the horizontal. Use as your generalized coordinate the cylinder's distance \(x\) measured down the plane from its starting point. Write down the Lagrange equation and solve it for the cylinder's acceleration \(\ddot{x}\). Remember that \(T=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2},\) where \(v\) is the velocity of the center of mass and \(\omega\) is the angular velocity.

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.

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.

Consider a bead that is threaded on a rigid circular hoop of radius \(R\) lying in the \(x y\) plane with its center at \(O,\) and use the angle \(\phi\) of two- dimensional polar coordinates as the one generalized coordinate to describe the bead's position. Write down the equations that give the Cartesian coordinates \((x, y)\) in terms of \(\phi\) and the equation that gives the generalized coordinate \(\phi\) in terms of \((x, y)\).

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