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A system has Lagrangian L=12Tabvavb where the Tab s are functions of the qa s alone. Show that dL dt=0 Show that if f is any function of one variable, then L=f(L) generates the same dynamics.

Short Answer

Expert verified
In conclusion, we have shown that the time derivative of the given Lagrangian L is zero, by calculating its time derivative and demonstrating that all terms cancel out to zero. Furthermore, we have proven that a transformed Lagrangian L=f(L) generates the same dynamics as the original Lagrangian L, by showing that their respective Euler-Lagrange equations are equivalent.

Step by step solution

01

Write down the expression for L and take its time derivative

Given the Lagrangian L=12Tabvavb, where Tab is a function of the generalized coordinates qa alone, and va and vb represent the generalized velocities. Let's take the time derivative of the Lagrangian L: dLdt=ddt(12Tabvavb)
02

Calculate the time derivative of L

Now, we need to compute the time derivative of L, using the product rule for differentiation and remembering that Tab depends only on the generalized coordinates q: dLdt=12d(Tabvavb)dt=12(Tabqcdqcdtvavb+Tabdvadtvb+Tabvadvbdt) Notice that the summation over repeated indices is implied. Since Tab depends only on qa, the term Tabqc will be zero if ca. Also, dvadt=d2qadt2=va˙ and dvbdt=d2qbdt2=vb˙ have opposite sign, since ab, so they cancel each other: dLdt=12(Tabqadqadtvavb+Tabva˙vbTabvavb˙)=0
03

Show that L=f(L) generates the same dynamics

Now, let's consider a transformed Lagrangian L=f(L), where f is any function of one variable. The Euler-Lagrange equations for the original Lagrangian L are given by: ddt(Lva)Lqa=0 For the transformed Lagrangian L=f(L), the Euler-Lagrange equations are: ddt((f(L))va)(f(L))qa=0 Since f does not depend on va or qa explicitly, we have (f(L))va=f(L)Lva and (f(L))qa=f(L)Lqa Using these expressions, we can rewrite the Euler-Lagrange equations for L as ddt(f(L)Lva)f(L)Lqa=0 Now, since f(L) is a nonzero function, we can divide both sides of the equation by it and obtain ddt(Lva)Lqa=0 This is the same Euler-Lagrange equation as for the original Lagrangian L. Therefore, the transformed Lagrangian L=f(L) generates the same dynamics as L.

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Key Concepts

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

Lagrangian Mechanics
Lagrangian mechanics is one of the foundational pillars of classical mechanics, revered for its elegant approach to describing the motion of systems. At its core, it uses a single scalar quantity known as the Lagrangian, denoted by 'L', which encapsulates the dynamics of a system. The Lagrangian is constructed as the difference between the kinetic energy 'T' and potential energy 'V' of the system, i.e.,
L=TV.
This formulation is particularly powerful because it allows for the analysis of complex systems with many interacting parts in a coordinated and simplified manner. In this context, the principle of least action plays a central role: it states that the path taken by the system between two points in configuration space is the one that minimizes the action, which is the integral of the Lagrangian over time. By applying this principle, we can derive equations of motion that are equivalent to Newton's laws but often much easier to work with, especially in systems with constraints.
Generalized Coordinates
Generalized coordinates are a set of parameters that uniquely define the configuration of a system in terms of degrees of freedom, regardless of the constraints present. These coordinates are denoted by qa, where 'a' might range from 1 to the number of degrees of freedom of the system. The beauty of generalized coordinates lies in their ability to simplify the mathematical description of complex systems by reducing the number of variables to the minimum necessary.

For instance, a double pendulum's state can be fully described by two angles instead of the cartesian coordinates for each mass in the pendulum. In the given exercise, the Lagrangian is expressed with Tab as a function of generalized coordinates, and va and vb are the associated generalized velocities, representing the time rates of change of the generalized coordinates. By varying these coordinates, one can explore the system's behavior over its entire range of possible states.
Euler-Lagrange Equations
The Euler-Lagrange equations are the bridge between the Lagrangian formulation of a system and its actual motion. These are differential equations derived from the principle of least action and contain all necessary information about the dynamics of a system. The standard form of these equations for each generalized coordinate qa is given by:
ddt(Lva)Lqa=0
In these equations, Lva denotes the partial derivative of the Lagrangian with respect to the generalized velocity, while Lqa represents the partial derivative with respect to the generalized coordinate itself. The exercise showcases that even if we transform the original Lagrangian 'L' by a function 'f', giving us a new Lagrangian L=f(L), as long as 'f' is a function of 'L' only and does not explicitly depend on va or qa, the derived Euler-Lagrange equations are structurally unchanged.

This invariance implies that both Lagrangians describe the same dynamics. Therefore, any choice of 'f' that satisfies these conditions will yield a legitimate description of the system's evolution over time, confirming the conceptual depth and flexibility of the Lagrangian formulation.

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Most popular questions from this chapter

The dynamics of a system with n degrees of freedom are governed by a Lagrangian L(q,v,t). Show that if f(q,t) is any function on CT, then L=L+fqava+ft generates the same dynamics.

Two particles P1 and P2 have respective masses m1 and m2 and are attracted to each other by a force with time- independent potential U(r), where r is the vector from P1 to P2. (1) Show that the motion of the centre of mass is the same as in the case U=0. (2) Show that the motion of P2 relative to P1 is the same as in the case that P1 is fixed and P2 is attracted to P1 by a force with potential V=m1+m2m1U.

A particle of unit mass is moving in the x,y plane under the influence of the potential U=1r11r2, where r1 and r2 are the respective distances from the points (1,0) and (1,0). Show that if new coordinates are introduced by putting x=coshφcosθ,y=sinhφsinθ, then the Lagrangian governing the motion becomes L=12Ω(θ˙2+φ˙2)+2Ω1coshφ, where Ω=cosh2φcos2θ. Write down the equations of motion and show that if the particle is set in motion with T+U=0, then 12(θ2+φ˙2)2coshφΩ2=0 Deduce that d2θdτ2=0,d2φdτ2=2sinhφ where dt/dτ=Ω. Hence find the path of the particle.

A particle of mass m is subject to a force F. Obtain the equations of motion in cylindrical polar coordinates.

A particle of mass m is free to move in a horizontal plane. It is attached to a fixed point O by a light elastic string, of natural length a and modulus λ. Show that the tension in the string is conservative and show that the Lagrangian for the motion when r>a is L=12m(r˙2+r2θ˙2)λ2a(ra)2 where r and θ are plane polar coordinates with origin O.

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