Derivation of Lagrange's eqs

In summary, the conversation revolves around deriving Lagrange's equations using Hamilton's principle and calculus of variations. The main question is whether to evaluate the variation at zero or at the actual curve. The expert explains that the correct approach is to set the variation equal to zero and use the fact that it is very small in the derivation of the equation. They also mention that the same approach can be used as in finding an extremum for a regular function, but with the understanding that the variation is small.
  • #1
lampCable
22
1

Homework Statement


So I'm deriving Lagrange's equations using Hamilton's principle which states that the motion of a dynamical system follows the path, consistent with any constraints, that minimise the time integral over the lagrangian [itex]L = T-U[/itex], where [itex]T[/itex] is the kinetic energy and [itex]U[/itex] is the potential energy.

We define the lagrangian as [itex]L = L(q_j,\dot{q}_j,t)[/itex]. Now I want to let [itex]q_j = q_j^{(0)}+\delta q_j[/itex], where [itex]\delta q_j[/itex] is the variation of [itex]q_j[/itex]. We also define
[tex]
S=\int_{t_1}^{t_2}L(q_j,\dot{q}_j,t)dt
[/tex]

So according to Hamilton's principle we would now like to minimise [itex]S[/itex]. At extremum we have [itex]\delta S = 0[/itex], i.e. the variation of S is zero.

Now, my problem:

My first experience with calculus of variations was to find Euler's equation. We considered then the functional
[tex]
J = \int_{x1}^{x2}f(y,y',x)dx
[/tex]
and our goal was to find the function [itex]y[/itex] that minimise S. To do this we let [itex]y(x,\alpha) = y(x)+\alpha\eta(x)[/itex], and set [itex]\frac{\partial J}{\partial\alpha}|_{\alpha=0}=0[/itex], where [itex]\eta[/itex] is some arbitrary function. This would give us an equation in [itex]y(x)[/itex] since we evaluate at [itex]\alpha=0[/itex].

So, returning to the derivation of Lagrange's equations. I set to find [itex]q_j[/itex] that minimise [itex]S[/itex] in similar fashion as we did for [itex]J[/itex]. But this time I do not have any [itex]\alpha[/itex] that I can put to zero. Should I instead take [itex]\delta S|_{\delta q_j = 0}=0[/itex]? For unless I have understood it wrong, it is actually [itex]q_j^{(0)}[/itex] that we want to find, right? I mean it seems strange to find [itex]q_j[/itex] to be some curve [itex]q_j^{(0)}[/itex] added by some arbitrary variations.

At the same time, I am not sure whether evaluating [itex]\delta q_j[/itex] at makes sense. What increase my doubts is that neither in my textbook (Marion and Thornton) nor at any lecture has it been emphasized that we evaluate with [itex]\delta q_j = 0[/itex].

Anyhow, I am thankful for answers.
 
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  • #2
Remember the definition of a derivative ? ##\displaystyle f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} ##. And to find an extremum of the function, you just set ##f'(x)=0 ## .
Its the same here, the only difference being that here we're talking about a functional, a function in the space of functions defined on a given interval with given boundary conditions. There is just one difference though. What we do here to derive the EL equations is, after doing the calculations and arriving at the following equation:
## \delta S=\int_{t_1}^{t_2} \left[ \frac{\partial L}{\partial q_j}-\frac{d}{dt} \frac{\partial L}{\partial \dot{q_j}} \right]\delta q_j dt=0##
We reason that this equation should be true for any non-zero ##\delta q_j ## so we should have ##\frac{\partial L}{\partial q_j}-\frac{d}{dt} \frac{\partial L}{\partial \dot{q_j}} =0##. The point is, although ##\delta q_j## is supposed to be very small, its not exactly zero. But there should be somewhere that we actually use the fact that ##\delta q_j## is very small, and its actually in the derivation of the above equation were we expand the Lagrangian in a Taylor series and treat ## \delta q_j ## as the small displacement from ## q_j^{(0)} ##.
Anyway, I think you still can use the ## y(x,\alpha)=y(x)+\alpha \eta(x) ## approach here. But even here, its actually ##\alpha\to 0## not ##\alpha=0##.
 

What is the concept behind Lagrange's equations?

Lagrange's equations are a set of equations used to describe the motion of systems with multiple degrees of freedom. They are derived from the principle of least action, which states that the motion of a system will follow the path that minimizes the action integral.

What is the difference between Lagrange's equations and Newton's laws of motion?

Lagrange's equations provide an alternative approach to describing the motion of a system compared to Newton's laws of motion. While Newton's laws focus on the forces acting on individual objects, Lagrange's equations consider the system as a whole and take into account all forces and constraints.

How are Lagrange's equations derived?

Lagrange's equations are derived from the principle of least action, which is based on the concept of virtual work. This principle states that the actual motion of a system will be the one that minimizes the action integral, which is the integral of the difference between kinetic and potential energy over time.

What types of systems can Lagrange's equations be applied to?

Lagrange's equations can be applied to a wide range of systems, including mechanical, electrical, and thermal systems. They can also be used to describe the motion of particles, rigid bodies, and fluids.

How are Lagrange's equations used in practical applications?

Lagrange's equations are used in various fields, including physics, engineering, and mathematics, to model and analyze the behavior of complex systems. They are particularly useful in solving problems involving multiple degrees of freedom and constraints, such as in celestial mechanics and control systems.

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