DEpartment of physics

In this lecture, we discuss how to generalize the Euler equation when there are are more than one dependent variabl. Another important thing we discuss is how to generalize the Euler equation when there are constrains involved in the minimization. For example, what is the shrtest path between two points on a cylindrical surface. There are two ways to approach this problem: Either by reducing the number of dependent variables or using the Lagrangean undertermined multiplier method. In this lecture, we will only do the first method. We will do the second method in the next chapter

Functions with more than one dependent variables In the last classes, we derived the condition to minimize $$S=\int f(y(x),y'(x);x) dx$$, where y(x) is the only unknown dependent function. The case more commonly encountered in Mechanics is that in which f is a functional of several dependent variables. $$f=f(y_1(x),y_1'(x)),y_2(x),y_2'(x), \cdots, x) $$ or we can simply write $$ f=f(y_i(x),y_i'(x),x) \,\,\,\,\,\, \text{for}\,\,\,\,\,\, i=1,2,\cdots,n $$ Compared to what we did in proving the Euler-Lagrangean Equations: $$ y_i(\alpha,x)=y_i(0,x)+\alpha \eta_i(x) $$ By writing the integral quantity and getting the derivative with respect to the variational parameter alpha, $$ \frac{\partial S}{\partial \alpha}=\int_{x_1}^{x_2} \sum_{i}\left(\frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y_i'}\right)\eta_i(x) dx $$ which gives $$ \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y_i'}=0 \,\,\,\,\,\,\,\, \text{for} \,\,\,\,\,\, i=1,2,\cdots, n $$ Notice that we assume the condition: all Y_i(x) 's are independent of each other. Lagrange Formulation in Classical Mechanics Experimental observation on classical Particle Dynamics has led to Physical laws, which says $$F = \frac{d \vec{P}}{ dt}$$ We have used this equation, along with other Newton?s laws to investigate how the dynamics of a particle evolve with time. If the particles are moving in a complicated manner, finding out the forces can become quite complicated. In order to solve the Newton's laws, (Well.... In order ti set up the Newton's laws) we need to know all the forces acting on the particle. Because of this difficulty of solving problem using the Newton's Laws, alternate methods are required. That is, we need alternate routes to obtain the result from Newton's laws. The new method we are going to discuss here is called the Hamilton's Principle. And the Resultant equations are called Lagrangean Equations. Hamilton's Principle The minimization conditions related to Classical Mechanics has a long history. Without going too much in to details, we start with the Hamilton's Principle. Of all possible paths along which a dynamical system may move from one point to another within a specified time interval (consistent with any constraints), the actual path followed is that which minimizes the time integral of the difference between the kinetic and potential energies. Therefore, in terms of classical mechanics, the Hamilton's Principle becomes: We want the quantity, $$ \int _{t_2}^{t_1} \left(T-U\right) dt $$ to be minimized (Least Action Principle) Notice that, the nature wants to have the time integral of (T -U ) becomes minimized. We only do the extermination, without checking if it is a maximum or a minimum. But in all important applications in dynamics minimizations happen. Now we know that the kinetic energy is a functional of velocities of each particle, Lets say $\dot{x}_i$, and the the potential in general is a function of the position $x_i$, The quantity whose integral needs to be minimized, $(T-U )$ is a function of $(\dot{x}_i , x_i )$. In this problem, the quantity, which needs to be minimized is called the Lagrangean. $(L=T-U)$ Let's write this mathematically: $$ \int_{x_1} ^{x_2} \delta (T-U)dt=\int_{x_1}^{x_2} L(\dot{x}_i,x_i;t)dt $$ should be minimized. Now if we think about the previous chapter on Calculus of variations, we can see the transformation of notation as: $$x \rightarrow t $$ $$y_i(x) \rightarrow x_i(t)$$ $$y'_{i}(x) \rightarrow \dot{x}_i(t) $$ $$f(y_i(x),y'_i(x),x) \rightarrow L(x_i,\dot{x}_i,t)$$ Example: Lagrange's Equations for Unconstrained Motion Consider a particle moving unconstrained in three dimensions, subject to a conservative net force $F(r)$. Let's first write the Kinetic Energy and the Potential Energy: $$ T=\frac{1}{2}\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right) $$ and the potential energy is written as: $$ U=U(x,y,z) $$ $$ L=\frac{1}{2}\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)-U(x,y,z) $$ Here independent variable is $t$, and we have three dependent variables. We apply three Euler-Lagrangean Equations: \begin{eqnarray} \frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=0 \\ \frac{\partial L}{\partial y}-\frac{d}{dt}\frac{\partial L}{\partial \dot{y}}=0 \\ \frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial \dot{z}}=0 \end{eqnarray} We have $\frac{\partial L}{\partial x}=-\frac{\partial U(x)}{\partial x}=F_x$. So by applying the Euler Lagrangean Equation w.r.t. $x$, we get \begin{equation*} F_x=m\dot{x} \end{equation} and we get similar equations for $x$ $y$ and $z$. We identify that $\frac{\partial L}{\partial x} $ is the $x$ component of the force. and $\frac{\partial L}{\partial \dot{x}}$ is the $x$ component of the momentum. We call these generalized force and generalized momentum. Why we call the generalize? It will be clear as we proceed through. Notes can be found here

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