2024-01-30 Calc. of Variations Continued

Last class

• Deriving Euler-Lagrange Equation using calculus of variations to minimize the action
  • The action is the integral over time of the Lagrangian

Deriving the Brachistachrone Problem curve with Calculus of variations

Given two points which are in a vertical plane , find the path $AmB$ which the movable particle m will traverse in the shortest time, assuming that acceleration is only due to gravity.

This is known as the Brachistachrone Problem. This is a also known as a Cycloid Curve.

Properties of Lagrange equations of motion in the case of coordinate transformation

We need a new set of coordinates

$$r_s=r_s(q_u,t)$$

With Jacobi determinant $|J|=\frac{d{r}_s}{d{q}_u}\ne 0$

Consider:

$${\dot{r}}_s=\frac{d}{dt}{r}_s(q_u(t),t)\text{\hspace{1em}(1)}$$ $$\frac{d{r}_s}{dt}+\frac{d{r}_s}{d{q}_u}{\dot{q}}_u\text{\hspace{1em}(2)}$$ $$L(r_s,{\dot{r}}_s,t)=L\left(r_s(q_u,t),\frac{d{r}_s}{dt}+\frac{d{r}_s}{d{q}_u}{\dot{q}}_u,t\right)$$ $$=\tilde{L}(q_u,{\dot{q}}_u,t)$$

Calculate

$$\frac{\partial \tilde{L}}{\partial q_u},\frac{\partial \tilde{L}}{\partial {\dot{q}}_u},\frac{d}{dt}\frac{\partial \tilde{L}}{\partial {\dot{q}}_u}$$ $$\frac{\partial \tilde{L}}{\partial q_u}=\frac{\partial L}{\partial r_s}\frac{\partial r_s}{\partial q_u}+\frac{\partial L}{\partial {\dot{r}}_s}\frac{\partial {\dot{r}}_s}{\partial q_u}$$ $$\frac{\partial \tilde{L}}{\partial {\dot{q}}_u}=\frac{\partial L}{\partial \dot{r_s}}\frac{\partial {\dot{r}}_s}{\partial {\dot{q}}_u}=\frac{\partial L}{\partial \dot{r_s}}\frac{\partial r_s}{\partial q_u}$$

Using equation 2

$$\frac{\partial {\dot{r}}_s}{\partial {\dot{q}}_u}=\frac{\partial r_s}{\partial q_u}$$ $$\frac{d}{dt}\left(\frac{\partial r_s}{\partial q_u}\right)=\frac{d}{dt}\left(\frac{\partial r_s(q_u(t),t)}{\partial q_u}\right)$$ $$=\frac{{\partial }^2{r}_s}{\partial q_{u}d{t}^2}+\frac{{\partial }^2{r}_s}{\partial q_{u}d{q}_u}\cdot {\dot{q}}_u$$ $$\frac{d}{d{q}_u}\left(\frac{\partial r_s}{\partial t}+\frac{\partial r_s}{\partial q_u}{\dot{q}}_u\right)=\frac{\partial {\dot{r}}_s}{\partial q_u}={\dot{r}}_s$$

From equation 2

Lagrange eqn is

$$\frac{\partial \tilde{L}}{\partial q_u}=\frac{d}{dt}\frac{\partial \tilde{L}}{\partial {\dot{q}}_u}=\frac{\partial L}{\partial r_s}\frac{\partial r_s}{\partial q_u}+\frac{\partial L}{\partial {\dot{r}}_s}\frac{\partial {\dot{r}}_s}{\partial q_u}-\frac{d}{dt}\left[\frac{\partial L}{\partial {\dot{r}}_s}\frac{\partial r_s}{\partial q_u}\right]$$ $$=\frac{\partial r_s}{\partial q_u}\left[\frac{\partial L}{\partial r_s}-\frac{d}{dt}\frac{\partial L}{\partial {\dot{r}}_s}\right]=0$$

Since $|\frac{\partial r_s}{\partial q_u}|\ne 0$

$$\frac{\partial \tilde{L}}{\partial q_n}-\frac{d}{dt}\frac{\partial \tilde{L}}{\partial {\dot{q}}_u}=0$$

and

$$\frac{\partial L}{\partial r_s}-\frac{d}{dt}\frac{\partial L}{\partial {\dot{r}}_s}=0$$

Lagrange equation is invariant under coordinate transforms!

Generalized Coordinates

• Given a system of $n$ objects, you have $3n$ coordinates required to describe position of n objects.

• However, depending on situation, not all $3n$ coordinates may be independent of each other! (uniform circular motion)

• The number of independent coordinates is equal to the degrees of freedom of the system.

• In a system with no constraints, then your D.O.F is equal to 3n (star example )

• If you have forces of constraint $s=3n$

this isn’t going to be on the first test

• With $m$ forces of constraint $s=3n-m$

Proper set of generalized coordinates

$$q_i⟹i=1,2,3,\dots$$

Set of independent coordinates $q_i$ is equal to the degrees of freedom $s_1$ of the system.

Generally speaking, this is an $s-$dimensional space aka configuration space.

2 questions:

1. What are generalized coordinates?
2. What is a coordinate transformation really?