What will happen if we take up Cartesian Coordinates in Lagrangian Formulation instead of Generalized Coordinates?

Question

Why do we actually need generalized coordinates? Is it a mathematical manipulation only or does it serve physical purpose? And will principle of stationary action stay valid if we use cartesian coordinates?

Answer 1

If you take a particle $m$ with coordinates $x,y,z$ in a field $U(x,y,z)$, then the Lagrangian $$ \mathcal L = \frac m2(\dot x^2+\dot y^2+\dot z^2) - U(x,y,z) $$ will lead you to the well-known equations: $$ m\ddot x=-\frac{\partial U}{\partial x},\qquad m\ddot y=-\frac{\partial U}{\partial y},\qquad m\ddot z=-\frac{\partial U}{\partial z}. $$

However, if you take a slightly complicated system: two balls $m$ connected with a rod of length $l=\sqrt{(x_1-x_2)^2+(y-y_2)^2}$, then the equations in Cartesian coordinates become complicated: $$ \mathcal L = \frac m2(\dot x_1^2+\dot y_1^2+\dot x_2^2+\dot y_2^2)+\lambda\Big((x_1-x_2)^2+(y_1-y_2)^2-l^2\Big),\\ m\ddot x_1 = \lambda(x_1-x_2),\qquad m\ddot x_2 = -\lambda(x_1-x_2),\\ m\ddot y_1 = \lambda(y_1-y_2),\qquad m\ddot y_2 = -\lambda(y_1-y_2) $$

You have a linear differential equation of 4th order you have to solve. It's not impossible but it will take a while.

Let's take generalized coordinates $x,y,\theta$: $x_1=x+\frac l2\cos\theta$, $y_1=y+\frac l2\sin\theta$, $x_2=x-\frac l2\cos\theta$, $y_2=y-\frac l2\sin\theta$: $$ \mathcal L = \frac m2\left(\left(\dot x-\dot\theta\frac l2\sin\theta\right)^2+\left(\dot y+\dot\theta\frac l2\cos\theta\right)^2+\left(\dot x+\dot\theta\frac l2\sin\theta\right)^2+\left(\dot x-\dot\theta\frac l2\cos\theta\right)^2\right) = \\ m(\dot x^2 + \dot y^2) + \frac{ml^2}4\dot\theta^2,\\ 2m\ddot x = 0,\qquad 2m\ddot y = 0,\qquad \frac{ml^2}2\ddot\theta = 0. $$ So much simpler. Thus, the ability to have generalized coordinates adds great flexibility to analyze and solve systems.

Answer 2

1. In the context of Newtonian mechanics of $N$ rectangular position coordinates ${\bf r}_1, \ldots, {\bf r}_N$, generalized coordinates $q^1,\ldots, q^n, $ are usually introduced as a way to eliminate $3N-n$ holonomic constraints $$f_m({\bf r},t)~=~0, \qquad m~\in~\{1,\ldots,3N-n \}.$$

2. Returning to OP's main question: A Lagrangian $$L({\bf r}(q,t),\dot{\bf r}(q,\dot{q},t),t)$$ of generalized coordinates $q^j$ may equivalently be replaced by a Lagrangian $$ L({\bf r},\dot{\bf r},t)+\sum_{m=1}^{3N-n}\lambda^m f_m({\bf r},t) $$ of rectangular position coordinates ${\bf r}_i$ and Lagrange multipliers $\lambda^m$.

3. Of course the larger picture is that the Lagrangian formalism is independent of the choice of coordinate system, cf. e.g. this Phys.SE post. So one may use the most convenient coordinate system in any given situation.

References:

1. H. Goldstein, Classical Mechanics; Chapter 1.