📂Classical Mechanics

What are Generalized Coordinates in Physics?

Definition¹

The coordinates of a system of particles with $n$ degrees of freedom, expressed in $n$ variables (1) that are independent of constraints and (2) mutually independent, are called generalized coordinates.

Generalized Coordinates

In a three-dimensional space where the degrees of freedom of a particle are $3$, the position of the particle can be expressed by the generalized coordinates $q_{1}, q_{2}, q_{3}$ as follows:

$$ \begin{align*} x &= x(q_{1}, q_{2}, q_{3}) \\ y &= y(q_{1}, q_{2}, q_{3}) \\ z &= z(q_{1}, q_{2}, q_{3}) \end{align*} $$

If the degrees of freedom are $2$ or $1$, they can be expressed as follows:

$$ \begin{align*} x &= x(q_{1}, q_{2}) \\ y &= y(q_{1}, q_{2}) \\ z &= z(q_{1}, q_{2}) \end{align*} \qquad\qquad \begin{align*} x &= x(q) \\ y &= y(q) \\ z &= z(q) \end{align*} $$

Circular Motion

Consider a particle performing circular motion on a unit circle in two-dimensional space. The position of the particle can be expressed in polar coordinates $r = (x,y)$. Here, since $y=\sqrt{1-x^{2}}$, the degrees of freedom of this system of particles is $2-1=1$.

$$ \begin{align*} x &= x\\ y &= \sqrt{1 - x^{2}} \end{align*} \quad \implies \quad r = (x, \sqrt{1-x^{2}}) $$

In this case, it is much more convenient to represent it by the angle $\theta$. $$ r = (\cos\theta, \sin\theta) $$ Here, the constraint is $x^{2} + y^{2} = 1$, and the generalized coordinate is $\theta$.

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Double Pendulum

Consider a double pendulum with a radius of $R, r (R \gt r)$. To represent the positions of the two pendulums in two dimensions, four coordinates $(x_{1}, y_{1})$, $(x_{2}, y_{2})$ are required, but the degrees of freedom are $2$. If the angles that the two pendulums form with the $x$ axis are called $\theta_{1}, \theta_{2}$, the positions of all particles in the system can be represented by these two variables.

$$ \begin{align*} x_{1} &= R\cos\theta_{1} \\ y_{1} &= R\sin\theta_{1} \\ x_{2} &= R\cos\theta_{1} + r\cos\theta_{2} \\ y_{2} &= R\sin\theta_{1} + r\sin\theta_{2} \end{align*} $$

Here, the generalized coordinates are $(\theta_{1}, \theta_{2})$.

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Generalized Velocity

The generalized velocity is expressed as follows according to the chain rule.

$$ \dot{x} = \sum\limits_{i=1}^{3} \dfrac{\partial x}{\partial q_{i}} \dot{q_{i}} \qquad \dot{y} = \sum\limits_{i=1}^{3} \dfrac{\partial y}{\partial q_{i}} \dot{q_{i}} \qquad \dot{z} = \sum\limits_{i=1}^{3} \dfrac{\partial z}{\partial q_{i}} \dot{q_{i}} $$

Energy Calculated by Generalized Coordinates

Example 1

Let’s look at a concrete example of calculating kinetic energy and potential energy using generalized coordinates. Consider a situation where an object of mass $M$ can move along the $x$ axis, and an object of mass $m$ is hanging and undergoing pendulum motion with a radius of $r$, as shown in the figure below.

To describe this system of particles, four variables, $(X, Y)$ for the position of $M$ and $(x, y)$ for the position of $m$, are required. However, due to the following constraints, the degrees of freedom are $2$, and the positions of all particles in the system can be expressed with only two variables. $$ Y = 0 \\ (x - X)^{2} + y^{2} = r^{2} $$

Since the system of particles can be represented by the two variables $X$ and $\theta$, the generalized coordinates are $(X, \theta)$. $$ \begin{align*} (X, Y) &= (X, 0) \\ (x, y) &= (X + r\sin\theta, -r\cos\theta) \end{align*} $$

The generalized velocity is as follows.

$$ \dot{X} = \dot{X}, \qquad \dot{x} = \dot{X} + r \dot{\theta} \cos\theta, \qquad \dot{y} = -r \dot{\theta} \sin\theta $$

Then, the kinetic energy $T$ and potential energy $V$ are as follows. $$ \begin{align*} T &= \dfrac{1}{2} M \dot{X}^{2} + \dfrac{1}{2} m \left( \dot{x}^{2} + \dot{y}^{2} \right) \\ &= \dfrac{1}{2} M \dot{X}^{2} + \dfrac{1}{2} m \left[ (\dot{X} + r \dot{\theta} \cos\theta)^{2} + (-r \dot{\theta} \sin\theta)^{2} \right] \\[1em] V &= MgY + mgy \\ &= -mgr\cos\theta \end{align*} $$

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Example 2

Consider the motion of a particle on a two-dimensional plane under a central force. If you choose polar coordinates as generalized coordinates, $q_{1} = r$ and $q_{2} = \theta$, the position and velocity are as follows.

$$ \begin{align*} x &= r\cos\theta &\quad y &= r\sin\theta \\ \dot{x} &= \dot{r}\cos\theta - r\dot{\theta}\sin\theta &\quad \dot{y} &= \dot{r}\sin\theta + r\dot{\theta}\cos\theta \end{align*} $$

Therefore, the kinetic energy and potential energy are:

$$ \begin{align*} T &= \dfrac{1}{2} m (\dot{x}^{2} + \dot{y}^{2}) = \dfrac{1}{2} m (\dot{r}^{2} + r^{2}\dot{\theta}^{2}) \\ V &= V(r) \end{align*} $$

The same result can be obtained using the velocity vector. The velocity in polar coordinates is as follows.

$$ \mathbf{v} = \dot{r}\mathbf{e}_{r} + r\dot{\theta}\mathbf{e}_{\theta} $$

Therefore, since $|\mathbf{v}|^{2} = \mathbf{v} \cdot \mathbf{v} = \dot{r}^{2} + r^{2}\dot{\theta}^{2}$, the kinetic energy is as follows.

$$ T = \dfrac{1}{2} m |\mathbf{v}|^{2} = \dfrac{1}{2} m (\dot{r}^{2} + r^{2}\dot{\theta}^{2}) $$

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