Generalized coordinates

In this post I’ll try to explain what Generalized coordinates are, how they came to be and finally where they are used with a simple.

To start with imagine a system of $k$ point masses present in a $d$ dimensional space. The $i^{th}$ point has a position vector $r_i = (x_{i1}, x_{i2}, \cdots, x_{id})$.

Note that if the system was unconstrained, then as any of the $k$ bodies can occupy any point in the space, we will need $kd$ number of scalars to completely define any particular configuration of the system.

For visualization purposes consider $d=3$, so we use $3$ scalars (the $x, y, z$ coordinates) to define the position of each point mass. As we have a total of $k$ such masses we need $3k$ scalars.

But often the systems we need to analyse are under some sort of constraints. In case of a robotic arm, each hinge point always maintains a fixed distance between them due to the rigid rod connecting them.

In such situations we don’t need all $kd$ scalars, rather we can use less number of scalars and still manage to uniquely represent all possible states that this constrained system can take. This is because the constraints decrease the degrees of freedom available to that system.

Holonomic constraints

We classify constraints as either being holonomic or nonholonomic.

If a constraint can be expressed purely as a relationship between the position variables ($r_1, r_2, \cdots, r_k$) and time $t$.
$${\displaystyle f(r_1, r_2, \cdots, r_k,\ t)=0,\,}$$
Then using each such constraint equation we can reduce the number of scalars needed by one. If we knew the time $t$ and all but one position variable, using the constraint equation that missing coordinate can be deduced.

When the constraint has inequalities or higher order derivative terms in them, they are classified as nonholonomic. Note that we have an equality symbol in the above expression. Also the function has to of the position variables themselves, not their derivatives.

All constraints which are unable to be expressed in the above form are called nonholonomic constraints.

Note that if there were velocity terms ($\dot r_i$) in the constraint equation, then integrating that constraint could in some cases yield an equation of the form shown above, if so then it’s a holonomic constraint. So what Wikipedia means when they say that if the constraint has velocity terms they “are not usually holonomic” is that it’s kinda hard to bring it to this form in that case.

When constraints are purely dependent on the positions at any time $t$ its a holonomic constraint, but in a nonholonomic constraint the constraint changes depending on how you reach that position! This seems to be analogous to path and state functions.

To visualise how a holonomic constraint reduces DOF, imagine a point mass is constrained to move along the line $y = mx+c$. So without this constraint we needed both position variables $(x,y)$ to describe its position anywhere in the XY plane, but now we just need $(x)$. As if we have $x$, its position is now known to be $(x,mx+c)$ due to the constraint.

Constrained to move in a circle of radius $r$

So a point mass is moving around in a plane (say the $XY$ plane). We can describe its position at any point of time using the vector,

$$\vec r = \left [ \begin{array}{cc} r_1(t) \\ r_2(t) \\ \end{array} \right ]= \left [ \begin{array}{cc} x(t) \\ y(t) \\ \end{array} \right ]$$

If we are given that this point mass is constrained to move in a circle of radius $c$, i.e. the constraint equation would like this,

$$|\vec r| = c^2$$
$$x(t)^2 + y(t)^2 = c^2$$

Note how the equation can be written as $x(t)^2 + y(t)^2 - c^2 + 0t = 0$, i.e. in the form $f(x,y,t)=0$. Therefore we conclude that this is a holonomic constraint.

So the “generalized coordinate” I will use is $\theta$, the angle between the $x$ axis and the position vector $\vec r$.

We know that $x = c \cos \theta$ and $y = c \sin \theta$.

Let the point mass move a little, it can of course only move in a small arc along the circle due to the constraint imposed on it. Such displacement is called Virtual displacement. In terms of $\theta$ it’s $\delta \theta$, while in terms of $\vec r$ it’s

$$\delta\vec r = \left [ \begin{array}{cc} \delta x \\ \delta y \\ \end{array} \right ]$$

The following formula relates this change in the $m$ generalized coordinates ($\delta q$) to that of the change in natural coordinates ($\delta\vec r$).
$$\delta\vec r_i=\sum _{{j=1}}^{m}{\frac {\partial {\vec r}_{i}}{\partial q_{j}}}\delta q_{j}\,$$

The term ${\frac {\partial {\vec r}_{i}}{\partial q_{j}}}$ is a measure of how much change in ${\vec r}_{i}$ happens for a unit change in $q_{j}$, keeping all other generalized coordinates $q_i \ \forall \ i \in [1,... , j-1, j+1, ... m]$ constant - that’s what partial differentiation is.

So by multiplying ${\frac {\partial {\vec r}_{i}}{\partial q_{j}}}$ with $\delta q_{j}$, you get the contributed of the $j_{th}$ generalized coordinate - $\ q_j$ in the displacement $\delta\vec r_i$.

Summing over all these contributions (from $1$ to $m$) we get the net virtual displacement $\delta\vec r_i$.

Coming back to our example, we can therefore note that

$$\delta x = \frac{\partial x}{\partial \theta}\delta \theta = c(-\sin \theta)\delta \theta$$

$$\delta y = \frac{\partial y}{\partial \theta}\delta \theta = c(\cos \theta)\delta \theta$$

Credits to Maschen - for this image, CC0, Link

This makes sense as for small displacement, the $\delta\vec r$ shown in the figure will be a straight line, thus instead of a sector, we have a right angled triangle with $\delta x$, $\delta y$ and $|\delta\vec r|$ as the sides. $|\delta\vec r|$ will be the length of the arc which is $c \ \delta \theta$.

So from pythagoras theorem we have $(\delta x)^2 + (\delta y)^2 = |\delta\vec r|^2 = (c \ \delta \theta)^2$.

And sure enough this fits in with our “derived” relationship between the generalized and natural coordinates.

$$(\delta x)^2 + (\delta y)^2 = (c(-\sin \theta)\delta \theta)^2 + (c(\cos \theta)\delta \theta)^2$$

$$=c^2(\delta \theta)^2((\sin \theta)^2+(\cos \theta)^2) = (c \ \delta \theta)^2$$

You can also try doing this with $q = s$ rather than $q = \theta$ like how I showed above. Where $s$ is the length of the curve traced out by the point mass from some reference point.

I thought of writing it up but it turns out that the length of a curve $y=f(x)$ from $(a,f(a)$ to $(b,f(b))$ is $S$. Which is kinda complicated,

$$S = \int_{a}^b\sqrt {1 + (f'(x))^2}$$

Well as always let me know if you have any thoughts on my post.

Written with StackEdit.