[NOTES/CM-09001] Degrees of Freedom of a Rigid Body

One can model a rigid body by a set of point particles, $\alpha=1,2\dots N,$ such that separation between any pairs remains constant when the body is in motion. The following chain of arguments show that a rigid body is completely described by six generalised coordinates.

1. Let a point $P$ move in such a way that its distance $d_A$ from a fixed point A remains constant. Then $P$ can move on a sphere with A as a center and $d_A$ as radius.
2. If we take two points $A,B$ which remain fixed and allow $P$ to move so that its distances from the two points $A$ and $B$ do not change, the point $P$ can move on a circle in plane perpendicular to the line $AB.$
3. Next suppose we are given three points $A,B,C$ which are not in a straight line and point $P$ whose distances from the three points are specified as $d_A,d_B,d_C$. Although three are two positions for a possible set of values of $d_A,d_B,d_C$, these being mirror images of each other in the plane $ABC,$ no motion of point $P$ with the constraints on distances $$PA=d_A,\qquad PB=d_B,\qquad PC=d_C.$$ is possible. The points in neighbour hood of $P$ violate the above constraints. A particle can jump from $P$ to $P^\prime$, but cannot reach from $P$ to $P^\prime$ by moving continuously along any path.

From the above discussion,  we conclude that a rigid body does not move if any three points of the rigid body, which are not in a straight line, are held fixed. The orientation of the moving body is completely determined by the position of three points.

Thus, if we fix three points of a rigid body and specify their coordinates, we completely specify the position and orientation of the rigid body. Specifying three points in three dimensions requires nine variables, or nine generalised coordinates, which must satisfy three constraints that their mutual distances $AB,BC,$ and $CA$ must remain constant in time. Therefore, the number of independent generalised coordinates needed for a rigid body is $9-3=6$.