Constraints in Rotation Matrix

[dontknow]

In Rigid body rotation, we need only 3 parameters to make a body rotate in any orientation. So to define a rotation matrix in 3d space we only need 3 parameters and we must have 6 constraint equation (6+3=9 no of elements in rotation matrix)

My doubt is if orthogonality conditions R.Transpose(R)=I must be satisfied for rotation matrix (no of constraints=6) and the determinant should be equal to +1 which makes it total 7 constraints, shouldn't be there only two parameters left( instead of 3). Let me know if I have to put some more details.
Refer: Pg no 138 from Goldstein

 

[PeroK]

In Rigid body rotation, we need only 3 parameters to make a body rotate in any orientation. So to define a rotation matrix in 3d space we only need 3 parameters and we must have 6 constraint equation (6+3=9 no of elements in rotation matrix)

My doubt is if orthogonality conditions R.Transpose(R)=I must be satisfied for rotation matrix (no of constraints=6) and the determinant should be equal to +1 which makes it total 7 constraints, shouldn't be there only two parameters left( instead of 3). Let me know if I have to put some more details.
Refer: Pg no 138 from Goldstein

The constraint that $\det R = 1$ does not reduce the dimension of the space. Both $O(3)$ and $SO(3)$ are three-dimensional. See here, for example:

http://www.physics.mcgill.ca/~yangob/groups.pdf

 

[vanhees71]

You restrict yourself to the part of the orthogonal group in 3D that is continuously connected to the identity. Then you have only the constraint $\hat{O}^{\text{T}} \hat{O}=\hat{1}$, because that implies that you necessarily have $\mathrm{det} \hat{O} \in \{-1,1 \}$. So the continuously connected part of O(3) is SO(3). So you have 6 constraints for 9 real matrix elements and thus 3 parameters. For rigid-body theory the usual choice are Euler angles.