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I searched on net for the question, " why are only small angular displacements considered vectors but not large angular displacement? " I saw an video where the reason was given that, this was because large angular displacements did'nt follow commutative property of vector addition but small angular displacements did follow the commutative property of vector addition. It was explained by giving an practical demonstartion how this was true, but in that demonstartion they used 3D space and as per their demonstation I think that in 2D space even large angular displacements follow commutative property of vector addition. So can we say that, even large angular displacement is a vector quantity in 2D space?

Qmechanic
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Sameer Nilkhan
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    A planar rotation is a fixed vector out of the plane. Since the direction of the vector does not change, then yes, standard vector algebra applies. – John Alexiou Jan 27 '20 at 17:23

2 Answers2

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I wouldn't call this a vector quantity. The reason for why small displacements are sometimes described as "vector quantities" (especially in introductory books and courses) is because of their relation with (Lie) group theory:

Rotations in $n$ dimensions form the Lie group SO$(n)$. Small angular displacements are approximately given by infinitesimal rotations. These infinitesimal transformations are given by the Lie algebra $\mathfrak{so}(n)$ (which is in particular a vector space). It can be shown that this Lie algebra is isomorphic to $\mathbb{R}^3$ with the cross product as multiplication. This explains why we often say that small rotations in 3D are "vector quantities".

Your example of 2 dimensions is also an interesting one. It can be shown that the Lie group SO$(2)$ of rotations in the plane is an Abelian group. This means that all rotations in the plane commute (as you said). However this does not mean that these rotations are described by vectors, this is again only the case for small rotations.

NDewolf
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  • angular displacements , may it be large or small in 2-D space have both magnitude , in terms of radian and direction , either clockwise or anticlockwise , then why are they not considered vectors ? Please explain with some example . – Sameer Nilkhan Jan 27 '20 at 10:05
  • You could indeed use 1D vectors the way you describe. This is a consequence of the fact that you can relate the circle with the interval $[0, 2\pi[$ (or when using vectors, the real line $\mathbb{R}$) using the exponential function.

    However this does not give you a useful technique to find out where points in the plane will go to under a rotation. (As was the case in 3D). How would you compose two rotations in the plane using these 1D vectors?

     – NDewolf Jan 27 '20 at 10:11
  • related https://physics.stackexchange.com/questions/146897/if-angular-velocity-angular-acceleration-are-vectors-why-not-angular-displace However, in a 2D (Clifford) space angular position is a bivector e.g. https://arxiv.org/pdf/gr-qc/9608052.pdf – Mr Anderson Jan 27 '20 at 11:17
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the criteria for the angular displacement to be a vector it must fulfill this equation.

$$\vec{\omega}=\vec{\dot{\phi}}\tag 1$$

where $\vec{\omega}$ is the angular velocity.

to obtain the angular velocity , you use this equation:

$$\dot{R}=R\,\tilde{\omega}\tag 2$$

where R is the rotation matrix and is function of the angular displacement $\phi=\phi_x,\phi_y,\phi_z$

only if you linearized equation (2) you get equation(1)

with: $\phi=\Delta\phi$ small angles

$$R\mapsto I_3+\Delta\phi\quad,\dot{R}\mapsto \dot{\Delta\phi} $$

$ \Rightarrow$

$$\widetilde{\dot{\Delta\phi}}=\tilde{\omega}\quad,\vec{\omega}=\frac{d}{dt}\vec{\Delta\phi} $$

equation (1) is fulfill thus $\Delta\phi$ is a vector with the components $\Delta\phi_x\,,\Delta\phi_y\,,\Delta\phi_z$

2D case

$$R= \begin{bmatrix} \cos(\phi_z) & -\sin(\phi_z) & 0 \\ \sin(\phi_z) & \cos(\phi_z) & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

you get

$$\vec{\omega}=\begin{bmatrix} 0 \\ 0 \\ \omega_z \\ \end{bmatrix}=\vec{\dot{\phi}}=\begin{bmatrix} 0 \\ 0 \\ \dot{\phi}_z \\ \end{bmatrix}$$

thus equation (1) is even for large angular displacement $\phi_z$ fulfill, your statement is correct!

$$\tilde{a}= \begin{bmatrix} 0 & -a_z & a_y \\ a_z & 0 & -a_x \\ -a_y & a_x & 0 \\ \end{bmatrix}$$

Eli
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