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Angular displacement isn't a vector quantity (according to some websites) then how angular velocity can be a vector? shouldn't it be a scalar?

Qmechanic
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Kaustav Guha Roy
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    Can you please link the website? In general the position vector for a circular motion is a vector and then so is the angular velocity. It is just often represented by a scalar, because the vector dependance is absorbed in the unit vectors. – Petroglyph Jun 22 '18 at 04:39
  • Possible duplicates: https://physics.stackexchange.com/q/389448/2451 , https://physics.stackexchange.com/q/69345/2451 , https://physics.stackexchange.com/q/286/2451 and links therein. – Qmechanic Jun 22 '18 at 10:21
  • Possibly the source you are looking at said "pseudo vector" instead of "not a vector?" It is a pseudo vector because it does things the other way under reflection. Other than that it has all the usual vector characteristics. –  Jun 22 '18 at 13:40
  • The way I was taught is that only infinitesimal angular displacements can be considered as vectors, which then means angular velocity is a vector since $\omega =\frac{d\theta}{dt}$ – BioPhysicist Jun 22 '18 at 13:43

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In three dimensions, any rotation can be expressed as an angular displacement about some axis. The angular velocity of a solid body is then defined to be the vector which points along the (instantaneous) axis of rotation, and whose magnitude is equal to the (instantaneous) rate of angular displacement about said axis.

In 2D, we only need $1$ parameter to uniquely specify a rotation, and so we can treat angular displacement and angular velocity as scalar quantities.

In any other number of dimensions, we run into problems. The reason for this is that rotations don't generally take place about an axis, but rather in a plane. In three dimensions, a plane through the origin is uniquely determined by a single, perpendicular axis, while in two dimensions, there's only one plane to speak of. In higher dimensions, things get more complicated.

A rotation matrix in $n$ dimensions is an orthogonal matrix with determinant $+1$ - in other words, an element of $SO(n)$. The dimensionality of $SO(n)$ is $\frac{n(n-1)}{2}$, which corresponds to the number of parameters needed to completely specify a rotation. When $n=3$, this corresponds to $3$ parameters, meaning that we can associate a given rotation with a unique 3D vector. When $n=2$, this corresponds to $1$ parameter, so we only need a scalar. However, when $n=4$, a rotation needs $6$ parameters to be completely determined, so neither a scalar nor a 4D vector is sufficient to get the job done.

J. Murray
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