Change in a vector upon rotation of the coordinate frame

[Jigyasa]

Homework Statement

Hi everyone. We were discussing conservation of angular momentum as a consequence of rotational invariance in class. There was one point where we needed to compute the change in a vector A when the coordinate frame is rotated by angle Δ(Φ).

Homework Equations

The teacher said that ΔA = ΔΦ x A (cross product), where ΔΦ is a vector perpendicular to the plane of rotation. I am not sure if I understand this equation correctly.

The Attempt at a Solution

This is how I understand it:
Consider rotation in the x-y plane. As Φ is an angle in the x-y plane, grad(Φ) will be in the direction of the outward normal (z direction). Now, (grad(Φ) x A) will give the direction of A

The problem is, I think I'm mixing up grad(Φ) and the vector ΔΦ. Can someone help?

 

[haruspex]

ΔA = ΔΦ x A (cross product), where ΔΦ is a vector perpendicular to the plane of rotation.

That is only valid for small rotations. For example, consider a radius vector length r in the plane and subjecting it to a small rotation dθ. The endpoint moves by rdθ at right angles to the radius vector. If we represent the rotation by a vector normal to the plane then we can write the new radius vector as $\vec r'=\vec r+\vec r\times\vec{d\theta}$.

grad(Φ)

No, grad is an entirely different beast. If you see grad(φ) written then almost surely the φ is very different too. It would be a scalar potential as a function of position, $\phi=\phi(\vec r)$. grad(φ) would be the vector formed by the partial derivatives of φ along the different coordinates.

 

[Jigyasa]

That is only valid for small rotations. For example, consider a radius vector length r in the plane and subjecting it to a small rotation dθ. The endpoint moves by rdθ at right angles to the radius vector. If we represent the rotation by a vector normal to the plane then we can write the new radius vector as $\vec r'=\vec r+\vec r\times\vec{d\theta}$.

No, grad is an entirely different beast. If you see grad(φ) written then almost surely the φ is very different too. It would be a scalar potential as a function of position, $\phi=\phi(\vec r)$. grad(φ) would be the vector formed by the partial derivatives of φ along the different coordinates.

How do you take the direction of dθ

I understand till the equation is in magnitude form. To convert it into vector form, I need to write dθ as a vector and then take its cross product with A

 

[haruspex]

How do you take the direction of dθ

A rotation can be represented by a vector with magnitude proportional to the extent of rotation and direction indicating the axis. It is not very useful for large rotations, but normally only used with infinitesimal ones and, thus, for derivatives - angular velocity and angular acceleration.
A convention is needed to decide which way along the axis the vector points. As far as I know, the right-hand rule is universal. If in your view of the rotation it is clockwise then the vector points away from you. This convention requires a corresponding convention for the cross product and the equations involved. E.g. tangential velocity is $\vec r\times\vec \omega$, not the other way around.

 

[Jigyasa]

Understood. Thanks a lot :)