Graduate Describing a Rotating Vector with Changing Angular-Velocity

[Jianbing_Shao]

I have a question about a rotating vector field:
if there is a vector $A^i(t_0)$ at the origin in coordinate space $IR^3$ , when $t≥t_0$, the vector rotates with a changing angular-velocity $ω^i(t)$. then we can get a rotating vector field $A^i(t)$. then how to describe $A^i(t)$ using angular-velocity $ω^i(t)$?
Further more, if $\theta^i(t_1)\doteq \int^{t_1}_{t_0}\omega^i(t)dt=0$, then can we assert that $A^i(t_1)=A^i(t_0)$?

 

[kent davidge]

You may use a coordinate system in which rotation is explicit. Example, in $\mathbb R^2$ let a position vector be represented by $\vec r = (x,y)$ in cartesian coordinates or $\vec r = x \partial_x + y \partial_y$ and $\vec r = r (\cos\theta, \sin\theta)$ in terms of polar coordinates. Now $(\cos\theta, \sin\theta)$ turns out to be the components of the radial basis vector in cartesian coordinates (a vector along the radial coordinate line in the polar coordinate system). So $\vec r = r \partial_r$. If you take the time derivative of this vector you get $\vec r ' = \omega \partial_\theta$ where $\omega = d\theta / dt$ is the "rotation" speed and $\partial_\theta$ is a vector lying along the angular coordinate line of the polar coordinate system, having components $(-r \sin\theta, r \cos\theta)$ in the cartesian coordinate system and I assumed that $r$ is constant because we want only rotation. So $\omega$ describes at what speed $\vec r$ rotates.

Because $\omega = d \theta / dt$ then $\theta(t_1) - \theta(t_0) = \int_{t_0}^{t_1} \omega dt$ and if this integral vanishes then $\theta(t_1) = \theta(t_0)$. So yes the vector was not rotating during the time period from $t_0$ to $t_1$.

 

[Jianbing_Shao]

Can we have a formula about this problem？
I am not so sure about your conclusion that the vector will invariant because the rotation around different axes is not commutative. So we can study a specific process: at first we rotate vector around $x^1$-axis by an angle $α$, then rotate it around $x^2$-axis by an angle $β$, next we rotate it around $x^1$-axis by an angle $-α$ , at last we rotate it around $x^2$-axis by an angle $-β$, Obviously this process can fulfill the condition $\int \omega(t) dt=0$. then the rotation can be written in such a formula:
$$\exp(-\beta X_2)\exp(-\alpha X_1)\exp(\beta X_2)\exp(\alpha X_1)$$
the rotation matrix equal to 1 only when $[\exp(\alpha X_1) , \exp(\beta X_2)]=0$. so in most cases this vector is not invariant if $\int \omega(t) dt=0$. This is the question that puzzles me.

 

[pervect]

I have a question about a rotating vector field:
if there is a vector $A^i(t_0)$ at the origin in coordinate space $IR^3$ , when $t≥t_0$, the vector rotates with a changing angular-velocity $ω^i(t)$. then we can get a rotating vector field $A^i(t)$. then how to describe $A^i(t)$ using angular-velocity $ω^i(t)$?
Further more, if $\theta^i(t_1)\doteq \int^{t_1}_{t_0}\omega^i(t)dt=0$, then can we assert that $A^i(t_1)=A^i(t_0)$?

I'm not sure I understand what you're asking in your first question. The usual formula non-tensor formula would be $\vec{v} = \vec{r} \times \vec{\omega}$

Here $\vec{r}$ is a displacement vector, so the formula gives a map from the displacement vector $\vec{r}$ to a velocity vector $\vec{v}$.

I believe the tensor version of the above formula, a map from a displacement r to a vector v, would be:

$$v^i = \epsilon^{ijk} \omega_j \, \wedge \, r_k$$

The $\wedge$ represents the wedge product, see for instance https://en.wikipedia.org/wiki/Exterior_algebra#Cross_and_triple_products

$\epsilon^{ijk}$ is the 3-d Levi-Civiti tensor.

In four dimensions, the cross product of two vectors is no longer a vector, but a bivector. (I'm not sure if you wanted to ask that, but I thought I'd mention it).

As far as your second answer goes, I'm not sure if I'm interpreting it correctly or not, but in general rotations do not commute, so if you rotate an object about different axes, even if the vectors $\omega^i$ representing the rotations sum to zero, the object will in general wind up being rotated.

 

[Jianbing_Shao]

I'm not sure I understand what you're asking in your first question. The usual formula non-tensor formula would be $\vec{v} = \vec{r} \times \vec{\omega}$

Here $\vec{r}$ is a displacement vector, so the formula gives a map from the displacement vector $\vec{r}$ to a velocity vector $\vec{v}$.

I believe the tensor version of the above formula, a map from a displacement r to a vector v, would be:

$$v^i = \epsilon^{ijk} \omega_j \, \wedge \, r_k$$

The $\wedge$ represents the wedge product, see for instance https://en.wikipedia.org/wiki/Exterior_algebra#Cross_and_triple_products

$\epsilon^{ijk}$ is the 3-d Levi-Civiti tensor.

In four dimensions, the cross product of two vectors is no longer a vector, but a bivector. (I'm not sure if you wanted to ask that, but I thought I'd mention it).

As far as your second answer goes, I'm not sure if I'm interpreting it correctly or not, but in general rotations do not commute, so if you rotate an object about different axes, even if the vectors $\omega^i$ representing the rotations sum to zero, the object will in general wind up being rotated.

You are right, I think the formula:
$$v^i = \epsilon^{ijk} \omega_j \, \wedge \, r_k$$
can also be written in such a form:
$$\partial_t A(t)=(\omega^k X_k)A(t)$$,
Here, $X_k$ represent the generators of rotation group.
and ,my question is if can we use a rotation matrix to describe the changing process of the rotating vector.A matrix such as $\exp(\theta^i(t)X_i)$, and $\theta^i(t)$ is determined by angular velocity $\omega^i(t)$.
If we can get such a formula, perhaps we can easily find out the answer of the second question.