Is $d\theta$ a vector and does the equation $d\vec{s}=d\theta \times \vec{r}$ make sense?

Question

I was reading about torque on Wikipedia and I found this passage (in the section "Relationship between torque, power, and energy")

The work done by a variable force acting over a finite linear displacement $s$ is given by integrating the force with respect to an elemental linear displacement $d\vec{s}$

$$W=\int_{s_1}^{s_2} \vec{F} \cdot \vec{ds}$$

Note that I inserted the vector arrows above, though in the Wikipedia article the variables are in bold (though it isn't very clear always whether there is a bold font or not so I may make a mistake in transcribing). In particular, I am not sure if $d\theta$ is a vector or not.

However, the infinitesimal linear displacement $d\vec{s}$ is related to a corresponding angular displacement $d\theta$ and the radius vector $\vec{r}$ as $$d\vec{s}=d\theta \times \vec{r}$$

Is the last equation correct? It seems like $d\theta$ is a scalar and $\vec{r}$ is a vector. My question is if I read the passage correctly, and if so, if it is correct as written.

Later in the passage, it seems like $d\theta$ is in fact a vector, because the $d\vec{s}$ expression is substituted into the work equation

$$W=\int_{s_1}^{s_2} \vec{F}\cdot d\theta \times \vec{r}$$

I am confused because usually I see arc length being $ds=rd\theta$.

Answer 1

The key is that $\text d\boldsymbol\theta$ is a differential, and thus is an infinitesimal quantity. This can be seen by doing the simple "experiment" below. While this answer is not mathematically rigorous, I think it shows how we get a vector from angular displacements. This experiment is from a "discussion question" exercise from Sears & Zemansky's "University Physics with Modern Physics" 13th edition:

Although angular velocity and angular acceleration can be treated as vectors, the angular displacement $\theta$, despite having a magnitude and direction, cannot. This is is because $\theta$ does not follow the commutative law of vector addition.

Prove this to yourself in the following way: Lay your physics textbook flat on the desk in front of you with the cover side up so that you can read the writing on it. Rotate it through $90^\circ$ about a horizontal axis so that farthest edge comes toward you. Call this angular displacement $\theta_1$. Then rotate by $90^\circ$ about a vertical axis so that the left edge comes toward you. Call this angular displacement $\theta_2$. The spine of the book should now face you, with the writing oriented so that you can read it.

Now start over again but carry out the two rotations in the reverse order. Do you get a different result? That is, does $\theta_1+\theta_2$ equal $\theta_2+\theta_1$?

Now repeat this experiment but this time with an angle of $1^\circ$ rather than $90^\circ$. Do you think that the infinitesimal displacement $\text d\boldsymbol\theta$ obeys the commutative law of addition and hence qualifies as a vector? If so, how is the direction of $\text d\boldsymbol\theta$ related to the direction of $\boldsymbol \omega$?

The infinitesimal angular displacement can be treated as a vector just fine, even if "larger" displacements cannot be.

Answer 2

In the wikipedia's article, it is written $d\boldsymbol \theta \times \mathbf r$, the cross product of a (pseudo)vector and a vector. As the direction of the pseudovector is perpendicular to the plane of movement, the resultant vector is tangent to the trajectory.

Answer 3

Lets start with this equation

$$\vec v=\vec\omega\times \vec r\tag 1$$

with : $$\vec v=v\,\vec e_v=\frac{ds}{dt}\,\vec e_v$$

and $$\vec\omega=\omega\,\vec e_\omega=\frac{d\theta}{dt}\,\vec e_\omega$$

you obtain that

$$ds\,\vec e_v=d\theta\,\vec e_\omega\times \vec r$$

hence

$$\vec{ds}=ds\,\vec e_v\quad,\vec{d\theta}=d\theta\,\vec e_\omega\\ \text{so that}\\ \vec{ds}=\vec{d\theta}\times\vec{r}$$

so

$$\vec F\cdot\vec{ds}\mapsto \vec F\cdot(\vec{d\theta}\times\vec{r})=\vec F\cdot\left(d\theta\,\vec e_\omega\times \vec r\right)=\vec F\cdot\left(\vec e_\omega\times \vec r\right)\,d\theta$$ this make sense