# Is torque always equal to the derivative of potential energy with respect to rotation angle?

For any three-dimensional rigid body, the applied torque on that body is defined as:

$\vec{\tau} = \vec{r} \times \vec{F}$

where $\vec{F}$ is the applied force on the object (i.e. $-\vec{\nabla} U$) and $\vec{r}$ is the vector pointing from the center of mass to the point at which the force is applied. My question is, does this always imply that:

$|\vec{\tau}| = \left|\frac{dU}{d\theta}\right|$,

where $d\theta$ represents an infinitesimal rotation of the body about an axis parallel to $\vec{r} \times \vec{F}$? If not, what would be one counterexample?

• Your second equation can be rewritten as $\tau \Delta \theta = \Delta U$; work done is equal to the change in potential energy. Sep 13, 2018 at 5:20

Firstly, the full version of your second equation, including the sign, and allowing an infinitesimal rotation by an angle $d\theta$ about any arbitrary axis $\hat{n}$ (a unit vector), is $$\hat{n}\cdot\vec{\tau} = -\frac{d U}{d\theta}$$ If you like, you can take this as the definition of the torque. Applying the chain rule allows us to relate this to the force, and to the angular derivative of the position vector $\vec{r}$: $$\hat{n}\cdot\vec{\tau} = -\frac{d U}{d\vec{r}} \cdot \frac{d\vec{r}}{d \theta} = -\vec{\nabla}U \cdot \frac{d\vec{r}}{d \theta} = -\frac{d\vec{r}}{d \theta} \cdot\vec{\nabla}U = \frac{d\vec{r}}{d \theta} \cdot \vec{F}$$ The general formula for actively rotating a vector by an angle $\theta$ according to the right-hand rule about an axis $\hat{n}$ is $$\vec{r}' = \vec{r} \cos\theta + \hat{n}\,(\hat{n}\cdot\vec{r})\,[1-\cos\theta] + (\hat{n}\times\vec{r}) \sin\theta$$ but in the limit of infinitesimal rotations $\theta\rightarrow d\theta$, $\cos d\theta\rightarrow 1$, $\sin d\theta\rightarrow d\theta$, this reduces to $$\vec{r}' - \vec{r} = d\vec{r} = (\hat{n}\times\vec{r})\, d\theta \quad \text{or} \quad \frac{d\vec{r}}{d \theta} = \hat{n}\times\vec{r}$$ [This formula is also familiar as $(d\vec{r}/d t) = \vec{\omega}\times\vec{r}$ where the angular velocity is $\vec{\omega}=\hat{n}\,(d \theta/dt)$].
So we end up with a scalar triple product on the right, which can be rearranged: $$\hat{n}\cdot\vec{\tau} = \hat{n}\times\vec{r} \cdot \vec{F} = \hat{n}\cdot\vec{r} \times \vec{F}$$ We can choose $\hat{n}$ to be the $x$, $y$, and $z$ axes in turn, to give each component of your first equation $$\vec{\tau}=\vec{r} \times \vec{F}.$$