classical mechanics - Is there an equivalent of a scalar potential for torques?

For a given scalar potential $V$, it is known that the corresponding force field $\mathbf{F}$ can be computed from

$$ \mathbf{F} = -\nabla V $$

Suppose a rigid body is placed inside this potential. The torque on the body $\mathbf{T}$ exerted by the scalar field will be

$$ \begin{align} \mathbf{T} &= \int_M \left( \mathbf{r} \times \mathbf{F}(\mathbf{r}) \right) dm \\ &= \int_M \left( \mathbf{r} \times -\nabla V(\mathbf{r}) \right) dm \\ &= \int_M \left( \nabla V(\mathbf{r}) \times \mathbf{r} \right) dm \end{align} $$

with $\mathbf{r}$ the position vector to the mass element $dm$, and the integration carried out over the entire body $M$.

So, being not too familiar with rigid body dynamics, I was wondering -- does something like a (vector/scalar) potential $P$ exist, such that the local torque induced by the potential $V$ can be expressed as

$$ \matrix{ \mathbf{T} = I \cdot \nabla P & & &\text{(or some similar form)} } $$

with $I$ the moment of inertia tensor of the rigid body?

If such a thing exists:

• what is its name?


• where should I start reading?


• What is the proper expression for the torque $\mathbf{T}$?


• how does $P$ relate to $V$?

If such a thing doesn't exist:

• why not? :)

Answer

Great question. A little background first.

Note that any force $\boldsymbol{F}$ moment $\boldsymbol{M}$ system on a point A can be equipollently translated into the screw axis S leaving only the components of $\boldsymbol{M}$ that are parallel to $\boldsymbol{F}$. The location is found by

$$ \boldsymbol{r} = \frac{\boldsymbol{F} \times \boldsymbol{M}}{\boldsymbol{F}\cdot\boldsymbol{F}} $$

Also the moment components parallel to $\boldsymbol{F}$ are described by a scalar pitch value $h$ found by

$$ h = \frac{ \boldsymbol{M} \cdot \boldsymbol{F}}{\boldsymbol{F} \cdot \boldsymbol{F}} $$

In reverse, a moment is defined by a force vector $\boldsymbol{F}$ passing through an axis located at $\boldsymbol{r}$ with pitch $h$

$$ \boldsymbol{M} = \boldsymbol{r} \times \boldsymbol{F} + h \boldsymbol{F} $$

Have you noticed how difficult it is to apply a pure moment on a rigid body, without applying a force? This is because you cannot have one without the other. A moment is really a result of the line of action of forces. So the scalar potential of a moment is really the same as the one for forces with

$$ \boldsymbol{M} = - \boldsymbol{r} \times \nabla V - h \nabla V = -\left( \left[1\right] h + \boldsymbol{r}\times \right) \nabla V $$

The problem is that in rigid body mechanics forces are not treated as scalar fields, but spatially constant, and temporally varying. Furthermore, I cannot think of a case where spatially varying moments arise that are NOT due to a force at a distance. I suppose you can come up with a tensor pitch $h$ instead of a scalar which is spatially varying for a definition like $\boldsymbol{M} = -\left( H + \boldsymbol{r}\times \right) \nabla V$, but then you will be making things up that do not have any physical meaning that I know of.