Is there an equivalent of a scalar potential for torques?

Question

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 1

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.

Answer 2

Nice question. Wish I were bright enough to think of things as out of left field as this one. So I'm sorry to say the answer is no, for two reasons:

1. The scalar field, call it $P(\mathbf{r})$ you seek has to be, by definition of a field, a function of the point in space $\mathbf{r}$ alone, whereas the torque on the body depends both on $\mathbf{r}$, the body's shape, mass distribution AND the body's orientation.

2. Let's suppose you try to overcome this first problem by defining a scalar field $P(\mathbf{r})$ that is meaningful only for a particular body (a "test standard") that is in a "standard" orientation that is always the same. Then, the torque "field" for this standard body and orientation as a function of its centre of mass (or some other "standard" point within the body defining its position) $\mathbf{r}$ will be $\mathbf{T}\left(\mathbf{r}\right) = \int_M \nabla V\left(\mathbf{r} + \mathbf{r}^\prime\right) \wedge \mathbf{r}^\prime \rho\left(\mathbf{r}^\prime\right) dV^\prime$ where the primed position vectors are the dummies for the integration and the position in the field $\mathbf{r}$ "offsets" them. $\rho$ is the body's mass distribution. Now, if this field is to be derivable from a scalar potential ($\mathbf{T}\left(\mathbf{r}\right) = \nabla P(\mathbf{r})$) then a necessary condition for this is that $\nabla \wedge \mathbf{T} = \mathbf{0}$ (the curl always annihilates the gradient where $\nabla \wedge \nabla$ exists). So, if you work out the curl we get $\nabla \wedge \mathbf{T}\left(\mathbf{r}\right) = \int_M \left[\mathbf{r}^\prime . \nabla\, \nabla V\left(\mathbf{r} + \mathbf{r}^\prime\right) - \mathbf{r}^\prime \nabla^2 V\left(\mathbf{r} + \mathbf{r}^\prime\right)\right]\rho\left(\mathbf{r}^\prime\right) dV^\prime$, which, unless I am very much mistaken, is not identically nought for a general scalar force potential $V$ and so, unfortunately, the necessary condition is not fulfilled and the torque field is not derivable from a potential.

Again, great idea and hope this helps.

Edit after Rody's question: "how about a vector potential".

A vector potential IS possible, but, as stated in 1. above, it is only meaningful for the particular body in question in a particular constant orientation.

To show this, we form the divergence of the torque "field":

$\nabla . \mathbf{T}\left(\mathbf{r}\right) = \int_M \left[\mathbf{r}^\prime . \nabla \wedge \nabla V\left(\mathbf{r} + \mathbf{r}^\prime\right) -\nabla V\left(\mathbf{r} + \mathbf{r}^\prime\right) . \nabla \wedge \mathbf{r}^\prime\right]\rho\left(\mathbf{r}^\prime\right) dV^\prime = 0$. So there is always a field $\mathbf{A}$ such that $\mathbf{T}\left(\mathbf{r}\right) = \nabla \wedge \mathbf{A}$. The easiest way to visualize this is in three dimensional Fourier space. In Fourier space the operations $\nabla . ()$ and $\nabla \wedge ()$ are replaced by $i\,\mathbf{k} . ()$ and $i\,\mathbf{k} \wedge ()$ where $\mathbf{k}$ is the "wavevector" (triple of the three Fourier transform variables $k_x$, $k_y$, $k_z$). So, a divergenceless (aka solenoidal) vector field in Fourier space is always orthogonal to $\mathbf{k}$ (i.e. orthogonal to the position vector in Fourier space); otherwise put, it is tangent to spheres centred at the origin. For such a vector field, in Fourier space we have $\mathbf{k} \wedge (\mathbf{k} \wedge \mathbf{\tilde{T}}) = k^2 \mathbf{\tilde{T}})$ and the curl is invertible for this special case. Therefore, to find the vector field $\mathbf{A}$ we transform $\mathbf{T}$ to Fourier space to get $\tilde{T}$, then the Fourier transform of $\mathbf{A}$ must be $\mathbf{\tilde{A}} = \frac{i}{k^2} \mathbf{k} \wedge \mathbf{\tilde{T}}$, then transform back to get the vector potential.