# Can a conservative field produce a torque?

I am asking whether the following Lagrangian for a point moving in a conservative field, can be correct :

$L(r, v, \omega) = \frac {mv^2}{2} + \frac {I \omega^2}{2} - U(r)$.

$r$ is the distance between the equipotential surface on which the movement begins and the equipotential surface on which the movement ends, $v = \text d r/ \text d t$, $\omega$ is angular velocity of rotation around some fix point in space (see example on the bottom of the text), $I = m\rho^2$, where $\vec \rho$ is the vector connecting the fix point in the space with the current position of the moving point (see the example).

What I am not sure on, is the presence of the term $I \omega^2/2$. I think that $\omega$ can vary only if the potential energy can produce a torque ($\vec F \ \text x \ \vec \rho$), and in that case $U$ should also depend on a variable $\theta$, indicating the angle between the vector $\vec \rho$ and a fix axis in the rotation plane.

But, if there is a torque, if $U$ depends not only on the distance between equipotential surfaces, but also on an angle $\theta$, is this anymore a conservative field? I know that in a conservative field the mechanical work doesn't depend on the path followed by the point, but on the distance between equipotential surfaces, however that doesn't help me in my question.

(As a simple example, one can think that the field is produced by an electric charge uniformly distributed on an ellipsoid. Then $d$ is the distance to the surface of the ellipsoid measured perpendicularly on the equipotential surfaces, and given a point $P$ in the field, $\vec \rho$ is the vector from the center of the ellipsoid to the pint $P$.)

• Strictly speaking, that $\frac{I\omega^2} 2$ term should be $\frac 1 2 \omega \cdot (I\omega)$. Moment of inertia is a tensor quantity. – David Hammen Jan 19 '15 at 0:25
• @DavidHammen but my moving object is a point, not an object with some volume. Well, I hope to "meet" you tomorrow. – Sofia Jan 19 '15 at 0:27
• Classically, the moment of inertia of a point mass is zero. That term vanishes for a point mass. – David Hammen Jan 19 '15 at 2:21
• @DavidHammen no, David, that term doesn't express spin of the point around itself, but around an external point, see the example with the ellipsoid. I also feel that the potential $U$ has to contain a variable $\theta$, after all $\omega$ is the derivative of $\theta$. – Sofia Jan 19 '15 at 11:11
• @DavidHammen my worry is that if $U$ depends on an angle $\theta$ and there is also a torque, then the field is no more conservative. What you say? – Sofia Jan 19 '15 at 11:50

• Something similar happens in fluid mechanics - e.g. a dumb-bell in a plane strain $\vec{u}\sim \vec{x}$ (Landau and Lifshitz 1959, $\S$ 11). – Nick P Jan 19 '15 at 1:43
I'm not sure about the notation, so there will be a bit of guessing here. I assume $v = \dot r$, so the kinetic term can be interpreted to be that of a point moving on a plane, described by polar coordinates. Now take any radial potential, which by the rotational symmetry generates a central field. Let us consider Kepler's problem to be definite here. It is well known that closed orbits are elliptic in general, hence $\omega$ is not constant throughout the motion, but $\mathbf F\times\mathbf r = 0$ at every time.
• "where ρ is the distance to some fix point in the space." Unless the fixed point is the origin, i.e. $\rho = 0$ implies $r = 0$, this fixed point breaks the rotational symmetry around the "origin". – Phoenix87 Jan 18 '15 at 23:20