On the geometric phase in the spatial equilibria of nonlinear rods

Abstract

Geometric phases have natural manifestations in large deformations of geometrically exact rods. The primary concerns of this article are the physical implications and observable consequences of geometric phases arising from the deformed patterns exhibited by a rod subjected to end moments. This mechanical problem is classical and has a long tradition dating back to Kirchhoff. However, the perspective from geometric phases seems to go more deeply into relations between local strain states and global geometry of shapes, and infuses genuinely new insights and better understanding, which enable one to describe this kind of deformation in a neat and elegant way. On the other hand, visual representations of these deformations provide beautiful illustrations of geometric phases and render the meaning of the abstract concept of holonomy more direct and transparent.

Appendix

From the view of mathematics, the exterior differential forms give unique insight into the geometry of the mechanical problem concerned with here. The concept of differential form, and the mathematical formalism for manipulating them, called exterior calculus , arise when concepts such as the work of a force along a path and the flux of a fluid through a surface are generalized to a curved manifold. Readers who have not been familiar with these mathematical tools should refer to the introductory materials presented in excellent books of Arnol’d [16] and Misner et al. [21].

The key to understanding this result is based upon the integration of the right invariant 1-form $m^{\flat }$ associated with the conserved moment vector $\varvec{m}$. Identifying the tangent bundle $T\mathrm {SO}(3)$ with the right trivialization $\mathrm {SO}(3)\times \mathbb {R}^3$ (physically, using spatial variable), the vector $\varvec{\theta }_{\varvec{\varLambda }} \in T_{\varvec{\varLambda }}\mathrm {SO}(3)$ has representation $\hat{\varvec{\theta }}\varvec{\varLambda }$. Then the 1-form $m^{\flat } |_{\varvec{\varLambda }} \in T^{*}_{\varvec{\varLambda }} \mathrm {SO}(3)$ at the point $\varvec{\varLambda }$ is naturally defined via

$$\begin{aligned} m^{\flat }(\varvec{\theta }_{\varvec{\varLambda }}) = \varvec{m} \cdot \varvec{\theta } \quad \text {for all } \varvec{\theta }_{\varvec{\varLambda }} \in T_{\varvec{\varLambda }}\mathrm {SO}(3). \end{aligned}$$

(A1)

Consider the following two curves lying on $\mathrm {SO}(3)$. The first is the physical curve defined by the rotation field of the rod (Fig. 9)

$$\begin{aligned} \gamma _p{:}\, s \mapsto \varvec{\varLambda }(s) \quad \text {for } s \in [0, S_{\sigma }]. \end{aligned}$$

(A2)

In order to constructed a closed curve, we define an auxiliary curve to connect the two end points of $\gamma _p$

$$\begin{aligned} \gamma _a{:}\, s \mapsto \exp (\alpha \varvec{\mu }) \varvec{\varLambda }(0) \quad \text {for } \alpha \in [0, \alpha ^{*}], \end{aligned}$$

(A3)

which satisfies $\gamma _p(0) = \gamma _a(0)$ and $\gamma _{p}(S_{\sigma }) = \gamma _{a}(\alpha ^*)$. Therefore, $\gamma = \gamma _p - \gamma _a$ (the curve obtained by first going along $\gamma _p$ and then backward along $\gamma _a$) is a closed curve on the $\mathrm {SO}(3)$. The mystery hidden in Eq. (40) can be revealed through the investigation of the integral of the 1-form $m^{\flat }$ along $\gamma$:

$$\begin{aligned} \int _{\gamma } m^{\flat } = \int _{\gamma _p} m^{\flat } - \int _{\gamma _a} m^{\flat }. \end{aligned}$$

(A4)

First, we consider the integral $\int _{\gamma _p} m^{\flat }$ along the physical curve. The tangent of $\gamma _p$ can be expressed as $\varvec{\omega }\varvec{\varLambda }$. Noting that $m^{\flat }(\varvec{\omega }\varvec{\varLambda }) = \varvec{m} \cdot \varvec{\omega } = 2 W$, we obtain

$$\begin{aligned} \int _{\gamma _p} m^{\flat } = \int _{0}^{S_{\sigma }} m^{\flat }(\varvec{\omega }\varvec{\varLambda }) \, \mathrm {d}s = 2 W S_{\sigma }. \end{aligned}$$

(A5)

Next, for the auxiliary curve $\gamma _{a}$, its tangent is simply the form $\varvec{\mu } \exp (\alpha \varvec{\mu })$. Since

$$\begin{aligned} m^{\flat }(\varvec{\mu }\exp (\alpha \varvec{\mu })) = \varvec{m} \cdot \varvec{\mu } = M, \end{aligned}$$

the value of the integral of $m^{\flat }$ along it can be obtained

$$\begin{aligned} \int _{\gamma _a} m^{\flat } = \int _{0}^{\alpha ^*} m^{\flat }(\varvec{\mu }\exp (\alpha \varvec{\mu })) \, \mathrm {d}\alpha = M \alpha ^{*}{.} \end{aligned}$$

(A6)

Finally, with previous results, the integral around the closed curve $\gamma$ takes the form

$$\begin{aligned} \int _{\gamma } m^{\flat } = 2 W S_{\sigma } - M \alpha ^{*}. \end{aligned}$$

(A7)

By applying the Stokes’ theorem, we can relate the integral $\int _{\gamma } m^{\flat }$ to the integral over an arbitrary surface $\sigma \subset \mathrm {SO}(3)$ encircled by $\gamma$, i.e., $\partial \sigma = \gamma$

$$\begin{aligned} \int _{\sigma } {\text {d}}m^{\flat } = \int _{\gamma } m^{\flat }. \end{aligned}$$

(A8)

In order to obtain the relation Eq. (40), we just need to prove that

$$\begin{aligned} \int _{\sigma } {\text {d}}m^{\flat } = M A. \end{aligned}$$

(A9)

We establish the linkage between quantities on the left and right hand side of Eq. (A9). To do so, we first need to compute the 2-form ${\text {d}}m^{\flat }$. This can be done by employing two arbitrary right invariant vector fields $\varvec{\xi }_{\varvec{\varLambda }}$ and $\varvec{\eta }_{\varvec{\varLambda }}$. Bearing in mind that $[\varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }}] = -{{\mathrm{skew}}}(\varvec{\xi } \times \varvec{\eta }) \varvec{\varLambda }$, we obtain

$$\begin{aligned} \begin{aligned} {\text {d}}m^{\flat }( \varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }})&= \varvec{\xi }_{\varvec{\varLambda }}\left( m^{\flat }( \varvec{\eta }_{\varvec{\varLambda }})\right) - \varvec{\eta }_{\varvec{\varLambda }}\left( m^{\flat }( \varvec{\xi }_{\varvec{\varLambda }})\right) - m^{\flat }( [\varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }}] ) \\&= - m^{\flat }( [\varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }}] ) \\&= \varvec{m} \cdot (\varvec{\xi } \times \varvec{\eta }) \ . \end{aligned} \end{aligned}$$

Note that although this calculation is carried out with help of the right invariant vector field, values of ${\text {d}}m^{\flat }$ just linearly depend on vectors at $T_{\varvec{\varLambda }} \mathrm {SO}(3)$. Therefore, ${\text {d}}m^{\flat }$ is determined via the relation

$$\begin{aligned} {\text {d}}m^{\flat }( \varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }}) = \varvec{m} \cdot (\varvec{\xi } \times \varvec{\eta }) \ \end{aligned}$$

(A10)

for all $\varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }} \in T_{\varvec{\varLambda }} \mathrm {SO}(3)$.

Fig. 9

Curves and surfaces in the proof

With an appropriate choice of a surface $\sigma \subset \mathrm {SO}(3)$, the map $\phi _{\varvec{m}}|_{\sigma }$, which is the restriction of $\phi _{\varvec{m}}$ on the surface $\sigma$, is an diffeomorphism to the surface $\varSigma \subset \mathscr {S}^2$ capping the reduced closed curve defined by image of the map $\phi _{\varvec{m}} \circ \gamma$. Let $\psi$ denote the inverse $(\phi _{\varvec{m}}|_{\sigma })^{-1} : \varSigma \mapsto \sigma$, according to the change of variables theorem, we have

$$\begin{aligned} \int _{\sigma } {\text {d}}m^{\flat } = \int _{\Sigma } \psi ^{*} {\text {d}}m^{\flat }. \end{aligned}$$

(A11)

The integral on the surface $\Sigma$ can be calculated by taking the following strategy. We note that every differential 2-form on the sphere can be written in the form $f \varepsilon$, where $\varepsilon$ is the standard area element of $\mathscr {S}^2$, and f is a scalar function. In particular, the pull back of ${\text {d}}m^{\flat }$ can be written into this form

$$\begin{aligned} \psi ^{*} {\text {d}}m^{\flat } = K \varepsilon , \end{aligned}$$

(A12)

where K is a smooth function on the subset $\varSigma$ of $\mathscr {S}^2$.

To determine the function K, let us first take an arbitrary point $\varvec{z} \in \varSigma$ and choose two unit vectors $\varvec{x}, \varvec{y} \in T_{\varvec{z}} \mathscr {S}^2$ tangent to $\mathscr {S}^2$ at the point $\varvec{z}$, which satisfy the relations

$$\begin{aligned} \varvec{x} \cdot \varvec{y} = 0 \quad \text{ and } \quad \varvec{x} \times \varvec{y} = \varvec{n}, \end{aligned}$$

where $\varvec{n}$ is the normal to $\mathscr {S}^2$ at $\varvec{z}$. Here two facts need to be keep in mind: the first is that the definition of standard volume form gives us $\varepsilon (\varvec{x}, \varvec{y}) = 1$; the second is that the matrix $\psi (\varvec{z})$ rotate the normal $\varvec{n}$ to the direction of $\varvec{m}$, i.e., $\varvec{\mu } = \psi (\varvec{z}) \varvec{n}$.

Let us denote $\varvec{\varLambda }$ to be the rotation matrix $\psi (\varvec{z})$, and take the vectors $\varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }} \in T_{\varvec{\varLambda }}\mathrm {SO}(3)$ which are the images of the vectors $\varvec{x}, \varvec{y}$ under the tangent map $T_{\varvec{z}} \psi$

$$\begin{aligned} \varvec{\xi }_{\varvec{\varLambda }} = T_{\varvec{z}} \psi \cdot \varvec{x}, \quad \varvec{\eta }_{\varvec{\varLambda }} = T_{\varvec{z}} \psi \cdot \varvec{y}. \end{aligned}$$

Since the value of $\psi ^{*} {\text {d}}m^{\flat }(\varvec{x}, \varvec{y})$ is defined by the equation

$$\begin{aligned} \psi ^{*} {\text {d}}m^{\flat }(\varvec{x}, \varvec{y}) = {\text {d}}m^{\flat }(\varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }}), \end{aligned}$$

and $\varepsilon (\varvec{x}, \varvec{y}) = 1$, the value K at the point z can be determined by the relation

$$\begin{aligned} K(\varvec{z}) = {\text {d}}m^{\flat }(\varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }}). \end{aligned}$$

By recalling that $\phi _{\varvec{m}}(\varvec{\varLambda }) = \varvec{\varLambda }^{\mathrm {T}}\varvec{m}$, for an arbitrary vector $\varvec{\theta }_{\varvec{\varLambda }} \in T_{\varvec{\varLambda }}\mathrm {SO}(3)$, the transformation of $\varvec{\theta }_{\varvec{\varLambda }}$ under the tangent map of $\phi _{\varvec{m}}$ at $\varvec{\varLambda }$ takes a simple form

$$\begin{aligned} T_{\varvec{\varLambda }}\phi _{\varvec{m}} \cdot \varvec{\theta }_{\varvec{\varLambda }} = \varvec{\varLambda }^{\mathrm {T}}(\varvec{m} \times \varvec{\theta }). \end{aligned}$$

(A13)

Therefore, we have

$$\begin{aligned} \varvec{m} \times \varvec{\xi } = \varvec{\varLambda } \varvec{x} \quad \text{ and } \quad \varvec{m} \times \varvec{\eta } = \varvec{\varLambda } \varvec{y}. \end{aligned}$$

By taking the cross product of $\varvec{m} \times \varvec{\xi }$ and $\varvec{m} \times \varvec{\eta }$, we find

$$\begin{aligned} (\varvec{m} \times \varvec{\xi }) \times (\varvec{m} \times \varvec{\eta }) = \varvec{\varLambda } (\varvec{x} \times \varvec{y}) = \varvec{\varLambda } \varvec{n} = \varvec{\mu }. \end{aligned}$$

However, with the help of the identity of the cross product and Eq. (A10) concerning ${\text {d}}m^{\flat }$, the left hand side can also be expressed in the form

$$\begin{aligned} (\varvec{m} \times \varvec{\xi }) \times (\varvec{m} \times \varvec{\eta }) = (\varvec{m} \cdot (\varvec{\xi } \times \varvec{\eta })) \varvec{m} = {\text {d}}m^{\flat }(\varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }}) \varvec{m} \ . \end{aligned}$$

Comparison of $\varvec{\mu }$ and ${\text {d}}m^{\flat }(\varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }}) \varvec{m}$ leads to

$$\begin{aligned} K(\varvec{z}) = {\text {d}}m^{\flat }(\varvec{\xi }_{\varvec{\varLambda }}, \varvec{\eta }_{\varvec{\varLambda }}) = 1/M \ . \end{aligned}$$

(A14)

Since the choice of $\varvec{z}$ is arbitrary, this result holds for all poinst $\varvec{z} \in \varSigma$. Hence, K is a constant function, and the 2-form $\psi ^{*} {\text {d}}m^{\flat }$ can be expressed in the form

$$\begin{aligned} \psi ^{*} {\text {d}}m^{\flat } = \frac{1}{M} \varepsilon \ . \end{aligned}$$

(A15)

With this result, the integral $\int _{\sigma } {\text {d}}m^{\flat }$ can be carried out in the following manner

$$\begin{aligned} \int _{\sigma } {\text {d}}m^{\flat } = \int _{\Sigma } \psi ^{*} {\text {d}}m^{\flat } = \frac{1}{M} \int _{\Sigma } \varepsilon = \frac{1}{M} {\text {area}}(\Sigma ) \ . \end{aligned}$$

(A16)

Remembering that the solid angle A can be expressed as $A = {\text {area}}(\Sigma )/M^2$, we arrive at the equation

$$\begin{aligned} \int _{\sigma } {\text {d}}m^{\flat } = MA \ . \end{aligned}$$

(A17)

Therefore, the demonstration of phase formula (40) is completed.

In summary, the phase formula (40) is established through a direct application of Stokes’ theorem on the integral of differential 1-form $m^{\flat }$ along the path $\gamma$ defined at the beginning of this appendix (see the Eq. (A8)). The surface integral is directly related to the global geometric property of the distribution of the stress field $\varvec{M}$ along the rod. After careful analysis, we find it could be expressed as the product of the magnitude of the moment and the solid angle enclosed by the closed orbit described by the stress vector $\varvec{M}$.