# Minimal distance transformations between links and polymers: Principles and
examples

Ali R. Mohazab† and Steven S. Plotkin†***e-mail<EMAIL_ADDRESS>†
Department of Physics and Astronomy, University of British Columbia, 6224
Agricultural Road, Vancouver, BC V6T1Z1, Canada

###### Abstract

The calculation of Euclidean distance between points is generalized to one-
dimensional objects such as strings or polymers. Necessary and sufficient
conditions for the minimal transformation between two polymer configurations
are derived. Transformations consist of piecewise rotations and translations
subject to Weierstrass-Erdmann corner conditions. Numerous examples are given
for the special cases of one and two links. The transition to a large number
of links is investigated, where the distance converges to the polymer length
times the mean root square distance (MRSD) between polymer configurations,
assuming curvature and non-crossing constraints can be neglected. Applications
of this metric to protein folding are investigated. Potential applications are
also discussed for structural alignment problems such as pharmacophore
identification, and inverse kinematic problems in motor learning and control.

###### pacs:

02.30.Xx, 45.10.Db, 45.40.-f , 45.40.Ln, 46.25.Cc , 64.70.Nd , 64.70.km ,
82.39.Rt , 87.10.-e , 87.10.Ed , 87.15.A- , 87.15.Cc , 87.15.hp

††: J. Phys.: Condens. Matter

##  1 Introduction

The standard variational definition of distance can be generalized to higher
dimensional objects such as strings or membranes. In a previous paper [1], one
of us has introduced the formalism for this calculation. Consider first zero-
dimensional objects (points). The distance between two points $A$ and $B$ is
defined through a transformation that takes $A$ to $B$, an object of dimension
one higher than the points themselves (here one-dimensional). The
transformation minimizing the arc-length travelled between $A$ and $B$ gives
the scalar distance $\mathcal{D}^{\ast}$. The differential increment of arc-
length may be defined as either $\sqrt{1+(dy/dx)^{2}+(dz/dx)^{2}}dx$, or
without the assumption that $y,z$ are functions of $x$, parametrically. To be
specific, introduce a “time” parameter $t$ such that $0\leq t\leq T$, and
${\bf r}(0)={\bf r}_{\mbox{\tiny{A}}}$, ${\bf r}(T)={\bf
r}_{\mbox{\tiny{B}}}$, and ${\bf r}(t)=(x(t),y(t),z(t))$. The distance between
${\bf r}_{\mbox{\tiny{A}}}$ and ${\bf r}_{\mbox{\tiny{B}}}$ can be found
variationally [2]:

$\displaystyle\mathcal{D}^{\ast}$ $\displaystyle=$
$\displaystyle\mathcal{D}\left[{\bf r}^{\ast}(t)\right]\mbox{where ${\bf
r}^{\ast}(t)$ satisfies}$ (1b)
$\displaystyle\delta\int_{0}^{T}\\!\\!\\!dt\>\left(g_{\mu\nu}\dot{x}^{\mu}(t)\dot{x}^{\nu}(t)\right)^{1/2}=0\>.$
or $\displaystyle\delta\int_{0}^{T}\\!\\!\\!dt\>\sqrt{\dot{{\bf
r}}^{2}}=0\;\;\;\;\mbox{(Euclidean metric)}$ (1c)

Here we have let $\dot{x}\equiv dx/dt$, and $\dot{{\bf r}}\equiv d{\bf r}/dt$.
The boundary conditions on the extremal path are ${\bf r}^{\ast}(0)={\bf
r}_{\mbox{\tiny{A}}}$ and ${\bf r}^{\ast}(T)={\bf r}_{\mbox{\tiny{B}}}$.

Taking the functional derivative in eq. (1c) gives Euler-Lagrange (EL)
equations for the Lagrangian $\mathcal{L}=\sqrt{\dot{{\bf r}}^{2}}$:

$\displaystyle\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{{\bf
r}}}\right)$ $\displaystyle=$ $\displaystyle 0$
$\displaystyle\mbox{or}\;\;\;\;\dot{\hat{{\bf v}}}$ $\displaystyle=$
$\displaystyle 0$ (2)

with $\hat{{\bf v}}$ the unit vector in the direction of the velocity.

Since the derivative of a unit vector is always orthogonal to that vector,
equation (2) says that the direction of the velocity cannot change, and
therefore straight line motion results. Applying the boundary conditions gives
$\hat{{\bf v}}=({\bf r}_{\mbox{\tiny{B}}}-{\bf
r}_{\mbox{\tiny{A}}})/\left|{\bf r}_{\mbox{\tiny{B}}}-{\bf
r}_{\mbox{\tiny{A}}}\right|$. However, any function ${\bf
v}(t)=\left|v_{o}(t)\right|\hat{{\bf v}}$ satisfying the boundary conditions
is a solution, so long as
$\int_{0}^{T}\\!\\!dt\>\left|v_{o}(t)\right|=\left|{\bf
r}_{\mbox{\tiny{B}}}-{\bf r}_{\mbox{\tiny{A}}}\right|$. The solution is
reparameterization-invariant. Then the extremal functional ${\bf r}^{\ast}(t)$
is given by

${\bf r}^{\ast}(t)={\bf r}_{\mbox{\tiny{A}}}+\frac{{\bf
r}_{\mbox{\tiny{B}}}-{\bf r}_{\mbox{\tiny{A}}}}{\left|{\bf
r}_{\mbox{\tiny{B}}}-{\bf
r}_{\mbox{\tiny{A}}}\right|}\int_{0}^{t}\\!\\!dt\>\left|v_{o}(t)\right|$ (3)

and the distance by

${\cal D}^{\ast}=\int_{0}^{T}\\!\\!dt\>\sqrt{\dot{{\bf
r}}^{\ast^{2}}}=\int_{0}^{T}\\!\\!dt\>\left|v_{o}(t)\right|=\left|{\bf
r}_{\mbox{\tiny{B}}}-{\bf r}_{\mbox{\tiny{A}}}\right|$ (4)

which represents the diagonal of a hypercube, as expected. At this point we
could fix the parameterization by choosing $\left|v_{o}(t)\right|=\left|{\bf
r}_{\mbox{\tiny{B}}}-{\bf r}_{\mbox{\tiny{A}}}\right|/T$ (constant speed), for
example.

The extremal transformation (3) is also a minimum. In section 2.4 we will give
the sufficient conditions for an extremum to be a (local) minimum, where we
will return to this example.

The above idea can be generalized to space curves, surfaces, or higher
dimensional manifolds [1]. The distance is defined through the transformation
between the objects that minimizes the cumulative amount of arc-length
travelled by all parts of the manifold.

##  2 Distance for polymers or strings

Describing the transformation ${\bf r}(s,t)$ between two space curves ${\bf
r}_{\mbox{\tiny A}}(s)$ and ${\bf r}_{\mbox{\tiny{B}}}(s)$ requires two scalar
parameters: $s$ the arc-length along the space curve, and $t$ the “time” as in
the above zero-dimensional case measuring progress during the transformation.
The boundary conditions are then ${\bf r}(s,0)={\bf r}_{\mbox{\tiny A}}(s)$
and ${\bf r}(s,T)={\bf r}_{\mbox{\tiny{B}}}(s)$. The minimal transformation
${\bf r}^{\ast}(s,t)$ is an object of dimension one higher than $A$ or $B$,
i.e. it yields a distance that is two-dimensional.

The distance $\mathcal{D}^{\ast}=\mathcal{D}[{\bf r}^{\ast}(s,t)]$, where the
functional $\mathcal{D}[{\bf r}]$ is given by

$\mathcal{D}[{\bf
r}]=\int_{0}^{L}\\!\\!\\!ds\\!\\!\int_{0}^{T}\\!\\!\\!dt\>\sqrt{\dot{{\bf
r}}^{2}}\>.$ (5)

Here we have used the shorthand ${\bf r}\equiv{\bf
r}(s,t)=(x(s,t),y(s,t),z(s,t))$ (a 3-vector), and $\dot{{\bf
r}}\equiv\partial{\bf r}/\partial t$.

It has been shown previously that the problem of distance does not map to a
simple soap film, nor to the minimal area of a world-sheet (which corresponds
to the action of a classical relativistic string) [1].

Formulated as above, the string can contract and expand arbitrarily in order
to minimize the distance travelled. The transforming object is akin to a
rubber band, and all points on ${\bf r}_{\mbox{\tiny A}}(s)$ will move in
straight lines to their partner points on ${\bf r}_{\mbox{\tiny{B}}}(s)$ to
minimize the distance. It is worth mentioning that protein chains for example
only change their length by about one percent at biological temperatures. To
accurately represent the transformation of a non-extensible string, a Lagrange
multiplier $\lambda(s,t)$ must be introduced into the effective Lagrangian,
weighting the constraint:

$\sqrt{{\bf r}^{\prime 2}}=1\>,$ (6)

where ${\bf r}^{\prime}\equiv\partial{\bf r}/\partial s$.

Under this constraint, points along the string can no longer move
independently of each other, but must always be a fixed (infinitesimal)
distance apart. The tangent vector $\hat{{\bf t}}={\bf r}^{\prime}$ is now a
unit vector, and the total length of the string is
$L=\int_{0}^{L}\\!\\!ds\>\sqrt{{\bf r}^{\prime 2}}=\int_{0}^{L}\\!\\!ds$.

Consider the minimal distance transformation between two configurations ${\bf
r}_{\mbox{\tiny{A}}}(s)$ and ${\bf r}_{\mbox{\tiny{B}}}(s)$ of an ideal
polymer of length $L$. Let us derive the EL equations for this case. From
equations (5) and (6), the effective action is

$\displaystyle{\cal D}$ $\displaystyle=$
$\displaystyle\int_{0}^{L}\\!\\!\int_{0}^{T}\\!\\!\\!ds\,dt\>\mathcal{L}\left(\dot{{\bf
r}},{\bf r}^{\prime}\right)$ (7a)
$\displaystyle\mbox{where}\;\;\;\;\mathcal{L}$ $\displaystyle=$
$\displaystyle\sqrt{\dot{{\bf r}}^{2}}-\lambda\left(\sqrt{{\bf r}^{\prime
2}}-1\right)$ (7b)

and the Lagrange multiplier $\lambda\equiv\lambda(s,t)$ is a function of both
$s$ and $t$. The extrema of the distance functional ${\cal D}$ in (7a) are
found from $\delta{\cal D}=0$. Taking the functional derivative gives EL
equations [1]:

$\dot{\hat{{\bf v}}}=\lambda\boldsymbol{\kappa}+\lambda^{\prime}\hat{{\bf
t}}\>.$ (8)

where $\hat{{\bf v}}$ is the unit velocity vector, $\hat{{\bf t}}$ is the unit
tangent vector, and $\boldsymbol{\kappa}$ is the curvature vector. In eq. (8)
we see explicitly that if the non-extensibility constraint is set to zero
($\lambda=0$), all points on ${\bf r}_{\mbox{\tiny{A}}}(s)$ move in straight
lines to ${\bf r}_{\mbox{\tiny{B}}}(s)$.

###  2.1 Discrete chains

To make the problem more amenable to solution, we can discretize the spatial
variables while letting the time variable remain continuous, i.e. we implement
the method of lines to solve eq. (8). Rather than directly discretizing eq.
(8) however, it is more natural to consider a discretized chain as shown in
figure 1 from the outset, and to calculate the EL equations for this system.
This recipe then gives the same result as properly discretizing eq. (8). For
the discretized chain, the constraint in eq. (6) becomes $|\Delta{\bf
r}|=\Delta s=L/(N-1)$, giving the length of each link. As the number of beads
$N\rightarrow\infty$ the system approaches a continuous chain. For finite $N$,
the Lagrangian becomes a function of the positions and velocities
$\left\\{{\bf r}_{i},\dot{{\bf r}}_{i}\right\\}$ of all beads $i$, $1\leq
i\leq N+1$. We use the shorthand notation $\mathcal{L}({\bf r}_{i},\dot{{\bf
r}}_{i})$.

|
---|---
a | b
Figure 1: Continuum (a) and discretized (b) polymer chain. The EL equation for
the continuum polymer is a nonlinear (vector) PDE, while the EL equations for
the discretized polymer are a set of nonlinear ODEs.

This recipe yields the distance metric for an ideal, freely-jointed chain,
which has no non-local interactions and no curvature constraints. While this
approximation is often used as a first step, real chains may behave quite
differently for several reasons. In many cases, the configuration which is an
energetic minimum is a straight line, or a single conformation dictated by the
chemistry of the polymeric bonds. At finite temperature, energy in the bath
induces conformational fluctuations. Real polymers also cannot cross
themselves, and because of their stereochemistry also take up volume. We leave
these interesting features for later analysis.

Equation (6) for the discretized chain becomes $N$ constraint equations added
to the effective Lagrangian:

$\sum_{i=1}^{N}\hat{\lambda}_{i,i+1}\left(\sqrt{({\bf r}_{i+1}-{\bf
r}_{i})^{2}}-\Delta s\right)$

where each $\hat{\lambda}_{i,i+1}\equiv\hat{\lambda}_{i,i+1}(t)$ is a function
of $t$, and $\hat{\lambda}_{N,N+1}=0$. Letting $\lambda\equiv
2\hat{\lambda}\,\Delta s$ and ${\bf r}_{i+1/i}\equiv{\bf r}_{i+1}-{\bf r}_{i}$
we rewrite this strictly for convenience as

$\sum\frac{\lambda_{i,i+1}}{2}\>\left({{\bf r}_{i+1/i}^{2}\over\Delta
s^{2}}-1\right)\>.$

We next convert to dimensionless variables by letting ${\bf r}=(\Delta
s)\hat{{\bf r}}$. To simplify the notation, from here on we simply refer to
$\hat{{\bf r}}$ as ${\bf r}$. The distance for the discretized chain becomes

${\cal D}[{\bf r}_{i},\dot{{\bf r}}_{i}]=\Delta
s^{2}\int_{0}^{T}\\!\\!\\!dt\>\mathcal{L}\left({\bf r}_{i},\dot{{\bf
r}}_{i}\right)$ (9)

with effective Lagrangian

$\mathcal{L}\left({\bf r}_{i},\dot{{\bf
r}}_{i}\right)=\sum_{i=1}^{N}\left(\sqrt{\dot{{\bf
r}}_{i}^{2}}-{\lambda_{i,i+1}\over 2}\left({\bf
r}_{i+1/i}^{2}-1\right)\right)\>.$ (10)

The derivatives $\dot{{\bf r}}$ and ${\bf r}_{i+1/i}$ are raised to different
powers in (10), however so long as ${\bf r}_{i+1/i}$ satisfies the constraint
$\left|{\bf r}_{i+1/i}\right|=1$, the EL equations for ${\bf r}_{i}(t)$ will
be the same whether the constraint $\sqrt{{\bf r}_{i+1/i}^{2}}=1$ or ${\bf
r}_{i+1/i}^{2}=1$ is used.

The reparameterization invariance present for point particles (c.f. section 1)
is still present for beads on the chain, but the parameterization of arclength
along the chain is taken to be fixed by the discretization.

###  2.2 General variation of the distance functional

For reasons that will become clear as we progress, we consider the general
variation of the functional ${\cal D}$, allowing for broken extremals. That
is, we allow the curves describing the particle trajectories to be non-smooth
in principle at one or more points in time. Consider the case of one such
point at time $t_{1}$. The distance can be written as

${\cal D}=\int_{0}^{t_{1}}\\!\\!\\!dt\>\mathcal{L}({\bf r}_{i},\dot{{\bf
r}}_{i})+\int_{t_{1}}^{T}\\!\\!\\!dt\>\mathcal{L}({\bf r}_{i},\dot{{\bf
r}}_{i})$ (11)

The space-trajectories of the particles must be continuous at time $t_{1}$, so
${\bf r}_{i}(t_{1}-\epsilon)={\bf r}_{i}(t_{1}+\epsilon)$, or in shorthand:

${\bf r}_{i}\left(t_{1}^{-}\right)={\bf r}_{i}\left(t_{1}^{+}\right)\>.$ (12)

Let ${\bf r}_{i}(t)$ and $\tilde{{\bf r}}_{i}(t)$ be two neighboring
trajectories from ${\bf r}_{i}(0)={\bf r}_{\mbox{\tiny{A}}i}$ to ${\bf
r}_{i}(T)={\bf r}_{\mbox{\tiny{B}}i}$ (see figure 2). Neighboring curves will
differ by the first order quantity ${\bf h}_{i}(t)=\tilde{{\bf r}}_{i}(t)-{\bf
r}_{i}(t)$. The fixed boundary conditions at $t=0,T$ dictate that ${\bf
h}_{i}(0)={\bf h}_{i}(T)=0$. The difference in distance between the two
trajectories is

$\displaystyle\Delta{\cal D}$ $\displaystyle=$ $\displaystyle{\cal D}[{\bf
r}_{i}+{\bf h}_{i}]-{\cal D}[{\bf r}_{i}]$ (13) $\displaystyle=$
$\displaystyle\int_{0}^{t_{1}+\delta t_{1}}\\!\\!\\!dt\>\mathcal{L}({\bf
r}_{i}+{\bf h}_{i},\dot{{\bf r}}_{i}+\dot{{\bf
h}}_{i})-\int_{0}^{t_{1}}\\!\\!\\!dt\>\mathcal{L}({\bf r}_{i},\dot{{\bf
r}}_{i})+\int_{t_{1}+\delta t_{1}}^{T}\\!\\!\\!dt\>\mathcal{L}({\bf
r}_{i}+{\bf h}_{i},\dot{{\bf r}}_{i}+\dot{{\bf
h}}_{i})-\int_{t_{1}}^{T}\\!\\!\\!dt\>\mathcal{L}({\bf r}_{i},\dot{{\bf
r}}_{i})$

Figure 2: General variations of a functional with fixed end points allow for
broken extremals. In the text we derive the extra “corner” conditions for a
piecewise continuous path to still be extremal for our distance functional.

Taylor expanding the Lagrangian to first order in ${\bf
h}_{i}$:‡**footnotetext: ‡ We use the notation $F_{{\bf r}}\equiv\partial
F/\partial{\bf r}$, $F_{\dot{{\bf r}}}\equiv\partial F/\partial\dot{{\bf r}}$.

$\mathcal{L}\approx\mathcal{L}({\bf r}_{i},\dot{{\bf
r}}_{i})+\sum_{i=1}^{N}\left(\mathcal{L}_{{\bf r}_{i}}\cdot{\bf
h}_{i}+\mathcal{L}_{\dot{{\bf r}}_{i}}\cdot\dot{{\bf h}}_{i}\right)$

and integrating by parts using the fixed boundary conditions at $t=0,T$, the
difference in distance up to first order in ${\bf h}_{i}$ is

$\displaystyle\Delta{\cal D}$ $\displaystyle\approx$
$\displaystyle\int_{0}^{t_{1}}\\!\\!\\!dt\>\sum_{i}\left(\mathcal{L}_{{\bf
r}_{i}}-{d\over dt}\mathcal{L}_{\dot{{\bf r}}_{i}}\right)\cdot{\bf
h}_{i}+\int_{t_{1}}^{T}\\!\\!\\!dt\>\sum_{i}\left(\mathcal{L}_{{\bf
r}_{i}}-{d\over dt}\mathcal{L}_{\dot{{\bf r}}_{i}}\right)\cdot{\bf h}_{i}$
(14) $\displaystyle+$ $\displaystyle\mathcal{L}(t_{1}^{-})\delta
t_{1}-\mathcal{L}(t_{1}^{+})\delta t_{1}+\sum_{i}\left.\mathcal{L}_{\dot{{\bf
r}}_{i}}\cdot{\bf
h}_{i}\right|_{t_{1}^{-}}-\sum_{i}\left.\mathcal{L}_{\dot{{\bf
r}}_{i}}\cdot{\bf h}_{i}\right|_{t_{1}^{+}}$

with the shorthand $\mathcal{L}(t)\equiv\mathcal{L}({\bf r}_{i}(t),\dot{{\bf
r}}_{i}(t))$.

###  2.3 Conditions for an extremum

The variation $\delta{\cal D}$ differs from $\Delta{\cal D}$ above only by
second order terms. Then for the transformation from $\\{{\bf
r}_{\mbox{\tiny{A}}i}\\}$ to $\\{{\bf r}_{\mbox{\tiny{B}}i}\\}$ to be an
extremum, $\delta{\cal D}=0$. Thus, the EL equations (in the top line of eq.
(14)) must vanish in each regime $[0,t_{1})$, $(t_{1},T]$. Using the form of
the Lagrangian in eq. (10), the EL equations become:

$\displaystyle\dot{\hat{{\bf v}}}_{1}+\lambda_{12}\,{\bf r}_{2/1}=0$ (15a)
$\displaystyle\dot{\hat{{\bf v}}}_{2}-\lambda_{12}\,{\bf
r}_{2/1}+\lambda_{23}\,{\bf r}_{3/2}=0$ (15c) $\displaystyle\hskip
28.45274pt\vdots$ $\displaystyle\dot{\hat{{\bf v}}}_{N}-\lambda_{N-1,N}\,{\bf
r}_{N/(N-1)}=0$

According to equation (14) there are additional conditions for the
transformation to be an extremum. To find these first note that up to first
order (see figure 2)

$\displaystyle{\bf h}_{i}(t_{1})\approx\delta{\bf r}_{i}(t_{1})-\dot{{\bf
r}}_{i}(t_{1})\,\delta t_{1}\>.$ (16)

Then the first variation in the distance is

$\displaystyle\delta{\cal D}$ $\displaystyle=$
$\displaystyle\left[\left.\left(\mathcal{L}-\sum_{i}\dot{{\bf
r}}_{i}\cdot\mathcal{L}_{\dot{{\bf
r}}_{i}}\right)\right|_{t_{1}^{-}}-\left.\left(\mathcal{L}-\sum_{i}\dot{{\bf
r}}_{i}\cdot\mathcal{L}_{\dot{{\bf
r}}_{i}}\right)\right|_{t_{1}^{+}}\right]\delta t_{1}$ (17) $\displaystyle+$
$\displaystyle\sum_{i}\left[\left.\mathcal{L}_{\dot{{\bf
r}}_{i}}\right|_{t_{1}^{-}}-\left.\mathcal{L}_{\dot{{\bf
r}}_{i}}\right|_{t_{1}^{+}}\right]\cdot\delta{\bf r}_{i}(t_{1})$

which must vanish at an extremum. Because the variations $\delta{\bf r}_{i}$
and $\delta t_{1}$ are all independent, the terms in square brackets in
equation (17) must vanish. Writing these expressions in terms of the conjugate
momenta ${\bf p}_{i}=\mathcal{L}_{\dot{{\bf r}}_{i}}$ and Hamiltonian
$\mathcal{H}=\sum_{i}\dot{{\bf r}}_{i}\cdot{\bf p}_{i}-\mathcal{L}$ gives the
conditions:

$\displaystyle\left.{\bf p}_{i}\right|_{{}_{t_{1}^{-}}}=\left.{\bf
p}_{i}\right|_{{}_{t_{1}^{+}}}$ (18a)
$\displaystyle\left.\mathcal{H}\right|_{{}_{t_{1}^{-}}}=\left.\mathcal{H}\right|_{{}_{t_{1}^{+}}}$
(18b)

These conditions are called the Weierstrass-Erdmann conditions or corner
conditions in the calculus of variations [2].

According to the Lagrangian in equation (10), the Hamiltonian is given by

$\mathcal{H}=-\sum_{i=1}^{N}{\lambda_{i,i+1}\over 2}\left({\bf
r}_{i+1/i}^{2}-1\right)$

which is identically zero, so corner condition (18b) provides no further
information.

The conjugate momenta according to (10) are given by

${\bf p}_{i}={\dot{{\bf r}}_{i}\over\left|\dot{{\bf r}}_{i}\right|}=\hat{{\bf
v}}_{i}\>.$ (19)

Therefore, according to corner condition (18a), extremal trajectories cannot
suddenly change direction: each ${\bf r}_{i}(t)$ follows a smooth path
continuous up to first derivatives in the spatial coordinates.

The fact that one corner condition provided no information due to the
vanishing of the Hamiltonian is related to our choice of parameterization in
formulating the problem. For example, in the case of the distance of the
single point particle mentioned in the introduction, the Lagrangian may be
defined either through independent variable $x$ as
$\mathcal{L}^{(x)}=\sqrt{1+y^{\prime 2}+z^{\prime 2}}$ (with e.g.
$y^{\prime}=dy/dx$), or parametrically through independent variable $t$ as
$\mathcal{L}^{(t)}=\sqrt{\dot{{\bf r}}^{2}}$. The conjugate momenta are then
either $\mathcal{L}^{(x)}_{y^{\prime}}=y^{\prime}/\sqrt{1+y^{\prime
2}+z^{\prime 2}}$ and
$\mathcal{L}^{(x)}_{z^{\prime}}=z^{\prime}/\sqrt{1+y^{\prime 2}+z^{\prime
2}}$, or $\mathcal{L}^{(t)}_{\dot{{\bf r}}}=\dot{{\bf r}}/|\dot{{\bf
r}}|\equiv\hat{{\bf v}}$. The Hamiltonia are either
$\mathcal{H}^{(x)}=1/\sqrt{1+y^{\prime 2}+z^{\prime 2}}$ or
$\mathcal{H}^{(t)}=\mathcal{L}^{(t)}-\dot{{\bf r}}\cdot(\dot{{\bf
r}}/|\dot{{\bf r}}|)=0$. The corner conditions can be shown to be equivalent
for both choices of independent variable: for $\mathcal{L}^{(t)}$ they give
$\hat{{\bf v}}(t_{1}^{-})=\hat{{\bf v}}(t_{1}^{+})$, so that the direction of
the tangent to the curve cannot have a discontinuity. Together, the
Hamiltonian and two conjugate momenta for $\mathcal{L}^{(x)}$ can be
interpreted as components of the unit tangent vector to the curve, i.e.
$\hat{{\bf t}}(x)=(\hat{{\bf i}}+y^{\prime}\hat{{\bf j}}+z^{\prime}\hat{{\bf
k}})/\sqrt{1+y^{\prime 2}+z^{\prime 2}}$, and so once again the corner
conditions enforce a continuous tangent vector, here $\hat{{\bf
t}}(x_{1}^{-})=\hat{{\bf t}}(x_{1}^{+})$.

####  2.3.1 Boundary conditions

In the continuum limit, the boundary conditions on ${\bf r}(s,t)$ are ${\bf
r}(s,0)={\bf r}_{\mbox{\tiny{A}}}(s)$, ${\bf r}(s,T)={\bf
r}_{\mbox{\tiny{B}}}(s)$ where ${\bf r}_{\mbox{\tiny{A}}}$ and ${\bf
r}_{\mbox{\tiny{B}}}$ are the two configurations of the polymer. For discrete
chains, these boundary conditions become

$\displaystyle\left\\{{\bf r}_{i}(0)\right\\}$ $\displaystyle=$
$\displaystyle\\{{\bf r}_{i}^{\left(A\right)}\\}$ (20a)
$\displaystyle\left\\{{\bf r}_{i}(T)\right\\}$ $\displaystyle=$
$\displaystyle\\{{\bf r}_{i}^{\left(B\right)}\\}\>.$ (20b)

There are also boundary conditions that hold for the end points of the chain
at all times. From equations (15a, 15c) we see that there are three solutions
for the end points of the chain:

1) If $\lambda\neq 0$, purely rotational motion results. This can be seen by
taking the dot product of eq. (15a) with ${\bf v}_{1}$, which yields
$\lambda_{12}{\bf v}_{1}\cdot{\bf r}_{2/1}=0$, so the velocity of the end
point is orthogonal to the link. The rotation must be about a point that is
internal to the link, i.e. on the line between points $1$ and $2$ for end
point $1$. This can be seen straightforwardly for the case of one link by
removing point $3$ from equations (15a) and (15c). Then the accelerations
$\dot{\hat{{\bf v}}}_{i}$ must be in opposite directions. This can only occur
if rotation is about a point on the line between points $1$ and $2$.

2) If $\lambda=0$, $\dot{\hat{{\bf v}}}_{i}=0$, and straight-line motion of
the end point results.

3) Writing out the time-derivative in (15a) yields

${\bf v}_{1}^{2}\dot{{\bf v}}_{1}-\left({\bf v}_{1}\cdot\dot{{\bf
v}}_{1}\right){\bf v}_{1}=-\lambda_{12}\left|{\bf v}_{1}\right|^{3}{\bf
r}_{2/1}$ (21)

which has the trivial solution ${\bf v}_{1}=0$. The end point can be at rest,
while other parts of the chain move.

###  2.4 Sufficient conditions for a minimum

For a transformation to be minimal, it is necessary, but not sufficient, that
it be an extremum. We now derive the sufficient conditions for a given
transformation to minimize the functional (9). We describe the formalism in
some detail because it is not typically taught to physicists- for further
reading see for example reference [2]. This section can be read independently
of the others, and might be skipped on first reading.

According to Sylvester’s criterion, a quadratic form
$\sum_{ij}A_{ij}x_{i}x_{j}$ is positive definite if and only if all descending
principle minors of the matrix $\|A_{ij}\|$ are positive, i.e.

$A_{11}>0\>,\hskip 14.22636pt\begin{vmatrix}A_{11}&A_{12}\\\
A_{21}&A_{22}\end{vmatrix}>0\>,\quad\begin{vmatrix}A_{11}&A_{12}&A_{13}\\\
A_{21}&A_{22}&A_{23}\\\
A_{31}&A_{32}&A_{33}\end{vmatrix}>0\>,\quad\ldots\quad,\mbox{det}\|A_{ij}\|>0\>,$
(22)

and a function $F$ of ${\bf x}\equiv(x_{1},x_{2},\ldots,x_{n})$ has a minimum
at ${\bf x}^{\star}$ if the Jacobian matrix $\|\partial^{2}F/\partial
x_{i}\partial x_{j}\|$ is positive definite at the position of the extremum
(where $\partial F/\partial x_{i}=0$).

For a function to be a minimum of a given functional, it must satisfy similar
sufficient conditions. Consider again the difference in distance between two
trajectories in (9)‡**footnotetext: ‡ We ignore corner conditions for purposes
of the derivation. It can be shown that they do not modify the result.. Taylor
expanding the Lagrangian to second order in ${\bf h}_{i}$:

$\displaystyle\Delta{\cal D}$ $\displaystyle=$ $\displaystyle{\cal
D}\left[{\bf r}_{i}+{\bf h}_{i}\right]-{\cal D}\left[{\bf r}_{i}\right]$ (23)
$\displaystyle=$ $\displaystyle\int_{0}^{T}\\!\\!\\!dt\>\mathcal{L}({\bf
r}_{i}+{\bf h}_{i},\dot{{\bf r}}_{i}+\dot{{\bf
h}}_{i})-\int_{0}^{T}\\!\\!\\!dt\>\mathcal{L}({\bf r}_{i},\dot{{\bf r}}_{i})$
$\displaystyle\approx$
$\displaystyle\int_{0}^{T}\\!\\!\\!dt\>\left[\sum_{i=1}^{N}\left(\mathcal{L}_{{\bf
r}_{i}}\cdot{\bf h}_{i}+\mathcal{L}_{\dot{{\bf r}}_{i}}\cdot\dot{{\bf
h}}_{i}\right)+{1\over
2}\sum_{i,j}^{3N}\left(\mathcal{L}_{x_{i}x_{j}}h_{i}h_{j}+2\mathcal{L}_{x_{i}\dot{x}_{j}}h_{i}\dot{h}_{j}+\mathcal{L}_{\dot{x}_{i}\dot{x}_{j}}\dot{h}_{i}\dot{h}_{j}\right)\right]$

At an extremum, the first order term in (23) is zero, and $\Delta{\cal
D}\approx\delta^{2}{\cal D}$, the second variation. For the extremum to be a
minimum, $\delta^{2}{\cal D}>0$. From eq. (10), the matrix
$\|\mathcal{L}_{x_{i}\dot{x}_{j}}\|=\|0\|$. Assuming
$\|\mathcal{L}_{x_{i}\dot{x}_{j}}\|$ is in general a symmetric matrix, i.e.
$\mathcal{L}_{x_{i}\dot{x}_{j}}=\mathcal{L}_{x_{j}\dot{x}_{i}}$, the second
term in the quadratic form of (23) may be integrated by parts to give:

$\delta^{2}{\cal D}={1\over
2}\int_{0}^{T}\\!\\!\\!dt\>\left[\langle\dot{h}|\mathrm{P}\dot{h}\rangle+\langle
h|\mathrm{Q}h\rangle\right]\>,$ (24)

where we have let $|h\rangle$ denote the vector $(h_{1},h_{2},\ldots,h_{3N})$,
and used the shorthand $\mathrm{P}$ and $\mathrm{Q}$ for the matrices:

$\displaystyle\mathrm{P}(t)=\|\mathrm{P}_{ij}\|=\|\mathcal{L}_{\dot{x}_{i}\dot{x}_{j}}\|$
$\displaystyle\mathrm{Q}(t)=\|\mathrm{Q}_{ij}\|=\left(\|\mathcal{L}_{x_{i}x_{j}}\|-{d\over
dx}\|\mathcal{L}_{x_{i}\dot{x}_{j}}\|\right)\>.$ (25)

From (10) the explicit form for these matrices may be calculated. $\mathrm{P}$
is block diagonal:

$\mathrm{P}=\begin{bmatrix}I_{ij}^{\left(1\right)}&0&\cdots&0\\\
0&I_{ij}^{\left(2\right)}&\cdots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
0&0&\cdots&I_{ij}^{\left(N\right)}\\\ \end{bmatrix}$ (26)

with each block matrix having elements

$\|\mathrm{I}_{ij}^{\left(J\right)}\|={1\over{\left|\dot{{\bf
r}}^{\,J}\right|}^{3}}\left(\delta_{ij}{\left.\dot{{\bf
r}}^{\left(J\right)}\right.}^{2}-\dot{x}^{\,\left(J\right)}_{i}\dot{x}^{\,\left(J\right)}_{j}\right)={1\over{\left|\dot{{\bf
r}}^{\,J}\right|}^{3}}\begin{bmatrix}\dot{y}^{2}+\dot{z}^{2}&-\dot{x}\dot{y}&-\dot{x}\dot{z}\\\
-\dot{x}\dot{y}&\dot{x}^{2}+\dot{z}^{2}&-\dot{y}\dot{z}\\\
-\dot{x}\dot{z}&-\dot{y}\dot{z}&\dot{x}^{2}+\dot{y}^{2}\\\
\end{bmatrix}_{\mbox{\footnotesize{(particle $J$)}}}$ (27)

Interestingly the numerator of (27) has the form of an inertia tensor for a
point particle in velocity-space. The matrix $\mathrm{Q}$ is block tri-
diagonal, because the spatial derivatives in (2.4) couple each bead to its two
neighbors. Using indices $I,J$ to enumerate beads and $i,j$ to enumerate
$x,y,z$ components for each bead:

$\displaystyle\|\mathrm{Q}_{IJ,ij}\|$
$\displaystyle=\delta_{ij}\left[\lambda_{J-1,J}\left(\delta_{IJ}-\delta_{I,J-1}\right)+\lambda_{J,J+1}\left(\delta_{IJ}-\delta_{I,J+1}\right)\right]$
or $\displaystyle\mathrm{Q}$
$\displaystyle=\begin{bmatrix}\lambda_{12}\mathbf{1}&-\lambda_{12}\mathbf{1}&0&0&\cdots&\\\
-\lambda_{12}\mathbf{1}&\left(\lambda_{12}+\lambda_{23}\right)\mathbf{1}&-\lambda_{23}\mathbf{1}&0&\cdots&\\\
0&-\lambda_{23}\mathbf{1}&\left(\lambda_{23}+\lambda_{34}\right)\mathbf{1}&-\lambda_{34}\mathbf{1}&&\\\
\vdots&\ddots&\ddots&\ddots&&\\\ &&&&&\\\
&&&&-\lambda_{N-1,N}\mathbf{1}&\lambda_{N-1,N}\mathbf{1}\\\ \end{bmatrix}$
(28)

For the transformation ${{\bf r}^{\ast}(t)}$ to be a minimum of ${\cal D}[{\bf
r}]$, the functional (24) must be positive definite for all $|h\rangle$. To
derive the conditions for this, we can temporarily ignore the fact that (24)
arose from the second variation of (9), and treat (24) as a new functional of
the function $|h(t)\rangle=|h_{1}(t),\ldots h_{3N}(t)\rangle$. We then ask
what $|h(t)\rangle$ extremizes (24). If $\delta^{2}{\cal D}>0$ we expect that
the only extremal solution would be the trivial one: $|h(t)\rangle=|0\rangle$,
at least for small variations of the $h_{i}(t)$. That is, changing the
transformation $\\{{\bf r}^{\ast}_{i}(t)\\}$ from that which extremized (9) to
a neighboring transformation $\\{{\bf r}^{\ast}_{i}(t)+{\bf h}_{i}(t)\\}$
would increase the distance travelled.

The system of $3N$ EL equations for $|h\rangle$ from (24) is

$-{d\over dt}|\mathrm{P}\dot{h}\rangle+|\mathrm{Q}h\rangle=|0\rangle$ (29)

with boundary conditions

$|h(0)\rangle=|h(T)\rangle=|0\rangle\>.$ (30)

Equation (29) is referred to as the Jacobi equation in the calculus of
variations.

First note that if $|h\rangle$ satisfies the system of equations in (29) as
well as the boundary conditions (30), then integration by parts gives

$\delta^{2}{\cal
D}=\int_{0}^{T}\\!\\!\\!dt\>\left(\langle\dot{h}|\mathrm{P}\dot{h}\rangle+\langle
h|\mathrm{Q}h\rangle\right)=\int_{0}^{T}\\!\\!\\!dt\>\langle h|-{d\over
dt}\left(\mathrm{P}\dot{h}\right)+\mathrm{Q}h\rangle=0\>.$ (31)

This means that for $\delta^{2}{\cal D}$ to be $>0$, any nontrivial
$|h(t)\rangle$ which satisfies the boundary conditions must not itself be an
extremal solution of the Jacobi equation, otherwise solutions $|{\bf
r}^{\ast}(t)\rangle$ perturbed by any constant times $|h(t)\rangle$ are
themselves extremals. One may think of this by analogy as the necessity for
the absence of any “Goldstone modes”, where excitations by various
$C|h(t)\rangle$ would lead to a family of curves with zero cost in action, and
thus zero effective restoring force, between them.

Alternatively we can ask what equation ${\bf h}\equiv|h\rangle$ must satisfy
if the EL equations are satisfied for both $\mathcal{L}({\bf r},\dot{{\bf
r}})$ and the neighboring extremal $\mathcal{L}({\bf r}+{\bf h},\dot{{\bf
r}}+\dot{{\bf h}})$. Taylor expanding $\mathcal{L}({\bf r}+{\bf h},\dot{{\bf
r}}+\dot{{\bf h}})$ in

$\mathcal{L}_{{\bf r}}({\bf r}+{\bf h},\dot{{\bf r}}+\dot{{\bf h}})-{d\over
dt}\mathcal{L}_{\dot{{\bf r}}}({\bf r}+{\bf h},\dot{{\bf r}}+\dot{{\bf h}})=0$

gives

$-{d\over dt}\left(\mathcal{L}_{\dot{{\bf r}}\dot{{\bf r}}}\cdot\dot{{\bf
h}}\right)+\left(\mathcal{L}_{{\bf r}{\bf r}}-{d\over dt}\mathcal{L}_{{\bf
r}\dot{{\bf r}}}\right)\cdot{\bf h}=0$

which is exactly Jacobi’s equation (29) with definitions (2.4).

From here on, it is much simpler to elucidate the central concepts for
sufficient conditions using the case of a single scalar function $h(t)$. The
analysis can be generalized to the multi-dimensional case with a bit more
effort, but the conclusions are essentially the same and so they will simply
be stated along with the conclusions for the ’1-D’ case. For further details
see [2].

We write equation (24) in 1-D as:

${1\over 2}\int_{0}^{T}\\!\\!\\!dt\>\left(P\dot{h}^{2}+Qh^{2}\right)$ (32)

It was realized originally by Legendre that the integral could be brought to
simpler form by adding zero to it in the form of a total derivative. Since

$\int_{0}^{T}\\!\\!\\!dt\>{d\over dt}\left(w(t)h^{2}\right)=0$

for any $w(t)$ so long as $h(t)$ satisfies the boundary conditions (30), we
can add it to the integral in (32) and seek a function $w(t)$ such that the
expression

$\delta^{2}{\cal D}={1\over
2}\int_{0}^{T}\\!\\!\\!dt\>\left(P\dot{h}^{2}+2wh\dot{h}+\left(Q+\dot{w}\right)h^{2}\right)$

may be written as a perfect square. This yields the differential equation

$P\left(Q+\dot{w}\right)=w^{2}$ (33)

for $w(t)$, and second variation

$\delta^{2}{\cal D}[h]={1\over
2}\int_{0}^{T}\\!\\!\\!dt\>P\left(\dot{h}+{w\over P}h\right)^{2}\>.$ (34)

Therefore a necessary condition for a minimum is for $P>0$. The analogous
condition in the multi-dimensional case is for the matrix $\|\mathrm{P}\|$ to
be positive definite.

If the differential term $\dot{h}+{w\over P}h$ in (34) were equal to zero for
some $h(t)$, the boundary condition $h(0)=0$ would then imply $\dot{h}(0)=0$
and thus $h(t)=0$ for all $t$ by the uniqueness theorem as applied to this
first order differential equation.

Therefore the functional (34) is positive definite if, and only if,
1.) $P>0$ ,
2.) A solution for eq. (33) exists for the whole interval $[0,T]$.
In general, there is no guarantee of condition (2) even if condition (1) is
valid. For example if $P=1$, $Q=-1$, (33) has solution $w(t)=\tan(t+c)$, which
has no finite solution if $|T|>\pi$. †**footnotetext: † Because
reparameterization invariance in our problem, the value of $T$ is adjustable,
however precisely because of this invariance, $\det\|\mathrm{P}\|=0$ and so is
no longer positive definite. We discuss this problem and its resolution below.

If (33) has a pole at say $\tilde{t}$, then for the integral (34) to remain
finite, $h(\tilde{t})\rightarrow 0$. This point is said to be conjugate to the
point $t_{o}=0$, i.e. it is a conjugate point.

Moreover, equation (33) is a Riccati equation, which may be brought to linear
form by the transformation $w(t)=-P\dot{H}/H$, with $H(t)$ an unknown
function. Substitution in (33) gives

$-{d\over dt}\left(P\dot{H}\right)+QH=0$ (35)

which is precisely equation (29)- the Jacobi equation for $h(t)$.

This means that for equation (33) to have a solution on $[0,T]$, $H(t)$, as
given by the solution to (35), must have no roots on $[0,T]$. But because
equation (35) holds for $h(t)$ as well, $h(t)$ must have no roots (conjugate
points) on $[0,T]$. Because $h(0)=h(T)=0$, the only way to extremize (32) is
to satisfy eq. (35) with the trivial solution $h(t)=0$. If $h(t)\neq 0$ for
$0<t<T$ then it would mean that there was a conjugate point at $\tilde{t}=T$.

In the multi-dimensional case an extremal $|h\rangle$ is one of $3N$ vectors
satisfying equations (29), i.e. $|h^{(\alpha)}\rangle=|h_{1}^{(\alpha)}\ldots
h_{3N}^{(\alpha)}\rangle$, $1\leq\alpha\leq 3N$. A conjugate point is defined
as a point where the determinant vanishes:

$\det\begin{vmatrix}h_{1}^{\left(1\right)}\left(t\right)&\cdots&h_{1}^{\left(3N\right)}\left(t\right)\\\
\vdots&&\vdots\\\
h_{1}^{\left(3N\right)}\left(t\right)&\cdots&h_{3N}^{\left(3N\right)}\left(t\right)\end{vmatrix}=0$

The conditions for a transformation to be minimal are then:
1.) The transformation $|{\bf r}^{\ast}(t)\rangle=\\{{\bf r}^{\ast}_{i}(t)\\}$
is extremal,
2.) Along $|{\bf r}^{\ast}(t)\rangle$, the matrix
$\mathrm{P}(t)=\mathcal{L}_{\dot{x}_{i}\dot{x}_{j}}$ is positive definite, and
3.) The interval $[0,T]$ contains no conjugate points to $t=0$.
The above ideas can be made clear with a few examples below.

####  2.4.1 Distance between points

From the effective Lagrangian $\mathcal{L}=\sqrt{\dot{{\bf r}}^{2}}$,
$\mathrm{P}=\|\mathcal{L}_{\dot{x}_{i}\dot{x}_{j}}\|$ is given in equation
(27), which has determinant $\det\mathrm{P}=0$, and so is not positive
definite. This is due to our choice of parameterization. If we break symmetry
by choosing one spatial direction as the independent variable,
$\mathcal{L}\left(x,y^{\prime},z^{\prime}\right)=\sqrt{1+y^{\prime
2}+z^{\prime 2}}$ (with e.g. $y^{\prime}\equiv dy/dx$ and $x_{0}\leq x\leq
x_{1}$). Then

$\mathrm{P}={1\over{\left(1+y^{\prime 2}+z^{\prime
2}\right)^{3/2}}}\begin{pmatrix}1+z^{\prime 2}&-y^{\prime}z^{\prime}\\\
-y^{\prime}z^{\prime}&1+y^{\prime 2}\end{pmatrix}$

with positive definite determinant $\det\|\mathrm{P}\|=\left(1+y^{\prime
2}+z^{\prime 2}\right)^{-1/2}>0$ for any trajectory. From eq (2.4),
$\|\mathrm{Q}(t)\|=\|0\|$. Along the extremal, where $y(x)=ax+y_{0}$,
$z(x)=bx+z_{0}$, equation (29) gives $\mathrm{P}\cdot{\bf h}^{\prime}={\bf
c}$, with ${\bf c}$ a constant vector and $\mathrm{P}$ a positive definite
matrix of constant values with respect to $x$. Solving this first-order
equation gives straight line solutions for ${\bf h}(x)$. Because ${\bf
h}(x_{0})=0$, there can be no conjugate points, and because ${\bf
h}(x_{1})=0$, the only solution to (29) is the trivial one, and the extremum
is a minimum.

####  2.4.2 Geodesics on the surface of a sphere

Taking the azimuthal angle $\phi$ as the independent variable, and polar angle
$\theta(\phi)$ as the dependent variable, the arc-length on the surface of a
unit sphere may be written as

${\cal
D}[\theta]=\int_{\phi_{0}}^{\phi_{1}}\\!\\!\\!d\phi\>\sqrt{\theta^{\prime
2}+\sin^{2}\theta}\>.$ (36)

The EL equations give the extremal trajectory as
$\cos\theta=A\sin\theta\cos\phi+B\sin\theta\sin\phi$ with $A,B$ constants.
This is the equation of a plane $z=Ax+By$, which intersects the surface of the
sphere to make a great circle. The scalar
$P=\mathcal{L}_{\theta^{\prime}\theta^{\prime}}=\sin^{2}\theta/\left(\theta^{\prime
2}+\sin^{2}\theta\right)^{3/2}$ which is always positive. To simplify the
problem, let $\phi_{0}=0$, and $\theta(\phi_{0})=\theta(\phi_{1})=\pi/2$, so
the great circle lies in the $z=0$ plane. Along this extremal $P$ is constant
and equal to $1$, while $Q=-1$. The second variation, eq. (32), is then
$(1/2)\int_{0}^{\phi_{1}}\\!\\!\\!d\phi\>\left(h^{\prime 2}-h^{2}\right)$. The
corresponding Jacobi equation, $h^{\prime\prime}+h=0$, must not have a root
between $[0,\phi_{1}]$. The nontrivial solution to the Jacobi equation
satisfying the initial condition $h(0)=0$ is $h(\phi)=C\sin\phi$, which has a
conjugate point at $\phi=\pi$. Thus for the extremal curve to be minimal,
$\phi_{1}$ must be $<\pi$, the location of the opposite pole on the sphere. If
$\phi_{1}<\pi$, there is no extremal solution for $h(\phi)$ other than the
trivial one which satisfies the boundary conditions. It is instructive to look
at the arc-length under sinusoidal variations around the extremal path which
satisfy the boundary conditions $h(0)=h(\phi_{1})=0$, so that
$\theta(\phi)=\pi/2+h(\phi)=\pi/2+\epsilon\sin\left(\pi\phi/\phi_{1}\right)$.
Inserting this into eq (36) above and expanding to second order in $\epsilon$,
we see that first order terms in $\epsilon$ vanish, and the difference in
distance from the extremal path is $\Delta{\cal
D}=\left(\epsilon^{2}/4\phi_{1}\right)\left(\pi^{2}-\phi_{1}^{2}\right)$. For
$\phi_{1}<\pi$ this is always greater than zero indicating the extremal is a
minimum. For $\phi_{1}>\pi$ this is always less than zero indicating the
extremal is a maximum with respect to these perturbations: the length may be
shortened. When $\phi_{1}=\pi$, $\Delta{\cal D}=0$ to second order. When
$h(\phi)$ represents the difference between great circles $\Delta{\cal D}$ is
precisely zero.

####  2.4.3 Harmonic oscillator

It is not widely appreciated that the classical action for a simple harmonic
oscillator is not always a minimum, and indeed in many cases can be a maximum
with respect to some perturbations. The action for a harmonic oscillator with
given spring constant is proportional to
$S[x]=\int_{0}^{T}\\!\\!\\!dt\>{1\over 2}(\dot{x}^{2}-x^{2})$, which has EL
equation $\ddot{x}+x=0$. Taking the specific initial conditions $x(0)=1$,
$\dot{x}(0)=0$, the extremal solution is $x(t)=\cos t$. The scalar
$P(t)=\mathcal{L}_{\dot{x}\dot{x}}=1$, which is always positive and satisfies
the necessary conditions for a minimum. The scalar $Q=\mathcal{L}_{xx}-{d\over
dt}\mathcal{L}_{x\dot{x}}=-1$. The second variation $\delta^{2}S[h]={1\over
2}\int_{0}^{T}\\!\\!\\!dt\>(\dot{h}^{2}-h^{2})$, which has Jacobi equation
$\ddot{h}+h=0$. This is the same Jacobi equation as that for geodesics on a
sphere, so the sufficient conditions will parallel those above. The boundary
condition $h(0)=0$ gives $h(t)=A\sin t$, with conjugate points at $t=n\pi$,
$n=1,2,\ldots$. This means that the action is a minimum only so long as
$T<\pi$, i.e. a half-period. If we let $x(t)$ be the extremal solution plus a
$\sin$ perturbation satisfying the Jacobi equation at the conjugate points:
$x(t)=\cos t+\epsilon\sin t$, then the difference in action from the extremal
path becomes $\Delta S=(\epsilon^{2}/4T)(\pi^{2}-T^{2})$. This result is exact
because the action for the oscillator is quadratic (as opposed to the action
for geodesics). When $T<\pi$, $\Delta S>0$ indicating the extremal is a
minimum. When $T$ is larger than a half-period, $\Delta S<0$ and the extremal
trajectory is a maximum (with respect to half-wavelength sinusoidal
perturbations), and when $T=\pi$, the end point is the conjugate point and
$\Delta S=0$.

We discuss sufficient conditions further below in the context of minimal
transformations for links.

##  3 Single Links

In the limit of one link, equations (15a-15c) reduce to:

$\displaystyle\dot{\hat{{\bf v}}}_{\mbox{\tiny{A}}}+\lambda\,{\bf
r}_{\mbox{\tiny{B/A}}}=0$ $\displaystyle\dot{\hat{{\bf
v}}}_{\mbox{\tiny{B}}}-\lambda\,{\bf r}_{\mbox{\tiny{B/A}}}=0$ (37)

where we have let $A$ represent point $1$, $B$ point $2$, and
$\lambda\equiv\lambda_{12}$. The link has length $1$ in our dimensionless
formulation, so the vector ${\bf r}_{\mbox{\tiny{B/A}}}$ could also have been
written as a unit vector $\hat{{\bf r}}_{\mbox{\tiny{B/A}}}$.

Both points $A$ and $B$ are end points and satisfy the boundary conditions of
section 2.3.1. This means that points $A$ and $B$ move by either pure
rotation, straight-line translation, or remain at rest. The initial and final
conditions may be written ${\bf r}_{\mbox{\tiny{A}}}(0)={\bf A}$, ${\bf
r}_{\mbox{\tiny{B}}}(0)={\bf B}$, ${\bf r}_{\mbox{\tiny{A}}}(T)={\bf
A}^{\prime}$, ${\bf r}_{\mbox{\tiny{B}}}(T)={\bf B}^{\prime}$.

The link in our problem has direction, so $A$ must transform to $A^{\prime}$
and $B$ to $B^{\prime}$. We will often use arrowheads in figures to denote
this direction.

###  3.1 Straight line transformations

As a first example, consider the two links shown in figure 3a. The four points
$A,B,A^{\prime},B^{\prime}$ need not lie in a plane (see for example fig 3b).
Let angle $\angle BAA^{\prime}\equiv a$ be obtuse. We draw straight lines from
$A$ to $A^{\prime}$ and $B$ to $B^{\prime}$, and ask whether such a
transformation is possible. We can thus derive the following rule:
$\bullet$ For a straight line transformation to exist between two links,
opposite angles of the quadrilateral made by $\overline{AB}$,
$\overline{A^{\prime}B^{\prime}}$, $\overline{AA^{\prime}}$,
$\overline{BB^{\prime}}$ must be obtuse.

|  |
---|---|---
a | b | c
Figure 3: Possible (a,b) and impossible (c) straight line transformations
between links $AB$ and $A^{\prime}B^{\prime}$. Figure $b$ shows a straight
line transformation where the initial and final states do not lie in the same
plane. In the text we derive the conditions for the possibility of a straight
line transformation between links.

Let the length that point $A$ travels be $x_{\mbox{\tiny A}}$, i.e. we imagine
the point $A^{\prime}$ and the distance $x_{\mbox{\tiny A}}=|AA^{\prime}|$ to
be variable. The length $r_{\mbox{\tiny B}}$ that point $B$ travels is then a
function of $x_{\mbox{\tiny A}}$ and the original angle $a$, $r_{\mbox{\tiny
B}}(x_{\mbox{\tiny A}},a)$. We can now find conditions on the angle
$b\equiv\angle BB^{\prime}A^{\prime}$ such that the transformation is
possible.

After some distance $x_{\mbox{\tiny A}}$ travelled by point $A$, the length of
the line from $B$ to $A^{\prime}$ is

$\displaystyle\overline{BA^{\prime}}$ $\displaystyle=$ $\displaystyle
x_{\mbox{\tiny A}}^{2}+1-2x_{\mbox{\tiny A}}\cos a$ $\displaystyle=$
$\displaystyle r_{\mbox{\tiny B}}^{2}+1-2r_{\mbox{\tiny B}}\cos b$

so that

$r_{\mbox{\tiny B}}(x_{\mbox{\tiny A}},a)=\cos
b\pm\sqrt{\cos^{2}b+f(x_{\mbox{\tiny A}},a)}$

with $f(x_{\mbox{\tiny A}},a)=x_{\mbox{\tiny A}}^{2}-2x_{\mbox{\tiny A}}\cos
a$. Since $a$ is obtuse, $f>0$ when $x_{\mbox{\tiny A}}>0$, and so the
positive root must be taken for $r_{\mbox{\tiny B}}$ to positive. When
$x_{\mbox{\tiny A}}=0$, $f(0,a)=0$, and

$r_{\mbox{\tiny B}}(0,a)=\cos b+\left|\cos b\right|=0$

Therefore $b$ must also be an obtuse angle. If two opposite angles are obtuse,
then the other two angles must be acute. This concludes the proof that the
above conditions are sufficient. An additional proof that they are necessary
is given in A.

We readily see that figure 3A is one pair of a larger set of straight line
transformations that can continue until one or both of the obtuse angles
reaches $90^{\circ}$. This collection forms a “bow tie” of admissible
configurations, as in figure 3.1. Note that straight lines in the
quadrilateral may cross as in the transformation from $A,B$ to
$A^{\prime},B^{\prime}$ in figure 3.1. Trivial translations of the link
without any concurrent rotation are a special case of general straight line
transformations.

|
---|---
A | B

Figure 4:
(A) An example of a set of link configurations connected by a straight-line
transformation. The link rotates clockwise as it translates to allow the end
points to move in straight lines. The translation can proceed no farther than
the end points $AB$ and $A^{\prime}B^{\prime}$, which have link vectors
$\overrightarrow{AB}$ or $\overrightarrow{A^{\prime}B^{\prime}}$ that are
perpendicular to one or the other of the vectors $\hat{{\bf v}}_{\mbox{\tiny
A}}$ or $\hat{{\bf v}}_{\mbox{\tiny B}}$. The totality of states thus
connected forms a “bowtie”. (B) A bowtie where the terminal states $AB$ and
$A^{\prime}B^{\prime}$ happen to cross each other.

###  3.2 Piece-wise extremal transformations: transformations with rotations

An immediate question is the nature of the transformation between $AB$ and
$A^{\prime}B^{\prime}$ in figure 3C, where opposite angles of the
quadrilateral are not obtuse. Recall our link has direction so $A$ cannot
transform to $B^{\prime}$. Then direct straight-line solution is not possible
due to the constraint of constant link length.

The only remaining solution is for the link to rotate as part of the
transformation. Consider first the rotation of link $AB$. The EL equations (3)
allow for pure rotations about $A$, $B$, or a common center along the link.
Likewise for link $A^{\prime}B^{\prime}$.

The rotation can occur from either link $AB$ (fig 5a) or link
$A^{\prime}B^{\prime}$ (fig 5b). After the link rotates to a critical angle,
it can then travel in a straight line. The extremals are broken in that they
involve matching up a piece consisting of pure rotation with a piece
consisting of pure translation of the end points of the link. Where the pieces
match they must satisfy the corner conditions (18a, 18b). This means that the
end points cannot suddenly change direction, a situation which is only
satisfied by a straight line trajectory that lies tangent to the circle of
rotation.

From figure 3.1, we see that a straight line transformation exists only when
an angle between a link and one of the straight line trajectories reaches
$\pi/2$.

The critical angle that link $AB$ must rotate is then determined by the point
where a line drawn from $B^{\prime}$ is just tangent to the unit sphere
centered at point $A$, point $B_{1}$ in figure 5a. There is generally a
different critical angle if the rotation occurs at link $A^{\prime}B^{\prime}$
as in fig 5B. It is shown in B that in general the critical angle is
determined by drawing the tangent to a circle or sphere about one of the link
ends.

a |  | b |
---|---|---|---
|  |  |
c |  | d |
|  |  |
e |  |  |
|  |  |
Figure 5: Transformations between two links involving broken extremals
consisting of rotation and translation. $(b)$ is the global minimum, with
shortest distance travelled during the transformation. $(a)$, $(c)$, and $(d)$
are local minima. $(e)$ is extremal, but not minimal as the trajectory of arc
$\stackrel{{\scriptstyle\frown}}{{B^{\prime}B_{1}}}$ passes through a
conjugate point.

If the rotation was about a common center, we see that one or another of the
link ends would violate a corner condition, so the rotation must be about one
of the link ends.

According to eqs. (26) and (27), the matrix $\mathrm{P}$ has a determinant of
zero due to the parametric formulation in the problem and so is not positive
definite. To show that the transformations in fig. 5a,b are indeed minimal, we
need to then express the problem in non-parametric form. To do this, let the
independent variable be the angle $\theta$ of the link with the vertical. Then
the displacement $x$ along the line $\overline{AA^{\prime}}$ is the unknown
function of $\theta$ to be determined by minimizing the total arc length
travelled. This distance can be written as

${\cal
D}[x]=\int_{\theta_{0}}^{\theta_{1}}\\!\\!\\!d\theta\>\left(\sqrt{x^{\prime
2}+2x^{\prime}\cos\theta+1}+\sqrt{x^{\prime 2}}\right)$

In this formulation, the scalar quantity
$\mathrm{P}(\theta)=\mathcal{L}_{x^{\prime}x^{\prime}}$ becomes

$\mathrm{P}(\theta)={\sin^{2}\theta\over\left(x^{\prime
2}+2x^{\prime}\cos\theta+1\right)^{3/2}}$

which is always $>0$ except for the isolated point $\theta=0$, in particular
it is positive along the extremal trajectory which is necessary for a minimum.
So we conclude that the transformation with the smaller angle of rotation in
fig 5b is here the global minimum, and the other transformation (fig 5a) is a
local minimum.

Figure 5e is also an extremal trajectory, satisfying corner conditions, and
with positive definite $P$. However it is not a local minimum because the
trajectory passes through a conjugate point (denoted by point $CP$, where the
dotted line along $\overline{A^{\prime}B^{\prime}}$ meets the great circle
about $A^{\prime}$). According to the results in section 2.4.2, if the
extremal trajectory (a great circle) traverses an angle larger than $\pi$
radians, it passes through a conjugate point and thus becomes unstable to
long-wavelength perturbations. Transformations involving rotations about
points $B$ or $B^{\prime}$ in figure 5 both have conjugate points and so are
not minimal.

The transformation in fig. 5c does not pass through a conjugate point and so
is in fact another local minimum. The part of the extremum along the straight
line section of the trajectory has no conjugate points as discussed above.

###  3.3 Systematically exploring transformations by varying link positions

We can investigate what happens to the minimal transformation when one of the
link positions or angles is varied with respect to the other. Let us start by
putting the two links head to tail as shown in figure 6A. The distance between
them is $2$ by simple translation of link end points.

We can now increase the angle between the two vectors by rotating the right
link for example, as in figures 6B-H. So long as the angle between the two
vectors is less than $90^{\circ}$, one link may slide along another and the
distance is unchanged (figs 6A-C). This is a special case of the
transformations shown in figure 3.1 (compare for example figure 6B with the
middle three unlabelled links in that figure).

Beyond $90^{\circ}$ however, the transformation must include rotation. Fig 6D
has an angle of $150^{\circ}$. The minimal transformation first rotates, for
example with the tail of the horizontal black arrow fixed, and the head
tracing out the blue arc, until the critical angle is reached, where a
straight line made from the final arrowhead (at the top of the figure) is just
tangent to the circle made by the blue arc. This state is indicated by a red
link in figure 6D. The link then translates to its reciprocal position at the
opposite end of the bowtie, denoted by a second red link (c.f. also figure
3.1B). At this point the arrowhead has completed the transformation. Finally
the tail rotates into its final position. The total distance travelled is
slightly larger than $2$.

When the angle between the vectors is $120^{\circ}$ as shown in 6E, the
transformation consists of pure rotations. Taking the initial state to be the
horizontal black vector, the link first rotates about its fixed tail, the head
tracing out the blue arc, until the link reaches the state shown in red, where
the position of the arrowhead has reached its final end point. Then the link
rotates about its head until the position of the tail reaches the final state.

When the angle between the links is larger than $120^{\circ}$ as shown in figs
6F-G, the transformation must involve rotation about an internal point along
the link. Let points $A$ and $B$ denote the tail and head of the link
respectively. If an infinitesimal rotation $\Delta\theta$ occurs about an
internal point $P$, the increment in distance travelled is

$\Delta{\cal D}=|{\bf r}_{\mbox{\tiny{B/P}}}|\Delta\theta+|{\bf
r}_{\mbox{\tiny{B/A}}}|\Delta\theta=\Delta\theta$

which is independent of the position of the instantaneous center of rotation
(ICR). This means that there are an infinity of transformations all giving the
same distance, depending on the time-dependence of the ICR. Two simple
alternatives with only two discrete positions of ICR are shown in figures
6F,G. Specifically, in figure 6F, the horizontal black vector first rotates
about its tail to the red configuration, which is a mirror image of the final
black vector. Then rotation is about an internal point determined by the
intercept of the red vector with the final black vector, with end points
tracing out the green arcs. In figure 6G the two ICRs are both internal and
determined by the intercepts of the initial and final states with the red
vector shown.

Figure 6H depicts the transformation for overlapping, opposite pointing
vectors. Rotation can now only occur about one point in the center of the
vectors.

a |  | b |  | c |
---|---|---|---|---|---
|  |  |  |  |
d |  | e |  | f |
|  |  |  |  |
g |  | h |  |  |
|  |  |  |  |
Figure 6: Successive transformations between two links made by rotating a link
so that there is a progressively larger angle between the links as vectors (or
smaller angle made between them as lines). The two boundary conditions (the
initial and final conditions) are shown as black links, and an intermediate
state is shown as a red link or links. The arcs traced out by the end points
are shown in blue or green, while straight line motions when they are not
along the links themselves are shown in grey. The distance travelled over the
course of the transformation is given below each figure.

Figure 7 illustrates what happens when one of the links is translated with
respect to another, starting from two different scenarios shown in 7A and 7E.
In 7A, the tail of the vertical link is displaced $(1/3,-1/3)$ with respect to
the tail of the horizontal link. The minimal transformation is a pure rotation
by $\pi/2$.

In figure 7B, the tail of the vertical link is now displaced to $(2/3,-1/3)$.
Pure rotations again give a distance of $\pi/2$. Rotation about a point on the
horizontal link that is equidistant from both arrowheads transforms the
initial arrowhead to the final (red intermediate state). Then rotation of the
tail about the arrowhead transforms to the final state.

In figure 7C, the minimal transformation first involves a translation by
sliding the arrowhead along the vertical, until the arrowheads overlap (red
intermediate state). The tail end of the link then rotates into place.

In figure 7D, straight lines from the end points will not satisfy the obtuse
condition in section 3.2, so the transformation must involve rotations. Here a
straight line transformation takes the link almost to the final state. It then
must undergo a small rotation to complete the transformation. Seen in reverse,
the vertical arrow must rotate to a critical angle determined by the criterion
in section 3.2, before the link can finish the transformation by pure
translation.

Figure 7E is figure 6F once again. The final condition (the tilted link) will
be systematically changed by translating it vertically away from the
horizontal link (which we choose arbitrarily as the initial configuration).

In figure 7F the tilted link is translated a distance $1/3$ vertically. The
transformation can be achieved by rotating the horizontal link about a point
equidistant from both arrowheads, to the red intermediate configuration. The
link then rotates about the arrowhead into the final configuration. The
distance is still the angle rotated for the reasons mentioned above in the
context of figures 6F-G, $\theta=(150/180)\pi$, which is unchanged from 7E. In
fact, so long as the arrowhead can be reached by rotation (the translated
distance is less than $d$ where $d$ is the solution to $d^{2}+d+1-\sqrt{3}=0$
for this angle), then the distance will be unchanged. The transformation at
the critical distance is shown in figure 7G. The rotations now occur about the
end-points: the tail and head of the link.

In figure 7H the translated distance is now equal to $1$. The transformation
first consists of a rotation about the tail to a critical angle (blue arc and
red intermediate state), then a translation much like that in figure 3.1 (grey
straight lines between red intermediate states), and finally a rotation about
the head (green arc) to the final configuration.

| a |  | e
---|---|---|---
|  |  |
| b |  | f
|  |  |
| c |  | g
|  |  |
| d |  | h
|  |  |
Figure 7: Successive transformations between two links made by translating one
link with respect to the other. In (a-d) the initial and final configurations
are perpendicular, while in (e-h) they are at an angle of $150^{\circ}$ to
each other. Note the distances in (e-g) are all the same, even though the end
points of the links are at varying distances from each other.

##  4 2-link chains

We now consider the next simplest case of $2$ links ($3$ beads). The
Lagrangian now reads:

$\mathcal{L}({\bf r}_{1},{\bf r}_{2},{\bf r}_{3},\dot{{\bf r}}_{1},\dot{{\bf
r}}_{2},\dot{{\bf r}}_{3})=\sqrt{\dot{{\bf r}}_{1}^{2}}+\sqrt{\dot{{\bf
r}}_{2}^{2}}+\sqrt{\dot{{\bf r}}_{3}^{2}}-{1\over
2}\lambda_{12}\left(\left({\bf r}_{2}-{\bf r}_{1}\right)^{2}-1\right)-{1\over
2}\lambda_{23}\left(\left({\bf r}_{3}-{\bf r}_{2}\right)^{2}-1\right)$ (38)

which has EL equations (c.f. eq.s 15a-15c) ‡**footnotetext: ‡ The links have
length $1$ in our dimensionless formulation, so the vectors ${\bf
r}_{\mbox{\tiny{B/A}}}$ and ${\bf r}_{\mbox{\tiny{C/B}}}$ could also have been
written as unit vectors $\hat{{\bf r}}_{\mbox{\tiny{B/A}}}$ and $\hat{{\bf
r}}_{\mbox{\tiny{C/B}}}$. :

$\displaystyle\dot{\hat{{\bf
v}}}_{\mbox{\tiny{A}}}+\lambda_{\mbox{\tiny{AB}}}\,{\bf
r}_{\mbox{\tiny{B/A}}}=0$ (39a) $\displaystyle\dot{\hat{{\bf
v}}}_{\mbox{\tiny{B}}}-\lambda_{\mbox{\tiny{AB}}}\,{\bf
r}_{\mbox{\tiny{B/A}}}+\lambda_{\mbox{\tiny{BC}}}\,{\bf
r}_{\mbox{\tiny{C/B}}}=0$ (39b) $\displaystyle\dot{\hat{{\bf
v}}}_{\mbox{\tiny{C}}}-\lambda_{\mbox{\tiny{BC}}}\,{\bf
r}_{\mbox{\tiny{C/B}}}=0\>.$ (39c)

The corner conditions (18a), (19) imply

$\hat{{\bf v}}_{i}\left(t^{-}\right)=\hat{{\bf v}}_{i}\left(t^{+}\right)$

so the direction of motion cannot suddenly change, unless along one part of
the extremal the velocity of point $i$ is zero (the point is at rest), where
its direction $\hat{{\bf v}}$ is then undefined.

The boundary conditions described in section 2.3.1 hold as well, so the end
points can either be at rest, move in straight lines, or purely rotate. This
gives $3\times 3=9$ possible scenarios to investigate here, many of which can
readily be ruled out. For example consider the states in figure 8a. Because
$A$ and $A^{\prime}$ are in the same position, rotation and translation of $A$
are ruled out and point $A$ remains at rest, leaving $3$ scenarios for the
other end point $C$. However since $C$ and $C^{\prime}$ are at different
positions and $ABC$ are along a straight line, $C$ cannot remain at rest
initially, leaving either translation or rotation for point $C$.

a |  | b |
---|---|---|---
|  |  |
c |  | d |
|  |  |
Figure 8: (a) Initial and final states for a chain of two links. The
transformation in (b) is non-extremal because it violates a corner condition
at $C^{\prime\prime}$. (c) and (d) are degenerate minima- rotations occurring
about $B^{\prime}$ or $B$ both have the same length. Intermediate states is
shown in red have opposite convexity in (c) and (d).

Suppose $C$ translates towards $C^{\prime}$ as in figure 8b. Then
$\dot{\hat{{\bf v}}}_{\mbox{\tiny{C}}}=0$ and from (39c,39b)
$\lambda_{\mbox{\tiny{BC}}}=0$ and $\dot{\hat{{\bf
v}}}_{\mbox{\tiny{B}}}=\lambda_{\mbox{\tiny{AB}}}\,{\bf
r}_{\mbox{\tiny{B/A}}}$. $B$ cannot move in a straight line without moving
point $A$, so $\lambda_{\mbox{\tiny{AB}}}\neq 0$ and thus $B$ must rotate
about point $A$. The transformation then proceeds as in figure 8b until $B$
reaches $B^{\prime}$ and $C$ reaches $C^{\prime\prime}$. Then however if
$C^{\prime\prime}$ were to rotate to $C^{\prime}$, the trajectory would
violate corner conditions at point $C^{\prime\prime}$. Therefore the direction
of translation of $C$ must not be directly to $C^{\prime}$ but must be
tangential to the arc
$\stackrel{{\scriptstyle\frown}}{{C^{\prime}C^{\prime\prime}}}$ as in figure
8c.

The reverse of this transformation is allowable as well, as can be seen by
swapping the labels $ABC\rightarrow A^{\prime}B^{\prime}C^{\prime}$. Here $C$
first rotates to the critical angle $\theta$ shown in fig 8d and then
translates to $C^{\prime}$.

In fact one can see that links $BC$ and $B^{\prime}C^{\prime}$ along with
lines $\overline{BB^{\prime}}$ and $\overline{CC^{\prime}}$ form a
quadrilateral as in figure 5, with the same consequences for rotation to a
critical angle. For the links in fig 8 the situation is symmetric so rotation
can occur at the beginning or end of the transformation. Figure 9a shows an
example with this symmetry broken, so that the distance is different depending
where the rotation occurs, as in figures 5a,b. In this case, the
transformation in fig 9c has the minimal distance, and that in fig 9b is
subminimal. Extensions of the transformation in figure 9 to large numbers of
links were explored in [1].

|  |
---|---|---
a | b | c
Figure 9: (a) Initial and final states for a polymer of $2$ links. The angle
between $\overline{AB}$ and $\overline{A^{\prime}B^{\prime}}$ is $\pi/4$. The
minimal transformations in (b) and (c) are now no longer degenerate. (c) is
the global minimum.

###  4.1 Transformations involving a change in convexity

Transformations between configurations with opposite convexity involve motion
out of the plane, even if the initial and final states lie in the plane. If
the transformation is constrained to lie in plane, the trajectories of some
points will be non-monotonic- those points must move farther away from their
final positions before approaching them. We illustrate these ideas with some
examples below.

Figure 10: A transformation between two states of opposite convexity: $ABC$
has convexity down and right, while $A^{\prime}B^{\prime}C^{\prime}$ has
convexity up and left. There is no extremal transformation in the plane that
can connect them, without some apparent violation of corner conditions.

Consider the initial and final states in figure 10. We again imagine $B$
rotating to $B^{\prime}$. If C were to translate to $C^{\prime}$ one would
have the intermediate configuration $A^{\prime}B^{\prime\prime}C^{\prime}$.
Now $C^{\prime}$ and $A^{\prime}$ must remain at rest to satisfy corner
conditions. Then the only way to finish the transformation is for
$B^{\prime\prime}$ to rotate about the axis $\overline{A^{\prime}C^{\prime}}$,
however then the trajectory of $B$ violates corner conditions and so is not
extremal. In C we take up the issue of minimal transformations for this case
when the links are constrained to lie in a plane.

We thus seek a point $B^{\prime\prime}$ and resulting trajectory
$\overrightarrow{BB^{\prime\prime}B^{\prime}}$ such that arc
$\stackrel{{\scriptstyle\frown}}{{BB^{\prime\prime}}}$ satisfies corner
conditions with arc
$\stackrel{{\scriptstyle\frown}}{{B^{\prime\prime}B^{\prime}}}$.

One solution is to effectively place $B^{\prime\prime}$ at position
$B^{\prime}$ by considering the boundary condition with $C$ at rest (and $A$
at rest). Then $B$ rotates to $B^{\prime}$ about axis $\overline{AC}$, and the
trajectory of $B$ lies on a circle defined by the intercept of two unit
spheres centered at $A$ and $C$. The sphere about $A$ is drawn in figure 11 as
a visual aid. Along arc $\stackrel{{\scriptstyle\frown}}{{BB^{\prime}}}$ both
$\lambda_{\mbox{\tiny{AB}}}\neq 0$ and $\lambda_{\mbox{\tiny{BC}}}\neq 0$.
Once in configuration $A^{\prime}B^{\prime}C$, $C$ can then undergo rotation
about $B^{\prime}$ to $C^{\prime}$, with $A^{\prime}$ and $B^{\prime}$
stationary.

The transformation in 11a is a local minimum in distance, however it is not
the global minimum. A shorter distance transformation can be seen by
considering the reverse transformation. Imagine $A^{\prime}$ and $C^{\prime}$
stationary while $B^{\prime}$ rotates about axis
$\overline{A^{\prime}C^{\prime}}$ in figure 11b. This rotation of $B^{\prime}$
follows a circular trajectory defined by the intercept of two unit spheres
centered at $A^{\prime}$ and $C^{\prime}$. The rotation occurs until point
$B^{\prime\prime}$, which is the point where above circle is tangent to a
great circle on the unit sphere about $A$ and passing through $B$. The arc
$\stackrel{{\scriptstyle\frown}}{{BB^{\prime\prime}}}$ is a great circle
because this is a geodesic for point $B$ given $A$ is fixed, which follows
from the Euler equations (39b,39c) when $\lambda_{\mbox{\tiny{BC}}}=0$. The
great circle is defined by the plane containing the points $A$, $B$, and
$B^{\prime\prime}$.

The angle between the (variable) vector $\overrightarrow{BC}$ of link $BC$ and
the tangent the the arc
$\stackrel{{\scriptstyle\frown}}{{B^{\prime}B^{\prime\prime}}}$ is always
$\pi/2$, so once the corner condition is met, point $C$ on link $BC$ can move
in straight line motion from $C^{\prime}$ to $C$ while $B$ moves on the great
circle from $B^{\prime\prime}$ to $B$. That is, the quadrilaterial criterion
of section 3.1 is met for $\Box BB^{\prime\prime}C^{\prime}C$.

To find point $B^{\prime\prime}$, let its position be ${\bf
r}_{\mbox{\tiny{B''}}}=(x_{o},y(x_{o}),z(x_{o}))$. The great circle is defined
by the plane passing through the points $A$, $B$, and $B^{\prime\prime}$. This
plane has normal ${\bf
n}\equiv\overrightarrow{AB}\times\overrightarrow{AB^{\prime\prime}}=(1,0,0)\times(x_{o},y(x_{o}),z(x_{o}))=(0,-z(x_{o}),y(x_{o}))$.
At the point $B^{\prime\prime}$ the normal is orthogonal to the tangent vector
of the circle defined by rotation about the $AC^{\prime}$ axis. This tangent
vector is $\hat{{\bf t}}=\partial{\bf r}/\partial s=x_{s}(1,y_{x},z_{x})$ by
the chain rule. At $B^{\prime\prime}$, $\hat{{\bf t}}\cdot{\bf n}=0$, or

$-z(x_{o})y_{x}(x_{o})+y(x_{o})z_{x}(x_{o})=0$ (40)

The functions $y(x)$ and $z(x)$ are defined by the intercept of two unit
spheres centered at $(0,0,0)$ and $(1/\sqrt{2},1+1/\sqrt{2},0)$, giving

$\displaystyle y(x)$ $\displaystyle=1-{\sqrt{2}\over{2+\sqrt{2}}}x$
$\displaystyle z(x)$ $\displaystyle=\sqrt{1-x^{2}-y(x)^{2}}\>.$ (41)

Together (40) and (41) give

$\displaystyle{\bf r}_{\mbox{\tiny{B''}}}=\begin{pmatrix}\sqrt{2}-1\\\
2(\sqrt{2}-1)\\\ \sqrt{2(5\sqrt{2}-7)}\end{pmatrix}$

The distance travelled along arc
$\stackrel{{\scriptstyle\frown}}{{BB^{\prime\prime}}}$ is
$\theta_{\mbox{\tiny{BB''}}}$, where
$\cos\theta_{\mbox{\tiny{BB''}}}=x_{o}=\sqrt{2}-1$. The distance travelled
along arc $\stackrel{{\scriptstyle\frown}}{{B^{\prime\prime}B^{\prime}}}$ can
similarly be shown to be
$r\theta_{\mbox{\tiny{B''B'}}}=\sin(\pi/8)\cos^{-1}(2\sqrt{2}-3)$. Adding the
distance $\overline{CC^{\prime}}$, the total (minimal) distance is thus ${\cal
D}=2.576$. There is of course a degenerate solution to the above with
$z\rightarrow-z$.

|
---|---
a | b
Figure 11: Subminimal (a) and minimal (b) transformations for the boundary
conditions in figure 10. The distances for each transformation are
approximately $3.007L^{2}$ and $2.576L^{2}$ respectively. Transformation (a)
proceeds from $ABC$ by first rotating $B$ to $B^{\prime}$ about axis
$\overline{AC}$, then rotating $C$ about point $B^{\prime}$. Transformation
(b) proceeds from $ABC$ by simultaneously translating $C$ to $C^{\prime}$
while rotating $B$ about $A$ on a great circle to point $B^{\prime\prime}$.
Finally point $B$ rotates from $B^{\prime\prime}$ to $B^{\prime}$ about axis
$A^{\prime}C^{\prime}$.

###  4.2 Transformations with initial and final states in 3-D

We now give a representative example where the initial and final
configurations do not lie in the same plane, as shown in figure 12. Because
$\overline{AB}\perp\overline{AA^{\prime}}$ and
$\overline{BC}\perp\overline{CC^{\prime}}$, neither $A$ nor $C$ will rotate
about $B$ as part of the transformation. Nor can $ABC$ simultaneously
translate directly to $A^{\prime}B^{\prime}C^{\prime}$, because for example
quadrilateral $\Box AA^{\prime}B^{\prime}B$ does not satisfy the rule of
opposite angles $\geq\pi/2$, so link $AB$ cannot slide (translate) to
$A^{\prime}B^{\prime}$.

This leaves $3$ options for the initial stages of the transformation:
1.) $A$ translates, $B$ rotates, $C$ remains fixed. $B$ then rotates about $C$
in the $CBB^{\prime}$ plane. The initial direction of motion of $B$ is then
$\hat{{\bf v}}_{\mbox{\tiny B}}=(-\hat{{\bf i}}+\hat{{\bf k}})/\sqrt{2}$,
however then $\hat{{\bf v}}_{\mbox{\tiny A}}$ can only move backward to
preserve link length ($\hat{{\bf v}}_{\mbox{\tiny A}}=-\hat{{\bf k}}$),
similar to figure 15. This rules out case (1).
2.) $A$ remains fixed, $B$ rotates, $C$ remains fixed. $B$ then rotates
towards $B^{\prime}$ about axis $\overline{AC}$ until it reaches a critical
angle where line $\overline{B^{\prime\prime}B^{\prime}}$ is tangent to its
circular trajectory (see fig. 12a). At this point the quadrilateral $\Box
B^{\prime\prime}CC^{\prime}B^{\prime}$ does not have opposite obtuse angles,
so a straight line transformation to $A^{\prime}B^{\prime}C^{\prime}$ is not
possible. It is possible to transform to a configuration
$A^{\prime}B^{\prime}C^{\prime\prime}$, where $C^{\prime\prime}$ is at
position $(1,1,1)$ and angle $\angle B^{\prime}C^{\prime\prime}C=\pi/2$, so
that $\hat{{\bf v}}_{\mbox{\tiny C}}=\hat{{\bf k}}$. Then the transformation
is completed by a $\pi/2$ rotation of $C^{\prime\prime}$ about $B^{\prime}$.
This transformation is subminimal.
3.) $A$ remains fixed, $B$ rotates, $C$ translates. In this case, $B$ rotates
toward $B^{\prime}$ in the $BAB^{\prime}$ plane, while $C$ translates to
$C^{\prime}$, until the state $AB^{\prime\prime}C^{\prime\prime}$ is reached
(see fig. 12b). State $AB^{\prime\prime}C^{\prime\prime}$ can be found as
follows. Because the rotation of $B$ is about the axis
$(0,-1/\sqrt{2},1/\sqrt{2})$, the position
$\overrightarrow{AB^{\prime\prime}}$ of $B^{\prime\prime}$ after rotation of
the (critical) angle $\theta$ is
$(\cos\theta,\sin\theta/\sqrt{2},\sin\theta/\sqrt{2})$. This angle is then
determined by the condition
$\overrightarrow{AB^{\prime\prime}}\cdot\overrightarrow{B^{\prime\prime}B^{\prime}}=0$,
where
$\overrightarrow{B^{\prime\prime}B^{\prime}}=\overrightarrow{AB^{\prime}}-\overrightarrow{AB^{\prime\prime}}$.
The solution to this condition is simply $\theta=\pi/4$. The location of
$C^{\prime\prime}$ is then determined from the condition that the link length
from $B^{\prime\prime}$ to $C^{\prime\prime}$ is one:
$|\overrightarrow{B^{\prime\prime}C^{\prime\prime}}|=1$, where
$\overrightarrow{B^{\prime\prime}C^{\prime\prime}}=\overrightarrow{AB^{\prime\prime}}+t\overrightarrow{CC^{\prime}}$.
Solving this condition for $t$ gives the position of $C^{\prime\prime}$ as
$({3+\sqrt{2}\over 5},1,{2(2-\sqrt{2})\over 5})$. At this point the
quadrilateral $\Box B^{\prime}B^{\prime\prime}C^{\prime\prime}C^{\prime}$ has
opposite obtuse angles, and quadrilateral $\Box
AB^{\prime\prime}B^{\prime}A^{\prime}$ has opposite angles $=\pi/2$, so it is
in a bowtie configuration as in the end point configurations in figure 3.1.
Therefore all points $AB^{\prime\prime}C^{\prime\prime}$ can translate from
this intermediate state to their final positions
$A^{\prime}B^{\prime}C^{\prime}$. The total distance travelled is
$\theta+|AA^{\prime}|+|CC^{\prime}|+|B^{\prime\prime}B^{\prime}|$ or ${\cal
D}=2+\pi/4+\sqrt{5}\approx 5.022$. The reverse of this transformation is also
possible, where point $B^{\prime}$ rotates about $A^{\prime}$ in the plane
$B^{\prime}AB$, while $C^{\prime}$ translates along
$\overrightarrow{C^{\prime}C}$. Inspection reveals the distance covered is the
same as the forward transformation.

|
---|---
a | b

Figure 12: (a) Subminimal transformation and (b) minimal transformations
between $ABC$ and $A^{\prime}B^{\prime}C^{\prime}$ (see text)

.

##  5 Limit of large link number

From the transformation discussed in section 4.1, we see that if both $\angle
ABC$ and $\angle A^{\prime}B^{\prime}C^{\prime}$ were $\pi/2$ as in figure
13a, then the transformations in figures 11a and 11b become degenerate, having
distance ${\cal D}=\pi/\sqrt{2}$. The transformation is completed by a single
rotation about axis $\overline{13}$.

|
---|---
a | b
|
c | d
Figure 13: Examples of transformations between initial and final states of
opposite convexity, for increasing numbers of links. (a) illustrates the
transformation for $N=2$ links. (b) $N=4$ and initial and final state form an
octagon. (c,d) $N=6$ and initial and final states form a dodecagon. (c) top
view. (d) view in perspective. Rotations are shown as solid color lines
(either green or blue). Translations are shown as dashed lines. The grey
dashed lines underneath $\overline{3^{\prime\prime}3^{\prime}}$ in (b) and
$\overline{4^{\prime\prime}4^{\prime}}$ in (d) are shown only to illustrate
that those lines are above the plane.

We can now examine the effect of increasing the link number. Let the number of
links increase to $4$, and let us preserve the symmetry that is present about
the horizontal axis in fig 13a, so the initial and final states become an
octagon (figure 13b). In the limit $N\rightarrow\infty$, the figure becomes a
circle.

If we separated the links in figure 13a by some distance in the $y$ direction
(perpendicular to axis $\overline{13}$), then the minimal transformation
involves the same rotation of $2$ about axis $\overline{13}$ until a critical
angle $\theta_{c}$, after which all three points $123$ can translate in
straight lines to $1^{\prime}2^{\prime}3^{\prime}$. In the same fashion, the
minimal transformation for the octagonal transformation in fig 13b involves a
rotation of point $3$ out of the plane about axis $\overline{24}$ to a
critical angle $\theta_{c}$ at which the point is located at position
$3^{\prime\prime}$. Once this critical angle is reached, point $3$ translates
in a straight line from $3^{\prime\prime}$ to $3^{\prime}$.

Because points $1$ and $5$ are stationary to satisfy corner conditions, points
$2$ and $4$ must move in great circles about points $1$ and $5$. However
points $2$ and $4$ cannot finish the transformation by moving on great
circles. At the configuration
$1^{\prime}2^{\prime\prime}3^{\prime}4^{\prime\prime}5^{\prime}$ in figure
13b, point $3$ has finished the transformation, but points $2$ and $4$ have
not. To satisfy corner conditions at the points $2^{\prime\prime}$ and
$4^{\prime\prime}$, the great circles must be out of plane as well. At points
$2^{\prime\prime}$ and $4^{\prime\prime}$, the transformation finishes with
rotations about axes $\overline{1^{\prime}3^{\prime}}$ and
$\overline{3^{\prime}5^{\prime}}$. The total distance ${\cal D}\approx 7.93$.

Of course the time reverse of this transformation (equivalent to swapping
primed and unprimed labels) is also a minimal transformation, as is the
transformation obtained by reflection about the $z=0$ plane.

Now consider increasing the chain to $6$ links, so the combination of ${\bf
r}_{i}(0)$ and ${\bf r}_{i}(T)$ becomes a dodecagon (12-sided polygon, see
figures 13c-d). As before the midpoint vertex (here ${\bf r}_{4}$) must rotate
out of the plane about axis $\overline{35}$ to a critical angle $\theta_{c}$
before translating in a straight line to ${\bf r}_{4^{\prime}}$. This critical
angle is where
$\overrightarrow{34^{\prime\prime}}\cdot\overrightarrow{4^{\prime\prime}4^{\prime}}=\overrightarrow{54^{\prime\prime}}\cdot\overrightarrow{4^{\prime\prime}4^{\prime}}=0$.
The quadrilaterals $\Box 22^{\prime}3^{\prime}3$ and $\Box
655^{\prime}6^{\prime}$ are of the type in figure 5, so point $3$ must rotate
about ${\bf r}_{2}(0)$ to a critical angle where
$\overrightarrow{23^{\prime\prime}}\cdot\overrightarrow{3^{\prime\prime}3^{\prime}}=0$,
and likewise for point $5$.

While point $3$ rotates to its critical angle, point $4$ translates along line
$\overline{4^{\prime\prime}4^{\prime}}$. Points ${\bf r}_{1}(0)$ and ${\bf
r}_{7}(0)$ overlap with ${\bf r}_{1}(T)$ and ${\bf r}_{7}(T)$ and so remain
fixed to satisfy corner conditions. After point $3$ has reached its critical
angle, it can translate along $\overline{3^{\prime\prime}3^{\prime}}$ as point
$2$ rotates about ${\bf r}_{1}$. However to satisfy corner conditions at point
$2^{\prime\prime}$, the rotation cannot remain in the $x-y$ plane. Point ${\bf
r}_{2^{\prime\prime}}$ is determined as the point where $\hat{{\bf
t}}\cdot{\bf n}_{plane}=0$, where $\hat{{\bf t}}$ is the tangent to the arc
$\stackrel{{\scriptstyle\frown}}{{22^{\prime\prime}}}$ defined by rotation
about axis $\overline{13^{\prime}}$, and ${\bf n}_{plane}$ is the normal to
the plane $122^{\prime\prime}$, i.e. ${\bf r}_{2/1}\times{\bf
r}_{2^{\prime\prime}/1}$. The same process holds for point $6$. These critical
points and some intermediate states for the transformation are shown in figure
13d. The total distance covered by the transformation is ${\cal D}\approx
16.3$.

It is sensible to consider the total length of chain as fixed to say $L=1$,
and to let the link length $ds_{\mbox{\tiny N}}$ for the chain of $N$ links be
determined by $Nds_{\mbox{\tiny N}}=L$. Because distances scale as
$ds_{\mbox{\tiny N}}^{2}$, the $N=2,4,6$ cases have ${\cal D}_{2}\approx
0.555L^{2}$, ${\cal D}_{4}\approx 0.496L^{2}$, ${\cal D}_{6}\approx
0.445L^{2}$. Note that this distance decreases with increasing number of
links: the constraints on the motion of the various beads during the
transformation are relaxed as the number of links is increased.

We can then imagine resting a piece of string on a table in the shape of a
semi-circular arc, and then asking how one can move this string to a facing
semicircle of opposite convexity. So long as the string has some non-zero
persistence length $\ell_{\mbox{\tiny P}}$, the transformation of minimal
distance must involve lifting the string off of the table to change its local
convexity. The vertical height the string must be lifted (see fig 13d) is of
order $\sim\sin(\pi\ell_{\mbox{\tiny P}}/L)\sim\ell_{\mbox{\tiny P}}/L$, which
goes to zero for an infinitely long chain.

As the number of links $N\rightarrow\infty$, some simplifications emerge. In
particular the contribution to the total distance due to rotations becomes
negligible, and the translational component dominates. To see this note that
the distance due to straight line motion scales as:

${\cal D}(\mbox{st. line})\sim ds\,NL\sim L^{2}$

while the distance travelled during rotations scales as

${\cal D}(\mbox{rot.})\sim ds\,N(\overline{\theta_{c}}ds)\sim L^{2}/N$

where we assume the worst case scenario where an extensive number of links
must rotate before translating. Because translation dominates the distance as
$N\rightarrow\infty$, the distance travelled converges to $L$ times the mean
root square distance (MRSD), i.e.

$\displaystyle{\cal D}_{\infty}$ $\displaystyle\rightarrow
ds\,\sum_{i=1}^{N+1}\left|{\bf r}_{i}(T)-{\bf r}_{i}(0)\right|$
$\displaystyle=L{1\over N}\sum_{i}\sqrt{\left({\bf r}_{\mbox{\tiny{B}}i}-{\bf
r}_{\mbox{\tiny{A}}i}\right)^{2}}$ $\displaystyle=L\left(\mbox{MRSD}\right)$
(42)

The MRSD for the examples in figures 13b,d are $0.394\,L$ and $0.400\,L$
respectively, which are both less than the actual distances travelled (in
units of $L$). In the limit $N\rightarrow\infty$, where the polygon becomes a
circle, the distance converges to ${\cal D}_{\infty}=4L^{2}/\pi^{2}\approx
0.4053L^{2}$. For large $N$ systems then, it is a good first approximation to
use $MRSD$ for the distance.

The MRSD is always less than the root mean square distance (RMSD), except in
special cases when they are equal. To see this, we can apply Hölder’s
inequality

$\sum_{k=1}^{N}\left(g_{k}\right)^{\alpha}\,\left(h_{k}\right)^{\beta}\leq\left(\sum_{k=1}^{N}g_{k}\right)^{\alpha}\left(\sum_{k=1}^{N}h_{k}\right)^{\beta}$

where $g_{k},h_{k}\geq 0$, $\alpha,\beta\geq 0$, and $\alpha+\beta=1$. With
the specific identifications $g_{k}=({\bf r}_{\mbox{\tiny{B}}k}-{\bf
r}_{\mbox{\tiny{A}}k})^{2}\equiv\Delta{\bf r}_{k}^{2}$, $h_{k}=1$, and
$\alpha=\beta=1/2$, we have directly

${1\over N}\sum_{k}\sqrt{\Delta{\bf r}_{k}^{2}}\>\leq\sqrt{{1\over
N}\sum_{k}\Delta{\bf r}_{k}^{2}}$

For example the RMSD for the circle configuration discussed above is
$\sqrt{2}L/\pi\approx 0.4502L$, which is greater than the MRSD.

The fact that the distance converges for large $N$ to MRSD rather than RMSD
suggests that RMSD may not be the best metric for determining similarity
between molecular structures, although it is ubiquitously used. This fact
warrants future investigation- it has implications in research areas from
structural alignment based pharmacophore identification [3, 4, 5] to protein
structure and function prediction [6, 7].

It was shown in [1] that chains with persistence length characterized by some
radius of curvature $R$ have extensive corrections to the MRSD-derived minimal
distance, which do not vanish as $N\rightarrow\infty$, but remain so long as
$R/L$ is nonzero. Likewise, chains that cannot cross themselves have nonlocal
EL equations and extensive corrections to the minimal distance. Nevertheless,
it is worthwhile to investigate some more complex polymers with MRSD as an
approximate distance metric. We pursue this in the next section.

###  5.1 MRSD as a metric for protein folding

Here we examine the use of MRSD as a metric or order parameter for protein
folding. To this end we adopt an unfrustrated $C_{\alpha}$ model of segment
$84-140$ of src tyrosine-protein kinease (src-SH3), by applying a Gō-like
Hamiltonian [8, 9, 10] to an off–lattice coarse-grained representation of the
src-SH3 native structure (pdb 1fmk). Amino acids are represented as single
beads centered at their $C_{\alpha}$ positions. The Gō-like energy of a
protein configuration $\alpha$ is given by the following Hamiltonian, which we
will explain term by term:

$\displaystyle\mathcal{H}(\alpha|N)$ $\displaystyle=$ $\displaystyle
k_{r}\sum_{bonds}\left(r_{\alpha}-r_{N}\right)^{2}+k_{\theta}\sum_{triples}\left(\theta_{\alpha}-\theta_{N}\right)^{2}$
(43) $\displaystyle+$
$\displaystyle\sum_{n=1,3}k_{\phi}^{\left(n\right)}\sum_{quads}\left[1-\cos\left(n\times(\phi_{\alpha}-\phi_{N})\right)\right]$
$\displaystyle+$ $\displaystyle\epsilon_{\mbox{\tiny N}}\sum_{j\geq
i+3}\left[6\left(\frac{\sigma_{ij}}{r_{ij}}\right)^{10}-5\left(\frac{\sigma_{ij}}{r_{ij}}\right)^{12}\right]+\epsilon_{\mbox{\tiny
NN}}\sum_{j\geq i+3}\left(\frac{\sigma_{ij}}{r_{ij}}\right)^{12}\>.$

Adjacent beads are strung together into a polymer through harmonic bond
interactions that preserve native bond distances between consecutive
$C_{\alpha}$ residues. Here $r_{\alpha}$ and $r_{N}$ represent the distances
between two subsequent residues in configurations $\alpha$ and the native
state $N$. As with other parameters in the Hamiltonian, the distances $r_{N}$
are based on the pdb structure and may vary pair to pair. The angles
$\theta_{N}$ represent the angles formed by three subsequent $C_{\alpha}$
residues in the pdb structure, and the angles $\phi_{N}$ represent the
dihedral angles defined by four subsequent residues. The dihedral potential
consists of a sum of two terms, one with period $2\pi$ and another with
$2\pi/3$, which give cis and trans conformations for angles between successive
planes of three amino acids, with a global dihedral potential minimum at
$\phi_{N}\in[-\pi,\pi]$.

The parameters $k_{r}$, $k_{\theta}$, and $k_{\phi}$, are taken to accurately
describe the energetics of the protein backbone: we used the values
$k_{r}=50\>\mbox{kcal/mol}$, $k_{\theta}=20\>\mbox{kcal/mol}$,
$k_{\phi}^{(1)}=1\>\mbox{kcal/mol}$ and $k_{\phi}^{(3)}=0.5\>\mbox{kcal/mol}$
for molecular dynamics (MD) simulations using the AMBER software package. For
MD simulations using LAMMPS, we had used slightly different values:
$k_{r}=80\>\mbox{kcal/mol}$, $k_{\theta}=16\>\mbox{kcal/mol}$,
$k_{\phi}^{(1)}=0.8\>\mbox{kcal/mol}$ and
$k_{\phi}^{(3)}=0.4\>\mbox{kcal/mol}$.

The last line in equation (43) deals with non-local interactions, both native
and non-native. If two amino acids are separated by $3$ more along the chain
($|i-j|\geq 3$), and have one or more pairs of heavy atoms within a cut-off
distance of $r_{c}=4.8$ Å in the pdb structure, the amino acids are said to
have a native contact. Then the respective coarse-grained $C_{\alpha}$
residues are given a Lennard-Jones-like 10-12 potential of depth
$\epsilon_{\mbox{\tiny N}}=-0.6\>\mbox{kcal/mol}$ ($-0.8\>\mbox{kcal/mol}$ for
LAMMPS simulations) and a position of the potential minimum equal to the
distance of the $C_{\alpha}$ atoms in the pdb structure. That is,
$\sigma_{ij}$ is taken equal to native distance between $C_{\alpha}$ residues
$i$ and $j$ if $i$–$j$ have a native contact.

If two amino acids are not in contact, their respective $C_{\alpha}$ residues
sterically repel each other ($\epsilon_{\mbox{\tiny
NN}}=+0.6\>\mbox{kcal/mol}$). Thus $\epsilon_{\mbox{\tiny NN}}=0$ if $i$-$j$
is a native residue pair, while $\epsilon_{\mbox{\tiny N}}=0$ if $i$-$j$ is a
non-native pair. For non-native residue pairs,
$\sigma_{ij}=4\>\mbox{Angstroms}$.

In an arbitrary configuration $\alpha$, two $C_{\alpha}$ residues $i$ and $j$
are considered to have formed a native contact if they have a distance
$r_{ij}\leq 1.2\sigma_{ij}$. The results do not strongly depend on the
specific value of this cutoff. The fraction of native contacts present in the
particular configuration $\alpha$ is then defined as $Q$ (or $Q_{\alpha}$).

The MRSD of configuration $\alpha$ is found by aligning this configuration to
the native structure, by minimizing MRSD over $3$ translational and $3$
rotational degrees of freedom.

Constant temperature molecular dynamics simulations were run for this system
using both AMBER and LAMMPS simulation packages. The probability for the
system to have given values of $Q$ and $MRSD$ within $(Q,Q+\Delta Q)$ and
$(MRSD,MRSD+\Delta MRSD)$ is proportional to the exponential of the free
energy $F(Q,MRSD)$. Thus the free energy can be directly obtained by sampling,
binning, and taking the logarithm:

$\displaystyle F(Q_{1},MRSD_{1})-F(Q_{2},MRSD_{2})=-k_{\mbox{\tiny
B}}T\log\left({p(Q_{1},MRSD_{1})\over p(Q_{2},MRSD_{2})}\right)$ (44)

with $F(1,0)=E_{\mbox{\tiny N}}$, the energy of the native structure.

Figure 14 shows the free energy surfaces obtained using the above recipe, for
the AMBER (fig 14a) and LAMMPS (fig 14b) molecular dynamics routines. The
temperature is taken to be the transition or folding temperature
$T_{\mbox{\tiny F}}$, where the unfolded and folded free energies are equal.

Notice that $F(Q)$ is comparable for both as it should be, moreover $F(MRSD)$
is as well. However the free energy surface plotted as a function of both $Q$
and $MRSD$ shows a marked difference. In addition to a native minimum, the
LAMMPS routine has an additional minimum at $Q\approx 0.95$ and $MRSD\approx
8.4$. The conformational states in this bin are closely related, with an
average MRSD between them of $1.8$Å. We can take the most representative state
in this bin as that which has a minimum MRSD from all the others in the bin
(at $Q\approx.95$, $MRSD\approx 8.4$):
$\leavevmode{\raisebox{-4.30554pt}{$\stackrel{{\scriptstyle\mbox{min}}}{{\mbox{\scriptsize{i}}}}$}}\left(\sum_{j\neq
i}^{{}^{\prime}}MRSD_{ij}/\sum_{j\neq i}^{{}^{\prime}}\right)\approx 1.6$Å.
Inspection reveals that this state is a mirror image of the pdb structure (see
fig 14b): If we reflect this structure about one plane, and subsequently align
this reflected structure to the pdb one, the MRSD is only $1.1$Å.

The discrepancy in free energy surfaces corresponding to the presence of a low
energy mirror-image structure arises because the COMPASS class 2 dihedral
potentials in the LAMMPS algorithm do not ascribe a sign to the angle $\phi$,
so the full range $[-\pi,\pi]$, is projected onto $[0,\pi]$. This gives the
set of actual dihedral angles $\\{\phi_{i}+\pi\\}$ the same energy as the set
$\\{\phi_{i}\\}$, so that the dihedral potentials have two minima rather than
one, and thus a protein chain of the opposite chirality (a mirror image) is
allowed and has the same energy as the pdb structure. We found that the CHARMM
and harmonic dihedral styles do not have this problem, however they have less
versatile function forms, so that we favored modifying the COMPASS dihedrals
to define $\phi$ over its full range.

a
---
b
Figure 14: Free energy surfaces for the folding of Gō-model src-SH3 using two
molecular dynamics simulation packages, AMBER (a) and LAMMPS (b). The contour
plots give $F(Q,MRSD)$. The projections $F(Q)$ and $F(MRSD)$ are also shown on
each side. The COMPASS class 2 dihedral potential in LAMMPS allows for a
mirror image of the folded structure (red color structure in inset) that is
not immediately evident from the $F(Q)$ or $F(MRSD)$ surfaces. Future
implementations of LAMMPS using COMPASS dihedrals for biomolecular simulations
must then correct for dihedral angles defined on the interval $[-\pi,\pi]$.

##  6 Conclusions

Analogously to the distance between two points, the distance between two
finite length space curves is a variational problem, and may be calculated by
minimizing a functional of $2$ independent variables $s$ and $t$, where $s$ is
the arc-length along the chain, and $t$ is the ’elapsed time’ during the
transformation.

We derived the Euler-Lagrange (EL) equation giving the solution to this
problem, which is a vector partial differential equation, with extremal
solution ${\bf r}^{\ast}(s,t)$. We also derived the sufficient conditions for
the extremal solution to be a minimum, through the Jacobi equation. Once the
minimal transformation ${\bf r}^{\ast}(s,t)$ is known, the distance ${\cal
D}^{\ast}\equiv{\cal D}[{\bf r}^{\ast}]$ follows.

We provided a general recipe for the solution to the EL equation using the
method of lines. The resulting $N+1$ EL equations for the discretized chain
are ODEs that can be interpreted geometrically and solved for minimal
solutions. Solutions consist generally of rotations and translations pieced
together so the direction of velocity of any link end point does not suddenly
change (the Weierstrass-Erdmann corner conditions).

We explored the minimal transformations for the simplest polymers, consisting
of $1$ or $2$ links, in depth. For transformations between $2$ links,
convexity becomes an issue (the analog to the direction of the radius of
curvature for a continuous string). For example, even if the initial and final
states lie in the same plane, if the convexities of these states are of
opposite sign the transformation must pass through intermediate states that
are out of the plane. Similarly, given a semicircular piece of string lying on
a table, to move it to a semicircle of opposite convexity using the minimal
amount of motion, the string must be lifted off the table.

The study of minimal transformations between small numbers of links has
applications to the inverse kinematic problem in robotics and movement
control. In the inverse kinematic problem, one is given the initial and final
positions of the end-effector (the hand of the robot), and asked for the
functional form of the joint variables for all intermediate states. Generally
there is no unique solution until some optimization functional is introduced,
such as minimizing the time rate of change of acceleration (the jerk), torque,
or muscle tension (see the review [11] and references therein). The minimal
distance transformation would be relevant if one sought the fastest
transformation between initial and final states, without explicit regard to
mechanical limitations. The indeterminate intermediate points can be handled
variationally as a free boundary value problem.

In the limit of a large number of links, some simplifications emerge. For
chains without curvature or non-crossing constraints, the distance converges
to $L$ times the mean root square distance ($MRSD$) of the initial and final
conformations. So for example the distance between 2 strings of length $L$
forming the top and bottom halves of a circle respectively is
$4L^{2}/\pi^{2}$, the distance between horizontal and vertical straight lines
of length $L$ which touch at one end is $L^{2}/\sqrt{2}$, and the distance to
fold a straight line upon itself (to form a hairpin) is $L^{2}/4$.

The fact that for large $N$ the distance (over $L$) converges to MRSD rather
than RMSD suggests that RMSD may not be the best metric for determining
similarity between molecular structures, although it is ubiquitously used.
Adopting MRSD may lead to improvements in structural alignment algorithms.

The MRSD was investigated as an approximate metric for protein folding. Free
energy surfaces for folding were constructed for two simulation packages,
AMBER and LAMMPS. It was found that including MRSD as an order parameter
uncovered discrepancies between the two molecular dynamics algorithms. Because
dihedral angles in LAMMPS (at least in COMPASS class 2 style) are only defined
on $[0,\pi]$, the potential admits a mirror image structure degenerate in
energy with the native structure. This is easily remedied and should not be
interpreted as a deficiency in the LAMMPS simulation package so long as one is
aware of it. It should be mentioned that the mirror-image structure would also
have been seen had RMSD been used as an additional order parameter.

It will be important for future studies to address the effects of persistence
length and non-crossing on the distance between biopolymer conformations [1].
Also important is the role of entropy of paths or transformations in
describing the accessibility of a particular biomolecular structure. Along
these lines it will be interesting to investigate whether the distance can be
a predictor of folding kinetics, or proximity to the native structure.

It is also an interesting question to ask whether the actual dynamics between
polymer configuraitons resembles the minimal transformation, after a suitable
averaging over trajectories. This question is linked with the role of the
entropy of transformations described above. It is also related to the problem
of finding the dominant pathway for a chemical reaction [12], which has
recently been applied to the problem of protein folding [13]. We have focused
here on the question of geometrical distance for complex systems, which can be
separated from the calculation of quantities such as reaction paths that
depend intrinsically on energetics, i.e. on the specific Hamiltonian of the
system. Quantifying the relationship between geometrical distance and the
dominant reaction path is an interesting future question worthy of
investigation.

The notion of distance and corresponding optimal transformation for a system
with many degrees of freedom is fundamental to a diverse array of research
subjects. Hence we saw potential applications for this metric in areas ranging
from drug design to robotics. It is not clear at present how useful the
calculation of the true Euclidean distance between high-dimensional objects
will be for practical applications, but we are optimistic.

##  7 Acknowledgements

We are thankful to Shirin Hadizadeh, Mike Prentiss, and Peter Wolynes for
helpful discussions. S.S.P. gratefully acknowledges support from the Natural
Sciences and Engineering Research Council and the A. P. Sloan Foundation.

##  Appendix A Necessary conditions for straight line transformations

It was shown in section 3.1 that to have straight line transformations between
links, it is sufficient to have facing obtuse angles on opposite sides of the
the quadrilateral defined by the transformation as shown in figure 3A. We now
show that it is a necessary condition as well, i.e. we show that a slide in
the correct direction is not possible in the absence of obtuse angles.

Figure 15:

Without loss of generality assume that the link is initially along the z axis.
The paths travelled by the link ends are shown in the figure. Note that the
end point trajectories of $A$ and $B$ are in 3D space so the paths travelled
by $A$ and $B$ need not cross or lie in the same plane. Let the unit vector
along $A$’s path be $\hat{{\bf v}}_{\mbox{\tiny A}}$ and the unit vector along
$B$’s path be $\hat{{\bf v}}_{\mbox{\tiny B}}$. Because the angles that the
path of A and the path of B make with the link are acute, the z-component of
$\hat{{\bf v}}_{\mbox{\tiny B}}$ ($\equiv z_{\mbox{\tiny{B}}}$) is negative
and the z-component of $\hat{{\bf v}}_{\mbox{\tiny A}}$
($z_{\mbox{\tiny{A}}}$) is positive. One can write $\hat{{\bf v}}_{\mbox{\tiny
A}}$ and $\hat{{\bf v}}_{\mbox{\tiny B}}$ as

$\displaystyle\hat{{\bf v}}_{\mbox{\tiny A}}$ $\displaystyle=$
$\displaystyle\boldsymbol{\rho}_{\mbox{\tiny A}}+z_{\mbox{\tiny{A}}}\hat{{\bf
z}}$ $\displaystyle\hat{{\bf v}}_{\mbox{\tiny B}}$ $\displaystyle=$
$\displaystyle\boldsymbol{\rho}_{\mbox{\tiny B}}+z_{\mbox{\tiny{B}}}\hat{{\bf
z}}$

where $\boldsymbol{\rho}_{\mbox{\tiny A}}$ and $\boldsymbol{\rho}_{\mbox{\tiny
B}}$ are vectors in xy plane and $z_{\mbox{\tiny{A}}}>0$ and
$z_{\mbox{\tiny{B}}}<0$.

Let ${\bf r}_{\mbox{\tiny{A}}}(t)$ and ${\bf r}_{\mbox{\tiny{B}}}(t)$ denote
the positions of the A and B ends at time $t$:

$\displaystyle{\bf r}_{\mbox{\tiny{A}}}$ $\displaystyle=$ $\displaystyle
t\hat{{\bf v}}_{\mbox{\tiny A}}$ $\displaystyle{\bf r}_{\mbox{\tiny{B}}}$
$\displaystyle=$ $\displaystyle g(t)\hat{{\bf v}}_{\mbox{\tiny B}}+\hat{{\bf
z}}$

The rigid link constraint dictates that

$({\bf r}_{\mbox{\tiny{A}}}-{\bf r}_{\mbox{\tiny{B}}})\cdot({\bf
r}_{\mbox{\tiny{A}}}-{\bf r}_{\mbox{\tiny{B}}})=1$

which translates to:

$g^{2}+2g\left(z_{\mbox{\tiny{B}}}-t\,(c+z_{\mbox{\tiny{A}}}\,z_{\mbox{\tiny{B}}})\right)-2tz_{\mbox{\tiny{A}}}+t^{2}+1=1$

with $c=\boldsymbol{\rho}_{\mbox{\tiny A}}\cdot\boldsymbol{\rho}_{\mbox{\tiny
B}}$. Solving for $g$ as a function of $t$, keeping in mind that $g(0)=0$:

$g(t)=-\left(z_{\mbox{\tiny{B}}}-t\,\left(c+z_{\mbox{\tiny{A}}}\,z_{\mbox{\tiny{B}}}\right)\right)+\sqrt{\left(z_{\mbox{\tiny{B}}}-t\,\left(c+z_{\mbox{\tiny{A}}}\,z_{\mbox{\tiny{B}}}\right)\right)^{2}-t^{2}+2tz_{\mbox{\tiny{A}}}}\>.$

Now if $g^{\prime}(t)>0$ it means that the B-end of the link is travelling in
the assumed direction, and if $g^{\prime}(t)<0$ it means that B-end is
travelling in the opposite direction (which means that the angle is not acute
anymore). Writing $g^{\prime}(0)$ we get:

$g^{\prime}(0)={{2\,z_{\mbox{\tiny{B}}}\,c+2\,z_{\mbox{\tiny{A}}}\,z_{\mbox{\tiny{B}}}^{2}-2\,z_{\mbox{\tiny{A}}}}\over{2\,\left|z_{\mbox{\tiny{B}}}\right|}}+c+z_{\mbox{\tiny{A}}}z_{\mbox{\tiny{B}}}={-z_{\mbox{\tiny{A}}}\over|z_{\mbox{\tiny{B}}}|}<0\>.$

Thus point $B$ can only travel in the opposite direction from what was
assumed, which in turn means an all-acute slide is not possible. We conclude
that the condition of “facing obtuse angles” is necessary and sufficient for
transformations consisting only of pure translations.

##  Appendix B Critical angles

The concept of critical angle was first introduced in 3.2. In order for a
straight-line slide of both ends to be possible, at some stage during the
transformation the link needs to rotate about one of the ends, with the other
end being stationary. In principle the rotation can be about either of the two
ends and it can happen at the beginning or the end of the transformation. The
conditions on the critical angle or orientation can be readily derived from
the broken extremal conditions. It was seen from 18a and 19, the non-trivial
corner conditions read:

$\hat{\bf v}_{i}|_{{}_{+}}=\hat{\bf v}_{i}|_{{}_{-}}\>.$ (B.45)

We know that the path travelled by the moving bead during the rotation is
circular and the path that is travelled during the slide part is a straight
line. Broken extremal condition forces these two paths to be patched smoothly,
which means that the straight-line path should be tangent to the circle. In
the 3D case, for the broken extremal condition to be satisfied, the straight
line slide path and the circular rotation path should lie in the same plane.
For example in figure 5 where $B$ is rotating about $A$ initially to $B_{1}$
and then slides to $B^{\prime}$, the rotation has to be in the plane formed by
the three points $ABB^{\prime}$.

Matching the directions of velocity as in (B.45) does not itself mean that a
link can subsequently slide in a straight line, however at the tangent point,
the tangent line to the circle is perpendicular to the radius, hence one
satisfies this second condition as well. Below we derive an analytical
expression for the critical angle for a particular case of single link
problem, as an example and illustration of the discussed concepts. Furthermore
the particular example will be used later in C to introduce minimal
transformations in $2$ dimensions.

Consider the single link action with the particular parametrization
$s=s(\theta)$, as discussed in section 3.2:

$\int(\sqrt{\dot{s}^{2}+1+2\dot{s}cos\theta}+\sqrt{\dot{s}^{2}})\ d\theta.$
(B.46)

where $s\equiv\overrightarrow{A(\theta)A}$ is the (signed) distance of $A$-end
from its initial position, and $\theta$ is the angle between the link and the
horizontal line (see figure 16).

The Euler Lagrange equation of motion reads:

${d\over
d\theta}({\dot{s}\over\sqrt{\dot{s}^{2}}}+{\dot{s}+cos\theta\over\sqrt{\dot{s}^{2}+1+2\dot{s}cos\theta}})=0$
(B.47)

We consider a transformation which is not (necessarily) a minimum:

$s=a\cos\theta-\sin\theta+b$ (B.48)

with $a$ and $b$ parameters to be determined.

Such a transformation in fact forces the two ends to travel on a straight line
(right from the beginning) , but the $A$ side may in fact retreat and then
move forward. We call such a transformation a “hyperextended transformation”.
A sample transformation of this kind is shown in figure 16. The parameters $a$
and $b$ in (B.48) can be tuned to meet the boundary conditions (see below).

Figure 16: Transformation in which both ends stay on a linear track

In fact it is seen that point $A$ on the link retreats backwards until it
reaches some critical angle, which is when link $\overline{AB}$ makes an angle
${\pi\over 2}$ with the straight line $\overline{BB^{\prime}}$ that point $B$
travels on. Subsequently $A$ then moves forward towards $A^{\prime}$.

Assume that $\theta$ runs from $\theta_{1}$ to $\theta_{2}$, where
$0<\theta_{2}<\pi/2$. For simplicity assume that both these angles are between
$0$ and ${\pi\over 2}$.

The boundary conditions dictate that:

$\displaystyle s(\theta_{1})$ $\displaystyle=$ $\displaystyle 0$ (B.49)
$\displaystyle s(\theta_{2})$ $\displaystyle=$ $\displaystyle l$ (B.50)

where $l$ is the distance between $A$ and $A^{\prime}$.

$a$ and $b$ can be explicitly solved to give:

$\displaystyle a$ $\displaystyle=$
$\displaystyle{{-\sin\theta_{2}+\sin\theta_{1}-l}\over{\cos\theta_{1}-\cos\theta_{2}}}$
(B.51) $\displaystyle b$ $\displaystyle=$
$\displaystyle-{{\cos\theta_{1}\,\left(-\sin\theta_{2}-l\right)+\sin\theta_{1}\,\cos\theta_{2}}\over{\cos\theta_{1}-\cos\theta_{2}}}$
(B.52)

For our purposes we only need to note that the critical angle occurs when
$\dot{s}\equiv{ds\over d\theta}$ becomes zero, that is when $A$ stops going
backward and starts moving forward:

$\dot{s}=-a\sin\theta-\cos\theta=0$ (B.53)

where $a$ is given in B.51.

We can now ask what should $\theta_{1}$ be so that there is no need for the
link to go backward, i.e. it moves forward from the beginning and the
transformation is monotonic. Equations (B.53) and (B.51) give:

$\cos\theta+{{-\sin\theta_{2}+\sin\theta-l}\over{\cos\theta-\cos\theta_{2}}}\sin\theta=0$
(B.54)

Figure 17: Geometric proof for critical angle condition

For pedagogical reasons we prove condition (B.54) using analytic geometry as
well. Looking at figure 17 we have the following:

$\displaystyle g^{2}+l_{1}^{2}$ $\displaystyle=$ $\displaystyle 1$ (B.55)
$\displaystyle g^{2}+l_{2}^{2}$ $\displaystyle=$ $\displaystyle a^{2}$ (B.56)
$\displaystyle{g\over a}$ $\displaystyle=$ $\displaystyle{1\over
l+l_{1}+l_{2}}$ (B.57)

We can solve $g=\sqrt{1-l_{1}^{2}}$ and $a=\sqrt{1-l_{1}^{2}+l_{2}^{2}}$ from
the first two equations and substitute in the third equation to give:

$l={\sqrt{1-l_{1}^{2}+l_{2}^{2}}\over\sqrt{1-l_{1}^{2}}}-l_{1}-l_{2}$ (B.58)

On the other hand based on our results for $g$ and $a$ we have:

$\displaystyle\sin\theta_{1}$ $\displaystyle=$
$\displaystyle{\sqrt{1-l_{1}^{2}}\over\sqrt{1-l_{1}^{2}+l_{2}^{2}}}$ (B.59)
$\displaystyle\cos\theta_{1}$ $\displaystyle=$
$\displaystyle{l_{2}\over\sqrt{1-l_{1}^{2}+l_{2}^{2}}}$ (B.60)
$\displaystyle\sin\theta_{2}$ $\displaystyle=$ $\displaystyle l_{1}$ (B.61)
$\displaystyle\cos\theta_{2}$ $\displaystyle=$
$\displaystyle\sqrt{1-l_{1}^{2}}$ (B.62)

Substitution of eqns (B.59-B.62) in equation (B.54) gives equation (B.58)
after some simplification.

For the particular case that we have discussed, the proposed transformation is
in fact a minimal solution if $\theta_{1}$ is greater than the critical angle,
because in that case a simple slide would be possible. If $\theta_{1}$ is less
than the critical angle a locally minimum solution as we know is pure rotation
to the critical angle and then straight line slide. Pure rotation has a nice
geometric interpretation in our parametrization. it corresponds to the null
solution $s=0$. Since at the critical angle $\dot{s}=0$ we see that $s=0$ will
be smoothly patched with $s=a\cos\theta-sin\theta+b$, as mandated by the
corner conditions in equation (18a).

Figure 18: A minimal transformation in $s(\theta)$ parametrization. The
horizontal segment corresponds to pure rotation and the curved section
corresponds to slide on straight paths. Here the corner conditions demand that
the derivative $\dot{s}$ be continuous at the critical angle.

##  Appendix C Minimal transformations in 2 dimensions

It was seen in section 4.1 that for the case of two links when one is confined
to moving in a plane, satisfying the constant link length constraints and
corner conditions do not seem to lead to solutions which are extremal. However
given the additional constraint that the links must lie in a plane, there must
be one or a set of minimal transformations. We need to look at other forms of
transformations, namely compound straight line transformations. We will
elaborate on the idea starting with single links.

The hyper extended solution that was discussed previously in B can be
considered as a very special example of compound straight line transformation.
These are transformations that are made strictly from straight line paths with
no pure rotation. A more general transformation is shown figure C beside the
old transformation.

Figure 19: The previous hyper extended solution is shown along with a more
general compound straight-line transformation, where
$\overrightarrow{AA^{\prime\prime}}$ travels in some general direction. Length
of each line segment is written beside it. For the hyper extended solution the
value of $AA^{\prime\prime}$ is multiplied by two because the path is
travelled twice.

Note that the corners do not technically violate the corner conditions because
the speed of “A” bead is zero at the corner point in any parametrization that
can simultaneously describe $A$ motion and $B$ motion: Since at the corner
point, the link makes an angle of 90 degrees with the path that B travels, the
speed of B at the critical angle in infinitely larger than the speed of A. In
fact one sees that we have an instantaneous pure rotation about $A$-bead, when
it is at the corner point. $\hat{v}_{a}$ is not clearly defined at the
corners, and everywhere else (when the speed of the bead(s) is not zero), the
two beads are travelling on a straight line. The two solutions depicted in the
figure come from two different parametrization of the most general form of the
action and result in different distances. But each of them is a local minimum
once the direction of $\overrightarrow{AA^{\prime\prime}}$ is picked, and
these local minima have different values for the distance.

We can then ask about the best position to put the corner point, to minimize
the distance travelled in the compound straight line transformation, with
respect to other compound straight line transformations. We assume the corner
occurs on one side and we take it to be the “A” side.

Note that at the corner, the link makes a $90^{\circ}$ angle with the B-bead
path $\overline{BB^{\prime}}$, meaning that the distance from the corner point
to B path is always the length of the link, i.e. unity here. Also note that
the total distance that the “A”-bead travels is the distance from the initial
point $A$ to the corner point $A^{\prime\prime}$, plus the distance from
$A^{\prime\prime}$ to the final position $A^{\prime}$.

The locus of points with equal sum of distances from two points $A$ and
$A^{\prime}$ defines an ellipse with foci at $A$ and $A^{\prime}$. Moreover
the length of the major axis of the ellipse equals the sum of the distances
from the foci. Thus the smaller the major axis of the ellipse with foci $A$
and $A^{\prime}$, the smaller the total distance travelled by the “A”-bead.
Moreover $A^{\prime\prime}$ should sit on a line parallel to B-path at a
distance of $1$ from the B-path line $\overline{BB^{\prime}}$. So in seeking
the shortest distance travelled the $A$ end of the link, we seek the point
$A^{\prime\prime}$ such that it lies on an ellipse with foci $A$ and
$A^{\prime}$, the ellipse shares at least one point with a line parallel to
$\overline{BB^{\prime}}$ and distance $1$ away from it, and lastly that the
ellipse has the smallest possible major axis (see figure 20). So the ellipse
giving the minimal distance is tangent to the parallel line, and
$A^{\prime\prime}$ is the tangent point. This is illustrated in figure 20.

Figure 20: Optimal Compound Straight Line Transformation

This solution can be straightforwardly extended to 2 links, as depicted in
figure 21. Consider then the example in figure 13a, where the links are no
longer allowed to move out of the plane (see figure 22). Here ${\bf
r}_{A}={\bf r}_{A^{\prime}}$ and ${\bf r}_{C}={\bf r}_{C^{\prime}}$ and the
above ellipses turns into a circles centered at $A$ and $C$. The circles have
radii $1-1/\sqrt{2}$, so that the perpendicular distance from line
$\overline{BB^{\prime}}$ to the farthest point on the circle is $1$ and a
fully extended intermediate state is allowed.

Figure 21: An optimal compound Straight line solution for 2 link. For this
particular class of solutions, the problem is divided into to disjoint
problems (one for each link) and solved separately. Figure 22: Minimal
transformation restricted to 2 dimensions, for 2 links of opposite convexity
which form opposite sides of a square.

References

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