Mechanical property of the helical configuration for a twisted intrinsically straight biopolymer

Abstract

We explore the effects of two typical torques on the mechanical property of the helical configuration for an intrinsically straight filament or biopolymer either in three-dimensional space or on a cylinder. One torque is parallel to the direction of a uniaxial applied force, and is coupled to the cross section of the filament. We obtain some algebraic equations for the helical configuration and find that the boundary conditions are crucial. In three-dimensional space, we show that the extension is always a monotonic function of the applied force. On the other hand, for a filament confined on a cylinder, the twisting rigidity and torque coupled to the cross section are irrelevant in forming a helix if the filament is isotropic and under free boundary condition. However, the twisting rigidity and the torque coupled to the cross section become crucial when the Euler angle at two ends of the filament are fixed. Particularly, the extension of a helix can subject to a first-order transition so that in such a condition a biopolymer can act as a switch or sensor in some biological processes. We also present several phase diagrams to provide the conditions to form a helix.

Appendix: Realization of two applied torques

In this appendix, we show how to realize external torques to obtain Eqs. (6) and (10). Meanwhile, we explain why $N_zs$ in Eqs. (6) and (10) are in fact different.

Relations between coordinates in different frames

First, note that in general the energy should be defined in a global fixed frame but the Euler angles are defined in local frames. We set the origin of the global fixed frame at $s=0$, denote the relevant position vector to be $\mathbf{r}_\mathrm{g}=(x_\mathrm{g}, y_\mathrm{g}, z_\mathrm{g})$ and the locus of the centerline to be $\mathbf{r}_\mathrm{c}(s)$. The origin of the Euler frame, a local body frame, is at $\mathbf{r}_\mathrm{c}(s)$. We denote further a position vector in Euler frame as $\mathbf{r}_\mathrm{E}=(x_\mathrm{E}, y_\mathrm{E}, z_\mathrm{E})$ and define a local ‘fixed’ frame which has the same origin as that of the Euler frame but its three axes have the same orientations as those of the global fixed frame. In other words, ‘fixed’ means to fix the orientations of axes. Consequently, a position vector in the local ‘fixed’ frame is $\mathbf{r}_f=(x_\mathrm{f}, y_\mathrm{f}, z_\mathrm{f})={\varvec{\lambda }}'\cdot \mathbf{r}_\mathrm{E}$, where ${\varvec{\lambda }}'$ is the transfer matrix of the rotation matrix ${\varvec{\lambda }}$ with (Goldstein 2002)

$$\begin{aligned} \lambda _{11}(s)& = \cos \phi \cos \psi -\cos \theta \sin \phi \sin \psi , \nonumber \\ \lambda _{12}(s)& = \sin \phi \cos \psi +\cos \theta \cos \phi \sin \psi , \nonumber \\ \lambda _{13}(s)& = \sin \psi \sin \theta , \nonumber \\ \lambda _{21}(s)& = -\cos \phi \sin \psi -\cos \theta \sin \phi \cos \psi , \nonumber \\ \lambda _{22}(s)& = -\sin \phi \sin \psi +\cos \theta \cos \phi \cos \psi , \nonumber \\ \lambda _{23}(s)& = \cos \psi \sin \theta , \ \lambda _{31}(s)= \sin \theta \sin \phi , \nonumber \\ \lambda _{32}(s)& = -\sin \theta \cos \phi , \ \lambda _{33}(s)= \cos \theta . \end{aligned}$$

(50)

Clearly, for a general space point $\mathbf{r}_\mathrm{g}=\mathbf{r}_\mathrm{c}+\mathbf{r}_f$.

A torque along the axis of the filament

Applying a pair of forces $\mathbf{F}_\mathrm{E}^\pm =(0, \pm F, 0)$ at $\mathbf{r}_\mathrm{E}^\pm =(\pm l, 0, 0)$, respectively, these forces yield a torque$=N_3\mathbf{t}_3=2Fl\mathbf{t}_3$ in Euler frame. In global fixed frame, forces become $\mathbf{F}_{g}^\pm ={\varvec{\lambda }}'(L)\cdot \mathbf{F}_\mathrm{E}^\pm =\pm F(\lambda _{21}(L), \lambda _{22}(L), \lambda _{23}(L))$ and they act at $\mathbf{r}_\mathrm{g}(L)^\pm =\mathbf{r}_\mathrm{c}(L)+{\varvec{\lambda }}'(L)\cdot (\pm l, 0, 0)=\mathbf{r}_\mathrm{c}(L) \pm l(\lambda _{11}(L), \lambda _{12}(L), \lambda _{13}(L))$. When the filament undergoes a small deformation, the work done by these two forces is

$$\begin{aligned} \mathrm{d}W& = \mathbf{F}_{g}^+ \mathrm{d} \mathbf{r}_\mathrm{g}^+(L) +\mathbf{F}_{g}^- \mathrm{d} \mathbf{r}_\mathrm{g}^-(L) \nonumber \\& = N_3 (\lambda _{21}\mathrm{d}\lambda _{11}+\lambda _{22}\mathrm{d}\lambda _{12}+\lambda _{23}\mathrm{d}\lambda _{13})\nonumber \\& = N_3 [\cos \theta (L)\mathrm{d}\phi (L)+\mathrm{d}\psi (L)]. \end{aligned}$$

(51)

For a finite deformation via a path P, the corresponding energy is therefore

$$\begin{aligned} E_t& = -N_3 \int _P [\cos \theta (L)\mathrm{d}\phi (L)+\mathrm{d}\psi (L)]. \end{aligned}$$

(52)

Since the energy is path-independent, we can choose the centerline of the filament as integral path, which leads to

$$\begin{aligned} E_t=-N_3 \int _0^L \omega _3 \mathrm{d}s. \end{aligned}$$

(53)

${\mathcal {E}}_t=-N_3 \omega _3$ is just the last term in Eqs. (6) and (10). It should be not too difficult to realize such a torque. If we fix $\phi (L)$, it becomes ${\mathcal {E}}_t=-N_3{{\dot{\psi }}}$. Moreover, from rotational symmetry, any torque along $\mathbf{t}_3$ yields the same energy term.

A torque along the fixed z-axis in 3D space

Now applying a pair of forces $\mathbf{F}_\mathrm{g}^\pm =\pm F(-\sin \alpha , \cos \alpha , 0)$ at $\mathbf{r}_\mathrm{g}(L)^\pm =\mathbf{r}_\mathrm{c}(L)\pm l(\cos \alpha , \sin \alpha , 0)$, respectively, $\alpha$ is an angle around z-axis. It follows that the energy $\mathrm{d}E_t=-[\mathbf{F}_\mathrm{g}^+\mathrm{d} \mathbf{r}_\mathrm{g}(L)^+ +\mathbf{F}_\mathrm{g}^-\mathrm{d} \mathbf{r}_\mathrm{g}(L)^-]=N_z\mathrm{d}\alpha =2Fl\mathrm{d}\alpha$. Clearly, $\mathrm{d}\alpha =\mathrm{d}\phi (L)+\cos \theta (L)\mathrm{d}\psi (L)$ since $\phi (L)$ is also around z-axis but $\psi (L)$ is around $\mathbf{t}_3$ and $\cos \theta$ is the z-component of $\mathbf{t}_3$. Therefore, $\mathrm{d}E_t$ gives the third term in Eq. (6). It is also not too difficult to realize such a torque. With a fixed $\psi (L)$, it becomes ${\mathcal {E}}_t=-N_z{{\dot{\phi }}}$. From rotational symmetry, any torque along z-axis also yields the same term. However, it is very difficult to obtain such a torque from a single force because in this case the energy from $\mathbf{r}_\mathrm{c}(L)$ becomes very complicate.

A torque along the cylinder axis

The third term in Eq. (10) has been used (Allard and Rutenberg 2009; Zhou et al. 2017) but it lacks a proper explanation on how to realize such a torque and it may be confused with the third term in Eq. (6) so that we clarify this point here.

On a cylinder, we can write $\mathbf{r}_\mathrm{g}=\mathbf{r}_\perp +z\mathbf{e}_z$, where $\mathbf{r}_\perp =R (1-\cos \varphi )\mathbf{e}_x - R \sin \varphi \mathbf{e}_y$ and $\varphi$ is yet unknown. Now we are applying a force F at $\mathbf{r}_\mathrm{g}(L)$ and let F parallel to $\mathbf{t}_\perp =\sin \phi \ \mathbf{e}_x-\cos \phi \mathbf{e}_y$ or $\mathbf{F}=F(\sin \phi \ \mathbf{e}_x-\cos \phi \mathbf{e}_y)$. It follows that $d\mathbf{r}_\perp =R (\sin \varphi \mathbf{e}_x - \cos \varphi \mathbf{e}_y)\mathrm{d}\varphi$ is also parallel to $\mathbf{t}_\perp$ so that $\varphi =\phi$. The work done by this force is $\mathrm{d}W=\mathbf{F}\cdot \mathrm{d} \mathbf{r}_\mathrm{g}(L)=\mathbf{F}\cdot \mathrm{d} \mathbf{r}_\perp (L)= F R \mathrm{d}\phi$. Therefore, the related energy density is ${\mathcal {E}}_t=-N_z{{\dot{\phi }}}(L)$ with $N_z=FR$ and it recovers the third term in Eq. (10).

Finally, note that this torque results from a single force so that it should be easier to realize than the $N_z$ in Eq. (6) since it requires a pair of forces.