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Positional Uncertainty

5.1. Overview

Positional uncertainty presents a fundamental challenge in nanomechanical engineering, determining many design decisions and limiting the range of workable systems. Positional uncertainty in compliant systems is a result of both thermal motion and quantum mechanical uncertainty, but thermal motion proves more important in almost all mechanical systems at room temperature. This chapter describes positional uncertainty in a set of structures characterized by a PES with a single potential well.

Section 5.2 discusses the concept of positional uncertainty and its importance in engineering. Section 5.3 presents classical and quantum mechanical analyses of displacements in thermally excited harmonic oscillators; Section 5.4 and 5.5 present similar analyses of longitudinal and transverse displacements in rods. These studies of elastic systems provide the basis for describing positional uncertainty in the structures considered in later chapters. Indeed, for analyzing mechanical systems at room temperature, Eq. (5.4) usually provides an adequate approximation. In practice, the rest of the discussion in these sections chiefly serves to quantify the errors in using Eq. (5.4), to show why its range of applicability is greater than its derivation would suggest (Section 5.8.1), and to provide more elaborate expressions that remain accurate at lower temperatures.

The following two sections discuss systems in which positional uncertainty results from fluctuations in entropic springs. The prototype of an entropic spring is a loaded piston in a gas-filled cylinder, analyzed from three perspectives in Section 5.6. A flexible, tensioned rod in a housing that subjects it to transverse elastic restoring forces is a more complex system, but provides a good description of devices useful in nanomechanical signal transmission; this is analyzed in Section 5.7. These results on entropic springs, though useful in analytical work of the sort presented in Part II, are not applied to any of the specific systems described there.

5.2. Positional uncertainty in engineering

In the design of nanomechanical systems, positional uncertainties stemming from thermal excitation and quantum mechanical principles are a fundamental concern. On the scale of conventional mechanical engineering, neither quantum uncertainties nor thermal excitations are significant; the closest macroscale analogues of these effects arise in systems excited by broad-band acoustic noise, yet issues arise there (e.g., fatigue and damping) that are alien to the molecular domain. Molecules are not subject to fatigue (for a more precise discussion, see Section 6.7.1a). Damping, which degrades mechanical energy to heat, cannot degrade the vibrations of heat itself.

In an engineering context, problems involving positional uncertainty can frequently be formulated in terms of a probability density function (PDF) for a coordinate describing a part of a system. Mechanical designs typically require that each part should, under specified conditions, occupy a particular position at a particular time to within specified tolerances. Errors occur at some finite rate owing to the finite probability mass in the tail of the PDF that extends beyond the tolerance band. Good approximations to the positional PDFs of typical systems are thus of fundamental value in nanomechanical engineering.

5.3. Thermally excited harmonic oscillators

Many parts of nanomechanical systems are adequately approximated by linear models, in which restoring forces are proportional to displacements. The prototype of such systems is the harmonic oscillator, consisting of a single mass with a single degree of freedom subject to a linear restoring force (measured by the stiffness ksk_{\mathrm{s}} ). Analytical results for the simple harmonic oscillator can be adapted and extended to systems with multiple degrees of freedom, as in the subsequent discussion of rods with internal vibrational modes.

5.3.1. Classical treatment

In classical statistical mechanics, the probability density function for the position coordinate xx of a particle subject to the potential energy function V(x)\mathscr{V}(x) is [from Eq. (4.10)]

fx(x)=exp[V(x)/kT]/exp[V(x)/kT]dx\begin{equation*} f_{x}(x)=\exp [-\mathscr{V}(x) / k T] / \int_{-\infty}^{\infty} \exp [-\mathscr{V}(x) / k T] d x \tag{5.1} \end{equation*}

For the harmonic potential,

V(x)=12ksx2\begin{equation*} \mathcal{V}(x)=\frac{1}{2} k_{\mathrm{s}} x^{2} \tag{5.2} \end{equation*}

the resulting probability density function is Gaussian:

fx(x)=exp(ksx2/2kT)exp(ksx2/2kT)dx=exp(ksx2/2kT)2πkskT=12πσclass exp(x2/2σclass 2)\begin{align*} f_{x}(x) & =\frac{\exp \left(-k_{\mathrm{s}} x^{2} / 2 k T\right)}{\int_{-\infty}^{\infty} \exp \left(-k_{\mathrm{s}} x^{2} / 2 k T\right) d x}=\frac{\exp \left(-k_{\mathrm{s}} x^{2} / 2 k T\right)}{\sqrt{2 \pi}} \sqrt{\frac{k_{\mathrm{s}}}{k T}} \tag{5.3}\\ & =\frac{1}{\sqrt{2 \pi} \sigma_{\text {class }}} \exp \left(-x^{2} / 2 \sigma_{\text {class }}^{2}\right) \end{align*}

yielding the classical value for the positional variance (= standard deviation squared = mean square displacement):

σclass 2=kT/ks\begin{equation*} \sigma_{\text {class }}^{2}=k T / k_{\mathrm{s}} \tag{5.4} \end{equation*}

This relationship has broader applicability than its derivation may suggest.

a. The irrelevance of external bombardment. The probability density function for the position and velocity of a harmonic oscillator is independent of the nature of its coupling to the surrounding thermal bath. This may occur through vibration of its attachment point, through absorption and emission of thermal radiation, or through bombardment by molecules in a surrounding gas. None of these influences alters the potential energy function, hence none alters the Boltzmann distribution for the oscillator.

This may seem counterintuitive, given a mental image of the dynamical effects of molecular bombardment. At equilibrium, however, an impinging gas molecule is as likely to absorb energy as to deliver it, and so molecular bombardment has no net effect on the amplitude of vibration. How a system is coupled to a thermal bath can affect its detailed dynamics (e.g., the smoothness or irregularity of its trajectory, the decay time for oscillations of unusual amplitude, etc.), but not the statistical distribution of dynamical quantities. This principle holds true for systems in general, and makes the study of positional uncertainty dependent only on potential energy functions.

5.3.2. Quantum mechanical treatment

In quantum statistical mechanics, the classical integral over xx (more generally, an integral over phase space, yielding a probability density function for both position and momentum) is replaced by a sum, and the probability density function is replaced by a probability distribution over a series of states ii :

P(i)=exp[E(i)/kT]/i=0exp[E(i)/kT]\begin{equation*} P(i)=\exp [-\mathscr{E}(i) / k T] / \sum_{i=0}^{\infty} \exp [-\mathscr{E}(i) / k T] \tag{5.5} \end{equation*}

From elementary quantum mechanics, the vibrational states n=0,1,2,3,n=0,1,2,3, \ldots of a harmonic oscillator have energies that depend on the frequency

E(n)=(n+12)ω;ω=ks/m\begin{equation*} \mathscr{E}(n)=\left(n+\frac{1}{2}\right) \hbar \omega ; \quad \omega=\sqrt{k_{\mathrm{s}} / m} \tag{5.6} \end{equation*}

The probability of finding the oscillator in vibrational state nn is

P(n)=exp[(n+12)ω/kT]/n=0exp[(n+12)ω/kT]\begin{equation*} P(n)=\exp \left[-\left(n+\frac{1}{2}\right) \hbar \omega / k T\right] / \sum_{n=0}^{\infty} \exp \left[-\left(n+\frac{1}{2}\right) \hbar \omega / k T\right] \tag{5.7} \end{equation*}

Rearranging and summing the series,

P(n)=exp(nω/kT)/n=0[exp(ω/kT)]nn=0yn=1+y1y,y<1P(n)=exp(nω/kT)[1exp(ω/kT)]\begin{gather*} P(n)=\exp (-n \hbar \omega / k T) / \sum_{n=0}^{\infty}[\exp (-\hbar \omega / k T)]^{n} \tag{5.8}\\ \sum_{n=0}^{\infty} y^{n}=1+\frac{y}{1-y}, \quad|y|<1 \tag{5.9}\\ P(n)=\exp (-n \hbar \omega / k T)[1-\exp (-\hbar \omega / k T)] \tag{5.10} \end{gather*}

The variance (mean square displacement) of the oscillator can be derived from its mean energy:

E=n=0P(n)En=n=0exp(nωkT)[1exp(ωkT)](n+12)ω\begin{equation*} \overline{\mathscr{E}}=\sum_{n=0}^{\infty} P(n) \mathscr{E}_{n}=\sum_{n=0}^{\infty} \exp \left(-n \frac{\hbar \omega}{k T}\right)\left[1-\exp \left(-\frac{\hbar \omega}{k T}\right)\right]\left(n+\frac{1}{2}\right) \hbar \omega \tag{5.11} \end{equation*}

Rearranging and summing both series,

E=ω[1exp(ωkT)]{12n=0[exp(ωkT)]n+n=0n[exp(ωkT)]n}n=0nyn=y(1y)2,y<1E=ω{12+[exp(ω/kT)1]1}\begin{gather*} \overline{\mathscr{E}}=\hbar \omega\left[1-\exp \left(-\frac{\hbar \omega}{k T}\right)\right]\left\{\frac{1}{2} \sum_{n=0}^{\infty}\left[\exp \left(-\frac{\hbar \omega}{k T}\right)\right]^{n}+\sum_{n=0}^{\infty} n\left[\exp \left(-\frac{\hbar \omega}{k T}\right)\right]^{n}\right\} \tag{5.12}\\ \sum_{n=0}^{\infty} n y^{n}=\frac{y}{(1-y)^{2}}, \quad y<1 \tag{5.13}\\ \overline{\mathscr{E}}=\hbar \omega\left\{\frac{1}{2}+[\exp (\hbar \omega / k T)-1]^{-1}\right\} \tag{5.14} \end{gather*}

In a harmonic oscillator, the total energy equals twice the mean potential energy, which is proportional to the mean square displacement:

12E=V=12ksx2=12ksσ2σ2=ωks{12+[exp(ω/kT)1]1}\begin{gather*} \frac{1}{2} \overline{\mathscr{E}}=\overline{\mathscr{V}}=\frac{1}{2} k_{\mathrm{s}} \overline{x^{2}}=\frac{1}{2} k_{\mathrm{s}} \sigma^{2} \tag{5.15}\\ \sigma^{2}=\frac{\hbar \omega}{k_{\mathrm{s}}}\left\{\frac{1}{2}+[\exp (\hbar \omega / k T)-1]^{-1}\right\} \tag{5.16} \end{gather*}

Describing the frequency in terms of the fundamental mechanical parameters and then rearranging yields an equation between dimensionless quantities (see Figure 5.1):

σ2ksm=12+[exp(ω/kT)1]1\begin{equation*} \sigma^{2} \frac{\sqrt{k_{\mathrm{s}} m}}{\hbar}=\frac{1}{2}+[\exp (\hbar \omega / k T)-1]^{-1} \tag{5.17} \end{equation*}

Figure 5.1. A dimensionless measure of variance vs. a dimensionless measure of temperature, Eq. (5.17).

Figure 5.2. The ratio of the actual variance to that predicted by a classical model, vs. a dimensionless measure of temperature, Eq. (5.18).

It often is desirable to determine whether the classical approximation is adequate for describing positional uncertainties in a system of engineering interest. A useful measure is the ratio of the total to the classical variance, which equals the ratio of the total to the classical energy:

σ2σclass 2=ωkT{12+[exp(ω/kT)1]1}=EkT\begin{equation*} \frac{\sigma^{2}}{\sigma_{\text {class }}^{2}}=\frac{\hbar \omega}{k T}\left\{\frac{1}{2}+[\exp (\hbar \omega / k T)-1]^{-1}\right\}=\frac{\overline{\mathscr{E}}}{k T} \tag{5.18} \end{equation*}

This function of the parameter kT/ωk T / \hbar \omega is graphed in Figure 5.2; where kT/ωk T / \hbar \omega substantially exceeds unity, the classical variance provides a good approximation.

With less elegance (but more immediately accessible physical significance), these dimensionless quantities can be unfolded into a set of graphs of root mean square displacement (σ)(\sigma) as a function of temperature, spring constant, and mass (Figure 5.3). For perspective, note that the stiffness of nonbonded interactions between atoms at their equilibrium separation typically is about 0.1 N/m0.1 \mathrm{~N} / \mathrm{m}; that of covalent bond angle-bending, about 30 N/m30 \mathrm{~N} / \mathrm{m}; and that of covalent bond stretching, about 400 N/m400 \mathrm{~N} / \mathrm{m}. Many small components in nanomechanisms have masses in the hundreds to thousands of daltons.

5.4. Elastic extension of thermally excited rods

Various molecular structures resemble rods; these include DNA helices, microtubules, ladder polymers, and several classes of nanomechanical components. Even flexible molecular chains resemble rods (for purposes of the present analysis) when held in an extended conformation by longitudinal tension or by lateral constraints. A rod is the simplest example of an extended object and can serve as a model for various mechanical systems.

Figure 5.3. This set of graphs plots the logarithm of the root mean square displacement of harmonic oscillators of varying mass and spring constant as a function of temperature, from Eq. (5.16). Note that the length at the top of the graphs corresponds roughly to an atomic diameter. Above the dashed lines, the classical approximation is accurate to within 10%10 \%; above the dotted lines, it is accurate to within 1%1 \%. (continued)

A rod clamped at one end and free at the other undergoes both transverse and longitudinal vibrations. This section analyzes the positional variance of the free end resulting from longitudinal vibrational modes; in rods of large aspect ratio, transverse vibrations can make a significant contribution to the longitudinal positional variance of the free end (analyzed in Section 5.6). The following first considers a classical model for a continuous elastic rod, then both an approximate and an exact quantum mechanical model for a rod consisting of a series of identical masses and springs, ending with an empirical approximation to this exact model.

Figure 5.3. (continued) A 1 nm31 \mathrm{~nm}^{3} volume containing 1000 daltons has ρ1661 kg/m3\rho \approx 1661 \mathrm{~kg} / \mathrm{m}^{3}.

5.4.1. Classical continuum treatment

A uniform rod of length \ell clamped at one end and free at the other supports longitudinal vibrational modes (n=0,1,2,3,)(n=0,1,2,3, \ldots) with wavelengths

λn=42n+1\begin{equation*} \lambda_{n}=\frac{4 \ell}{2 n+1} \tag{5.19} \end{equation*}

The speed of sound vsv_{\mathrm{s}} in the rod can be calculated from the linear modulus EE_{\ell} and linear density ρ\rho_{\ell} yielding the modal frequencies ωn\omega_{n}.

vs=E/ρ;ω0=π2E/ρ;ωn=ω0(2n+1)\begin{equation*} v_{\mathrm{s}}=\sqrt{E_{\ell} / \rho_{\ell}} ; \quad \omega_{0}=\frac{\pi}{2 \ell} \sqrt{E_{\ell} / \rho_{\ell}} ; \quad \omega_{n}=\omega_{0}(2 n+1) \tag{5.20} \end{equation*}

Each longitudinal mode can be regarded as a harmonic oscillator with a certain frequency, { }^{\circ}effective stiffness, and { }^{\circ}effective mass. The effective stiffness relates the square of the amplitude (at the rod end, where the variance is to be computed) to the potential energy at maximum displacement during a vibrational cycle, which equals the maximum kinetic energy:

12knAn2=max(V)=max(T)=012ρ[v(x)]2dx=012ρ(ωnAn)2sin2[(2n+1)2πx]dx=14ρ(ωnAn)2\begin{align*} \frac{1}{2} k_{n} A_{n}^{2} & =\max (\mathcal{V})=\max (\mathscr{T}) \\ & =\int_{0}^{\ell} \frac{1}{2} \rho_{\ell}[v(x)]^{2} d x \\ & =\int_{0}^{\ell} \frac{1}{2} \rho_{\ell}\left(\omega_{n} A_{n}\right)^{2} \sin ^{2}\left[\frac{(2 n+1)}{2 \ell} \pi x\right] d x \\ & =\frac{1}{4} \rho_{\ell} \ell\left(\omega_{n} A_{n}\right)^{2} \tag{5.21} \end{align*}

This yields the effective stiffness knk_{n} of mode nn :

kn=Eπ28(2n+1)2\begin{equation*} k_{n}=\frac{E_{\ell}}{\ell} \frac{\pi^{2}}{8}(2 n+1)^{2} \tag{5.22} \end{equation*}

Combining this with the classical expression for positional variance in a harmonic oscillator as a function of temperature and stiffness, Eq. (5.4), the positional variance at the free end associated with mode nn is:

σn, class 2=kTE8π21(2n+1)2\begin{equation*} \sigma_{n, \text { class }}^{2}=k T \frac{\ell}{E_{\ell}} \frac{8}{\pi^{2}} \frac{1}{(2 n+1)^{2}} \tag{5.23} \end{equation*}

The total variance is the sum of the modal variances:

σclass 2=n=0σn, class 2=kTE8π2n=01(2n+1)2\begin{equation*} \sigma_{\text {class }}^{2}=\sum_{n=0}^{\infty} \sigma_{n, \text { class }}^{2}=k T \frac{\ell}{E_{\ell}} \frac{8}{\pi^{2}} \sum_{n=0}^{\infty} \frac{1}{(2 n+1)^{2}} \tag{5.24} \end{equation*}

Applying the identity

n=01(2n+1)2=π28\begin{equation*} \sum_{n=0}^{\infty} \frac{1}{(2 n+1)^{2}}=\frac{\pi^{2}}{8} \tag{5.25} \end{equation*}

yields the classical variance in the position of the free end for a rod of uniformly distributed mass and elasticity:

σclass 2=kT/E\begin{equation*} \sigma_{\text {class }}^{2}=k T \ell / E_{\ell} \tag{5.26} \end{equation*}

This is exactly kT/ksk T / k_{\mathrm{s}}, where ksk_{\mathrm{s}} is the stretching stiffness of the rod as a whole. This is the variance for a simple harmonic oscillator of the same stiffness, and hence of a "rod" having a single mass and a single mode-the opposite extreme from a rod of uniformly distributed mass and elasticity. The significance of this equality is discussed in Section 5.8.1.

5.4.2. Quantum mechanical treatments

a. Discrete rod models. For a continuous rod, a quantum mechanical treatment yields a divergent series for the positional variance, owing to the zero-point vibrations of an infinity of high-frequency modes; the continuum model is thus unacceptable even as an approximation. The following will work with a more realistic rod model (Figure 5.4) consisting of a series of NN identical springs and masses, supporting NN longitudinal vibrational modes. Introducing complexities

Figure 5.4. Diagram of a discrete rod, showing masses, springs, and the length coordinate (measuring between atomic centers).

such as differing masses would alter the detailed dynamics of rod vibrations, but would usually have little effect on the positional variance. More drastic, however, is the approximation that entire planes of atoms perpendicular to the rod axis can be lumped together and treated as single masses, neglecting the degrees of freedom introduced by the physical extent and flexibility of each plane.

A limiting-case analysis illustrates the essential physics. In one case (Figure 5.5), the atoms in each plane are tightly coupled to one another, each plane shares a single longitudinal degree of freedom, and the approximation under consideration is correct (by construction). In the other limiting case, a "rod" consists of an uncoupled bundle of jj component rods, each one atom wide and mm atoms long; this increases the total number of modes is by a factor of jj. If a component rod has a positional standard deviation σc\sigma_{c}, then the position of the end of the bundle (interpreted as the mean position of the component ends) has a standard deviation

σb=σc/j\begin{equation*} \sigma_{\mathrm{b}}=\sigma_{\mathrm{c}} / \sqrt{j} \tag{5.27} \end{equation*}

and the variance of the bundle end position is thus inversely proportional to jj. As we will see, this is exactly the variance that results from treating the bundle as a single unit with N=mN=m, hence the uncoupled and the tightly coupled results are identical. NN may thus be taken as the number of atomic planes along the length of the rod; a conservatively generous estimate (given that real interatomic spacings are greater than 0.1 nm0.1 \mathrm{~nm} ) is

N=1010 (for  in meters) \begin{equation*} N=10^{10} \ell \quad \text { (for } \ell \text { in meters) } \tag{5.28} \end{equation*}

Figure 5.5. Limiting-case rods. Part (a) diagrams a rod in the limiting case in which masses and spring constants are lumped on a plane-by-plane basis; (b) diagrams the limiting case in which the rod is treated as a bundle of decoupled simple rods.

b. Semicontinuum quantum mechanical approximation. A simple approximation to a discrete system uses the continuum results but truncates the sum of the modal variances at n=N1n=N-1. The continuum expressions for the modal effective stiffnesses and frequencies, together with Eq. (5.17), yield the positional variance associated with mode nn in the continuum approximation:

σn2Eρ=4π(2n+1){12+[exp(ω0kT(2n+1))1]1}\begin{equation*} \sigma_{n}^{2} \frac{\sqrt{E_{\ell} \rho_{\ell}}}{\hbar}=\frac{4}{\pi(2 n+1)}\left\{\frac{1}{2}+\left[\exp \left(\frac{\hbar \omega_{0}}{k T}(2 n+1)\right)-1\right]^{-1}\right\} \tag{5.29} \end{equation*}

Summing over the NN modes in a discrete rod yields the semicontinuum approximation to the total positional variance of the free end of a clamped rod:

σ2Eρ=n=0N14π(2n+1){12+[exp(ω0kT(2n+1))1]1}\begin{equation*} \sigma^{2} \frac{\sqrt{E_{\ell} \rho_{\ell}}}{\hbar}=\sum_{n=0}^{N-1} \frac{4}{\pi(2 n+1)}\left\{\frac{1}{2}+\left[\exp \left(\frac{\hbar \omega_{0}}{k T}(2 n+1)\right)-1\right]^{-1}\right\} \tag{5.30} \end{equation*}

c. Exact quantum mechanical treatment. The simple model above has serious defects when vibrational wavelengths approach the interatomic spacing of the rod(λda)\operatorname{rod}\left(\lambda \approx d_{\mathrm{a}}\right), because the dispersive wave-propagation characteristics of the discrete structure become important. These effects are large for the single mode of a "rod" with N=1N=1, and continue to be large for a substantial fraction of the modes in rods where N1N \gg 1. Defining the discrete-mass variables in terms of the continuum variables,

da=N;ma=ρN;ks=EN\begin{equation*} d_{\mathrm{a}}=\frac{\ell}{N} ; \quad m_{\mathrm{a}}=\frac{\ell \rho_{\ell}}{N} ; \quad k_{\mathrm{s}}=\frac{E_{\ell} N}{\ell} \tag{5.31} \end{equation*}

Keeping the intuitive definition of the rod length as the distance from the attachment point to the last mass,

λn=4(2n+1)(1+12N)\begin{equation*} \lambda_{n}=\frac{4 \ell}{(2 n+1)}\left(1+\frac{1}{2 N}\right) \tag{5.32} \end{equation*}

The correct, dispersive relationship of frequency to wavelength (Ashcroft and Mermin, 1976) is

ωn=2ksmasin(πdλn)=2EρNsin(2n+12N+1π2)\begin{equation*} \omega_{n}=2 \sqrt{\frac{k_{\mathrm{s}}}{m_{\mathrm{a}}}} \sin \left(\frac{\pi d}{\lambda_{n}}\right)=\frac{2}{\ell} \sqrt{\frac{E_{\ell}}{\rho_{\ell}}} N \sin \left(\frac{2 n+1}{2 N+1} \frac{\pi}{2}\right) \tag{5.33} \end{equation*}

The effective mass mnm_{n} of a mode nn is the sum over the (equal) masses of the square of the local modal amplitude, divided by the square of the modal amplitude at the free end of the rod:

mn=ma[sin2(2n+12N+1πN)]1i=1Nsin2(2n+12N+1πi)\begin{equation*} m_{n}=m_{\mathrm{a}}\left[\sin ^{2}\left(\frac{2 n+1}{2 N+1} \pi N\right)\right]^{-1} \sum_{i=1}^{N} \sin ^{2}\left(\frac{2 n+1}{2 N+1} \pi i\right) \tag{5.34} \end{equation*}

which simplifies to

mn=ρ[sin2(2n+12N+1πN)]1(12+14N)\begin{equation*} m_{n}=\rho_{\ell} \ell\left[\sin ^{2}\left(\frac{2 n+1}{2 N+1} \pi N\right)\right]^{-1}\left(\frac{1}{2}+\frac{1}{4 N}\right) \tag{5.35} \end{equation*}

Given Eq. (5.16) and the above values for frequency and effective mass, the modal variance

σn2=mnωn{12+[exp(ωnkT)1]1}\begin{equation*} \sigma_{n}^{2}=\frac{\hbar}{m_{n} \omega_{n}}\left\{\frac{1}{2}+\left[\exp \left(\frac{\hbar \omega_{n}}{k T}\right)-1\right]^{-1}\right\} \tag{5.36} \end{equation*}

which leads to the following dimensionless expression for the exact total variance expressed in terms of ω0\omega_{0} (the frequency of the fundamental mode given by the continuum approximation):

σ2Eρ=22N+1n=0N1[sin2(2n+12N+1πN)/sin(2n+12N+1π2)×(12+{exp[4Nπ(ω0kT)sin(2n+12N+1π2)]1}1)]\begin{align*} \sigma^{2} \frac{\sqrt{E_{\ell} \rho_{\ell}}}{\hbar}=\frac{2}{2 N+1} \sum_{n=0}^{N-1} & {\left[\sin ^{2}\left(\frac{2 n+1}{2 N+1} \pi N\right) / \sin \left(\frac{2 n+1}{2 N+1} \frac{\pi}{2}\right)\right.} \tag{5.37}\\ & \left.\times\left(\frac{1}{2}+\left\{\exp \left[\frac{4 N}{\pi}\left(\frac{\hbar \omega_{0}}{k T}\right) \sin \left(\frac{2 n+1}{2 N+1} \frac{\pi}{2}\right)\right]-1\right\}^{-1}\right)\right] \end{align*}

This equation is graphed in Figure 5.6; note that the result for n=0n=0, while identical to that given for the harmonic oscillator in Figure 5.1, is shifted to the left on this graph by the expression of results in terms of ω0\omega_{0} [Eq. (5.20)] rather than in terms of the true frequency ω\omega.

Note that division of a rod into jj component rods with

E=E/j;ρ=ρ/j\begin{equation*} E_{\ell}^{\prime}=E_{\ell} / j ; \quad \rho_{\ell}^{\prime}=\rho_{\ell} / j \tag{5.38} \end{equation*}

yields

σ=σj\begin{equation*} \sigma^{\prime}=\sigma \sqrt{j} \tag{5.39} \end{equation*}

Figure 5.6. A dimensionless measure of variance vs. a dimensionless measure of temperature, for rods supporting varying numbers of longitudinal modes, Eq. (5.37).

and the standard deviation of the bundle end position (i.e., of the mean displacement of the component ends) is

σb=jσ2/j=σ\begin{equation*} \sigma_{\mathrm{b}}=\sqrt{j \sigma^{\prime 2}} / j=\sigma \tag{5.40} \end{equation*}

d. An engineering approximation. For engineering purposes, it is advantageous to have a simple analytical expression rather than a sum over a variable number of terms. In seeking such an expression, we can begin with the relatively simple semicontinuum approximation, Eq. (5.30). This sum can be rearranged into two terms

σ2Eρ=2πn=0N1(2n+1)1+4πn=0N1((2n+1){exp[ω0kT(2n+1)]1})1\begin{equation*} \sigma^{2} \frac{\sqrt{E_{\ell} \rho_{\ell}}}{\hbar}=\frac{2}{\pi} \sum_{n=0}^{N-1}(2 n+1)^{-1}+\frac{4}{\pi} \sum_{n=0}^{N-1}\left((2 n+1)\left\{\exp \left[\frac{\hbar \omega_{0}}{k T}(2 n+1)\right]-1\right\}\right)^{-1} \tag{5.41} \end{equation*}

and the second term, which dominates in the classical limit, can be simplified by considering the classical limit (ω0/kT1)\left(\hbar \omega_{0} / k T \ll 1\right) :

σ2Eρ=2πn=0N1(2n+1)1+4πkTω0n=0N1(2n+1)2\begin{equation*} \sigma^{2} \frac{\sqrt{E_{\ell} \rho_{\ell}}}{\hbar}=\frac{2}{\pi} \sum_{n=0}^{N-1}(2 n+1)^{-1}+\frac{4}{\pi} \frac{k T}{\hbar \omega_{0}} \sum_{n=0}^{N-1}(2 n+1)^{-2} \tag{5.42} \end{equation*}

The first, logarithmically divergent sum can be replaced by an integral approximation corrected by a constant. Summing the second series to N=N=\infty yields a simple expression (also transformed to eliminate ω0\omega_{0} ) exhibiting better classical-limit behavior than the original expression:

σ2=πEρ[0.54+log(2N+1)]+kTE\begin{equation*} \sigma^{2}=\frac{\hbar}{\pi \sqrt{E_{\ell} \rho_{\ell}}}[0.54+\log (2 N+1)]+k T \frac{\ell}{E_{\ell}} \tag{5.43} \end{equation*}

This expression is accurate in the classical limit (ω0/kT0)\left(\hbar \omega_{0} / k T \rightarrow 0\right), and provides a good approximation in the quantum limit (kT/ω00)\left(k T / \hbar \omega_{0} \rightarrow 0\right). Its deviations are strictly in a conservative direction (overestimating variance), but can amount to tens of percent for values of kT/ω01k T / \hbar \omega_{0} \approx 1. A similar expression giving a more accurate result in this middle range can be obtained by multiplying the classical term by an empirically chosen function:

σ2=[0.54+log(2N+1)]πEρ+kTEexp[(0.70.39N)π2kTEρ]\begin{equation*} \sigma^{2}=\frac{\hbar[0.54+\log (2 N+1)]}{\pi \sqrt{E_{\ell} \rho_{\ell}}}+k T \frac{\ell}{E_{\ell}} \exp \left[-\left(0.7-\frac{0.39}{\sqrt{N}}\right) \frac{\pi \hbar}{2 \ell k T} \sqrt{\frac{E_{\ell}}{\rho_{\ell}}}\right] \tag{5.44} \end{equation*}

Figure 5.7 compares the results of these two approximations to the results given by Eq. (5.37). Both are shown to give conservative estimates of the variance, and the latter equation is shown to give estimates that are within a few percent of the correct values. It can serve as an adequate approximation for most engineering purposes.

Figures 5.8 and 5.9 plot quantum mechanical variances in forms that are useful for making quick estimates. Figure 5.8 plots positional standard deviations for rods with diamondlike properties at 300 K300 \mathrm{~K}; Figure 5.9 plots ratios of quantum to classical results for a range of properties and temperatures.

Figure 5.7. A plot of exact variances as in Figure 5.6, with the approximations of Eqs. (5.43) and (5.44) shown for comparison.

Figure 5.8. Standard deviation in elastic longitudinal displacement for the end of a thermally excited rod, plotted vs. diameter and length, assuming mechanical properties like those of bulk diamond (ρ=3500 kg/m3,E=1012 N/m2),N/=1010 m1\left(\rho=3500 \mathrm{~kg} / \mathrm{m}^{3}, E=10^{12} \mathrm{~N} / \mathrm{m}^{2}\right), N / \ell=10^{10} \mathrm{~m}^{-1}, and T=T= 300 K. Entropic effects, neglected here, are included in Figure 5.16. Note that quantum effects make a major contribution to the positional uncertainty only for 1 nm\ell \leq 1 \mathrm{~nm}. At sufficiently small dimensions, neglect of atomic-scale structural detail becomes unacceptable even as an approximation (e.g., the dark gray region to the left describes nonexistent subatomic diameters); at larger dimensions, this approximation is excellent.

Figure 5.9. A plot of ratios of exact [Eq. (5.37)] to classical [Eq. (5.26)] variances. The final graph in the sequence assumes an unrealistically high acoustic speed (40 km/s(40 \mathrm{~km} / \mathrm{s}, vs. 17 km/s\sim 17 \mathrm{~km} / \mathrm{s} for diamond), and thus represents an upper bound on the magnitude of quantum effects in real structures under familiar physical conditions.

Figure 5.9. (Continued)

5.5. Elastic bending of thermally excited rods

The analysis of the transverse positional variance resulting from transverse vibrational modes substantially parallels the longitudinal case. The following first considers a classical model for a continuous elastic rod, then a semicontinuum quantum mechanical model, a fit to the continuum limit of this model, and a conservative approximation to the combined effects of bending and { }^{\circ}shear deformation.

5.5.1. Classical treatment

a. Continuum model. Knowledge of modal frequencies and modal effective masses suffices to characterize the system. Following Timoshenko et al. (1974), the bending vibrations in a plane of symmetry of a uniform rod clamped at one end and free at the other (i.e., of a cantilever beam) have normal modes with shapes described by

yn(x)=C1sin(κnx)+C2cos(κnx)+C3sinh(κnx)+C4cosh(κnx)\begin{equation*} \mathrm{y}_{n}(x)=C_{1} \sin \left(\kappa_{n} x\right)+C_{2} \cos \left(\kappa_{n} x\right)+C_{3} \sinh \left(\kappa_{n} x\right)+C_{4} \cosh \left(\kappa_{n} x\right) \tag{5.45} \end{equation*}

where y(x)y(x) is the transverse displacement from the equilibrium position at point xx. The constants C1,C2,C3C_{1}, C_{2}, C_{3}, and C4C_{4} are determined by boundary and normalization conditions. The boundary conditions require that

cosRncoshRn=1;Rn=κn(n+12)π\begin{equation*} \cos \mathscr{R}_{n} \cosh \mathscr{R}_{n}=-1 ; \quad \mathscr{R}_{n}=\kappa_{n} \ell \approx\left(n+\frac{1}{2}\right) \pi \tag{5.46} \end{equation*}

and the modal frequencies are given by the relationship

ωn=(Rn/)2kb/ρ\begin{equation*} \omega_{n}=\left(\mathscr{R}_{n} / \ell\right)^{2} \sqrt{k_{\mathrm{b}} / \rho_{\ell}} \tag{5.47} \end{equation*}

where kbk_{\mathrm{b}} is the bending stiffness of the rod\operatorname{rod} in Jm/rad2\mathrm{J} \cdot \mathrm{m} / \mathrm{rad}^{2}, and the elastic energy per unit length resulting from bending

Ebx=12kb(2yx2)2\begin{equation*} \frac{\partial \mathscr{E}_{\mathrm{b}}}{\partial x}=\frac{1}{2} k_{\mathrm{b}}\left(\frac{\partial^{2} y}{\partial x^{2}}\right)^{2} \tag{5.48} \end{equation*}

The modal effective mass (taking the free-end displacement as the generalized position coordinate) can be found by integrating the square of the local normalized amplitude [requiring y()=1y(\ell)=1 ] with respect to the mass:

A=(cosRn+coshRn)[cos(Rnx/)cosh(Rnx/)]B=(sinRnsinhRn)[sin(Rnx/)sinh(Rnx/)]mn=0(A+B)2(sinRnsinhRn)2ρdx=ρ4\begin{align*} A & =\left(\cos \mathscr{R}_{n}+\cosh \mathscr{R}_{n}\right)\left[\cos \left(\mathscr{R}_{n} x / \ell\right)-\cosh \left(\mathscr{R}_{n} x / \ell\right)\right] \\ B & =\left(\sin \mathscr{R}_{n}-\sinh \mathscr{R}_{n}\right)\left[\sin \left(\mathscr{R}_{n} x / \ell\right)-\sinh \left(\mathscr{R}_{n} x / \ell\right)\right] \\ m_{n} & =\int_{0}^{\ell} \frac{(A+B)^{2}}{\left(\sin \mathscr{R}_{n} \sinh \mathscr{R}_{n}\right)^{2}} \rho_{\ell} d x=\frac{\rho_{\ell} \ell}{4} \tag{5.49} \end{align*}

for all values of nn. The modal frequency and the modal effective mass define the modal effective transverse stiffness resulting from bending

kt,b,n=kb43Rn4\begin{equation*} k_{\mathrm{t}, \mathrm{b}, n}=\frac{k_{\mathrm{b}}}{4 \ell^{3}} \mathscr{R}_{n}^{4} \tag{5.50} \end{equation*}

which (as in the longitudinal analysis) yields an expression for the total classical transverse variance resulting from bending { }^{\circ}compliance:

σt,b,class2=4kT3kbn=01Rn4\begin{equation*} \sigma_{\mathrm{t}, \mathrm{b}, \mathrm{class}}^{2}=4 k T \frac{\ell^{3}}{k_{\mathrm{b}}} \sum_{n=0}^{\infty} \frac{1}{\mathscr{R}_{n}^{4}} \tag{5.51} \end{equation*}

Applying the identity

n=01Rn4=112\begin{equation*} \sum_{n=0}^{\infty} \frac{1}{\mathscr{R}_{n}^{4}}=\frac{1}{12} \tag{5.52} \end{equation*}

yields the classical expression for the transverse variance in the position of the free end of a rod of uniformly distributed mass and bending stiffness, neglecting compliance owing to shear and the effects of discreteness (see Section 5.5.1b):

σt,b,class2=kT3/3kb\begin{equation*} \sigma_{\mathrm{t}, \mathrm{b}, \mathrm{class}}^{2}=k T \ell^{3} / 3 k_{\mathrm{b}} \tag{5.53} \end{equation*}

But elementary theory of flexure shows that the bending stiffness of the end of a continuum cantilever rod is

kt,b=3kb/3\begin{equation*} k_{\mathrm{t}, \mathrm{b}}=3 k_{\mathrm{b}} / \ell^{3} \tag{5.54} \end{equation*}

hence the classical variance is simply kT/kt,bk T / k_{\mathrm{t}, \mathrm{b}}. As in the longitudinal case, the vibrational modes prove irrelevant to the classical analysis (see Section 5.4.1); knowledge of the overall stiffness suffices.

For a tube, the bending stiffness is

kb=Eπ4(r24r14)\begin{equation*} k_{\mathrm{b}}=E \frac{\pi}{4}\left(r_{2}^{4}-r_{1}^{4}\right) \tag{5.55} \end{equation*}

and the transverse stiffness at the far end of an anchored tube is

kt,b=E3π43(r24r14)\begin{equation*} k_{\mathrm{t}, \mathrm{b}}=E \frac{3 \pi}{4 \ell^{3}}\left(r_{2}^{4}-r_{1}^{4}\right) \tag{5.56} \end{equation*}

where r1r_{1} and r2r_{2} are the inner and outer tube radii, and EE is { }^{\circ}Young's modulus.

b. Discrete model. In bending (unlike stretching) the transition from a continuum to a lumped system changes the overall stiffness (not merely the modal stiffnesses). For a rod considered as a series of masses and angular springs, as shown in Figure 5.10, the rod-bending stiffness is related to the intermass spacings and angular stiffnesses:

kb=kθda\begin{equation*} k_{\mathrm{b}}=k_{\theta} d_{\mathrm{a}} \tag{5.57} \end{equation*}

For rods in which the number of segments, NN, is finite, the compliance of the rod can be described as the sum of contributions from a series of bending joints

1kt,b=3kbn=0N11N(1nN)23kb(3N+49N2)\begin{equation*} \frac{1}{k_{\mathrm{t}, \mathrm{b}}}=\frac{\ell^{3}}{k_{\mathrm{b}}} \sum_{n=0}^{N-1} \frac{1}{N}\left(1-\frac{n}{N}\right)^{2} \approx \frac{\ell^{3}}{k_{\mathrm{b}}}\left(\frac{3 N+4}{9 N-2}\right) \tag{5.58} \end{equation*}

where the approximation is exact for N=1,2N=1,2, and \infty; for intermediate values it is always high (i.e., conservative) and in error by less than 1%1 \%.

The preceding analysis takes account of transverse compliance caused by bending, but rods of multiatom width also exhibit compliance through shear.

Figure 5.10. A discrete rod, illustrating the angles entering into calculations of bending stiffness effects.

Figure 5.11. Modal deformation in pure bending (a) vs. pure shear (b). Bending dominates in high aspect-ratio rods; shear dominates in rods of low aspect ratio. Both deformations are here shown in idealized form.

For typical materials, shear makes a substantial contribution to the total transverse compliance, ktk_{\mathrm{t}}, when rod diameter dd \approx \ell; in vibration, it has significant effects on modal shapes and stiffnesses when dd is of the same order as the associated wavelength. Figure 5.11 illustrates normal modes from pure bending (zero shear compliance) and pure shear (zero bending compliance) for n=2n=2, with the idealization that plane cross sections remain plane despite shear. A standard engineering approximation treats shear as an independent, additive source of compliance, with transverse stiffness effects from shear taking the same analytical form as longitudinal stiffness effects but substituting the (linear) shear modulus GG_{\ell} for the (linear) stretching modulus EE_{\ell}. Combining the resulting shear compliance, 1/kt,s=/G1 / k_{\mathrm{t}, \mathrm{s}}=\ell / G_{\ell}, with Eq. (5.58) yields the engineering approximation:

σt, class 2=kT[3kb(3N+49N2)+G]\begin{equation*} \sigma_{\mathrm{t}, \text { class }}^{2}=k T\left[\frac{\ell^{3}}{k_{\mathrm{b}}}\left(\frac{3 N+4}{9 N-2}\right)+\frac{\ell}{G_{\ell}}\right] \tag{5.59} \end{equation*}

Aside from quantum corrections, this approximation breaks down when /d\ell / d becomes small, yet bending compliance remains important. In this regime, outof-plane deformation of the free surface becomes a significant source of additional compliance. To treat objects of such low aspect-ratio as rods is, at best, a crude approximation.

5.5.2. Semicontinuum quantum mechanical treatment

The preceding classical treatment can serve as the basis for a quantum mechanical analysis. As in the earlier analysis of longitudinal vibrations, use of the continuum approximation neglects the dispersive properties of a discrete medium, which become significant as the characteristic wavelengths of normal modes approach interatomic distances. As will be seen, however, zero-point vibrations in the higher transverse modes (unlike those of the higher longitudinal modes) make only a modest contribution to the positional variance. Accordingly, dispersion is neglected in the following analysis; this neglect should have little effect for N10N \geq 10.

We begin with a model including only the effects of bending compliance, deferring discussion of shear compliance. (Figure 5.11 illustrates the difference between these forms of compliance.) The above values of modal effective mass

Figure 5.12. Dimensionless transverse variance for rods, neglecting shear compliance. The simple approximation is based on Eq. (5.64); the exponential-factor approximation on Eq. (5.65), and the exact semicontinuum sum on Eq. (5.60).

and effective stiffness yield the semicontinuum approximation to the transverse bending variance for a uniform rod. In dimensionless terms analogous to those of the longitudinal-mode analysis:

σt,b2kbρ=n=0N14Rn2(12+{exp[ω0kT(RnR0)2]1}1)\begin{equation*} \sigma_{\mathrm{t}, \mathrm{b}}^{2} \frac{\sqrt{k_{b} \rho_{\ell}}}{\hbar \ell}=\sum_{n=0}^{N-1} \frac{4}{\mathscr{R}_{n}^{2}}\left(\frac{1}{2}+\left\{\exp \left[\frac{\hbar \omega_{0}}{k T}\left(\frac{\mathscr{R}_{n}}{\mathscr{R}_{0}}\right)^{2}\right]-1\right\}^{-1}\right) \tag{5.60} \end{equation*}

(see Figure 5.12).

5.5.3. Engineering approximations

Expressing the variance directly,

σt,b2=kbρn=0N14Rn2{12+[exp(kTRn22kbρ)1]1}\begin{equation*} \sigma_{\mathrm{t}, \mathrm{b}}^{2}=\frac{\hbar \ell}{\sqrt{k_{b} \rho_{\ell}}} \sum_{n=0}^{N-1} \frac{4}{\mathscr{R}_{n}^{2}}\left\{\frac{1}{2}+\left[\exp \left(\frac{\hbar}{k T} \frac{\mathscr{R}_{n}^{2}}{\ell^{2}} \sqrt{\frac{k_{b}}{\rho_{\ell}}}\right)-1\right]^{-1}\right\} \tag{5.61} \end{equation*}

Multiplying out and taking the classical limit of the second term yields

σt,b2=kbρn=0N12Rn2+n=0N14Rn4kT3kb\begin{equation*} \sigma_{\mathrm{t}, \mathrm{b}}^{2}=\frac{\hbar \ell}{\sqrt{k_{b} \rho_{\ell}}} \sum_{n=0}^{N-1} \frac{2}{\mathscr{R}_{n}^{2}}+\sum_{n=0}^{N-1} \frac{4}{\mathscr{R}_{n}^{4}} k T \frac{\ell^{3}}{k_{b}} \tag{5.62} \end{equation*}

In the limit of large NN, the second term yields the classical continuum result, and may be replaced by the modified classical expression derived above to take account of the reduced bending stiffness of discrete rods where NN is small. For the first term (describing zero-point contributions to the variance) solving the equation that defines values of Rn\mathscr{R}_{n} yields the convergent series

n=02Rn2=0.5688+0.0908+0.0324+0.0165+0.7588\begin{equation*} \sum_{n=0}^{\infty} \frac{2}{\mathscr{R}_{n}^{2}}=0.5688+0.0908+0.0324+0.0165+\cdots \approx 0.7588 \tag{5.63} \end{equation*}

Substituting the infinite series limit yields the approximation

σt,b2=0.76kbρ+kT3kb(3N+49N2)\begin{equation*} \sigma_{\mathrm{t}, \mathrm{b}}^{2}=0.76 \frac{\hbar \ell}{\sqrt{k_{b} \rho_{\ell}}}+k T \frac{\ell^{3}}{k_{\mathrm{b}}}\left(\frac{3 N+4}{9 N-2}\right) \tag{5.64} \end{equation*}

Given the size of the first term in the above series and the shortcomings of the continuum model for rods of low NN, it is useful to consider the case N=1N=1. In a rod consisting of a single mass and a single point of bending, the continuum model overestimates the effective stiffness by a factor of three and underestimates the effective mass by a factor of four, making its estimate of the quantum mechanical positional variance conservative by a factor of (4/3)1/21.15(4 / 3)^{1 / 2} \approx 1.15.

This approximation, like that of Eq. (5.43), significantly overestimates the positional variance in the transition region between the quantum and classical limits (Figure 5.12). A more accurate result for the important case of large NN may be obtained by evaluating the limit of the original sum, Eq. (5.60), as NN \rightarrow \infty and fitting an empirical expression to the results:

σt,b2=0.76kbρ+kT33kbexp(1.972kbρkT)\begin{equation*} \sigma_{\mathrm{t}, \mathrm{b}}^{2}=0.76 \frac{\hbar \ell}{\sqrt{k_{b} \rho_{\ell}}}+k T \frac{\ell^{3}}{3 k_{\mathrm{b}}} \exp \left(-\frac{1.97}{\ell^{2}} \sqrt{\frac{k_{\mathrm{b}}}{\rho_{\ell}}} \frac{\hbar}{k T}\right) \tag{5.65} \end{equation*}

This approximation is always high, but never by more than 1%1 \%. Figure 5.12 compares these approximations to Eq. (5.60).

5.5.4. Shear and bending in the quantum limit

The approximations developed for longitudinal positional variance, Eq. (5.43) and Eq. (5.44), have direct analogs for the positional variance that would occur in hypothetical rods having shear but no bending compliance. The more accurate of the two takes the form

σt,s2=[0.54+log(2N+1)]πGρ+kTGexp[(0.70.39N)π2kTGρ]\begin{equation*} \sigma_{\mathrm{t}, \mathrm{s}}^{2}=\frac{\hbar[0.54+\log (2 N+1)]}{\pi \sqrt{G_{\ell} \rho_{\ell}}}+k T \frac{\ell}{G_{\ell}} \exp \left[-\left(0.7-\frac{0.39}{\sqrt{N}}\right) \frac{\pi \hbar}{2 \ell k T} \sqrt{\frac{G_{\ell}}{\rho_{\ell}}}\right] \tag{5.66} \end{equation*}

In the classical limit, the effects of bending and shear compliance are simply additive, yielding expressions like Eq. (5.59). In the quantum limit, effective modal masses and frequencies play a role, and a precise analysis would have to include the effects of bending and shear on a mode-by-mode basis. An upper bound on their combined effect can be had more simply. Consider the variance of a harmonic oscillator in the quantum limit

σ2=/(2ksm)\begin{equation*} \sigma^{2}=\hbar /\left(2 \sqrt{k_{\mathrm{s}} m}\right) \tag{5.67} \end{equation*}

If we consider its stiffness to have two sources, ks1k_{\mathrm{s} 1} and ks2k_{\mathrm{s} 2}, the variance becomes

σ2=2mks11+ks21\begin{equation*} \sigma^{2}=\frac{\hbar}{2 \sqrt{m}} \sqrt{k_{\mathrm{s} 1}^{-1}+k_{\mathrm{s} 2}^{-1}} \tag{5.68} \end{equation*}

The expression

σest 2=σ12+σ22=2ks1m+2ks2m\begin{equation*} \sigma_{\text {est }}^{2}=\sigma_{1}^{2}+\sigma_{2}^{2}=\frac{\hbar}{2 \sqrt{k_{\mathrm{s} 1} m}}+\frac{\hbar}{2 \sqrt{k_{\mathrm{s} 2} m}} \tag{5.69} \end{equation*}

overestimates the actual variance by a factor

1σest2σ2=ks1/ks2+1(ks1/ks2)+12\begin{equation*} 1 \leq \frac{\sigma_{\mathrm{est}}^{2}}{\sigma^{2}}=\frac{\sqrt{k_{\mathrm{s} 1} / k_{\mathrm{s} 2}}+1}{\sqrt{\left(k_{\mathrm{s} 1} / k_{\mathrm{s} 2}\right)+1}} \leq \sqrt{2} \tag{5.70} \end{equation*}

Accordingly, it should be conservative to estimate the variance resulting from a vibrational mode subject to both bending and shear as the sum of the variances of a hypothetical mode constrained purely by bending forces and of one constrained by pure shear forces. (The differences between modal shapes in pure bending and those in pure shear would complicate a more precise analysis.) Thus, treating both classical and quantum variances as additive, expressions for the total transverse variance at the end of a rod take the form

σt2=σt,b2+σt,s2\begin{equation*} \sigma_{t}^{2}=\sigma_{t, b}^{2}+\sigma_{t, s}^{2} \tag{5.71} \end{equation*}

with the choice of expressions for the shear and bending contributions depending on the desired accuracy, the magnitude of quantum effects, and the value of NN. Figure 5.13 uses Eqs. (5.64) and (5.66) to graph the standard deviation of the transverse displacement at room temperature for rods with mechanical properties approximating those of bulk diamond. As can be seen, under these conditions, in the regime where shear and bending compliance are both important, quantum effects on positional uncertainty are minor for rods of nanometer or greater size; accordingly, Eq. (5.59) provides a good approximation.

Figure 5.13. Transverse positional standard deviation for rods at 300 K300 \mathrm{~K}, including shear compliance. The bending component is based on Eq. (5.64); the shear component is based on Eq. (5.66), assuming mechanical properties resembling those of bulk diamond, as in Figure 5.8; similar remarks apply.

Figure 5.14. Piston, cylinder, and gas molecules. Note that the position coordinate xx measures not the distance of the piston from the bottom of the cylinder, but the range of motion available to a gas molecule. A hardsphere, hard-surface model is assumed here and in the text.

5.6. Piston displacement in a gas-filled cylinder

Earlier sections have considered linear, elastic systems in which the motion can be divided into normal modes, treated as independent harmonic oscillators. A different approach is necessary for nonlinear systems in which the displacement of one component affects the range of motion possible to another, that is, for entropic springs. Here, the simplest example is not a mass and spring, but a loaded piston in a cylinder containing an ideal gas.

Figure 5.14 illustrates the system and the defining coordinates. The diameter of the cylinder proves irrelevant, and the displacement of the piston is chosen such that a zero displacement corresponds to zero freedom of movement for the gas molecule in the xx direction. The assumption of an ideal gas entails ignoring all forces between gas molecules, and accordingly ignoring the reduction in available volume that a molecule experiences as a result of the bulk of other molecules that may be present. Classical positional uncertainty in this system will be analyzed from three (ultimately equivalent) perspectives.

5.6.1. Weighting in terms of potential energy and available states

Let the compressing force on the piston be a constant, FcF_{\mathrm{c}}. For an empty cylinder, the potential energy of the piston is then FcxF_{\mathrm{c}} x, and applying the Boltzmann weighting to this potential yields the exponential PDF

fx(x)=exp(Fcx/kT)0exp(Fcx/kT)dx=FckTexp(FcxkT);x0\begin{equation*} f_{x}(x)=\frac{\exp \left(-F_{\mathrm{c}} x / k T\right)}{\int_{0}^{\infty} \exp \left(-F_{\mathrm{c}} x / k T\right) d x}=\frac{F_{\mathrm{c}}}{k T} \exp \left(-\frac{F_{\mathrm{c}} x}{k T}\right) ; \quad x \geq 0 \tag{5.72} \end{equation*}

The addition of NN gas molecules does not change the potential energy, but does increase the number of states associated with each piston position by a factor proportional to the space available to each molecule, that is, by a factor proportional to xNx^{N}. Introducing this factor to account for the number of states yields the Erlang PDF

fx(x)=xNexp(Fcx/kT)0xNexp(Fcx/kT)dx=1N!(FckT)N+1xnexp(FcxkT);x0\begin{align*} f_{x}(x) & =\frac{x^{N} \exp \left(-F_{\mathrm{c}} x / k T\right)}{\int_{0}^{\infty} x^{N} \exp \left(-F_{\mathrm{c}} x / k T\right) d x} \tag{5.73}\\ & =\frac{1}{N !}\left(\frac{F_{\mathrm{c}}}{k T}\right)^{N+1} x^{n} \exp \left(-\frac{F_{\mathrm{c}} x}{k T}\right) ; \quad x \geq 0 \end{align*}

with mean and variance

xˉ=(N+1)kTFc;σx2=(N+1)(kTFc)2\begin{equation*} \bar{x}=(N+1) \frac{k T}{F_{\mathrm{c}}} ; \quad \sigma_{x}^{2}=(N+1)\left(\frac{k T}{F_{\mathrm{c}}}\right)^{2} \tag{5.74} \end{equation*}

In this approach, the configurational states of the system are treated as known, and the Boltzmann-weighted probabilities are then integrated over these states.

5.6.2. Weighting in terms of a mean-force potential

Alternatively, one can treat the gas as a nonlinear spring described by the ideal gas equation (here in molecular rather than the more familiar molar units)

pV=NkT\begin{equation*} p V=N k T \tag{5.75} \end{equation*}

yielding the time-average force owing to pressure

Fp=NkT/x\begin{equation*} F_{\mathrm{p}}=-N k T / x \tag{5.76} \end{equation*}

Since the gas is now treated as a spring external to the piston, its configurational states are now ignored (as discussed in Section 4.3.7). Treating FpF_{\mathrm{p}} on the same basis as FcF_{c}, the work-energy as a function of position is

Fc(NkT/x)dx=FcxNkTlogx+C\begin{equation*} \int F_{\mathrm{c}}-(N k T / x) d x=F_{\mathrm{c}} x-N k T \log x+C \tag{5.77} \end{equation*}

hence the Boltzmann-weighted positional PDF is

fx(x)=exp(FcxkTnlogx+CkT)0exp(FcxkTnlogx+CkT)dx=xNexp(Fcx/kT)0xNexp(Fcx/kT)dx\begin{equation*} f_{x}(x)=\frac{\exp \left(-\frac{F_{\mathrm{c}} x}{k T}-n \log x+\frac{C}{k T}\right)}{\int_{0}^{\infty} \exp \left(-\frac{F_{\mathrm{c}} x}{k T}-n \log x+\frac{C}{k T}\right) d x}=\frac{x^{N} \exp \left(-F_{\mathrm{c}} x / k T\right)}{\int_{0}^{\infty} x^{N} \exp \left(-F_{\mathrm{c}} x / k T\right) d x} \tag{5.78} \end{equation*}

Note that the nonlinearity of the stiffness invalidates the simple relationship

σx2=kT/ks\begin{equation*} \sigma_{x}^{2}=k T / k_{\mathrm{s}} \tag{5.79} \end{equation*}

Evaluating ksk_{\mathrm{s}} at the mean and most probable values of xx

kTks, mean =(N+1)2N(kTFc)2;kTks, max-prob =N(kTFc)2\begin{equation*} \frac{k T}{k_{\mathrm{s}, \text { mean }}}=\frac{(N+1)^{2}}{N}\left(\frac{k T}{F_{\mathrm{c}}}\right)^{2} ; \quad \frac{k T}{k_{\mathrm{s}, \text { max-prob }}}=N\left(\frac{k T}{F_{\mathrm{c}}}\right)^{2} \tag{5.80} \end{equation*}

yields differing results, both of which differ from the true variance, Eq. (5.74). Both expressions, however, approach the correct value in the limit of large NN, where the standard deviation in position is small enough for a linear approximation to hold for fluctuations of ordinary magnitude.

5.6.3. Weighting in terms of the Helmholtz free energy

In a third approach, we begin with the Helmholtz free energy,

F=ETS\begin{equation*} \mathscr{F}=\mathscr{E}-T \mathscr{S} \tag{5.81} \end{equation*}

The internal energy E\mathscr{E} is the sum of the potential energy FcxF_{\mathrm{c}} x, the kinetic energy of the gas ( 3/2NkT3 / 2 N k T, assuming a monatomic gas), and of the piston ( 3kT3 k T, assuming freedom to slide, rotate, and rattle). The translational entropy of an ideal gas

(Knox, 1971) is

Strans =Nk(52+32lnmkT2π2lnNV)=Nk(ClnNx)\begin{equation*} \mathscr{S}_{\text {trans }}=N k\left(\frac{5}{2}+\frac{3}{2} \ln \frac{m k T}{2 \pi \hbar^{2}}-\ln \frac{N}{V}\right)=N k\left(C-\ln \frac{N}{x}\right) \tag{5.82} \end{equation*}

hence the free energy of the system as a function of piston position is

F=ETS=Fcx+(3+32N)kTNkT(ClnNx)\begin{equation*} \mathscr{F}=\mathscr{E}-T \mathscr{S}=F_{\mathrm{c}} x+\left(3+\frac{3}{2} N\right) k T-N k T\left(C-\ln \frac{N}{x}\right) \tag{5.83} \end{equation*}

Taking the Boltzmann-weighted PDF in terms of the free energy again yields

fx(x)=exp(FcxkT+3+32NNC+NlnXN)0exp(FcxkT+3+32NNC+NlnXN)dx=xNexp(Fcx/kT)0xNexp(Fcx/kT)dx\begin{align*} f_{x}(x) & =\frac{\exp \left(-\frac{F_{\mathrm{c}} x}{k T}+3+\frac{3}{2} N-N C+N \ln \frac{X}{N}\right)}{\int_{0}^{\infty} \exp \left(-\frac{F_{\mathrm{c}} x}{k T}+3+\frac{3}{2} N-N C+N \ln \frac{X}{N}\right) d x} \\ & =\frac{x^{N} \exp \left(-F_{\mathrm{c}} x / k T\right)}{\int_{0}^{\infty} x^{N} \exp \left(-F_{\mathrm{c}} x / k T\right) d x} \tag{5.84} \end{align*}

5.6.4. Comparison and quantum effects

These three perspectives on the problem are closely related: The variation in available states as a function of piston position considered in the first corresponds to the variation in entropy considered in the third. The second approach considers movement of the piston as doing work on the gas (under reversible, isothermal conditions); work done under these conditions equals the change in FF considered in the third. The choice of perspective in such problems is a matter of convenience.

Physical displacements frequently couple to changes in the range of motion available in some degree of freedom, hence altering its entropy and doing work against an entropic spring. These systems share the basic properties of the piston and cylinder system just considered, including a (kT)2(k T)^{2} dependence of positional uncertainty in the classical regime, as opposed to the kTk T dependence of classical uncertainty stemming from conventional springs. In the quantum limit, systems that would exhibit entropic spring effects at higher temperatures will display a small residual compliance stemming from the compression of zero-point probability distributions. This is not an entropic effect, since (in the quantum limit) all modes are consistently in their ground state with an invariant entropy of zero; nonetheless, it is a compressive effect that is a smooth extension of the behavior in the classical regime.

5.7. Longitudinal variance from transverse deformation

Transverse vibrations in a rod cause longitudinal shortening by forcing it to deviate from a straight line. This longitudinal-transverse coupling causes longitudinal positional variance, and provides another example of an entropic spring. Although the following analysis is not used in this volume, sheathed flexible rods like those described have applications in a wide range of nanomechanical systems.

5.7.1. General approach

Rods sliding in channels can be used to couple mechanical displacements occurring at one location to displacements at a relatively distant location. Thus, it is of interest to consider the longitudinal positional variance of rods in systems where rod motion occurs in a channel that imposes transverse restoring forces (modeled as a stiffness per unit length, kk_{\ell} ) via overlap repulsion, and where boundary conditions may impose a mean tension γ\gamma_{\ell} on the rod.

The effect of shear compliance on transverse vibration is usually small in systems where longitudinal-transverse coupling is significant. Further, in the approximation that the rod behaves as an isotropic, elastically linear material, shear deformation has no effect on length. Shear is accordingly neglected.

The discrete structure of the rod imposes a limit to the number of modes and modifies the coupling constants as λn\lambda_{n} approaches 2Δ(=λmin)2 \Delta \ell\left(=\lambda_{\min }\right). The following analysis adopts a semicontinuum model that takes account of limitations on modes while neglecting changes in coupling constants. This approximation can break down in systems where modes with λλmin\lambda \approx \lambda_{\min } dominate the variance. Where the restoring force for these modes is dominated by kk_{\ell}, the approximation is conservative; where the restoring force is dominated by γ\gamma_{\ell}, the approximation is accurate; where the restoring force is dominated by kbk_{\mathrm{b}}, the approximation is too low. In this case, the maximum correction factor for the variance (for λ=λmin\lambda=\lambda_{\min } ) is (π/2)46.09(\pi / 2)^{4} \approx 6.09, falling to 1.52 for λ=2λmin\lambda=2 \lambda_{\min } and to 1.02 for λ=10λmin \lambda=10 \lambda_{\text {min }}.

The nonlinearity of overlap repulsions makes the use of the constant stiffness kk_{\ell} a rough approximation. Since stiffness increases with displacement, this approximation is conservative, underestimating the constraining forces. Discussion of quantum mechanical effects is deferred to Section 5.7.3.

5.7.2. Coupling and variance

The following analysis considers a weighting in terms of potential energy and available states for each of a set of normal-mode deformations, summing the resulting variances to yield the total variance. The use of normal modes here implies nothing about vibrations, but merely provides a convenient set of orthogonal functions with which to describe all possible rod configurations.

For the sinusoidal deformation characteristic of mode nn, there exists a constant Cn\mathscr{C}_{n} (dependent on rod parameters) relating the contraction in the length of the rod to the potential energy of the deformation

Δn=CnEn\begin{equation*} \Delta \ell_{n}=\mathscr{C}_{n} \mathscr{E}_{n} \tag{5.85} \end{equation*}

The classical variance in rod length resulting from modal longitudinal-transverse coupling can be determined from the PDF for the potential energy (which in turn can be derived from the Gaussian PDF for the amplitude),

fEn(En)=exp(En/kT)/πkTEn\begin{equation*} f_{\mathscr{E}_{n}}\left(\mathscr{E}_{n}\right)=\exp \left(-\mathscr{E}_{n} / k T\right) / \sqrt{\pi k T \mathscr{E}_{n}} \tag{5.86} \end{equation*}

and for the contraction in length

fΔn(Δn)=exp(Δn/CnkT)/πCnkTΔn\begin{equation*} f_{\Delta \ell_{n}}\left(\Delta \ell_{n}\right)=\exp \left(-\Delta \ell_{n} / \mathscr{C}_{n} k T\right) / \sqrt{\pi \mathscr{C}_{n} k T \Delta \ell_{n}} \tag{5.87} \end{equation*}

From this one can derive the variance in the potential energy

σEn2=En2(En)2=0En2exp(En/kT)πkTEndEn[0Enexp(En/kT)πkTEndEn]2=12(kT)2\begin{align*} \sigma_{\mathscr{E}_{n}}^{2} & =\overline{\mathscr{E}_{n}^{2}}-\left(\overline{\mathscr{E}_{n}}\right)^{2} \\ & =\int_{0}^{\infty} \frac{\mathscr{E}_{n}^{2} \exp \left(-\mathscr{E}_{n} / k T\right)}{\sqrt{\pi k T \mathscr{E}_{n}}} d \mathscr{E}_{n}-\left[\int_{0}^{\infty} \frac{\mathscr{E}_{n} \exp \left(-\mathscr{E}_{n} / k T\right)}{\sqrt{\pi k T \mathscr{E}_{n}}} d \mathscr{E}_{n}\right]^{2} \\ & =\frac{1}{2}(k T)^{2} \tag{5.88} \end{align*}

and hence the variance in length resulting from longitudinal-transverse coupling in mode nn.

σ,t,n2=12(CnkT)2\begin{equation*} \sigma_{\ell, t, n}^{2}=\frac{1}{2}\left(\mathscr{C}_{n} k T\right)^{2} \tag{5.89} \end{equation*}

The mean contraction is

Δn=12CnkT\begin{equation*} \overline{\Delta \ell_{n}}=\frac{1}{2} \mathscr{C}_{n} k T \tag{5.90} \end{equation*}

5.7.3. Rods with tension and transverse constraints

Consider a long, continuous rod with a bending modulus kbk_{\mathrm{b}}, subject to a mean tension γ\gamma_{\ell} and a transverse stiffness per unit length kk_{\ell}. The energy per unit length associated with a sinusoidal deformation can be derived by integrating the contributions from each of these restoring-force terms. For a rod of length \ell supporting modes with amplitudes AnA_{n} and λ=2/n(n=1,2,3,)\lambda=2 \ell / n(n=1,2,3, \ldots),

En=An24[kb(nπ)4+γ(nπ)2+k]\begin{equation*} \frac{\mathscr{E}_{n}}{\ell}=\frac{A_{n}^{2}}{4}\left[k_{\mathrm{b}}\left(\frac{n \pi}{\ell}\right)^{4}+\gamma_{\ell}\left(\frac{n \pi}{\ell}\right)^{2}+k_{\ell}\right] \tag{5.91} \end{equation*}

The fractional change in length is

Δn=(nπAn2)2\begin{equation*} \frac{\Delta \ell_{n}}{\ell}=\left(\frac{n \pi A_{n}}{2 \ell}\right)^{2} \tag{5.92} \end{equation*}

hence

Cn=ΔnEn=[kb(nπ)2+γ+k(nπ)2]1\begin{equation*} \mathscr{C}_{n}=\frac{\Delta \ell_{n}}{\mathscr{E}_{n}}=\left[k_{\mathrm{b}}\left(\frac{n \pi}{\ell}\right)^{2}+\gamma_{\ell}+k_{\ell}\left(\frac{n \pi}{\ell}\right)^{-2}\right]^{-1} \tag{5.93} \end{equation*}

and the total (classical) variance resulting from transverse vibrations in one of the two possible polarizations equals the sum of the modal variances

σ,t2=n=1N12(CnkT)2=n=1N12(kT)2[kb(nπ)2+γ+k(nπ)2]2;N=/Δ\begin{align*} \sigma_{\ell, \mathrm{t}}^{2} & =\sum_{n=1}^{N} \frac{1}{2}\left(\mathscr{C}_{n} k T\right)^{2} \\ & =\sum_{n=1}^{N} \frac{1}{2}(k T)^{2}\left[k_{\mathrm{b}}\left(\frac{n \pi}{\ell}\right)^{2}+\gamma_{\ell}+k_{\ell}\left(\frac{n \pi}{\ell}\right)^{-2}\right]^{-2} ; \quad N=\ell / \Delta \ell \tag{5.94} \end{align*}

The total variance is the sum of contributions from both polarizations; if these are equivalent, the total is twice the above value.

Stretching the rod compresses the range of motion of the transverse modes, doing work and reducing modal entropies; from a mean-force potential perspective, transverse modes introduce a source of (nonlinear) longitudinal compli- ance. This perspective clarifies the nature of the quantum effects: Each transverse mode can be regarded as a harmonic oscillator with a restoring force modulated by the degree of rod extension. In the quantum-mechanical limit, the transverse positional variance is proportional not to the transverse compliance (as in the classical regime)

σclassical limit 2=kT/ks\begin{e