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Problem set-up

I am looking at two adjacent elastic cantilever beams. They are micron/nanoscale size and so they vibrate due to thermal noise. The problem I am trying to solve is whether the vibrations are enough for the tips of the cantilevers to touch or not.

The beams are the same identical height $H$, separated by distance $a$, and have identical Youngs modulus $E$, area moment of inertia $I$, and mass-per-length density of $\mu$. The deflection of the tip of each cantilever is $\delta_1(t)$ and $\delta_2(t)$.

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According to the Equipartition Theorem, the vibrations due to thermal noise will be such that

$$\frac{1}{2}k_B T = \frac{1}{2}K \sigma^2$$

where $k_B$ is the Boltzmann constant, $T$ is the absolute temperature, $K$ is the spring constant of the elastic cantilever $K=3EI/H^3$, and $\sigma^2$ is the mean square deflection of the cantilever caused by thermal vibrations.

Solving for the standard deviation of the displacement $\sigma$ we get:

$$\sigma = \sqrt{\frac{k_B T H^3}{3EI}}$$

From this it's trivial to solve for the when the average deflection of the tip is equal to half the gap width $a/2$. However $\sigma$ is just the standard deviation of the deflection, at any point in time the amplitude of the vibration may be much larger or smaller.

Assumption: the probability of the amplitude of the deflection of either cantilever being $z$ follows a Gaussian distribution:

$$P(z) = \frac{1}{\sqrt{2\sigma^2 \pi}}\exp{\left(-\frac{z^2}{2\sigma^2}\right)} $$

That means that at any time the amplitude of the deflection could be any value from $-\infty$ to $+\infty$ (though physically it's of course impossible for it to be greater than the total length of the cantilever $H$) but since $\sigma \ll H$ it should be a close enough of an approximation.

So now we have two vibrating cantilevers, the amplitude of each being random but having a known probability distribution. For simplicity we'll ignore all the vibrational modes except the first one.

Assumption: the two vibrating cantilevers are not necessarily in phase, with each one having a random phase angle. The phase angle of each cantilever has a uniform probability distribution:

$$P_{phase}(\theta_i) = 1/2\pi; \quad \theta_i\in[0,2\pi]$$

However since we are only worried about the difference between the phase angle of the two cantilevers, we can do a coordinate transform so that the phase angle of cantilever 1 is zero. Therefore for the position of the ends of each of the cantilevers as a function of time we have the following expressions:

$$\delta_1(t)=z_1 \cos (\omega t)$$ $$\delta_2(t)=z_2 \cos (\omega t + \theta)$$

where $\omega$ is the vibrating frequency, $t$ is time, and $\theta$ the phase angle difference between the two cantilevers. The frequency $\omega$ is assumed to be the 1st natural frequency which is

$$ \omega_1 = {\beta_1}^2\sqrt{\frac{EI}{\mu}} $$

where $\beta_1$ is the first Eigenvalue, determined by the Eigencondition $\cosh(\beta_1 H) \cos(\beta_1 H) +1=0$, which becomes $\beta_1 = 0.59686 \pi / H$.

Touching Condition

The tips of the two vibrating cantilevers will touch if at any time the following condition is met:

$$a-\delta_1+\delta_2 \leq 0$$

The $-\delta_1 + \delta_2$ can be evaluated by using phasor addition:

$$-z_1 \cos(\omega t) + z_2 \cos(\omega t + \theta) = z_3 \cos(\omega t + \theta_3)$$

where:

$$z_3= \sqrt{(z_2 \cos \theta - z_1)^2 + (z_2 \sin \theta)^2}$$

$$ \theta_3 = \arctan \left( \frac{z_2 \sin \theta}{z_2 \cos \theta - z_1} \right) $$

So the touching condition becomes

$$ a + z_3 \cos(\omega t + \theta_3) \leq 0 $$

For the sinusoidal term, we don't really care where it specifically is in the phase angle, all we care is whether or not the inequality is true at any time. So we can simply take the limits of that term, which is $\pm z_3$. Since we want the LHS to be less than zero and $a$ is always positive, we can simplify the inequality to this:

$$ a- z_3 \leq 0 $$

$$ a - \sqrt{(z_2 \cos \theta - z_1)^2 + (z_2 \sin \theta)^2} $$

and rearranging slightly:

$$ (z_2 \cos \theta - z_1)^2 + (z_2 \sin \theta)^2 - a^2 \geq 0 $$

Problem

Ultimately I want to determine the probability of the two cantilevers touching. That comes to finding the probability that inequality above is true. I know the probability distributions for the two displacements $z_1$ and $z_2$, and the phase angle $\theta$ as shown above. But I don't know how I'm supposed to use those to evaluate the probability of the inequality being true. I assume it will involve integrating $z_1$ and $z_2$ from $-\infty$ to $+\infty$ and $\theta$ from $0$ to $2\pi$, but beyond that I have no idea.

Edit: I cross-posted the final mathematical question at math.stackexchange here.

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  • 1
    $\begingroup$ The difficulty in the question seems to be mathematical, rather than physical. Have you considered posting on Mathematics SE? ... You could consider running a simulation. Probably the solution will require numerical computation anyway. $\endgroup$ – sammy gerbil Jan 26 '17 at 22:27
  • $\begingroup$ Part of the reason I posted it here was to see if others here thought my analysis of the problem made sense, see if perhaps there is another way to go about solving this, etc. I have some further questions about this that relate to the actual physics of it, but I need to be able to solve this first to even be able to get to those. But submitting it to math.overflow might be a good idea, if just phrased in terms of the mathematical problem at the end. $\endgroup$ – Derek Jan 27 '17 at 1:40
  • $\begingroup$ I could certainly just brute-force it and solve it using a Monte Carlo approach, but I would like to have some kind of analytical expression/integral, even if I have to evaluate that numerically. That will let me understand scaling trends, etc. on how touching probability is affected by parameters like length, stiffness, gap width, etc. $\endgroup$ – Derek Jan 27 '17 at 1:44
  • $\begingroup$ I'm not convinced this is right. Don't you need to take into account the frequency response function of the cantilevers to the random excitation? For example if they are lightly damped, their motion will be far from "random" - it will mostly be a superposition of their normal modes of vibration. $\endgroup$ – alephzero Jan 27 '17 at 5:03
  • $\begingroup$ alphzero, yes, the motion will be a superposition of the normal modes. I'm simplifying the problem by just using the 1st mode and ignoring the others, since the contribution of the total deflection of the tip from the $n>1$ modes is much smaller than that of the $n=1$ mode. Since they both have the same $E$, $I$, and $\mu$, then their characteristic frequency $\omega$ will be the same. $\endgroup$ – Derek Jan 27 '17 at 5:23
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Consider the coordinate system with origin at midpoint between the two cantilevers. Let $z$ be the horizontal coordinate, positive rightwards. Therefore cantilevers are located at $z=\pm a/2$. Probability distribution for displacement $z_1$ and $z_2$ of cantilever tips are $N_{z_1}(+a/2,\sigma)$ and $N_{z_2}(-a/2,\sigma)$ respectively for the right and left cantilevers, where $N$ is normal distribution. You are assuming that the motion of the two cantilevers is independent. Therefore the joint probability distribution that right cantilever-tip shall be displaced to $z_1$ and left cantilever-tip shall be displaced to $z_2$ is simply \begin{align} f(z_1,z_2)=f(z_1)f(z_2)=N_{z_1}(+a/2,\sigma)N_{z_2}(-a/2,\sigma) \end{align} For the cantilever-tips to touch we must have $z_1\leq z_2$ and $\lvert z_{1,2}\rvert\leq a/2$ (assuming that contact happens always in the region between the two cantilevers). That is, for any given $\lvert z_1\rvert\leq a/2$ for right cantilever-tip, the probability that the cantilever-tips shall touch is equal to the probability that left cantilever-tip is displaced to $z_2$ such that $z_1\leq z_2\leq a/2$. This probability is simply \begin{align} \int_{-a/2}^{+a/2}dz_1\int_{z_1}^{+a/2}dz_2~f(z_1,z_2) \end{align}

Of course a big assumption is being made here, which is that if the cantilevers touch then they shall do so only within a small neighborhood of their tips, and that the probability that tip of one cantilever shall touch some point outside of this neighborhood of the other cantilever is negligible. Geometrically thinking, this is a dubious assumption indeed.

-----------------Update------------------

Although I am not convinced that probability of $(z_2\cos\theta-z_1)^2+(z_2\sin\theta)^2-a^2\geq 0$ gives you the probability you are seeking, I shall show how to calculate it.

I take it that $z_1,z_2,\theta$ are all independent random variables. Let $f()$ denote p.d.f. of its argument random-variable. Then joint p.d.f. is \begin{align} f(z_1,z_2,\theta)=f(z_1)f(z_2)f(\theta) \end{align} Now \begin{align} & (z_2\cos\theta-z_1)^2+(z_2\sin\theta)^2-a^2\geq 0 \\ \Rightarrow & (z_2\cos\theta-z_1)^2 \geq a^2-(z_2\sin\theta)^2 \\ \Rightarrow & (\alpha-z_1)^2 \geq \beta^2 \\ & \alpha\equiv z_2\cos\theta,\beta\equiv\sqrt{ a^2-(z_2\sin\theta)^2} \end{align} If we fix $z_2,\theta,$ then $\alpha,\beta,$ are constants. Then $h(z_1)\equiv (\alpha-z_1)^2$ is a parabola and $h(z_1)=\beta^2$ is a horizontal line. The two intersect at $\alpha\pm\beta$. The inequality $(\alpha-z_1)^2 \geq \beta^2$ is satisfied if $z_1\leq \alpha-\beta$ or $z_1\geq \alpha+\beta$. Further we require that $\beta$ to be real, which implies \begin{align} \beta^2 & \geq 0 \\ a^2-(z_2\sin\theta)^2 & \geq 0 \\ (z_2\sin\theta)^2 & \leq a^2 \\ z_2^2 & \leq \left( \frac{a}{\sin\theta} \right)^2 \\ \therefore \quad \lvert z_2 \rvert \leq \frac{a}{\lvert \sin\theta \rvert} \end{align} Therefore required probability is \begin{align} & \int_0^{2\pi}d\theta \int_{-a/\lvert \sin\theta\rvert}^{+a/\lvert \sin\theta\rvert}dz_2 \left[ \int_{-\infty}^{\alpha-\beta} dz_1~f(z_1,z_2,\theta)+\int_{\alpha+\beta}^{+\infty} dz_1~f(z_1,z_2,\theta) \right] \\ = & \int_0^{2\pi}d\theta \int_{-a/\lvert \sin\theta\rvert}^{+a/\lvert \sin\theta\rvert}dz_2 \left[ 1-\int_{\alpha-\beta}^{\alpha+\beta}dz_1~f(z_1,z_2,\theta) \right] \end{align}

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  • $\begingroup$ Does this assume that the phase of each of the cantilevers is strictly antiphase, i.e. the left cantilever reaches its right-most position at the same time the right cantilever reaches its left-most position? $\endgroup$ – Derek Jan 27 '17 at 17:07
  • $\begingroup$ As I understand it, the amplitudes of oscillation are distributed Normally (and the phases are distributed uniformly), but what distribution results for the displacements of the ends? $\endgroup$ – sammy gerbil Jan 28 '17 at 0:18
  • $\begingroup$ @sammygerbil In the question I interpreted $z$ to be tip displacement. If that is not the case then we shall need to find the p.d.f. for tip displacement. $\endgroup$ – Deep Jan 28 '17 at 3:34
  • $\begingroup$ @Derek I formulated the answer in terms of a given p.d.f. for tip displacement. In this case phase considerations are redundant. Also time does not enter the picture because all probability distributions are time-independent (i.e. stationary), therefore so is the answer. You may think of it this way: Consider a large ensemble of such a pair of cantilevers. Then at any time instant the fraction of the ensemble in which cantilevers are touching is the same. $\endgroup$ – Deep Jan 28 '17 at 3:40
  • $\begingroup$ Deep, thanks for showing this analysis. This is great. I think I may have made some incorrect assumptions in setting up the problem though, as DanielSank points out in his comments above. I think you were expressing similar concerns where you stated that you weren't convinced that the probability question I stated gives the actual probability that I'm seeking. $\endgroup$ – Derek Jan 30 '17 at 20:27

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