# Electron vs Optical microscopes

Scanning electron microscopes have up to 1 nm resolution  and up to 1,000,000 X magnification which is much better than optical microscopes that have up to 200 nm resolution and 1000 X magnification. That’s cool but it turns out that they’re also a bit more complicated to operate.

A month ago I wanted images of a polymer sample using a scanning electron microscope but then realized that the samples had to be specially prepared by coating them with an electrically conductive material. I didn’t have any expertise in this art so the images were never made. But, it’s interesting to figure out why such preparation is necessary.

From this website I learned that unlike an optical microscope, the electron microscope works in the following manner:

1. The light source is replaced by a beam of very fast moving electrons.
2. The specimen usually has to be specially prepared(i.e. conductive coating) and held inside a vacuum chamber from which the air has been pumped out (because electrons do not travel very far in air).
3. The lenses are replaced by a series of coil-shaped electromagnets through which the electron beam travels.

The reason why the magnification allowed by an SEM is much more important than that of an optical microscope is due to the wavelength of an electron(1eV), 1.23 nm, which is much shorter than the wavelength of a photon(1eV),1240 nm. This may be computed using the relation found by De Broglie: $\lambda = \frac{h}{p}$ where $h$ is planck’s constant and $p$ is the momentum.

$\frac{\lambda_{photon}}{\lambda_{electron}} \approx 1000$ which is the main reason why the magnification of SEM’s is approximately 1000X more important than that of optical microscopes.

# The Poisson Ratio and conservation of volume

This summer, I had the chance to work on modelling robot skin at Hanson Robotics and I’ve learned quite a bit of continuum mechanics in the process. But, one of the most important lessons I’ve learned so far turned out to be quite simple yet counter-intuitive.

Prior to reading any texts on elasticity, I would assume that volume was always conserved as I thought this would be useful for estimating elastic properties. In particular, I used this assumption when estimating the Poisson Ratio and Young’s Modulus of the rectangular sample I was modelling. However, this led to really bad estimates for the Poisson Ratio in particular which is defined to be the negative ratio of transverse to axial strain.

After reflecting on the source of my error, I wondered whether the the volume of an elastic material with a prismatic geometry wasn’t necessarily conserved when subject to small uni-axial strains.

I realised that this fact can be easily demonstrated for Poisson Ratios in the range $-1 \leq \nu < 0.5$:

1. We are given $\nu=-\frac{\Delta{W}/W_0}{\Delta{L}/L_0}=-\frac{\Delta{T}/T_0}{\Delta{L}/L_0}$ where $-1 \leq \nu < 0.5$

2. Let $\frac{\Delta{L}}{L_0}=\alpha > 0$ then $latex \begin{cases} \Delta{W}=-W_0*\nu*\alpha \\ \Delta{T}=-T_0*\nu*\alpha\\ \end{cases}$
3. $V'=L*T*W=(L_0+\alpha L_0)*T_0*(1-\nu \alpha)*W_0*(1-\nu \alpha)=L_0*T_0*W_0*(1+\alpha)*(1-\nu \alpha)^2=V*(1+\alpha)*(1-\nu \alpha)^2$

4. The equation in part 3 actually reduces to:

$\frac{\Delta{V}}{V}=(1+\alpha)*(1-\nu \alpha)^2-1$

Clearly, for $\nu \leq 0$, $\frac{V'}{V}>1$ so the volume has increased. And for $\nu >0$, the volume is constant only when $\nu = \frac{1}{\alpha}*(1-\sqrt{\frac{1}{1+\alpha}})$. Since $\nu$ is supposed to be constant and $\alpha$ is allowed to vary we must conclude that the volume is not constant in this case either.

If I should trust my arguments then for any given non-negative poisson ratio, there is a single non-trivial value for $\alpha=\Delta{L}/L_0$ such that $V'=V$. The graph of the function $f(\alpha)= \frac{1}{\alpha}*(1-\sqrt{\frac{1}{1+\alpha}})$ is given below:

Now, we must deal with the case $\nu = \frac{1}{2}$ separately:

1. Let’s suppose, without loss of generality, that our material has the geometry of a cylinder. Then $V=\pi r^2 L$.
2. If its volume is constant(i.e. it’s incompressible), then: $\quad dV= 2\pi r L dr +\pi r^2 dL = 0$$\nu \equiv \frac{dr/r}{dL/L} = \frac{1}{2}$

So the moral of the story is that for most materials that we interact with in everyday life, their volume isn’t conserved when stretched or compressed unless they happen to have a poisson ratio of $\nu = \frac{1}{2}$.