## 28 November 2012

### The Carvallo paradox and femtosecond lasers - part I

Fourier analysis amounts to writing any signal (including those limited in time) as a sum of infinitely extended harmonic functions. One way of framing this apparent contradiction is the Carvallo paradox:

Since a spectrograph only selects one component of the signal to be analyzed (and this component is infinite in time), it should detect that component both before and after receiving the signal.

The standard answer is that spectrographs have a finite resolution: when selecting light with a given wavelength $$\lambda$$, the result is in practice a finite interval $$( \lambda - \Delta \lambda, \lambda + \Delta \lambda)$$, defining the spectral resolution $$R = \lambda / \Delta \lambda$$. Let us assume a (respectable) value $$R = 10^5$$ in the visible range, at $$\lambda = 500 \, \text{nm}$$.

The uncertainty relation connects the resolution to the lenth of the pulse by: $\Delta \omega \Delta t \geq \frac{1}{2} \Rightarrow \Delta t = \frac{R}{2 \omega} \geq 8 \; 10^{-11} \, \text{s} \simeq 100 \, \text{ps}$ where in the second step I used $$\left | \Delta \omega / \omega \right | = \left | \Delta \lambda / \lambda \right |$$. A pulse with the spectral sharpness $$\Delta \lambda$$ must then be at least 100 ps long.

This is a very short interval for classical spectroscopy measurements. However, using modern optical techniques one can create ultrashort pulses, down to $$\Delta t \simeq 10 \, \text{fs}$$. When observing such a pulse at a resolution $$R$$ we should then see it as spread out to 100 ps. Where is the error?

More about in in the next part, where I'll also try to give a version of the paradox that is not affected by resolution.

## 27 November 2012

### Gaussian integral of an error function

In Surely You're Joking, Mr. Feynman!, Richard Feynman mentions a useful technique he used for evaluating integrals, namely taking the derivative under the integral sign. I will show here how this trick works in calculating the Gaussian integral of an error function. Averages over Gaussian distributions are omnipresent in physics, and the error function is just the primitive of the Gaussian, making the calculations relatively easy (and the result quite elegant.) Nevertheless, Mathematica (version 8) cannot perform this integral, and I could not find it in Gradshteyn & Ryzhik. I needed it to describe the interaction of a phase front with an external field, see the paper here.

Let us define: $I(\alpha, \beta, \gamma ) = \int_{-\infty}^{\infty} \text{d}x \exp (-\alpha x^2) \,\text{erf}(\beta x + \gamma)$ with $$\alpha, \beta \, \text{and}\, \gamma$$ real and $$\alpha$$ positive. For $$\gamma = 0$$ the integrand is an odd function, so $$I(\alpha, \beta, 0 ) = 0$$. We can also estimate $I' (\gamma) = \frac{\partial}{\partial \gamma} I(\alpha, \beta, \gamma) = \frac{2}{\sqrt{\pi}} \int_{-\infty}^{\infty} \text{d}x \exp (-\alpha x^2)\, \exp \left [-(\beta x + \gamma)^2\right ]$ which is a simple Gaussian integral: $$\displaystyle I' (\gamma) = \frac{2}{\sqrt{\alpha + \beta^2}} \exp \left ( - \frac{\alpha \gamma ^2}{\alpha + \beta^2}\right )$$

Finally, $I(\alpha, \beta, \gamma ) = \int_{0}^{\gamma} \text{d}u \, I' (u) = \sqrt{\vphantom{\beta}\frac{\pi}{\alpha}} \,\text{erf} \; \left ( \gamma \sqrt{\frac{\vphantom{\beta} \alpha}{\alpha + \beta^2}} \right )$ The reader can check that all derivatives exist and all integrals converge. What happens if we replace the linear term in the error function by a quadratic one?

## 26 November 2012

### An aesthetic argument against solipsism

In his short story The Other, from the volume The Book of Sand, Borges uses an artistic proof to convince his interlocutor (his younger self, actually) that their encounter is real and not a mere dream. He quotes a verse from Hugo (that the young Borges had not yet read), which is so striking that it could not have been dreamed up; the two then communicate across half a century.

This being a Borges story, things are more complicated than the summary above. In particular, the older Borges concludes that, although they had met, the younger one had in fact been dreaming. What I am interested in here is whether the story puts forward a successful argument against solipsism.

Such an argument requires demonstrating the existence of another person, who is:
• essentially similar, enough to achieve meaningful communication, and
• sufficiently different, so that it cannot be a figment of the subject's imagination
Does a successful encounter with a work of art fulfill these two conditions?

## 25 November 2012

### Power laws in small-angle scattering - part II

In the first part I showed that the SAXS intensity scattered by a platelet system goes like $$I(q) \sim q^{-2}$$, at least in some intermediate (but as yet unspecified) q range. Here I will show that for thin rods this dependence becomes $$q^{-1}$$, I will then derive the terminal (Porod) behaviour $$q^{-4}$$ and briefly consider the transition between these two regimes.

### Rods: α = 1

For a rod, the electron density profile:
$\rho (\mathbf{r}) = \delta(x) \delta(y) \mathrm{Cst}(z)$ with the same notations as in part I. Its Fourier transform $$\tilde{\rho} (\mathbf{q}) = \mathrm{Cst}(q_x) \mathrm{Cst}(q_y) \delta(q_z)$$ and, as in part I, I will assume the same form for the intensity $$I(\mathbf{q})$$. The "dual" object of a rod under Fourier transform is a platelet:
When spreading this intensity over reciprocal space we must keep in mind that the intersection of the plane with a sphere of constant $$q$$ is a great circle (shown in red in the figure above). Thus, the total contribution $$2 \pi q I_0$$ increases with $$q$$, but it must also be divided by the surface area of the sphere, yielding for the two spheres: $I(q_0) = \frac{2 \pi q_0 I_0}{4 \pi q_0^2} = \frac{I_0}{2 q_0} \quad \mathrm{and} \quad I(2 q_0) = \frac{4 \pi q_0 I_0}{4 \pi (2 q_0)^2} = \frac{I_0}{4 q_0}$ so that $$I(2 q_0) = I(q_0) /2$$ and $$\alpha = 1$$.

### Interfaces: α = 4 (Porod)

The density profile of an interface is invariant under x and y translations and is a Heaviside (step) function along z: $\rho (\mathbf{r}) = \mathrm{Cst}(x) \mathrm{Cst}(y) H(z)$ The Fourier transform of $$H(z)$$ is $$1/q_z$$, as one can see either by direct evaluation or by noting that the derivative of $$H(z)$$ is the Dirac delta etc. Finally, $\left | \tilde{\rho}(\mathbf{q}) \right |^2 = \delta(q_x) \delta(q_y) {q_z}^{-2}$ As for the platelet in part I, the scattering is confined along a rod perpendicular to the interface, but its intensity, instead of being constant, decreases along its length. Further spreading this signal over the sphere adds an additional $${q}^{-2}$$ factor, for a final $${q}^{-4}$$ dependence.

### Crossover

Although both are infinitely extended in the plane, a single interface (Porod) and two interfaces very close together (thin plate) exhibit very different scattering laws. However, at high enough scattering vector, all objects reach the Porod regime. We will discuss this crossover for the case of a plate with finite thickness. The Fourier transform of this object along its normal is easily shown to be a cardinal sinus, so that: $$\left | \tilde{\rho}(q_z) \right |^2 = \left [ \frac{\sin(q_z a)}{q_z a} \right ] ^2$$. Close to the origin, ie. for $$q_z a \ll 1$$, this function is flat, hence the approximation $$\sim \mathrm{Cst}(q_z)$$ we used for the platelet in part I. At high $$q$$, on the other hand, it behaves as $$q^{-2}$$, yielding the Porod law. To put it differently, when the typical scale $$L = 2\pi/q$$ over which we observe the object is much larger than its thickness, we are in the "platelet" regime and do not resolve the two interfaces. When $$L \ll 2a$$, on the other hand, we only observe "one interface at a time", justifying the treatment above and thus the $$q^{-4}$$ law. This regime is "terminal" insofar there is no smaller typical length scale in the system, until that of the composing atoms or molecules.

## 24 November 2012

### DIC microscopy image

The edge of a spin-coated droplet (reflection DIC image).

## 23 November 2012

### Power laws in small-angle scattering - part I

The small-angle X-ray scattering (SAXS) spectrum of particles with a well-defined shape (such as rods or platelets) is often characterized by a power-law dependence: $$I(q) \sim q^{-\alpha}$$, where the exponent $$\alpha$$ is directly related to the particle geometry. For "compact" particles, the large-$$q$$ intensity scales as $$q^{-4}$$ (Porod regime). Below, I'll give the most compact and yet -hopefully- understandable derivation I can think of for these power laws.

To simplify the derivation, we'll consider these objects as infinitely thin and infinitely large, meaning that we'll be looking at them on length scales much larger than their thickness and much smaller than their lateral extension. The approximation is legitimate, since it is in this range of length (or, conversely, scattering vector) that the power-law regimes are encountered.

### Platelets: α = 2

For a platelet, this simplification yields for the electron density profile:
$\rho (\mathbf{r}) = \mathrm{Cst}(x) \mathrm{Cst}(y) \delta(z)$
where "Cst" (constant) means that the density does not vary as a function of $$x$$ and $$y$$. Of course, a constant does not need an argument, but we will specify it in order to keep track of the space dimensions.

We now need the Fourier transform of $$\rho (\mathbf{r})$$, $$\tilde{\rho} (\mathbf{q})$$. Since the Fourier transform of a constant is a Dirac delta and viceversa, we simply have:
$\tilde{\rho} (\mathbf{q}) = \delta(q_x) \delta(q_y) \mathrm{Cst}(q_z)$
The intensity scattered at a given wave vector $$I(\mathbf{q}) = | \tilde{\rho} (\mathbf{q})|^2$$, and we'll admit that the latter function has the same form as $$\tilde{\rho} (\mathbf{q})$$. This is plausible: if a distribution is perfectly localized, we expect its square to share this property. Rigorously speaking, however, the square of a Dirac delta makes no sense (within classical distribution theory) so a proper derivation involves considering a finite size for the object, taking the modulus square of its Fourier transform and only letting the size go to infinity in the last step. Finally, the intensity scattered by a platelet perpendicular to the $$z$$ axis is a thin rod parallel to $$q_z$$, as shown in the figure below.

In solution, colloidal particles assume all possible orientations, so that this intensity is spread evenly over the sphere of constant $$q$$. Consider two such spheres, with radii $$q_0$$ and $$2 q_0$$. The rod contributes to each sphere the same amount, namely twice its (constant) intensity $$2 I_0$$, shown as red dots at the poles. This signal must however be divided by the surface area of the sphere, an area that increases as the square of the radius: $I(q_0) = \frac{2 I_0}{4 \pi q_0^2} \quad \mathrm{and} \quad I(2 q_0) = \frac{2 I_0}{4 \pi (2 q_0)^2}$ so that $$I(2 q_0) = I(q_0) /4$$ and, finally, $$\alpha = 2$$.

More power laws coming in part II of this post...

## 22 November 2012

### Physics and Photography

Short review of a short but very well-written introduction to the principles of photography: Science for the Curious Photographer: An Introduction to the Science of Photography by Charles S. Johnson Jr (author's blog is here). This slim (180 p.) volume covers a lot of ground, not only the obvious topics such as image formation or the physiology of vision but also, for instance, the operation of CMOS and CCD detectors and a discussion of polarization. The presentation is fairly technical, so an adequate background in physics is necessary.

The only critique would be that, for an introduction, it is somewhat light on the references (many of which are simply Wikipedia articles).

## 21 November 2012

### Electrical circuits and Euler's polyhedron formula

When solving an electrical circuit consisting of impedances and voltage sources, one needs to apply Kirchhoff's laws:
• The current law, yielding N-1 equations (with N the number of nodes)
•  The voltage law, for an additional L equations (where L is the number of elementary loops)
The unknowns are the currents flowing in each branch (B of them). We also assume that the branches do not cross. Note that writing the current law for the N-th node or the voltage law for a composite loop (consisting of several adjacent elementary loops) does not provide any further information, the resulting equations being linear combinations of the previous ones.

For the problem to be well-posed the number of unknowns and equations is equal, which we can write as:
$N + L - B = 1$ This relation is easily proven in plane geometry, but here I would like to show its intimate connection with Euler's formula, which states that, for a convex polyhedron, $V + F - E = 2$ where V, F and E are the numbers of vertices, faces and edges, respectively.

Let us start by establishing a correspondence between circuits and polyhedra, as shown in the figure below. Place a sphere on top of the (planar) circuit diagram and connect each node to the North pole by a line segment (this is known as a stereographic projection.) We define the vertices as the intersections of these segments with the sphere; the result is a convex polyhedron.

It is easily seen that, with the notations above, we have the straightforward equivalences V = N and E = B. The number of faces, however, F = L + 1, since the "topmost" face corresponds to the open area surrounding the circuit. Substitution in either of the equations above yields the other one.

## 18 November 2012

### Polarized microscopy image

Schlieren texture in nematic 5CB.