de Gennes narrowing

 In colloidal solutions, a widely-used relation connects the scale-dependent collective diffusion constant and the structure factor:
\begin{equation}
D_c(q) =\frac{D_0}{S(q)}
\label{eq:dGn}
\end{equation} and is generally known as de Gennes narrowing since its use by de Gennes in the context of quasi-elastic neutron scattering from liquids [1].

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. 

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.
As discussed above, the Patterson function is similar to the density and thus we will apply the same approximation to \(P(\mathbf{r})\), which is the natural descriptor of the system, due to its intimate relation with the intensity \(I(\mathbf{q}) = \left | \tilde{\rho}(\mathbf{q}) \right |^2\). 

Patterson functions

Fourier transforms

We will use the following convention for the Fourier transforms:\begin{equation} \begin{split} \rho(\mathbf{q}) = \mathcal{F} [\rho(\mathbf{r})](\mathbf{q}) & \triangleq \int_{\mathcal{V}} \rho(\mathbf{r}) \exp(-i \mathbf{q} \mathbf{r}) {\textrm d} \mathbf{r} \\ \rho(\mathbf{r}) = \mathcal{F}^{-1} [\tilde{\rho}(\mathbf{q})](\mathbf{r}) & \triangleq \dfrac{1}{(2\pi)^3}\int_{\mathbb{R}^3} \tilde{\rho}(\mathbf{q}) \exp(i \mathbf{q} \mathbf{r}) {\textrm d} \mathbf{q} \end{split} \label{eq:Fourierdef} \end{equation} where we integrate in real space over the (as yet unspecified) volume of interest \(\mathcal{V}\) and in reciprocal space over the entire \(\mathbb{R}^3\).

Wiener-Khinchine theorem

The autocorrelation of the real-space density function is \(\Gamma_{\rho \rho} = \int_{\mathcal{V}} \rho(\mathbf{r}') \rho(\mathbf{r}'+\mathbf{r}) {\textrm d} \mathbf{r}'\), which can be developed (using the second line of \eqref{eq:Fourierdef}) into:\begin{equation} \begin{split} \Gamma_{\rho \rho}(\mathbf{r}) & = \dfrac{1}{(2\pi)^6} \int_{\mathcal{V}} {\textrm d} \mathbf{r}' \rho(\mathbf{r}') \int_{\mathbb{R}^3} {\textrm d} \mathbf{q} \, \tilde{\rho}(\mathbf{q}) \exp(i \mathbf{q} \mathbf{r}') \int_{\mathbb{R}^3} {\textrm d} \mathbf{q}' \tilde{\rho}(\mathbf{q}') \exp[i \mathbf{q}' (\mathbf{r}' + \mathbf{r})] \\ & = \dfrac{1}{(2\pi)^6} \int_{\mathbb{R}^3} {\textrm d} \mathbf{q} \int_{\mathbb{R}^3} {\textrm d} \mathbf{q}' \tilde{\rho}(\mathbf{q}) \tilde{\rho}(\mathbf{q}') \exp(i \mathbf{q}' \mathbf{r}) \underbrace{\int_{\mathcal{V}} {\textrm d} \mathbf{r}' \exp[i (\mathbf{q} + \mathbf{q}') \mathbf{r} ]}_{(2\pi)^3 \delta (\mathbf{q} + \mathbf{q}')} \\ & = \dfrac{1}{(2\pi)^3} \int_{\mathbb{R}^3} {\textrm d} \mathbf{q}' \exp(i \mathbf{q}' \mathbf{r}) \tilde{\rho}(\mathbf{q}') \underbrace{\int_{\mathbb{R}^3} {\textrm d} \mathbf{q} \, \tilde{\rho}(\mathbf{q}) \delta (\mathbf{q} + \mathbf{q}')}_{\tilde{\rho}(-\mathbf{q}')} \end{split} \end{equation} where we assumed that everything converges, and thus we can interchange the integration order at will. Dropping the prime and noting that \(\tilde{\rho}(-\mathbf{q}) = \overline{\tilde{\rho}(\mathbf{q})}\) (Friedel's law) we finally prove the Wiener-Khinchine theorem: the autocorrelation function of the scattering length density is the inverse Fourier transform of its spectral density: \begin{equation} \Gamma_{\rho \rho}(\mathbf{r}) = \dfrac{1}{(2\pi)^3} \int_{\mathbb{R}^3} {\textrm d} \mathbf{q} \exp(i \mathbf{q} \mathbf{r}) \left | \tilde{\rho}(\mathbf{q}) \right |^2 = \mathcal{F}^{-1} [|\tilde{\rho}(\mathbf{q})|^2] \label{eq:WK} \end{equation}

The Patterson function

As discussed during the lecture, the scattered intensity is precisely the spectral density of \(\rho(\mathbf{r})\): \(I(\mathbf{q}) = \left | \tilde{\rho}(\mathbf{q}) \right |^2\). Unlike \(\rho(\mathbf{r})\) itself, its autocorrelation \(\Gamma_{\rho \rho}(\mathbf{r})\) is directly accessible via Fourier transform from the experimental data, provided their quality and \(q\)-range are sufficient. Since it is frequently used in crystallography, it has a specific name: the Patterson function, denoted by \(P(\mathbf{r})\).

Correlation and convolution

In reciprocal space, the signal recorded by the detector at position \(\mathbf{q}\) is characterized by the electric field amplitude \(E(\mathbf{q})\), but the experimentally accessible quantity is its modulus squared, the intensity \(I(\mathbf{q}) = |E(\mathbf{q})|^2\). In real space, the structure is described by the density function \(\rho(\mathbf{r})\) but, as we will see in the next post, it is useful to define two new types of functions "of the type of the square", but where the two instances of \(\rho\) are evaluated in different space points.

Curvature of a planar curve

I have done this calculation several times over the years, so I might as well write it down in detail, in case it may be of use to someone else.

We are interested in the curvature \(C = 1/R\) of a planar curve \(y=f(x)\) at a given point A, where \(R\) is the curvature radius at that particular point, defined with respect to the curvature center \(O\) (intersection of the normals raised to the curve in A and its infinitesimal neighbor B.)

The angle subtending AB is: \(\displaystyle \mathrm{d}\alpha = \mathrm{d}s/R \Rightarrow C = \frac{\mathrm{d}\alpha}{\mathrm{d}s}\)

The length of the curve element AB is: \(\displaystyle \mathrm{d}s = \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2} \Rightarrow \frac{\mathrm{d}s}{\mathrm{d}x } = \sqrt{1+ f'(x)^2}\)

The derivative of \(f\) is directly related to the angle \(\alpha\): \(\displaystyle f'(x) = \frac{\mathrm{d}y}{\mathrm{d}x} = \tan \alpha \Rightarrow \alpha = \arctan \frac{\mathrm{d}y}{\mathrm{d}x} = \arctan [f'(x)] \Rightarrow \frac{\mathrm{d}\alpha}{\mathrm{d}x} = \frac{1}{1+f'(x)^2} f''(x)\)

Putting together the three relations above yields:

\[C = \frac{\mathrm{d}\alpha}{\mathrm{d}s} = \frac{f''(x)}{\left [ 1 + f'(x)^2\right ]^{3/2}}\]

The Dirac delta "function" - part II

The relation between δ(x) and dx

After introducing the Dirac delta "function" \(\delta(x)\) in the previous post, I'll try now to explain the relation between it and the differential element \(\mathrm{d}x\). In the process, I'll through all mathematical rigour out the window and aim for an intuitive understanding.

The Dirac delta "function" - part I

When describing a physical system, one would often like to describe some of its components as pointlike particles, i.e. with no spatial extent. Obviously, all objects have a finite size, but this size can be much smaller than the length scales relevant to the problem at hand. All the attributes of the particle (mass, charge, etc.) can then be assigned to a space point.

How to read an equation

The mere formal expression of an equation is not very useful, unless complemented by a more or less intuitive understanding. Different people may have different intuitions of a given formula or different mental images of one physical systems (more on that later).

The interesting part is that putting together two such different intuitions of a relation can yield non-trivial results with almost no algebraic manipulation, as I'll show below. What is the meaning of the following formula ?

\[\frac{1}{\sqrt{2\pi} \sigma} \int_{-\infty}^{\infty} \text{d}x \exp (i q x) \exp \left (- \frac{x^2}{2 \sigma ^2} \right ) \tag{1}\]

Power-law distribution of war magnitudes

While reading Steven Pinker's The better angels of our nature I stumbled upon the following argument for the magnitude of war (number of casualties) following a power-law distribution (page 220):

Recall the mathematical law that a variable will fall into a power-law distribution if it is an exponential function of a second variable that is distributed exponentially. My own guess is that the combination of escalation and attrition is the best explanation for the power-law distribution of war magnitudes.

The Kramers-Kronig relations - part 2

In part 1, we had stopped before going to the frequency domain  because we needed the Fourier transform of the sign function. This is where the technical difficulty appears, because we cannot simply write:

\begin{equation}
\label{eq:sgnTF}
\operatorname{sgn}(\omega) = \int_{-\infty}^{\infty} \text{d} t \exp (-i \omega t) \operatorname{sgn}(t) \tag{5}
\end{equation} as the integral does not converge. One can however define\begin{align}
\label{eq:sgnvp}
&\operatorname{sgn}(\omega) = \lim_{\epsilon \to 0} \int_{-\infty}^{\infty} \text{d} t \exp (-i \omega t - \epsilon |t|) \operatorname{sgn}(t) = \nonumber \\
&- \lim_{\epsilon \to 0} \left [ \frac{1}{i \omega + \epsilon} + \frac{1}{i \omega - \epsilon} \right ]= \lim_{\epsilon \to 0} \frac{2i \omega}{\omega^2 + \epsilon ^2}= \mathcal{P} \left ( \frac{2i}{\omega}\right ) \tag{6}
\end{align}

The Kramers-Kronig relations - part 1

[See part 2 for some technical aspects]
Very nice derivation of the Kramers-Kronig relations (on Wikipedia, of all places), exploiting the relation between the even and odd components of a function \(\chi (t)\) and the real and imaginary parts of its Fourier transform \(\chi (\omega) = \chi ' (\omega) + i \, \chi '' (\omega)\).

One usually invokes the analyticity of \(\chi (\omega)\) in the upper half-plane, which must first be derived from the causality: \(\chi (t) = 0\) for \(t < 0\). Complex integration along a well-chosen contour then yields the Kramers-Kronig relations in their standard form [1]:

\begin{align}
\label{eq:KK}  
\chi (\omega) &= \frac{1}{i \, \pi} \mathcal{P} \int_{-\infty}^{\infty} \text{d} \omega ' \frac{\chi (\omega ')}{\omega ' - \omega} \quad \text{or, for the components:} \nonumber \\
\chi '(\omega) &= \frac{1}{\pi} \mathcal{P} \int_{-\infty}^{\infty} \text{d} \omega ' \frac{\chi ''(\omega ')}{\omega ' - \omega} \\
\chi ''(\omega) &= -\frac{1}{\pi} \mathcal{P} \int_{-\infty}^{\infty} \text{d} \omega ' \frac{\chi '(\omega ')}{\omega ' - \omega} \nonumber
\end{align}
where \(\mathcal{P}\) denotes Cauchy's principal value.

Torsion constant of a rod. Dimensional derivation

The torsion constant of a circular rod (the torque needed to twist it by a given angle) is easily calculated from the equations of elasticity. Up to a numerical constant, it can also be derived by dimensional analysis, as shown below.

Consider the rod in Figure a), with radius \(r\), length \(L\) and shear modulus \(G\). Its upper end A is clamped. The torque \(T\) needed to turn the free end B is linear¹ in the twist angle \(\theta\):

\begin{equation}
T = \kappa \theta
\label{twist}
\end{equation}

Clearly, \(\theta\) depends on \(r\), \(L\) and \(G\). Are these the only relevant parameters? How about the bulk modulus \(B\), for instance? A non-rigorous way of showing its irrelevance is by considering a material with finite \(B\) and \(G = 0\) (e.g. a liquid): in this case the torsion constant is clearly zero². We can then write:

\begin{equation}
\kappa = K \, G^a r^b L^c
\label{kappa}
\end{equation}

Gibbs-Duhem and Euler relations

In a previous post I discussed the derivation of Euler's equation in thermodynamics and proposed a geometrical illustration. Here, I will continue the discussion by including the Gibbs-Duhem relation. We will need the following ingredients:

The Euler equation:
\begin{equation}
U = TS -pV + \sum_i \mu_i N_i
\label{eq:Euler}
\end{equation}
First-order homogeneity:
\begin{equation}
U(\lambda S, \lambda V, \lambda N_i) = \lambda U(S, V, N_i)
\label{eq:homog}
\end{equation}
The fundamental relation:
\begin{equation}
\text{d}U = T\text{d}S -p\text{d}V + \sum_i \mu_i  \text{d}N_i
\label{eq:fund}
 \end{equation}
Taking the derivative of \eqref{eq:Euler} and subtracting \eqref{eq:fund} yields
The Gibbs-Duhem relation:
\begin{equation}
S\text{d}T -V\text{d}p + \sum_i N_i  \text{d}\mu_i =0
\label{eq:GD}
\end{equation}

 

To get some insight into the Gibbs-Duhem relation, let us start from the observation that the space dimension is \(k+2\), where \(k\) is the number of components. When working with the energy, we use the extensive variables \( S,V,N_i\), as shown in the Figure, but there are also \(k+2\) intensive parameters \( T, P, \mu _i\). As already discussed, the transformation that takes the system along the dashed line connecting the current point and the origin (the "diagonal") occurs at constant intensive parameters, so that all terms in \eqref{eq:GD} are identically zero. We used this path to prove the Euler relation, so we label it as such.

The intensive parameters do not "feel" a transformation along the Euler path. For them to change, the system must evolve in the "perpendicular" subspace¹ (shown as a grey surface) of dimension \(k+1\). The Gibbs-Duhem relation \eqref{eq:GD} states that the variation of the intensive parameters is constrained to this subspace (that we duly label). An intuitive image is then that Euler is perpendicular to Gibbs-Duhem.

[Update 13/03/2013] A similar (but much simpler) situation occurs when mixing \(n\) different components. We can manipulate \(n\) parameters (the mass of each component) but can only set independently \(n-1\) intensive parameters (the mass concentrations) because the sum of all concentrations is always 1. The "missing" \(n\)-th variable is simply the total mass of the solution.

---

^{1. I use the quotes around the term perpendicular because \( (S,V,N_i)\) is not a metric space and we cannot define a scalar product (and hence an angle) on it. We can however define an affine transformation (scaling all parameters by the same amount \(\lambda\)) and say more properly that a change in the intensive parameters requires a non-affine transformation in the \( (S,V,N_i)\) space. ↩}

The Euler equation in thermodynamics

The internal energy \(U\) and its natural variables \(S, V, N_i\) are extensive quantities. It is then -mathematically- very easy (see Callen [1], section 3.1 for the canonical derivation) to prove the Euler equation:

\begin{equation}
U = TS -pV + \sum_i \mu_i N_i
\label{eq:Euler}
\end{equation}Briefly, one only needs to write the definition of first-order homogeneity:
\begin{equation}
U(\lambda S, \lambda V, \lambda N_i) = \lambda U(S, V, N_i)
\label{eq:homog}
\end{equation}take the derivative with respect to \(\lambda\) and set \(\lambda = 1\).

This demonstration is very elegant but it can hide the physical meaning of the relation \eqref{eq:Euler}. It is always a good idea to consider a system undergoing a precise transformation. Unless we can clearly identify the latter, we have not really understood the problem.

In our case, the transformation can be described as follows: all variables increase at the same rate. Since they are extensive, we can consider a system with length \(L \) (see Figure 1) and a cursor that can slide along the \(x\) axis.

Figure 1

At the position of the cursor we can introduce in the medium a wall that defines a new system, of length \(\lambda L \). Clearly, all extensive variables \(U, S, V, N_i\) have been multiplied by \(\lambda\), while the intensive parameters \(T, p, \mu_i\) remain unchanged. We are thus moving along the "diagonal" of parameter space, i.e. the line connecting the current point to the origin, as shown in Figure 2 for a one-component system.

Figure 2

As on any path, we can of course write the fundamental relation for an infinitesimal displacement:

\begin{equation}
\text{d}U = T\text{d}S -p\text{d}V + \sum_i \mu_i  \text{d}N_i
\label{eq:fund}
\end{equation} On this particular path (and only on this one [2]) the derivatives \(T, p, \mu_i\) are all constant so we can extend \eqref{eq:fund} to arbitrary displacements, yielding precisely \eqref{eq:Euler}.

In a future post I will try to show how the Gibbs-Duhem relation fits into this geometrical picture. UPDATE: here it is!

---

[1] H. B. Callen, Thermodynamics and an Introduction to Thermostatistics (2nd ed.), New York: John Wiley & Sons, 1985.

[2] Since we derive \eqref{eq:Euler} by moving along one particular path, one might wonder why it holds for all parameter values, even outside the given path. It should be noted that \eqref{eq:Euler} applies to a given state (a point in parameter space), and for each point we can draw its "diagonal". In contrast, \eqref{eq:fund} is written for a transformation.