## 30 March 2013

### Pont Neuf

It is a little-known fact that the oldest bridge in every French city is called Pont Neuf.

## 6 March 2013

### de Gennes narrowing

In colloidal solutions, a widely-used relation connects the scale-dependent collective diffusion constant and the structure factor:

D_c(q) =\frac{D_0}{S(q)}
\label{eq:dGn}
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].

Intuitively, equation \eqref{eq:dGn} makes sense: if $$S(q)$$ is high for a certain value of $$q$$ then fluctuations with that particular $$q$$ are frequent, meaning that their energetic cost is low and that they will decay slowly. However, I have not yet found in the literature a simple yet rigorous derivation. This is what I will attempt below.

Consider a system characterized by a conserved scalar parameter $$\phi(\mathbf{r})$$ (for instance, the local particle concentration in a suspension). The free energy only depends on this field: $$\mathcal{F}[\phi(\mathbf{r})]$$ or, equvalently, on its Fourier components: $$\mathcal{F}[\phi(\mathbf{q})]$$.

For an isotropic system in the absence of applied fields, all fluctuations $$\phi(\mathbf{q})$$ with $$\left |\mathbf{q} \right | > 0$$ will eventually decay to zero. To fix the ideas, we will consider an overdamped relaxation (model B in the Hohenberg-Halperin classification [2], but of course far from criticality).

Let us write Fick's laws, with $$\mathbf{j}$$ and $$\mu$$ the current and chemical potential associated to $$\phi$$ (this is similar to the presentation in [2], Eqs. (2.2)-(2.9)):
$\frac{\partial \phi(\mathbf{r},t)}{\partial t} = - \nabla \mathbf{j} \, ; \quad \mathbf{j} = - \lambda \nabla \mu \, ; \quad \mu = \frac{\delta \mathcal{F}}{\delta \phi(\mathbf{r},t)} \, \Rightarrow \, \frac{\partial \phi(\mathbf{r},t)}{\partial t} = \lambda \nabla ^2 \frac{\delta \mathcal{F}}{\delta \phi(\mathbf{r},t)}$ where $$\delta$$ denotes the functional derivative. Introducing the Fourier components yields:
\label{eq:relax}
\frac{\partial \phi(\mathbf{r},t)}{\partial t}= \lambda \nabla ^2 \frac{\delta \mathcal{F}}{\delta \phi(\mathbf{r},t)}  \Rightarrow \frac{\partial \phi(\mathbf{q},t)}{\partial t} = - \lambda q^2 \frac{\text{d} \mathcal{F}}{\text{d}\phi(\mathbf{q},t)}
If the different Fourier modes are uncoupled, we can write the equipartition relation:

\label{eq:equi}
\mathcal{F} = \sum_{\mathbf{q}}  \frac{A(\mathbf{q})}{2} \phi(\mathbf{q}) ^2 \Rightarrow \frac{\text{d} \mathcal{F}}{\text{d}\phi(\mathbf{q},t) } = A(\mathbf{q}) \phi(\mathbf{q})
From \eqref{eq:relax} and \eqref{eq:equi}, the evolution of mode $$\mathbf{q}$$ is given by:
\label{eq:evol}
\frac{\partial \phi(\mathbf{q},t)}{\partial t}=- \lambda q^2 A(\mathbf{q}) \phi(\mathbf{q,t}) \Rightarrow \frac{\phi(q,t)}{\phi(\mathbf{q},0)} = \exp \left \lbrace - D_c(q) q^2 t \right \rbrace
where we invoked the isotropy of the system. The collective diffusion coefficient is given by:
\label{eq:Dc}
D_c(q) = \lambda A(q)
On the other hand,equipartition also implies:
S(\mathbf{q}) = \left \langle \phi(\mathbf{q}) \phi(\mathbf{-q}) \right \rangle = \frac{k_B T}{A(\mathbf{q})}
\label{eq:Sq}
with $$\left \langle \cdot \right \rangle$$ the ensemble average. From \eqref{eq:Dc} and \eqref{eq:Sq} we finally obtain:

\label{eq:final}
D_c(q) = \frac{\lambda k_B T}{S(q)}

For simplicity's sake I glossed over some technical details, such as the proper use of $$\phi(\mathbf{q})$$ and $$\phi(\mathbf{-q})$$ and the rigorous form of the time evolution in Eq. \eqref{eq:evol}.

[1] P. G. de Gennes, Liquid dynamics and inelastic scattering of neutrons, Physica A 25, 825-839 (1959).
[2] P. C. Hohenberg and B. Halperin, Theory of dynamic critical phenomena, Rev. Mod. Phys. 49, 435-479 (1977).

### 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:

U = TS -pV + \sum_i \mu_i N_i
\label{eq:Euler}

First-order homogeneity:

U(\lambda S, \lambda V, \lambda N_i) = \lambda U(S, V, N_i)
\label{eq:homog}

The fundamental relation:

\text{d}U = T\text{d}S -p\text{d}V + \sum_i \mu_i  \text{d}N_i
\label{eq:fund}

Taking the derivative of \eqref{eq:Euler} and subtracting \eqref{eq:fund} yields
The Gibbs-Duhem relation:

S\text{d}T -V\text{d}p + \sum_i N_i  \text{d}\mu_i =0
\label{eq:GD}

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.

## 4 March 2013

### 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:

U = TS -pV + \sum_i \mu_i N_i
\label{eq:Euler}
Briefly, one only needs to write the definition of first-order homogeneity:

U(\lambda S, \lambda V, \lambda N_i) = \lambda U(S, V, N_i)
\label{eq:homog}
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:

\text{d}U = T\text{d}S -p\text{d}V + \sum_i \mu_i  \text{d}N_i
\label{eq:fund}
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.

### Was Heidegger a Cratylist?

Some thoughts on:

Right Names: On Heidegger’s Closet Cratylism
by Christopher Eagle

The author claims that Heidegger practised a covert form of Cratylism, defined as the belief that some words are more appropriate than others for designating certain realities.

(a) In Plato's dialogue this adequacy should be understood as similarity; succinctly stated, "that rightness must be founded on mimesis." (p. 58).

Such an interpretation is however incompatible with Heidegger's thought, as the author admits:

"At a glance, Cratylism would appear to be necessarily at odds with Heidegger’s conception of language, due to his well-known antipathy for correspondence theory." (p. 59).

so the adequacy criterion must be modified:

(b) "If Heidegger allows us therefore to rewrite the terms of Cratylism in a way that situates it outside of the correspondence theory of truth, he has made this possible by shifting the sense of rightness away from an epistemological ground towards an ontological one." (p. 71).

While Cratylism fits naturally within framework (a), its meaning in relation with (b) is unclear. There is some mystique of language in Heidegger (see Poetically man dwells), but I am not sure that Cratylism is the proper term for it.