Thermal diffusion and the 2nd law of thermodynamics

Question

If we look at the definition of heat flux,

$$\stackrel{\to }{J}=-\kappa\stackrel{\to }{\nabla}T \, ,$$

we may notice that it's defined to be a vector showing in the opposite direction of the temperature gradient, in accordance with Clausius's statement of the second law of thermodynamics. From there, by using the continuity equation we may derive the thermal diffusion equation

$${a}{\stackrel{\to }{\nabla }}^{2}T=\frac{\partial T}{\partial t} \, .$$

My question is: can we view this equation as yet another statement of the second law of thermodynamics? Yes, we did derive it using Clausius's statement and the continuity equation (a way of interpreting conservation of energy, since we have no other heat sources), but the famous statement of the second law $d{S}_{universe}\ge 0$ isn't a postulate either (it's derived from Clausius's inequality, which comes from Kelvin's statement of the 2nd law, at least in my book), so that leaves me confused regarding what we can, and what we can't define as forms of the 2nd law.

Answer 1

Fourier's law $\vec\jmath=-\kappa\vec\nabla T$ implies $$ \frac{dS}{dt}=\int d^3x \frac{\kappa}{T^2}(\vec\nabla T)^2 + \ldots $$ where $\ldots$ is other sources of dissipation (viscosity etc.). This result requires a little effort, but it is explained in standard fluid dynamics text books (chapter V of Landau, for example). Then the second law of thermodynamics requires $$ \kappa\geq 0. $$ Obviously, Fourier's law is just the linear response approximation for the heat current, and not an equivalent form of the second law. It has nothing to say other forms of dissipation, and about heat currents beyond linear gradients. However, Fourier's law is consistent with the second law provided $\kappa\geq0$.

Answer 2

When we talk about there being multiple statements of the second law what we mean is that the different statements are logically equivalent, i.e. we may postulate any valid statement of the second law we choose and derive all the others. So we can deduce the increase of entropy by assuming Clausius' statement, but we can also assume the increase in entropy and deduce Clausius' statement (or for that matter Kelvin's statement, Caratheodory's principle or any of the other statements)

There are a number of problems with taking $\mathbf{J} = - k\nabla T$ as a statement of the second law. Firstly it is a statement about systems which are continuous in space, where other statements of the second law do not place such a restriction. It may be possible to get around this, though probably not without postulating additional information.

Secondly Fourier's Law is, in some senses, stricter than the second law. A law $\mathbf{J} = - k\nabla T |\nabla T|^2$ would also be consistent, as would any Law of the form $$ \mathbf{J} = - \nabla T \sum_{n=0}^\infty k_n |\nabla T |^{2n} $$ so Fourier's Law is not equivalent to the second law of thermodynamics

Answer 3

The 2nd law is a restriction on the class of constitutive relations that would specify/describe the material behavior, here heat conduction. Fourier's law of heat conduction is not a "law" or part of a "law", it is rather a constitutive relation describing material properties for which the 2nd law applies. The relationship to the 2nd law is manifested in the a priori inequality

$$\vec q \cdot \frac {1}{\theta} \vec \nabla \theta \ge 0$$.

($$\text{Notations}:\begin{cases} \vec q: \text{heating flux}, \\ \theta: \text{temperature}, \\ \eta: \text{specific entropy},\\ \rho: \text{mass density}, \\ \delta: \text{dissipated specific energy}.\end{cases} $$)

In fact, this is a special case. The more general one is obtained by combining Planck's inequality for internal dissipation

$$\delta = \theta \dot \eta - \frac{1}{\rho}\vec \nabla \cdot\vec q \ge 0 \\\\\; \text{with the Fourier inequality, and then one gets},\\ \rho \delta - \vec q \cdot \frac{1}{\theta} \vec\nabla \theta \ge 0, \\ \text{or after rearrangement} \\ \rho \dot \eta - \vec \nabla \left(\frac{\vec q}{\theta}\right) \ge 0 \\\ \text{Upon integration},\\ \dot H \ge \int {\frac {\vec q}{\theta}}.$$, which is the time rate of the conventional form of the Clausius inequality.

One cannot obtain the Fourier and Planck inequalities individually from the Clausius inequalty. All this is described in Truesdell: Rational Thermodynamics, chapter 2. (For simplicity I left out the volume heating supply term from the Clausius inequality)