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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.

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  • $\begingroup$ A thought that came to mind is that the second law actually gives direction to the arrow of time (towards a global increase in entropy), but this equation governs just local cases (i.e. works on subspaces of the ${ℝ}^{3}$ domain with no sources of heat or possibilities to do work), but those cases can be amended into the equation. $\endgroup$ Sep 7, 2015 at 0:49

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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 $\kappa\geq0$.

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  • $\begingroup$ That's fine, but I wonder if it can get more general than this... $\endgroup$ Sep 7, 2015 at 1:05
  • $\begingroup$ Not sure wht your question is. Obviously Fourier's law is only a special case of the second law. Added a short comment. $\endgroup$
    – Thomas
    Sep 7, 2015 at 1:08
  • $\begingroup$ Yes, that comment (and By Symmetry's answer) is the direction of reasoning I wanted to see to answer my question. Anyway, thanks for the contribution, +1. $\endgroup$ Sep 7, 2015 at 1:11
  • $\begingroup$ It's actually much worse than this because one can not continue much past the linear approximation for small gradients. If we heat systems so fas that we get large gradients then the heat transport soon becomes convective with convection cells first developing in nice geometric patterns that soon become chaotic and eventually we will cause phase transitions in the material aka "boiling". Every cook is familiar with this phenomenon. In effect there is thermodynamics which is really "near-thermostatics", a narrow linear transport range and then full chaos. $\endgroup$
    – CuriousOne
    Sep 7, 2015 at 2:26
  • $\begingroup$ I don't quite agree. Convection, turbulence, bubbling, boiling, detonation, etc is perfectly well described by fluid dynamics at the Navier-Stokes-Fourier level. To actually see a breakdown of the gradient expansion you usually have to look at more subtle things, like extremely dilute gases. $\endgroup$
    – Thomas
    Sep 7, 2015 at 2:39
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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

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  • $\begingroup$ Yes, this was the type of answer I was looking for, thanks. $\endgroup$ Sep 7, 2015 at 1:12
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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)

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  • $\begingroup$ Hi, sir; I've (superficially) modified the mathjax to make it look better without altering the main body. You can check this & if there is any mistake, then pardon me; you can rollback it BTW:D $\endgroup$
    – user36790
    Sep 7, 2015 at 14:14

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