Is Fourier's law of conduction a consequence of the second principle?

Question

In classical thermodynamics courses, entropy is often motivated by the need to justify that heat flows from high temperatures zones to lower temperatures zones: this is seen as a consequence of maximizing entropy. However it could also be seen to be a consequence of the law of heat conduction. Looking up in the non-equuilibrium thermodynqmics literature I can see a connection with Onsager's reciprocal relations but from what I can tell they are introduced as phenomenological laws.

Can the law of conduction be derived from the assumptions of classical thermodynamics?

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user Whelp

Comments

there is thermodynamics of non-equilibrium

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user Nikos M.

Answer 1

When we speak about heat conduction we must use the laws of continuum thermomechanics, where quantities like temperature, internal energy, etc. can vary from place to place.

The second law of thermodynamics for a body $B$ when the body is at rest and the only form of heating is by contact (as opposed to bulk heating, as in a microwave oven for example) can be written this way: $$\frac{\mathrm{d}}{\mathrm{d}t}\int_B s \,\mathrm{d}v \geqslant - \int_{\partial B} \frac{\boldsymbol{q}}{T} \cdot \mathrm{d}\boldsymbol{a}\tag{*}\label{2ndlaw}$$ where $\partial B$ denotes the surface of the body, $s$ is the entropy per volume, $\boldsymbol{q}$ is the heat outflux (energy/time/area), $\mathrm{d}v$ is the volume element and $\mathrm{d}\boldsymbol{a}$ the area element at the surface, with outward-pointing normal. Note that $\boldsymbol{q}\cdot\mathrm{d}\boldsymbol{a} < 0$ means that the body is heated. The inequality above is the "Clausius-Duhem inequality" mentioned in hyportnex's answer.

If the temperature always stays uniform throughout the body, the inequality above reduces to $\mathrm{d}S/\mathrm{d}t \geqslant Q/T$, typical for uniform processes, where $S$ is the total entropy of the body and $Q = -\int_{\partial B} \boldsymbol{q}\cdot\mathrm{d}\boldsymbol{a}$ the total heating (energy/time) of the body through its surface.

We can rewrite the inequality $\eqref{2ndlaw}$, which also holds for any part of the body, in local form using Gauss's theorem: $$\frac{\partial s}{\partial t} \geqslant -\nabla\cdot\frac{\boldsymbol{q}}{T} \equiv -\frac{1}{T}\nabla\cdot \boldsymbol{q} + \frac{1}{T^2}\boldsymbol{q} \cdot\nabla T\tag{**}\label{2ndlawloc}$$ (again, this is a special form valid when the body is at rest and the only form of heating is by contact).

In steady-state conditions the entropy and internal energy don't change with time, so the term on the left side and the first term on the right side (which by the first law equals the increase in internal energy divided by $T$) vanish. Considering that absolute temperature is positive, we are left with $$\boldsymbol{q} \cdot \nabla T \leqslant 0,\tag{***}\label{heatcond}$$ which says that the heat flux must form an obtuse angle with the temperature gradient – in other words, "heat flows from hot to cold", which I suppose is what the question refers to as "the law of conduction". So: yes, this law can be derived from the second and first laws for continua – under a steady-state condition in a rigid body at least, for example in an iron bar which has been heated at one end and cooled at the other at a constant rate for some time.

But in more general situations the inequality $\eqref{heatcond}$ needs not hold.

That it needs not hold in general is clear from the local form $\eqref{2ndlawloc}$. For example, let me quote from Astarita (1990), § 7.1:

the second law reduces to the requirement in equation $\eqref{heatcond}$ only for steady-state phenomena. In other words, for unsteady-state phenomena, the second law does not forbid heat to flow in the direction of increasing temperature, if only for short intervals of time.

He continues in § 7.5:

Even if the isotropic form of Fourier's law has been established experimentally for steady-state conditions, it need not hold also under unsteady-state conditions. Indeed, consider the following constitutive equation for the heat flux [...] $$\boldsymbol{q} + \theta \frac{\partial\boldsymbol{q}}{\partial t} = -k \nabla T\tag{7.5.3}\label{astarita}$$ with $\theta$, the relaxation time for heat flux, a positive constant. This equation is guaranteed to deliver a heat flux vector which, in steady state, forms an obtuse angle with the temperature gradient vector, and its validity (or lack thereof) cannot be ascertained by steady-state experiments.

At unsteady state, equation $\eqref{astarita}$ does allow the heat flux vector to form an acute angle with $\nabla T$, as the following simple example shows. Suppose $\nabla T$ has been held constant at some value, and correspondingly the heat flux forms an obtuse angle with it. At some time $t = 0$, the temperature gradient is suddenly reversed. The heat flux will also reverse in a time scale of order $\theta$, but at $t$ larger than $0$ by an amount negligibly small as compared to $\theta$ it will still have the direction it had at negative times–and hence it will form an acute angle with $\nabla T$.

This, however, does not contradict the second law, since equation $\eqref{heatcond}$ requires the heat flux to form an obtuse angle with $\nabla T$ only at steady state. If a finite relaxation time $\theta$ is allowed for, the usual Maxwell relations do not hold at unsteady state, and hence the other terms appearing in equation $\eqref{2ndlawloc}$ may well compensate for a positive value of the last term on the left-hand side.

Indeed, some experimental results on the rate of crystallization in polymers suggest that a constitutive equation for the heat flux of the type of equation $\eqref{astarita}$ is needed in order to model the data.

What Astarita says also implies that Fourier's law of conduction, $\boldsymbol{q} = -k(V, T) \nabla T$, cannot be derived from the second law only: it is a constitutive equation, that is, an equation specifying the heat-conduction properties of particular bodies only. Other bodies may satisfy different laws (cf. Astarita's $\eqref{astarita}$). Fourier's law $\boldsymbol{q} = -k(V, T) \nabla T$ can be derived by assuming, besides the second law, also a dependence of the fluid properties on particular variables and a form of linearity; see for example Samohýl & Pekař (2014), §§ 3.5–7, and the references there.

For the history and other comments about the second law for continua $\eqref{2ndlaw}$ see Truesdell (1984).

References

• Astarita, G (2000): Thermodynamics: An Advanced Textbook for Chemical Engineers (Springer).

• Samohýl, I., Pekař, M. (2014): The Thermodynamics of Linear Fluids and Fluid Mixtures (2nd ed., Springer).

• Truesdell, C. A., editor (1984): Rational Thermodynamics (2nd ed., Springer).

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user pglpm

Comments

You set $\textrm{div} \textbf{q}=0$ for steady state but that is not necessary. For fluid motion Truesdell derives the energy conservation laws as (page 110, 2.18) $\rho \dot {\epsilon} = w + \textrm{div} \textbf{h}+ \rho s$, so that in steady state for $\textrm{div} \textbf{h}=0$ to hold even if you assume $s=0$ (volume supply) that you also need the stress power be zero (eq 2.14) $w=0$.

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user hyportnex

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Also, on page 117 "The derivation of the Clausius-Duhem inequality makes it clear that, conversely, the Planck inequality and the Fourier inequality do not follow from it in general. By (2.48), it is equivalent to the local inequality (2.47). A little reflection shows that the Fourier inequality, correct as it is for the case envisaged by the classical theory of heat conduction, cannot be general. "

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user hyportnex

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continued "... By placing a sufficiently strong local source of energy at a cold spot, we certainly ought to be able to force a neighboring hot spot to grow hotter rather than colder, especially if we help the process by putting a sink of energy at the hot spot. That is, heat can flow from cold to hot, just as water can flow uphill. In view of the definition (2.25), the internal dissipation $\delta$ is influenced by the local heating, be it through the supply s or the conduction div h. "

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user hyportnex

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"....According to the Clausius-Duhem inequality in the form (2.47), the heating flux — h must lie on or within a certain cone whose axis points along the gradient of coldness. If there is no internal dissipation, that cone is a plane normal to the gradient. If the internal dissipation is positive, the cone is obtuse, having an aperture which depends upon the magnitudes of the heating flux and the gradient of coldness as well as upon the coldness itself and the internal dissipation. "

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user hyportnex

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"...Therefore, in the context of a theory based on the Clausius-Duhem inequality, the Planck inequality serves to exclude certain constitutive equations. A theory that forces — h to lie in or on the cone is consistent with the Clausius-Duhem inequality, while one that permits — h to lie outside the cone is not."

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user hyportnex

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The inequality (eq 2.47) containing internal dissipation (eq. 2.25) and Fourier's conduction and also follows from the Clausius-Duhem inequality is $\rho\delta + \textbf{h} \cdot \textrm{grad} \theta/ \theta \ge 0$, where $\delta = \theta \dot{\eta} - \textrm{div}\textbf{h} - s/\rho$ (eq 2.25)

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user hyportnex

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Thank you @hyportnex. I'm familiar with the passage you mention, I was thinking of including it in my answer but wanted to keep it simpler. Sorry I don't fully get your point. Are you saying that (***) in my answer doesn't follow from the Clausius-Duhem inequality? That's correct, but it follows if we also assume steady-state and that the body is rigid (no work), as I wrote.

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user pglpm

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I added some text to make this more precise after eqn (***).

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user pglpm

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Regarding (***), even if you assume rigidity could a heat sink attached to the boundary while $\textrm{div} \textbf{q} \ne 0$ assure that $\dot{\epsilon} = 0$?

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user hyportnex

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Sure, but I assumed that "the only form of heating is by contact ", before eqn (*). My first point was to show that the statement "heat flows from hot to cold", in a common situation in which it's envisaged – no body sources or sink, stationarity, body at rest – does follow from the Clausius-Duhem inequality. My second point was to stress with an example that that statement only holds under particular conditions. There are all sorts of way by which heat can flow from cold to hot: mechanical or electromagnetic work, non-stationarity, heat body sources or sinks, mass transport...

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user pglpm

Answer 2

Short answer is no but even then it depends on what you mean by "the 2nd law of thermodynamics". In conventional treatments of so-called equilibrium thermodynamics Fourier's law of heat conduction is completely independent of the rest. In what is called "rational thermodynamic" where the 2nd law is formulated as the "Clausius-Duhem inequality" it is in fact becomes part of the "2nd law" and a generalization of it as well. From the Clausius-Duhem inequality it can be shown that for heat conduction in the linear regime the conductivity must be positive or if in an anisotropic crystal a positive definite tensor. The symmetry of the tensor would follow from the so-called Onsager's reciprocity principle but Truesdell claims it has never been verified experimentally for all crystalline classes, but his, C. A. Truesdell: Rational Thermodynamics where you can read quite a lot on this subject is an "old" book, so there maybe newer experimental results on that. In fact, Truesdell uses the paucity of experiments on the symmetry of the heat conduction tensor to denounce "Onsagerism" as a quasi-religious movement that has never produced much of anything. The same formalism is used to introduce the "rational thermodynamics" of diffusion.

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user hyportnex

Comments

Is Principles of General Thermodynamics better than Callen's Thermodynamics and an Introduction to Thermostatistics?

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user Geremia

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@Geremia, havent read the last book, but the outlook of the "General Thermodynamics" book on thermodynamics and entropy (as basic and not as merely statistical) is sth i agree with (and more). Here is an arxiv paper which summarises the approach of the book (plus there are links to further study along the same lines). Note i have stopped answering to this site (Phys.SE) but wanted to respond to this comment

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user Nikos M.

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@Geremia, if you dont put "@ + NikosM" i dont get a notification of your comment (only the author of this answer gets one)

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user Nikos M.

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@Geremia Callen's book has been probably the most popular undergraduate text on thermodynamics for physics majors in the US for at least 40 years. It introduces entropy as a fundamental macroscopic descriptor of the state of matter just as energy, mass or charge is. The idea goes back to Tisza who references Gibbs in this approach to describe equilibrium in thermostatics. It is truly an excellent book.

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user hyportnex

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@Whelp: to start with, you can check Truesdell (ed.): Rational Thermodynamics (2nd ed., Springer, 1984), and for a more applied point of view Astarita: Thermodynamics: An Advanced Textbook for Chemical Engineers (Springer 1990).

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user pglpm

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@Geremia, well i download it once some time ago, in pdf format. If you search with the authors and the title (+ .pdf) you should find it. Here is the TOC after a quick search (i remember it took a while to find it but it was there)

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user Nikos M.

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@Geremia, here one can read it online (from here)

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user Nikos M.

Answer 3

One cannot "derive" any non-equilibrium rate law from thermodynamics, simply because they are beyond the scope of the theory. Thermodynamics simply does not deal with such phenomena and hence cannot tell you how such processes occur (in this case heat conduction). All that thermodynamics does is relate mean values of certain properties of systems amongst each other (which also implies that second order quantities like specific heat which depend on property fluctuations cannot be derived in thermodynamics and are instead an external input to the theory). Instead, a so called derivation of Fourier's law is still primarily phenomenological and can be done, for which one must have a look at Chaikin and Lubensky's book on Principles of Condensed Matter, under the topic of Hydrodynamics.

I shall just briefly mention some of the points as given in the book. The Fourier law can be thought as the lowest order contribution in a gradient expansion, thereby considering only small thermal gradients. There is no reason for linear response theory to be valid in a large temperature range, if it is, then it is a peculiarity of the material. Considering the slow spatial modulation in temperature as a long wavelength thermal excitation and the conservation of energy, we can systematically write down the linearized equation for the energy current, i.e heat. This would directly give you Fourier's law in the process. The associated transport coefficient is labelled the thermal conductivity and whose sign is fixed by the second law of thermodynamics, for stability reasons.

A detailed procedure is provided in the book mentioned above.

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user surajshankar

Comments

there is thermodynamics of non-equilibrium (look it up), plus almost all the laws of thermodynamics can be (and indeed have been) generalised to non-equilibrium

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user Nikos M.

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There is an enormous literature on non-equilibrium thermodynamics and your statement "Thermodynamics simply does not deal with such phenomena and hence cannot tell you how such processes occur (in this case heat conduction)" is just plainly wrong. Just because the subject is not finished as classical "thermostatics" is, does not mean it does not exist. Look up the works by Bridgman, Eckart, Truesdell, Coleman, Noll, Prigogine, Onsager, Glansdorff, Casas-Vazquez, Lebon, Jou, etc. Maybe not as sexy or fashionable as quantum gravity is but it is a very much alive research area.

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user hyportnex

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Yes, non-equilibrium methods do exist, though I personally like to classify that separately along with linear response theory. Looking at thermodynamics as the entirety of the content of the four laws limits us only to equilibrium systems, which is kind of how I like to see and use thermodynamics. In no way do I mean to espouse a belief that thermodynamics in the broadest sense is a "dead" field.

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user surajshankar

Answer 4

Can the law of conduction be derived from the assumptions of classical thermodynamics?

The answer should be no, because classical thermodynamics does not deal with description of irreversible processes in time; it only deals with equilibrium states. Second law of thermodynamics does not assert that entropy decreases in time, only that after irreversible process ends up in a new equilibrium state, entropy could not have decreased.

The Fourier law of heat conduction is a way to describe what happens to temperature of an object in time. In one way, it is more general description, because it describes non-equilibrium state, but in other way it is also less general, because it does not apply to all heat conduction processes.

It seems strange, but these two theories of heat are largely separate - they deal with different kind of questions about heat and temperature. Perhaps this means we do not have satisfactory theory of them (irreversible thermodynamics) yet.

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user Ján Lalinský

Comments

see my comment on 1st answer

This post imported from StackExchange Physics at 2025-01-31 21:49 (UTC), posted by SE-user Nikos M.