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Take, for simplicity, a Lennard-Jones fluid below the critical temperature, which is to say that there is a phase separation into fluid and gas and thus an interface is formed. The macroscale picture is that there's tension acting tangent to this interface. But of course at the microscale things are not as pretty: The interface is diffuse and therefore the interfacial tension, too, should be distributed accordingly.

My question is, then: how does one define stress (in classical systems) with long-range interactions?


I have listed a couple of options below. As I understand it, the differences in the approaches are very much related to how things are handled in general relativity: Hilbert vs. Canonical vs. Belinfante-Rosenfeld stress-energy tensors. I am assuming the debate, if there ever was one, has been resolved in GR, but not being an expert on the subject matter, I would greatly appreciate if the lessons drawn from those could be explained in simple terms.

  1. Now, we could define the stress as the tensor whose divergence is force density. Now obviously this definition is not unique, and equating the time derivative of momentum density with that of force, we end up with $$\sigma^{mn}(r) = \left\langle- \sum_i \frac{p^m_ip^n_i}{m}\delta(\mathbf{r}-\mathbf{r}_i) + \sum_{i>j}\nabla_i^m\varphi(r_{ij})\int_i^j\delta(\mathbf{r}-\mathbf{\ell})\mathrm{d}\ell^n\right\rangle$$ where the integral is over any contour from the particle $i$ to $j$ [Schofield & Henderson, Proc. R. Soc. Lond. A 379, 231 (1982)]. That this contour is arbitrary is at the heart of the problem: With different choices one gets different results, which is problematic. It is often advocated that the stress tensor itself is a non-physical entity, and one should only be concerned with measurable thermodynamical quantities such as interfacial tension (which ought to be an integral over the stress), but this seems odd to me. While the latter can be calculated from the former, if the stress tensor is nonunique, so would the interfacial tension be (this actually does not happen in planar geometry, but does in more general systems).

  2. Another option might be to use, much like the aforementioned Hilbert stress-energy tensor, $$\sigma^{mn} = \frac{2}{\sqrt{g}}\frac{\delta \mathcal{F}}{\delta g_{mn}}$$ where the variation of free energy $\delta \mathcal{F} = \int \sqrt{g}\sigma^{mn}\delta\varepsilon_{mn}\mathrm{d}^3x$; $\varepsilon_{mn}$ being the strain, interpreted through metric changes for the above formula [Mistura, Int. J. Thermophys. 8, 397 (1987)]. This form is automatically symmetric and unique. Moreover, it is equivalent to the previous definition should one choose the integration contours to be straight lines from one atom to the next. This all sounds great, but will this always give the correct results for measurable quantities i.e. is it fully consistent with thermodynamics?

On top of these issues, one likely has to worry about non-extensibility.


To summarize: Can one generalize the Hilbert stress-energy tensor to classical thermodynamics by calling the variation of free energy w.r.t. the metric (multiplied by $2/\sqrt{g}$) the stress tensor? Or must one use the canonical stress(-energy) tensor? The latter would lead to problems with gauge invariance; How does one interpret these? Is there an obvious proof? And finally: Are the suggested approaches doomed to fail due to the long range nature of the problem, and thus the non-extensibility of the thermodynamical quantities?

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  • $\begingroup$ I can't answer your question, but of all the questions I've seen today on PSE this is the one that deserves an answer, I just hope it isn't ignored because of its difficulty level. $\endgroup$ – Zach466920 Aug 19 '15 at 4:31
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Only the symmetric stress tensor is physical, since thermodynamics demands a symmetric stress tensor. Since the symmetric stress tensor is unique, your option 2 is the correct one. (Canonical versions may be simpler but need not be physical; cf. the canonical momentum, which is often different from the physical momenetum.)

This is completely unrelated to the question of long-range forces as the stress tensor is a fully local concept. In particular, the stress tensor is already determined on the most fundamental level (quantum field theory) and remains the same on every level of approximation.

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  • $\begingroup$ For thermodynamics "to work" (extensibility), do you not need to always assume that the forces are short range? How does the stress defined as above correspond to a thermodynamic quantity? So suppose I want to compute the surface tension of vdW: Can I compute the stress from the formula above and then merely integrate it over the surface? Or would this give some other "stress"? e.g. Rossi et al., J. Chem. Phys. 132, 074902 (2010) have only established the connection with short-range forces. $\endgroup$ – alarge Aug 26 '15 at 0:34
  • $\begingroup$ Integrals over the stress tensor are extensive quantities and have in a thermodynamic setting integrals over the metric tensor as conjugate intensive quantities. See the book By Oettinger, Beyond equilibrium thermodynamics. - So probably the answer to your surface tension question is yes, after some slight smoothing of the stress tensor in a thin shell around the surface. $\endgroup$ – Arnold Neumaier Aug 26 '15 at 8:20
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Short answer: The major difficulty lies with the definitions themselves, and none of the possibilities given has a real physical meaning which can be univoquely related to stress in non extensive systems in its conventional mechanical original meaning.

The long one: What kind of systems does this apply to? This is not answered by referring to systems with long range interactions, because that same concept is not clearly defined, or better said, it is not completely clear what effect it has on the macroscopical behavior of the system.

For example molecules interacting with Lennard-Jones potential exhibit Ideal Gas behavior as well as Van der Waals gas behavior, and even liquid behavior, under the appropriate conditions. Has the potential changed? No, just the average kinetic energy of the molecules which impacts average separation, so apparently, whatever we want to classify this potential as, it clearly can give rise to different macroscopic behavior, and even extensivity.

The same happens with nuclei: whatever our definition of long range is, clearly the potential due to one of the nucleon has a range of comparable size to the nucleus, and you would expect this to be long enough range. Yet the nucleus exhibits saturation like liquids, with constant density for all stable nuclei.

Moreover, the approach taken on this topic determines a lot what you can expect. Stress is a concept from macroscopic-like descriptions, where you don't care for the details of the microscopic description. It also requires that the system which you want to characterize with stress has some degree of rigidity. I mean with this that the different parts of the system need to be linked in such a way that stress does not change the system completely, but offers an opposition to it.

This is definitely the case of solids, for which you could even zoom and still find microscopic parts compressing or extending or twisting under stress, but never slipping through each other like layers.

The caso of liquids is interesting though, because although microscopically particles are not really connected, and they will pass through each other almost unconnected, at a higher level you can define stress based on the amount of matter in average on each portion of its volume, which will echibit this constancy and kind of rigidity needed for a description in terms of mechanical stress. Gases under certain conditions can be treated as such but again, at certain macroscopical enough scale.

The cons of describing a fluid in this terms is thus, this is only valid for a scale in which the parts of the system are under thermodynamical equilibrium, at least locally. If this does not hold, then it is a mess to describe in thermodynamical terms, and still subject of research.

So for system of interacting flying particles like the one on deffinition one, to me it makes no sense to have a detail account of energy at such microscopical scale, when making a time derivative. It makes no sense to me because is a theoretical deffinition as useful as defining the path of brownian motion by the number of hits you get an averaged of time: it provides no insight on the nature of the phenomenon, and is basically saying: ``if we knew all the momenta and forces for every instant, we could calculate the macroscopic behavior of the system'', which is unpractical even for theoretical standards. Plus, we know that collective behavior, raised from a chaotic microscopic behavior, can often be very simple. This to me is call for theoretitians to find a different approach, maybe meso-scale like, in which the relevant behavior is explained, where the ab initio description is not a mechanicist one.

This is already the case of the second option you give. And the fact that simplifying the paths of integration considerably in the first one gives you the second one already proves the above point. This does not mean that microscopically particles behave so simplistic, it just means that if the macroscopic system behaves simply (i.e. local thermodynamical equilibrium, isotropic) then you obtain the same average behavior as if the system were simpler.

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  • $\begingroup$ This answer seems to be in conflict with the other one: Could you elaborate? As for your comment on not knowing all the forces at all instances of time: Definition 1 is actually regularly used to compute stresses from molecular dynamics simulations, and having an exact definition for microscopic stress which could be integrated over to give the value that can be macroscopically measured would in fact be of great benefit and directly applicable. $\endgroup$ – alarge Aug 26 '15 at 0:25
  • $\begingroup$ Because the other doesn't mention the problem with long range interactions. The problem with the first expression is that is not unique, and for integrations some simplifications are made that produce the effect of observed averaging over time and over many particles. The second one is good provided the system exhibit equilibrium in some macroscopic degree, but when the system shows signs of non extensibility like long range interactions are said to cause, then its unclear if it really represents physical behaviour of the system. $\endgroup$ – rmhleo Aug 26 '15 at 7:16

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