Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can you think of examples where a macroscopic law coexists with a fully known microscopic law, but the former hasn't been derived from the latter (yet)? Or maybe a rule of thumb, which works but hasn't been understood yet. Or some emergence which is hard to see from the individual interacting parts. The microscopic arrangement should be simple. So plain complexity like in biology is something different. Most thermodynamic results and entropy-related results you can derive from microscopic assumptions.

I heard apparently Fourier's law isn't that clear

Maybe there are more examples from other physical areas (quantum mechanics, chemistry etc.)? Basically something where there are fully understood microscopic laws and also a macroscopically observed law which theoretically should be derivable.

share|cite|improve this question
The emergence of physical laws is a long-running theme in condensed matter. Phil Anderson wrote a famous and influential essay on this:… – genneth Apr 12 '12 at 10:51
To be clear, full understanding only occurs when we can connect the microscopic and macroscopic laws, but often one finds the macroscopic laws and only later find the microscopic ones; and much later, how to connect them. Thus historically, and more importantly logically, neither comes first. – genneth Apr 12 '12 at 10:52
Very good point! In fact I'm exactly refering to something like "more is different" and asking where such a high level law has been found, which was inaccessible from microscopic ones. – Gerenuk Apr 16 '12 at 8:59

There are thousands of such examples, it is basically all situations in condensed matter physics. You see a lot of regularities that have no explanation.

Here's one of the most annoying ones for me: Moseley's law--- you can knock out one of the two electrons most tightly bound to a heavy atom (in the K-shell). This leaves a hole orbiting the nucleus. The energy of this hole can be calculated from the Bohr model, except that you need to use a nuclear charge reduced by exactly 1 unit. This is due to electron screening.

But why is this exactly one unit? Measurements in heavy atoms show that the K-shell Moseley screening is one electron charge. But the other K-shell electron is orbiting at the same r, and the far-away electrons contribute different amounts, and yet somehow when you sum up all their screening contributions, no matter what the atom, you end up reducing the nuclear charge by one unit. This is not understood. I will ask it as a question.

share|cite|improve this answer
Nice example, could you please provide a reference for this – richard Feb 20 '14 at 13:17

At present, the Navier-Stokes equations for the dynamics of water haven't yet been derived from microscopic principles.

share|cite|improve this answer
David Tong recently gave a course on kinetic theory in which he also derived (admittedly with a few approximations etc.) the Navier-Stokes equations from a simple interacting Hamiltonian (cf. It is however true that some parametres in that equation can only be computed for dilute gases and not liquids. – Claudius Dec 4 '12 at 23:52
@Claudius: Yes. He says on p.50: ''Our derivation of the Navier-Stokes equation relied on the dilute gas approximation. However, the equation is more general than that'', which is the time-honored wisdom. No microscopic argument is given (anywhere that I'd know of) for why the equation is more general than that. Instead, one resorts to arguments of symmetry (rotation or Galilei invariance) and simplicity. Thus there is no knwon way to compute from first principles the parameters in Navier-Stokes for dense fluids. – Arnold Neumaier Dec 5 '12 at 10:14
Moreover, Tong's interaction is assumed to be rotationally invariant, corresponding to spherical particles, not water molecules, which have even in the simplest sensible approximation far more complex interactions.. – Arnold Neumaier Dec 5 '12 at 10:16
You certainly have a point there, yes. Apologies for the noise. – Claudius Dec 5 '12 at 13:12

This is an example from hydrodynamics. When the effects of viscosity can be ignored (inviscid flow), a uniform incident flow can exert on immersed bodies only lift forces perpendicular to the asymptotic flow velocity. However, there exist an infinite number of solutions of the flow equations of motion satisfying the asymptotic conditions at infinity and the requirement that the local velocities are tangent to the body boundaries. The solutions differ by the values of the flow circulation around the body. The theory does not give preference to any of these solutions. Each solution is associated with a different lift force.

However, in the case of airfoils having a sharp trailing edge, nature prefers the solution with the exact circulation giving a vanishing velocity at the trailing edge. This is called the Kutta condition which states that the airfoil generates just enough circulation that the air speed at the airfoil (sharp) trailing edge is exactly zero. There are arguments that this condition stems from the full theory (probably expressed through the Navier-Stokes equation),but it hasen't been derived.

share|cite|improve this answer

It is curious that no one has answered the most obvious one:

Non-perturvative Einstein(-Hilbert) general relativity.

share|cite|improve this answer
+1. Certainly good observation there! – namehere Dec 4 '12 at 11:59

As Ron noted, there are many, many examples within condensed matter; they often share a very similar story where the microscopic laws are known well (exactly, for the case of simulations), but the macroscopic laws are derived by symmetry concerns.

Take for example, liquid crystals. We could simulate a collection of hard rods or ellipsoids - this is our perfect "microscopic" model of a nematic. On a larger scale, we could describe this object in terms of a continuum vector field $\mathbf{n}(r)$ representing the direction of these rods - but what is the free energy associated with a configuration $\mathbf{n}(r)$? What we can do is to write down the most general free energy possible that obeys the symmetry of the system we want to describe. For instance, for the nematic phase of a liquid crystal, the free energy must be even in $\mathbf{n}(r)$ since the rods have inversion symmetry. There are also other requirements, like rotational and translational invariance, etc. It turns out that the most general form possible (up to second order in gradients of $\mathbf{n}$) is given by the Frank free energy, $\mathcal{F}_{d}=\frac{1}{2}K_1(\nabla\cdot\mathbf{\hat{n}})^2+\frac{1}{2}K_2(\mathbf{\hat{n}}\cdot\nabla\times\mathbf{\hat{n}})^2+\frac{1}{2}K_3(\mathbf{\hat{n}}\times\nabla\times\mathbf{\hat{n}})^2$ ( The derivation of this is shown in the book by de Gennes and Prost (The Physics of Liquid Crystals).

We do not know from the microscopic theory what the parameters $K_{1,2,3}$ should be - but we know what the form of the free energy should be; we have "hidden our ignorance" in these parameters. (It may be possible to get from a microscopic theory to $K_{1,2,3}$ in some special cases - but in general, I don't think it is). This symmetry-based approach is very powerful, and is used all over condensed matter physics. A cute example is the dynamics of bird flocks, which can be described by a generalization of the Navier-Stokes equations: Hydrodynamics and phases of flocks (PDF)

share|cite|improve this answer

Any problem that requires solving of non-trivial Schroedinger equations. For example, protein folding problem. It is known what equations the system should satisfy and those equations can be written down. Yet they cannot be solved with modern computers which would take millions of years tor that.

share|cite|improve this answer
How about quantum computing? – user26143 Jun 10 '14 at 16:08
@user26143 on quantum computers they can be solved. – Anixx Jun 10 '14 at 16:45

Another favorite: It's remarkably difficult to compute the nucleation rate of water molecules during a phase transation from microscopic equations. Water molecules have a dipole moment of order 1, so most of the usual approximation tricks don't work.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.