Is there a rigorous derivation of the limits for continuum properties in solid mechanics? For instance, the stress-strain relationship may be linear for large samples (the slope being the Young's Modulus) but at what limit does that break down?

Even if a linear relationship exists for the stress-strain curve between two atoms, or two crystals, the slope likely isn't the same as for the large sample. What spatial scale separates the two regimes and is it similar to fluid dynamics where there is the Boltzmann equation at the non-continuum regime, the Navier-Stokes equations in the continuum regime, and a bridge between the two regimes (Chapman-Enskog equations)?

  • $\begingroup$ @Danu Sorry sorry, it is the Planck constant n (en.wikipedia.org/wiki/Planck_constant )Some linguistic confusion with the uncertainty principle which depends on it and is denoted by h. $\endgroup$
    – anna v
    Feb 16, 2014 at 7:18
  • $\begingroup$ I will make an answer out of it. $\endgroup$
    – anna v
    Feb 16, 2014 at 7:31

3 Answers 3


The limit to which you refer is known as the thermodynamic limit in statistical mechanics. It consists in taking the limit of infinite particles ($N\rightarrow \infty$) and infinite volume ($V\rightarrow \infty$) while keeping a finite density $N/V$.

In a solid, both electrons and atomic nuclei contribute to the thermodynamical and elastic quantities, such as the Young modulus. Electronic contributions, which typically aren't negligible, are usually considered from a quantum mechanical point of view, while nuclei contributions can be considered classically. You can see how the thermodynamical limits plays a role by considering non-interacting electrons in a periodical lattice (I hope you are familiar with quantum mechanics, because that's at the heart of it). If you solve the Schrödinger equation for the hamiltonian of the non-interacting electrons in the potential of the lattice, you will find a relation between the energy E and the wave-number k of the electrons. For instance, in the one dimensional tight-binding model, you find:

\begin{equation} E(k)=-2tcos(ka) \end{equation}

where a is the separation between atoms in the lattice, and t describes the probability of an electron tunneling from one atom to a neighboring one.

In order to fully solve the problem, since the Schrödinger equation is a differential equation, you need to set boundary conditions. The choice of boundary conditions is important for "small" systems, but it plays no significant role for "big" systems (that is, in the thermodynamic limit). The imposition of boundary conditions results in the quantization of momentum, and you'll usually find that you have as many allowed quantum states as sites in the lattice (if you allow only one electron per site, and neglecting the spin). For periodic boundary conditions, the allowed states are given by:

\begin{equation} k_n=\frac{2\pi n}{L} \end{equation}

where L is the length of the system and n an arbitrary integer. Note that this implies that the electrons can't take any value of energy, but only those corresponding to a $k_n$ (energy is quantized). One can notice that the separation in wave-number between two "adjacent" states (that is, a state for a given n and the state for n+1) will be then \begin{equation} \Delta k=k_{n+1}-k_n=\frac{2\pi}{L} \end{equation} To count the number of states N, per unit length, you can do the following sum: \begin{equation} \frac{N}{L}=\frac{1}{L}\sum_{k} 1=\frac{1}{L}\sum_{k} \frac{\Delta k}{\frac{2\pi}{L}}=\frac{1}{2\pi}\sum_{k} \Delta k \end{equation} Being able to do sums over the states is really useful in solid state physics to calculate average values and thermodynamical quantities through a statistical mechanics approach.

Now comes the important part: when you take the thermodynamic limit ($L\rightarrow \infty$), the separation $\Delta k$ becomes infinitely small, which implies that the possible values for the energy become continuous, and that you have to replace the sum over discrete states by an integral over the (now continuous) values of k:

\begin{equation} \sum_{k} \Delta k \rightarrow \int dk \end{equation}

so: \begin{equation} \frac{1}{L}\sum_{k} 1=\frac{1}{2\pi}\int dk \end{equation}

If you do the same analysis in the three dimensional case (by the way, you can do it for d dimensions), you have to replace L by V and more $2\pi$ factors appear. The systematic way of taking this limit is then by replacing the sums by integrals in your calculations in the following way:

\begin{equation} \frac{1}{V}\sum_{k} \rightarrow \frac{1}{(2\pi)^3}\int dk \end{equation}

While it is true than in reality you will never have $L\rightarrow \infty$, you can consider that you have entered this regime when $\frac{L}{a}>>1$. For two atoms, this will never be the case since L=a by definition (the size L of your system is the distance between the atoms a), but considering the typical interatomic distances, which are of the order of Angstroms or nanometers, a sample of a few micrometers can already be treated by this approach.


Is there a rigorous derivation of the limits for continuum properties in solid mechanics

Continuum properties stop holding whenever we reach the quantum mechanical regime, i.e. where atoms and molecules become distinct and follow quantum mechanical solutions rather than classical collective ones.

What spatial scale separates the two regimes

Quantum mechanics is a probabilistic theory and at the microlevel ruled by the Heisenberg uncertainty principle, the HUP.


Planck's constant is a very small number which for classical dimensions in space and energy can be be considered equal to zero.

The uncertainty principle has to be taken into account when looking at the limits of validity of the classical going to quantum mechanical, because it constrains what variables can be measured accurately together. Thus a general rule cannot be devised but it will depend on the units of the variables under consideration : space and momentum , energy and time, plus other more esoteric ones.

A rule of thumb though is to compare what one is interested in with the Planck constant h . One has to get the correct units for the quantity in question. If the quantity/variable in question is larger within experimental errors and accuracy requirements than h , h can be considered 0 and the quantity treated classically.

People have studied how the quantum mechanical formulations at their limits give rise to the classical but it is not a simple mathematical problem and each case has to be studied separately. To get an idea of the complexity you could have a look at how classical electromagnetic waves emerge from photons, for example.

  • 1
    $\begingroup$ So if I have a simulation using a continuum approach, say finite elements, of a material with lattice dimensions of ~10 Angstroms and my discretization has a grid spacing of 20 Angstroms, I should expect the results to be accurate? It seems like the long-range effects at scales comparable to the lattice would be non-negligible and the traditionally continuum properties which are local would fail to hold true anymore. I guess another way to phrase the question -- how does one determine the representative volume element for a material where subdivision of the element results in diff. properties? $\endgroup$
    – tpg2114
    Feb 16, 2014 at 15:55
  • 1
    $\begingroup$ And also, this answer seems to imply that solids are different from fluids where the Knudsen number determines whether continuum approaches will work. Is there an equivalent ratio of length scale of interest (say, the length of a rod) to length scale of physical property (say, the lattice dimension) in a solids? $\endgroup$
    – tpg2114
    Feb 16, 2014 at 15:59
  • $\begingroup$ This needs a solid state expert for an answer. For example stresses and strains that you ask about have to be related with the lattice dimensions through the electromagnetic forces which are finally responsible. Hand waving: a simulation should use much higher distances than the lattice distance for continuum, a classical idea because if you simulate the em forces the discreteness will be important combined with the size of the force. lets say the effective momentum of the exchanged virtual photons times the dimensions in space in the simulation should be larger than h_bar/2 $\endgroup$
    – anna v
    Feb 16, 2014 at 16:28

When you're asking practically important questions about the mechanical strength of materials, size matters for many reasons.

  • Let's talk about dislocations, although the same discussion also applies to other defects and impurities. The density and motion of dislocations are critical factors in determining the mechanical strength of materials. A large crystal will have zillions of dislocations, but a small enough crystal is likely to have exactly one dislocation or zero dislocations. (Actually, very small crystals may be especially likely to be perfect, because dislocations are so close to the surface that they can just slide out.) If there are dislocations, they might be able to move more easily or less easily in a small crystal versus a large one (after all, their length is different).
  • Surface layers have different mechanical properties than bulk layers (after all, there is surface reconstruction, interface strain, etc.) A small crystal has different surface-to-volume ratio than a large one, which affects mechanical properties.
  • How much strain is enough to crack a macroscopic material? It depends on how nano-cracks propagate and grow into micro-cracks and stuff like that. In a very small sample, the cracks do not need to grow as much, there is relatively more surface area to seed them, etc. -- the parameters are all different so we can expect the strength to be different too.

So for all these reasons, you should not be surprised to find that mechanical properties can change at all different length scales, depending on the material and how it's prepared. I expect that there are general rules-of-thumb about what length scales you can expect different effects to happen, but I don't know them. I usually hear people discuss these kinds of things in the context of nanotechnology (less than 50nm thick objects) but I don't really know.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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