The continuum approximation

Start with one mass and its neighbours

Imagine equal masses along a straight line, with a spring between each neighbouring pair. Supply m = 0.5 kg for each mass, rest spacing a = 0.25 m and spring constant K = 8 N/m. The line is infinite in both directions. We assume longitudinal motion, linear spring forces, an inertial frame and no friction or other applied forces. These are model stipulations, not measured material data.

Label each mass by an integer j. Its reference position is x = ja. Its signed longitudinal displacement is u_j(t), positive along increasing x; its actual position is ja + u_j(t). The continuum approximation will replace the separate u_j values by one smooth field u(x,t). The reference coordinate remains the label used to compare them.

For preparation, use Units, standards, and physical constants, Algebra, functions, graphs, and logarithms, and Geometry, trigonometry, and coordinate systems. Begin with the neighbour forces, wavelength comparison and figure. Derivatives, Taylor series, and local approximation is an optional return path for the remainder estimate; The classical wave equation and Newton’s laws and free-body diagrams provide optional background. The needed force and wave relations are introduced here.

Add the two signed spring forces

The right spring changes its length by u_(j+1) − u_j, so its force on mass j is +K[u_(j+1) − u_j]. The left spring changes its length by u_j − u_(j−1), and pulls in the opposite signed direction: −K[u_j − u_(j−1)]. Adding gives F_j = K[u_(j+1) − 2u_j + u_(j−1)].

Define acceleration as the second time derivative d²u_j/dt². The supplied mass law is m d²u_j/dt² = F_j. Thus d²u_j/dt² = (K/m)[u_(j+1) − 2u_j + u_(j−1)]. The bracket is a second difference: it compares one displacement with its two neighbours, rather than measuring the displacement alone.

If all three displacements are equal, the second difference is zero. It is also zero for a local affine pattern u(x) = sx + b at equally spaced positions. A larger positive middle displacement than both neighbours instead gives a negative force. These checks explain the sign before a continuous derivative is introduced.

Supply two travelling-mode families

Choose an amplitude U = 0.001 m. A discrete travelling mode is u_j(t) = U cos(qj − ω_D t), where q is the phase change between neighbouring reference positions. Define the angular spatial wavenumber k = q/a in m⁻¹. Both qj and ω_D t are numerical radian phases; ω_D has unit s⁻¹, so the accumulated angle is (ω_D t) rad.

The cosine addition formula gives u_(j+1) − 2u_j + u_(j−1) = −4 sin²(q/2)u_j. Its second time derivative is −ω_D²u_j. The force law therefore gives ω_D = 2√(K/m) sin(q/2) on our chosen positive branch 0 < q ≤ π. Here √(K/m) = 4 s⁻¹.

Supply the continuum mode u_C(x,t) = U cos(kx − ω_C t), with ω_C = q√(K/m). Its initial displacement at x = ja agrees with the discrete mode: both are U cos(qj). Their initial velocities differ: Uω_D sin(qj) for the chain and Uω_C sin(qj) for the continuum. We compare their travelling-mode relations, not two evolutions with identical full initial data.

At integer j, replacing q by q + 2πn for integer n leaves the nodal spatial phase unchanged. We retain the positive first-zone choice above; at q = π the nodal mode alone does not distinguish the two travel directions. The q = 0 limit is a static uniform displacement in this family, separate from the positive-frequency comparisons.

Replace the difference under a smoothness condition

For a field that changes slowly over one spacing, the second difference can be approximated by a²u_xx, where u_xx is its second spatial derivative. Substitution gives u_tt = c²u_xx, with c = a√(K/m) = 1 m/s. The subscripts indicate derivatives with respect to time or reference position. This is a continuum wave equation for this uniform linear model.

The approximation has a checkable condition. If u has a continuous fourth spatial derivative over [x−a,x+a], let M₄ bound its absolute value there at the chosen time. Taylor expansion on each side cancels the odd terms and gives |u(x+a) − 2u(x) + u(x−a) − a²u_xx| ≤ a⁴M₄/12. Multiplying by K/m bounds the acceleration residual; it does not by itself bound a complete later trajectory.

For a harmonic shape, M₄ = Uk⁴ is a global bound. The residual bound becomes (K/m)Uq⁴/12, while the continuum acceleration amplitude is (K/m)Uq². Their bound ratio is q²/12 for q ≠ 0. A smaller q controls this particular approximation. Return to the Taylor lesson for the derivative proof; the exact frequency comparison below gives a second way to see the effect.

The shared amplitude ratio U/a = 0.004 keeps the stipulated motions small relative to the spacing, and adjacent masses remain separated by at least a − 2U = 0.248 m. That amplitude check does not make every wavelength long. Both chosen waves have the same small U but different q.

Compare two wavelengths at the same time

Wavelength is λ = 2π/k = 2πa/q. For q = π/10 it is 20 spacings, or 5 m. For q = π/2 it is four spacings, or 1 m. The figure compares the two mode families at t = 0.5 s: a blue continuous curve for the continuum mode, unjoined circles for the longer chain mode and unjoined diamonds for the shorter chain mode.

Each panel uses the same reference-position and displacement scales. Although displacement is drawn vertically, the physical motion is longitudinal along the chain. The plot is a graph of signed displacement against a reference label, not a transverse picture or a path of a mass. The finite 0 ≤ x ≤ 5 m window creates no ends on the infinite chain.

The ratio ω_D/ω_C = 2 sin(q/2)/q also equals the discrete phase speed divided by c. The longer-wave frequency is about 0.41% below its continuum value; the shorter-wave frequency is about 9.97% below. These are model frequency differences, not measured material-error percentages. The exact table retains the expressions behind the rounded descriptions.

Read the marks without inventing a discrete curve

At the reference nodes the chain values are exact for the supplied discrete mode. The chain provides displacement values at integer labels j; we have not defined a discrete displacement field at a non-integer reference label. Only the continuum solution is connected in the figure. Its finite line segments approximate an analytic curve.

The table uses u/U, so its displacement entries are dimensionless. Multiply by U = 0.001 m to recover metres. It selects the first five reference nodes in both panels; the figure also shows the other nodes through j = 20. Expressions with zero are exact trigonometric values, even though finite numerical sine or cosine evaluation can leave a tiny roundoff remainder.

For the shorter case at j = 1, the continuum table gives zero but the exact chain gives sin(2√2), about 0.308. Their displacements therefore differ by about 0.308 mm there. This illustrates the chosen mode comparison; it is not a universal error bound for arbitrary chain initial conditions.

Separate a phase bound from a trajectory claim

For a fixed positive q, the phase difference δ(t) = (ω_C − ω_D)t grows linearly with time. At 0.5 s it is about 0.002581 rad for the longer wave and 0.313166 rad for the shorter wave. Doubling the time doubles this phase difference, even though the amplitude U stays the same.

The cosine subtraction identity gives a nodal displacement bound |u_C(ja,t) − u_j(t)| ≤ 2U|sin[δ(t)/2]| ≤ min(2U,Uδ(t)) for t ≥ 0. The bound need not be reached at a displayed node. It applies to these two mode choices, including their different initial velocities. It does not describe all chain solutions.

For 0 < q ≤ π, the sine remainder also gives 0 ≤ 1 − ω_D/ω_C ≤ q²/24. This supports the small-q frequency comparison at fixed parameters. It does not make closeness uniform for arbitrarily long time. If a is refined while keeping a finite c fixed, K/m must change accordingly; keeping K and m fixed instead makes c = a√(K/m) tend to zero.

Keep the material assumptions separate

A continuum description treats fields such as displacement or density as values varying over position. Local averaging requires a scale large enough to avoid individual-particle fluctuations yet small enough to retain the macroscopic variation of interest. Whether such a useful separation exists is a physical question; this toy chain supplies no universal numerical threshold.

Our coefficients are uniform because all m, a and K were chosen equal. Homogeneous means that material properties do not vary with position; that is an additional assumption, not the definition of a continuum. A continuum model may instead use smoothly varying properties.

Nor does a continuum imply incompressibility. In this one-dimensional smooth reference map, actual position is x + u(x,t). If 1 + u_x > 0, mass conservation gives current line density μ = μ₀/(1 + u_x), where μ₀ = m/a = 2 kg/m. A nonzero changing strain u_x can change that density. This is a line-density relation, not a three-dimensional fluid-density model.

The corresponding axial rigidity Ka = 2 N is not a Young modulus in pascals without additional geometry. A real material needs an appropriate constitutive relation and calibrated parameters. The exact chain calculation illustrates a controlled approximation; it does not establish accuracy for heterogeneous media, crystals, thermal motion or every microscopic solution.

Worked examples

Try it yourself

1. Use the worked local state u_(j−1) = 0.0004 m, u_j = 0.0007 m, u_(j+1) = 0.0002 m. Does reversing all three displacement signs reverse the force and acceleration?

   Show the answer and reasoning

   Yes. The reversed state gives F_j = +0.0064 N and acceleration = +0.0128 m/s².

   The second difference is linear in the three displacements, so it changes from −0.0008 m to +0.0008 m. The stipulated positive K and m preserve that reversal.

2. With a = 0.25 m and q = π/10, find λ/a and λ. Does the small amplitude alone establish the long-wave condition?

   Show the answer and reasoning

   λ/a = 20 and λ = 5 m. Small amplitude alone does not establish a long wavelength.

   λ = 2πa/q. Both figure cases use U/a = 0.004, but their wavelength ratios are 20 and 4.

3. For q = π/2, find ω_D/ω_C. Does a mass necessarily move along the line at that phase speed?

   Show the answer and reasoning

   ω_D/ω_C = 2√2/π ≈ 0.900316. A mass need not move at the phase speed.

   The frequencies are 4√2 s⁻¹ and 2π s⁻¹. Phase speed follows a fixed oscillation phase; a mass velocity is the time derivative of its own displacement.

4. At equally spaced reference nodes near one point, supply u(x) = 0.002x with x and u in metres. What second difference and net spring force does this local affine pattern give?

   Show the answer and reasoning

   The second difference is 0 m, giving a net spring force of 0 N at that node.

   The offsets at x+a and x−a cancel: 0.002(x+a) − 2(0.002x) + 0.002(x−a) = 0. Individual spring forces need not vanish. This local pattern is not asserted to be a bounded harmonic mode on the whole infinite chain.

5. For the longer mode, does doubling time from 0.5 s to 1 s double δ? Does that prove the actual displacement difference at every node doubles?

   Show the answer and reasoning

   It doubles δ, but does not prove that every nodal displacement difference doubles.

   δ = (ω_C − ω_D)t is linear in time. The exact cosine difference also depends on the oscillation phase; Uδ is an upper bound for the absolute displacement difference, which is also bounded by 2U. These bounds need not be attained at every node.

6. In a different uniform model, change every mass to 1 kg while keeping a = 0.25 m and K = 8 N/m. What continuum speed results? Does the continuum assumption require constant density during motion?

   Show the answer and reasoning

   c = (√2/2) m/s ≈ 0.707107 m/s. A continuum assumption does not require constant density during motion.

   c = a√(K/m) = 0.25√8 m/s. This changes all masses uniformly, not one heterogeneous defect. A smooth continuum may have changing line density μ₀/(1+u_x) when its reference deformation changes.

What to carry forward

• A continuum approximation replaces discrete values by a field under declared scale and smoothness assumptions.
• The second difference supplies the signed neighbour force; its smooth approximation produces a wave equation in this model.
• Small amplitude and long wavelength are independent checks, and mode phase speed differs from material velocity.
• These two travelling-mode choices share initial displacement but have different initial velocities; their phase discrepancy can grow with time.
• Reference positions, finite viewing windows and unjoined discrete marks keep the diagram honest. Continuum does not by itself mean homogeneous or incompressible.