research35 min studyLesson 186 in Physics

Turbulent energy cascade and Kolmogorov scaling

Relate energy flux across three-dimensional turbulent scales to inertial-range estimates and the viscous cutoff, with assumptions made explicit.

On this page

What you’ll understand

  • Distinguish energy stored at a scale from energy flux through scales.
  • Derive inertial-range velocity and spectral dimensions.
  • Estimate Kolmogorov microscales and state the limits of their interpretation.

An irregular velocity field contains motion over many length scales. In ordinary three-dimensional turbulence forced at large scales, nonlinear interactions transfer energy toward smaller scales until viscosity removes it. The transfer need not resemble literal large vortices breaking neatly into smaller ones. Kolmogorov reasoning instead asks which quantities can control statistics sufficiently far from both forcing and viscous effects. Its predictions are useful precisely when their scale-separation and statistical assumptions are tested.

Track a flux rather than a photograph

Let ε\varepsilon be mean kinetic-energy dissipation per unit mass, with dimensions length squared divided by time cubed. In statistically steady turbulence, mean injection balances mean dissipation after accounting for boundary transport. An inertial interval is a set of scales where direct forcing and molecular dissipation are both small in the scale-by-scale budget, so the mean transfer rate is approximately constant. A spectrum alone shows energy distribution; establishing a cascade also requires information about transfer direction and the budget.

The forward energy cascade is a characteristic of many three-dimensional turbulent regimes, not a theorem that every instantaneous transfer must be downscale. Backscatter and spatially intermittent transfer can occur while the mean flux remains forward. Rotation, stratification, magnetic fields, walls, and restricted dimensionality may alter the dynamics and statistical assumptions. A picture with several small swirls cannot by itself establish an inertial range. Scale separation and a resolved statistical measurement are needed before fitting a universal-looking slope.

τℓ∼ℓδuℓ,ε∼(δuℓ)2τℓ⟹δuℓ∼(εℓ)1/3\tau_\ell\sim\frac{\ell}{\delta u_\ell},\qquad \varepsilon\sim\frac{(\delta u_\ell)^2}{\tau_\ell}\quad\Longrightarrow\quad \delta u_\ell\sim(\varepsilon\ell)^{1/3}
The turnover-time estimate assumes an inertial-range local-interaction picture. The velocity increment δu_ℓ is characteristic of separation ℓ; the relations are scaling estimates rather than exact pointwise equalities.↔ Scroll wide equations horizontally.

Derive a spectrum with a declared normalization

For homogeneous isotropic turbulence, define the one-dimensional shell-integrated spectrum E(k) so that its integral over positive wavenumber equals mean kinetic energy per unit mass. It therefore has dimensions length cubed divided by time squared. If the inertial-range spectrum depends only on ε\varepsilon and k, dimensional consistency gives the familiar minus-five-thirds exponent. A dimensionless coefficient remains undetermined by this argument. It depends on the precise spectrum convention and is not obtained by dimensional analysis.

This reasoning is a statistical similarity hypothesis. It does not make every instantaneous velocity increment equal to its estimated characteristic value, nor does it prove that all higher-order statistics share simple self-similarity. The exact longitudinal third-order law, under its homogeneous, isotropic, incompressible high-Reynolds-number inertial-range conditions, provides a stronger flux-related result than the dimensional spectrum argument. Keep these different logical statuses visible: an exact limiting relation and a phenomenological dimensional estimate are not interchangeable evidence.

12⟨∣u′∣2⟩=∫0∞E(k) dk,E(k)∼CKε2/3k−5/3\frac12\langle |\mathbf u'|^2\rangle=\int_0^\infty E(k)\,dk,\qquad E(k)\sim C_K\varepsilon^{2/3}k^{-5/3}
This shell-integrated spectrum convention is for homogeneous isotropic three-dimensional turbulence. The inertial-range estimate requires separation from forcing and dissipative scales; C_K is not fixed by dimensional reasoning.↔ Scroll wide equations horizontally.

Find the scales where viscosity matters

At sufficiently small scale, the local Reynolds number δuℓℓ/ν\delta u_\ell\ell/\nu becomes order one. Combining this criterion with the inertial estimate gives the Kolmogorov length η=(ν3/ε)1/4\eta=(\nu^3/\varepsilon)^{1/4}, time τη=(ν/ε)1/2\tau_\eta=(\nu/\varepsilon)^{1/2}, and velocity uη=(νε)1/4u_\eta=(\nu\varepsilon)^{1/4}. These are characteristic scales constructed from mean dissipation. Strong local fluctuations can require finer resolution for particular observables, especially high derivatives or intense events. Resolving an averaged length estimate is not an automatic accuracy certificate.

A direct numerical simulation must resolve the energy-containing region as well as the dissipative scales, and must run long enough for its desired statistics. Increasing Reynolds number widens the separation and rapidly raises spatial and temporal costs. A large-eddy simulation instead resolves a range of larger scales and models the influence of smaller ones. Neither a fitted spectral line nor a visually complicated animation settles whether a numerical calculation has the correct energy budget, adequate resolution, or statistically converged rare-event measurements.

Estimate a dissipative scale

An illustrative homogeneous turbulent flow has ν=10−6 m2 s−1\nu=10^{-6}\,\mathrm{m^2\,s^{-1}} and mean ε=10−2 m2 s−3\varepsilon=10^{-2}\,\mathrm{m^2\,s^{-3}}. Estimate its Kolmogorov length, time, and velocity, then check their local Reynolds number.

  1. The length is η=(10−18/10−2)1/4=10−4 m\eta=(10^{-18}/10^{-2})^{1/4}=10^{-4}\,\mathrm m, or 0.1 mm. This is a mean-dissipation-based characteristic scale, not a statement that no smaller structure can occur.

  2. The time is τη=(10−6/10−2)1/2=10−2 s\tau_\eta=(10^{-6}/10^{-2})^{1/2}=10^{-2}\,\mathrm s. The velocity is uη=(10−610−2)1/4=10−2 m s−1u_\eta=(10^{-6}10^{-2})^{1/4}=10^{-2}\,\mathrm{m\,s^{-1}}.

  3. Their local Reynolds number is uηη/ν=(10−2)(10−4)/10−6=1u_\eta\eta/\nu=(10^{-2})(10^{-4})/10^{-6}=1. Also η/uη=10−2 s\eta/u_\eta=10^{-2}\,\mathrm s, consistent with the independently constructed time.

The result

The estimates meet the order-one local Reynolds criterion by construction. Actual resolution choices still depend on the numerical method, domain, and target observable.

Explore the model

Interactive lab

Interactive lab · spectral

Stored small-scale content

Predict whether lowering coefficient decay p makes high modes more visible in the finite sum Σ sin(nx−t)/nᵖ. Increase mode count N at fixed p and compare fine-scale structure while paused. Use the controls to distinguish adding available scales from changing their assigned amplitude.

Stored small-scale contentA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe prescribed sine sum illustrates spectral content and truncation. Its coefficients do not exchange energy, so it has no turbulent cascade, computed flux, Kolmogorov dissipation, or dynamically generated minus-five-thirds spectrum.

Check your understanding

A velocity spectrum shows a short interval close to a minus-five-thirds slope. What further evidence is needed before claiming a forward inertial energy cascade?

Try it yourself

At fixed viscosity, mean dissipation increases by a factor of sixteen. By what factors do the Kolmogorov length, time, and velocity change? Explain one implication for measuring the most rapidly varying motion.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Read the powers of dissipation in each definition separately. A shorter characteristic time calls for a faster measurement response.

Compare with the solution

The length changes by 16^(−1/4)=1/2, the time by 16^(−1/2)=1/4, and velocity by 16^(1/4)=2. A sensor or calculation that resolved the original dissipative time may no longer resolve the new one; its time step or bandwidth must be reconsidered. These are characteristic mean-based scalings, so convergence of an intermittent observable still requires a direct resolution check.

Take it with you

  • A turbulent cascade concerns mean energy transfer across scales, not merely a complex-looking velocity field.
  • Kolmogorov inertial scaling requires a statistical regime and separation from forcing and dissipation.
  • Mean-dissipation microscales guide resolution but do not replace observable-specific convergence checks.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.