Kolmogorov Turbulent Energy Cascade

Equivalent forms

E(k) \propto k^{-5/3}

\eta = \left(\frac{\nu^3}{\varepsilon}\right)^{1/4}\ (\text{Kolmogorov scale})

u_\ell \sim (\varepsilon\,\ell)^{1/3}

Dimensional analysis alone pins the spectrum of chaos to a single exponent: −5/3.

Symbol	Name	SI unit	Dimension	Range
$C$	Kolmogorov Constant Universal proportionality constant (~1.5)	dimensionless	1	1.2 – 2
$\varepsilon$	Dissipation Rate Turbulent kinetic-energy dissipation per unit mass	m²/s³	L²·T⁻³	0.001 – 100
$k$	Wavenumber Spatial wavenumber (2π/eddy size)	rad/m	L⁻¹	0.1 – 1000
$E$	Energy Spectrumoutput Energy per unit wavenumber	m³/s²	L³·T⁻²	0 – 100

Unit systems

Symbol	SI	CGS	Imperial
$C$	—	—	—
$\varepsilon$	m²/s³	cm²/s³	ft²/s³
$k$	rad/m	rad/cm	rad/ft
$E$	m³/s²	cm³/s²	ft³/s²

Where it holds

High-Reynolds-number, locally isotropic turbulence in the inertial range

\eta

≪ ℓ ≪ L, far from forcing and viscous scales. Real flows show small intermittency corrections to the exponent

(\sim -1.71

measured

) but -5/3

is the leading law.

Dimensional analysis

[\varepsilon^ ${2/3} k^{-5/3}] = (L^{2}\cdot T^{-3})^{2/3}(L^{-1})^{-5/3} = (L^{4/3}T^{-2})(L^{5/3}) = L^{3}\cdot T^{-2} \checkmark$ (energy spectrum)

Discovery

Andrey Kolmogorov · 1941

Kolmogorov argued that at high Reynolds number the small scales of turbulence are statistically universal, set only by the energy-transfer rate ε and viscosity ν. Pure dimensional analysis then forced E(k) ∝ ε^(2/3)k^(−5/3) — the celebrated K41 law, later confirmed in tidal channels and wind tunnels.

Try this

Big whirls make little whirls — turbulence shreds energy across scales by one universal power law.

In the inertial range of turbulence the energy spectrum is E(k) = C·ε^(2/3)·k^(−5/3). For dissipation rate ε = 1 m²/s³ and Kolmogorov constant C ≈ 1.5, evaluate E at wavenumber k = 10 rad/m.

Research status: stable

Real-world applications

Subgrid models in weather and climate simulation (LES)
Aerodynamic and combustion turbulence modeling
Astrophysical and interstellar turbulence
Mixing and dispersion of pollutants

Common misconceptions

Turbulence is purely random with no structure — its energy spectrum is a precise, universal power law
Viscosity sets the inertial-range slope — it does not appear there; only $\varepsilon$ does
The exponent is exactly $-5/3$ — intermittency gives a tiny correction, but K41 is the leading-order law

Experimental verification

Grant, Stewart & Moilliet's 1962 tidal-channel measurements first showed

the -5/3

slope over several decades of k; countless wind-tunnel, atmospheric and oceanic spectra since confirm it, with

C \approx 1.5

across flows.

Derivation

Assume that in the inertial range E(k) depends only on the wavenumber k and the energy-transfer rate

\varepsilon

.

Dimensionally

[E] = L^{3}T^{-2}

,

[\varepsilon ] = L^{2}T^{-3}

,

[k] = L^{-1}

.

The only combination giving

L^{3}T^{-2} is \varepsilon ^(2/3)k^(-5/3)

; inserting a dimensionless constant C gives

E(k) = C\varepsilon ^(2/3)k^(-5/3)

.

Limiting cases

Inertial range⟶ E \propto k^{-5/3}Between forcing and dissipation scales the spectrum follows the universal five-thirds law set only

by \varepsilon

.

k \to k_\

\eta = 1

/\eta⟶ Viscous cutoffNear the Kolmogorov scale

\eta

viscosity dissipates the energy and the spectrum rolls off exponentially.

k \to k_{forcing}⟶ Energy injectionAt the largest scales the spectrum departs from

-5/3

where energy is fed into the flow.

What if…

What if you stir twice as hard (raise

\varepsilon )

?

Since $E \propto \varepsilon ^(2/3)$ , doubling $\varepsilon$ lifts the whole inertial-range spectrum by $2^(2/3) \approx 1.59\times$ — more energy at every scale.

What if the Reynolds number is low?

The inertial range shrinks to nothing — forcing and viscous scales overlap and the clean $-5/3$ law never appears.

1

Spectrum at a given scale

Given ·

C:: 1.5
\varepsilon:: 1
k:: 10

Find · E

Steps

Compute $k^(-5/3)$ : $10^(-5/3) = 10^(-1.667) \approx 0.0215$
$\varepsilon ^(2/3) = 1^(2/3) = 1$
$E = 1.5 \times 1 \times 0.0215 \approx 0.0323\,\mathrm{m}^{3}/s^{2}$

Result ·

E = C\varepsilon ^(2/3)k^(-5/3) = 1.5\times 1^(2/3)\times 10^(-5/3) = 1.5\times 0.0215 \approx 0.0323\,\mathrm{m}^{3}/s^{2}

.

Reynolds Number→Navier-Stokes Equation→Newton's Law of Viscosity→