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Derivation

Kolmogorov's Four-Fifths Law

Statement

For statistically homogeneous, isotropic, incompressible turbulence at high Reynolds number, the third-order longitudinal velocity structure function S_3(r)=\langle(\delta u_L)^3\rangle is exactly proportional to the scale separation r throughout the inertial range \eta\ll r\ll L: S_3(r)=-\tfrac{4}{5}\,\varepsilon\, r, where \varepsilon is the mean rate of turbulent kinetic energy dissipation per unit mass and \delta u_L=u_L(\mathbf{x}+\mathbf{r})-u_L(\mathbf{x}) is the longitudinal velocity increment.

Why it matters

The four-fifths law is one of a mere handful of exact, non-trivial results in the theory of turbulence — derived directly from the Navier–Stokes equations with no closure assumption, no free constant, and no dimensional guesswork. It fixes both the sign (negative, encoding the forward cascade of energy from large to small scales) and the exact numerical prefactor 4/5.

It is the anchor of Kolmogorov's 1941 theory: the associated dimensional argument S_p\sim(\varepsilon r)^{p/3} is only a scaling hypothesis, but the p=3 case is a theorem. Because it survives intermittency, the 4/5 law serves as the benchmark against which every closure model, subgrid model, and experimental cascade measurement is calibrated.

Assumptions
Incompressibility, \(\nabla\!\cdot\!\mathbf{u}=0\).The pressure–velocity and dilatation terms would no longer drop out of the two-point budget; the isotropic-divergence operator \(\partial_r+4/r\) that generates the exact integral loses its conservative form.
Statistical homogeneity.Single-point statistics would depend on position, so \(\langle u_L^3(\mathbf{x})\rangle\) need not vanish and the reflection identity \(B_{LL,L}(-r)=-B_{LL,L}(r)\) fails, breaking the kinematic link \(S_3=6B_{LL,L}\).
Statistical isotropy.The correlation tensors would carry independent transverse components; \(S_3\) would no longer be reconstructible from a single scalar longitudinal function and the closed scalar equation could not be written.
High Reynolds number with a wide inertial range \(\eta\ll r\ll L\).Without scale separation there is no window in which both the unsteady/forcing term and the viscous term are simultaneously negligible; the clean \(-\tfrac45\varepsilon r\) is replaced by a Reynolds-number-dependent expression with finite-\(Re\) corrections.
Local stationarity of the small scales (spectral equilibrium).If the small scales cannot adjust instantaneously to the energy flux, \(\partial_t S_2\) is not negligible against \(\varepsilon\); an unsteady term \(\propto\int r^4\partial_t S_2\,dr\) survives and corrupts the prefactor.
Derivation
1
\[ \frac{\partial B_{LL}}{\partial t}=\frac{1}{r^4}\frac{\partial}{\partial r}\!\left(r^4\,B_{LL,L}\right)+\frac{2\nu}{r^4}\frac{\partial}{\partial r}\!\left(r^4\,\frac{\partial B_{LL}}{\partial r}\right) \]
Kármán–Howarth equation: the exact two-point energy budget obtained by writing Navier–Stokes at \(\mathbf{x}\) and \(\mathbf{x}+\mathbf{r}\), correlating, and reducing the tensors with isotropy. Here \(B_{LL}(r,t)=\langle u_L(\mathbf{x})u_L(\mathbf{x}+\mathbf{r})\rangle\), \(B_{LL,L}(r,t)=\langle u_L^2(\mathbf{x})\,u_L(\mathbf{x}+\mathbf{r})\rangle\), and \(\tfrac1{r^4}\partial_r(r^4\,\cdot)=(\partial_r+\tfrac4r)\,\cdot\) is the radial part of the divergence in the isotropic sector. Built on the Newtonian constitutive relation. A
2
\[ B_{LL}=u'^2-\tfrac{1}{2}S_2,\qquad B_{LL,L}=\tfrac{1}{6}S_3 \]
Kinematic identities. Expanding \(\langle(\delta u_L)^2\rangle\) and using homogeneity gives \(S_2=2u'^2-2B_{LL}\). Expanding \((\delta u_L)^3\): the single-point odd moments \(\langle u_L^3\rangle\) vanish by isotropy, and homogeneity plus the isotropic parity \(B_{LL,L}(-r)=-B_{LL,L}(r)\) collapse the cross terms to \(S_3=6B_{LL,L}\). Here \(u'^2=\langle u_L^2\rangle\). B
3
\[ -\tfrac{2}{3}\varepsilon-\tfrac{1}{2}\frac{\partial S_2}{\partial t}=\frac{1}{6r^4}\frac{\partial}{\partial r}\!\left(r^4 S_3\right)-\frac{\nu}{r^4}\frac{\partial}{\partial r}\!\left(r^4\frac{\partial S_2}{\partial r}\right) \]
Substitute Step 2 into Step 1. The \(r\)-independent constant \(u'^2\) drops under \(\partial_r\). At a point the isotropic kinetic energy is \(\tfrac32 u'^2\), and its decay defines dissipation: \(\tfrac32\,d u'^2/dt=-\varepsilon\Rightarrow d u'^2/dt=-\tfrac23\varepsilon\); this supplies the constant on the left. B
4
\[ -\tfrac{2}{3}\varepsilon=\frac{1}{6r^4}\frac{\partial}{\partial r}\!\left(r^4 S_3\right)-\frac{\nu}{r^4}\frac{\partial}{\partial r}\!\left(r^4\frac{\partial S_2}{\partial r}\right) \]
Drop \(\partial_t S_2\). At inertial scales the small-scale field is in statistical equilibrium with the cascade, so the local time rate of change of \(S_2(r)\) is of order \(u'^2/T_{\text{large}}\), negligible against \(\varepsilon\) by a factor \(\sim(r/L)^{2/3}\) as \(Re\to\infty\). This is the one genuine physical (not kinematic) assumption. C
5
\[ -\tfrac{2}{3}\varepsilon\,\frac{r^5}{5}=\tfrac{1}{6}\,r^4 S_3-\nu\,r^4\frac{\partial S_2}{\partial r} \]
Multiply by \(r^4\) and integrate from \(0\) to \(r\). The boundary terms at the origin vanish: analyticity of the increments gives \(S_3\sim r^3\) and \(\partial_r S_2\sim r\) as \(r\to0\), so \(r^4S_3\sim r^7\) and \(r^4\partial_r S_2\sim r^5\) both \(\to0\). The left side integrates to \(-\tfrac23\varepsilon\,r^5/5\). B
6
\[ S_3(r)=-\tfrac{4}{5}\,\varepsilon\, r+6\nu\,\frac{d S_2}{d r} \]
Divide by \(r^4\), giving \(-\tfrac{2}{15}\varepsilon r=\tfrac16 S_3-\nu\,dS_2/dr\), then multiply through by \(6\). This is the exact Kolmogorov equation, valid across the entire range of scales (dissipation and inertial alike). A
7
\[ \frac{6\nu\,|dS_2/dr|}{\tfrac45\varepsilon r}\sim\frac{\nu\,S_2/r}{\varepsilon r}\sim\left(\frac{\eta}{r}\right)^{4/3}\ll1 \]
In the inertial range the viscous term is negligible. Using \(S_2\sim(\varepsilon r)^{2/3}\) and \(\eta=(\nu^3/\varepsilon)^{1/4}\), the ratio of the viscous term to the cascade term scales as \((\eta/r)^{4/3}\), which vanishes for \(r\gg\eta\). Dropping it isolates the pure cascade law. C
Result
\[ \boxed{\,S_3(r)=\langle(\delta u_L)^3\rangle=-\tfrac{4}{5}\,\varepsilon\, r\,}\qquad \eta\ll r\ll L \]

Reading. The mean cube of the longitudinal velocity increment is negative and grows linearly with separation. The negative sign is the signature of the direct energy cascade: on average larger eddies push energy toward smaller scales, biasing the increment distribution toward negative skewness. The magnitude divided by \(r\) is a direct, model-free measurement of the dissipation rate \(\varepsilon\). The full (unabridged) relation including the viscous term, \(S_3=-\tfrac45\varepsilon r+6\nu\,dS_2/dr\), holds at every scale.

Units check. \([S_3]=(\text{m/s})^3=\text{m}^3\,\text{s}^{-3}\). Dissipation per unit mass \([\varepsilon]=\text{m}^2\,\text{s}^{-3}\), so \([\varepsilon r]=\text{m}^2\,\text{s}^{-3}\cdot\text{m}=\text{m}^3\,\text{s}^{-3}\). Both sides match; \(4/5\) is dimensionless.

Limiting cases
  • Dissipation range, \(r\to0\): analyticity forces \(S_3\sim r^3\to0\). In the full Kolmogorov equation the viscous term \(6\nu\,dS_2/dr\) rises to cancel \(-\tfrac45\varepsilon r\); with \(S_2\to\tfrac{1}{15}(\varepsilon/\nu)r^2\) one finds \(6\nu\,dS_2/dr\to\tfrac45\varepsilon r\), exactly annihilating the cascade term as required.
  • Inertial midrange, \(\eta\ll r\ll L\): viscous and unsteady terms both negligible; the pure \(-\tfrac45\varepsilon r\) plateau appears when \(-S_3/(\varepsilon r)\) is plotted, a hallmark used to confirm a resolved inertial range.
  • Energy-containing range, \(r\to L\) and beyond: two-point increments decorrelate, \(\langle(\delta u_L)^3\rangle\to\langle u_L^3\rangle_{+}-\langle u_L^3\rangle_{-}\to0\); the linear law is cut off by the finite integral scale.
Breaks when
  • Finite Reynolds number. When \(L/\eta\) is only moderate the inertial window shrinks; the retained unsteady/forcing term contributes a correction \(\sim r/L\) and the viscous term a correction \(\sim(\eta/r)^{4/3}\), so measured \(-S_3/(\varepsilon r)\) never quite reaches \(4/5\). One must extrapolate \(Re\to\infty\).
  • Anisotropy or inhomogeneity. In shear flows, near walls, or with large-scale gradients the scalar reduction of Step 2 fails; additional production terms enter the budget and the prefactor and even the linear form no longer hold (the Monin/Antonia generalisations must be used instead).
  • Compressible or forced-at-small-scale flows. Dilatational and pressure–dilatation terms re-enter the budget; injecting energy directly into the inertial range destroys the constant-flux structure that produces the exact \(4/5\).
Failure modes
  • Confusing the exponent with a scaling exponent. Students write \(S_3\sim r^{-4/5}\); the \(4/5\) is a coefficient, the \(r\)-exponent is exactly \(1\).
  • Sign errors. Dropping the minus sign hides the physics — a positive \(S_3\) would mean an inverse cascade. The forward 3-D cascade requires \(S_3<0\).
  • Using \(\varepsilon\) as a dissipation with viscosity in it. Here \(\varepsilon\) is the flux/mean dissipation rate per unit mass, independent of \(\nu\) at high \(Re\); it is not \(\nu\langle|\nabla u|^2\rangle\) evaluated with a specific molecular \(\nu\) as a free knob.
  • Applying it to the transverse increment. The law is specifically longitudinal (increment component along \(\mathbf{r}\)); \(\langle(\delta u_T)^3\rangle\) has its own (different, mixed) exact relation.
  • Assuming it needs K41 self-similarity. The 4/5 law is exact even though K41's \(S_2\sim r^{2/3}\) is only approximate; students wrongly discard it once told intermittency corrects the other exponents.
  • Forgetting scale separation. Reading off \(\varepsilon\) from \(S_3\) at a single \(r\) inside the dissipation range gives a badly underestimated \(\varepsilon\) because the viscous term has not yet become negligible.
Discussion

The four-fifths law is remarkable precisely because it is derived without closure. Every statistical theory of turbulence eventually confronts the closure problem — the equation for the \(n\)-point moment always involves the \((n+1)\)-point moment. The Kármán–Howarth equation is no exception: it relates the second-order \(B_{LL}\) to the third-order \(B_{LL,L}\). What makes Kolmogorov's manoeuvre work is that we do not try to solve the hierarchy; we simply integrate the exact budget across a range where two of its three terms are asymptotically negligible, leaving one algebraic relation between \(S_3\) and \(\varepsilon\). The unknown third-order object is thereby pinned exactly.

Physically, the constancy of the energy flux is the heart of the matter. In the inertial range energy is neither injected nor dissipated; it merely transits from large to small scales at the constant rate \(\varepsilon\). The Kolmogorov equation is a statement of scale-space flux conservation, and \(-\tfrac45\varepsilon r\) is the flux written in the natural third-order observable. This is why the result is so robust to the details of the large-scale forcing: only the through-flux \(\varepsilon\) appears.

The deepest point is the law's immunity to intermittency. Real turbulence is spatially spotty: dissipation concentrates on a thin, fractal set, and this makes the high-order structure-function exponents \(\zeta_p\) deviate from the K41 line \(p/3\) (anomalous scaling). Yet \(\zeta_3=1\) exactly, with the prefactor \(4/5\) unrenormalised. The reason is that \(S_3\) is not an arbitrary moment but the specific combination fixed by the exact conservation law; intermittency reshuffles the probability distribution of \(\delta u_L\) but cannot alter the constraint that the mean energy flux equals \(\varepsilon\). In modern language the 4/5 law is an exact consequence of the mean scale-space flux, whereas anomalous scaling lives in the fluctuations of that flux.

Common misconceptions. The 4/5 law does not "prove" the 5/3 spectrum — that spectrum follows from an additional dimensional hypothesis about \(S_2\). Nor is the law approximate: unlike almost everything else in K41 theory, it is a theorem, exact in the limit \(Re\to\infty\). And the negative sign is not a convention; it is the empirical fingerprint that three-dimensional turbulence cascades energy forward, in contrast to the inverse cascade of two-dimensional turbulence where the analogous relation reverses sign.

Worked examples

Example 1 — Predicting \(S_3\) from a known dissipation rate.

1
\[ S_3(r)=-\tfrac45\,\varepsilon\,r \]
Inertial-range four-fifths law; we are told \(r=0.10\ \text{m}\) lies inside the inertial range and \(\varepsilon=1.0\times10^{-2}\ \text{m}^2\,\text{s}^{-3}\) (a typical laboratory/atmospheric value). A
2
\[ S_3=-\tfrac45\,(1.0\times10^{-2}\ \text{m}^2\text{s}^{-3})(0.10\ \text{m}) \]
Insert numbers with units carried explicitly. A
3
\[ S_3=-8.0\times10^{-4}\ \text{m}^3\,\text{s}^{-3} \]
Evaluate: \(0.8\times10^{-2}\times0.10=8.0\times10^{-4}\). A
\[ S_3(0.10\ \text{m})=-8.0\times10^{-4}\ \text{m}^3\,\text{s}^{-3} \]

Reading. A negative value confirms the forward cascade; its magnitude sets the scale of the (skewed) third moment of velocity increments at that separation.

Units check. \(\text{m}^2\text{s}^{-3}\times\text{m}=\text{m}^3\text{s}^{-3}\), as required for a cube of velocity.

Example 2 — Measuring \(\varepsilon\) from a structure-function experiment.

1
\[ \varepsilon=-\frac{5}{4}\,\frac{S_3(r)}{r} \]
Invert the four-fifths law for \(\varepsilon\). A hot-wire probe in a wind-tunnel inertial range yields \(S_3(r)=-2.4\times10^{-3}\ \text{m}^3\text{s}^{-3}\) at \(r=0.050\ \text{m}\). A
2
\[ \varepsilon=-\frac{5}{4}\,\frac{-2.4\times10^{-3}\ \text{m}^3\text{s}^{-3}}{0.050\ \text{m}} \]
Substitute measured values; the double negative makes \(\varepsilon>0\) as it must. A
3
\[ \varepsilon=\frac{5}{4}\times(4.8\times10^{-2}\ \text{m}^2\text{s}^{-3})=6.0\times10^{-2}\ \text{m}^2\,\text{s}^{-3} \]
Compute \(2.4\times10^{-3}/0.050=4.8\times10^{-2}\), then multiply by \(5/4\). A
\[ \varepsilon=6.0\times10^{-2}\ \text{m}^2\,\text{s}^{-3} \]

Reading. The 4/5 law gives a probe-independent, closure-free dissipation estimate — the preferred experimental method, since it avoids needing to resolve the full dissipation-range gradients.

Units check. \(\text{m}^3\text{s}^{-3}/\text{m}=\text{m}^2\text{s}^{-3}\), the units of dissipation per unit mass.

Problems
  1. Confirm dimensionally that the full Kolmogorov equation \(S_3=-\tfrac45\varepsilon r+6\nu\,dS_2/dr\) is homogeneous in units.
    Solution\([S_3]=\text{m}^3\text{s}^{-3}\). First term: \([\varepsilon r]=\text{m}^2\text{s}^{-3}\cdot\text{m}=\text{m}^3\text{s}^{-3}\). Second term: \([\nu\,dS_2/dr]=(\text{m}^2\text{s}^{-1})\cdot(\text{m}^2\text{s}^{-2}/\text{m})=(\text{m}^2\text{s}^{-1})(\text{m}\,\text{s}^{-2})=\text{m}^3\text{s}^{-3}\). All three terms are \(\text{m}^3\text{s}^{-3}\); the equation is dimensionally consistent, and the numerical coefficients \(4/5\), \(6\) are dimensionless.
  2. In an atmospheric boundary layer \(\varepsilon=5.0\times10^{-3}\ \text{m}^2\text{s}^{-3}\). Find the longitudinal separation \(r\) at which \(S_3=-2.0\times10^{-3}\ \text{m}^3\text{s}^{-3}\), assuming it lies in the inertial range.
    SolutionFrom \(S_3=-\tfrac45\varepsilon r\), \(r=-\tfrac{5}{4}S_3/\varepsilon=-\tfrac54\,(-2.0\times10^{-3})/(5.0\times10^{-3})=\tfrac54\times0.40=0.50\ \text{m}\). So \(r=0.50\ \text{m}\).
  3. A DNS reports \(-S_3/( \varepsilon r)=0.71\) at its best-resolved scale. Express the shortfall from the exact value as a percentage, and name the two correction terms responsible.
    SolutionExact value is \(4/5=0.80\). Shortfall \(=(0.80-0.71)/0.80=0.1125\), i.e. about \(11\%\) below the asymptotic plateau. The two responsible terms are (i) the retained unsteady/forcing (large-scale) contribution, giving an \(\mathcal{O}(r/L)\) correction that lowers the ratio toward large \(r\), and (ii) the viscous term \(6\nu\,dS_2/dr\), giving an \(\mathcal{O}((\eta/r)^{4/3})\) correction toward small \(r\). Finite Reynolds number keeps the curve below \(0.80\) everywhere.
  4. Using the small-\(r\) result \(S_2(r)\to\tfrac{1}{15}(\varepsilon/\nu)\,r^2\), show that the full Kolmogorov equation forces \(S_3\to0\) faster than linearly as \(r\to0\).
    SolutionDifferentiate: \(dS_2/dr=\tfrac{2}{15}(\varepsilon/\nu)r\). Then \(6\nu\,dS_2/dr=6\nu\cdot\tfrac{2}{15}(\varepsilon/\nu)r=\tfrac{12}{15}\varepsilon r=\tfrac45\varepsilon r\). Substituting into \(S_3=-\tfrac45\varepsilon r+6\nu\,dS_2/dr\) gives \(S_3=-\tfrac45\varepsilon r+\tfrac45\varepsilon r=0\) at leading (linear) order. The linear parts cancel exactly, so the leading non-zero behaviour is the cubic \(S_3\sim r^3\) set by the next term in the analytic expansion — consistent with the smoothness of the velocity field at scales below \(\eta\).
  5. Grid turbulence has \(\varepsilon=2.0\times10^{-2}\ \text{m}^2\text{s}^{-3}\), \(\nu=1.5\times10^{-5}\ \text{m}^2\text{s}^{-1}\), and integral scale \(L=0.20\ \text{m}\). (a) Compute the Kolmogorov length \(\eta=(\nu^3/\varepsilon)^{1/4}\). (b) State whether \(r=1.0\times10^{-2}\ \text{m}\) plausibly lies in the inertial range. (c) If so, compute \(S_3(r)\).
    Solution(a) \(\nu^3=(1.5\times10^{-5})^3=3.375\times10^{-15}\ \text{m}^6\text{s}^{-3}\); \(\nu^3/\varepsilon=3.375\times10^{-15}/2.0\times10^{-2}=1.69\times10^{-13}\ \text{m}^4\); \(\eta=(1.69\times10^{-13})^{1/4}\approx2.0\times10^{-4}\ \text{m}=0.20\ \text{mm}\). (b) The chosen \(r=1.0\times10^{-2}\ \text{m}=10\ \text{mm}\) satisfies \(\eta=0.20\ \text{mm}\ll r\ll L=200\ \text{mm}\): it sits about \(50\eta\) above the dissipation scale and one-twentieth of \(L\), comfortably inside the inertial range. (c) \(S_3=-\tfrac45\varepsilon r=-\tfrac45(2.0\times10^{-2})(1.0\times10^{-2})=-1.6\times10^{-4}\ \text{m}^3\text{s}^{-3}\).