Consider a sufficiently smooth 3D turbulent fluid flow occupying a periodic cubic box of side length $L$. Any sufficiently smooth periodic field variable $f(\boldsymbol{x})$, where $\boldsymbol{x}=(x_l)$ and $x_l\in[0,L)$ for $l=1,2,3$, can be expanded in a Fourier series
\[f(\boldsymbol{x})=\sum_{\boldsymbol{k}}\hat{f}(\boldsymbol{k})e^{i\boldsymbol{k}\cdot\boldsymbol{x}},\quad \boldsymbol{k}\in\frac{2\pi}{L}\mathbb{Z}^3.\]It follows that
\[\nabla f=\sum_{\boldsymbol{k}}(i \boldsymbol{k})\hat{f}(\boldsymbol{k})e^{i\boldsymbol{k}\cdot\boldsymbol{x}},\quad \nabla^2 f=\sum_{\boldsymbol{k}}(-k^2)\hat{f}(\boldsymbol{k})e^{i\boldsymbol{k}\cdot\boldsymbol{x}},\]where $k \equiv \lvert \boldsymbol{k} \rvert = \sqrt{k_l k_l}$, and repeated Cartesian indices are summed from 1 to 3. For a real-valued field, the Fourier coefficients satisfy the Hermitian symmetry $\hat{f}(-\boldsymbol{k})=\hat{f}^*(\boldsymbol{k})$.
The incompressible Navier–Stokes equations (INSE) governing the turbulent velocity field $\boldsymbol{u}(\boldsymbol{x},t)$ are
\[\partial_t\boldsymbol{u}+\boldsymbol{u}\cdot\nabla\boldsymbol{u}=-\nabla p+\nu \nabla^2\boldsymbol{u}+\boldsymbol{f}, \quad \nabla\cdot\boldsymbol{u}=0.\]It can also be written in index form as
\[\frac{\partial u_l}{\partial t}+u_m\frac{\partial u_l}{\partial x_m}=-\frac{\partial p}{\partial x_l}+\nu \frac{\partial^2 u_l}{\partial x_m\partial x_m}+f_l,\] \[\frac{\partial u_m}{\partial x_m}=0.\]Here $p$ denotes the pressure divided by the constant density. Expanding $\boldsymbol{u}(\boldsymbol{x},t)$, $p(\boldsymbol{x},t)$, and $\boldsymbol{f}(\boldsymbol{x},t)$ into Fourier series, the equations in wavenumber space become
\[\begin{aligned} &\sum_{\boldsymbol{k}}\frac{\partial \hat{u}_l(\boldsymbol{k},t)}{\partial t}e^{i\boldsymbol{k}\cdot\boldsymbol{x}} +\bigg(\sum_{\boldsymbol{p}}\hat{u}_m(\boldsymbol{p},t)e^{i\boldsymbol{p}\cdot\boldsymbol{x}}\bigg)\bigg(\sum_{\boldsymbol{q}}(i q_m)\hat{u}_l(\boldsymbol{q},t)e^{i\boldsymbol{q}\cdot\boldsymbol{x}}\bigg) \\ &={} -\sum_{\boldsymbol{k}}(ik_l)\hat{p}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}} +\nu\sum_{\boldsymbol{k}}(-k^2)\hat{u}_l(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}} +\sum_{\boldsymbol{k}}\hat{f}_l(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}, \end{aligned}\]and
\[\sum_{\boldsymbol{k}}(ik_m)\hat{u}_m(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}=0.\]The nonlinear term can be rewritten as
\[\sum_{\boldsymbol{p}}\sum_{\boldsymbol{q}}(iq_m)\hat{u}_m(\boldsymbol{p},t)\hat{u}_l(\boldsymbol{q},t)e^{i(\boldsymbol{p}+\boldsymbol{q})\cdot\boldsymbol{x}}.\]We now focus on the coefficient of $e^{i\boldsymbol{k}\cdot\boldsymbol{x}}$ on both sides of the equation, which corresponds to the evolution equation for a single Fourier mode $\boldsymbol{k}$. To obtain the evolution for a single Fourier mode $\boldsymbol{k}$, apply the operator
\[\mathcal{F}\{\,\cdot\,\}_{\boldsymbol{k}}\equiv\langle(\,\cdot\,)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\rangle_L\equiv \frac{1}{L^3}\int_0^L\int_0^L\int_0^L(\,\cdot\,)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\,\mathrm{d}\boldsymbol{x}.\]Using orthogonality of the modes, one finds
\[\bigg(\frac{\partial }{\partial t}+\nu k^2\bigg)\hat{u}_l(\boldsymbol{k},t) = -ik_m\sum_{\boldsymbol{k}'}\hat{u}_m(\boldsymbol{k}-\boldsymbol{k}',t)\hat{u}_l(\boldsymbol{k}',t) -ik_l\hat{p}(\boldsymbol{k},t) +\hat{f}_l(\boldsymbol{k},t),\]and
\[ik_m\hat{u}_m(\boldsymbol{k},t)=0.\]Here, the derivation of the quadratic nonlinear term has utilized the continuity equation and a variable substitution, $\boldsymbol{p}+\boldsymbol{k}’=\boldsymbol{k}$.
The projection below applies to $\boldsymbol{k}\ne\boldsymbol{0}$. The zero mode instead obeys $\partial_t\hat{u}_l(\boldsymbol{0},t)=\hat{f}_l(\boldsymbol{0},t)$, while $\hat{p}(\boldsymbol{0},t)$ is arbitrary. In a zero-mean formulation, the zero velocity mode is absent.
Intuitively, in wavenumber space, the velocity $\hat{\boldsymbol{u}}$ should be normal to the wavevector $\boldsymbol{k}$ due to the incompressibility constraint. Hence, it’s natural to ask what will happen if we decompose INSE into two components – one parallel to and one normal to $\boldsymbol{k}$.
Projecting the modal momentum equation along $\boldsymbol{k}$ gives its longitudinal component, equivalent to the pressure Poisson equation in wavenumber space,
\[0 = -i\bigg(\frac{k_lk_n}{k^2}\bigg)k_m\sum_{\boldsymbol{k}'}\hat{u}_m(\boldsymbol{k}-\boldsymbol{k}',t)\hat{u}_n(\boldsymbol{k}',t) -ik_l\hat{p}(\boldsymbol{k},t) +\bigg(\frac{k_lk_n}{k^2}\bigg)\hat{f}_n(\boldsymbol{k},t),\]For $\boldsymbol{k}\ne\boldsymbol{0}$, imposing $k_n\hat{f}_n(\boldsymbol{k},t)=0$, equivalently $({k_lk_n}/{k^2})\hat{f}_n(\boldsymbol{k},t)=0$, ensures divergence-free forcing for numerical simulations of forced incompressible homogeneous isotropic turbulence.
Subtracting this longitudinal component from the modal momentum equation gives
\[\bigg(\frac{\partial }{\partial t}+\nu k^2\bigg)\hat{u}_l(\boldsymbol{k},t) = -P_{ln}(\boldsymbol{k})(ik_m)\sum_{\boldsymbol{k}'}\hat{u}_m(\boldsymbol{k}-\boldsymbol{k}',t)\hat{u}_n(\boldsymbol{k}',t) +P_{ln}(\boldsymbol{k})\hat{f}_n(\boldsymbol{k},t),\]Here $P_{ln}(\boldsymbol{k})\equiv\delta_{ln}-{(k_lk_n)}/{k^2}$ is the projection tensor, and the pressure is canceled in this equation.
Energy balance equation
We then drop the forcing term in the projected equation, focusing on the intrinsic properties of the turbulent flows governed by INSE. In particular, we are concerned with how the kinetic energy is transferred in this system.
\[\bigg(\frac{\partial }{\partial t}+2\nu k^2\bigg)\frac{1}{2}\hat{u}_l^*(\boldsymbol{k},t)\hat{u}_l(\boldsymbol{k},t) = \operatorname{Re}\bigg\{-P_{ln}(\boldsymbol{k})(ik_m)\sum_{\boldsymbol{k}'}\hat{u}_m(\boldsymbol{k}-\boldsymbol{k}',t)\hat{u}_n(\boldsymbol{k}',t)\hat{u}_l^*(\boldsymbol{k},t)\bigg\}.\]Here the superscript * denotes the complex conjugate.
Taking the ensemble average of the above equation, we obtain the balance equation of kinetic energy, $\hat{E}(\boldsymbol{k},t)\equiv\langle\frac{1}{2}\hat{u}_l^*(\boldsymbol{k},t)\hat{u}_l(\boldsymbol{k},t)\rangle$.
\[\bigg(\frac{\partial }{\partial t}+2\nu k^2\bigg)\hat{E}(\boldsymbol{k},t) = -k_mP_{ln}(\boldsymbol{k})\,\Im\bigg\{\sum_{\boldsymbol{k}'}\big\langle{\hat{u}_l(\boldsymbol{k},t)\hat{u}_m^*(\boldsymbol{k}-\boldsymbol{k}',t)\hat{u}_n^*(\boldsymbol{k}',t)}\big\rangle\bigg\}.\]The right-hand side follows from $\operatorname{Re}(iz)=-\Im(z)$ after taking the complex conjugate of the triple product. Here $\Im$ denotes the imaginary part.
We denote the right-hand side as $\hat{T}(\boldsymbol{k},t)$, and – using the Hermitian symmetry of the real-valued velocity field, and a variable substitution – it can also be expressed as
\[\hat{T}(\boldsymbol{k},t)=-k_mP_{ln}(\boldsymbol{k})\,\Im\bigg\{\sum_{\substack{\boldsymbol{p},\,\boldsymbol{q} \\ \boldsymbol{k}+\boldsymbol{p}+\boldsymbol{q}=0}}\big\langle{\hat{u}_l(\boldsymbol{k},t)\hat{u}_m(\boldsymbol{p},t)\hat{u}_n(\boldsymbol{q},t)}\big\rangle\bigg\} \\ =-\Im\bigg\{\sum_{\substack{\boldsymbol{p},\,\boldsymbol{q} \\ \boldsymbol{k}+\boldsymbol{p}+\boldsymbol{q}=0}}\big\langle{k_m\hat{u}_n(\boldsymbol{k},t)\hat{u}_m(\boldsymbol{p},t)\hat{u}_n(\boldsymbol{q},t)}\big\rangle\bigg\}.\]The second equality uses $P_{ln}(\boldsymbol{k})\hat{u}_l(\boldsymbol{k},t)=\hat{u}_n(\boldsymbol{k},t)$, which follows from incompressibility.
When it is summed over all $\boldsymbol{k}$, we have
\[\begin{aligned} &\sum_{\boldsymbol{k}}\hat{T}(\boldsymbol{k},t)=-\Im\bigg\{\sum_{\substack{\boldsymbol{k},\,\boldsymbol{p},\,\boldsymbol{q} \\ \boldsymbol{k}+\boldsymbol{p}+\boldsymbol{q}=0}}\big\langle{k_m\hat{u}_n(\boldsymbol{k},t)\hat{u}_m(\boldsymbol{p},t)\hat{u}_n(\boldsymbol{q},t)}\big\rangle\bigg\} \\ &=\Im\bigg\{\sum_{\substack{\boldsymbol{k},\,\boldsymbol{p},\,\boldsymbol{q} \\ \boldsymbol{k}+\boldsymbol{p}+\boldsymbol{q}=0}}\big\langle{(p_m+q_m)\hat{u}_n(\boldsymbol{k},t)\hat{u}_m(\boldsymbol{p},t)\hat{u}_n(\boldsymbol{q},t)}\big\rangle\bigg\} \\ &=\Im\bigg\{\sum_{\substack{\boldsymbol{k},\,\boldsymbol{p},\,\boldsymbol{q} \\ \boldsymbol{k}+\boldsymbol{p}+\boldsymbol{q}=0}}\big\langle{q_m\hat{u}_n(\boldsymbol{q},t)\hat{u}_m(\boldsymbol{p},t)\hat{u}_n(\boldsymbol{k},t)}\big\rangle\bigg\} \\ &=-\sum_{\boldsymbol{q}}\hat{T}(\boldsymbol{q},t). \end{aligned}\]The second equality uses $k_m=-(p_m+q_m)$, the third uses incompressibility $p_m\hat{u}_m(\boldsymbol{p},t)=0$, and the final equality relabels the summed wavevectors.
Therefore the sum is equal to its own negative and must vanish:
\[\sum_{\boldsymbol{k}}\hat{T}(\boldsymbol{k},t)=0,\]namely, this term merely represents a transfer of energy between modes.
References
Batchelor, G. K. (1953). The Theory of Homogeneous Turbulence, Cambridge University Press.
McComb, W. D. (1990). The Physics of Fluid Turbulence, Clarendon Press.
Frisch, U. (1995). Turbulence: The Legacy of A. N. Kolmogorov, Cambridge University Press.
Pope, S. B. (2000). Turbulent Flows, Cambridge University Press.