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Take zero mean flow locally and incompressible fluctuations. Subtracting the averaged induction equation givesThe first-order smoothing approximation omits the second term on the right when solving for $\mathbf{b}$, but retains $⟨\mathbf{u}\times \mathbf{b}⟩$ in the mean equation. For a locally uniform, slowly changing mean magnetic field the forcing becomes $(\overline{\mathbf{B}}\cdot \nabla )\mathbf{u}$.

There are two controlled ways to make the omitted fluctuation term small. At small eddy magnetic Reynolds number $R_m=u_{\mathrm{rms}}\ell /\eta$, diffusion gives $b/B\sim R_m$, so the omitted $ub/\ell$ is smaller than the retained $uB/\ell$ by $R_m$. Alternatively, if $u_{\mathrm{rms}}\tau /\ell \ll 1$ for the fluctuation response time $\tau$, then $b/B\sim u_{\mathrm{rms}}\tau /\ell$ and the short-memory response is perturbative. Mean-field scale separation is additionally required. Merely calling the turbulence weak or assuming that the mean magnetic field is small does not justify the omission.

With homogeneous initial transients removed, the causal linear solution isInsert this into the electromotive correlation. Isotropy gives ${\alpha }_{ij}=\alpha {\delta }_{ij}$, so taking the trace yields the helicity formula for isotropic first-order smoothing:Using the alternating-tensor sign givesThe minus sign follows from ${ϵ}_{imn}=-{ϵ}_{min}$. This derivation specifies the diffusion and correlation weighting, rather than asserting that alpha is proportional to instantaneous helicity for arbitrary turbulence.

If $\eta \tau /{\ell }^2\ll 1$, diffusion over a correlation time is negligible. Define the helicity correlation time byFor an ordinary decaying correlation with nonzero mean kinetic helicity, this gives the familiar estimateIt is the kinetic helicity, not magnetic helicity. Opposite flow handedness reverses the alpha effect. The time-integral formula remains meaningful if the equal-time helicity vanishes or its correlation changes sign, when replacing it by a single positive correlation time may fail. In the same weak-diffusion short-memory approximation the gradient forcing $-(\mathbf{u}\cdot \nabla )\overline{\mathbf{B}}$ gives $\beta =\frac{1}{3}{\int }_0^{\infty }⟨\mathbf{u}(t)\cdot \mathbf{u}(t-\tau )⟩d\tau \simeq {\tau }_U{u}_{\mathrm{rms}}^2/3$; the two correlation times need not be equal.

In the small-$R_m$, slowly varying quasistatic regime, diffusion is instead essential. The response givesFor velocity concentrated at wavenumber $k_0$, this becomes $\alpha =-⟨\mathbf{u}\cdot \mathbf{\varpi }⟩/(3\eta k_0^2)$. Its effective memory is the magnetic diffusion time on the eddy scale. The controlled low-$R_m$ setting is developed in Moffatt's analysis of turbulent dynamo action; the causal-response calculation above also covers short-memory turbulence without assuming that diffusion is always negligible.

Write the total velocity as $\mathbf{U}=\Omega y\hat{\mathbf{x}}+\mathbf{u}$ and the magnetic field as $B_0\hat{\mathbf{x}}+\mathbf{b}$. For solenoidal fields the resistive induction equation is $({\partial }_t+\mathbf{U}\cdot \nabla )\mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{U}+\eta {\nabla }^2\mathbf{B}$. The first-order smoothing approximation drops the fluctuating terms $\mathbf{b}\cdot \nabla \mathbf{u}-\mathbf{u}\cdot \nabla \mathbf{b}$, retaining shear exactly. The uniform field is not stretched by the background shear because ${\partial }_x(\Omega y)=0$. The other retained stretching terms are $\mathbf{b}\cdot \nabla (\Omega y\hat{\mathbf{x}})=\Omega b_y\hat{\mathbf{x}}$ and $B_0{\partial }_x\mathbf{u}$. HenceThe same phase cancellation as in part (i), together with ${\nabla }^2{e}^{i\mathbf{k}\cdot \mathbf{x}}=-K^2{e}^{i\mathbf{k}\cdot \mathbf{x}}$, proves that a response can be taken as $\mathrm{Re}(\mathbf{c}(t)e^{i\mathbf{k}(t)\cdot \mathbf{x}})$, whereIf $\mathbf{k}\cdot \mathbf{c}=0$ initially, its derivative is $-\eta K^2\mathbf{k}\cdot \mathbf{c}$, so magnetic solenoidality is preserved.

For $j=y,z$ the response equation is ${\dot{c}}_j+\eta K^2{c}_j=i{k}_x{B}_0{v}_j$. Its integrating-factor solution isThe velocity amplitudes and wavevector vary on the shear timescale. In the fast-diffusion limit $\eta (k_x^2+k_y^2)\gg |\Omega |$, and after the free transient, the integral is concentrated within time $(\eta K^2{)}^{-1}$ of $t$. Replacing the slowly varying factors there by their current values gives the quasistatic magnetic response of a helical shearing wave:Complex conjugation introduces a minus sign in $c_j^{\ast }$. With precisely the EMF normalization specified in the question,A literal spatial average of the product of the two real Fourier waves would be half this quantity; no extra factor of $1/2$ belongs in the question's defined $\mathcal{E}$.

Use $A=k_x^2+k_z^2$ and part (i) to obtain $\mathcal{E}(t)\simeq -B_0{H}_{0}A/[\eta (A+{\Omega }^2{k}_x^2{t}^2{)}^2]$. For the time integral, take the driven response without a free magnetic transient and suppose the fast-diffusion approximation holds through the contributing interval, in particular at $t=0$; $\eta k_x^2\gg |\Omega |$ is a sufficient uniform condition. For $\Omega k_x\ne 0$, substitute $z=|\Omega k_x|t/\sqrt{A}$. Since ${\int }_{-\infty }^{\infty }(1+z^2{)}^{-2}dz=\pi /2$, this givesThe integral identity follows immediately from $z=\tan \chi$, reducing it to ${\int }_{-\pi /2}^{\pi /2}{\cos }^2\chi d\chi$. If $k_x=0$, the forcing and $H$ vanish and the driven EMF is zero. If $\Omega =0$ but $H_0\ne 0$, the leading EMF is constant rather than time-integrable; the finite integral relies on nonzero shear.

Under the first-order smoothing approximation, neglect the fluctuating product in the resistive induction equation while retaining its contribution to the mean electromotive force. Since the test field ${\mathbf{B}}_0$ is constant and the velocity is a solenoidal vector field, the fluctuation equation isLet $a=\eta k^2>0$, $d=a^2+{\Omega }^2$, and ${\mathbf{C}}_j=i({\mathbf{B}}_0\cdot {\mathbf{k}}_j){\hat{\mathbf{u}}}_j$. For each Fourier mode, the diffusion equation becomes a scalar linear relaxation equation applied to each vector component. Take its long-time periodic response, with $\Omega \ne 0$. For the first mode,while for the second,Thus the complex response coefficients areThey giveAn arbitrary initial fluctuation also contains a homogeneous diffusive transient. In the forced modes this is $\mathrm{Re}{\sum }_j{\mathbf{r}}_j{e}^{i{\mathbf{k}}_j\cdot \mathbf{x}-at}$; more general initial modes also diffuse. The displayed harmonic expression is the periodic particular solution after those transients, not the most general solution at finite time. The constants $a,d$ and the factors of $i$ are unchanged by taking the final real part.

To justify the helicity relation, write ${\hat{\mathbf{u}}}_j={\mathbf{a}}_j+i{\mathbf{c}}_j$ with real vectors. Solenoidality gives ${\mathbf{k}}_j\cdot {\mathbf{a}}_j={\mathbf{k}}_j\cdot {\mathbf{c}}_j=0$. Their cross product is parallel to ${\mathbf{k}}_j$, andThis is the helicity vector of a solenoidal Fourier mode. In these conventions the spatially averaged kinetic helicity density of $\mathrm{Re}({\hat{\mathbf{u}}}_j{e}^{i{\mathbf{k}}_j\cdot \mathbf{x}})$ is $-H_j{k}^2$; the sign is set by the specified cross-product order.

For equal wavevectors, spatial averaging of two real harmonic fields givesTime averaging gives another factor $1/2$, because $⟨{\cos }^2\Omega t{⟩}_t=⟨{\sin }^2\Omega t{⟩}_t=1/2$ and the mixed temporal average is zero. Consequently each diagonal mode contributes to the mean-field electromotive forceDistinct modes with ${\mathbf{k}}_1\ne \pm {\mathbf{k}}_2$ have no spatially averaged cross term. The printed condition also permits the opposite-wavevector case. The possible cross terms still cancel after time averaging: the out-of-phase response coefficients are ${\mathbf{p}}_2=-\Omega {\mathbf{C}}_2/d$ and ${\mathbf{q}}_1=\Omega {\mathbf{C}}_1/d$, while integration by parts gives $⟨{\mathbf{u}}_1\times ({\mathbf{B}}_0\cdot \nabla ){\mathbf{u}}_2{⟩}_x=⟨{\mathbf{u}}_2\times ({\mathbf{B}}_0\cdot \nabla ){\mathbf{u}}_1{⟩}_x$ for the real spatial fields. This handles all the distinct wavevectors allowed by the question, assuming the usual periodic-cell or whole-space spatial average.

Thus the alpha tensor isBoth averaging factors matter. The alpha tensor is real and symmetric, and this contribution vanishes when the mode kinetic helicity densities vanish.

For the two perpendicular wavevectors and common $H_1=H_2=H$, put ${\alpha }_0=\eta H{k}^4/[2({\Omega }^2+{\eta }^2{k}^4)]$. ThenMean-field dynamo action is possible when ${\alpha }_0\ne 0$ and sufficiently long mean-field wavelengths are allowed. The missing $zz$ entry does not prevent an alpha-squared dynamo. Indeed a slowly varying mean magnetic field transverse to $z$, with dependence $e^{iKz+\lambda t}$, obeysThe two growth rates are $-\eta K^2\pm {\alpha }_{0}K$, so the larger is positive if $0<|K|<|{\alpha }_0|/\eta$. This supplies a brief constructive reason; scale separation also requires $|K|\ll k$. A spatially uniform test field itself has zero curl of its mean-field electromotive force and does not grow. If $H=0$, or if boundary conditions exclude all unstable long wavelengths, this particular alpha effect does not yield growth. The conclusion concerns the stated first-order smoothing approximation; no uncomputed turbulent-diffusion correction is assumed.