= Solution
An explicit example avoids an unspecified <correlation time>. For each coordinate axis $\mathbf e_j$, choose transverse unit vectors $\mathbf a_j,\mathbf b_j$ with $\mathbf e_j\times\mathbf a_j=\mathbf b_j$, and take the three real <circular polarizations>
$$
\mathbf u_j=U(\mathbf a_j\cos\phi_j-\sigma\mathbf b_j\sin\phi_j),\qquad \phi_j=kx_j-\Omega t,\qquad \sigma=\pm1.
$$
The propagation directions are mutually perpendicular and all three waves have the same handedness. Direct differentiation gives
$$
\nabla\times\mathbf u_j=\sigma k\mathbf u_j,\qquad \nabla^2\mathbf u_j=-k^2\mathbf u_j,\qquad \mathbf u_j\times\partial_{\phi_j}\mathbf u_j=-\sigma U^2\mathbf e_j.
$$
A spatial average over the periodic cell removes cross terms between different waves. Hence the mean <kinetic helicity> of their sum is
$$
\mathcal H=\langle\mathbf u\cdot\nabla\times\mathbf u\rangle=3\sigma kU^2.
$$
The sum has isotropic second moments, $\langle u_i u_j\rangle=U^2\delta_{ij}$.
For a locally uniform test <magnetic field> $\overline{\mathbf B}$, the periodic <first-order smoothing> response obeys
$$
(\partial_t-\eta\nabla^2)\mathbf b_j=k\overline B_j\partial_{\phi_j}\mathbf u_j.
$$
Set $a=\eta k^2>0$. Since $\partial_t=-\Omega\partial_\phi$ and $\partial_\phi^2\mathbf u_j=-\mathbf u_j$, direct substitution gives the response after transients decay:
$$
\mathbf b_j=\frac{k\overline B_j}{a^2+\Omega^2}(a\partial_{\phi_j}\mathbf u_j-\Omega\mathbf u_j).
$$
The term proportional to $\mathbf u_j$ contributes no <cross product> with that same wave. Cross terms between distinct waves average to zero. Therefore
$$
\boldsymbol{\mathcal E}=\sum_j\langle\mathbf u_j\times\mathbf b_j\rangle=-\frac{\sigma kU^2a}{a^2+\Omega^2}\overline{\mathbf B},
$$
and the <isotropic alpha effect of three helical traveling waves> is
$$
\boxed{\alpha=-\frac{a}{3(a^2+\Omega^2)}\mathcal H,\qquad a=\eta k^2.}
$$
This is the alpha-helicity relation with the exact response time $\tau_{\rm eff}=a/(a^2+\Omega^2)$ for these monochromatic waves. In the quasistatic limit $\Omega=0$, $\alpha=-\mathcal H/(3\eta k^2)$. A single wave would give an anisotropic response along its propagation direction; the three equal waves make the response tensor a scalar multiple of the identity. The calculation retains the validity assumptions of <first-order smoothing>, rather than extrapolating the formula to arbitrary fluctuation amplitude.
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