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Let's discuss stability of exact one-mode solution of Maxwell equations with presense of chiral anomaly and in hydrodynamical approximation. Corresponding system of equations has the form

\ [\nabla \times \mathbf E] = -\partial_{0}\mathbf B \qquad (1),

\ [\nabla \times \mathbf B] = \sigma \mathbf E + \frac{\alpha}{\pi}\mu_{5}\mathbf B \qquad (2),

\ \partial_{0}\mu_{5} = \frac{\alpha }{\pi f_{5}^{2}}(\mathbf E \cdot \mathbf B) \qquad (3).

For one-mode solution,

\ \mathbf B(t) = b(t)\mathbf b_{k}, \quad \mathbf b_{k} = (sin(kz), cos(kz), 0), \quad  b(0) = B_{0} .

Then we have from \ (1)-(3)

\ \mathbf E = -\frac{1}{k}\dot{b}(t)\mathbf b_{k}, \quad \partial_{0}\mu_{5} = -\frac{\alpha }{2\pi f_{5}^{2}k}\dot{b}^{2}(t) \Rightarrow \mu_{5} = \mu_{5}(0) + \frac{\alpha B_{0}^{2}}{2 \pi kf_{5}^{2}} - \frac{\alpha b^{2}(t)}{2 \pi kf_{5}^{2}},

\ kb(t) = -\frac{\sigma}{k}\dot{b}(t) + \frac{\alpha}{\pi}\left( \mu_{5}(0) + \frac{\alpha B_{0}^{2}}{2 \pi kf_{5}^{2}} - \frac{\alpha b^{2}(t)}{2 \pi kf_{5}^{2}}\right)b(t).

For \ \frac{\alpha}{\pi}\mu_{5}(0) \neq k we have solution

\ b(t) = \frac{B_{0}\sqrt{\gamma B_{0}^{2} + \delta }e^{(B_{0}^{2}\gamma + \delta ) t}}{\sqrt{ \gamma B_{0}^{2}e^{2(B_{0}^{2}\gamma + \delta ) t} + \delta}} = \frac{\tilde{B}e^{(B_{0}^{2}\gamma + \delta ) t}}{\sqrt{ \gamma B_{0}^{2}e^{2(B_{0}^{2}\gamma + \delta ) t} + \delta}}, \quad \delta = \frac{\alpha \mu_{5}^{0}}{\pi} - k , \quad \gamma = \frac{\alpha^{2}}{2\pi^{2}\sigma f_{5}^{2}}, \quad \delta + B_{0}^{2}\gamma > 0 \qquad (4).

From \ (4) we see that

\ b(\infty ) \to B_{0}\sqrt{1 + \frac{\delta }{\gamma B_{0}^{2}}} , \quad \mathbf E (\infty ) \to 0, \quad \mu_{5} (\infty ) \to \frac{\pi k }{\alpha} .

PerturbationsEdit

Let's perturb solution of \ (1)-(3) and then linearize system on perturbations: the result is

\ \partial_{0}\delta \mu_{5} = \gamma \left( (\delta \mathbf E \cdot \mathbf B ) + (\delta \mathbf B \cdot \mathbf E )\right) \qquad (5),

\ [\nabla \times \delta \mathbf E] = -\partial_{0}\delta \mathbf B \qquad (6),

\ [\nabla \times \delta \mathbf B ] = \sigma \delta \mathbf E + \frac{\alpha}{\pi}\left( \mu_{5}\delta \mathbf B + \mathbf B \delta \mu_{5}\right) \qquad (7).

Let's take curl and divergence of \ (6)-(7):

\ (\nabla \cdot \delta \mathbf E ) = -\frac{\alpha}{\pi \sigma}(\mathbf B \cdot \nabla \delta \mu_{5}) \qquad (8),

\ \nabla (\nabla \cdot \delta \mathbf E) - \Delta \delta \mathbf E = - \partial_{0}[\nabla \times \delta \mathbf B] \qquad (9)

\ -\Delta \delta \mathbf B = -\sigma \partial_{0}\delta \mathbf B + \frac{\alpha}{\pi}\left( \mu_{5} [\nabla \times \delta \mathbf B ]+ [\nabla \delta \mu_{5} \times \mathbf B] + \delta \mu_{5}[\nabla \times \mathbf B]\right) \qquad (10).

From eq. \ (8) it follows that

\ \delta \mathbf E = -\frac{\alpha }{\pi \sigma }\delta \mu_{5}\mathbf B + \delta \tilde {\mathbf E}, \quad \nabla \cdot \delta \tilde{\mathbf E} = 0 \qquad (11),

from which after substituting eq. \ (11) to eq. \ (7) in particular follows (for simplicity \ \tilde{\mathbf E} is set to zero) that

\ [\nabla \times \delta \mathbf B] = \frac{\alpha}{\pi}\mu_{5} \delta \mathbf B \Rightarrow \Delta \delta \mathbf B = -\left( \frac{\alpha \mu_{5}}{\pi }\right)^{2}\delta \mathbf B \Rightarrow \delta \mathbf B = \mathbf A cos (\frac{\alpha \mu_{5}}{\pi }(\mathbf n \cdot \mathbf r)) \qquad (12),

where

\ (\mathbf n \cdot \mathbf A ) = 0, \quad \mathbf A = const, \quad \mathbf n^{2} = 1.

In general,

\ \mathbf B = \frac{\pi \alpha }{\mu_{5}}[\nabla \times [\nabla \times (\mathbf a \varphi ) ]] + [\nabla \times (\mathbf a \varphi )], \quad \Delta \varphi + \left( \frac{\alpha \mu_{5}}{\pi }\right)^{2}\varphi = 0, \quad \mathbf a^{2} = 1 \qquad (13).

\ (12) is oscillating solution, so it doesn't grow. (general case is still analyzed)

Let's now derive the expressions for \ \delta \mathbf E , \delta \mu_{5}. Let's use \ (5), (11), (12) and set \ \delta \tilde{\mathbf E} to zero:

\ \partial_{0}\delta \mu_{5} = \gamma \left( -\frac{b^{2}(t)\alpha}{\pi \sigma}\delta \mu_{5} + (\mathbf E \cdot \mathbf A )cos\left( \frac{\alpha \mu_{5}}{\pi}(\mathbf n \cdot \mathbf r)\right)\right) \qquad (14).

We know that \ \mathbf E \sim \dot{b}(t), while \ \mu_{5} = const + b^{2}(t). We may try to solve it in quadratures.

\ \delta \mu_{5} = \delta \mu_{5}^{h} + \delta \mu_{5}^{inh.}, \quad \delta \mu_{5}^{h} = \int \frac{dt}{1 + \frac{\gamma\alpha}{\pi \sigma }b^{2}(t)}, \quad \delta \mu_{5}^{inh.} = \int dt (\mathbf E \cdot \mathbf A)cos\left(\frac{\mu_{5}\alpha}{\sigma \pi}(\mathbf n \cdot \mathbf r)\right).

The second summand contains Fresnel integrals which are converge for all arguments. For \ t \to \infty this summand doesn't depend on \ t. Formally it doesn't break stability of \ \delta \mu_{5} trivial solution, but this solution isn't asymptotically stable.

Absolutely analogical conclusion about \ \delta \mathbf E is also true (I've used eq. \ (11)).

Stability of stationary one-mode solutionEdit

Let's now assume stationary case of one-mode solution, \ \mu_{5}^{0} = \frac{\pi k}{\alpha }. Then \ \mathbf B = B_{0}\mathbf b_{k}, \mathbf E = 0, \mu_{5} = \mu_{5}^{0}. Expressions \ (12)-(13) are still true. As for expressions for \ \delta \mathbf E , \delta \mu_{5}, we have (instread of eq. \ (14) and below)

\ \partial_{0}\delta \mu_{5} = -\frac{\gamma \alpha }{\pi \sigma }\delta \mu_{5}\mathbf B^{2} = -\frac{B_{0}^{2}\gamma \alpha }{\pi \sigma }\delta \mu_{5} \Rightarrow \delta \mu_{5} = Ce^{-\frac{B_{0}^{2}\gamma \alpha}{\pi \sigma}t} \qquad (15).

Then,

\ \delta \mathbf E = -\frac{\alpha }{\pi \sigma }Ce^{-\frac{B_{0}^{2}\gamma \alpha}{\pi \sigma}t}\mathbf B \qquad (16).


The other analisysEdit

\ \sigma \partial_{0}\delta A_{+} = \partial_{z}^{2}\delta A_{+} +ik_{0} \partial_{z}\delta A_{+} - \gamma \left(\frac{1}{2}\delta A_{+}- \frac{\delta A_{-}}{2}e^{-2ikz} \right) \qquad (A1),

\ \sigma \partial_{0}\delta A_{-} = \partial_{z}^{2}\delta A_{-} - ik_{0}\partial_{z}\delta A_{-} - \gamma \left(\frac{1}{2}\delta A_{-} - \frac{\delta A_{+}}{2}e^{2ikz} \right), \quad \gamma = \frac{\alpha^{2}B_{0}^{2}}{\pi^{2}f_{5}^{2}} \qquad (A2).

In the simplest case \ \delta A_{-}(z, t) = \delta A(t)e^{ikz}. Then

\ \sigma \partial_{0}\delta A = -k^{2}\delta A - k_{0}k\delta A .

Then we have as condition of stability

\ -k^{2} + k_{0}k < 0 \Rightarrow k = \left(k_{0}, \infty \right).

Because of \ A(1)^{\dagger} = A(2), we have in general case that \ A_{-}(z, t) = a_{-}(z, t)e^{-ikz}. Then

\ \sigma \partial_{0}\delta A_{+} = \partial_{z}^{2}\delta A_{+} +ik_{0} \partial_{z}\delta A_{+} + \gamma \delta a_{-}e^{ikz} \qquad (A3),

\ \sigma \partial_{0}\delta A_{-} = \partial_{z}^{2}\delta A_{-} - ik_{0}\partial_{z}\delta A_{-} + \gamma \delta a_{+}e^{-ikz}, \quad \gamma = \frac{\alpha^{2}B_{0}^{2}}{\pi^{2}f_{5}^{2}} \qquad (A4),

or, by proceeding to \ A(z, t),

\ \sigma \partial_{0}a_{+} = -k^{2}\delta a_{+} + \partial_{z}^{2}\delta a_{+} - kk_{0}\delta a_{+} + ik_{0}\partial_{z}\delta a_{+} + \gamma \delta a_{-},

\ \sigma \partial_{0}a_{-} = -k^{2}\delta a_{-} + \partial_{z}^{2}\delta a_{-} - kk_{0}\delta a_{-} - ik_{0}\partial_{z}\delta a_{-} + \gamma \delta a_{+}.

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