Stability for one-mode solution of Maxwell MHD equation with chiral anomaly

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

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

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

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

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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