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This question is related to nonlinear Hall effect proposed in this paper. The Boltzmann equation in the electric field under relaxation time approximation is:

$$-e E_a \partial_a f+\partial_t f=\dfrac{f_0-f}{\tau} \tag{1} $$

in which $\partial_a=\dfrac{\partial}{\partial_{k_a}}$ and $f_0$ is the equilibrium distribution in the absence of external fields. Note that we have adopted the Einstein convention for vector analysis. To solve Eq. $(1)$ we take the following two assumpations:

  • The electric field is given as: $$E_a(t)=Re\{\mathcal{E}_ae^{i\omega t}\} \qquad \mathcal{E}_a \in C \tag{2}$$

  • The distribution is expanded up to second order: $$f=Re\{f_0+f_1+f_2\} \tag{3}$$ in which $f_n\propto |\mathcal{E}|^n \tag{4} $

Under these conditions, the authors obtain the following analytic solutions: $$f_1=f_1^\omega e^{i\omega t} \quad f_1^\omega=\dfrac{e\tau\mathcal{E}_a\partial_a f_0}{1+i\omega\tau} \tag{5}$$ $$f_2=f_2^0+f_2^{2\omega}e^{2i\omega t} \quad f_2^0=\dfrac{(e\tau)^2\mathcal{E}_a^* \mathcal{E}_b\partial_{ab}f_0}{2(1+i\omega \tau)} \quad f_2^{2\omega}=\dfrac{(e\tau)^2\mathcal{E}_a\mathcal{E}_b\partial_{ab}f_0}{2(1+i\omega\tau)(1+2i\omega\tau)} \tag{6} $$

My question is how to obtain Eqs. $(5)$ and $(6)$?

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1 Answer 1

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  • Finally, I figure out this question on my own. Put Eq.$(2)$ and $(3)$ into Eq.$(1)$: \begin{equation} -e \Re\{\mathcal{E}_a e^{i\omega t}\} (\partial_a f_0+\partial_a f_1)+(\partial_tf_1+\partial_tf_2)=\dfrac{-f_1-f_2}{\tau} \tag{7} \end{equation} in which $\Re\{\mathcal{E}_a e^{i\omega t}\}f_2\propto |\mathcal{E}|^3$ is droped and $\partial_t f_0=0$. Furthermore, we can investigate Eq$(7)$ by the order of $|\mathcal{E}|$.

  • For $|\mathcal{E}|^1$ term: \begin{equation} -e \mathcal{E}_a e^{i\omega t} \partial_a f_0+\partial_t f_1=-\dfrac{f_1}{\tau} \tag{8} \end{equation} in which $\partial_t f_0$ is real.

  • For $|\mathcal{E}|^2$ term:

\begin{equation} -e \Re\{ \mathcal{E}_a e^{i\omega t} \}\partial_af_1+\partial_t f_2=-\dfrac{f_2}{\tau} \tag{9} \end{equation} in which $\Re\{ \mathcal{E}_a e^{i\omega t} \}$ is real.

  • Eqs.$(8)$ and $(9)$ are linear inhomogenious differential equation of order one, which can be solved easily. \begin{equation} \dfrac{dx(t)}{dt}+p(t)x(t)=q(t) \Rightarrow x(t)=e^{-\int p(t) dt}[c+\int q(t) e^{\int p(t) dt}dt]\tag{10} \end{equation}
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  • $\begingroup$ +10. Great work figuring it out and taking the time to type out an answer in case someone in the future might find it helpful! $\endgroup$ Commented Aug 10, 2021 at 4:41
  • $\begingroup$ @NikeDattani Thanks. $\endgroup$
    – Jack
    Commented Aug 10, 2021 at 5:09

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