relaxors (e.g., BZT).

Susceptibility of BZT \[sec:lead-free-relaxor\]
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For the static susceptibility of lead-free relaxor BZT[@BZT-susceptibility; @Prosandeev; @dielectric-susceptibility; @Petzelt2014; @Wang2016], we assume (i) dipoles with kinetic energy that overcomes potential well can be treated as free dipoles, subject only to thermal excitation; (ii) dipoles inside the potential well only contribute a constant susceptibility, $\chi_{2}$. The total susceptibility is thus given by $$\begin{aligned}
\chi(T)= & \chi_{1}\mathcal{L}\left(\frac{\theta}{T}\right)P_{1}\left(E_{b},T\right)+\chi_{2}P_{2}\left(E_{b},T\right)\label{eq:BZT-susceptibility}\end{aligned}$$ where $\mathcal{L}\left(x\right)=\coth\left(x\right)-1/x$ is the Langevin function, known for depicting orientational polarization under thermal fluctuations [@Kasap2006; @CurieLaw]. $E_{b}$, $\chi_{1}$, $\chi_{2}$ and $\theta$ are constants, which will be determined by fitting experimental or numerical data. It can be inferred from equation \[eq:BZT-susceptibility\] that $\chi_{1}$ is the susceptibility of the material at zero Kevin, $\chi_{1}\mathcal{L}\left(\frac{\theta}{T}\right)$ is essentially the Curie law at high temperature, and $\theta$ is proportional to the magnitude of the low-frequency electric field used in experimental measurements.

![Fitting the static susceptibility of Ba(Zr$_{0.5}$,Ti$_{0.5}$)O$_{3}$ obtained from Monte-Carlo simulation using Eq. (\[eq:BZT-susceptibility\]). The blue dots are from Monte-Carlo simulation [@Akbarzadeh2012] and the red solid line is the fitting curve using Eq. (\[eq:BZT-susceptibility\]). \[fig:Fitting-susceptibility-of\]](Fig1){width="8cm"}

We first test Eq. (\[eq:BZT-susceptibility\]) against the static susceptibility versus temperature obtained with Monte-Carlo (MC) simulation in a previous work [@Akbarzadeh2012]. \[fig:Fitting-susceptibility-of\] shows the overall 