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删除172字节 、 2025年3月17日 (一) 14:49
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:<math> \frac{d \theta_i}{d t} = \omega_i - K r \sin(\theta_i) </math>。
 
:<math> \frac{d \theta_i}{d t} = \omega_i - K r \sin(\theta_i) </math>。
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== Large ''N'' limit ==
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== ''N''很大时的极限情况 ==
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Now consider the case as ''N'' tends to infinity. Take the distribution of intrinsic natural frequencies as ''g''(''ω'') (assumed [[normalization constant|normalized]]).  Then assume that the density of oscillators at a given phase ''θ'', with given natural frequency ''ω'', at time ''t'' is <math>\rho(\theta, \omega, t)</math>.  Normalization requires that
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当振荡器数量趋于无穷大时,考虑此情况。假设固有自然频率的分布为''g''(''ω''(已归一化)。然后假设在给定相位''θ''和给定自然频率''ω''时,振荡器的密度在时间''t''为<math>\rho(\theta, \omega, t)</math>。归一化要求
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:<math> \int_{-\pi}^{\pi} \rho(\theta, \omega, t) \, d \theta = 1. </math>
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:<math> \int_{-\pi}^{\pi} \rho(\theta, \omega, t) , d \theta = 1. </math>
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The [[continuity equation]] for oscillator density will be
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振荡器密度的[[连续性方程]]为
 
:<math> \frac{\partial \rho}{\partial t} + \frac{\partial}{\partial \theta}[\rho v] = 0, </math>
 
:<math> \frac{\partial \rho}{\partial t} + \frac{\partial}{\partial \theta}[\rho v] = 0, </math>
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where ''v'' is the drift velocity of the oscillators given by taking the infinite-''N'' limit in the transformed governing equation, such that
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其中''v''是振荡器的漂移速度,通过在变换后的控制方程中取无穷-''N''极限得到,使得
 
:<math> \frac{\partial \rho}{\partial t} + \frac{\partial}{\partial \theta}[\rho \omega + \rho K r \sin(\psi-\theta)] = 0. </math>
 
:<math> \frac{\partial \rho}{\partial t} + \frac{\partial}{\partial \theta}[\rho \omega + \rho K r \sin(\psi-\theta)] = 0. </math>
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Finally, the definition of the order parameters must be rewritten for the continuum (infinite ''N'') limit. <math>\theta_i</math> must be replaced by its ensemble average (over all <math>\omega</math>) and the sum must be replaced by an integral, to give
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最后,必须将有序参数的定义改写为连续(无穷''N'')极限。<math>\theta_i</math>必须替换为其在所有<math>\omega</math>上的合集平均值,并用积分替换求和,得到
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:<math>  
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:<math>
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r e^{i \psi} = \int_{-\pi}^{\pi} e^{i \theta} \int_{-\infty}^{\infty} \rho(\theta, \omega, t) g(\omega) \, d \omega \, d \theta.
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r e^{i \psi} = \int_{-\pi}^{\pi} e^{i \theta} \int_{-\infty}^{\infty} \rho(\theta, \omega, t) g(\omega) , d \omega , d \theta.
 
</math>
 
</math>
  

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