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"微观连接要解释什么",
"先选连接律再求演化",
"十二条连接怎样给出连续主部",
"精确色散与长波截断",
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---
## 微观连接要解释什么
最终书稿采用 FCC 十二近邻连接作为空间的首选微观基态模型，并用具体连接律推导连续主部与完整离散色散。第四章小结同时明确：最密堆积不能单独排除其他实现，FCC 的选择与纵横同速仍分别需要动力学和本构依据。
连续模型能在给定恢复系数与动态权重后求传播，但没有因此解释这些参数的来源。微观连接模型要说明空间状态怎样相邻耦合、怎样刷新，哪种长波极限产生连续变量，以及何时出现方向色散或截止。第四、第五章以 FCC 给出具体的微观实现，使邻接关系可以直接计算；空间基元本身不是处于空间中的普通物质原子。
第四章以常规立方晶胞边长 $a_0$ 表示离散尺度，最近邻距离为 $L_0=a_0/\sqrt2$。常规晶胞的四个代表位置属于同一 FCC Bravais 晶格描述，不是四种独立基元。这里 $a_0$、近邻步长 $L_0$、物质核半径与相位周期长度各有定义，不能未经动力学匹配直接等同。
十二近邻、立方对称以及四面体和八面体通道，提供组织密度、剪切和取向响应的可能性。它们并没有单独给出基态选择、正确纵横波速或粒子谱；最密堆积也不能代替相互作用能和候选基态比较。
## 先选连接律再求演化
第五章先取一个已经选定标量或偏振的实分量 $q_{\mathbf R}(t)$，格点位置为 $\mathbf R$。对反向配对、互易且位置无关的连接，离散作用量为
$$
L_d=\sum_{\mathbf R}\left[\frac{Z_d}{2}\dot q_{\mathbf R}^{\,2}-\frac{Z_d}{4}\sum_{n=1}^{12}\Gamma_n(q_{\mathbf R+\mathbf d_n}-q_{\mathbf R})^2+J_{\mathbf R}q_{\mathbf R}\right].
$$
$Z_d>0$ 是时间导数项的归一化系数，$[\Gamma_n]=\mathrm{s^{-2}}$，$J_{\mathbf R}$ 与该分量耦合。若 $q$ 为长度单位，则 $Z_d$ 为质量单位，各格点括号为能量；这一单位选择没有额外证明格点是物质小球。系数 $1/4$ 包括连接在全格点有向求和中被计数两次的修正。
对 $q_{\mathbf R}$ 变分，来自相邻键的项要一起计入。互易条件下得到
$$
\ddot q_{\mathbf R}=\sum_{n=1}^{12}\Gamma_n(q_{\mathbf R+\mathbf d_n}-q_{\mathbf R})+s_{\mathbf R},\qquad s_{\mathbf R}=J_{\mathbf R}/Z_d.
$$
这一式子才是选定连接律的演化。邻接几何与恢复系数在其中分别出现；对完整矢量或内部取向体系，连接还带内部指标，不能把本标量式直接当作全部空间自由度。
## 十二条连接怎样给出连续主部
最近邻矢量与二阶矩为
每个平面有四个独立符号组合。以 $x$ 分量为例，$xy$ 和 $xz$ 组各贡献 $a_0^2$，其余组的 $x$ 分量为零；交叉乘积因符号配对相消。反向键还满足 $\sum_n\mathbf d_n=0$。
在 $ka_0\ll1$，邻点变化可展开为
$$
q(\mathbf r+\mathbf d_n)-q(\mathbf r)=d_{n,i}\partial_iq+\frac12d_{n,i}d_{n,j}\partial_i\partial_jq+\frac16d_{n,i}d_{n,j}d_{n,l}\partial_i\partial_j\partial_lq+\cdots.
$$
反向连接具有相同系数时，奇数阶在求和中消失。于是连续主部的速度平方张量为
$$
C_{ij}=\frac12\sum_n\Gamma_n d_{n,i}d_{n,j},\qquad
\partial_t^2q-C_{ij}\partial_i\partial_jq=s+O(c_q^2a_0^2\nabla^4q).
$$
$[C_{ij}]=\mathrm{m^2s^{-2}}$。等连接 $\Gamma_n=\Gamma_0$ 时，$C_{ij}=\Gamma_0a_0^2\delta_{ij}$，$c_q^2=\Gamma_0a_0^2$。把 $c_q$ 匹配到 $c$ 还需要参数条件，而完整纵横同速还要检查各自的连接响应。
对缓慢变化的互易连接，应从相同连接能得到带权散度形式，见[连续动力学](/theory/space-dynamics/continuum-dynamics)。其粗粒化同时要保留时间与源的归一化，不能只拿空间二阶矩就宣布两个作用量等价。
## 精确色散与长波截断
对等连接无源方程取 $q_{\mathbf R}\propto e^{i(\mathbf k\cdot\mathbf R-\omega t)}$，时间项给 $-\omega^2$，每条连接给 $e^{i\mathbf k\cdot\mathbf d_n}-1$。正反配对消去虚部，故 $\omega^2=\Gamma_0\sum_n[1-\cos(\mathbf k\cdot\mathbf d_n)]$。逐组求和得到
**来源公式核对。** 最终书稿 (5.1.12) 前的半角平方和系数印为 $2c^2/a_0^2$，与 (5.1.11) 不相等：按 $1-\cos x=2\sin^2(x/2)$，对三个 $i0$ 时完整式在离散波数域中非负。取 $c_q^2=\Gamma_0a_0^2$，展开为
$$
\omega^2=c_q^2k^2-c_q^2a_0^2\left[\frac{\sum_i k_i^4}{48}+\frac{\sum_{i