M03.1 递归Riemann张量的数学涌现

定理M03.1.1 (观察者度规的φ-递归构造)

陈述：观察者坐标系的度量张量从φ-权重结构涌现： $$g_{\mu \nu }^{(obs)}=\sum _{k=0}^{n_{obs}}{\eta }^{(\varphi )}(k;m_{obs}){\varphi }^{-k}{g}_{\mu \nu }^{(k)}$$

其中$g_{\mu \nu }^{(k)}$是第$k$层的基础度量分量。

证明： 步骤1：基础度量的递归分层定义 第$k$层度量： $$g_{\mu \nu }^{(k)}=\{\begin{array}{ll}\text{diag}(-1,+1,+1,+1)& k=0\text{（基础Minkowski）}\\ \delta g_{\mu \nu }^{(k)}& k\ge 1\text{（递归修正）}\end{array}$$

步骤2：φ-权重的度量调制机制 $$g_{00}^{(obs)}=-\sum _{k=0}^{n_{obs}}{\eta }^{(\varphi )}(k;m_{obs}){\varphi }^{-k}=-\sum _{k=0}^{n_{obs}}{F}_k{\varphi }^{-k}$$

主导项：$k=0,1,2$贡献 $$g_{00}^{(obs)}\approx -(F_0+F_1{\varphi }^{-1}+F_2{\varphi }^{-2})=-(0+1\cdot {\varphi }^{-1}+1\cdot {\varphi }^{-2})$$ $$=-({\varphi }^{-1}+{\varphi }^{-2})=-{\varphi }^{-2}(\varphi +1)=-{\varphi }^{-2}{\varphi }^2=-1$$

步骤3：空间度量分量的类似计算 $$g_{ii}^{(obs)}=+\sum _{k=0}^{n_{obs}}{F}_k{\varphi }^{-k}\approx +1$$

步骤4：Minkowski度量的数学涌现 $$g_{\mu \nu }^{(obs)}=\text{diag}(-1,+1,+1,+1)+O({\varphi }^{-n_{obs}})$$

φ-递归结构自然涌现标准时空度量。$\square$

定理M03.1.2 (Christoffel符号的递归几何涌现)

陈述：Christoffel符号从度量的φ-递归结构涌现： $${\Gamma }_{\mu \nu }^{(\varphi )\lambda }=\sum _{k=0}^{n_{obs}}{\eta }^{(\varphi )}(k;m_{obs}){\varphi }^{-k}{\Gamma }_{\mu \nu }^{(k)\lambda }$$

证明： 步骤1：第9章递归微分几何的Christoffel计算 $${\Gamma }_{\mu \nu }^{(\varphi )\lambda }=\frac{1}{2}{g}^{(\varphi )\lambda \rho }\left(\frac{\partial g_{\mu \rho }^{(\varphi )}}{\partial x^{\nu }}+\frac{\partial g_{\nu \rho }^{(\varphi )}}{\partial x^{\mu }}-\frac{\partial g_{\mu \nu }^{(\varphi )}}{\partial x^{\rho }}\right)$$

步骤2：度量导数的φ-递归分解 $$\frac{\partial g_{\mu \nu }^{(\varphi )}}{\partial x^{\rho }}=\sum _{k=0}^{n_{obs}}{\eta }^{(\varphi )}(k;m_{obs}){\varphi }^{-k}\frac{\partial g_{\mu \nu }^{(k)}}{\partial x^{\rho }}$$

步骤3：逆度量的φ-递归展开 $$g^{(\varphi )\lambda \rho }=\sum _{k=0}^{n_{obs}}{\eta }^{(\varphi )}(k;m_{obs}){\varphi }^{-k}{g}^{(k)\lambda \rho }$$

步骤4：Christoffel符号的层级贡献 $${\Gamma }_{\mu \nu }^{(\varphi )\lambda }=\sum _{k=0}^{n_{obs}}{\varphi }^{-k}{\Gamma }_{\mu \nu }^{(k)\lambda }+\text{cross-terms}$$

交叉项来源于不同层级的耦合。$\square$

定理M03.1.3 (Riemann曲率张量的递归分解)

陈述：Riemann曲率张量具有φ-递归的层级分解： $$R_{\mu \nu \rho \sigma }^{(\varphi )}=\sum _{k=0}^{n_{obs}}{\varphi }^{-k}{R}_{\mu \nu \rho \sigma }^{(k)}+\sum _{i<j}{\varphi }^{-(i+j)}{R}_{\mu \nu \rho \sigma }^{(i,j)}$$

其中$R^{(k)}$是第$k$层贡献，$R^{(i,j)}$是层间相互作用项。

证明： 步骤1：Riemann张量的定义在φ-递归框架 $$R_{\mu \nu \rho \sigma }^{(\varphi )}=g_{\sigma \lambda }^{(\varphi )}{R}_{\mu \nu \rho }^{(\varphi )\lambda }$$

其中 $$R_{\mu \nu \rho }^{(\varphi )\lambda }=\frac{\partial {\Gamma }_{\mu \rho }^{(\varphi )\lambda }}{\partial x^{\nu }}-\frac{\partial {\Gamma }_{\mu \nu }^{(\varphi )\lambda }}{\partial x^{\rho }}+{\Gamma }_{\mu \rho }^{(\varphi )\sigma }{\Gamma }_{\sigma \nu }^{(\varphi )\lambda }-{\Gamma }_{\mu \nu }^{(\varphi )\sigma }{\Gamma }_{\sigma \rho }^{(\varphi )\lambda }$$

步骤2：导数项的φ-递归分解 $$\frac{\partial {\Gamma }_{\mu \rho }^{(\varphi )\lambda }}{\partial x^{\nu }}=\sum _{k=0}^{n_{obs}}{\varphi }^{-k}\frac{\partial {\Gamma }_{\mu \rho }^{(k)\lambda }}{\partial x^{\nu }}$$

步骤3：二次项的交叉贡献分析 $${\Gamma }_{\mu \rho }^{(\varphi )\sigma }{\Gamma }_{\sigma \nu }^{(\varphi )\lambda }=\sum _{i,j=0}^{n_{obs}}{\varphi }^{-(i+j)}{\Gamma }_{\mu \rho }^{(i)\sigma }{\Gamma }_{\sigma \nu }^{(j)\lambda }$$

步骤4：曲率分解的收敛性验证 主导项：$k=0$（平坦时空） 修正项：$k\ge 1$（φ-权重${\varphi }^{-k}$指数衰减）

总曲率：$R_{\mu \nu \rho \sigma }^{(\varphi )}=R_{\mu \nu \rho \sigma }^{(flat)}+{\sum }_{k=1}^{n_{obs}}{\varphi }^{-k}{R}_{\mu \nu \rho \sigma }^{(k)}+O({\varphi }^{-n_{obs}})$

收敛性通过φ-权重保证。$\square$

定理M03.1.4 (Einstein张量的递归涌现)

陈述：Einstein张量从Riemann曲率的递归结构自然涌现： $$G_{\mu \nu }^{(\varphi )}=R_{\mu \nu }^{(\varphi )}-\frac{1}{2}{R}^{(\varphi )}{g}_{\mu \nu }^{(\varphi )}=\sum _{k=0}^{n_{obs}}{\varphi }^{-k}{G}_{\mu \nu }^{(k)}$$

证明： 步骤1：Ricci张量的φ-递归缩并 $$R_{\mu \nu }^{(\varphi )}=g^{(\varphi )\rho \sigma }{R}_{\mu \rho \nu \sigma }^{(\varphi )}=\sum _{k=0}^{n_{obs}}{\varphi }^{-k}{R}_{\mu \nu }^{(k)}$$

步骤2：标量曲率的递归计算 $$R^{(\varphi )}=g^{(\varphi )\mu \nu }{R}_{\mu \nu }^{(\varphi )}=\sum _{k=0}^{n_{obs}}{\varphi }^{-k}{R}^{(k)}$$

步骤3：Einstein张量的层级分解 $$G_{\mu \nu }^{(k)}=R_{\mu \nu }^{(k)}-\frac{1}{2}{R}^{(k)}{g}_{\mu \nu }^{(k)}$$

步骤4：Einstein张量的递归性质验证 $${\nabla }_{\mu }^{(\varphi )}{G}^{(\varphi )\mu \nu }=\sum _{k=0}^{n_{obs}}{\varphi }^{-k}{\nabla }_{\mu }^{(k)}{G}^{(k)\mu \nu }=0$$

Bianchi恒等式在每个递归层都成立，保证Einstein张量的几何一致性。$\square$