菲兹克斯喵

菲兹克斯喵

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Lesson 10 多极展开

2025年10月21日 15:52

对于 Coulomb 特解做展开：

φ (x) = \int_{C} ρ (x^{'...剩余内容已隐藏 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x'}\\left(\\frac{1}{r}\\right)\\right|_{\\vec{x}'=0}+y'\\left.\\frac{\\partial}{\\partial y'}\\left(\\frac{1}{r}\\right)\\right|_{\\vec{x}'=0}+z'\\left.\\frac{\\partial}{\\partial z'}\\left(\\frac{1}{r}\\right)\\right|_{\\vec{z}'=0}\\right]+\\cdots\\\\\\\\\n\u0026\\overset{r_{\\vec{x}'=0}=R}{=}\\frac{1}{R}+(\\vec{x}'\\cdot\\nabla)\\frac{1}{R}+\\frac{1}{2}\\vec{x}'\\vec{x}':\\nabla\\nabla\\frac{1}{R}+\\cdots\n\\end{aligned}\nr1​​=r1​\n​x\n'=0​+[x'\partialx'\partial​(r1​)\n​x\n'=0​+y'\partialy'\partial​(r1​)\n​x\n'=0​+z'\partialz'\partial​(r1​)\n​z\n'=0​]+\dots=rx\n'=0​=RR1​+(x\n'\cdot\nabla)R1​+21​x\n'x\n':\nabla\nablaR1​+\dots​\n偶极项：\nφ=14πεp⃗\cdotx⃗R3\\varphi = \\frac{1}{4\\pi\\varepsilon}\\frac{\\vec{p}\\cdot\\vec{x}}{R^3}\nφ=4πε1​R3p\n​\cdotx\n​\n四极项：\nφ=q4πε(x⃗0'x⃗1'+x⃗1'x⃗0')(3x⃗x⃗-R2I)2R5\\varphi=\\frac{q}{4\\pi\\varepsilon}\\frac{(\\vec{x}_0'\\vec{x}_1'+\\vec{x}_1'\\vec{x}_0')(3\\vec{x}\\vec{x}-R^2\\bold{I})}{2R^5}\nφ=4πεq​2R5(x\n0'​x\n1'​+x\n1'​x\n0'​)(3x\nx\n-R2I)​\n这是双偶极子共线对顶的这种情况的领头阶，实际上也可以用四电荷方框的电荷分布来计算领头阶.\n如果定义四极矩 D=\summ,nqnx⃗nx⃗mD = \\sum_{m,n}q_{n}\\vec{x}_n\\vec{x}_mD=\summ,n​qn​x\nn​x\nm​，可以改写四极项.\n对于连续电荷分布，有\nφ=14πε\intV'ρ(x⃗')[1R-x⃗'\cdot\nabla1R+12!x⃗x⃗:\nabla\nabla1R+\dots ]dτ'\\varphi=\\frac{1}{4\\pi\\varepsilon}\\int_{V'}\\rho(\\vec{x}')\\left[\\frac{1}{R}-\\vec{x}'\\cdot\\nabla\\frac{1}{R}+\\frac{1}{2!}\\vec{x}\\vec{x}:\\nabla\\nabla\\frac{1}{R}+\\cdots\\right]\\text{d}\\tau'\nφ=4πε1​\intV'​ρ(x\n')[R1​-x\n'\cdot\nablaR1​+2!1​x\nx\n:\nabla\nablaR1​+\dots]dτ'\n令：\nQ=\intV'ρ(x⃗')dτ' ,p⃗=\intV'ρ(x⃗')x⃗'dτ' ,D=\intV'3x⃗'x⃗'ρ(x⃗')dτ'Q=\\int_{V'}\\rho(\\vec{x}')\\text{d}\\tau'\\,,\\quad\\vec{p}=\\int_{V'}\\rho(\\vec{x}')\\vec{x}'\\text{d}\\tau'\\,,\\quad\\bold{D}=\\int_{V'}3\\vec{x}'\\vec{x}'\\rho(\\vec{x}')\\text{d}\\tau'\nQ=\intV'​ρ(x\n')dτ',p\n​=\intV'​ρ(x\n')x\n'dτ',D=\intV'​3x\n'x\n'ρ(x\n')dτ'\n上式改写为\nφ=14πε\int[QR-p⃗\cdot\nabla1R+16D:\nabla\nabla1R+\dots ]dτ'\\varphi=\\frac{1}{4\\pi\\varepsilon}\\int\\left[\\frac{Q}{R}-\\vec{p}\\cdot\\nabla\\frac{1}{R}+\\frac{1}{6}\\bold{D}:\\nabla\\nabla\\frac{1}{R}+\\cdots\\right]\\text{d}\\tau'\nφ=4πε1​\int[RQ​-p\n​\cdot\nablaR1​+61​D:\nabla\nablaR1​+\dots]dτ'\n一个小区域内连续分布的电荷在远处激发的电场等于一系列多极子在远处激发的场的叠加."])self.__next_f.push([1,"18:Tc70,"])self.__next_f.push([1,"订阅更新(RSS/Atom): 对于 Coulomb 特解做展开：\nφ(x)=\intCρ(x')4πε∣x'-x∣d3x'\\varphi(x)=\\int_C\\frac{\\rho(x')}{4\\pi\\varepsilon|x'-x|}\\text{d}^3x'\nφ(x)=\intC​4πε∣x'-x∣ρ(x')​d3x'\n展开考虑的是 ∣x'∣/∣x∣|x'|/|x|∣x'∣/∣x∣ 为小量. 令 r'=∣x⃗-x⃗'∣r'=|\\vec{x}-\\vec{x}'|r'=∣x\n-x\n'∣，固定 xxx (场点) 不变，把 x'x'x' 在原点附近展开，则\n1r=1r∣x⃗'=0+[x'\partial\partialx'(1r)∣x⃗'=0+y'\partial\partialy'(1r)∣x⃗'=0+z'\partial\partialz'(1r)∣z⃗'=0]+\dots=rx⃗'=0=R1R+(x⃗'\cdot\nabla)1R+12x⃗'x⃗':\nabla\nabla1R+\dots\\begin{aligned}\n\\frac{1}{r}\u0026=\\left.\\frac{1}{r}\\right|_{\\vec{x}'=0}+\\left[x'\\left.\\frac{\\partial}{\\partial x'}\\left(\\frac{1}{r}\\right)\\right|_{\\vec{x}'=0}+y'\\left.\\frac{\\partial}{\\partial y'}\\left(\\frac{1}{r}\\right)\\right|_{\\vec{x}'=0}+z'\\left.\\frac{\\partial}{\\partial z'}\\left(\\frac{1}{r}\\right)\\right|_{\\vec{z}'=0}\\right]+\\cdots\\\\\\\\\n\u0026\\overset{r_{\\vec{x}'=0}=R}{=}\\frac{1}{R}+(\\vec{x}'\\cdot\\nabla)\\frac{1}{R}+\\frac{1}{2}\\vec{x}'\\vec{x}':\\nabla\\nabla\\frac{1}{R}+\\cdots\n\\end{aligned}\nr1​​=r1​\n​x\n'=0​+[x'\partialx'\partial​(r1​)\n​x\n'=0​+y'\partialy'\partial​(r1​)\n​x\n'=0​+z'\partialz'\partial​(r1​)\n​z\n'=0​]+\dots=rx\n'=0​=RR1​+(x\n'\cdot\nabla)R1​+21​x\n'x\n':\nabla\nablaR1​+\dots​\n偶极项：\nφ=14πεp⃗\cdotx⃗R3\\varphi = \\frac{1}{4\\pi\\varepsilon}\\frac{\\vec{p}\\cdot\\vec{x}}{R^3}\nφ=4πε1​R3p\n​\cdotx\n​\n四极项：\nφ=q4πε(x⃗0'x⃗1'+x⃗1'x⃗0')(3x⃗x⃗-R2I)2R5\\varphi=\\frac{q}{4\\pi\\varepsilon}\\frac{(\\vec{x}_0'\\vec{x}_1'+\\vec{x}_1'\\vec{x}_0')(3\\vec{x}\\vec{x}-R^2\\bold{I})}{2R^5}\nφ=4πεq​2R5(x\n0'​x\n1'​+x\n1'​x\n0'​)(3x\nx\n-R2I)​\n这是双偶极子共线对顶的这种情况的领头阶，实际上也可以用四电荷方框的电荷分布来计算领头阶.\n如果定义四极矩 D=\summ,nqnx⃗nx⃗mD = \\sum_{m,n}q_{n}\\vec{x}_n\\vec{x}_mD=\summ,n​qn​x\nn​x\nm​，可以改写四极项.\n对于连续电荷分布，有\nφ=14πε\intV'ρ(x⃗')[1R-x⃗'\cdot\nabla1R+12!x⃗x⃗:\nabla\nabla1R+\dots ]dτ'\\varphi=\\frac{1}{4\\pi\\varepsilon}\\int_{V'}\\rho(\\vec{x}')\\left[\\frac{1}{R}-\\vec{x}'\\cdot\\nabla\\frac{1}{R}+\\frac{1}{2!}\\vec{x}\\vec{x}:\\nabla\\nabla\\frac{1}{R}+\\cdots\\right]\\text{d}\\tau'\nφ=4πε1​\intV'​ρ(x\n')[R1​-x\n'\cdot\nablaR1​+2!1​x\nx\n:\nabla\nablaR1​+\dots]dτ'\n令：\nQ=\intV'ρ(x⃗')dτ' ,p⃗=\intV'ρ(x⃗')x⃗'dτ' ,D=\intV'3x⃗'x⃗'ρ(x⃗')dτ'Q=\\int_{V'}\\rho(\\vec{x}')\\text{d}\\tau'\\,,\\quad\\vec{p}=\\int_{V'}\\rho(\\vec{x}')\\vec{x}'\\text{d}\\tau'\\,,\\quad\\bold{D}=\\int_{V'}3\\vec{x}'\\vec{x}'\\rho(\\vec{x}')\\text{d}\\tau'\nQ=\intV'​ρ(x\n')dτ',p\n​=\intV'​ρ(x\n')x\n'dτ',D=\intV'​3x\n'x\n'ρ(x\n')dτ'\n上式改写为\nφ=14πε\int[QR-p⃗\cdot\nabla1R+16D:\nabla\nabla1R+\dots ]dτ'\\varphi=\\frac{1}{4\\pi\\varepsilon}\\int\\left[\\frac{Q}{R}-\\vec{p}\\cdot\\nabla\\frac{1}{R}+\\frac{1}{6}\\bold{D}:\\nabla\\nabla\\frac{1}{R}+\\cdots\\right]\\text{d}\\tau'\nφ=4πε1​\int[RQ​-p\n​\cdot\nablaR1​+61​D:\nabla\nablaR1​+\dots]dτ'\n一个小区域内连续分布的电荷在远处激发的电场等于一系列多极子在远处激发的场的叠加."])self.__next_f.push([1,"9:{\"metadata\":[[\"$\",\"title\",\"0\",{\"children\":\"Lesson 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特解做展开：\nφ(x)=\intCρ(x')4πε∣x'-x∣d3x'\\varphi(x)=\\int_C\\frac{\\rho(x')}{4\\pi\\varepsilon|x'-x|}\\text{d}^3x'\nφ(x)=\intC​4πε∣x'-x∣ρ(x')​d3x'\n展开考虑的是 ∣x'∣/∣x∣|x'|/|x|∣x'∣/∣x∣ 为小量. 令 r'=∣x⃗-x⃗'∣r'=|\\vec{x}-\\vec{x}'|r'=∣x\n-x\n'∣，固定 xxx (场点) 不变，把 x'x'x' 在原点附近展开，则\n1r=1r∣x⃗'=0+[x'\partial\partialx'(1r)∣x⃗'=0+y'\partial\partialy'(1r)∣x⃗'=0+z'\partial\partialz'(1r)∣z⃗'=0]+\dots=rx⃗'=0=R1R+(x⃗'\cdot\nabla)1R+12x⃗'x⃗':\nabla\nabla1R+\dots\\begin{aligned}\n\\frac{1}{r}\u0026=\\left.\\frac{1}{r}\\right|_{\\vec{x}'=0}+\\left[x'\\left.\\frac{\\partial}{\\partial x'}\\left(\\frac{1}{r}\\right)\\right|_{\\vec{x}'=0}+y'\\left.\\frac{\\partial}{\\partial y'}\\left(\\frac{1}{r}\\right)\\right|_{\\vec{x}'=0}+z'\\left.\\frac{\\partial}{\\partial z'}\\left(\\frac{1}{r}\\right)\\right|_{\\vec{z}'=0}\\right]+\\cdots\\\\\\\\\n\u0026\\overset{r_{\\vec{x}'=0}=R}{=}\\frac{1}{R}+(\\vec{x}'\\cdot\\nabla)\\frac{1}{R}+\\frac{1}{2}\\vec{x}'\\vec{x}':\\nabla\\nabla\\frac{1}{R}+\\cdots\n\\end{aligned}\nr1​​=r1​\n​x\n'=0​+[x'\partialx'\partial​(r1​)\n​x\n'=0​+y'\partialy'\partial​(r1​)\n​x\n'=0​+z'\partialz'\partial​(r1​)\n​z\n'=0​]+\dots=rx\n'=0​=RR1​+(x\n'\cdot\nabla)R1​+21​x\n'x\n':\nabla\nablaR1​+\dots​\n偶极项：\nφ=14πεp⃗\cdotx⃗R3\\varphi = \\frac{1}{4\\pi\\varepsilon}\\frac{\\vec{p}\\cdot\\vec{x}}{R^3}\nφ=4πε1​R3p\n​\cdotx\n​\n四极项：\nφ=q4πε(x⃗0'x⃗1'+x⃗1'x⃗0')(3x⃗x⃗-R2I)2R5\\varphi=\\frac{q}{4\\pi\\varepsilon}\\frac{(\\vec{x}_0'\\vec{x}_1'+\\vec{x}_1'\\vec{x}_0')(3\\vec{x}\\vec{x}-R^2\\bold{I})}{2R^5}\nφ=4πεq​2R5(x\n0'​x\n1'​+x\n1'​x\n0'​)(3x\nx\n-R2I)​\n这是双偶极子共线对顶的这种情况的领头阶，实际上也可以用四电荷方框的电荷分布来计算领头阶.\n如果定义四极矩 D=\summ,nqnx⃗nx⃗mD = \\sum_{m,n}q_{n}\\vec{x}_n\\vec{x}_mD=\summ,n​qn​x\nn​x\nm​，可以改写四极项.\n对于连续电荷分布，有\nφ=14πε\intV'ρ(x⃗')[1R-x⃗'\cdot\nabla1R+12!x⃗x⃗:\nabla\nabla1R+\dots ]dτ'\\varphi=\\frac{1}{4\\pi\\varepsilon}\\int_{V'}\\rho(\\vec{x}')\\left[\\frac{1}{R}-\\vec{x}'\\cdot\\nabla\\frac{1}{R}+\\frac{1}{2!}\\vec{x}\\vec{x}:\\nabla\\nabla\\frac{1}{R}+\\cdots\\right]\\text{d}\\tau'\nφ=4πε1​\intV'​ρ(x\n')[R1​-x\n'\cdot\nablaR1​+2!1​x\nx\n:\nabla\nablaR1​+\dots]dτ'\n令：\nQ=\intV'ρ(x⃗')dτ' ,p⃗=\intV'ρ(x⃗')x⃗'dτ' ,D=\intV'3x⃗'x⃗'ρ(x⃗')dτ'Q=\\int_{V'}\\rho(\\vec{x}')\\text{d}\\tau'\\,,\\quad\\vec{p}=\\int_{V'}\\rho(\\vec{x}')\\vec{x}'\\text{d}\\tau'\\,,\\quad\\bold{D}=\\int_{V'}3\\vec{x}'\\vec{x}'\\rho(\\vec{x}')\\text{d}\\tau'\nQ=\intV'​ρ(x\n')dτ',p\n​=\intV'​ρ(x\n')x\n'dτ',D=\intV'​3x\n'x\n'ρ(x\n')dτ'\n上式改写为\nφ=14πε\int[QR-p⃗\cdot\nabla1R+16D:\nabla\nabla1R+\dots ]dτ'\\varphi=\\frac{1}{4\\pi\\varepsilon}\\int\\left[\\frac{Q}{R}-\\vec{p}\\cdot\\nabla\\frac{1}{R}+\\frac{1}{6}\\bold{D}:\\nabla\\nabla\\frac{1}{R}+\\cdots\\right]\\text{d}\\tau'\nφ=4πε1​\int[RQ​-p\n​\cdot\nablaR1​+61​D:\nabla\nablaR1​+\dots]dτ'\n一个小区域内连续分布的电荷在远处激发的电场等于一系列多极子在远处激发的场的叠加."])self.__next_f.push([1,"1e:[\"$\",\"meta\",\"11\",{\"name\":\"twitter:description\",\"content\":\"$21\"}]\n1f:[\"$\",\"link\",\"12\",{\"rel\":\"icon\",\"href\":\"/favicon.ico?favicon.0b3bf435.ico\",\"sizes\":\"256x256\",\"type\":\"image/x-icon\"}]\n20:[\"$\",\"$L22\",\"13\",{}]\n"])}