\u671d\u6c38\u541b\u3078\u306e\u79c1\u4fe1\u3002
\n
\n\u4e00\u822c\u306b\uff0c\u7121\u9650\u5c0f\u5ea7\u6a19\u5909\u63db<\/p>\n
$$\\bar{x}^{\\mu} = x^{\\mu} + \\xi^{\\mu}$$<\/p>\n
\u306b\u3088\u308b\u30e1\u30c8\u30ea\u30c3\u30af\u306e\u5909\u63db\u306f<\/p>\n
\\begin{eqnarray}
\n\\bar{g}_{\\alpha\\beta}(\\bar{x}) &=& {g}_{\\mu\\nu} ({x}) \\frac{\\partial {x}^{\\mu}}{\\partial \\bar{x}^{\\alpha}}\\frac{\\partial {x}^{\\nu}}{\\partial \\bar{x}^{\\beta}} \\\\
\n&=& {g}_{\\mu\\nu}(x)
\n\\left( \\delta^{\\mu}_{\\ \\ \\alpha} – \\xi^{\\mu}_{\\ \\ ,\\alpha}\\right)
\n\\left( \\delta^{\\nu}_{\\ \\ \\beta} – \\xi^{\\nu}_{\\ \\ ,\\beta}\\right) \\\\
\n&=& {g}_{\\alpha\\beta}(x) – g_{\\alpha\\nu} \\xi^{\\nu}_{\\ \\ ,\\beta}
\n– g_{\\mu\\beta} \\xi^{\\mu}_{\\ \\ ,\\alpha}
\n\\end{eqnarray}<\/p>\n
\u3053\u308c\u306b\u4f34\u3046\u30b2\u30fc\u30b8\u5909\u63db\u306f\uff0c\u5de6\u8fba\u3092<\/p>\n
$$\\bar{g}_{\\alpha\\beta}(\\bar{x}) = \\bar{g}_{\\alpha\\beta}(x) + g_{\\alpha\\beta, \\lambda}\\xi^{\\lambda}$$<\/p>\n
\u3068\u3057\u3066<\/p>\n
$$\\bar{g}_{\\alpha\\beta}(x) = {g}_{\\alpha\\beta}(x) – g_{\\alpha\\nu} \\xi^{\\nu}_{\\ \\ ,\\beta} – g_{\\mu\\beta} \\xi^{\\mu}_{\\ \\ ,\\alpha}\u00a0-g_{\\alpha\\beta, \\lambda}\\xi^{\\lambda}$$<\/p>\n
$$ \\xi^0 = \\alpha, \\quad \\xi^i\u00a0 = {\\color{red}{\\delta^{ij} \\beta_{, j}}}$$<\/p>\n
\u3068\u3057\u3066<\/p>\n
\\begin{eqnarray}
\n\\bar{g}_{0i} = –\u00a0 \\bar{B}_{,i} &=& g_{0i} – g_{00} \\xi^0_{\\ \\ ,i} – g_{ji} \\xi^{j}_{\\ \\ , 0} – g_{0i, \\lambda}\\xi^{\\lambda} \\\\
\n&=& – B_{, i} + \\alpha_{, i} – a^2 \\delta_{ji} \\delta^{jk} \\dot{\\beta}_{, k} \\\\
\n\\therefore\\ \\ \\bar{B}_{,i} &=& B_{, i}\\, -\\, \\alpha_{, i} + a^2\\dot{\\beta}_{, i}\u00a0 \\\\
\n\\therefore\\ \\ \\bar{B} &=& B\\, -\\, \\alpha + a^2 \\dot{\\beta}
\n\\end{eqnarray}<\/p>\n
\u307b\u304b\u3082\u307e\u3068\u3081\u308b\u3068\uff0c<\/p>\n
\\begin{eqnarray}
\n\\bar{A} &=& A – \\dot{\\alpha} \\\\
\n\\bar{B} &=& B\\, -\\, \\alpha + a^2 \\dot{\\beta}\\\\
\n\\bar{E} &=& E – \\beta \\\\
\n\\bar{F} &=& F – \\frac{\\dot{a}}{a} \\alpha \\\\
\n\\bar{\\delta} &=& \\delta + 3\\frac{\\dot{a}}{a} \\alpha \\\\
\n\\bar{v} &=& v + a^2 \\dot{\\beta}
\n\\end{eqnarray}<\/p>\n
\u3053\u308c\u3088\u308a<\/p>\n
\\begin{eqnarray}
\n\\bar{B} + a^2 \\dot{\\bar{E}} &=& B + a^2 \\dot{E} – \\alpha \\\\
\n\\bar{v} – \\bar{B} &=& v – B + \\alpha
\n\\end{eqnarray}<\/p>\n
\u3067\u3042\u308b\u304b\u3089<\/p>\n
\\begin{eqnarray}
\n\\varPhi &\\equiv& A – \\left(B + a^2 \\dot{E} \\right)^{\\dot{}}\\\\
\n\\varPsi &\\equiv& F – \\frac{\\dot{a}}{a} \\left(B + a^2 \\dot{E} \\right) \\\\
\n\\varDelta &\\equiv& \\delta – 3\\frac{\\dot{a}}{a}\\left(v – B\\right) \\\\
\nV &\\equiv& v + a^2 \\dot{E}
\n\\end{eqnarray}<\/p>\n","protected":false},"excerpt":{"rendered":"
\u671d\u6c38\u541b\u3078\u306e\u79c1\u4fe1\u3002<\/p>