{"id":7916,"date":"2024-03-06T14:02:43","date_gmt":"2024-03-06T05:02:43","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?p=7916"},"modified":"2024-03-19T15:10:54","modified_gmt":"2024-03-19T06:10:54","slug":"%e7%9c%9f%e8%bf%91%e7%82%b9%e9%9b%a2%e8%a7%92%e3%81%a8%e9%9b%a2%e5%bf%83%e8%bf%91%e7%82%b9%e9%9b%a2%e8%a7%92%e3%81%a8%e3%81%ae%e9%96%a2%e4%bf%82%e3%81%ab%e3%81%a4%e3%81%84%e3%81%a6%e3%82%82%e3%81%86","status":"publish","type":"post","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/7916\/","title":{"rendered":"\u771f\u8fd1\u70b9\u96e2\u89d2\u3068\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2\u3068\u306e\u95a2\u4fc2\u306b\u3064\u3044\u3066\u3082\u3046\u5c11\u3057"},"content":{"rendered":"

\u30b1\u30d7\u30e9\u30fc\u904b\u52d5\u306e\u771f\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span> $\\phi$ \u3068\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span> $u$ \u3068\u306e\u95a2\u4fc2\u306b\u3064\u3044\u3066\u306f\uff0cMemo\u300c\u30b1\u30d7\u30e9\u30fc\u904b\u52d5\u306e\u6642\u9593\u5e73\u5747\u3092\u771f\u8fd1\u70b9\u96e2\u89d2\u306e\u7a4d\u5206\u3067\u6c42\u3081\u308b<\/a>\u300d\u306b\u307e\u3068\u3081\u305f\u3002\u6728\u4e0b\u5b99\u8457\u300c\u5929\u4f53\u3068\u8ecc\u9053\u306e\u529b\u5b66\u300d\u306b\u3088\u308c\u3070\uff0c<\/p>\n

\\begin{eqnarray}
\n\\tan \\phi
\n&=& \\frac{\\sqrt{1-e^2} \\sin u}{\\cos u \u2013 e}
\n\\end{eqnarray}<\/p>\n

\u3042\u308b\u3044\u306f\uff0c\u534a\u89d2\u8868\u793a\u3067<\/p>\n

\\begin{eqnarray}
\n\\tan \\frac{\\phi}{2} &=& \\sqrt{\\frac{1+e}{1-e}} \\tan \\frac{u}{2}
\n\\end{eqnarray}<\/p>\n

\u3053\u308c\u3092\u3082\u3046\u5c11\u3057\u5225\u306e\u89d2\u5ea6\u304b\u3089\u898b\u3066\u307f\u3088\u3046\u3068\u3044\u3046\u8a71\u3002<\/p>\n

<\/p>\n

\u52d5\u5f84\u5ea7\u6a19\u306e\u8868\u8a18\u306b\u3088\u308b\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2\u306e\u5b9a\u7fa9<\/h3>\n

\u6955\u5186\u306e\u7126\u70b9\u3092\u539f\u70b9\u3068\u3057\u305f\u6975\u5ea7\u6a19 $r, \\, \\phi$ \u3067\u8868\u3057\u305f\u6955\u5186\u306e\u5f0f\u306f<\/p>\n

$$r(\\phi) = \\frac{a (1-e^2)}{1 + e \\cos\\phi}$$<\/p>\n

\u3053\u306e\u89d2\u5ea6\u5ea7\u6a19 $\\phi$ \u3092\u3042\u3089\u305f\u307e\u3063\u305f\u8a00\u3044\u65b9\u3067\u771f\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span>\u3068\u547c\u3076\u306e\u3067\u3042\u3063\u305f\u3002\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2 $u$ \u306e\u5c0e\u5165\u306f\u3000Memo\u300c\u30b1\u30d7\u30e9\u30fc\u904b\u52d5\u306e\u6642\u9593\u5e73\u5747\u3092\u771f\u8fd1\u70b9\u96e2\u89d2\u306e\u7a4d\u5206\u3067\u6c42\u3081\u308b<\/a>\u300d\u306b\u307e\u3068\u3081\u3066\u3044\u308b\u306e\u3067\u305d\u308c\u3092\u898b\u3066\u3044\u305f\u3060\u304f\u3068\u3057\u3066\uff0c\u3053\u306e $u$ \u3092\u4f7f\u3046\u3068\u52d5\u5f84\u5ea7\u6a19 $r$ \u306f\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u66f8\u3051\u308b\u3053\u3068\u304c\u308f\u304b\u3063\u3066\u3044\u308b\u3002<\/p>\n

$$r = a (1 -e \\cos u)$$<\/p>\n

\u3067\u3042\u308c\u3070\uff0c\u3044\u3063\u305d\u306e\u3053\u3068\uff0c<\/p>\n

$$ \\frac{r}{a} = \\frac{1-e^2}{1 + e \\cos\\phi} \\equiv 1 -e \\cos u$$<\/p>\n

\u3092\u771f\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span> $\\phi$ \u3068\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span> $u$ \u306e\u95a2\u4fc2\u306e\u5b9a\u7fa9\u3067\u3042\u308b\u3068\u3057\u3066\uff0c\u5168\u3066\u306f\u3053\u308c\u304b\u3089\u59cb\u3081\u308b\u3068\u3057\u305f\u3089\u3069\u3046\u304b\uff0c\u3068\u3044\u3046\u8a71\u3002<\/p>\n

\u771f\u8fd1\u70b9\u96e2\u89d2 $\\phi$ \u3068\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2 $u$ \u306e\u95a2\u4fc2<\/h3>\n

\u3053\u306e\u5b9a\u7fa9\u304b\u3089\uff0c\u4ee5\u4e0b\u306e\u3088\u3046\u306b $\\phi$ \u306e\u95a2\u6570\u3092 $u$ \u3092\u4f7f\u3063\u3066\u8868\u3059\u3053\u3068\u304c\u3067\u304d\u308b\u3002<\/p>\n

\\begin{eqnarray}
\n\\frac{1 -e^2}{1 + e \\cos\\phi} &\\equiv& 1 -e \\cos u \\\\
\n\\cos \\phi &=& \\frac{\\cos u -e}{1 -e \\cos u} \\\\
\n\\sin \\phi &=&\\frac{ \\sqrt{1 -e^2}\u00a0 \\,\\sin u}{1 -e \\cos u} \\\\
\nd\\phi &=& \\frac{\\sqrt{1 -e^2}}{1 -e \\cos u} \\,du
\n\\end{eqnarray}<\/p>\n

\u6700\u5f8c\u306e $d\\phi = \\dots$ \u306e\u5f0f\u306f\uff0cMemo\u300c\u30b1\u30d7\u30e9\u30fc\u904b\u52d5\u306e\u6642\u9593\u5e73\u5747\u3092\u771f\u8fd1\u70b9\u96e2\u89d2\u306e\u7a4d\u5206\u3067\u6c42\u3081\u308b<\/a>\u300d\u3067\u6c42\u3081\u305f\uff0c\u6642\u9593\u7a4d\u5206\u3068\u89d2\u5ea6\u7a4d\u5206\u306e\u9593\u306e\u95a2\u4fc2<\/p>\n

$$ \\frac{dt}{T} = \\frac{ r\\, du}{2 \\pi a} = \\frac{r^2 \\, d\\phi}{2 \\pi a^2 \\sqrt{1 -e^2}}$$<\/p>\n

\u304b\u3089\u3082\u76f4\u63a5\u6c42\u3081\u308b\u3053\u3068\u304c\u3067\u304d\u308b\u3002<\/p>\n

\u4e00\u5fdc\uff0c\u8a08\u7b97\u3092\u307e\u3068\u3081\u3066\u304a\u304f\u3068\uff0c$\\cos\\phi$ \u306b\u3064\u3044\u3066\u306f\uff0c<\/p>\n

\\begin{eqnarray}
\n\\frac{1 -e^2}{1 + e \\cos\\phi} &\\equiv& 1 -e \\cos u \\\\
\n1 + e \\cos\\phi &=& \\frac{1 -e^2}{1 -e \\cos u} \\\\
\ne \\cos\\phi &=& \\frac{1 -e^2}{1 -e \\cos u} -1 \\\\
\n&=& \\frac{1 -e^2 – (1 -e \\cos u)}{1 -e \\cos u} \\\\
\n&=& \\frac{e (\\cos u -e)}{1 -e \\cos u} \\\\
\n\\therefore\\ \\ \\cos\\phi &=&\\frac{\\cos u -e}{1 -e \\cos u}
\n\\end{eqnarray}<\/p>\n

$\\cos\\phi$ \u304c\u308f\u304b\u308b\u3068\uff0c$\\sin \\phi$ \u306f<\/p>\n

\\begin{eqnarray}
\n\\sin^2 \\phi &=& 1 – \\cos^2 \\phi \\\\
\n&=& \\frac{(1 -e\\cos u)^2 – (\\cos u -e)^2}{(1 -e\\cos u)^2} \\\\
\n&=& \\frac{1 -2 e \\cos u + e^2 \\cos^2 u – \\cos^2 u + 2 e \\cos u -e^2}{(1 -e\\cos u)^2} \\\\
\n&=& \\frac{(1 -e^2) (1 – \\cos^2 u)}{(1 -e\\cos u)^2} \\\\
\n&=& \\frac{(1 -e^2) \\sin^2 u}{(1 -e\\cos u)^2} \\\\
\n\\therefore\\ \\ \\sin \\phi &=& \\frac{ \\sqrt{1 -e^2}\u00a0 \\,\\sin u}{1 -e \\cos u}
\n\\end{eqnarray}<\/p>\n

\u307e\u305f\uff0c<\/p>\n

\\begin{eqnarray}
\nd\\left(\\frac{1 -e^2}{1 + e \\cos\\phi}\\right) &=& d\\left(1 -e \\cos u\\right) \\\\
\n\\frac{1 -e^2}{(1 + e \\cos\\phi)^2} \\, e \\sin\\phi\\, d\\phi &=& e \\sin u \\, du \\\\
\n\\therefore \\ \\ d \\phi &=& \\frac{(1 + e \\cos\\phi)^2}{1 -e^2} \\, \\frac{\\sin u}{\\sin \\phi} \\, du \\\\
\n&=& (1 -e^2) \\frac{1}{(1 -e\\cos u)^2} \\frac{1 -e \\cos u}{\\sqrt{1 -e^2}} \\, du \\\\
\n&=& \\frac{\\sqrt{1 -e^2}}{1 -e \\cos u} \\,du
\n\\end{eqnarray}<\/p>\n

\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2 $u$ \u3092\u4f7f\u3063\u305f\u7f6e\u63db\u7a4d\u5206\u306e\u4f8b<\/h3>\n

\u306a\u306e\u3067\uff0c\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2\u306e\u5c0e\u5165\u7d4c\u7def\u306f\u4e00\u65e6\u5fd8\u308c\u3066\uff0c\u7f6e\u63db\u7a4d\u5206\u306e\u969b\u306e\u5909\u6570\u5909\u63db\u3068\u3057\u3066 $u$ \u3092\u7528\u3044\u308b\u3068\u3044\u3046\uff0c\u9053\u5177\u3068\u3057\u3066\u306e\u610f\u7fa9\u3060\u3051\u3067\u3082\u3051\u3063\u3053\u3046\u4fbf\u5229\u3067\u3042\u308b\u3002\u4ee5\u4e0b\u306f\u305d\u306e\u4f8b\u3002<\/p>\n

\u306a\u3093\u3067\u3053\u3093\u306a\u7a4d\u5206\u3092\u66f8\u3044\u3066\u304a\u304f\u304b\u3068\u3044\u3046\u3068\uff0c\u5352\u8ad6\u30cd\u30bf\u3068\u3057\u3066\u305d\u306e\u3046\u3061\u306b\u5fc5\u8981\u306b\u306a\u308b\u304b\u3089\u3067\u3059\u3088\uff0c\u305f\u3076\u3093\u3002<\/p>\n

\u4f8b 1.<\/h4>\n

\\begin{eqnarray}
\n\\int_0^{2 \\pi} \\frac{1}{1 + e \\cos \\phi}\\, d \\phi
\n&=& \\frac{1}{1 -e^2} \\int_0^{2 \\pi} (1 -e \\cos u) \\cdot \\frac{\\sqrt{1 -e^2}}{1 -e \\cos u} du \\\\
\n&=& \\frac{2 \\pi}{\\sqrt{1 -e^2}}
\n\\end{eqnarray}<\/p>\n

\u3042\u3063\u3061<\/strong><\/span><\/a>\u3067\u3082\u3053\u306e\u7a4d\u5206\u3092\u8a08\u7b97\u3057\u3066\u3044\u305f\u304c\uff0c\u3084\u305f\u3089\u5927\u5909\u306a\u3053\u3068\u306b\u306a\u3063\u3066\u3044\u305f\u3002<\/p>\n

\u3061\u306a\u307f\u306b\uff0c\u88ab\u7a4d\u5206\u95a2\u6570\u306e\u5206\u6bcd\u306e $e$ \u306e\u524d\u306e\u7b26\u53f7\u3092\u30de\u30a4\u30ca\u30b9\u306b\u5909\u3048\u3066\u3082\u7b54\u3048\u306f\u540c\u3058\u306b\u306a\u308b\u3002<\/p>\n

\\begin{eqnarray}
\n\\int_0^{2 \\pi} \\frac{1}{1 -e \\cos \\phi}\\,d \\phi
\n&=& \\frac{2 \\pi}{\\sqrt{1 -e^2}}
\n\\end{eqnarray}<\/p>\n

\u3053\u306e\u3053\u3068\u306f\uff0c$e \\rightarrow -e $ \u3068\u3057\u3066\u8a08\u7b97\u3092\u8ffd\u3063\u3066\u3044\u3051\u3070\u7406\u89e3\u3067\u304d\u308b\u3060\u308d\u3046\u3002\u307e\u305f\uff0c$\\phi = \\pi -x$\u00a0 \u3068\u3044\u3046\u5909\u6570\u5909\u63db\u3092\u3057\u3066\uff0c\u88ab\u7a4d\u5206\u95a2\u6570\u304c\u5468\u671f $2 \\pi$ \u306e\u5468\u671f\u95a2\u6570\u3067\u3042\u308b\u3053\u3068\u3092\u4f7f\u3048\u3070\uff0c<\/p>\n

\\begin{eqnarray}
\n\\int_0^{2 \\pi} \\frac{1}{1 + e \\cos \\phi}\\,d \\phi
\n&=& \\int_0^{ \\pi} \\frac{d \\phi}{1 + e \\cos \\phi} + \\int_{\\pi}^{2 \\pi} \\frac{d \\phi}{1 + e \\cos \\phi} \\\\
\n&=& \\int^0_{ \\pi} \\frac{-dx}{1 -e \\cos x} + \\int_{0}^{- \\pi} \\frac{-dx}{1 -e \\cos x} \\\\
\n&=& \\int_0^{ \\pi} \\frac{dx}{1 -e \\cos x} + \\int^{0+2\\pi}_{- \\pi+2\\pi} \\frac{dx}{1 -e \\cos x} \\\\
\n&=& \\int_0^{ 2 \\pi} \\frac{dx}{1 -e \\cos x}
\n\\end{eqnarray}<\/p>\n

\u3068\u3057\u3066\u3082\u8a3c\u660e\u3067\u304d\u308b\u3067\u3042\u308d\u3046\u3002\u3053\u306e\u7d50\u679c\u306f\u4ee5\u4e0b\u306e\u4f8b 4.<\/strong> <\/span>\u3067\u4f7f\u3046\u3002<\/p>\n

\u4f8b 2.<\/h4>\n

\\begin{eqnarray}
\n\\int_0^{2 \\pi} \\frac{1}{(1 + e \\cos \\phi)^2}\\,d \\phi
\n&=& \\frac{1}{(1 -e^2)^2} \\int_0^{2 \\pi} (1 -e \\cos u)^2 \\cdot \\frac{\\sqrt{1 -e^2}}{1 -e \\cos u} \\,du \\\\
\n&=& \\frac{\\sqrt{1 -e^2}}{(1 -e^2)^2} \\int_0^{2 \\pi} (1 -e \\cos u)\\, du \\\\
\n&=& \\frac{\\sqrt{1 -e^2}}{(1 -e^2)^2} \\Bigl[ u -e \\sin u \\Bigr]_0^{2 \\pi} \\\\
\n&=& \\frac{2 \\pi \\sqrt{1 -e^2}}{(1 -e^2)^2}
\n\\end{eqnarray}<\/p>\n

\u3053\u306e\u7a4d\u5206\u306f\u6955\u5186\u306e\u9762\u7a4d\u3092\u6c42\u3081\u308b\u969b\u306b\u73fe\u308c\u308b\u3002<\/p>\n