{"id":7891,"date":"2024-03-05T14:12:13","date_gmt":"2024-03-05T05:12:13","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?p=7891"},"modified":"2024-03-06T10:23:17","modified_gmt":"2024-03-06T01:23:17","slug":"%e3%82%b1%e3%83%97%e3%83%a9%e3%83%bc%e9%81%8b%e5%8b%95%e3%81%ae%e6%99%82%e9%96%93%e5%b9%b3%e5%9d%87%e3%82%92%e7%9c%9f%e8%bf%91%e7%82%b9%e9%9b%a2%e8%a7%92%e3%81%ae%e3%81%bf%e3%81%ae%e7%a9%8d%e5%88%86","status":"publish","type":"post","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/7891\/","title":{"rendered":"\u30b1\u30d7\u30e9\u30fc\u904b\u52d5\u306e\u6642\u9593\u5e73\u5747\u3092\u771f\u8fd1\u70b9\u96e2\u89d2\u306e\u7a4d\u5206\u3067\u6c42\u3081\u308b"},"content":{"rendered":"

\u6728\u4e0b\u5b99\u8457\u300c\u5929\u4f53\u3068\u8ecc\u9053\u306e\u529b\u5b66\u300d2.6\u7bc0\u3067\u306f\uff0c\u30b1\u30d7\u30e9\u30fc\u904b\u52d5\u3057\u3066\u3044\u308b\u5929\u4f53\u306e\u8ecc\u9053\u91cf\u306e\u6642\u9593\u5e73\u5747\uff08\u3059\u306a\u308f\u3061\u6642\u9593\u7a4d\u5206\uff09\u3092\uff0c\u30b1\u30d7\u30e9\u30fc\u65b9\u7a0b\u5f0f\u3092\u6e80\u305f\u3059\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span> $u$ \uff08\u3084\u5e73\u5747\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span> $l$\uff09 \u306e\u7a4d\u5206\u306b\u7f6e\u304d\u63db\u3048\u3066\u8a08\u7b97\u3057\u3066\u3044\u308b\u3002<\/p>\n

\u3053\u308c\u3092\uff08\u4e00\u822c\u76f8\u5bfe\u8ad6\u7684\u306a\u904b\u52d5\u3078\u306e\u9069\u7528\u3092\u5ff5\u982d\u306b\u304a\u3044\u3066\uff09\u771f\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span> $\\phi$ \u306e\u307f\u306e\u7a4d\u5206\u3067\u3084\u3063\u3066\u307f\u308b\u3002<\/p>\n

<\/p>\n

\u6955\u5186\u8ecc\u9053\u3068\u771f\u8fd1\u70b9\u96e2\u89d2<\/h3>\n

\u6955\u5186\u306e\u5f0f\u3092\u7126\u70b9\u3092\u539f\u70b9\u3068\u3059\u308b\u6975\u5ea7\u6a19 $r, \\phi$ \u3067\u66f8\u304f\u3068\uff0c<\/p>\n

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

\u3053\u3053\u3067 $a$ \u306f\u8ecc\u9053\u9577\u534a\u5f84\uff0c$e$ \u306f\u96e2\u5fc3\u7387\u3002\u89d2\u5ea6\u5ea7\u6a19 $\\phi$ \u306f\u771f\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span>\u3068\u547c\u3070\u308c\u308b\u3002\u771f\u8fd1\u70b9\u96e2\u89d2<\/strong>\u306f\u6975\u5ea7\u6a19\u7cfb\u306e\u89d2\u5ea6\u5ea7\u6a19\u305d\u306e\u3082\u306e\u306e\u3053\u3068\u3002<\/p>\n

\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2\u306e\u5b9a\u7fa9\u3068\u30b1\u30d7\u30e9\u30fc\u65b9\u7a0b\u5f0f<\/h3>\n

\u6955\u5186\u306e\u4e2d\u5fc3\u3092\u539f\u70b9\u3068\u3057\u305f\u30c7\u30ab\u30eb\u30c8\u5ea7\u6a19\u7cfb $X, Y$ \u3067\u306f\uff0c\u6955\u5186\u306e\u5f0f\u306e\u6a19\u6e96\u5f62\u306f<\/p>\n

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

\u3053\u308c\u3092\u8868\u3059\u5a92\u4ecb\u5909\u6570 $u$ \u3092\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u3057\u3066\u5c0e\u5165\u3059\u308b\u3002<\/p>\n

\\begin{eqnarray}
\nX\u00a0 &\\equiv& a \\cos u =ae + r \\cos\\phi\u00a0 \\\\
\nY &\\equiv& a \\sqrt{1-e^2}\u00a0 \\sin u = r \\sin\\phi
\n\\end{eqnarray}<\/p>\n

\u3053\u306e $u$ \u3092\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span>\u3068\u3044\u3046\u3002\u3053\u306e\u96e2\u5fc3\u8fd1\u70b9\u96e2\u89d2<\/strong><\/span>\u3068\u6975\u5ea7\u6a19 $r, \\phi$ \u306e\u95a2\u4fc2\u306f\u4ee5\u4e0b\u306e\u901a\u308a\u3002<\/p>\n

\\begin{eqnarray}
\nr^2 &=& \\left( a \\cos u -a e\\right)^2 + \\left(a \\sqrt{1-e^2}\u00a0 \\sin u \\right)^2 \\\\
\n&=& a^2 \\left( 1 -e \\cos u\\right)^2 \\\\
\n\\therefore\\ \\ r &=& a \\left( 1 -e \\cos u\\right)
\n\\end{eqnarray}<\/p>\n

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

\u534a\u89d2\u8868\u793a\u3092\u597d\u3080\u4eba\u3082\u3044\u308b\u306e\u3067\uff0c<\/p>\n

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

\u534a\u89d2\u306e $\\tan$ \u3092\u5168\u89d2\u306e $\\sin, \\ \\cos$ \u3067\u8868\u3059\u306e\u306f1\u5e74\u751f\u306e\u6388\u696d\u3067\u3082\u3084\u3063\u3066\u307e\u3059\u3002\u4ee5\u4e0b\u306e\u30da\u30fc\u30b8\u3092\u53c2\u7167\uff1a<\/p>\n