{"id":4828,"date":"2023-01-06T11:41:54","date_gmt":"2023-01-06T02:41:54","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?page_id=4828"},"modified":"2025-01-20T14:34:08","modified_gmt":"2025-01-20T05:34:08","slug":"%e5%8f%82%e8%80%83%ef%bc%9a%e9%9d%99%e7%a3%81%e5%a0%b4%e3%82%92%e6%b1%82%e3%82%81%e3%82%8b%e9%9a%9b%e3%81%ab%e4%bd%bf%e3%81%a3%e3%81%9f%e7%a9%8d%e5%88%86%e3%82%92%e4%ba%ba%e5%8a%9b%e3%81%a7%e7%a2%ba","status":"publish","type":"page","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/%e9%9b%bb%e7%a3%81%e6%b0%97%e5%ad%a6-i\/%e9%9d%99%e7%a3%81%e5%a0%b4%ef%bc%9a%e9%9b%bb%e6%b5%81%e5%af%86%e5%ba%a6%e3%81%8b%e3%82%89%e7%9b%b4%e6%8e%a5%e9%9d%99%e7%a3%81%e5%a0%b4%e3%82%92%e6%b1%82%e3%82%81%e3%82%8b\/%e5%8f%82%e8%80%83%ef%bc%9a%e9%9d%99%e7%a3%81%e5%a0%b4%e3%82%92%e6%b1%82%e3%82%81%e3%82%8b%e9%9a%9b%e3%81%ab%e4%bd%bf%e3%81%a3%e3%81%9f%e7%a9%8d%e5%88%86%e3%82%92%e4%ba%ba%e5%8a%9b%e3%81%a7%e7%a2%ba\/","title":{"rendered":"\u53c2\u8003\uff1a\u9759\u78c1\u5834\u3092\u6c42\u3081\u308b\u969b\u306b\u4f7f\u3063\u305f\u7a4d\u5206\u3092\u4eba\u529b\u3067\u6c42\u3081\u3066\u307f\u308b"},"content":{"rendered":"

\u9759\u78c1\u5834\u3092\u6c42\u3081\u308b\u969b\u306b\u4f7f\u3063\u305f\u7a4d\u5206\u306f Maxima \u3092\u6570\u5b66\u516c\u5f0f\u96c6\u3068\u3057\u3066\u4f7f\u3046\u3053\u3068\u3067\u78ba\u8a8d\u3067\u304d\u3066\u3044\u308b<\/a>\u304c\uff0cMaxima \u3067\u89e3\u6790\u7684\u306b\u7a4d\u5206\u3067\u304d\u308b\u3053\u3068\u304c\u308f\u304b\u308c\u3070\uff0c\u4eba\u529b\u3067\u3082\u89e3\u3044\u3066\u307f\u305f\u304f\u306a\u308b\u3082\u306e\u3002\u3067\u3082\u307b\u3068\u3093\u3069\u306f\u300c\u9759\u96fb\u5834\u3092\u6c42\u3081\u308b\u969b\u306b\u4f7f\u3063\u305f\u7a4d\u5206\u3092\u4eba\u529b\u3067\u6c42\u3081\u3066\u307f\u308b<\/a>\u300d\u30b7\u30ea\u30fc\u30ba\u3067\u3059\u3067\u306b\u3084\u3063\u3066\u3044\u308b\u3002<\/p>\n

<\/p>\n

\u76f4\u7dda\u96fb\u6d41\u306b\u3088\u308b\u78c1\u5834<\/span>\u3067\u4f7f\u3063\u305f\u7a4d\u5206<\/h3>\n

$$\\int_{-\\infty}^{\\infty} \\frac{1}{\\left\\{x^2 + y^2 + (z-z\u2019)^2 \\right\\}^{\\frac{3}{2}}}\\,dz\u2019 = \\frac{2}{x^2 + y^2}$$<\/p>\n

\u3053\u308c\u306f\u300c\u9759\u96fb\u5834\u3092\u6c42\u3081\u308b\u969b\u306b\u4f7f\u3063\u305f\u7a4d\u5206\u3092\u4eba\u529b\u3067\u6c42\u3081\u3066\u307f\u308b\uff1a\u7b2c1\u8a71<\/a>\u300d\u3067<\/p>\n

$$\\int_{-\\infty}^{\\infty} \\frac{1}{(a^2 + Z^2)^{\\frac{3}{2}}} dZ = \\frac{2}{a^2}$$<\/p>\n

\u3068\u306a\u308b\u3053\u3068\u3092\u8aac\u660e\u6e08\u307f\u3002<\/p>\n

\u30bd\u30ec\u30ce\u30a4\u30c9\u3092\u6d41\u308c\u308b\u96fb\u6d41\u306b\u3088\u308b\u78c1\u5834\u3067\u4f7f\u3063\u305f\u7a4d\u5206<\/h3>\n

$$\\int_{-\\infty}^{\\infty}\\frac{z -z\u2019}{\\left((x-x\u2019)^2 + (y-y\u2019)^2 + (z-z\u2019)^2\\right)^{3\/2}} dz\u2019 =0$$<\/p>\n

\u3053\u308c\u306f\uff0c<\/p>\n

\\begin{eqnarray}
\n&& \\int_{-\\infty}^{\\infty}\\frac{z -z\u2019}{\\left((x-x\u2019)^2 + (y-y\u2019)^2 + (z-z\u2019)^2\\right)^{3\/2}} dz\u2019 \\\\
\n&=& \\int_{-\\infty}^{\\infty}\\frac{Z}{\\left((x-x\u2019)^2 + (y-y\u2019)^2 + Z^2\\right)^{3\/2}} dZ
\n\\end{eqnarray}<\/p>\n

\u3068\u3059\u308c\u3070\uff0c\u88ab\u7a4d\u5206\u95a2\u6570\u306f $Z$ \u306b\u3064\u3044\u3066\u5947\u95a2\u6570\u3060\u304b\u3089 $\\displaystyle \\int_{-b}^{b} dZ$ \u306f\u30bc\u30ed\u3068\u306a\u308b\u3053\u3068\u306f\u660e\u3089\u304b\u3002<\/p>\n

\u3082\u3046\u4e00\u3064\u306e\u7a4d\u5206<\/p>\n

\\begin{eqnarray}
\n\\int_0^{2\\pi}\u00a0 \\,d\\phi\u2019\u00a0 \\frac{ a^2\u00a0 -a\u00a0 y\\sin\\phi\u2019 -a x\\cos\\phi\u2019\u00a0 }{x^2 + y^2 + a^2 -2 a x \\cos\\phi\u2019 -2 a y \\sin\\phi\u2019 } &=& \\left\\{
\n\\begin{array}{ll}
\n2\\pi\u00a0 & (a > \\sqrt{x^2 + y^2})\\\\
\n0 & (a < \\sqrt{x^2 + y^2})
\n\\end{array}
\n\\right.
\n\\end{eqnarray}<\/p>\n

\u3082\u300c\u9759\u96fb\u5834\u3092\u6c42\u3081\u308b\u969b\u306b\u4f7f\u3063\u305f\u7a4d\u5206\u3092\u4eba\u529b\u3067\u6c42\u3081\u3066\u307f\u308b\uff1a\u7b2c2\u8a71<\/a>\u300d\u306e\u300c$y\\neq 0$ \u306e\u4e00\u822c\u306e\u5834\u5408\u306e\u8003\u5bdf<\/a>\u300d\u306e\u9805\u3092\u307f\u308c\u3070\uff08$\\sqrt{x^2+y^2}$ \u306f\u305d\u306e\u307e\u307e\u306b\u3057\uff0c$r’ \\rightarrow a$ \u3068\u7f6e\u304d\u63db\u3048\u3066\uff09<\/p>\n

\\begin{eqnarray}
\n\\int_0^{2\\pi} d\\phi\u2019\\frac{x^2 + y^2 -(x a \\cos\\phi\u2019 + y a \\sin\\phi\u2019)}{x^2 + y^2 + a^2 -2(x a \\cos\\phi\u2019 + y a \\sin\\phi\u2019)}
\n&=& \\left\\{
\n\\begin{array}{ll}
\n0 & ( \\sqrt{x^2 + y^2} < a)\\\\
\n2\\pi\u00a0 & (\u00a0 \\sqrt{x^2 + y^2} > a)
\n\\end{array}
\n\\right.
\n\\end{eqnarray}<\/p>\n

\u3068\u306a\u308b\u3053\u3068\u304c\u308f\u304b\u3063\u3066\u3044\u308b\u306e\u3067\uff0c<\/p>\n

\\begin{eqnarray}
\n&&\\int_0^{2\\pi}\u00a0 \\,d\\phi\u2019\u00a0 \\frac{ a^2\u00a0 -a\u00a0 y\\sin\\phi\u2019 -a x\\cos\\phi\u2019\u00a0 }{x^2 + y^2 + a^2 -2 a x \\cos\\phi\u2019 -2 a y \\sin\\phi\u2019 }\u00a0 \\\\
\n&=& \\int_0^{2\\pi}\u00a0 \\,d\\phi\u2019 \\left\\{ 1 -\\frac{x^2 + y^2 -(x a \\cos\\phi\u2019 + y a \\sin\\phi\u2019)}{x^2 + y^2 + a^2 -2(x a \\cos\\phi\u2019 + y a \\sin\\phi\u2019)}\\right\\} \\\\
\n&=& 2\\pi -\\int_0^{2\\pi} d\\phi\u2019\\frac{x^2 + y^2 -(x a \\cos\\phi\u2019 + y a \\sin\\phi\u2019)}{x^2 + y^2 + a^2 -2(x a \\cos\\phi\u2019 + y a \\sin\\phi\u2019)} \\\\
\n&=& \\left\\{
\n\\begin{array}{ll}
\n2 \\pi -0\u00a0 = 2 \\pi & (\\sqrt{x^2 + y^2} < a)\\\\
\n2 \\pi -2\\pi = 0 & (\u00a0 \\sqrt{x^2 + y^2} > a)
\n\\end{array}
\n\\right.
\n\\end{eqnarray}<\/p>\n

\u3064\u307e\u308a\uff0c\u3081\u3067\u305f\u304f<\/p>\n

$$\\int_0^{2\\pi}\u00a0 \\,d\\phi\u2019\u00a0 \\frac{ a^2\u00a0 -a\u00a0 y\\sin\\phi\u2019 -a x\\cos\\phi\u2019\u00a0 }{x^2 + y^2 + a^2 -2 a x \\cos\\phi\u2019 -2 a y \\sin\\phi\u2019 } = \\left\\{
\n\\begin{array}{ll}
\n2 \\pi & ( \\sqrt{x^2 + y^2} < a )\\\\
\n0 & ( \\sqrt{x^2 + y^2} > a)
\n\\end{array}
\n\\right.$$<\/p>\n

\u3068\u306a\u308b\u3002<\/p>\n","protected":false},"excerpt":{"rendered":"

\u9759\u78c1\u5834\u3092\u6c42\u3081\u308b\u969b\u306b\u4f7f\u3063\u305f\u7a4d\u5206\u306f Maxima \u3092\u6570\u5b66\u516c\u5f0f\u96c6\u3068\u3057\u3066\u4f7f\u3046\u3053\u3068\u3067\u78ba\u8a8d\u3067\u304d\u3066\u3044\u308b\u304c\uff0cMaxima \u3067\u89e3\u6790\u7684\u306b\u7a4d\u5206\u3067\u304d\u308b\u3053\u3068\u304c\u308f\u304b\u308c\u3070\uff0c\u4eba\u529b\u3067\u3082\u89e3\u3044\u3066\u307f\u305f\u304f\u306a\u308b\u3082\u306e\u3002\u3067\u3082\u307b\u3068\u3093\u3069\u306f\u300c\u9759\u96fb\u5834\u3092\u6c42\u3081\u308b\u969b\u306b\u4f7f\u3063\u305f\u7a4d\u5206\u3092\u4eba\u529b\u3067\u6c42\u3081\u3066\u307f\u308b\u300d\u30b7\u30ea\u30fc\u30ba\u3067\u3059\u3067\u306b\u3084\u3063\u3066\u3044\u308b\u3002<\/p>

\u7d9a\u304d\u3092\u8aad\u3080<\/a><\/p>\n","protected":false},"author":33,"featured_media":0,"parent":2687,"menu_order":20,"comment_status":"closed","ping_status":"closed","template":"","meta":{"inline_featured_image":false,"footnotes":""},"class_list":["post-4828","page","type-page","status-publish","hentry","nodate","item-wrap"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t