{"id":4779,"date":"2023-01-05T10:08:20","date_gmt":"2023-01-05T01:08:20","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?p=4779"},"modified":"2024-12-16T15:52:47","modified_gmt":"2024-12-16T06:52:47","slug":"%e9%9d%99%e9%9b%bb%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%e3%82%84%e3%81%a3%e3%81%a6%e3%81%bf","status":"publish","type":"post","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/4779\/","title":{"rendered":"\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"},"content":{"rendered":"

\u9759\u96fb\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\u305d\u306e\u30b7\u30ea\u30fc\u30ba\u7b2c1\u8a71\u306f\uff0c\u4e00\u69d8\u306a\u7dda\u96fb\u8377\u306b\u3088\u308b\u96fb\u5834<\/a>\u3092\u6c42\u3081\u308b\u969b\u306b\u4f7f\u3063\u305f\u7a4d\u5206\u3002\u96fb\u78c1\u6c17\u5b66\u3068\u3044\u3046\u3088\u308a\u306f\u7406\u5de5\u7cfb\u306e\u6570\u5b66B\uff08\u5fae\u5206\u7a4d\u5206\uff09\u306e\u6f14\u7fd2\u554f\u984c\u7528\u306b\u3002<\/p>\n

<\/p>\n

\u4e00\u69d8\u306a\u7dda\u96fb\u8377\u306b\u3088\u308b\u96fb\u5834<\/a>\u3067\u4f7f\u3063\u305f\u7a4d\u5206<\/h3>\n

$\\displaystyle \\int_{-\\infty}^{\\infty}
\n\\frac{1}{\\left(x^2 + y^2 + (z-z’)^2\\right)^{3\/2}} dz’ =\u00a0 \\frac{2}{x^2 + y^2}$<\/p>\n

\u3053\u308c\u306f\u57fa\u672c\u7684\u306b<\/p>\n

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

\u304c\u308f\u304b\u308c\u3070\u3088\u3044\u3002$a^2 \\Rightarrow x^2 + y^2$ \u3068\u7f6e\u304d\u63db\u3048\u308c\u3070\u6c42\u3081\u308b\u7b54\u3048\u306b\u306a\u308b\u3002<\/p>\n

$Z = a \\tan\\theta$ \u3068\u304a\u3051\u3070\uff0c$\\displaystyle dZ = \\frac{a}{\\cos^2\\theta} d\\theta$\u3002<\/p>\n

\u4e0d\u5b9a\u7a4d\u5206<\/h4>\n

\u307e\u305a\u4e0d\u5b9a\u7a4d\u5206\u306f<\/p>\n

\\begin{eqnarray}
\n\\int \\frac{1}{(a^2 + Z^2)^{\\frac{3}{2}}} dZ &=&
\n\\int\\frac{1}{(a^2 + a^2 \\tan^2\\theta)^{\\frac{3}{2}}}\\frac{a}{\\cos^2\\theta} d\\theta \\\\
\n&=&
\n\\int\\frac{1}{a^3 (1 + \\tan^2\\theta)^{\\frac{3}{2}}}\\frac{a}{\\cos^2\\theta} d\\theta \\\\
\n&=& \\frac{1}{a^2} \\int |\\cos\\theta|\\, d\\theta \\quad(\\mbox{we assume} \\ \\cos\\theta > 0) \\\\
\n&=& \\frac{1}{a^2} \\sin\\theta \\\\
\n&=& \\frac{1}{a^2} \\tan\\theta \\times \\cos\\theta \\\\
\n&=& \\frac{1}{a^2} \\frac{\\tan\\theta}{\\sqrt{1 + \\tan^2\\theta}} \\\\
\n&=& \\frac{1}{a^2} \\frac{a \\tan\\theta}{\\sqrt{a^2 + a^2\\tan^2\\theta}} \\\\
\n&=& \\frac{1}{a^2} \\frac{Z}{\\sqrt{a^2 + Z^2}} \\\\
\n\\end{eqnarray}<\/p>\n

\u4e0d\u5b9a\u7a4d\u5206\u306e\u5225\u89e3<\/h4>\n

\u4e00\u65e6\u7b54\u3048\u304c\u308f\u304b\u308c\u3070\uff0c\u3044\u304b\u3088\u3046\u306b\u3082\u3053\u3058\u3064\u3051\u306f\u3067\u304d\u308b\u308f\u3051\u3067<\/p>\n

\\begin{eqnarray}
\n\\frac{1}{(a^2 + Z^2)^{\\frac{3}{2}}} &=& \\frac{1}{\\sqrt{a^2 + Z^2}} \\frac{1}{a^2 + Z^2} \\\\
\n&=& \\frac{1}{\\sqrt{a^2 + Z^2}} \\frac{1}{a^2} \\left( 1 – \\frac{Z^2}{a^2 + Z^2}\\right) \\\\
\n&=& \\frac{1}{a^2} \\left( \\frac{1}{\\sqrt{a^2 + Z^2}} – Z \\frac{Z}{(a^2 + Z^2)^{\\frac{3}{2}}}\\right) \\\\
\n&=& \\frac{1}{a^2} \\frac{d}{dZ} \\frac{Z}{\\sqrt{a^2 + Z^2}} \\\\
\n\\therefore \\ \\ \\int \\frac{1}{(a^2 + Z^2)^{\\frac{3}{2}}} dZ &=& \\int \\frac{1}{a^2} \\frac{d}{dZ} \\frac{Z}{\\sqrt{a^2 + Z^2}} dZ \\\\
\n&=& \\frac{1}{a^2} \\frac{Z}{\\sqrt{a^2 + Z^2}}
\n\\end{eqnarray}<\/p>\n

\u5b9a\u7a4d\u5206<\/h4>\n

\u5b9a\u7a4d\u5206\u306f<\/p>\n

\\begin{eqnarray}
\n\\int_{-\\infty}^{\\infty} \\frac{1}{(a^2 + Z^2)^{\\frac{3}{2}}} dZ
\n&=& \\frac{1}{a^2} \\Bigl[ \\sin\\theta\\Bigr]_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\\\
\n&=& \\frac{2}{a^2}
\n\\end{eqnarray}<\/p>\n

\u307e\u305f\u306f<\/p>\n

\\begin{eqnarray}
\n\\int_{-\\infty}^{\\infty} \\frac{1}{(a^2 + Z^2)^{\\frac{3}{2}}} dZ
\n&=& \\frac{1}{a^2} \\Bigl[ \\frac{Z}{\\sqrt{a^2 + Z^2}}\\Bigr]_{-\\infty}^{\\infty} \\\\
\n&=& \\frac{2}{a^2}
\n\\end{eqnarray}<\/p>\n","protected":false},"excerpt":{"rendered":"

\u9759\u96fb\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\u305d\u306e\u30b7\u30ea\u30fc\u30ba\u7b2c1\u8a71\u306f\uff0c\u4e00\u69d8\u306a\u7dda\u96fb\u8377\u306b\u3088\u308b\u96fb\u5834\u3092\u6c42\u3081\u308b\u969b\u306b\u4f7f\u3063\u305f\u7a4d\u5206\u3002\u96fb\u78c1\u6c17\u5b66\u3068\u3044\u3046\u3088\u308a\u306f\u7406\u5de5\u7cfb\u306e\u6570\u5b66B\uff08\u5fae\u5206\u7a4d\u5206\uff09\u306e\u6f14\u7fd2\u554f\u984c\u7528\u306b\u3002<\/p>

\u7d9a\u304d\u3092\u8aad\u3080<\/a><\/p>\n","protected":false},"author":33,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false,"footnotes":""},"categories":[19],"tags":[],"class_list":["post-4779","post","type-post","status-publish","format-standard","hentry","category-19","nodate","item-wrap"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\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