{"id":4811,"date":"2023-01-05T13:24:50","date_gmt":"2023-01-05T04:24:50","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?p=4811"},"modified":"2023-03-14T16:47:36","modified_gmt":"2023-03-14T07:47:36","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%e6%b1%82%e3%82%81%e3%81%a6%e3%81%bf-2","status":"publish","type":"post","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/4811\/","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\u7b2c4\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\u7b2c4\u8a71\u306f\uff0c\u7403\u5bfe\u79f0\u306a\u96fb\u8377\u5206\u5e03\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

\u7403\u5bfe\u79f0\u306a\u96fb\u8377\u5206\u5e03\u306b\u3088\u308b\u96fb\u5834<\/a>\u3067\u4f7f\u3063\u305f\u7a4d\u5206<\/h3>\n

$\\boldsymbol{r} = (0, 0, r)$ \u306e\u5834\u5408<\/h4>\n

$$ \\int_0^{\\pi} \\sin\\theta’ d\\theta’
\n\\frac{(r-r’\\cos\\theta’)}{\\left\\{r^2 + (r’)^2 -2 r r’ \\cos\\theta’\\right\\}^{3\/2}}
\n= \\frac{2}{r^2} H(r-r’) = \\left\\{
\n\\begin{array}{ll}
\n{\\displaystyle \\frac{2}{r^2}} & (r \\geq r’)\\\\ \\ \\\\
\n0 & (r < r’)
\n\\end{array}
\n\\right.$$<\/p>\n

\u306a\u305c\u3053\u306e\u7a4d\u5206\u306b\u30d8\u30d3\u30b5\u30a4\u30c9\u306e\u968e\u6bb5\u95a2\u6570 $H(r-r’)$ \u304c\u51fa\u3066\u304f\u308b\u306e\u304b\u3092\u5c11\u3057\u8003\u5bdf\u3002<\/p>\n

$t = \\cos\\theta’$ \u3068\u304a\u304f\u3068\uff0c$\\sin\\theta’ \\,d\\theta’ = – dt$<\/p>\n

\\begin{eqnarray}
\n&&\\int_0^{\\pi}
\n\\frac{(r-r’\\cos\\theta’)\\sin\\theta’ d\\theta’}{\\left\\{r^2 + (r’)^2 -2 r r’ \\cos\\theta’\\right\\}^{3\/2}} \\\\
\n&=& \\int_{-1}^1\u00a0 \\frac{(r-r’ t)}{\\left\\{r^2 + (r’)^2 -2 r r’ t\\right\\}^{3\/2}} dt \\\\
\n&=& \\frac{1}{r^2} \\Biggl[\\frac{r t – r’}{\\sqrt{r^2 + (r’)^2 – 2 r r’ t}} \\Biggr]_{-1}^1 \\\\
\n&=& \\frac{1}{r^2} \\left( \\frac{r-r’}{\\sqrt{(r – r’)^2}} – \\frac{-r-r’}{\\sqrt{(r + r’)^2}}\\right) \\\\
\n&=& \\frac{1}{r^2} \\left( \\frac{r-r’}{|r – r’|} + \\frac{r +r’}{|r + r’|}\\right) \\\\
\n&=& \\frac{1}{r^2} \\left( 1 + \\frac{r-r’}{|r – r’|} \\right) \\\\
\n&=&\\left\\{
\n\\begin{array}{ll}
\n{\\displaystyle \\frac{2}{r^2}} & (r \\geq r’)\\\\ \\ \\\\
\n0 & (r < r’)
\n\\end{array}
\n\\right.
\n\\end{eqnarray}<\/p>\n

$\\boldsymbol{r} = (x, y, z)$ \u306e\u4e00\u822c\u306e\u5834\u5408<\/h4>\n

\u305d\u3082\u305d\u3082\u7403\u5bfe\u79f0\u306a\u96fb\u8377\u5bc6\u5ea6 $\\rho(r)$ \u306b\u3064\u3044\u3066<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{E} &=&
\n\\frac{1}{4\\pi\\varepsilon_0}\\iiint \\frac{\\rho(r’)(\\boldsymbol{r} – \\boldsymbol{r}’)}{|\\boldsymbol{r} – \\boldsymbol{r}’|^3} dV’ \\\\
\n&=& \\frac{1}{4\\pi\\varepsilon_0} \\int (r’)^2 dr’ \\rho(r’) \\iint \\sin\\theta’\u00a0 d\\theta’ \\,d\\phi’
\n\\frac{\\boldsymbol{r} – \\boldsymbol{r}’}{\\left(r^2 + (r’)^2 – 2\\boldsymbol{r}\\cdot\\boldsymbol{r}’\\right)^{\\frac{3}{2}} }
\n\\end{eqnarray}<\/p>\n

\u3092\u6c42\u3081\u308b\u306e\u304c\u672c\u984c\u3067\u3042\u3063\u305f\u3002\u3053\u306e\u7a4d\u5206\u304c\u96e3\u3057\u305d\u3046\u3060\u304b\u3089\uff0c$\\boldsymbol{r} = (0, 0, r)$ \u306e\u5834\u5408\u306b\u3084\u3063\u3066\u307f\u305f\u308f\u3051\u3060\u3002\u4e00\u822c\u306e $\\boldsymbol{r} = (x, y, z)$ \u306e\u5834\u5408\u306f\uff0c\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u3057\u3066\u307f\u305f\u3089\u3069\u3046\u3060\u308d\u3046\u304b\u3002<\/p>\n

\u307e\u305a\uff0c<\/p>\n

$$\\boldsymbol{r}’ = r’
\n\\sin\\theta’ \\cos\\phi’ \\boldsymbol{i}
\n+ r’\u00a0 \\sin\\theta’ \\sin\\phi’ \\boldsymbol{j} +r’ \u00a0 \\cos\\theta’ \\boldsymbol{k}
\n$$<\/p>\n

\u306e\u304b\u308f\u308a\u306b\uff0c$\\displaystyle \\boldsymbol{e}_r \\equiv \\frac{\\boldsymbol{r}}{r}$ \u3068\u3057\uff0c$\\boldsymbol{e}_r$ \u306b\u76f4\u4ea4\u3059\u308b\u57fa\u672c\u30d9\u30af\u30c8\u30eb $\\boldsymbol{e}_1, \\boldsymbol{e}_2$ \u3092\u6b21\u306e\u3088\u3046\u306b\u3057\u3066\u5c0e\u5165\u3059\u308b\u3002<\/p>\n

\u307e\u305a\uff0c\u6b63\u898f\u76f4\u4ea4\u95a2\u4fc2\u306f<\/p>\n

$$\\boldsymbol{e}_r \\cdot \\boldsymbol{e}_r = 1, \\ \\
\n\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_1= 1, \\ \\
\n\\boldsymbol{e}_2 \\cdot \\boldsymbol{e}_2 = 1$$<\/p>\n

$$\\boldsymbol{e}_r \\cdot \\boldsymbol{e}_1 = 0, \\ \\
\n\\boldsymbol{e}_r \\cdot \\boldsymbol{e}_2= 0, \\ \\
\n\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2 = 0$$<\/p>\n

\u307e\u305f\uff0c\u5916\u7a4d\u306b\u3064\u3044\u3066\u306f<\/p>\n

$$\\boldsymbol{e}_r \\times \\boldsymbol{e}_1 = \\boldsymbol{e}_2, \\ \\
\n\\boldsymbol{e}_r \\times \\boldsymbol{e}_2= – \\boldsymbol{e}_1$$<\/p>\n

$\\boldsymbol{i} , \\boldsymbol{j} , \\boldsymbol{k} $ \u7cfb\u304b\u3089 $\\boldsymbol{e}_r,\u00a0 \\boldsymbol{e}_1, \\boldsymbol{e}_2$ \u7cfb\u3078\u306f\u539f\u70b9\u306e\u307e\u308f\u308a\u306e\u5ea7\u6a19\u7cfb\u306e\u56de\u8ee2\u306b\u3088\u3063\u3066\u5909\u63db\u3067\u304d\u308b\u3002<\/p>\n

\u4eca\u5f8c\uff0c\u8a08\u7b97\u306e\u969b\u306f $\\boldsymbol{e}_r,\u00a0 \\boldsymbol{e}_1, \\boldsymbol{e}_2$ \u7cfb\u3092\u4f7f\u3046\u3053\u3068\u306b\u3057\u3066\uff0c<\/p>\n

$$\\boldsymbol{r}’ =
\nr’\u00a0 \\sin\\vartheta’ \\cos\\varphi’ \\boldsymbol{e}_1
\n+ r’\u00a0 \\sin\\vartheta’ \\sin\\varphi’ \\boldsymbol{e}_2 +r’ \u00a0 \\cos\\vartheta’ \\boldsymbol{e}_r
\n$$<\/p>\n

\u3059\u3050\u4f7f\u3046\u91cf\u3092\u3042\u3089\u304b\u3058\u3081\u8a08\u7b97\u3057\u3066\u304a\u304f\u3068\uff0c<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{r}\\cdot\\boldsymbol{r}’ &=& r r’ \\cos\\vartheta’ \\\\
\n\\boldsymbol{r}\\times\\boldsymbol{r}’ &=& r r’ \\left(
\n\\sin\\vartheta’ \\cos\\varphi’ \\boldsymbol{e}_r\\times\\boldsymbol{e}_1
\n+ \\sin\\vartheta’ \\sin\\varphi’ \\boldsymbol{e}_r\\times\\boldsymbol{e}_2\\right) \\\\
\n&=& r r’ \\left(
\n\\sin\\vartheta’ \\cos\\varphi’ \\boldsymbol{e}_2
\n-\\sin\\vartheta’ \\sin\\varphi’ \\boldsymbol{e}_1\\right)
\n\\end{eqnarray}<\/p>\n

\u96fb\u5834\u3092\u6c42\u3081\u308b\u7a4d\u5206\u5f0f\u306f<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{E} &=&
\n\\frac{1}{4\\pi\\varepsilon_0}\\iiint \\frac{\\rho(r’)(\\boldsymbol{r} – \\boldsymbol{r}’)}{|\\boldsymbol{r} – \\boldsymbol{r}’|^3} dV’ \\\\
\n&=& \\frac{1}{4\\pi\\varepsilon_0} \\int (r’)^2 dr’ \\rho(r’) \\iint \\sin\\vartheta’\u00a0 d\\vartheta’ \\,d\\varphi’
\n\\frac{\\boldsymbol{r} – \\boldsymbol{r}’}{\\left(r^2 + (r’)^2 – 2\\boldsymbol{r}\\cdot\\boldsymbol{r}’\\right)^{\\frac{3}{2}} }
\n\\end{eqnarray}<\/p>\n

\u3053\u3053\u3067\uff0c<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{F} &\\equiv&
\n\\iint \\sin\\vartheta’\u00a0 d\\vartheta’ \\,d\\varphi’
\n\\frac{\\boldsymbol{r} – \\boldsymbol{r}’}{\\left(r^2 + (r’)^2 – 2\\boldsymbol{r}\\cdot\\boldsymbol{r}’\\right)^{\\frac{3}{2}} } \\\\
\n&=& \\iint \\sin\\vartheta’\u00a0 d\\vartheta’ \\,d\\varphi’
\n\\frac{\\boldsymbol{r} – \\boldsymbol{r}’}{\\left(r^2 + (r’)^2 – 2r r’ \\cos\\vartheta’\\right)^{\\frac{3}{2}} }
\n\\end{eqnarray}<\/p>\n

\u3068\u5b9a\u7fa9\u3059\u308b\u3068\uff0c<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{r} \\times\\boldsymbol{F}
\n&=& \\iint \\sin\\vartheta’\u00a0 d\\vartheta’ \\,d\\varphi’
\n\\frac{- \\boldsymbol{r}\\times\\boldsymbol{r}’}{\\left(r^2 + (r’)^2 – 2r r’ \\cos\\vartheta’\\right)^{\\frac{3}{2}} } \\\\
\n&=&\\int \\sin\\vartheta’\u00a0 d\\vartheta’ {\\color{red}{\\int_0^{2\\pi}\\,d\\varphi’}}
\n\\frac{- r r’ \\left(
\n\\sin\\vartheta’ {\\color{red}{\\cos\\varphi’}} \\boldsymbol{e}_2
\n-\\sin\\vartheta’ {\\color{red}{\\sin\\varphi’}} \\boldsymbol{e}_1\\right)}{\\left(r^2 + (r’)^2 – 2r r’ \\cos\\vartheta’\\right)^{\\frac{3}{2}} } \\\\
\n&=& \\boldsymbol{0}
\n\\end{eqnarray}<\/p>\n

\u3068\u306a\u308a\uff0c$\\boldsymbol{F}$ \u306f $\\boldsymbol{r}$ \u306b\u5e73\u884c\u306a\u6210\u5206\u3057\u304b\u3082\u305f\u306a\u3044\u3053\u3068\u304c\u7c21\u5358\u306b\u308f\u304b\u308b\u3002<\/p>\n

\u3057\u305f\u304c\u3063\u3066<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{F} &=& \\left(\\boldsymbol{e}_r\\cdot\\boldsymbol{F}\\right)\\, \\boldsymbol{e}_r \\\\
\n&=& \\frac{\\boldsymbol{r}}{r^2} \\left(\\boldsymbol{r}\\cdot\\boldsymbol{F}\\right) \\\\
\n&=& \\frac{\\boldsymbol{r}}{r^2} \\iint \\sin\\vartheta’\u00a0 d\\vartheta’ \\,d\\varphi’
\n\\frac{\\boldsymbol{r}\\cdot(\\boldsymbol{r} – \\boldsymbol{r}’)}{\\left(r^2 + (r’)^2 – 2r r’ \\cos\\vartheta’\\right)^{\\frac{3}{2}} }\u00a0 \\\\
\n&=& \\frac{\\boldsymbol{r}}{r} \\int_0^{2\\pi} \\,d\\varphi’ \\int_0^{\\pi} \\sin\\vartheta’\u00a0 d\\vartheta’
\n\\frac{r –\u00a0 r’\\cos\\vartheta’}{\\left(r^2 + (r’)^2 – 2r r’ \\cos\\vartheta’\\right)^{\\frac{3}{2}} }\u00a0 \\\\
\n&=& 2\\pi \\frac{\\boldsymbol{r}}{r}\\times \\left\\{
\n\\begin{array}{ll}
\n{\\displaystyle \\frac{2}{r^2}} & (r \\geq r’)\\\\ \\ \\\\
\n0 & (r < r’)
\n\\end{array}
\n\\right.\\\\
\n&=&\\left\\{
\n\\begin{array}{ll}
\n{\\displaystyle \\frac{4\\pi \\boldsymbol{r}}{r^3}} & (r \\geq r’)\\\\ \\ \\\\
\n0 & (r < r’)
\n\\end{array}
\n\\right.
\n\\end{eqnarray}<\/p>\n

\u3088\u3063\u3066\u3081\u3067\u305f\u304f\uff0c<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{E} &=& \\frac{1}{4\\pi\\varepsilon_0} \\int_0^{\\infty} (r’)^2 d r’ \\rho(r’) \\boldsymbol{F} \\\\
\n&=& \\frac{1}{4\\pi\\varepsilon_0} \\int_0^{r} (r’)^2 d r’\\rho(r’)\u00a0 \\frac{4\\pi \\boldsymbol{r} }{r^3} \\\\
\n&=& \\frac{1}{4\\pi\\varepsilon_0} \\frac{\\boldsymbol{r} }{r^3}\\int_0^{r} 4\\pi(r’)^2 \\rho(r’)\u00a0 d r’\\\\
\n&=& \\frac{Q_r}{4\\pi\\varepsilon_0} \\frac{\\boldsymbol{r} }{r^3}
\n\\end{eqnarray}<\/p>\n

\u3068\u306a\u308b\u3002<\/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\u7b2c4\u8a71\u306f\uff0c\u7403\u5bfe\u79f0\u306a\u96fb\u8377\u5206\u5e03\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-4811","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