{"id":4787,"date":"2023-01-09T12:39:54","date_gmt":"2023-01-09T03:39:54","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?p=4787"},"modified":"2023-03-14T16:47:17","modified_gmt":"2023-03-14T07:47:17","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-2","status":"publish","type":"post","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/4787\/","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\u7b2c2\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\u7b2c2\u8a71\u306f\uff0c\u8ef8\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

\u8ef8\u5bfe\u79f0\u306a\u96fb\u8377\u5206\u5e03\u306b\u3088\u308b\u96fb\u5834<\/a>\u3067\u4f7f\u3063\u305f\u7a4d\u5206<\/h3>\n

$$\\int_0^{2\\pi} d\\phi’ \\frac{x-r’ \\cos\\phi’}{(x-r’ \\cos \\phi’)^2 + (y-r’ \\sin \\phi’)^2} = \\frac{2\\pi x}{x^2 + y^2} H\\left(\\sqrt{x^2 + y^2} – r’\\right) \\tag{1}$$<\/p>\n

$$\\int_0^{2\\pi} d\\phi’ \\frac{y-r’ \\sin\\phi’}{(x-r’ \\cos \\phi’)^2 + (y-r’ \\sin \\phi’)^2} = \\frac{2\\pi y}{x^2 + y^2} H\\left(\\sqrt{x^2 + y^2} – r’\\right) \\tag{2}$$<\/p>\n

\u3053\u3053\u3067 $H\\left(\\sqrt{x^2 + y^2} – r’\\right)$ \u306f\u30d8\u30d3\u30b5\u30a4\u30c9\u306e\u968e\u6bb5\u95a2\u6570\u3067\u3042\u308a\uff0c<\/p>\n

$$H\\left(\\sqrt{x^2 + y^2} – r’\\right)=
\n\\left\\{
\n\\begin{array}{ll}
\n1 & (\\sqrt{x^2 + y^2} \\geq r’)\\\\ \\ \\\\
\n0 & (\\sqrt{x^2 + y^2} < r’)
\n\\end{array}
\n\\right.$$<\/p>\n

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

$y=0$ \u306e\u5834\u5408\u306e\u8003\u5bdf<\/h3>\n

$y=0$ \u3068\u3059\u308b\u3068 $(\\mbox{1})$ \u5f0f\u306f<\/p>\n

\\begin{eqnarray}
\n\\int_0^{2\\pi} d\\phi’ \\frac{x-r’ \\cos\\phi’}{(x-r’ \\cos \\phi’)^2 + (y-r’ \\sin \\phi’)^2}
\n&\\Rightarrow& \\int_0^{2\\pi} d\\phi’ \\frac{x-r’ \\cos\\phi’}{x^2 + (r’)^2 -2 x r’ \\cos \\phi’}
\n\\end{eqnarray}<\/p>\n

\u307e\u305a\uff0c\u89e3\u304f\u3079\u304d\u7a4d\u5206\u3092\uff08\u5909\u6570\u3092\u3061\u3087\u3063\u3068\u5909\u3048\u3066\uff0c\u3088\u308a\u4e00\u822c\u7684\u30b7\u30c1\u30e5\u30a8\u30fc\u30b7\u30e7\u30f3\u306b\u3082\u4f7f\u3048\u308b\u3088\u3046\u306b\u3057\u3066\uff09<\/p>\n

\\begin{eqnarray}
\n\\int_0^{2\\pi} \\frac{a – b \\cos\\phi}{a^2 + b^2 -2 a b \\cos \\phi} d\\phi
\n\\end{eqnarray}<\/p>\n

\u306e\u5f62\u306b\u3057\uff0c\u6388\u696d\u3067\u3084\u3063\u305f\u3088\u3046\u306b<\/a> $\\sin \\phi, \\cos \\phi$ \u306e\u6709\u7406\u95a2\u6570\u306e\u7a4d\u5206\u306f\uff0c\u4ee5\u4e0b\u306e\u3088\u3046\u306a\u5909\u63db\u3092\u3057\u3066\u7f6e\u63db\u7a4d\u5206\u3059\u308c\u3070\u3088\u3044\u3068\u3044\u3046\u3053\u3068\u304c\u77e5\u3089\u308c\u3066\u3044\u308b\u306e\u3067\uff0c\u30bb\u30aa\u30ea\u30fc\u306b\u305d\u3063\u3066\u3084\u3063\u3066\u307f\u308b\u3002<\/p>\n

\\begin{eqnarray}
\nt &\\equiv& \\tan\\frac{\\phi}{2} \\\\
\n\\sin\\phi &=& \\frac{2t}{1+t^2} \\\\
\n\\cos\\phi &=& \\frac{1-t^2}{1+t^2} \\\\
\nd\\phi &=& \\frac{2}{1+t^2} dt
\n\\end{eqnarray}<\/p>\n

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

\u307e\u305a\u4e0d\u5b9a\u7a4d\u5206\u306f\u5909\u6570\u5909\u63db $\\displaystyle t \\equiv \\tan\\frac{\\phi}{2}$ \u306e\u5f8c\uff0c\u90e8\u5206\u5206\u6570\u306b\u5206\u89e3\u3057\u3066\u4ee5\u4e0b\u306e\u3088\u3046\u306b…<\/p>\n

\\begin{eqnarray}{}&&
\n\\int \\frac{a – b \\cos\\phi}{a^2 + b^2 -2 a b \\cos \\phi} d\\phi \\\\
\n&&\\quad=
\n\\int \\frac{a – b \\frac{1-t^2}{1+t^2}}{a^2 + b^2 -2 a b \\frac{1-t^2}{1+t^2}}\\frac{2}{1+t^2} dt \\\\
\n&&\\quad= \\int \\frac{a(1+t^2)-b (1-t^2)}{\\left(a^2 + b^2\\right)(1+t^2) -2 a b (1-t^2)}\\frac{2}{1+t^2} dt \\\\
\n&&\\quad= \\int \\frac{(a+b)t^2 -(a-b)}{(a+b)^2 t^2 + (a-b)^2}\\frac{2}{1+t^2} dt \\\\
\n&&\\quad= \\frac{1}{a} \\int \\left\\{\\frac{1}{1+t^2} + \\frac{a+b}{a-b}\\frac{1}{1+\\left(\\frac{a+b}{a-b}t\\right)^2} \\right\\} dt \\\\
\n&&\\quad= \\frac{1}{a} \\left\\{\\tan^{-1} t + \\frac{|a-b|}{a-b}\\tan^{-1} \\left(\\frac{a+b}{|a-b|} t\\right) \\right\\} \\\\
\n&&\\quad= \\frac{1}{a} \\left\\{\\frac{\\phi}{2} + \\frac{|a-b|}{a-b}\\tan^{-1} \\left(\\frac{a+b}{|a-b|} \\tan\\frac{\\phi}{2}\\right) \\right\\}
\n\\end{eqnarray}<\/p>\n

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

\u5b9a\u7a4d\u5206\u3082\uff08\u7a4d\u5206\u533a\u9593\u5185\u3067 $\\tan$ \u304c\u767a\u6563\u3059\u308b\u3068\u3053\u308d\u304c\u3042\u308b\u306e\u3067\uff09\u614e\u91cd\u306b\u3002<\/p>\n

\u307e\u305a $a > b$ \u306e\u5834\u5408\u306f $\\displaystyle \\frac{|a-b|}{a-b} = 1$ \u3060\u304b\u3089<\/p>\n

\\begin{eqnarray}&&
\n\\int_0^{2\\pi} \\frac{a – b \\cos\\phi}{a^2 + b^2 -2 a b \\cos \\phi} d\\phi \\\\
\n&&\\quad=
\n\\frac{1}{a}\\Bigl[\\frac{\\phi}{2} \\Bigr]_0^{2\\pi} \\ + \\frac{1}{a}\\Biggl[\\tan^{-1}\\left(\\frac{a+b}{|a-b|} \\tan\\frac{\\phi}{2}\\right) \\Biggr]_0^{\\pi-0}
\n+ \\frac{1}{a}\\Biggl[\\tan^{-1}\\left(\\frac{a+b}{|a-b|} \\tan\\frac{\\phi}{2}\\right) \\Biggr]_{\\pi+0}^{2\\pi} \\\\
\n&&\\quad= \\frac{\\pi}{a} + \\frac{1}{a}\\left\\{\\frac{\\pi}{2}-0 \\right\\}
\n+\\frac{1}{a}\\left\\{0-\\left(-\\frac{\\pi}{2}\\right) \\right\\} \\\\
\n&&\\quad= \\frac{\\pi}{a} + \\frac{\\pi}{a} \\\\
\n&&\\quad= \\frac{2\\pi}{a}
\n\\end{eqnarray}<\/p>\n

$a\\\u00a0 {\\color{red}{<}} \\ b$ \u306e\u5834\u5408\u306f $\\displaystyle \\frac{|a-b|}{a-b} = {\\color{red}{-}}\u00a0 1$ \u3060\u304b\u3089<\/p>\n

\\begin{eqnarray}&&
\n\\int_0^{2\\pi} \\frac{a – b \\cos\\phi}{a^2 + b^2 -2 a b \\cos \\phi} d\\phi \\\\
\n&&\\quad=
\n\\frac{1}{a}\\Bigl[\\frac{\\phi}{2} \\Bigr]_0^{2\\pi} {\\color{red}{-}} \\frac{1}{a}\\Biggl[\\tan^{-1}\\left(\\frac{a+b}{|a-b|} \\tan\\frac{\\phi}{2}\\right) \\Biggr]_0^{\\pi-0}
\n{\\color{red}{-}} \\frac{1}{a}\\Biggl[\\tan^{-1}\\left(\\frac{a+b}{|a-b|} \\tan\\frac{\\phi}{2}\\right) \\Biggr]_{\\pi+0}^{2\\pi} \\\\
\n&&\\quad= \\frac{\\pi}{a} {\\color{red}{-}} \\frac{1}{a}\\left\\{\\frac{\\pi}{2}-0 \\right\\}
\n{\\color{red}{-}}\\frac{1}{a}\\left\\{0-\\left(-\\frac{\\pi}{2}\\right) \\right\\} \\\\
\n&&\\quad= \\frac{\\pi}{a} {\\color{red}{-}} \\frac{\\pi}{a} \\\\
\n&&\\quad= 0
\n\\end{eqnarray}<\/p>\n

\u307e\u3068\u3081\u308b\u3068<\/p>\n

\\begin{eqnarray}
\n\\int_0^{2\\pi} \\frac{a – b \\cos\\phi}{a^2 + b^2 -2 a b \\cos \\phi} d\\phi\u00a0 &=& \\left\\{
\n\\begin{array}{ll}
\n{\\displaystyle\\frac{2\\pi}{a}} & (a > b)\\\\ \\ \\\\
\n0 & (a \\ {\\color{red}{<}} \\ b)
\n\\end{array}
\n\\right.
\n\\end{eqnarray}<\/p>\n

\u5143\u306e\u5909\u6570\u306b\u76f4\u3059\u3068<\/p>\n

\\begin{eqnarray}
\n\\int_0^{2\\pi} \\frac{x – r’ \\cos\\phi’}{x^2 + (r’)^2 -2 x r’ \\cos \\phi} d\\phi’ \\\\ &=& \\left\\{
\n\\begin{array}{ll}
\n{\\displaystyle \\frac{2\\pi}{x}} & (x > r’)\\\\ \\ \\\\
\n0 & (x \\ {\\color{red}{<}} \\ r’)
\n\\end{array}
\n\\right.
\n\\end{eqnarray}<\/p>\n

$y \\neq 0$ \u306e\u4e00\u822c\u306e\u5834\u5408\u306e\u8003\u5bdf<\/h3>\n

$$F_1 \\equiv \\int_0^{2\\pi} d\\phi’ \\frac{x-r’ \\cos\\phi’}{(x-r’ \\cos \\phi’)^2 + (y-r’ \\sin \\phi’)^2} $$<\/p>\n

$$F_2 \\equiv \\int_0^{2\\pi} d\\phi’ \\frac{y-r’ \\sin\\phi’}{(x-r’ \\cos \\phi’)^2 + (y-r’ \\sin \\phi’)^2}$$<\/p>\n

\u3068\u304a\u304f\u3068\uff0c<\/p>\n

\\begin{eqnarray}
\ny F_1 – x F_2 &=& \\int_0^{2\\pi} d\\phi’
\n\\frac{y(x-r’ \\cos\\phi’) – x (y-r’ \\sin\\phi’)}{(x-r’ \\cos \\phi’)^2 + (y-r’ \\sin \\phi’)^2} \\\\
\n&=& \\int_0^{2\\pi} d\\phi’
\n\\frac{x r’ \\sin\\phi’ – y r’ \\cos\\phi’}{x^2 + y^2 + (r’)^2 –\u00a0 2(x r’ \\cos\\phi’ + y r’ \\sin\\phi’)} \\\\
\n&=& \\frac{1}{2} \\Bigl[ \\ln \\left|x^2 + y^2 + (r’)^2 –\u00a0 2(x r’ \\cos\\phi’ + y r’ \\sin\\phi’)\\right| \\Bigr]_0^{2\\pi} \\\\
\n&=& 0 \\\\
\n\\therefore\\ \\ y F_1 &=& x F_2 \\\\
\nF_2 &=& \\frac{y}{x} F_1
\n\\end{eqnarray}<\/p>\n

\u6b21\u306b\uff0c<\/p>\n

\\begin{eqnarray}
\nx F_1 + y F_ 2 &=& x F_1 + y \\frac{y}{x} F_1 \\\\
\n&=& \\frac{x^2 + y^2}{x} F_1 \\\\
\n&=& \\int_0^{2\\pi} d\\phi’
\n\\frac{x (x-r’ \\cos\\phi’) + y (y-r’ \\sin\\phi’)}{(x-r’ \\cos \\phi’)^2 + (y-r’ \\sin \\phi’)^2} \\\\
\n&=& \\int_0^{2\\pi} d\\phi’
\n\\frac{x^2 + y^2 – (x r’ \\cos\\phi’ + y r’ \\sin\\phi’)}{x^2 + y^2 + (r’)^2 –\u00a0 2(x r’ \\cos\\phi’ + y r’ \\sin\\phi’)} \\\\
\n&=& \\int_0^{2\\pi} d\\phi’
\n\\frac{x^2 + y^2 – \\sqrt{x^2 + y^2}\\, r’ \\cos(\\phi’-\\alpha)}{x^2 + y^2 + (r’)^2 –\u00a0 2\\sqrt{x^2 + y^2}\\, r’ \\cos(\\phi’-\\alpha)} \\\\ \\ \\\\
\n&&\\quad \\alpha \\equiv \\tan^{-1} \\frac{y}{x}
\n\\end{eqnarray}<\/p>\n

$\\sqrt{x^2 + y^2} \\equiv a, r’ \\equiv b, \\phi’-\\alpha \\equiv \\phi$ \u3068\u7f6e\u3044\u3066\u307f\u308b\u3068<\/p>\n

\\begin{eqnarray}
\n\\frac{x^2 + y^2}{x} F_1 &=& a \\int_{-\\alpha}^{2\\pi-\\alpha} d\\phi
\n\\frac{a – b \\cos\\phi}{a^2 +b^2 –\u00a0 2a b \\cos\\phi} \\\\
\n&=&
\na \\int_{-\\alpha}^{0} d\\phi
\n\\frac{a – b \\cos\\phi}{a^2 +b^2 –\u00a0 2a b \\cos\\phi} \\\\
\n&& + a \\int_{0}^{2\\pi} d\\phi
\n\\frac{a – b \\cos\\phi}{a^2 +b^2 –\u00a0 2a b \\cos\\phi} \\\\
\n&& + a \\int_{2\\pi}^{2\\pi-\\alpha} d\\phi
\n\\frac{a – b \\cos\\phi}{a^2 +b^2 –\u00a0 2a b \\cos\\phi} \\\\
\n&=& a \\int_{0}^{2\\pi} d\\phi
\n\\frac{a – b \\cos\\phi}{a^2 +b^2 –\u00a0 2a b \\cos\\phi} \\\\
\n&& + a \\int_{-\\alpha}^{0} d\\phi
\n\\frac{a – b \\cos\\phi}{a^2 +b^2 –\u00a0 2a b \\cos\\phi} + a \\int_{0}^{-\\alpha} d\\phi
\n\\frac{a – b \\cos\\phi}{a^2 +b^2 –\u00a0 2a b \\cos\\phi} \\\\
\n&=& a \\int_{0}^{2\\pi} d\\phi
\n\\frac{a – b \\cos\\phi}{a^2 +b^2 –\u00a0 2a b \\cos\\phi} \\\\
\n&=& \\left\\{
\n\\begin{array}{ll}
\n2\\pi & (a \\geq b)\\\\
\n0 & (a <\u00a0 b)
\n\\end{array}
\n\\right. \\\\ \\ \\\\
\n\\therefore \\ \\ F_1 &=&\\left\\{
\n\\begin{array}{ll}
\n{\\displaystyle \\frac{2\\pi x}{x^2 + y^2}} & (\\sqrt{x^2 + y^2} \\geq r’)\\\\
\n0 & (\\sqrt{x^2 + y^2} < r’)
\n\\end{array}
\n\\right.\u00a0 \\\\ \\ \\\\
\nF_2 &=& \\frac{y}{x} F_1 \\\\
\n&=& \\left\\{
\n\\begin{array}{ll}
\n{\\displaystyle \\frac{2\\pi y}{x^2 + y^2}} & (\\sqrt{x^2 + y^2} \\geq r’)\\\\
\n0 & (\\sqrt{x^2 + y^2} < r’)
\n\\end{array}
\n\\right.
\n\\end{eqnarray}<\/p>\n

$y \\neq 0$ \u306e\u4e00\u822c\u306e\u5834\u5408\u3092\u30d9\u30af\u30c8\u30eb\u5f62\u3067\u8003\u5bdf<\/h3>\n

\u305d\u3082\u305d\u3082\uff0c$z$ \u8ef8\u306b\u3064\u3044\u3066\u8ef8\u5bfe\u79f0\u306a\u96fb\u8377\u5bc6\u5ea6 $\\rho(\\boldsymbol{r}) \\Rightarrow \\rho(\\sqrt{x^2+y^2})$ \u306b\u3064\u3044\u3066\uff0c<\/p>\n

$$\\boldsymbol{E} = \\frac{1}{4\\pi\\varepsilon_0}
\n\\iiint \\frac{\\rho(\\boldsymbol{r}’) \\left(\\boldsymbol{r}-\\boldsymbol{r}’\\right)}{|\\boldsymbol{r}-\\boldsymbol{r}’|^3} dV’$$<\/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$y=0$ \u306e\u5834\u5408\u306b\u3084\u3063\u3066\u307f\u305f\u308f\u3051\u3060\u3002<\/p>\n

\u4e00\u822c\u306e\u5834\u5408\u306f\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u3084\u3063\u3066\u307f\u305f\u3089\u3069\u3046\u3060\u308d\u3046\u304b\u3002<\/p>\n

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

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

\u306e\u304b\u308f\u308a\u306b\uff0c$z$ \u8ef8\u306b\u76f4\u4ea4\u3059\u308b\uff0c$z$ \u8ef8\u304b\u3089\u306e\u300c\u52d5\u5f84\u300d\u30d9\u30af\u30c8\u30eb\u3092<\/p>\n

$$\\boldsymbol{r}_{\\!\\perp} \\equiv (x, y, 0), \\quad r_{\\!\\perp} \\equiv \\sqrt{\\boldsymbol{r}_{\\!\\perp}\\cdot\\boldsymbol{r}_{\\!\\perp}} = \\sqrt{x^2 + y^2}$$<\/p>\n

\u3068\u3057\uff0c$\\boldsymbol{r}_{\\!\\perp} $ \u65b9\u5411\u306e\u5358\u4f4d\u30d9\u30af\u30c8\u30eb $\\boldsymbol{e}_{\\!\\perp}$ \u3092<\/p>\n

$$\\boldsymbol{e}_{\\!\\perp} \\equiv \\frac{\\boldsymbol{r}_{\\!\\perp}}{r_{\\!\\perp}}$$<\/p>\n

$\\boldsymbol{e}_{\\!\\perp}$ \u306b\u76f4\u4ea4\u3059\u308b\u5358\u4f4d\u30d9\u30af\u30c8\u30eb $\\boldsymbol{e}_1$ \u3092\u6b21\u306e\u3088\u3046\u306b\u3057\u3066\u5c0e\u5165\u3059\u308b\u3002<\/p>\n

$$\\boldsymbol{e}_{\\!\\perp}\\cdot \\boldsymbol{e}_1 = 0, \\quad \\boldsymbol{k}\\cdot \\boldsymbol{e}_1 = 0$$<\/p>\n

$$\\boldsymbol{e}_{\\!\\perp}\\times \\boldsymbol{e}_1 = \\boldsymbol{k}$$<\/p>\n

$\\boldsymbol{i}, \\boldsymbol{j}, \\boldsymbol{k}$ \u7cfb\u304b\u3089 $\\boldsymbol{e}_{\\!\\perp}, \\boldsymbol{e}_1, \\boldsymbol{k}$ \u7cfb\u3078\u306f $z$ \u8ef8\u306e\u307e\u308f\u308a\u306e\u5ea7\u6a19\u7cfb\u306e\u56de\u8ee2\u306b\u3088\u3063\u3066\u5909\u63db\u3067\u304d\u308b\u3002<\/p>\n

\u4ee5\u5f8c\uff0c\u8a08\u7b97\u306e\u969b\u306f $\\boldsymbol{e}_{\\!\\perp}, \\boldsymbol{e}_1, \\boldsymbol{k}$ \u7cfb\u3092\u4f7f\u3046\u3053\u3068\u306b\u3057\u3066<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{r} &=& \\boldsymbol{r}_{\\!\\perp} + z \\boldsymbol{k} \\\\
\n&=& r _{\\!\\perp} \\boldsymbol{e}_{\\!\\perp} + z \\boldsymbol{k} \\\\
\n\\boldsymbol{r}’ &=& \\boldsymbol{r}’_{\\!\\perp} + z’ \\boldsymbol{k} \\\\
\n&=& r’_{\\!\\perp} \\cos\\varphi’ \\boldsymbol{e}_{\\!\\perp} + r’_{\\!\\perp} \\sin\\varphi’ \\boldsymbol{e}_{1} +z’ \\boldsymbol{k}
\n\\end{eqnarray}<\/p>\n

\u3055\u3066\uff0c\u96fb\u78c1\u5834\u3092\u6c42\u3081\u308b\u7a4d\u5206\u5f0f\u306f<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{E} &=& \\frac{1}{4\\pi\\varepsilon_0}
\n\\iiint \\frac{\\rho(r’_{\\!\\perp}) \\left(\\boldsymbol{r} –\u00a0 \\boldsymbol{r}’ \\right)}{|\\boldsymbol{r} –\u00a0 \\boldsymbol{r}’ |^3} dV’ \\\\
\n&\\equiv& \\frac{1}{4\\pi\\varepsilon_0} \\iint r’_{\\!\\perp}\\,dr’_{\\!\\perp} \\,d\\varphi’ \\rho(r’_{\\!\\perp}) \\,\\boldsymbol{F}
\n\\end{eqnarray}<\/p>\n

\u3053\u3053\u3067 $\\boldsymbol{F}$ \u306f<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{F} &\\equiv& \\int_{-\\infty}^{\\infty} dz’ \\frac{\\boldsymbol{r} –\u00a0 \\boldsymbol{r}’ }{|\\boldsymbol{r} –\u00a0 \\boldsymbol{r}’ |^3} \\\\
\n&=& \\int_{-\\infty}^{\\infty} dz’ \\frac{\\boldsymbol{r}_{\\!\\perp} –\u00a0 \\boldsymbol{r}’_{\\!\\perp} + (z-z’) \\boldsymbol{k}}{\\left( r_{\\!\\perp}^2 + (r’_{\\!\\perp})^2 – 2 \\boldsymbol{r}\\cdot\\boldsymbol{r}’ + (z-z’)^2\\right)^{\\frac{3}{2}}}\\\\
\n&=& \\frac{2 (\\boldsymbol{r}_{\\!\\perp} –\u00a0 \\boldsymbol{r}’_{\\!\\perp})}{r_{\\!\\perp}^2 + (r’_{\\!\\perp})^2 – 2 \\boldsymbol{r}\\cdot\\boldsymbol{r}’}
\n\\end{eqnarray}<\/p>\n

\u3053\u3053\u3067\uff0c\u65e2\u306b\u793a\u3055\u308c\u3066\u3044\u308b\u4ee5\u4e0b\u306e\u7d50\u679c\u3092\u4f7f\u3063\u305f\u3002<\/p>\n

$$\\int_{-\\infty}^{\\infty} \\frac{1}{\\left(a^2 + Z^2 \\right)^{\\frac{3}{2}}} dZ = \\frac{2}{a^2}, \\quad
\n\\int_{-\\infty}^{\\infty} \\frac{Z}{\\left(a^2 + Z^2 \\right)^{\\frac{3}{2}}} dZ = 0$$<\/p>\n

\u3068\u3044\u3046\u3053\u3068\u3067\uff0c<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{E}
\n&=& \\frac{1}{4\\pi\\varepsilon_0} \\iint r’_{\\!\\perp}\\,dr’_{\\!\\perp} \\,d\\varphi’ \\rho(r’_{\\!\\perp}) \\,\\boldsymbol{F} \\\\
\n&=& \\frac{1}{2\\pi\\varepsilon_0} \\int r’_{\\!\\perp} \\,dr’_{\\!\\perp} \\rho(r’_{\\!\\perp}) \\int_0^{2\\pi}\u00a0 d\\varphi’
\n\\frac{\\boldsymbol{r}_{\\!\\perp} –\u00a0 \\boldsymbol{r}’_{\\!\\perp}}{r_{\\!\\perp}^2 + (r’_{\\!\\perp})^2 – 2 \\boldsymbol{r}\\cdot\\boldsymbol{r}’}
\n\\end{eqnarray}<\/p>\n

\u3042\u3068\u306f\uff0c$\\boldsymbol{E} $ \u304c $\\boldsymbol{r}_{\\!\\perp} $ \u306b\u5e73\u884c\u306a\u6210\u5206\u3057\u304b\u3082\u305f\u306a\u3044\u3053\u3068\u3092\u6b21\u306e\u3088\u3046\u306b\u3057\u3066\u793a\u3059\u3002<\/p>\n

\u307e\u305a\uff0c$\\boldsymbol{r}_{\\!\\perp} $ \u3068\u306e\u5916\u7a4d\u3092\u3068\u308b\u3068<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{r}_{\\!\\perp}\\times\\boldsymbol{E}
\n&=& \\frac{1}{2\\pi\\varepsilon_0} \\int r’_{\\!\\perp} \\,dr’_{\\!\\perp} \\rho(r’_{\\!\\perp}) \\int_0^{2\\pi}\u00a0 d\\varphi’
\n\\frac{- \\boldsymbol{r}_{\\!\\perp}\\times\u00a0 \\boldsymbol{r}’_{\\!\\perp}}{r_{\\!\\perp}^2 + (r’_{\\!\\perp})^2 – 2 \\boldsymbol{r}\\cdot\\boldsymbol{r}’} \\\\
\n&=& \\frac{1}{2\\pi\\varepsilon_0} \\int r’_{\\!\\perp} \\,dr’_{\\!\\perp} \\rho(r’_{\\!\\perp})
\n\\int_0^{2\\pi}\u00a0 d\\varphi’ \\frac{- r_{\\!\\perp}\u00a0 r_{\\!\\perp}’ \\sin\\varphi’}{r_{\\!\\perp}^2 + (r’_{\\!\\perp})^2 – 2r_{\\!\\perp} r’_{\\!\\perp}\u00a0 \\cos\\varphi’ } \\boldsymbol{k} \\\\
\n&=& \\frac{1}{2\\pi\\varepsilon_0} \\int r’_{\\!\\perp} \\,dr’_{\\!\\perp} \\rho(r’_{\\!\\perp})
\n\\Bigl[\\ln\\left| r_{\\!\\perp}^2 + (r’_{\\!\\perp})^2 – 2r_{\\!\\perp} r’_{\\!\\perp}\u00a0 \\cos\\varphi’ \\right|\\Bigr]^0_{2\\pi}\\, \\boldsymbol{k} \\\\
\n&=& \\boldsymbol{0}
\n\\end{eqnarray}<\/p>\n

\u3057\u305f\u304c\u3063\u3066\uff0c$\\boldsymbol{E} $ \u306f $\\boldsymbol{r}_{\\!\\perp}$ \u306b\u5e73\u884c\u306a\u6210\u5206\u3057\u304b\u3082\u305f\u306a\u3044\u304b\u3089<\/p>\n

\\begin{eqnarray}
\n\\boldsymbol{E} &=& (\\boldsymbol{e}_{\\!\\perp} \\cdot\\boldsymbol{E} ) \\,\\boldsymbol{e}_{\\!\\perp}\u00a0 \\\\
\n&=& \\frac{\\boldsymbol{r}_{\\!\\perp}}{r^2_{\\!\\perp}} (\\boldsymbol{r}_{\\!\\perp} \\cdot\\boldsymbol{E} ) \\\\
\n&=& \\frac{\\boldsymbol{r}_{\\!\\perp}}{r^2_{\\!\\perp}}\\frac{1}{2\\pi\\varepsilon_0}
\n\\int_0^{\\infty}\u00a0 r’_{\\!\\perp} \\,dr’_{\\!\\perp} \\rho(r’_{\\!\\perp})
\n\\int_0^{2\\pi}\u00a0 d\\varphi’ \\frac{r_{\\!\\perp}^2 – r_{\\!\\perp} r’_{\\!\\perp}\u00a0 \\cos\\varphi’ }{r_{\\!\\perp}^2 + (r’_{\\!\\perp})^2 – 2r_{\\!\\perp} r’_{\\!\\perp}\u00a0 \\cos\\varphi’ } \\\\
\n&=& \\frac{\\boldsymbol{r}_{\\!\\perp}}{r^2_{\\!\\perp}}\\frac{1}{2\\pi\\varepsilon_0}
\n\\int_0^{r}\u00a0 r’_{\\!\\perp} \\,dr’_{\\!\\perp} \\rho(r’_{\\!\\perp}) \\times 2\\pi \\\\
\n&=& \\frac{Q_{r_{\\perp}}}{2\\pi\\varepsilon_0} \\frac{\\boldsymbol{r}_{\\!\\perp}}{r^2_{\\!\\perp}} \\\\ \\ \\\\
\nQ_{r_{\\perp}} &\\equiv& 2\\pi \\int_0^{r_{\\perp}}\u00a0 r’_{\\!\\perp} \\,dr’_{\\!\\perp} \\rho(r’_{\\!\\perp})
\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\u7b2c2\u8a71\u306f\uff0c\u8ef8\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-4787","post","type-post","status-publish","format-standard","hentry","category-19","nodate","item-wrap"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/wp-json\/wp\/v2\/posts\/4787","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/wp-json\/wp\/v2\/users\/33"}],"replies":[{"embeddable":true,"href":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/wp-json\/wp\/v2\/comments?post=4787"}],"version-history":[{"count":30,"href":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/wp-json\/wp\/v2\/posts\/4787\/revisions"}],"predecessor-version":[{"id":4935,"href":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/wp-json\/wp\/v2\/posts\/4787\/revisions\/4935"}],"wp:attachment":[{"href":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/wp-json\/wp\/v2\/media?parent=4787"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/wp-json\/wp\/v2\/categories?post=4787"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/wp-json\/wp\/v2\/tags?post=4787"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}