{"id":7414,"date":"2025-01-30T11:30:05","date_gmt":"2025-01-30T02:30:05","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?p=7414"},"modified":"2025-06-07T10:22:19","modified_gmt":"2025-06-07T01:22:19","slug":"%e9%ab%98%e3%81%95-h-%e3%81%8b%e3%82%89%e3%81%ae%e6%96%9c%e6%96%b9%e6%8a%95%e5%b0%84%e3%81%ae%e6%9c%80%e5%a4%a7%e6%b0%b4%e5%b9%b3%e5%88%b0%e9%81%94%e8%b7%9d%e9%9b%a2%e3%82%92%e6%b1%82%e3%82%81","status":"publish","type":"post","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/7414\/","title":{"rendered":"\u9ad8\u3055 h \u304b\u3089\u306e\u659c\u65b9\u6295\u5c04\u306e\u6700\u5927\u6c34\u5e73\u5230\u9054\u8ddd\u96e2\u3092\u89e3\u6790\u7684\u306b\u6c42\u3081\u308b"},"content":{"rendered":"

\u300c\u9ad8\u3055 h \u304b\u3089\u306e\u659c\u65b9\u6295\u5c04\u306e\u6700\u5927\u5230\u9054\u8ddd\u96e2\u3092\u6c42\u3081\u308b\u6e96\u5099<\/a>\u300d\u306e\u30da\u30fc\u30b8\u3092\u53c2\u7167\u3002<\/p>\n

\u6700\u5927\u6c34\u5e73\u5230\u9054\u8ddd\u96e2\u3068\u306a\u308b\u6253\u3061\u51fa\u3057\u89d2\u5ea6 $\\theta$ \u3092\u6c42\u3081\u308b\u306b\u306f\uff0c\u5e73\u65b9\u6839\u3092\u542b\u3080\u975e\u7dda\u5f62\u65b9\u7a0b\u5f0f\u3092\u89e3\u304f\u5fc5\u8981\u304c\u3042\u308b\u306e\u3067\uff0cSymPy \u3067\u3084\u3063\u3066\u307f\u305f\u3002<\/p>\n

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\u30e9\u30a4\u30d6\u30e9\u30ea\u306e import<\/h3>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[1]:<\/div>\n
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# 1\u6587\u5b57\u5909\u6570\u306e Symbol \u306e\u5ba3\u8a00\u304c\u7701\u7565\u3067\u304d\u308b<\/span>\r\nfrom<\/span> sympy.abc<\/span> import<\/span> *<\/span>\r\n\r\nfrom<\/span> sympy<\/span> import<\/span> *<\/span>\r\n\r\ninit_printing<\/span>()<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u6700\u5927\u6c34\u5e73\u5230\u9054\u8ddd\u96e2\u3068\u306a\u308b $\\sin\\theta_{\\rm max}$ \u3092\u89e3\u6790\u7684\u306b\u6c42\u3081\u308b<\/h3>\n
\u6c34\u5e73\u5230\u9054\u8ddd\u96e2 $L$ \u3092 $s \\equiv \\sin \\theta$ \u3067\u8868\u3059<\/h4>\n
\u6c34\u5e73\u5230\u9054\u8ddd\u96e2 $L(\\theta)$ \u3092 $s \\equiv \\sin \\theta$ \u3092\u4f7f\u3063\u3066\u8868\u3057\u3066\u307f\u307e\u3059\u3002<\/p>\n
\\begin{eqnarray}
\nL(\\theta) &=& \\left( \\sin{\\left(\\theta \\right)} + \\sqrt{2 H + \\sin^{2}{\\left(\\theta \\right)}}\\right) \\cos{\\left(\\theta \\right)} \\\\
\n&=& \\left(s + \\sqrt{s^2 + 2H}\\right) \\, \\sqrt{1 -s^2}
\n\\end{eqnarray}<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[2]:<\/div>\n
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var<\/span>(<\/span>'H s'<\/span>,<\/span> positive<\/span> =<\/span> True<\/span>)<\/span> # H, s > 0<\/span>\r\n\r\ndef<\/span> Ls<\/span>(<\/span>s<\/span>):<\/span>\r\n    return<\/span> (<\/span>s<\/span> +<\/span> sqrt<\/span>(<\/span>2<\/span>*<\/span>H<\/span> +<\/span> s<\/span>**<\/span>2<\/span>))<\/span> *<\/span> sqrt<\/span>(<\/span>1<\/span> -<\/span> s<\/span>**<\/span>2<\/span>)<\/span>\r\nLs<\/span>(<\/span>s<\/span>)<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[2]:<\/div>\n
$\\displaystyle \\sqrt{1 -s^{2}} \\left(s + \\sqrt{2 H + s^{2}}\\right)$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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$L$ \u304c\u6700\u5927\u3068\u306a\u308b $\\sin\\theta$ \u3092\u6c42\u3081\u308b<\/h4>\n
dLs<\/code>$\\displaystyle \\equiv\\frac{d L(s)}{d s} = 0$ \u3068\u306a\u308b $s \\equiv s_{\\rm max}$ \u3092 solve()<\/code> \u3067\u6c42\u3081\u307e\u3059\u3002<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[3]:<\/div>\n
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dLs<\/span> =<\/span> diff<\/span>(<\/span>Ls<\/span>(<\/span>s<\/span>),<\/span> s<\/span>)<\/span>\r\ndLs<\/span> =<\/span> dLs<\/span>.<\/span>simplify<\/span>()<\/span>\r\ndLs<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[3]:<\/div>\n
$\\displaystyle \\frac{\\left(s + \\sqrt{2 H + s^{2}}\\right) \\left(-s^{2} -s \\sqrt{2 H + s^{2}} + 1\\right)}{\\sqrt{1 -s^{2}} \\sqrt{2 H + s^{2}}}$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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dLs = 0<\/code> \u3092 s<\/code> \u306b\u3064\u3044\u3066\u89e3\u304f\u3002<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[4]:<\/div>\n
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solve<\/span>(<\/span>dLs<\/span>,<\/span> s<\/span>)<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[4]:<\/div>\n
$\\displaystyle \\left[ \\frac{\\sqrt{2}}{2 \\sqrt{H + 1}}\\right]$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u3068\u3044\u3046\u3053\u3068\u3067\uff0c\u6700\u5927\u6c34\u5e73\u5230\u9054\u8ddd\u96e2\u3068\u306a\u308b $\\sin\\theta_{\\rm max}$ \u306f<\/p>\n
\\begin{eqnarray}
\n\\sin\\theta_{\\rm max} &=& s_{\\rm max} = \\frac{1}{\\sqrt{2 (1 + H)}} \\\\
\n\\therefore\\ \\ \\theta_{\\rm max} &=& \\arcsin\\left(\\frac{1}{\\sqrt{2 (1 + H)}}\\right)
\n\\end{eqnarray}<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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\u6700\u5927\u6c34\u5e73\u5230\u9054\u8ddd\u96e2<\/h4>\n
\u305d\u306e\u3068\u304d\u306e\u6700\u5927\u6c34\u5e73\u5230\u9054\u8ddd\u96e2 $L_{\\rm max} = L(\\theta_{\\rm max}, H)$ \u306f…<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[5]:<\/div>\n
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smax<\/span> =<\/span> 1<\/span>\/<\/span>sqrt<\/span>(<\/span>2<\/span>*<\/span>(<\/span>1<\/span>+<\/span>H<\/span>))<\/span>\r\nLs<\/span>(<\/span>smax<\/span>)<\/span>.<\/span>simplify<\/span>()<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[5]:<\/div>\n
$\\displaystyle \\frac{\\sqrt{2 H + 1} \\left(\\sqrt{4 H \\left(H + 1\\right) + 1} + 1\\right)}{2 \\left(H + 1\\right)}$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u7c21\u5358\u5316\u3057\u3066\u304f\u308c\u306a\u3044\u306a\u3041…<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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\u6700\u5927\u6c34\u5e73\u5230\u9054\u8ddd\u96e2\u3068\u306a\u308b $\\tan\\theta_{\\rm max}$ \u3092\u89e3\u6790\u7684\u306b\u6c42\u3081\u308b<\/h3>\n
\u6c34\u5e73\u5230\u9054\u8ddd\u96e2 $L(\\theta)$ \u3092 $t \\equiv \\tan \\theta$ \u3092\u4f7f\u3063\u3066\u8868\u3057\u3066\u307f\u307e\u3059\u3002<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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\u6c34\u5e73\u5230\u9054\u8ddd\u96e2 $L$ \u3092 $t \\equiv \\tan\\theta$ \u3067\u8868\u3059<\/h4>\n
$t \\equiv \\tan\\theta$ \u3068\u3057\u3066\u898f\u683c\u5316\u3055\u308c\u305f\u6c34\u5e73\u5230\u9054\u8ddd\u96e2 $L$ \u3092 $t$ \u3067\u8868\u3059\u3068<\/p>\n
\\begin{eqnarray}
\nL(t)
\n&=& \\frac{1}{1+t^2} \\left\\{t + \\sqrt{(1+2H)\\, t^2 + 2H} \\right\\}
\n\\end{eqnarray}<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[6]:<\/div>\n
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var<\/span>(<\/span>'t H'<\/span>,<\/span> positive<\/span> =<\/span> True<\/span>)<\/span> # positive = True \u3068\u3059\u308b\u3068\u5409<\/span>\r\n\r\ndef<\/span> Lt<\/span>(<\/span>t<\/span>):<\/span>\r\n    return<\/span> (<\/span>t<\/span>+<\/span>sqrt<\/span>((<\/span>1<\/span>+<\/span>2<\/span>*<\/span>H<\/span>)<\/span>*<\/span>t<\/span>**<\/span>2<\/span> +<\/span> 2<\/span>*<\/span>H<\/span>))<\/span>\/<\/span>(<\/span>1<\/span> +<\/span> t<\/span>**<\/span>2<\/span>)<\/span>\r\nLt<\/span>(<\/span>t<\/span>)<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[6]:<\/div>\n
$\\displaystyle \\frac{t + \\sqrt{2 H + t^{2} \\left(2 H + 1\\right)}}{t^{2} + 1}$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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$L$ \u304c\u6700\u5927\u3068\u306a\u308b $\\tan\\theta$ \u3092\u6c42\u3081\u308b<\/h4>\n<\/div>\n<\/div>\n<\/div>\n
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$\\displaystyle \\frac{d L(t)}{dt} = 0$ \u3068\u306a\u308b $t \\equiv t_{\\rm max}$ \u3092 solve()<\/code> \u3067\u6c42\u3081\u307e\u3059\u3002<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[7]:<\/div>\n
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dLt<\/span> =<\/span> diff<\/span>(<\/span>Lt<\/span>(<\/span>t<\/span>),<\/span> t<\/span>)<\/span>\r\nsols<\/span> =<\/span> solve<\/span>(<\/span>dLt<\/span>,<\/span> t<\/span>)<\/span>\r\nsols<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[7]:<\/div>\n
$\\displaystyle \\left[ \\frac{1}{\\sqrt{2 H + 1}}\\right]$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u5ff5\u306e\u305f\u3081\uff0c\u672c\u5f53\u306b\u89e3\u306b\u306a\u3063\u3066\u3044\u308b\u304b\uff0c\u78ba\u8a8d\u3057\u307e\u3059\u3002<\/p>\n
dLt.subs(t, sols[0])<\/code> \u3068\u3057\u3066\uff0cdLt<\/code> \u306e t<\/code> \u306b sols[0]<\/code> \u3092\u4ee3\u5165\u3057\uff0csimplify()<\/code> \u3067\u7c21\u5358\u5316\u3057\u3066\u307f\u307e\u3059\u3002<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[8]:<\/div>\n
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simplify<\/span>(<\/span>dLt<\/span>.<\/span>subs<\/span>(<\/span>t<\/span>,<\/span> sols<\/span>[<\/span>0<\/span>]))<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[8]:<\/div>\n
$\\displaystyle 0$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u3088\u3063\u3066\uff0c$L(t)$ \u304c\u6700\u5927\u3068\u306a\u308b $t$ \u306e\u5024 $t_{\\rm max}$ \u306f…<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[9]:<\/div>\n
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tmax<\/span> =<\/span>  1<\/span>\/<\/span>sqrt<\/span>(<\/span>1<\/span> +<\/span> 2<\/span>*<\/span>H<\/span>)<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u6700\u5927\u6c34\u5e73\u5230\u9054\u8ddd\u96e2<\/h4>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[10]:<\/div>\n
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simplify<\/span>(<\/span>Lt<\/span>(<\/span>tmax<\/span>))<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[10]:<\/div>\n
$\\displaystyle \\sqrt{2 H + 1}$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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\u3064\u307e\u308a\uff0c\u9ad8\u3055 $H$ \u306e\u3068\u304d\u306e\u6700\u5927\u6c34\u5e73\u5230\u9054\u8ddd\u96e2\u306f\uff0c<\/p>\n
$$L_{\\rm max} = \\sqrt{1 + 2H}$$<\/p>\n
\u3068\u306a\u308b\u3053\u3068\u304c\u308f\u304b\u308a\u307e\u3059\u3002<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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\u3053\u306e\u3068\u304d\u306e\u6ede\u7a7a\u6642\u9593 $T_1(\\theta)$ \u306f…<\/h4>\n
$$T_1(\\theta) = \\sin\\theta + \\sqrt{\\sin^2\\theta + 2 H}$$<\/p>\n
\u306b<\/p>\n
\\begin{eqnarray}
\nt_{\\rm max} &=& \\tan\\theta_{\\rm max} = \\frac{1}{\\sqrt{1 + 2 H}} \\\\
\n\\therefore\\ \\ \\theta_{\\rm max} &=& \\arctan \\left(\\frac{1}{\\sqrt{1 + 2 H}}\\right)
\n\\end{eqnarray}<\/p>\n
\u3067\u6c7a\u307e\u308b $\\theta_{\\rm max}$ \u3092\u4ee3\u5165\u3057\u3066…<\/p>\n
\u3042\u308b\u3044\u306f\uff0c<\/p>\n
\\begin{eqnarray}
\ns_{\\rm max} &=& \\sin\\theta_{\\rm max} = \\frac{1}{\\sqrt{2(1 + H)}}
\n\\end{eqnarray}<\/p>\n
\u3092\u4f7f\u3063\u3066…<\/p>\n<\/div>\n<\/div>\n<\/div>\n
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In\u00a0[11]:<\/div>\n
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T1<\/span> =<\/span> smax<\/span> +<\/span> sqrt<\/span>(<\/span>factor<\/span>(<\/span>smax<\/span>**<\/span>2<\/span> +<\/span> 2<\/span>*<\/span>H<\/span>))<\/span>\r\nT1<\/span>.<\/span>factor<\/span>()<\/span>\r\n<\/pre>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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Out[11]:<\/div>\n
$\\displaystyle \\sqrt{2} \\sqrt{H + 1}$<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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$$T_1 = \\sin\\theta_{\\rm max} + \\sqrt{\\sin^2\\theta_{\\rm max} + 2H} = \\sqrt{2 (1 + H)}$$<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"
\u300c\u9ad8\u3055 h \u304b\u3089\u306e\u659c\u65b9\u6295\u5c04\u306e\u6700\u5927\u5230\u9054\u8ddd\u96e2\u3092\u6c42\u3081\u308b\u6e96\u5099\u300d\u306e\u30da\u30fc\u30b8\u3092\u53c2\u7167\u3002<\/p>\n
\u6700\u5927\u6c34\u5e73\u5230\u9054\u8ddd\u96e2\u3068\u306a\u308b\u6253\u3061\u51fa\u3057\u89d2\u5ea6 $\\theta$ \u3092\u6c42\u3081\u308b\u306b\u306f\uff0c\u5e73\u65b9\u6839\u3092\u542b\u3080\u975e\u7dda\u5f62\u65b9\u7a0b\u5f0f\u3092\u89e3\u304f\u5fc5\u8981\u304c\u3042\u308b\u306e\u3067\uff0cSymPy \u3067\u3084\u3063\u3066\u307f\u305f\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":[12,25],"tags":[],"class_list":["post-7414","post","type-post","status-publish","format-standard","hentry","category-sympy","category-25","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