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磁場中における荷電粒子の軌跡

磁場中における荷電粒子の軌跡 (書きかけ)

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  • 極座標の運動方程式は以下のようになる。
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  • \newcommand{\dfrac}{\displaystyle\frac}
\begin{eqnarray*}
\left\{
\begin{array}{lll}
m \left[\dfrac{d^2r}{dt^2}-r\left(\dfrac{d\theta}{dt}\right)^2\right] = F_r \\
m \left(2\dfrac{dr}{dt}\dfrac{d\theta}{dt}+r\dfrac{d^2\theta}{dt^2}\right) = \dfrac{m}{r} \dfrac{d}{dt}\left(r^2\frac{d\theta}{dt}\right) = F_\theta \\
\end{array}
\right.
\end{eqnarray*}

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  • \newcommand{\dfrac}{\displaystyle\frac}
\begin{eqnarray*}
\left\{
\begin{array}{l}
m \left[\dfrac{d^2r}{dt^2}-r\left(\dfrac{d\theta}{dt}\right)^2\right]= qr\dfrac{d\theta}{dt}B(r) \\
m \left(2\dfrac{dr}{dt}\dfrac{d\theta}{dt}+r\dfrac{d^2\theta}{dt^2}\right) =\dfrac{m}{r} \dfrac{d}{dt}\left(r^2\dfrac{d\theta}{dt}\right) = -q\dfrac{dr}{dt}B(r)
\end{array}
\right.
\end{eqnarray*}

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  • \begin{eqnarray*}
m\frac{d}{dt}\left(r^2\frac{d\theta}{dt}\right) = -q\frac{dr}{dt}rB(r) =-q\frac{d}{dt}\left(\int_\infty^r r'B(r')dr'\right)\\
\frac{d}{dt}\left(mr^2\frac{d\theta}{dt}+q\int_\infty^r r'B(r')dr'\right) = 0\\
mr^2\frac{d\theta}{dt}+q\int_\infty^r r'B(r')dr' = L\\
\frac{d\theta}{dt} = \frac{1}{mr^2}\left(L-q\int_\infty^r r'B(r')dr'\right)
\end{eqnarray*}

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  • また運動エネルギーの式より
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  • \begin{eqnarray*}
\frac{1}{2}m\left[\left(\frac{dr}{dt}\right)^2 + r^2\left(\frac{d\theta}{dt}\right)^2 \right]= E\\
\frac{1}{2}m\left[\left(\frac{d\theta}{dt}\frac{dr}{d\theta}\right)^2 + r^2\left(\frac{d\theta}{dt}\right)^2 \right]= E\\
\frac{1}{2}m\left[\left(\frac{dr}{d\theta}\right)^2 + r^2 \right]\left(\frac{d\theta}{dt}\right)^2= E\\
\frac{1}{2}m\left[\left(\frac{dr}{d\theta}\right)^2 + r^2 \right]\left[\frac{1}{mr^2}\left(L-q\int_\infty^r r'B(r')dr'\right)\right]^2= E\\
\left(\frac{dr}{d\theta}\right)^2 + r^2 = \frac{2mr^4E}{\left(L-q\int_\infty^r r'B(r')dr'\right)^2}\\
\frac{dr}{d\theta} = r\sqrt{\frac{2mr^2E}{\left(L-q\int_\infty^r r'B(r')dr'\right)^2} - 1}\\
\end{eqnarray*}

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  • これより
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  • \begin{eqnarray*}
\frac{d\theta}{dr}
&=& \frac{1}{r\displaystyle\sqrt{\frac{2mr^2E}{\left(L-q\int_\infty^r r'B(r')dr'\right)^2} - 1}}\\
&=& \frac{L-q\int_\infty^r r'B(r')dr'}{r\sqrt{2mr^2E - \left(L-q\int_\infty^r r'B(r')dr'\right)^2}}\\
&=& \frac{L-q\int_\infty^r r'B(r')dr'}{\sqrt{2mE}}\frac{1}{r\sqrt{r^2 - \left(\frac{L-q\int_\infty^r r'B(r')dr'}{\sqrt{2mE}}\right)^2}}\\
\end{eqnarray*}

メモ

  • 準備
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    • \begin{eqnarray*}
\frac{dr}{dt} &=& \frac{dr}{d\theta}\frac{d\theta}{dt}\\
\frac{d^2r}{dt^2} &=& \frac{d}{dt}\left(\frac{dr}{d\theta}\frac{d\theta}{dt}\right)\\
 &=& \frac{d\theta}{dt}\frac{d}{dt}\left(\frac{dr}{d\theta}\right) + \frac{dr}{d\theta}\frac{d}{dt}\left(\frac{d\theta}{dt}\right)\\
 &=& \frac{d\theta}{dt}\frac{d}{d\theta}\left(\frac{dr}{d\theta}\right)\frac{d\theta}{dt} + \frac{dr}{d\theta}\frac{d^2\theta}{dt^2}\\
 &=& \frac{d^2r}{d\theta^2}\left(\frac{d\theta}{dt}\right)^2 + \frac{dr}{d\theta}\frac{d^2\theta}{dt^2}\\
\end{eqnarray*}

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  • 極座標の運動方程式は以下のようになる。
    •  

    • \begin{eqnarray*}
m \left[\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2\right] &=& F_r \\
m \left(2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}\right) &=& \frac{m}{r} \frac{d}{dt}\left(r^2\frac{d\theta}{dt}\right) = F_\theta \\
\end{eqnarray*}

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    • \begin{eqnarray*}
m \left[\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2\right] &=& qr\frac{d\theta}{dt}B(r) \\
m \left(2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}\right) &=& \frac{m}{r} \frac{d}{dt}\left(r^2\frac{d\theta}{dt}\right) = -q\frac{dr}{dt}B(r) \\
\end{eqnarray*}

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    • \begin{eqnarray*}
\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2 &=& r\frac{d\theta}{dt}b(r) \\
2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2} &=& \frac{1}{r} \frac{d}{dt}\left(r^2\frac{d\theta}{dt}\right) = -\frac{dr}{dt}b(r) \\
\end{eqnarray*}

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    • \begin{eqnarray*}
r\frac{d^2\theta}{dt^2} &=& - 2\frac{dr}{dt}\frac{d\theta}{dt}-\frac{dr}{dt}b(r) \\
\frac{d^2\theta}{dt^2} &=& - \frac{2}{r}\frac{dr}{dt}\frac{d\theta}{dt}-\frac{1}{r}\frac{dr}{dt}b(r) \\
&=& -\frac{dr}{dt}\left(\frac{2}{r}\frac{d\theta}{dt}+\frac{1}{r}b(r) \right)\\
&=& -\frac{dr}{d\theta}\frac{d\theta}{dt}\left(\frac{2}{r}\frac{d\theta}{dt}+\frac{1}{r}b(r) \right)\\
\end{eqnarray*}

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    • \begin{eqnarray*}
\frac{d^2r}{dt^2}
 &=& \frac{d^2r}{d\theta^2}\left(\frac{d\theta}{dt}\right)^2 + \frac{dr}{d\theta}\frac{d^2\theta}{dt^2}\\
 &=& \frac{d^2r}{d\theta^2}\left(\frac{d\theta}{dt}\right)^2 - \left(\frac{dr}{d\theta}\right)^2\frac{d\theta}{dt}\left(\frac{2}{r}\frac{d\theta}{dt}+\frac{1}{r}b(r) \right)\\
\end{eqnarray*}

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    • \begin{eqnarray*}
\frac{d^2r}{d\theta^2}\left(\frac{d\theta}{dt}\right)^2 - \left(\frac{dr}{d\theta}\right)^2\frac{d\theta}{dt}\left(\frac{2}{r}\frac{d\theta}{dt}+\frac{1}{r}b(r) \right)-r\left(\frac{d\theta}{dt}\right)^2 &=& r\frac{d\theta}{dt}b(r) \\
\frac{d^2r}{d\theta^2}\frac{d\theta}{dt} - \left(\frac{dr}{d\theta}\right)^2\left(\frac{2}{r}\frac{d\theta}{dt}+\frac{1}{r}b(r) \right)-r\frac{d\theta}{dt} &=& rb(r) \\
\frac{d^2r}{d\theta^2}\frac{d\theta}{dt} - \left(\frac{dr}{d\theta}\right)^2\left(\frac{2}{r}\frac{d\theta}{dt}+\frac{1}{r}b(r) \right)+r\frac{d\theta}{dt} - r^2\left(\frac{2}{r}\frac{d\theta}{dt} + \frac{1}{r}b(r)\right)&=& 0 \\
\frac{d^2r}{d\theta^2}\frac{d\theta}{dt} - \left(\left(\frac{dr}{d\theta}\right)^2+r^2\right)\left(\frac{2}{r}\frac{d\theta}{dt}+\frac{1}{r}b(r) \right)+r\frac{d\theta}{dt}&=& 0 \\
\frac{d^2r}{d\theta^2} - \left(\left(\frac{dr}{d\theta}\right)^2+r^2\right)\left(\frac{2}{r}+\left.\frac{b(r)}{r}\right/\frac{d\theta}{dt} \right)+r&=& 0 \\
\end{eqnarray*}

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    • \begin{eqnarray*}
m\frac{d}{dt}\left(r^2\frac{d\theta}{dt}\right) = -qr\frac{dr}{dt}B(r) =-\frac{1}{2}q\frac{dr^2}{dt}B(r)\\
\frac{d}{dt}\left(mr^2\frac{d\theta}{dt}+\frac{1}{2}qr^2B(r)\right) = 0 \\
\end{eqnarray*}

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    • \begin{eqnarray*}
mr^2\frac{d\theta}{dt}+\frac{1}{2}qr^2B(r) = L \\
\frac{d\theta}{dt} = \frac{L}{mr^2}-\frac{q}{2m}B(r)
\end{eqnarray*}

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    • \begin{eqnarray*}
\frac{dr}{dt} &=& \frac{dr}{d\theta}\frac{d\theta}{dt} = -r^2\frac{du}{d\theta} \left( \frac{L}{mr^2}-\frac{q}{2m}B(r)\right)= \frac{du}{d\theta} \left( -\frac{L}{m}+\frac{qr^2}{2m}B(r)\right) \\
\frac{d^2r}{dt^2} &=& \frac{d}{dt} \left[\frac{du}{d\theta} \left( -\frac{L}{m}+\frac{qr^2}{2m}B(r)\right)\right]= \frac{d}{d\theta}\left(\frac{du}{d\theta}\right)\frac{d\theta}{dt}\left(-\frac{L}{m}+\frac{qr^2}{2m}B(r)\right)+\frac{du}{d\theta}\frac{q}{2m}B(r)\frac{dr^2}{dt}\\
&=& -\frac{d^2u}{d\theta^2}\frac{1}{u^2}\left(-\frac{L}{m}+\frac{q}{2mu^2}B(r)\right)^2+\frac{du}{d\theta}\frac{q}{2m}B(r)\frac{dr^2}{d\theta}\frac{d\theta}{dt}\\
&=& -\frac{d^2u}{d\theta^2}\frac{1}{u^2}\left(-\frac{L}{m}+\frac{q}{2mu^2}B(r)\right)^2-2\frac{1}{u^3}\left(\frac{du}{d\theta}\right)^2\frac{q}{2m}B(r)\frac{d\theta}{dt}\\
\end{eqnarray*}