물리학 부록

물리학 부록

기초

부록A-1

$$ \begin{align*} \frac{ d |x| } {d x} &= \frac{d \sqrt{x^2} }{d x} \\ &= \frac{d \sqrt{x^2}}{d x^2} \frac{d x^2}{dx} \\ &= \frac{1}{2}\frac{1}{\sqrt{x^2}} \cdot 2x \\ &= \dfrac{1}{|x|}x \end{align*} $$

전자기학

부록E-1

원 글로 가기 $$ \delta \big( f(x) \big) =\sum \limits_{x_0} \frac{\delta(x-x_0)}{ \frac{\partial f}{\partial x}\Big|_{x=x_0} } $$ 이때 $x_0$는 $f(x)$의 해이다. 위의 사실을 이용하면 $$ \delta \left( t'-t+\frac{ | \mathbf{r} -\mathbf{w}(t') | }{c} \right) = \frac{ \delta(t'-t_r) }{ 1+\frac{\partial}{\partial t'}\left( \frac{|\mathbf{r}-\mathbf{w}(t')|}{c} \right)\Big| _{t'=t_r} } $$ $\frac{ d}{dx} f(g(x))^{\frac{1}{2}} =\dfrac{1}{2\sqrt{ f(g(x)) }}g'(x)$이므로 우변의 분모를 풀어보면 $$ \begin{align*} 1+\frac{\partial}{\partial t'}\left( \frac{|\mathbf{r}-\mathbf{w}(t')|}{c} \right) &= 1+\frac{1}{c} \dfrac{ \partial }{\partial t'} \left( \left[\big( \mathbf{r}-\mathbf{w}(t') \big) \cdot \big( \mathbf{r}-\mathbf{w}(t') \big) \right]^{\frac{1}{2}} \right) \\ &= 1+\frac{1}{c} \dfrac{1}{2 \sqrt{ \big( \mathbf{r}-\mathbf{w}(t') \big) \cdot \big( \mathbf{r}-\mathbf{w}(t') \big)} } \dfrac{\partial}{\partial t'} \bigg( \big( \mathbf{r}-\mathbf{w}(t') \big)\cdot \big( \mathbf{r}-\mathbf{w}(t') \big) \bigg) \\ &= 1+\frac{1}{c} \dfrac{1}{2 |\mathbf{r}-\mathbf{w}(t')|} 2\big(\mathbf{r}-\mathbf{w}(t') \big)\cdot\dfrac{\partial}{\partial t'} \big( \mathbf{r}-\mathbf{w}(t') \big) \\ &= 1 +\frac{ \mathbf{r}-\mathbf{w}(t') }{c | \mathbf{r}-\mathbf{w}(t')|} \cdot \big(-\mathbf{v}(t') \big) \\ &= 1-\frac{ [ \mathbf{r}-\mathbf{w}(t') ]\cdot \mathbf{v}(t') }{c | \mathbf{r}-\mathbf{w}(t')|} \end{align*} $$ 여기에 $t'=t_r$을 대입하면 $$ 1-\frac{ [\mathbf{r}-\mathbf{w}(t_r)] \cdot \mathbf{v}(t_r) }{c | \mathbf{r}-\mathbf{w}(t_r)|} =1-\frac{ \boldsymbol{\eta} \cdot \mathbf{v}(t_r) }{c \eta} = 1-\frac{ \hat{\boldsymbol{\eta}} \cdot \mathbf{v}(t_r) }{c } $$ 따라서 $$ \delta \left( t'-t+\frac{ | \mathbf{r} -\mathbf{w}(t') | }{c} \right) =\frac{ \delta(t'-t_r) } {1-\frac{ \hat{\boldsymbol{\eta}} \cdot \mathbf{v}(t_r) }{c } } $$

양자역학

부록Q-1

원 글로 가기

$$ \mathbb{A} = \begin{pmatrix} e^{-ika} & e^{ika} \\ ike^{-ika} & -ike^{ika} \end{pmatrix},\quad \mathbb{B} = \begin{pmatrix} e^{\kappa a} & e^{-\kappa a} \\ \kappa e^{\kappa a} & -\kappa e^{-\kappa a} \end{pmatrix} $$ 여인수 전개로 위 두 행렬의 역행렬을 구하면 $$ \begin{align*} \mathbb{A}^{-1} &= \frac{1}{|\mathbb{A}|} \mathrm{adj}(A) \\ &= \frac{1}{-ik -ik} \begin{pmatrix} -ike^{ika} & -ike^{-ika} \\ -e^{ika} & e^{-ika} \end{pmatrix}^T \\ &= \frac{1}{-2ik} \begin{pmatrix} -ike^{ika} & -e^{ika} \\ -ike^{-ika} & e^{-ika} \end{pmatrix} \\ &= \frac{1}{2} \begin{pmatrix} e^{ika} & \frac{1}{ik}e^{ika} \\ e^{-ika} & \frac{-1}{ik} e^{-ika} \end{pmatrix} \end{align*} $$

$$ \begin{align*} \mathbb{B}^{-1} &= \frac{1}{|\mathbb{B}|} \mathrm{adj}(B) \\ &= \frac{1}{-\kappa -\kappa } \begin{pmatrix} -\kappa e^{-\kappa a} & -\kappa e^{\kappa a} \\ -e^{-\kappa a} & e^{\kappa a} \end{pmatrix}^T \\ &= \frac{1}{-2\kappa } \begin{pmatrix} -\kappa e^{-\kappa a} & -e^{-\kappa a} \\ -\kappa e^{\kappa a} & e^{\kappa a} \end{pmatrix} \\ &= \frac{1}{2} \begin{pmatrix} e^{-\kappa a} & \frac{1}{\kappa }e^{-\kappa a} \\ e^{\kappa a} & \frac{-1}{\kappa } e^{\kappa a} \end{pmatrix} \end{align*} $$ 그러면 $$ \begin{align*} \begin{pmatrix} A_{+} \\ A_{-} \end{pmatrix} &= \mathbb{A}^{-1} \begin{pmatrix} e^{-\kappa a} & e^{\kappa a} \\ \kappa e^{-\kappa a} & -\kappa e^{\kappa a} \end{pmatrix} \mathbb{B}^{-1} \begin{pmatrix} e^{ik a} & e^{-ik a} \\ ik e^{ik a} & -ik e^{-ika } \end{pmatrix} \begin{pmatrix} C_{+} \\ 0 \end{pmatrix} \\ &= \frac{1}{2} \begin{pmatrix} e^{ika} & \frac{1}{ik}e^{ika} \\ e^{-ika} & \frac{-1}{ik} e^{-ika} \end{pmatrix} \begin{pmatrix} e^{-\kappa a} & e^{\kappa a} \\ \kappa e^{-\kappa a} & -\kappa e^{\kappa a} \end{pmatrix} \frac{1}{2} \begin{pmatrix} e^{-\kappa a} & \frac{1}{\kappa }e^{-\kappa a} \\ e^{\kappa a} & \frac{-1}{\kappa } e^{\kappa a} \end{pmatrix} \begin{pmatrix} e^{ik a} & e^{-ik a} \\ ik e^{ik a} & -ik e^{-ika } \end{pmatrix} \begin{pmatrix} C_{+} \\ 0 \end{pmatrix} \\ &= \frac{1}{2^2} \begin{pmatrix} e^{ika} & \frac{1}{ik}e^{ika} \\ e^{-ika} & \frac{-1}{ik} e^{-ika} \end{pmatrix} \begin{pmatrix} e^{-\kappa a} & e^{\kappa a} \\ \kappa e^{-\kappa a} & -\kappa e^{\kappa a} \end{pmatrix} \begin{pmatrix} e^{-\kappa a} & \frac{1}{\kappa }e^{-\kappa a} \\ e^{\kappa a} & \frac{-1}{\kappa } e^{\kappa a} \end{pmatrix} \begin{pmatrix} C_{+}e^{ika} \\ C_{+}ike^{ika} \end{pmatrix} \end{align*} $$ 두번째, 세번째 행렬을 먼저 계산해보자. $$ \begin{align*} \begin{pmatrix} e^{-\kappa a} & e^{\kappa a} \\ \kappa e^{-\kappa a} & -\kappa e^{\kappa a} \end{pmatrix} \begin{pmatrix} e^{-\kappa a} & \frac{1}{\kappa }e^{-\kappa a} \\ e^{\kappa a} & \frac{-1}{\kappa } e^{\kappa a} \end{pmatrix} &= \begin{pmatrix} e^{-2\kappa a} +e^{2\kappa } & \frac{1}{\kappa} (e^{-2\kappa a } - e^{\kappa a} ) \\ \kappa (e^{-2\kappa a } - e^{2\kappa a } ) & e^{-2\kappa a} + e^{2\kappa a} \end{pmatrix} \\ &= \begin{pmatrix} 2\cosh (2\kappa a) & -\frac{1}{\kappa} 2\sinh (2\kappa a) \\ -2\kappa \sinh(2\kappa a ) & 2\cosh (2\kappa a) \end{pmatrix} \end{align*} $$ 이를 원래 식에 대입해서 계산하면 $$ \begin{align*} \begin{pmatrix} A_{+} \\ A_{-} \end{pmatrix} &= \frac{1}{2} \begin{pmatrix} e^{ika} & \frac{1}{ik}e^{ika} \\ e^{-ika} & \frac{-1}{ik} e^{-ika} \end{pmatrix} \begin{pmatrix} \cosh (2\kappa a) & -\frac{1}{\kappa} \sinh (2\kappa a) \\ -\kappa \sinh(2\kappa a ) & \cosh (2\kappa a) \end{pmatrix} \begin{pmatrix} C_{+}e^{ika} \\ C_{+}ike^{ika} \end{pmatrix} \\ &= \frac{1}{2} \begin{pmatrix} e^{ika} \big(\cosh (2\kappa a) -\frac{\kappa}{ik}\sinh (2\kappa a) \big) & e^{ika} \big( -\frac{1}{\kappa} \sinh (2\kappa a) + \frac{1}{ik}\cosh (2\kappa a) \big) \\ e^{-ika} \big( \cosh (2\kappa a ) + \frac{\kappa}{ik} \sinh (2\kappa a ) \big) & e^{-ika} \big( -\frac{1}{\kappa } \sinh (2\kappa a) - \frac{1}{ik}\cosh (2\kappa a) \big) \end{pmatrix} \begin{pmatrix} e^{ika} \\ ik e^{ika} \end{pmatrix} C_+ \end{align*} $$ 그러면 $$ \begin{align*} A_{+} &= \frac{1}{2} \left[ e^{i2ka} \big(\cosh (2\kappa a) -\frac{\kappa}{ik}\sinh (2\kappa a) \big) + ike^{i2ka} \big( -\frac{1}{\kappa} \sinh (2\kappa a) + \frac{1}{ik}\cosh (2\kappa a) \big) \right]C_{+} \\ &= \frac{1}{2} e^{i2ka} \bigg[\cosh (2\kappa a) -\frac{\kappa}{ik}\sinh (2\kappa a) -\frac{ik}{\kappa} \sinh (2\kappa a) + \cosh (2\kappa a) \bigg] C_{+} \\ &= \frac{1}{2} e^{i2ka} \left[ 2\cosh (2\kappa a ) + i\left( \frac{\kappa}{k} -\frac{k}{\kappa} \right) \sinh (2\kappa a) \right] \\ &= \frac{1}{2} e^{i2ka} \Big( 2\cosh (2\kappa a) +i2\eta \sinh (2\kappa a) \Big) C_+ \\ &= e^{i2ka} \Big( \cosh (2\kappa a ) + i\eta \sinh (2\kappa a) \Big) C_+ \end{align*} $$ 이 때 $\dfrac{\kappa}{k}-\dfrac{k}{\kappa}=2\eta$로 치환하였다. 따라서 $$ \dfrac{C_{+}} {A_{+}} = \frac{e^{-i2ka}} {\cosh (2\kappa a) + i\eta \sinh (2\kappa a) } $$ 또한 $$ \begin{align*} A_{-} &= \frac{1}{2} \left[ \cosh (2\kappa a) +\frac{\kappa}{ik}\sinh (2\kappa a) -\frac{ik}{\kappa} \sinh (2\kappa a)-\cosh (2\kappa a) \right]C_{+} \\ &= \dfrac{1}{2} \sinh (2\kappa a) \left( \dfrac{\kappa}{ik}-\dfrac{ik}{\kappa} \right) C_+ \\ &= \dfrac{-i}{2} \sinh (2\kappa a) \left( \dfrac{\kappa}{k} + \frac{k}{\kappa} \right) C_+ \\ &= -i\xi \sinh (2\kappa a) C_{+} \end{align*} $$ 이때 $\dfrac{\kappa}{k}+\dfrac{k}{\kappa}=2\xi$로 치환하였다. 따라서 $$ \dfrac{A_{-}}{A_{+}} = \dfrac{-i\xi \sinh (2\kappa a) C_{+} } {e^{i2ka} \Big( \cosh (2\kappa a ) + i\eta \sinh (2\kappa a) \Big) C_+} = \dfrac{-i\xi \sinh (2\kappa a) e^{-i2ka} } { \cosh (2\kappa a ) + i\eta \sinh (2\kappa a)} $$

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