Problem 6.9#
Starting with Equation 6.74:
\[\begin{equation*}
\sin\left(\frac{1}{2}\,\omega\,\Delta t\right) = c\,\Delta t \:\left[ \left( \frac{\sin\left( \frac{1}{2}k_x\,\Delta x \right)}{\Delta x}\right)^2 + \left( \frac{\sin\left( \frac{1}{2}k_y\,\Delta y \right)}{\Delta y}\right)^2 \right]^{1/2}
\end{equation*}\]
Taking the derivative of this with respect to the wavevector in the x direction, \(k_x\), remembering that \(\omega \sim \omega(k_x, \, k_y)\) we get
\[\begin{equation*}
\frac{1}{2} \, \Delta t \cos\left(\frac{1}{2}\,\omega\,\Delta t\right) \frac{\partial \omega}{\partial k_x} = \frac{1}{2} \, c\,\Delta t \:\left[ \left( \frac{\sin\left( \frac{1}{2}k_x\,\Delta x \right)}{\Delta x}\right)^2 + \left( \frac{\sin\left( \frac{1}{2}k_y\,\Delta y \right)}{\Delta y}\right)^2 \right]^{\:- 1/2} \, \frac{\cos \left( \frac{1}{2}k_x\,\Delta x \right) \sin \left( \frac{1}{2}k_x\,\Delta x \right)}{\Delta x}
\end{equation*}\]
Noting here that we can rearrange Eqn. 6.74 to
\[\begin{equation*}
\left[ \left( \frac{\sin\left( \frac{1}{2}k_x\,\Delta x \right)}{\Delta x}\right)^2 + \left( \frac{\sin\left( \frac{1}{2}k_y\,\Delta y \right)}{\Delta y}\right)^2 \right]^{\:-1/2} = \frac{ c\,\Delta t } {\sin\left(\frac{1}{2}\,\omega\,\Delta t\right)}
\end{equation*}\]
and substitute it in to yield
\[\begin{equation*}
\frac{1}{2} \, \Delta t \cos\left(\frac{1}{2}\,\omega\,\Delta t\right) \frac{\partial \omega}{\partial k_x} = \frac{1}{2} \: \frac{ c^2 \,(\Delta t)^2 } {\sin\left(\frac{1}{2}\,\omega\,\Delta t\right)} \, \frac{\cos \left( \frac{1}{2}k_x\,\Delta x \right) \sin \left( \frac{1}{2}k_x\,\Delta x \right)}{\Delta x}
\end{equation*}\]
Rearranging this slightly to get
\[\begin{equation*}
\frac{\partial \omega}{\partial k_x} = \frac{ c^2 \Delta t }{\Delta x} \, \frac{\cos \left( \frac{1}{2}k_x\,\Delta x \right) \sin \left( \frac{1}{2}k_x\,\Delta x \right)}{ \cos\left(\frac{1}{2}\,\omega\,\Delta t\right) \sin\left(\frac{1}{2}\,\omega\,\Delta t\right) }
\end{equation*}\]
Finally the double angle formula \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\) allows both the numerator and denomenator to be converted:
\[\begin{align*}
\cos \left( \frac{1}{2}k_x\,\Delta x \right) \sin \left( \frac{1}{2}k_x\,\Delta x \right) &= \frac{1}{2} \sin \left(k_x\,\Delta x \right) \\
\cos\left(\frac{1}{2}\,\omega\,\Delta t\right) \sin\left(\frac{1}{2}\,\omega\,\Delta t \right) &= \frac{1}{2} \sin\left(\omega\,\Delta t \right)
\end{align*}\]
such that derivative can be written as
\[\begin{equation*}
\frac{\partial \omega}{\partial k_x} = \frac{ c^2\Delta t }{\Delta x} \, \frac{ \sin \left(k_x\,\Delta x \right)}{ \sin\left(\omega\,\Delta t \right) }
\end{equation*}\]
which is the desired solution for Eqn 6.76. There is is no difference in deriving 6.77 other than to take the derivative with respect to \(k_y\) and so this is left to the reader.