Continued writing experiment 4B report.

diff --git a/year2/4b/4b.pdf b/year2/4b/4b.pdf
index 343ceb4a56e537f141a5e7a612ddd1b0b146d7a4..b12f13c24487156a63a61c70288ba89f0a921e06 100644 (file)
Binary files a/year2/4b/4b.pdf and b/year2/4b/4b.pdf differ

diff --git a/year2/4b/4b.tex b/year2/4b/4b.tex
index b54e224c9598bf1f47be2b3af11adcb98a23302e..6e6fe136bcd0389697da8e19e7c133f310c9c017 100644 (file)
--- a/year2/4b/4b.tex
+++ b/year2/4b/4b.tex
@@ -224,15 +224,61 @@ Furthermore the calculus-based approximation\autocite{hughes-hase-uncertainties}
     \alpha_r = \sqrt{\frac{\pi c \hbar \mu}{\tilde{B_{\nu}}}} \alpha_B
 \end{equation}
 
+\subsection{Determination of Vibrational Constants}
+
 The values for the harmonic constant $\tilde{\nu_e}$ and the dimensionless anharmonicity constant $x_e$ in equation
 \ref{eq:vib-energy} (which gives the expected vibrational energy levels if the rotational spectrum is
-ignored) can be determined by 
+ignored).
 
 \begin{equation}\label{eq:vib-energy}
     \tilde{E_{\nu}} = \tilde{\nu_e} \left( \nu + \frac{1}{2} \right) -
     \tilde{\nu_e} x_e \left( \nu + \frac{1}{2} \right)^2
 \end{equation}
 
+From equation \ref{eq:vib-energy} we can obtain the energy related to the pure vibrational transition
+$\tilde{E}(\nu_f \leftarrow 0)$ (the $\nu = 0$ to $\nu = \nu_f$ transition) as equation \ref{eq:delta-e-coeff}.
+
+\begin{align}
+    \tilde{E}\left( \nu_f \leftarrow 0 \right) &= \tilde{E}_{\nu_f} - \tilde{E}_0 \nonumber \\
+        &= \tilde{\nu_e}\left( \nu_f + \frac{1}{2} - \frac{1}{2} \right) - \tilde{\nu_e} x_e \left(
+            \frac{1}{4} - \left( \nu_f + \frac{1}{2} \right)^2 \right) \nonumber \\
+        \label{eq:delta-e-coeff}
+        &= \tilde{\nu_e} \left( \nu_f - \left( \nu_f^2 + \nu_f \right) x_e \right)
+\end{align}
+
+The energy related to the R branch transitions can be determined to yield
+equation \ref{eq:r-trans-energy} where $J$ is the rotational state adopted in
+the lower vibrational state ($\nu = 0$).
+
+\begin{equation}\label{eq:r-trans-energy}
+    R(J) = \Delta \tilde{E}(\nu_f \leftarrow 0) + (\tilde{B_{\nu_f}} + \tilde{B_0}) (J + 1) + (\tilde{B_{\nu_f}} -
+    \tilde{B_0}) ( J + 1)^2
+\end{equation}
+
+Setting $J = 0$ hence gives equation \ref{eq:r-trans-energy-j0}.
+
+\begin{equation}\label{eq:r-trans-energy-j0}
+    R(J) - 2 \tilde{B_{\nu_f}} =  \Delta \tilde{E}(\nu_f \leftarrow 0)
+\end{equation}
+
+Subsituting equation \ref{eq:delta-e-coeff} into equation \ref{eq:r-trans-energy-j0} yields equation
+\ref{eq:r-trans-energy-sub}.
+
+\begin{equation}\label{eq:r-trans-energy-sub}
+    R(0) - 2 \tilde{B_{\nu_f}} = \tilde{\nu_e} \left( \nu_f - \left( \nu_f^2 + \nu_f \right) x_e \right)
+\end{equation}
+
+Now from equation \ref{eq:r-trans-energy-sub} a system of linear equations can be
+obtained by setting $\nu_f = 1$ and $\nu_f = 2$ (equations xx and xx) for which we can solve to give equations xx and xx.
+
+
+
+
+
+
+for which setting $J = 0$ and $\nu_f = 1$ and subsequently $\nu_f = 2$ gives the
+system of linear equations (equations xx and xx) for which we can solve to give equations xx and xx.
+
 %TODO: Total energy equation, determination of \nu_e x_e, calculation of values, determination of k with error
 %propagation and table.