Generated by the Physics Derivation Graph.
https://arxiv.org/pdf/2004.04818.pdf
Eq. \ref{eq:1719451} is an initial equation. \begin{equation} v = \sqrt{ \frac{K + (4/3) G}{\rho} } \label{eq:1719451} \end{equation} Eq. \ref{eq:2178289} is an initial equation. \begin{equation} K = f \frac{E}{a^3} \label{eq:2178289} \end{equation} Based on the assumption \(K >> G$, drop non-dominant term in Eq. \ref{1719451}; yeilds Eq. \ref{9155336} \begin{equation} v = \sqrt{ \frac{K}{\rho} } \label{eq:9155336} \end{equation} Substitute LHS of Eq. \ref{eq:2178289} into Eq. \ref{eq:9155336}; yields Eq. \ref{eq:5077893}. \begin{equation} v = \sqrt{ \left( f\frac{E}{a^3} \right) \frac{1}{\rho} } \label{eq:5077893} \end{equation} Eq. \ref{eq:2438445} is an initial equation. \begin{equation} \rho = \frac{m}{a^3} \label{eq:2438445} \end{equation} Multiply both sides of Eq. \ref{eq:2438445} by \(a^3\); yields Eq. \ref{eq:7834577}. \begin{equation} a^3 \rho = m \label{eq:7834577} \end{equation} Substitute RHS of Eq. \ref{eq:7834577} into Eq. \ref{eq:5077893}; yields Eq. \ref{eq:5020923}. \begin{equation} v = \sqrt{f} \sqrt{\frac{E}{m}} \label{eq:5020923} \end{equation} Based on the assumption \(\sqrt{f} \approx 2$, drop non-dominant term in Eq. \ref{5020923}; yeilds Eq. \ref{4534919} \begin{equation} v = \sqrt{\frac{E}{m}} \label{eq:4534919} \end{equation} Eq. \ref{eq:5961293} is an initial equation. \begin{equation} E_{\rm Rydberg} = \frac{ m_e e^4 }{ 32 \pi^2 \epsilon_0^2 \hbar^2} \label{eq:5961293} \end{equation} Eq. \ref{eq:6901924} is an assumption. \begin{equation} E_{\rm Rydberg} = E \label{eq:6901924} \end{equation} Eq. \ref{eq:9431422} is an initial equation. \begin{equation} \alpha = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{\hbar c} \label{eq:9431422} \end{equation} Multiply both sides of Eq. \ref{eq:9431422} by \(c\); yields Eq. \ref{eq:6181437}. \begin{equation} \alpha c = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{\hbar} \label{eq:6181437} \end{equation} Substitute LHS of Eq. \ref{eq:6901924} into Eq. \ref{eq:5961293}; yields Eq. \ref{eq:3642765}. \begin{equation} E = \frac{ m_e e^4 }{ 32 \pi^2 \epsilon_0^2 \hbar^2} \label{eq:3642765} \end{equation} Substitute LHS of Eq. \ref{eq:3642765} into Eq. \ref{eq:4534919}; yields Eq. \ref{eq:2063484}. \begin{equation} v = \sqrt{ \frac{m_e}{m} \frac{e^4}{32 \pi^2 \epsilon_0^2 \hbar^2} } \label{eq:2063484} \end{equation} Simplify Eq. \ref{eq:2063484}; yields Eq. \ref{eq:4586348}. \begin{equation} v = \frac{e^2}{4 \pi \epsilon_0 \hbar} \sqrt{\frac{m_e}{2 m}} \label{eq:4586348} \end{equation} Substitute LHS of Eq. \ref{eq:6181437} into Eq. \ref{eq:4586348}; yields Eq. \ref{eq:5883117}. \begin{equation} v = \alpha c \sqrt{ \frac{m_e}{2 m} } \label{eq:5883117} \end{equation} Eq. \ref{eq:6979804} is an initial equation. \begin{equation} m = A m_p \label{eq:6979804} \end{equation} Substitute LHS of Eq. \ref{eq:6979804} into Eq. \ref{eq:5883117}; yields Eq. \ref{eq:8323044}. \begin{equation} v = \alpha c \sqrt{ \frac{m_e}{A m_p} } \label{eq:8323044} \end{equation} The maximum of Eq. \ref{eq:8323044} with respect to \(A\) is Eq. \ref{eq:9568206} \begin{equation} v_u = \alpha c \sqrt{ \frac{m_e}{m_p} } \label{eq:9568206} \end{equation} Eq. \ref{eq:9568206} is one of the final equations.