- #1
luinthoron
- 14
- 1
Hello, I am trying to work out this exercise for my personal research connected with my bachelor thesis. The task is to compare equations (25.42) and (25.47) and express $u_0$ in terms of \tilde{L}. I have so far put the two equations together getting
\begin{equation}
12u^2u_0\tilde{L}^2-18uu_0^2\tilde{L}^2-u_0^2\tilde{L}^2+2uu_0\tilde{L}^2-\tilde{E}_0^2=2u-1
\end{equation}
After this I tried putting some terms together but I think I am missing another equation since there are in fact two unknowns: $u_0$ and $\tilde{E}_0$ or is there some trick I am missing?
For those without access to MTW, here are the equations: \\
(25.42)
\begin{equation}
\left(\frac{\mathrm{d}u}{\mathrm{d}\varphi}\right)^2=\frac{\tilde{E}^2}{\tilde{L}^2}-\frac{1}{\tilde{L}^2}\left(1-2u\right)\left(1+\tilde{L}^2u^2\right)
\end{equation}
and (25.47)
\begin{equation}
\left(\frac{\mathrm{d}u}{\mathrm{d}\varphi}\right)^2+\left(1-6u_0\right)\left(u-u_0\right)^2-2\left(u-u_0\right)^3=\frac{\tilde{E}^2-\tilde{E}_0^2}{\tilde{L}^2}
\end{equation}
Thank you.
\begin{equation}
12u^2u_0\tilde{L}^2-18uu_0^2\tilde{L}^2-u_0^2\tilde{L}^2+2uu_0\tilde{L}^2-\tilde{E}_0^2=2u-1
\end{equation}
After this I tried putting some terms together but I think I am missing another equation since there are in fact two unknowns: $u_0$ and $\tilde{E}_0$ or is there some trick I am missing?
For those without access to MTW, here are the equations: \\
(25.42)
\begin{equation}
\left(\frac{\mathrm{d}u}{\mathrm{d}\varphi}\right)^2=\frac{\tilde{E}^2}{\tilde{L}^2}-\frac{1}{\tilde{L}^2}\left(1-2u\right)\left(1+\tilde{L}^2u^2\right)
\end{equation}
and (25.47)
\begin{equation}
\left(\frac{\mathrm{d}u}{\mathrm{d}\varphi}\right)^2+\left(1-6u_0\right)\left(u-u_0\right)^2-2\left(u-u_0\right)^3=\frac{\tilde{E}^2-\tilde{E}_0^2}{\tilde{L}^2}
\end{equation}
Thank you.