MTW Gravitation Ex. 25.16: Comparing Equations & Expressing u_0 in \tilde{L}

  • Context: Graduate 
  • Thread starter Thread starter luinthoron
  • Start date Start date
  • Tags Tags
    Gravitation Shift
Click For Summary
SUMMARY

The discussion focuses on solving exercise 25.16 from MTW, specifically comparing equations (25.42) and (25.47) to express \( u_0 \) in terms of \( \tilde{L} \). The user derives a cubic equation for \( \tilde{E}_0^2 \) and a quadratic equation for \( u_0 \), leading to the solutions \( {u_0}_{1,2}=\frac{1\pm\sqrt{1-\frac{12}{\tilde{L}^2}}}{6} \). The user expresses concern about the implications of the derived condition for \( \tilde{L} \) and seeks clarification on selecting the correct root for \( u_0 \).

PREREQUISITES
  • Understanding of general relativity concepts as presented in MTW.
  • Familiarity with differential equations and their applications in physics.
  • Knowledge of cubic and quadratic equations and their solutions.
  • Ability to manipulate algebraic expressions and coefficients.
NEXT STEPS
  • Study the derivation of equations (25.42) and (25.47) in MTW for deeper context.
  • Explore methods for solving cubic equations in the context of physics problems.
  • Investigate the implications of the conditions derived from \( u_0 \) and \( \tilde{L} \) on physical models.
  • Review techniques for determining the validity of roots in quadratic equations.
USEFUL FOR

Students and researchers in theoretical physics, particularly those focusing on general relativity and mathematical modeling of gravitational systems.

luinthoron
Messages
14
Reaction score
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.
 
Physics news on Phys.org
Each equation expresses (du/dφ)2 in terms of a cubic in u. Expand out the cubics and equate the coefficients of each power of u.
 
Thank you for the advice. I ended up with:

\begin{equation}
\tilde{E}_0^2=4\tilde{L}^2u_0^3-\tilde{L}^2u_0^2+1
\end{equation}

\begin{equation}
3u_0^2-u_0=-\frac{1}{\tilde{L}^2}
\end{equation}

The second one is a quadratic equation, so I can write the solution

\begin{equation}
{u_0}_{1,2}=\frac{1\pm\sqrt{1-\frac{12}{\tilde{L}^2}}}{6} .
\end{equation}

But this would give a condition for \tilde{L}, which I find suspicious. Also which of the two roots is correct? It stays possitive in both cases and I can't come with any other clue to help me choose. Any additional hints, please?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
535
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K