Radial Null Geodesic: Solving the Equation

  • Context: Graduate 
  • Thread starter Thread starter Ichimaru
  • Start date Start date
  • Tags Tags
    Geodesic Radial
Click For Summary
SUMMARY

The discussion focuses on solving the radial null geodesic equations derived from the metric ds² = -c²αdt² + (1/α)dr² + dω², where α = 1 - (r²/rₛ²). The user successfully establishes the relationships for a light ray and attempts to integrate the equations to derive the geodesic path. The integration leads to the expression r = rₛ coth(ct/rₛ), but the user encounters difficulties with the integral ∫(1/(1 - (r²/rₛ²)))dr = ct, which results in a logarithmic form. The discussion emphasizes the importance of considering the conditions r > rₛ or r < rₛ during integration.

PREREQUISITES
  • Understanding of general relativity and geodesic equations
  • Familiarity with differential equations and integration techniques
  • Knowledge of hyperbolic functions and their properties
  • Basic concepts of metric tensors and spacetime geometry
NEXT STEPS
  • Study the derivation of geodesic equations in general relativity
  • Learn about hyperbolic functions and their applications in physics
  • Explore integration techniques for solving differential equations
  • Investigate the implications of light paths in curved spacetime
USEFUL FOR

Physicists, mathematicians, and students studying general relativity, particularly those interested in the behavior of light in curved spacetime and the mathematical formulation of geodesics.

Ichimaru
Messages
9
Reaction score
0
Hi,

I've found geodesic equations for the metric:

\begin{equation}

ds^{2} = -c^{2} \alpha dt^{2} + \frac{1}{ \alpha } dr^{2} + d \omega ^{2}

\end{equation}

where

\begin{equation}

\alpha = 1 - \frac{r^{2}}{r_{s}^{2}}

\end{equation}

I have found that for a light ray:

\begin{equation}

\alpha \frac{dt}{d \lambda} = 1 ;
\frac{dr}{d \lambda} = c

\end{equation}

Where lambda is an affine parameter and i have appliedthe conditions:

\begin{equation} \frac{dt}{d \lambda} = 1 , r = 0

\end{equation}

I am then told to integrate these equations to get:

\begin{equation}

r = r_{S}tanh(\frac{ct}{r_{s}})

\end{equation}

However when I try to integrate I get:

\begin{equation}

\int_{0}^{r} \frac{1}{1-\frac{r^2}{r_{s}^{2}}}dr = ct

\end{equation}

goes to:

\begin{equation}

\frac{r_{s}}{2} ln \frac{r+r_{s}}{r-r_{s}} = ct

\end{equation}

Which rearranges to:

\begin{equation}

r = r_{s}coth( \frac{ct}{r_s})

\end{equation}

Any help would be appreciated, thanks!
 
Physics news on Phys.org
The same integral can have several different forms. You need to think about whether r > rs or r < rs.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K