Radial Null Geodesic: Solving the Equation

In summary, the conversation discusses the geodesic equations for the metric, specifically for a light ray. The equations involve the constant alpha, which is dependent on r and rs. The conditions for integrating the equations are given, and the resulting integration yields two different forms depending on whether r is greater than or less than rs.
  • #1
Ichimaru
9
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!
 
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  • #2
The same integral can have several different forms. You need to think about whether r > rs or r < rs.
 

FAQ: Radial Null Geodesic: Solving the Equation

What is a radial null geodesic?

A radial null geodesic is a path in spacetime that is taken by a massless particle, such as a photon, that moves in a straight line towards or away from a central point of gravitational attraction, without changing its direction.

How is the equation for a radial null geodesic solved?

The equation for a radial null geodesic can be solved using the geodesic equation, which takes into account the curvature of spacetime and the mass of the object traveling along the geodesic. This equation can be solved using mathematical techniques such as integration and differentiation.

What is the significance of solving the equation for a radial null geodesic?

Solving the equation for a radial null geodesic allows us to understand the path of massless particles in the presence of strong gravitational fields. This is important in fields such as astrophysics, where we study the behavior of light and other forms of radiation in the vicinity of massive objects like black holes.

Are there any practical applications of solving the equation for a radial null geodesic?

Yes, there are practical applications of solving the equation for a radial null geodesic. For example, the equation can be used to calculate the bending of light around massive objects, which is crucial for accurate observations and measurements in astrophysics.

Can the equation for a radial null geodesic be applied to other types of particles?

Yes, the equation for a radial null geodesic can be applied to any type of massless particle, not just photons. This includes particles like gravitons, which are hypothetical particles that transmit the force of gravity. However, the equation may need to be modified for particles with mass, as they will follow a different type of geodesic.

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