• Support PF! Buy your school textbooks, materials and every day products Here!

Null geodesics of light from a black hole accretion disk

  • Thread starter Favicon
  • Start date
  • #1
14
0
Sorry I don't know latex so this may look a little messy.

Homework Statement



I'm trying to solve the equation for null geodesics of light travelling from a rotating black hole accretion disk to an observer at r = infinity. The point of emission for each photon is given by co-ordinates r, phi where r is radial distance from centre of the black hole, phi is azimuthal angle around the accretion disk (phi = 0 is defined to be the tangent point). The problem is stated as follows:

"Light travels on null geodesics given by the solution of the equation

d2u/dphi2 = 3u2 - u

where u = 1/r. The full paths can be found by integrating this from u=1/rem, phiem to u = 0 (r=infinity), phi = 0. This requires varying the initial gradient (du/dphi)em = - uemtanE until the correct solution is found for an angle E = E' + theta, where E' is the 'straight line' angle, and theta is the additional deflection from lightbending as the photon travels from rem to infinity. Explore the size of theta to estimate where the straight line approximation may break down."

I've also been told that to solve the equation I need to split it into two 1st order ODEs, but I'm not sure how to do that.

Homework Equations




The Attempt at a Solution



I'm really struggling just to try and understand the description, let alone solve the equation. Please could someone explain to me what this means and how I can extract the light paths from the given equations?
 
Last edited:

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,261
790
Sorry I don't know latex so this may look a little messy.
I learned latex when I joined Physics Forums. See

https://www.physicsforums.com/showthread.php?t=8997.
I'm trying to solve the equation for null geodesics of light travelling from a rotating black hole
"Light travels on null geodesics given by the solution of the equation

d2u/dphi2 = 3u2 - u

where u = 1/r.
But this equation is for null geodesics of non-rotating black holes. Is this what you want to do?
I've also been told that to solve the equation I need to split it into two 1st order ODEs, but I'm not sure how to do that.
This second-order equation can be reduced to a pair of first-order equations by setting [itex]p = du/d\phi[/itex], so that [itex]dp/d\phi = d^2 u/d\phi^2[/itex]. Consequently,
the second-order equation is eqiuvalent to

[tex]
\begin{equation*}
\begin{split}
\frac{du}{d\phi} &= p \\
\frac{dp}{d \phi} &= 3u^2 - u.\\
\end{split}
\end{equation*}
[/tex]
 
  • #3
14
0
Thanks for the latex tip.

But this equation is for null geodesics of non-rotating black holes. Is this what you want to do?
Yes it probably is a non-rotating black hole. I'm actually writing a program to produce the expected line spectrum from a black hole, but the description I've been given (to explain the physics of relativistic line smearing) isn't very clear so when it talked about the motion of the accretion disk I assumed it meant the black hole itself was rotating.

[tex]\frac{du}{d\phi} = p[/tex]
[tex]\frac{dp}{d\phi} = 3u^{2} - u[/tex]
So now I have
[tex]\frac{du}{d\phi} = p_{em} = -u_{em}tan(E)[/tex]
and have to vary E? But I still can't see how I'll know when I've found the correct value for E.
 

Related Threads for: Null geodesics of light from a black hole accretion disk

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
8
Views
1K
Replies
4
Views
6K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
1
Views
361
Replies
0
Views
353
Replies
1
Views
899
Top