Deflection angles of photons passing by black holes

[tcw]

In trying to compute deflection angles for photons given their impact parameter (closest distance of trajectory to centre of black hole if unaffected) I am trying to numerically integrate the following equation (d^2/d(phi)^2)(u)+u=3Mu^2. However I am stuck as to how to work out the initial condition for du/d(phi) given the impact parameter of the photon. I have a feeling I'm missing something simple geometrically...

 

[bcrowell]

Hi, tcw,

You'll be more likely to get helpful replies if you mark up your math in LaTeX so it's readable. Here's an example of how to do that with your diffeq: $(d^2/d\phi^2)u+u=3mu^2$. To see how I did that, click on the QUOTE button on my post.

Could you explain your notation? What is u?

Here is a calculation like the one you're talking about: http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 (see subsection 6.2.7)

I don't think it will be easy to find the initial conditions given the impact parameter. I would think that in the strong-field case, the impact parameter would depend in some complicated way on the initial trajectory. Given the initial conditions, the results of your numerical calculation should tell you the impact parameter.

-Ben

 

[Bill_K]

tcw, Do you realize that your equation can be easily solved? Multiplying both sides by 2 du/dφ and integrating leads to a first integral, the particle's energy:

(du/dφ)² + u² = 2mu³ + C

Now separate the variables:

(C - u² + 2mu³)^-½ du = dφ

and integrate again:

φ = ∫(C - u² + 2mu³)^-½ du

Of course you still have to evaluate the integral numerically!

 

[tcw]

Thank you both for your help.

Ben, I have read through the article linked and it has given me some thoughts on how to approach the initial conditions problem; namely, instead of using the initial conditions from far off we use the impact parameter (b, say) as an initial condition starting close to the black hole. I tried to implement their code given but it didn't quite work for me in Matlab.

Bill, I had a look through your suggestion. The idea is interesting. For the first constant of integration we can deduce C=E^2/L^2=1/b^2. Now as for the numerical integration, I'm trying to work out the limits of integration for u. Since we start from far off and end far away, it seems like we should u=0 and u=0 for the two limits but of course that's nonsense. I thought about taking one of the limits as 1/b 1/distance of closest approach. However what puzzles me is that as we increase b, the numerical value for the integration gets larger which is counter intuitive as we expect less deflection for a larger b.

 

[pervect]

It sounds like you are doing what MTW does in exercise 25.24 pg 679 - except for an extraneous M in your equation

MTW defines u = M/r, and u_b = M/b, and the equations of motion for an infalling light ray are then

(du/dphi)^2 + (1-2u)u^2 = u_b ^2

Differentiating this with respect to phi, one gets

2 (du/dphi) (d^2 u / dphi^2) + 2u (du/dphi) -6u^2 (du/dphi) = 0, or

(d^2 u / dphi^2) + u = 3 u^2

which is your equation - except for that pesky M. You are correct to note u(phi) starts out at zero when phi has its initial value (probably easiest to take that initial value as zero), this corresponds to r=infinity, and ends up there too.

The impact parameter is given by the first, undifferentiated equation simply as

(du/phi) = u_b = M/b, evaluated at u=0, using MTW's notation.

I don't know what to do about your extra "M".

 

[tcw]

Thanks pervect, your post has helped me out. For the initial conditions in my original numerical integration, it turns out using 2M/b for (du/dphi) works.