Calculating a Deflection of Light Integral from Einstein's 1907 paper on Gravity and Light

Question

$\frac{1}{c^2}\int_{\theta=-\pi/2}^{\theta=\pi/2} \frac{kM}{r^2}cos\theta ds=2\frac{kM}{c^2\Delta}$

An observation: $\cos\theta=\frac{\Delta}{r}$

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The entire calculation is done in an early section of the answere here: http://physics.stackexchange.com/q/14056/ — Ron Maimon, May 31 '12 at 19:42
Thanks! I've read quite a bit of the other sections as well. — MadScientist, Jun 01 '12 at 02:01

score 1 · Accepted Answer · answered May 31 '12 at 18:12

Use

$s = \Delta \tan\theta \quad \Rightarrow \quad ds = \frac{\Delta}{\cos^2\theta}d\theta $

and

$r = \frac{\Delta }{\cos \theta }$

and insert it into your integral. What you get is

$\frac{1}{c^2} \intop_{-\pi/2}^{\pi/2}\frac{kM}{\Delta } \cos\theta d\theta = 2 \frac{kM}{c^2 \Delta} $.

1 Answers1