View Single Post
Feb24-12, 01:49 PM
P: 1,555
Quote Quote by yuiop View Post
You may well be right, but can you support that with an equation?
The Doppler factor wrt to far away light for a radially free falling from infinity test observer in a Schwarzschild solution depends on two factors:

1. The gravitational based Doppler effect:

[tex]{\Large \frac {1}{\sqrt {1-{\frac {{\it rs}}{r}}}}}[/tex]

2. The velocity based Doppler effect, in case for free falling from infinity we have:

[tex]\Large \sqrt {1-{\frac {{\it rs}}{r}}} \left( 1+\sqrt {{\frac {{\it rs}}{r}}}
\right) ^{-1}[/tex]

If we multiply the two factors and play around with the math to simplify the formula and normalize to rs=1 we get:
[tex] \Large \left( 1+\sqrt {{r}^{-1}} \right) ^{-1}[/tex]

Which is 0.5 if r goes to 1.

(rs is the Schwarzschild radius)