Quote by yuiop
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)