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 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:

$${\Large \frac {1}{\sqrt {1-{\frac {{\it rs}}{r}}}}}$$

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

$$\Large \sqrt {1-{\frac {{\it rs}}{r}}} \left( 1+\sqrt {{\frac {{\it rs}}{r}}} \right) ^{-1}$$

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

Which is 0.5 if r goes to 1.