Black hole falling observer blueshifted light

Question

As per chapter 7, figure 8 Link from Exploring Black Holes, 2nd Edition for a falling (rain frame) observer dt(reception) of light pulses sent from far away one second apart is greater than one second. So outside light should appear red shifted to a rain observer, which is not the real case. Light appears blue shifted to a rain observer after he crosses the horizon. So, what am I missing?

Answer 1

Suppose you are falling towards a black hole and the light is coming from behind you i.e. you and the light are moving in the same direction towards the black hole. Since you are moving away from the source of the light there is a Doppler redshift, and when you take into account all the effects the you find the red shift dominates. This isn't a hard calculation and I described it in my answer to Do free falling observers see gravitational blueshift? The final result is that the falling observer sees a red shift given by:

$$ \frac{f}{f_0} = \frac{1}{1 + \sqrt{\frac{r_s}{r}}} $$

But remember that black holes can deflect light beams, so a light beam from behind can overtake you, be bent round 180° by the black hole and then move back towards you again. Now the light is moving towards you so it is blue shifted, and in principle this blue shift can become infinitely large. This is what Taylor et al are referring to in their figure 7 in chapter 12:

The value $b$ on the graph is the impact parameter. When $b = 0$ the light is coming from directly behind you and you and the light just fall in a straight line directly into the black hole. In this case you see the light redshifted all the way to your final doom at the singularity at the centre. Value of $b \ne 0$ means the light is not coming from directly behind you so it can be deflected by the black hole and meet you from a different direction. Whether you see this light as blue or red shifted depends on the direction in which it meets you.