Observers at rest in gravitational fields will see infalling light signals as blueshifted. Do inertial observers in free fall in a gravitational field see the same shift? If someone is standing on a cliff at rest in the gravitational field and then steps off, do they instantly see a change in the blueshift as a result of being in free fall in an inertial reference frame?

  • $\begingroup$ Do you assume the light source is falling with them or that it is fixed in the cliff frame? $\endgroup$ Commented Oct 19, 2015 at 17:44
  • $\begingroup$ I'm thinking of light emitted from a distant galaxy. I know the expansion of the Universe will make that signal redshifted, so I am interested in the difference between the signal measured from that galaxy by an observer at rest on the cliff and an observer who has stepped off) $\endgroup$ Commented Oct 19, 2015 at 17:56
  • $\begingroup$ the amount of blueshift changes with the velocity as described by the doppler shift $\endgroup$
    – Jaywalker
    Commented Oct 19, 2015 at 19:32
  • $\begingroup$ The accepted answer is wrong, and this question is a duplicate. $\endgroup$
    – user4552
    Commented Dec 13, 2019 at 14:31

1 Answer 1


There are two effects that will change the frequency of the light:

  1. the gravitational time dilation

  2. the Doppler shift due to the velocity of the infalling observer

To a first approximation we can simply multiply these together to work out what the falling observer sees.

The gravitational time dilation for a freely falling observer is calculated in my answer to A clock in freefall, and it turns out to be:

$$ \frac{d\tau}{dt} = 1 - \frac{r_s}{r} \tag{1} $$

Note that this is greater than the time dilation for an observer hovering near the black hole because there is time dilation due to the velocity as well as due to the gravitational field.

The Doppler shift is given by:

$$ \frac{f}{f_0} = 1 - \frac{v_{obs}}{v_{light}} \tag{2} $$

The velocity of the free falling observer is

$$ \frac{v_{obs}}{c} = \left( 1 - \frac{r_s}{r} \right) \sqrt{\frac{r_s}{r}} $$

and the velocity of light near the black hole is:

$$ \frac{v_{light}}{c} = 1 - \frac{r_s}{r} $$

Substituting these into our expression (2) for the Doppler shift gives:

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

and it just remains to convert this to the frequency observed by the freely falling observer by dividing by the factor calculated in equation (1):

$$\begin{align} \frac{f}{f_0} &= \frac{1 - \sqrt{\frac{r_s}{r}}}{1 - \frac{r_s}{r}} \\ &= \frac{1}{1 + \sqrt{\frac{r_s}{r}}} \end{align}$$

And there's your answer. The value of $f/f_0$ is less than one so an observer falling freely into a black hole sees the light as red shifted not blue shifted.

  • 1
    $\begingroup$ I tried hard to convince myself this was wrong; but it isn't. Nice that the redshift is exactly 2 at the event horizon. See also iopscience.iop.org/article/10.1088/1742-6596/104/1/012008/pdf $\endgroup$
    – ProfRob
    Commented Oct 20, 2015 at 15:47
  • $\begingroup$ @Rob Jefferies - +1 for the reference, this is very helpful $\endgroup$ Commented Oct 20, 2015 at 15:53
  • $\begingroup$ This is correct for radially infalling light, but not for light coming from other angles. See physics.stackexchange.com/questions/26185/… , which this question duplicates. $\endgroup$
    – user4552
    Commented Dec 13, 2019 at 14:30

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