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I have read that to an observer at rest outside a black hole, they will see the light from the free faller get redshifted and dimmer. What is the mathematical explanation for the light actually getting dimmer?

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    $\begingroup$ "I have read" - please provide references for claims if possible. $\endgroup$ – ACuriousMind Oct 27 '15 at 13:26
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Suppose I am a long way from a black hole watching you are hovering near the event horizon, then your time is dilated with respect to mine. I won't go into the details since lots of questions hereabouts involve this calculation. I'll just mention the result:

$$ \frac{d\tau}{dt} = \sqrt{1 - \frac{2GM}{c^2r}} \tag{1} $$

In this equation $d\tau$ is the number of seconds you measure while $dt$ is the number of seconds I measure, and $r$ is your distance from the black hole.

Suppose you are emitting EM radiation with a frequency on one hertz i.e. one cycle per second. One of you seconds corresponds to more than one second of my time. Using equation (1) we find the time I measure corresponding to your one second is:

$$ \Delta t = \frac{1}{\sqrt{1 - \frac{2GM}{c^2r}}} $$

For me the one cycle of EM radiation you emit every second lasts for $\Delta t$ seconds, so the period has increased and therefore the original 1 Hertz frequency has decreased to:

$$ \nu = \sqrt{1 - \frac{2GM}{c^2r}} $$

which is less than one so the radiation I receive has been red shifted. I used the example of 1Hz radiation to make the calculating simple. More generally, if you emit radiation with a frquency $\nu_0$ the frequency I observe is:

$$ \nu = \nu_0 \sqrt{1 - \frac{2GM}{c^2r}} $$

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  • $\begingroup$ But why would the light also get dimmer while falling. Does the decrease in frequency necessarily mean the signal will also be less intense? $\endgroup$ – Chris L. Oct 27 '15 at 13:41
  • $\begingroup$ @PetTaxi: the power, i.e. energy per second, of the received light will be a factor of $\sqrt{1 - 2GM/c^2r}$ less than the emitted light due to the time dilation. There will be an additional dimming due to loss of energy to the gravitational field, but I must admit I'm not sure how to calculate that. I'll need to go away and think about it. $\endgroup$ – John Rennie Oct 28 '15 at 6:05
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It is not dimmer in the sense that one photon is dimmer after climbing out of the gravitational well of the black hole. It is dimmer in the sense that fewer photons arrive per second.

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Assuming you are stationary with respect to the black hole, the light source appears dimmer due to the fact that the distance between you and it is increasing. This is described by the the equation for luminous intensity.

see https://en.wikipedia.org/wiki/Intensity_(physics)

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A black hole's gravity has so much force that light doesn't have enough speed to escape the gravitational pull being emitted by the black hole.

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  • $\begingroup$ I'm considering the time before the free faller reaches the horizon. In this region, the light still escapes, but gets dimmer over time. $\endgroup$ – Chris L. Oct 27 '15 at 13:17
  • $\begingroup$ @PetTaxi Ok I'll do some research. $\endgroup$ – DubGamer87 Oct 27 '15 at 13:18
  • $\begingroup$ @PetTaxi it's because it shows the light getting sucked in. $\endgroup$ – DubGamer87 Oct 27 '15 at 13:20
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    $\begingroup$ This answer is totally out of the question. $\endgroup$ – Fabrice NEYRET Oct 27 '15 at 13:21
  • $\begingroup$ I believe it does attempt to answer the question, it's just very wrong. $\endgroup$ – David Z Oct 27 '15 at 14:15

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