Alice is in a rocket hovering outside a BH event horizon when her rocket runs out of fuel, and thus begins a free-fall decent. Bob is also in an inertial frame but at a very large distance.

It is commonly said that Bob will see:

a. Alice forever approaching the event horizon but never crossing it

b. Alice's clock tick ever more slowly

c. light emitted by Alice as becoming ever more red-shifted

My question: Is the observed time dilation due to relative velocity (i.e. reciprocal) or is it due to gravitational time dilation (non reciprocal)?

In other words: Does Alice see Bob aging quickly (non reciprocal dilation) or does she see Bob aging less than herself (reciprocal dilation)?

Edit: Yes, Schwarzschild geometry. Alice and Bob are radial. Mass of BH and Alice's initial distance from the horizon are unspecified (free parameters) which will impact the evolution of the time dilation of Bob from Alice's frame.


3 Answers 3


Doppler shifts (red/blue) are the same effect as changes in the time intervals between signals and changes in the rate at which one person sees the other aging.

You don't say whether Bob and Alice are along the same radial line, which can make a difference. Let's say they are.

Alice is initially at rest, so she sees Bob's signals as blueshifted due to gravitational time dilation. As time goes on and she accelerates, she will see this change to a redshift. (The case where she starts fairly far away from the black hole is particularly simple. In that situation, the Doppler shift is $1/(1+r^{-1/2})$, where $r$ is her radial coordinate in units of the Schwarzschild radius.

  • $\begingroup$ thanks for the related link to 26185 - a very similar question. I'm trying to stay focused on the time dilation rather than the red/blue shift which brings in Doppler effects. $\endgroup$
    – Hope Case
    Mar 26, 2019 at 16:02
  • 1
    $\begingroup$ @HopeCase: I'm trying to stay focused on the time dilation rather than the red/blue shift which brings in Doppler effects. They're not different effects. As explained in he first paragraph of my answer, they're identical. $\endgroup$
    – user4552
    Mar 26, 2019 at 16:16
  • $\begingroup$ Of course, time dilation and frequency shift are equivalent. Thanks for correcting my brain fart. Your answer mentions Alice will see a change to redshift - which I assume comes from the increasing relative velocity. At the same time, she is coming closer to the event horizon and the gravitational dilation increases - which is in the blue direction. I'm curious as to how these effects play out as she falls. $\endgroup$
    – Hope Case
    Mar 28, 2019 at 16:47

It is a misconception that relative velocity will itself cause time dilation. Black holes do have stress-energy and the difference between the stress-energy (strong effect) at the event horizon and far away from it (lesser effect), will cause the time dilation. This is what you call gravitational time dilation.

Because of the equivalence principle, acceleration (which is absolute in the universe) will have the same effects as the gravitational zone. So yes, acceleration will (can) cause time dilation. But in your case, if the movements of Alice and Bob are constant speed, that fact itself will not cause time dilation.

Now in your c. you say that light emitted will be seen more red-shifted. That is actually not true always. It depends on the actual position of Alice and Bob and their relative motion. Sometimes it will acutally be blueshift. Please see here:

Communication near a black hole

You are asking about the aging, and if you only take the gravitational stress-energy difference into account (and disregard their position and motion) then Alice will see Bob age quickly. If you consider their position and motion, then the answer is again depending on the situation.

  • 1
    $\begingroup$ It is a misconception that relative velocity will itself cause time dilation. The OP hasn't expressed any such idea, so this isn't particularly relevant. $\endgroup$
    – user4552
    Mar 25, 2019 at 21:48

The time dilation of Alice in Bob's frame is

$$ \rm \frac{d t}{d \tau} = 1 \div \sqrt{1-r_s/r} \ \div \sqrt{1-v^2/c^2}$$

while the time dilation of Bob in Alice's frame is

$$ \rm \frac{d \tau}{d t} = \sqrt{1-r_s/r} \ \div \sqrt{1-v^2/c^2}$$

where t is the time elapsed on Bob's clock, τ on Alice's and v the local velocity of Alice relative to a stationary probe. Bob is far away from the black hole in this scenario. The gravitative component of the time dilation is absolute, while the kinematic component is relative, see here.

Here we are assuming Schwarzschild geometry, in the vicinity of rotating or charged black holes the split of the components is little bit more complicated but following the same principle.

So if Alice would be free falling from infinity (then v would be the negative escape velocity) both components would cancel in her frame, and Bob would age with the same rate as herself. In Bob's frame Alice then would age with the squared rate of locally stationary observers at the same height above the black hole.

If you are not only asking about the time dilation but also what one sees with his eyes, you also have to take the regular Doppler shift into account, which makes the received signals redder when they move away from each other, and bluer if they move towards each other. Which effect is the strongest depends not only on her exact position but also on her velocity and vector relative to Bob.

For example, if Alice falls from rest position at r=5rs, and Bob stays fixed at this position, Alice reaches the horizon after a proper time of τ=33.7GM/c³. If she watches Bob, the time she sees on his clock in the moment when she crosses the horizon would be t=27.328GM/c³, so Bob's signal would visually appear redshifted in her frame.

Since her free fall velocity is smaller than the escape velocity (because she fell from rest at finite height) Bob's clock would still be ticking faster in her frame though, but in this scenario the effect of the regular Doppler shift would trump the time dilation visually.

  • $\begingroup$ So the net time dilation is a combination of the two, with the reciprocal part increasing as the relative velocities increase. Thanks for the link to Wikipedia - and the introduction to celerity. $\endgroup$
    – Hope Case
    Mar 26, 2019 at 15:08
  • $\begingroup$ Frequencies and clock ticks have to be dilated in exactly the same ratio, because frequencies are just clock ticks of Bob's signaling mechanism. You can't have frequencies redshift and clock ticks speed up in Alice's frame. $\endgroup$ Mar 26, 2019 at 17:28
  • $\begingroup$ @Peter Shor: the received frequency does not have to be the same as the time dilation, that is only the case if the motion is completely transversal. For more details on the subject google "radial and transverse doppler", and consider that you already have redshift in the Newtonian limit where the time dilation becomes neglible. $\endgroup$
    – Yukterez
    Mar 26, 2019 at 18:14
  • $\begingroup$ @Yukterez: I didn't say "redshift" and "time dilation". I said "frequencies" and "clock ticks". The frequency depends on the intervals between when Bob's radio waves are at two successive maxima. Clock ticks are the intervals between when Bob's clock goes "tick". These have to be slowed down by the same amount, because they're both constant in Bob's frame. $\endgroup$ Mar 26, 2019 at 18:54
  • $\begingroup$ How fast the clock ticks depends on the time dilation, and how fast you see it tick (the received frequency) depends both on the time dilation and the velocity vector. $\endgroup$
    – Yukterez
    Mar 26, 2019 at 19:10

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