Imagine two identical black holes in a circular orbit, and Alice is smack-dab in the middle of the system (at the barycenter). Bob is at infinity.

Let's assume that Alice and Bob are stationary relative to each other. Alice does not experience any net acceleration (from being at the barycenter of the system).

Does Alice experience any time dilation relative to Bob?

Here are some arguments for why she should not:

  1. Alice is stationary relative to Bob - so no special relativistic effects.

  2. Alice is not accelerating - so by the equivalence principle, she is not experiencing gravitational effects and therefore, no time dilation.

However, this feels like the wrong conclusion because she's still stuck in the gravitational well of this binary system and requires a non-zero escape velocity to exit the system and meet with Bob.

In literature, lot of discussion about time dilation talks about escape velocity which is straightforward when talking about single spherically symmetric masses - but I am not sure how it applies to this system.

Of course, the resolution here might just be they both don't experience any relative time dilation, but any signals they try to send each other will always be gravitationally redshifted. And there is no way for them to meet up and compare clocks that show the same passage of time.

  • 1
    $\begingroup$ You didn't say how close the black holes orbit each other. Consider that, whether a black hole or whether a star, it really doesn't make a difference. The mass of the body and the orbital distance are all that really matter. For example, a black hole with one solar mass could replace our sun and the mechanics of the solar system would be largely unchanged. It would be very cold and dark, but time and space around Earth would be no different than they are now. $\endgroup$
    – J...
    Commented Feb 1, 2022 at 21:47
  • 2
    $\begingroup$ As many, you seem to be confusing the gravitational acceleration with potential. The time dilation effectively is the gravitational potential, so acceleration is proportional to the gradient of the time dilation, but not to the value of the time dilation. For example, the time dilation in the center of the Earth is the highest while the gravitational acceleration there is zero. $\endgroup$
    – safesphere
    Commented Feb 2, 2022 at 1:10

3 Answers 3


Lets suppose that the orbit is very large, so we could apply the linearized theory at the center. The spacetime metric is then $$\tag{1} ds^2 \approx (1 + 2 \phi) \, dt^2 - (1 - 2 \phi) (dx^2 + dy^2 + dz^2), $$ where $\phi$ is the newtonian potential. For two black holes on the circular orbit: $$\tag{2} \phi = \sum_k \phi_k = -\, \frac{2 G M}{r}. $$ The time dilation is defined by the following formula (for stationary observers at the center and at infinity): $$\tag{3} d\tau = \sqrt{g_{00}} \, dt \approx (1 + \phi) \, dt. $$ So the time retardation of the central observer would be $$\tag{4} \Delta \tau \approx \frac{2 G M}{r} \, \Delta t. $$ Notice that if the orbit is very large so that $r \gg 2 G M$, then $\Delta \tau \approx 0$. The time dilation would be negligible.

  • $\begingroup$ So, in the spirit of the original question: If the orbit is small, the central observer will experience time dilation? $\endgroup$
    – XYZT
    Commented Feb 1, 2022 at 17:20
  • $\begingroup$ @XYZT, yes, there should be a time dilation. $\endgroup$
    – Cham
    Commented Feb 1, 2022 at 18:53
  • $\begingroup$ @XYZT this answer’s first sentence suggests that it is not applicable to small orbits at all: “Lets suppose that the orbit is very large…”. $\endgroup$
    – Holger
    Commented Feb 2, 2022 at 13:00

For your particular example, the question is: "is spacetime curved more at Alice's location than at Bob's", or equivalently "is the gravitational potential higher for Alice than for Bob"? And the answer is "yes": it would in fact take energy for Alice to go visit Bob. She is located at a local maximum of potential, but not a global maximum. So she experiences time dilation relative to Bob.

Time dilation is more about gravitational potential than about acceleration (to be very precise it's about the length of the path through spacetime). Acceleration has no effect on time dilation, except in so far as it changes the instantaneous speed of the object; this is the clock postulate and has been experimentally verified in particle accelerators, where particles experience literally millions of g's of acceleration but their time dilation relative to the laboratory frame is entirely due to their speed.

The equivalence principle says that there's a kind of pseudo-potential created in an accelerating coordinate system (such as a coordinate system co-moving with an accelerating rocket). That is, it does indeed take energy for someone at the bottom of an accelerating rocket to move to the top. But from an outside observer's perspective, the different times counted by clocks at the top and bottom of the rocket are due to their different paths through spacetime, which is only indirectly due to acceleration.


A partial answer / extended comment.

Think about how a time dilated observer must observe distant light sources to be blueshifted to the amount of her time dilation.

For example: Suppose Bob sees Alice get into a blue space ship and fly into position, while Alice sees Bob ready a red laser. Bob sees Alice's space ship gradually turn red as she gets there. When Alice gets to position, Bob shines his laser at her space ship.

Alice sees a blue laser which reflects off of her ship back to Bob. Bob sees a red laser reflect off of a red space ship. But Alice knows that Bob's laser is red in Bob's frame and Bob knows that Alice's space ship is blue in Alice's frame. If either of them does the calculation, they will find that the red laser was shifted in frequency by $+\Delta \nu$ as it traveled from Bob to Alice, then by $- \Delta \nu$ as it traveled back from Alice to Bob. The reason for the change in frequency is Work: the gravitational field does work on the beam as it goes from Bob to Alice, and the beam does work on the gravitational field as it goes from Alice to Bob.

Change in photon energy is proportional to change in frequency: $\Delta T = h \Delta \nu$. Energy is conserved. Therefore, if we know the work per unit mass-energy required to move a parcel of mass-energy from one point to another, we know the time dilation factor between those points, regardless of the value of the gravitational (pseudo)force vector at either point.

  • $\begingroup$ This is incorrect. Photons don’t change frequency in flight and the gravitational field does no work on the beam. The red and blue shift effects are due exclusively to the difference in the frames of reference. When measured in the same frame, there is no gravitational redshift or blueshift. For example, in Bob’s frame, the photons of his laser have the same frequency, energy, and color when they hit Alice as they had at the emission near Bob. You seem to be confusing General Relativity with Newtonian gravity. In GR, gravity is not a force, so it does no work. $\endgroup$
    – safesphere
    Commented Feb 2, 2022 at 5:41
  • $\begingroup$ I am not confusing anything. Pseudoforces do work equal to their distance integral as measured by a frame for which the psuedoforce is measured. Alice and Bob both measure the pseudoforce of gravity, and although Alice measures a radially length-contracted universe, her time dilated frame measures a reciprocally faster pseudoforce such that her integral and Bob's are identical. $\endgroup$
    – g s
    Commented Feb 2, 2022 at 17:28
  • $\begingroup$ If, for instance, you drop a giant meteor onto a planet, the dinosaurs will find scant comfort in the fact that the giant meteor was really just travelling at $c$ in a straight line towards a future spacetime in which the meteor and the dinosaurs were in the same place at the same time. The work done by the gravitational field as measured by the dinosaurs really vaporizes the dinosaurs, whether it's real in a metaphysical sense or not. $\endgroup$
    – g s
    Commented Feb 2, 2022 at 17:34
  • $\begingroup$ You are still confusing General Relativity with Newtonian gravity. The source of the kinetic energy of the meteor in GR is not the work done by the field like in Newtonian gravity, but a mass defect as a result of the time dilation. I am not sure what you mean by the the same integrals for Bob and Alice, but they measure a different energy of the same photon. The energy of a photon is conserved in the Schwarzschild spacetime when properly measured in the same frame, not in different frames (e.g. first Bob, then Alice). $\endgroup$
    – safesphere
    Commented Feb 3, 2022 at 8:56

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