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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.

Qmechanic
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XYZT
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3 Answers3

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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.

Cham
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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.

Eric Smith
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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.

g s
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