# When you are in a gravitational field, do object far away get physically closer to you as you get closer to the mass?

An observer A is close to a black hole and an observer B one light year away. They are both remaining at constant radial distance from the black hole. A is at 2 Rs away from the center of the black hole, meaning his clock ticks at about half the rate of B's clock. This means that the two way travel of light signals between the two observers takes about 2 years to far away B, but only about 1 year to A. If you keep placing A closer to Rs and repeat the experiment, the difference in clock tick rates increases, reaching infinity at Rs. The time the 2 way signal takes in A's frame to reach B and return decreases towards 0 close to the event horizon. In B's frame, the time the 2 way signal takes to go to A and return increases toward infinity as A is placed closer to the event horizon. Eventually there will be a point where the light signal only takes a fraction of a second to go to B and back from A's frame.

a) In A's frame, does B affect A gravitationally as if it was at the distance A measures with the time light takes to travel between them?

b) If the answer to a) is yes, what are the consequences of this effect to an infalling observer?

Edit: Re-worded parts of the question for clarifications and removed text related to my own theory.

• What is a shell observer supposed to be? Commented Jun 17 at 20:44
• @Triatticus A shell observer is an observer staying at a constant radial distance from the black hole.
– Zach
Commented Jun 18 at 14:44
• @safesphere I should have clarified in the question that A and B are separated by 1 lightyear "before you add the black hole", because as you mentioned distances measured in each frame changes after it is "placed". The only thing that truly matters is clock ticks counted by each observer. If A’s time is dilated twice, he will count half as many clock tick for the round trip of light as B would on his clock, no matter the chosen distance.
– Zach
Commented Jun 20 at 17:55
• @safesphere I found that the increase in the travel time for light due to it's slowing down gets overpowered by the clock rate slowing down near the event horizon. I am curious to see what you find.
– Zach
Commented Jun 21 at 18:23

I should have clarified in the question [...] The only thing that truly matters is clock ticks counted by each observer. If A’s time is dilated twice, he will count half as many clock tick for the round trip of light as B would on his clock, no matter the chosen distance.

I found that the increase in the travel time for light due to it's slowing down gets overpowered by the clock rate slowing down near the event horizon.

Based on the radial null geodesic equation in the Schwarzschild spacetime outside the horizon, the coordinate time of light passing between $$r$$ and $$R$$ (the blue curve on the chart below) is:

$$t=r_s \ln{\frac{R-r_s}{r-r_s}} + R - r \tag{1}$$

The time dilation at $$r$$ (the green curve, not to scale) is:

$$\tau=t\,\sqrt{1-\dfrac{r_s}{r}}$$

Combining both formulas and multiplying by $$2$$ gives the round trip time of light between $$r$$ and $$R$$ by the clock of the shell observer hovering at $$r$$ (the red curve):

$$\tau=2\,\sqrt{1-\dfrac{r_s}{r}}\left( r_s \ln{\frac{r_s-R}{r_s-r}} + R - r \right )$$

On the chart below, $$M=1$$, the horizon is at $$r_s=2$$, the observer B is at $$R=10\,r_s=20$$, the observer A is hovering anywhere in between. A holds a flashlight; B holds a mirror. A sends a flash of light out and measures the time the light takes to return. The red curve shows that the round trip time of light by A's clock tends to zero at the horizon.

If you are (hypothetically) hovering asymptotically close to the horizon and send a flash of light to a remote mirror, no matter how far, you see your light returning almost instantly, because in your coordinates, the remote speed of light becomes arbitrarily high.

Please note that the same conclusion does not apply to a free falling observer.

EDIT at the request of the OP to answer the original questions:

a) In A's frame, does B affect A gravitationally as if it was at the distance A measures with the time light takes to travel between them?

For simplicity, I will stay clear of the "distance" part of the question and focus only on the "travel time". Gravitational interactions propagate at the speed of light. If B creates a gravitational wave moving in the direction of A, then this wave would reach A at the same time as a flash of light from B. By the clock of A, this time tends to zero at the horizon. Therefore A would receive any gravitational wave from anywhere in the universe almost instantly while hovering arbitrarily close to the horizon. Also note that both light and gravitational waves received by A would be severely blueshifted and thus would instantly shred A into pieces and vaporize.

These dramatic and unintuitive effects illustrate that hovering close to the horizon is an unnatural behavior for matter. And hovering too close to the horizon is practically impossible due to prohibitively unrealistic values of acceleration, rigidity, and energy required.

b) If the answer to a) is yes, what are the consequences of this effect to an infalling observer?

Unlike hovering, free falling is a natural behavior for matter, an inertial frame, in which matter simply moves forward along its timeline. So a free falling observer experiences none of the discussed effects.

The coordinate time of falling from $$R$$ to $$r$$ is:

$$t=\sqrt{\dfrac{R}{2M}-1}\cdot\left(\left(\dfrac{R}{2}+2M\right)\cdot\arccos\left(\dfrac{2r}{R}-1\right)+\dfrac{R}{2}\sin\left(\arccos\left(\dfrac{2r}{R}-1\right)\right)\right)+\, 2M\ln\left(\left|\dfrac{\sqrt{\dfrac{R}{2M}-1}+\tan\left(\dfrac{1}{2}\arccos\left(\dfrac{2r}{R}-1\right)\right)}{\sqrt{\dfrac{R}{2M}-1}-\tan\left(\dfrac{1}{2}\arccos\left(\dfrac{2r}{R}-1\right)\right)}\right|\right)$$

The coordinate time for light to move from $$R$$ to $$r$$ is given by formula $$(1)$$ above, but let us consider that light is moving from a larger distance $$R^*>R$$:

$$t=r_s \ln{\frac{R^*-r_s}{r-r_s}} + R^* - r$$

Equating these formulas to eliminate $$t$$, we obtain the relation between $$r$$, $$R$$, and $$R^*$$ ahown on the second chart below. There $$M=1$$, the horizon is at $$r_s=2$$, A is in a free fall from $$R=10\,r_s=20$$, B from a larger distance $$R^*$$ sends a flash of light that catches with A before the singularity. The vertical axis shows $$R^*$$, the location of B. The horizontal axis shows $$r$$, at which the light from B catches with A.

For example, the blue dot shows that light of B from the distance of $$R^*=104.6$$ catches up at the horizon $$r=2$$ with A falling from rest at $$R=20$$. The chart shows that A never sees the light emitted at the time of his fall from distances above $$R^*=105$$. In other words, a free falling observer does not see the infinite future of the external universe.

Finally, the proper time of A falling from rest at $$R$$ to $$r$$ is:

$$\tau=\dfrac{R}{2}\sqrt{\dfrac{R}{2M}}\left(\arccos\left(\dfrac{2r}{R}-1\right)+\sin\left(\arccos\left(\dfrac{2r}{R}-1\right)\right)\right)$$

The third chart below shows that the clock of A (the vertical axis) shows $$\tau=98$$ at the horizon, the time that takes A to fall from $$R=20$$ and for the light of B from $$R^*=104.6$$ to catch up with A. As you can see, it is not at all nearly instant like in the case of the shell observer.

• Thank you so much for taking the time to calculate and show this. It solves part of the question, but I cannot mark the question as solved because we still need answers to questions a) and b). a) needs to be answered first. I was looking to find out if for observer A the gravitational effects of B would feel as if B was a fraction of a light second away, and therefore A would feel the gravity of other (possibly all) mass and energy "far away" as if they were arbitrarily close.
– Zach
Commented Jun 24 at 14:47
• Once again thank you for taking the time. As you now have answered the question I will accept the answer. But the conclusion of a) seems to cause a paradox. I might have to open a new question. As you place A closer and closer to the event horizon, it seems there is a frame outside the horizon where the energy density outside the black hole equalizes with the energy density of the black hole as EM signals become almost instantaneous to all energy outside. The black hole would not exist in that frame, which causes the paradox.
– Zach
Commented Jun 25 at 14:52
• @Zach There is no paradox, because the energy you describe is kinetic, which is frame dependent and doesn’t bend spacetime. Nothing physically changes, because you move to another frame. Physics is not frame dependent. The stress-energy of a photon is its 4-momentum, which is a null vector with a zero magnitude. Therefore light doesn’t bend spacetime despite widespread misconceptions. In other words, all energy of a photon is kinetic, which doesn’t bend spacetime. Energy of everything that falls also becomes kinetic-only at the horizon. This is why the stress-energy of a black hole is zero. Commented Jun 26 at 5:45
• I now see there is no paradox because the effects would be seen from all frames, although harder to see from an observer far away. Just following from the answer to a), I don't see how there wouldn't be a tipping point close to the horizon where the gravitational pull of matter in the outside universe does not overpower the gravity of the black hole as the gravity of matter from all the stars, planets, etc. from the outside universe would gets arbitrarily close to the observer. As it was concluded that the gravity A would feel is as if the matter was physically that arbitrarily close.
– Zach
Commented Jun 27 at 13:45
• @Zach A shorter travel time means a higher speed, not a smaller distance. The distance is actually larger. A’s time is severely dilated, so he sees light moving enormously fast like in a fast forwarded movie. In this movie, leaking faucets pour a lot of water, but a mattress doesn’t yield under your weight any more than at a normal frame rate. Commented Jun 28 at 5:45