Gravitational time dilation calculation near a Black Hole

[Antonio]

Hello to everyone!

I am trying to understand in practice the Gravitational Time Dilation by calculating the time distortion near the supermassive black hole Sgr A* in our Galactic Center.

According to Einstein's Theory of General Relativity, the time distortion near a black hole is calculated with the following formula:

T^2 = t^2 / 1 - ( Rs / r ), where:

T= Observer A time, far away from the black hole
t = Observer B close to the black hole
r = Distance between the black hole and Observer B
Rs= Schwarzschild Radius

For the Sgr A* black hole I found that the Schwarzschild Radius is Rs=13 * 10^16 km. Based on that, I tried to calculate how close to the Event Horizon of the black hole does Observer B have to be so that Observer A sees 60 seconds pass for every 12 seconds on Observer B clock. I calculated that the distance should be r=541.666,67 km

Could you tell me please if my reasoning is correct and if not what is the proper answer?

Thank you in advance!

 

[stevendaryl]

According to Einstein's Theory of General Relativity, the time distortion near a black hole is calculated with the following formula:

T^2 = t^2 / 1 - ( Rs - r ), where:

T= Observer A time, far away from the black hole
t = Observer B close to the black hole
r = Distance between the black hole and Observer B
Rs= Schwarzschild Radius

That's not quite right. It should be:

T^2 = t^2 \frac{1}{1-(R_s/r)}

But actually, that formula is only valid for a special case: when Observer B is hovering in place outside the black hole. A more realistic case would be for observer B to be traveling in a circular orbit around the black hole.

 

[Antonio]

Initially, thank you very much for your answer. I am really sorry, I made a typo in the formula, which I have just corrected it.

1. So the distance from the event horizon [ r=541.666,67 km ] that I have calculated is true only for a hypothetical scenario in which the Observer B is hovering in place outside the black hole.
2. In order to calculate a more realistic case, in which the Observer B is traveling in a circular orbit around the black hole should I calculate also the time distortion due to the relative velocity?

 

[Nugatory]

In order to calculate a more realistic case, in which the Observer B is traveling in a circular orbit around the black hole should I calculate also the time distortion due to the relative velocity?

There's more to it than that because the gravitational time dilation is different for a hovering observer and one in a freefall orbit (as you would expect, because they are following very different paths through spacetime). The wikipedia page on "Gravitational time dilation" has the formula you need, and then you will have to add the effects of relative velocity on top of that.

 

[stevendaryl]

Initially, thank you very much for your answer. I am really sorry, I made a typo in the formula, which I have just corrected it.

1. So the distance from the event horizon [ r=541.666,67 km ] that I have calculated is true only for a hypothetical scenario in which the Observer B is hovering in place outside the black hole.
2. In order to calculate a more realistic case, in which the Observer B is traveling in a circular orbit around the black hole should I calculate also the time distortion due to the relative velocity?

Yes, but in Schwarzschild coordinates, the time-dilation effect of tangential motion (constant r) is different from that of radial motion.

For circular motion, we have:

dt^2 = [(1-R_s/r) - r^2 \dot{\phi}^2] dT^2

where \dot{\phi} is the angular velocity, as measured in terms of the time variable T.

 

[PeterDonis]

A more realistic case would be for observer B to be traveling in a circular orbit around the black hole.

It is worth noting, though, that free-fall circular orbits are only possible outside $r = 3M$ (3/2 the Schwarzschild radius), and those orbits are unstable until you get outside $r = 6M$ (3 times the Schwarzschild radius). The formula for time dilation in a free-fall circular orbit is

$$
T = \frac{t}{\sqrt{1 - \frac{3}{2} \frac{R_s}{r}}}
$$

The orbital velocity doesn't appear directly because for a free-fall orbit it has a fixed relationship to $R_s$ and $r$.

 

[PeterDonis]

It's also worth noting that all of the formulas we are discussing are for a non-rotating black hole. AFAIK the hole at the center of our galaxy is not rotating (or at least not significantly).

 

[Antonio]

It's also worth noting that all of the formulas we are discussing are for a non-rotating black hole. AFAIK the hole at the centre of our galaxy is not rotating (or at least not significantly).

Thank you all for the very informative answers.

Finally and after exploring your comments, I have understood much better the way that time distorts near a black hole. Especially, the Wikipedia page on "Gravitational Time Dilation" includes all the necessary formulas regarding the time dilation for an Observer either in a "hovering situation" - rest frame or in a Circular Orbit.

However, the only unclear point to me now is whether the Sgr A* Black Hole is indeed a non-rotating black hole and if we can use these formulas. After a search on the Web, I found that Sgr A* Black Hole is rotating, but I couldn't find if its rotation is significant or not regarding the application of these formulas.

So, can we assume that Sgr A* Black Hole is a non-rotating spherical body and finally to apply these formulas for simple Gravitational Time Dilation calculations? Or our results would be completely wrong if the rotation of the Sgr A* Black Hole is significant?

Thank you.

 

[PeterDonis]

can we assume that Sgr A* Black Hole is a non-rotating spherical body and finally to apply these formulas for simple Gravitational Time Dilation calculations? Or our results would be completely wrong if the rotation of the Sgr A* Black Hole is significant?

If the rotation is significant, the correct formulas would be different (and more complicated).

I don't know how much information we have on the actual rotation rate of the Sgr A black hole.