<p>Not at all a dumb question. As you have heard, it is true that time is affected by gravity. The stronger the gravitational field, the slower time passes. If you're far from any gravitating matter, time passes "normally".</p>

<p>But to answer your question, we must specify what is meant by "the black holes's time" (let's call the black hole <span class="math-container">$\mathrm{BH}_\mathrm{Sgr\,A^*}$</span>; see note below on the nomenclature), since it depends on how far from Sgr A* we are talking. The time pace at a distance <span class="math-container">$r$</span> from the center of a BH is given by
<span class="math-container">$$t = t_\infty \sqrt{1 - \frac{r_\mathrm{S}}{r}},$$</span>
where <span class="math-container">$t_\infty$</span> is the time "at infinity", i.e. far from the BH, and
<span class="math-container">$$r_\mathrm{S} \equiv \frac{2GM}{c^2} \simeq 3\,\mathrm{km}\,\times \left( \frac{M}{M_\odot}\right)$$</span>
is the so-called Schwarzschild radius (the "surface" of the BH), which is where not even light can escape. Here, <span class="math-container">$G$</span> is the gravitational constant, <span class="math-container">$M$</span> is the mass of the BH, <span class="math-container">$c$</span> is the speed of light, and <span class="math-container">$M_\odot$</span> is the mass of the Sun.</p>

<p>The last equality shows that a BH with the mass of the Sun would have a radius of 3 km. The mass of <span class="math-container">$\mathrm{BH}_\mathrm{Sgr\,A^*}$</span> is some 4.1 million Solar masses, so its radius is <span class="math-container">$r_\mathrm{S} = 12.1$</span> million km.</p>

<p>Plugging in the other numbers, we can see that at a distance from <span class="math-container">$\mathrm{BH}_\mathrm{Sgr\,A^*}$</span> of</p>

<ol>
<li><strong>1 lightyear</strong>, time runs slower by a factor of 1.00000064, i.e. unnoticeably.</li>
<li><strong>1 astronomical unit</strong> (the distance from Earth to the Sun), time runs 4% slower.</li>
<li><strong>1 million km from the surface</strong>, time runs slower by a factor of 3.6.</li>
<li><strong>1000 km from the surface</strong>, time runs slower by a factor of 110.</li>
<li><strong>1 km from the surface</strong>, time runs slower by a factor of ~3500.</li>
<li><strong>1 m from the surface</strong>, time runs more than a 100,000 times slower.</li>
<li><strong>At the surface</strong>, time stops.</li>
</ol>

<p>Note that this time dilation is <em>what a distant observer</em> (i.e. the guy with the <span class="math-container">$t_\infty$</span> time) would measure for an observer at the distance <span class="math-container">$r$</span>. The person at <span class="math-container">$r$</span> would just measure his/her own time as usual. For instance, according to point 5 above, if you were hovering 1 km from the surface, waving your hand every second, then I, choosing to stay at a safe distance of 1 lightyear but with a magically powerful telescope, would see you wave approximately once every hour. And when you run out of fuel and plummet into the BH, then when you cross the surface you wouldn't notice anything particular, but I would see you frozen in time. This is the concept of relativity.</p>

<p>Finally, let me use this chance to clarify something that people, including myself, often have gotten wrong: <strong>Sagittarius A</strong> (without an asterisk) is a radio source in the center of the Milky Way. It consists of three parts: <strong>Sagittarius A East</strong> (a supernova remnant), <strong>Sagittarius A West</strong> (dust and gas clouds), and <strong>Sagittarius A*</strong>, or <strong>Sgr A*</strong>, which is a very bright and compact radio source believed to be formed by a supermassive BH. Sgr A* isn't actually the BH <em>itself</em>. I think the BH doesn't really have a name, so I'll call it <span class="math-container">$\mathrm{BH}_\mathrm{Sgr\,A^*}$</span>. Maybe that's a bad name…</p>

CC BY-SA 3.0
2015-04-14T13:08:37.100Z
@user6760: Time stops only for an "outside" observer. The person close to the BH doesn't notice anything. I clarified in the text. Thanks for prompting me to do this.
pela
CC BY-SA 3.0
2015-04-14T13:51:25.277Z
Great answer and very illustrating numbers about the gradual time [dilatation](http://www.etymonline.com/index.php?term=dilation&allowed_in_frame=0) with distance from a SMBH. Could throw in the effect for a GPS satellite there too. As for the naming issue, maybe we should call it the "A-Star Sagittarius Hole", or for short the AS... no I won't type that. I'm afraid that the next IAU meeting might buy it, though.
LocalFluff
CC BY-SA 3.0
2015-04-14T19:24:25.120Z
:D @LocalFluff. And thanks for the link, I didn't know about the "mis"-etymology of dilation. Also thanks for the encouragement about the GPS, but I think I'll stick with the time stuff.
pela
CC BY-SA 3.0
2015-04-15T10:31:30.743Z
For a hypothetical Sgr A*-massed Schwarzschild black hole, the tidal forces across $1.8\,\mathrm{m}$-tall human near the horizon should be on the scale of $10^{-4}\, \mathrm{gee}$ or so. Supermassive black holes don't spaghettify until well past the horizon. The time dilation calculations are misleading because the Schwarzschild radial coordinate does not straightforwardly correspond to a radial distance. For example, if $r_\text{ft} = r_\text{S}+1\,\mathrm{m}$, then $r_\text{hd} = r_\text{ft} + 16\,\mathrm{\mu m}$ for the human. That's one way to think about why the tidal forces are small.
Stan Liou
CC BY-SA 3.0
2015-04-15T11:57:19.253Z
Excellent comment, @StanLiou, I hadn't thought of that. I'll remove the part about time dilation difference.
pela
CC BY-SA 3.0
2016-07-30T20:54:37.323Z
*"then ( A ), choosing to stay at a safe distance of 1 lightyear but with a magically powerful telescope, would see ( B ) wave approximately once every hour..."*  What if B has the same model telescope, what does B see looking at A?  Does B see A waving A's hand very fast?  (Say, about 60 times/second)
Fattie
CC BY-SA 3.0
2016-08-05T13:46:03.577Z
@JoeBlow: Yes, if B hovers above the horizon (as in this example), the difference in the rate of time is the opposite, so B sees A wave fast. But if B is freely falling, then I think — but I'm not exactly sure of this, to be honest — that B would see A's time run normally. At least B wouldn't see A's time run as fast, as in the case of hovering.
pela
CC BY-SA 4.0
2019-10-20T05:09:15.130Z
Thank You for answering with this detail. I am curious will the gravitational Time Dialation 1 meter from the event horizon would "stack" onto the time dialation caused by the "velocity" of the plazma traveling around the EH? So Time slows down for both reasons. 1) proximity to the gravitational field,.. and 2) the speed at which the accretion disk travels around SGR A EH. Is this correct? And.. If Time Stops at the surface of the EH solely as a result of gravitational time dilation, then any matter traveling at the extreme high velocity around the BH would be subject to further Time Dialation
Richard
CC BY-SA 4.0
2019-10-20T05:09:15.130Z
Would that reverse time for that particle; and if so "when" would the "observer" see that particle?
Richard
CC BY-SA 4.0
2019-10-20T20:10:26.410Z
@Richard Yes, a particle traveling at high speeds in a gravitational field is subject to two "cumulative" time dilations, an SR and a GR time dilation, whether it be close to a BH, or — as has actually been experimented with — in an airplane flying here at Earth. Both effect slow down time. But in our reference frame, the particle approaching the EH moves slower and slower, so the SR time dilation approaches zero. No time is reversed outside the BH.
pela
CC BY-SA 4.0
2023-04-08T06:42:54.670Z
Is it possible to stay or orbit at 1 astronomical unit near BH with current technology?
Pasupathi Rajamanickam
CC BY-SA 4.0
2023-04-08T14:52:54.913Z
@PasupathiRajamanickam Staying at such an orbit doesn't really require any technology; you will just orbit. Of course _getting there_, from Earth, requires technology that we don't (yet) have, since it's some 25,000 lightyears away, but I assume that's not what you mean. Note though that you cannot have a _stable_ orbit inside $3r_\mathrm{S}$, the so-called [innermost stable circular orbit](https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit), i.e. ~36 million kilometers, or 0.24 AU. Moreover, your orbit may also be disturbed by other matter plunging into the BH.
pela
CC BY-SA 4.0
2023-04-12T17:41:46.737Z
@pela Thank you for calling out ISCO. QQ, what about temperature at .24AU? A normal space shuttle can withstand temperature at this location. 
Let me worry about getting there ;) wanted to visualize.
Pasupathi Rajamanickam
CC BY-SA 4.0
2023-04-12T20:46:45.840Z
@PasupathiRajamanickam The accretion onto the Sgr A* black hole is very small, so I'd think gas temperature wouldn't be a problem, but I don't know on top of my head. But you're right that accretion disks in general are extremely hot, up to several millions of degrees. Perhaps you should ask a new, separate question about this?
pela
CC BY-SA 4.0
2015-04-14T11:35:25.207Z
2019-11-01T13:18:32.240Z
2019-11-01T13:18:32.240Z
pela
<p>This may be a really dumb question (I'm more of a Biologist than an Astronomer) so I apologize in advance for my little knowledge pertaining to Astronomy, but, if I'm not mistaken, time is effected by gravity, right? So what is Sagittarius A*'s time compared to ours since it has so much stronger gravity? Do we know specifically the difference?</p>

<space-time><supermassive-black-hole><time-dilation>
Time according to the gravity of Sagittarius A*?
CDB
# Time according to the gravity of Sagittarius A*?

(Asked by: username_0 on 2015-04-14)

This may be a really dumb question (I'm more of a Biologist than an Astronomer) so I apologize in advance for my little knowledge pertaining to Astronomy, but, if I'm not mistaken, time is effected by gravity, right? So what is Sagittarius A\*'s time compared to ours since it has so much stronger gravity? Do we know specifically the difference?

**Tags:** space-time, supermassive-black-hole, time-dilation


---

## 1 answers

### Answer by username_1 on 2015-04-14. Score: 25 | Accepted answer

Not at all a dumb question. As you have heard, it is true that time is affected by gravity. The stronger the gravitational field, the slower time passes. If you're far from any gravitating matter, time passes "normally".

But to answer your question, we must specify what is meant by "the black holes's time" (let's call the black hole $\mathrm{BH}\_\mathrm{Sgr\,A^\*}$; see note below on the nomenclature), since it depends on how far from Sgr A\* we are talking. The time pace at a distance $r$ from the center of a BH is given by
$$t = t\_\infty \sqrt{1 - \frac{r\_\mathrm{S}}{r}},$$
where $t\_\infty$ is the time "at infinity", i.e. far from the BH, and
$$r\_\mathrm{S} \equiv \frac{2GM}{c^2} \simeq 3\,\mathrm{km}\,\times \left( \frac{M}{M\_\odot}\right)$$
is the so-called Schwarzschild radius (the "surface" of the BH), which is where not even light can escape. Here, $G$ is the gravitational constant, $M$ is the mass of the BH, $c$ is the speed of light, and $M\_\odot$ is the mass of the Sun.

The last equality shows that a BH with the mass of the Sun would have a radius of 3 km. The mass of $\mathrm{BH}\_\mathrm{Sgr\,A^\*}$ is some 4.1 million Solar masses, so its radius is $r\_\mathrm{S} = 12.1$ million km.

Plugging in the other numbers, we can see that at a distance from $\mathrm{BH}\_\mathrm{Sgr\,A^\*}$ of

1. **1 lightyear**, time runs slower by a factor of 1.00000064, i.e. unnoticeably.
2. **1 astronomical unit** (the distance from Earth to the Sun), time runs 4% slower.
3. **1 million km from the surface**, time runs slower by a factor of 3.6.
4. **1000 km from the surface**, time runs slower by a factor of 110.
5. **1 km from the surface**, time runs slower by a factor of ~3500.
6. **1 m from the surface**, time runs more than a 100,000 times slower.
7. **At the surface**, time stops.

Note that this time dilation is *what a distant observer* (i.e. the guy with the $t\_\infty$ time) would measure for an observer at the distance $r$. The person at $r$ would just measure his/her own time as usual. For instance, according to point 5 above, if you were hovering 1 km from the surface, waving your hand every second, then I, choosing to stay at a safe distance of 1 lightyear but with a magically powerful telescope, would see you wave approximately once every hour. And when you run out of fuel and plummet into the BH, then when you cross the surface you wouldn't notice anything particular, but I would see you frozen in time. This is the concept of relativity.

Finally, let me use this chance to clarify something that people, including myself, often have gotten wrong: **Sagittarius A** (without an asterisk) is a radio source in the center of the Milky Way. It consists of three parts: **Sagittarius A East** (a supernova remnant), **Sagittarius A West** (dust and gas clouds), and **Sagittarius A\***, or **Sgr A\***, which is a very bright and compact radio source believed to be formed by a supermassive BH. Sgr A\* isn't actually the BH *itself*. I think the BH doesn't really have a name, so I'll call it $\mathrm{BH}\_\mathrm{Sgr\,A^\*}$. Maybe that's a bad name…

> **Comments:**
> **username_1**: @user6760: Time stops only for an "outside" observer. The person close to the BH doesn't notice anything. I clarified in the text. Thanks for prompting me to do this.  
> **username_2**: Great answer and very illustrating numbers about the gradual time [dilatation](http://www.etymonline.com/index.php?term=dilation&allowed_in_frame=0) with distance from a SMBH. Could throw in the effect for a GPS satellite there too. As for the naming issue, maybe we should call it the "A-Star Sagittarius Hole", or for short the AS... no I won't type that. I'm afraid that the next IAU meeting might buy it, though.  
> **username_1**: :D @username_2. And thanks for the link, I didn't know about the "mis"-etymology of dilation. Also thanks for the encouragement about the GPS, but I think I'll stick with the time stuff.  
> **username_3**: For a hypothetical Sgr A*-massed Schwarzschild black hole, the tidal forces across $1.8\,\mathrm{m}$-tall human near the horizon should be on the scale of $10^{-4}\, \mathrm{gee}$ or so. Supermassive black holes don't spaghettify until well past the horizon. The time dilation calculations are misleading because the Schwarzschild radial coordinate does not straightforwardly correspond to a radial distance. For example, if $r_\text{ft} = r_\text{S}+1\,\mathrm{m}$, then $r_\text{hd} = r_\text{ft} + 16\,\mathrm{\mu m}$ for the human. That's one way to think about why the tidal forces are small.  
> **username_1**: Excellent comment, @username_3, I hadn't thought of that. I'll remove the part about time dilation difference.  
> **username_4**: *"then ( A ), choosing to stay at a safe distance of 1 lightyear but with a magically powerful telescope, would see ( B ) wave approximately once every hour..."*  What if B has the same model telescope, what does B see looking at A?  Does B see A waving A's hand very fast?  (Say, about 60 times/second)  
> **username_1**: @JoeBlow: Yes, if B hovers above the horizon (as in this example), the difference in the rate of time is the opposite, so B sees A wave fast. But if B is freely falling, then I think — but I'm not exactly sure of this, to be honest — that B would see A's time run normally. At least B wouldn't see A's time run as fast, as in the case of hovering.  
> **username_5**: Thank You for answering with this detail. I am curious will the gravitational Time Dialation 1 meter from the event horizon would "stack" onto the time dialation caused by the "velocity" of the plazma traveling around the EH? So Time slows down for both reasons. 1) proximity to the gravitational field,.. and 2) the speed at which the accretion disk travels around SGR A EH. Is this correct? And.. If Time Stops at the surface of the EH solely as a result of gravitational time dilation, then any matter traveling at the extreme high velocity around the BH would be subject to further Time Dialation  
> **username_5**: Would that reverse time for that particle; and if so "when" would the "observer" see that particle?  
> **username_1**: @username_5 Yes, a particle traveling at high speeds in a gravitational field is subject to two "cumulative" time dilations, an SR and a GR time dilation, whether it be close to a BH, or — as has actually been experimented with — in an airplane flying here at Earth. Both effect slow down time. But in our reference frame, the particle approaching the EH moves slower and slower, so the SR time dilation approaches zero. No time is reversed outside the BH.  
> **username_6**: Is it possible to stay or orbit at 1 astronomical unit near BH with current technology?  
> **username_1**: @username_6 Staying at such an orbit doesn't really require any technology; you will just orbit. Of course _getting there_, from Earth, requires technology that we don't (yet) have, since it's some 25,000 lightyears away, but I assume that's not what you mean. Note though that you cannot have a _stable_ orbit inside $3r_\mathrm{S}$, the so-called [innermost stable circular orbit](https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit), i.e. ~36 million kilometers, or 0.24 AU. Moreover, your orbit may also be disturbed by other matter plunging into the BH.  
> **username_6**: @username_1 Thank you for calling out ISCO. QQ, what about temperature at .24AU? A normal space shuttle can withstand temperature at this location.  Let me worry about getting there ;) wanted to visualize.  
> **username_1**: @username_6 The accretion onto the Sgr A* black hole is very small, so I'd think gas temperature wouldn't be a problem, but I don't know on top of my head. But you're right that accretion disks in general are extremely hot, up to several millions of degrees. Perhaps you should ask a new, separate question about this?  

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