<blockquote>
  <p>I am curious on how to turn those factors into relatable terms such as: At 1 km from the event horizon ____ time passes on earth in an hour.</p>
</blockquote>

<p>The key here is to understand what the formula relates to.  The (all important) <em>observer</em> is in this case a notional observer so distant that curvature (and hence gravitational effects) are negligible.</p>

<p>That observer is the "base clock" we reference.  Everything is dilated relative to <em>that</em> clock.</p>

<p>That means that the dilation on Earth (not strictly speaking an ideal distant observer) has to be taken into account to get an answer in the form you mean.  So starting with the formula you had :</p>

<p><span class="math-container">$$t = t_\infty \sqrt{1 - \frac{r_s}{r}}$$</span></p>

<p>We want this relative to Earth so we need :</p>

<p><span class="math-container">$$t_e = t_\infty \sqrt{1 - \frac{r_s}{r_e}}$$</span></p>

<p>And combining youy get :</p>

<p><span class="math-container">$$t = t_e \sqrt{ \frac {1 - \frac {r_s} r} {1 - \frac {r_s} {r_e}} }
= t_e \sqrt { \frac { r_e(r-r_s) } { r(r_e-r_s) } }$$</span></p>

<p>Now in the specific case of Earth and Sagitarius A*, <span class="math-container">$r_e$</span> is <em>huge</em> so the dilation factor for Earth is as close to one as makes no difference in practice, so you don't <em>really</em> need to go to all that trouble and you can use <span class="math-container">$t_\infty \approx t_e$</span>.</p>

CC BY-SA 4.0
2018-11-20T22:35:18.647Z
2018-11-20T22:35:18.647Z
StephenG - Help Ukraine
<p>On this post,</p>

<p><a href="https://astronomy.stackexchange.com/questions/10467/time-according-to-the-gravity-of-sagittarius-a/10470#10470">Time according to the gravity of Sagittarius A?</a></p>

<p>They found the factor of time on the varying distances from the SMBH, I am curious on how to turn those factors into relatable terms such as:
At 1 km from the event horizon ____ time passes on earth in an hour.</p>

<p>I'm just not sure how to figure this out.</p>

CC BY-SA 4.0
2018-11-20T04:36:37.427Z
Hi, Blake, welcome to Astronomy Stack Exchange! I'm not sure what you mean by "reliable"; did you mean "relatable" instead - referring to more common units understandable on a human scale? Please add a bit more detail to your question to address that. Thanks!
HDE 226868
CC BY-SA 4.0
2018-11-20T05:53:58.417Z
Thank you and yes, common units. I just want to be able to take the factor of time and explain how much time would pass on earth at a certain distance from the event horizon of a black hole, in hours, days and/or years. I'm sure this is a simple question but I can't remember what to do with the factors. I'm not a mathematician. Haha
Blake Frazier
CC BY-SA 4.0
2018-11-20T15:43:03.880Z
So what are you asking?  Your link gives a very simple formula for time dilation as a function of radius (relative to the Schwartzchild radius).  So look up the $r_s$ for the black hole of interest and you're done.
Carl Witthoft
CC BY-SA 4.0
2018-11-21T03:17:55.993Z
@JamesK So 3500 hours pass on Earth as 1 hour passes 1km from the horizon.
PM 2Ring
CC BY-SA 4.0
2018-11-21T07:42:09.187Z
I think @JamesK has it backwards and PM 2Ring has it right.
Phil N DeBlanc
CC BY-SA 4.0
2018-11-21T07:54:52.243Z
Yes I wrote that backwards
James K
<black-hole><space-time><space><supermassive-black-hole><time-dilation>
How to convert factors of time dilation into something relatable?
Blake Frazier
# How to convert factors of time dilation into something relatable?

(Asked by: username_0 on 2018-11-20)

On this post,

[Time according to the gravity of Sagittarius A?](https://astronomy.stackexchange.com/questions/10467/time-according-to-the-gravity-of-sagittarius-a/10470#10470)

They found the factor of time on the varying distances from the SMBH, I am curious on how to turn those factors into relatable terms such as:
At 1 km from the event horizon \_\_\_\_ time passes on earth in an hour.

I'm just not sure how to figure this out.

**Tags:** black-hole, space-time, space, supermassive-black-hole, time-dilation
> **Comments:**
> **username_1**: Hi, Blake, welcome to Astronomy Stack Exchange! I'm not sure what you mean by "reliable"; did you mean "relatable" instead - referring to more common units understandable on a human scale? Please add a bit more detail to your question to address that. Thanks!  
> **username_0**: Thank you and yes, common units. I just want to be able to take the factor of time and explain how much time would pass on earth at a certain distance from the event horizon of a black hole, in hours, days and/or years. I'm sure this is a simple question but I can't remember what to do with the factors. I'm not a mathematician. Haha  
> **username_2**: So what are you asking?  Your link gives a very simple formula for time dilation as a function of radius (relative to the Schwartzchild radius).  So look up the $r_s$ for the black hole of interest and you're done.  
> **username_3**: @JamesK So 3500 hours pass on Earth as 1 hour passes 1km from the horizon.  
> **username_4**: I think @JamesK has it backwards and PM 2Ring has it right.  
> **username_5**: Yes I wrote that backwards  

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## 1 answers

### Answer by username_6 on 2018-11-20. Score: 1

> I am curious on how to turn those factors into relatable terms such as: At 1 km from the event horizon \_\_\_\_ time passes on earth in an hour.

The key here is to understand what the formula relates to. The (all important) *observer* is in this case a notional observer so distant that curvature (and hence gravitational effects) are negligible.

That observer is the "base clock" we reference. Everything is dilated relative to *that* clock.

That means that the dilation on Earth (not strictly speaking an ideal distant observer) has to be taken into account to get an answer in the form you mean. So starting with the formula you had :

$$t = t\_\infty \sqrt{1 - \frac{r\_s}{r}}$$

We want this relative to Earth so we need :

$$t\_e = t\_\infty \sqrt{1 - \frac{r\_s}{r\_e}}$$

And combining youy get :

$$t = t\_e \sqrt{ \frac {1 - \frac {r\_s} r} {1 - \frac {r\_s} {r\_e}} }
= t\_e \sqrt { \frac { r\_e(r-r\_s) } { r(r\_e-r\_s) } }$$

Now in the specific case of Earth and Sagitarius A\*, $r\_e$ is *huge* so the dilation factor for Earth is as close to one as makes no difference in practice, so you don't *really* need to go to all that trouble and you can use $t\_\infty \approx t\_e$.


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