<p>Let's start with a description of sidereal time. From my knowledge, sidereal time at any location can be described as:</p>
<p><code>the right ascension currently at the zenith</code></p>
<p>Here are the definitions of <a href="https://en.wikipedia.org/wiki/Zenith" rel="nofollow noreferrer">zenith</a> and <a href="https://en.wikipedia.org/wiki/Right_ascension" rel="nofollow noreferrer">right ascension</a>, if you need them.</p>
<p>So, if the sidereal time at some location is 12 hours, 0 minutes, and 0 seconds, the right ascension directly above that location at that time will be 12:00:00. So, one sidereal day is the time it takes for the sidereal time at any location to &quot;repeat itself&quot; - i.e., the time between one 00:00:00 and the next 00:00:00. This means that one sidereal day is the time it takes for Earth to rotate 360 degrees.</p>
<p>The time that full rotation takes - i.e., one <a href="https://en.wikipedia.org/wiki/Sidereal_time" rel="nofollow noreferrer">sidereal day</a> - has been calculated to be roughly 23 hours and 56 minutes, or 1436 minutes. Since you're measuring your time in <em>solar</em> time, you would expect to see this 4-minute discrepancy in a lot of places, because, well, 23 hours and 56 minutes is 4 minutes less than a full solar day of 24 hours. From one day to the next, most objects (the Moon is a notable exception) will rise 4 minutes earlier.</p>
<p><strong>So, yes, your analysis is correct - this daily 4 minute change is not really &quot;significant&quot; since it is a consequence of the Earth's rotation around the Sun, and you should expect to see that.</strong></p>
<hr />
<p>A side note:</p>
<p>Additionally, we can see that the time it takes Earth to rotate one full degree is:</p>
<p><span class="math-container">$\dfrac{1436\: \rm{minutes}}{360 \: \rm{degrees}} = 3.99 \dfrac{\rm{minutes}}{\rm{degree}}$</span></p>
<p>So, this 4-minute daily change is also a nearly 1-degree daily change.</p>

CC BY-SA 4.0
2020-11-19T00:10:18.143Z
2020-11-19T00:10:18.143Z
sforsingh
<p>My understanding of sidereal time is that the position of a star will be in the same place 4 minutes earlier the next day. So a 20 days later the star will be 1hr 20 mins earlier. So if the star is in the same place it must be a multiple of 4 minutes earlier (and the seconds that I have ignored). So it is not possible for the star to be in the same place and not be that time multiple apart.</p>
<p>Its just that I have some data points where the locations are identical and they are all multiples of 4 mins apart but I am worried that that is not significant because it is inevitable. Is that correct?</p>

CC BY-SA 4.0
2020-11-19T00:04:00.383Z
I don't really get the confusion. I don't understand why you are "worried" about this.  Suppose it is not significant? why would that be a problem and something to worry about?
James K
CC BY-SA 4.0
2020-11-19T20:14:41.623Z
@ James K I suppose you are right. The fact that multiple data points map to the same location is more informative.
user36093
<positional-astronomy>
Confused about the significance of sidereal time
user36093
# Confused about the significance of sidereal time

(Asked by: username_0 on 2020-11-18)

My understanding of sidereal time is that the position of a star will be in the same place 4 minutes earlier the next day. So a 20 days later the star will be 1hr 20 mins earlier. So if the star is in the same place it must be a multiple of 4 minutes earlier (and the seconds that I have ignored). So it is not possible for the star to be in the same place and not be that time multiple apart.

Its just that I have some data points where the locations are identical and they are all multiples of 4 mins apart but I am worried that that is not significant because it is inevitable. Is that correct?

**Tags:** positional-astronomy

> **Comments:**
> **username_1**: I don't really get the confusion. I don't understand why you are "worried" about this.  Suppose it is not significant? why would that be a problem and something to worry about?  
> **username_0**: @ username_1 I suppose you are right. The fact that multiple data points map to the same location is more informative.  

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## 1 answers

### Answer by username_2 on 2020-11-19. Score: 1 | Accepted answer

Let's start with a description of sidereal time. From my knowledge, sidereal time at any location can be described as:

`the right ascension currently at the zenith`

Here are the definitions of [zenith](https://en.wikipedia.org/wiki/Zenith) and [right ascension](https://en.wikipedia.org/wiki/Right_ascension), if you need them.

So, if the sidereal time at some location is 12 hours, 0 minutes, and 0 seconds, the right ascension directly above that location at that time will be 12:00:00. So, one sidereal day is the time it takes for the sidereal time at any location to "repeat itself" - i.e., the time between one 00:00:00 and the next 00:00:00. This means that one sidereal day is the time it takes for Earth to rotate 360 degrees.

The time that full rotation takes - i.e., one [sidereal day](https://en.wikipedia.org/wiki/Sidereal_time) - has been calculated to be roughly 23 hours and 56 minutes, or 1436 minutes. Since you're measuring your time in *solar* time, you would expect to see this 4-minute discrepancy in a lot of places, because, well, 23 hours and 56 minutes is 4 minutes less than a full solar day of 24 hours. From one day to the next, most objects (the Moon is a notable exception) will rise 4 minutes earlier.

**So, yes, your analysis is correct - this daily 4 minute change is not really "significant" since it is a consequence of the Earth's rotation around the Sun, and you should expect to see that.**

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A side note:

Additionally, we can see that the time it takes Earth to rotate one full degree is:

$\dfrac{1436\: \rm{minutes}}{360 \: \rm{degrees}} = 3.99 \dfrac{\rm{minutes}}{\rm{degree}}$

So, this 4-minute daily change is also a nearly 1-degree daily change.


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