<p>The conversion between local and equatorial coordinates is
<span class="math-container">$$\sin\text{Az}=-\frac{\sin\text{HA}\cos\delta}{\cos\text{El}},\quad \sin\text{El}=\sin\delta\sin\phi+\cos\delta\cos\phi\cos\text{HA}$$</span>
with <span class="math-container">$\delta$</span> declination, <span class="math-container">$\phi$</span> your latitude and <span class="math-container">$\text{HA}$</span> the hour angle at the time of the observation, which is the local sidereal time minus right ascension. You can then solve these two equations for declination and hour angle, and convert the hour angle to right ascension by <span class="math-container">$\alpha=\text{LST}-\text{HA}$</span>, with <span class="math-container">$\text{LST}$</span> being the local sidereal time at the second observation.</p>
<p>In your case, the elevation and azimuth aren't changing, so the declination and hour angle are the same. What <em>has</em> changed is the local sidereal time, which - for two observations spaced an hour apart - has changed by approximately one hour, meaning that in turn the right ascension will have changed by approximately one hour.</p>

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2021-06-22T14:59:40.730Z
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HDE 226868
<p>The declination won't change. The right ascension will increase by about 15 degrees per hour or 0.25 degrees per minute.
The actual number is closer to 15.04 degrees per hour of time: (366.25 * 360)/(365.25 * 24). For RA expressed as time it would be about 1.002 hours in 1 hour of time: 366.25/365.25. 366.25 is the number of apparent revolutions of the celestial sphere in a year, and 365.25 is the number of apparent revolutions of the Sun - days. 24 is the number of hours in a day and 360 the number of degrees in a circle.</p>

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stretch
<p>Assuming I was observing a source at a given right ascension <span class="math-container">$\alpha_0$</span> declination <span class="math-container">$\delta_0$</span> at a given time <span class="math-container">$t_0$</span> under a certain azimuth <span class="math-container">$A$</span> and elevation <span class="math-container">$a$</span> angle.</p>
<p>If I now change my time to <span class="math-container">$t_1$</span> (e.g. 1 hour later), how can I determine which <span class="math-container">$\alpha_1$</span> and <span class="math-container">$\delta_1$</span> will be observed based on the same azimuth <span class="math-container">$A$</span> and altitude <span class="math-container">$a$</span> values?</p>
<p>I suppose one way would be to really go the route and transform <span class="math-container">$(\alpha_0, \delta_0)→(t_0,A, a)→(t_1,A, a)→(\alpha_1, \delta_1)$</span>.</p>
<p>My question is: Is there a simpler, more elegant solution to the problem?
Can I rotate <span class="math-container">$(\alpha_0, \delta_0)$</span> directly to <span class="math-container">$(\alpha_1, \delta_1)$</span>?
If so, what steps/rotations are necessary for it (and is this supported in <em>astropy</em>?).
I have a bigger source list thus, going via azimuth/elevation is a bit time-consuming and I think it should be possible to simply rotate the sky-sphere based on a known earth rotation (expressed through delta time)?</p>
<p>The &quot;source&quot; is an extragalactic radio source -&gt; I do not have to bother about its distance, movement or about refraction etc.
Currently, I was only applying a nutation model but maybe this is even unnecessary...</p>

<coordinate>
rotate right ascension, declination based on delta time
user7431005
# rotate right ascension, declination based on delta time

(Asked by: username_0 on 2021-06-22)

Assuming I was observing a source at a given right ascension $\alpha\_0$ declination $\delta\_0$ at a given time $t\_0$ under a certain azimuth $A$ and elevation $a$ angle.

If I now change my time to $t\_1$ (e.g. 1 hour later), how can I determine which $\alpha\_1$ and $\delta\_1$ will be observed based on the same azimuth $A$ and altitude $a$ values?

I suppose one way would be to really go the route and transform $(\alpha\_0, \delta\_0)→(t\_0,A, a)→(t\_1,A, a)→(\alpha\_1, \delta\_1)$.

My question is: Is there a simpler, more elegant solution to the problem?
Can I rotate $(\alpha\_0, \delta\_0)$ directly to $(\alpha\_1, \delta\_1)$?
If so, what steps/rotations are necessary for it (and is this supported in *astropy*?).
I have a bigger source list thus, going via azimuth/elevation is a bit time-consuming and I think it should be possible to simply rotate the sky-sphere based on a known earth rotation (expressed through delta time)?

The "source" is an extragalactic radio source -> I do not have to bother about its distance, movement or about refraction etc.
Currently, I was only applying a nutation model but maybe this is even unnecessary...

**Tags:** coordinate

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## 2 answers

### Answer by username_1 on 2021-06-22. Score: 2 | Accepted answer

The conversion between local and equatorial coordinates is
$$\sin\text{Az}=-\frac{\sin\text{HA}\cos\delta}{\cos\text{El}},\quad \sin\text{El}=\sin\delta\sin\phi+\cos\delta\cos\phi\cos\text{HA}$$
with $\delta$ declination, $\phi$ your latitude and $\text{HA}$ the hour angle at the time of the observation, which is the local sidereal time minus right ascension. You can then solve these two equations for declination and hour angle, and convert the hour angle to right ascension by $\alpha=\text{LST}-\text{HA}$, with $\text{LST}$ being the local sidereal time at the second observation.

In your case, the elevation and azimuth aren't changing, so the declination and hour angle are the same. What *has* changed is the local sidereal time, which - for two observations spaced an hour apart - has changed by approximately one hour, meaning that in turn the right ascension will have changed by approximately one hour.


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### Answer by username_2 on 2021-06-22. Score: 1

The declination won't change. The right ascension will increase by about 15 degrees per hour or 0.25 degrees per minute.
The actual number is closer to 15.04 degrees per hour of time: (366.25 \* 360)/(365.25 \* 24). For RA expressed as time it would be about 1.002 hours in 1 hour of time: 366.25/365.25. 366.25 is the number of apparent revolutions of the celestial sphere in a year, and 365.25 is the number of apparent revolutions of the Sun - days. 24 is the number of hours in a day and 360 the number of degrees in a circle.


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