# rotate right ascension, declination based on delta time

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...

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.