<p>At some geocentric latitude $\phi$ and longitude $\lambda$, the elevation or altitude $a$ of a star with right ascension $\alpha$ and declination $\delta$ is given by
$$\begin{aligned}
\sin a = \sin\phi \sin \delta + \cos \phi \cos\delta \cos h&amp;&amp;&amp;&amp;&amp;&amp;(1)
\end{aligned}$$
where
$$\begin{aligned}
h=\theta_\phi - \alpha\qquad&amp;&amp;\qquad&amp;&amp;\qquad&amp;&amp;\quad\,&amp;&amp;(2)
\end{aligned}$$
is the <a href="http://en.wikipedia.org/wiki/Hour_angle" rel="nofollow">hour angle</a> to the star, expressed in terms of the local sidereal time $\theta_{\phi}$ and the right ascension $\alpha$.</p>

<p>Differentiating (1) with respect to time yields
$$\begin{aligned}
\cos a \frac{da}{dt} = &amp;
\phantom{+}\,(\cos\phi\sin\delta - \sin\phi \cos\delta \cos h)\,\frac{d\phi}{dt} \\
&amp; + \,(\sin\phi \cos\delta - \cos\phi \sin\delta \cos h)\,\frac{d\delta}{dt} \\
&amp; - \cos\phi \cos\delta \sin h\,\frac{dh}{dt} &amp;\!\!\!(3)
\end{aligned}
$$
While not quite zero, the time derivatives of latitude $\phi$ and declination $\delta$ are negligibly small compared to the $2\pi$ radians per sidereal day time derivative of hour angle $h$. For all practical purposes, the above thus reduces to
$$\begin{aligned}
\cos a \frac{da}{dt} = - \cos\phi \cos\delta \sin h\,\frac{dh}{dt} \quad\!&amp;&amp;&amp;&amp;&amp;&amp;(4)
\end{aligned}
$$
The left hand side of (4) is zero at extrema of elevation angle. Thus we're looking for conditions that make the right hand side of (4) equal zero. Since $dh/dt$ is nonzero, the right hand  side is zero only if one or more of $\cos \phi$, $\cos \delta$, or $\sin h$ is zero. The first two cases ($\cos \phi = 0$ and $\cos \delta = 0$) represent conditions where elevation is constant. The only condition of interest is $\sin h = 0$, which means an hour angle of 0° or 180° (hour angle is constrained to lie between 0° (inclusive) and 360° (exclusive)).</p>

<p>Since latitude $\phi$ and declination $\delta$ both lie between -90° and +90°, the condition $h=180^{\circ}$ represents the minimum possible elevation, while $h=0$ represents the maximum possible elevation. From (2), $h=0$ means $\theta=\alpha$, or local sidereal time being equal to right ascension.</p>

CC BY-SA 3.0
2014-12-28T19:03:13.743Z
David, could you recommend me a book that introduces positional astronomy and develops all these concepts?
Vladimir Vargas
CC BY-SA 3.0
2014-12-28T14:05:15.827Z
2014-12-28T14:05:15.827Z
David Hammen
<p>I'm currently taking a course called Introduction to Astronomy at Coursera. It's not very formal so they didn't present a proof of the statement in the title. I understand why logically one could argue that when a star lies on the projection of the observer's local meridean onto the celestial sphere (I understand that this projection defines the observer's local sidereal time) then this star is at its highest point on the sky.</p>

<p>I haven't found any mathematical proof of this statement and also do not know how to approach the problem geometrically (either euclidean geometry or analytic geometry).</p>

<p>I appreciate any hint or direct answer on how to approach the problem. Thanks.</p>

CC BY-SA 3.0
2014-12-29T22:10:26.507Z
@barrycarter I understood that my sidereal time is the right ascension of the star that is on the projection of my meridean onto the celestial sphere.
Vladimir Vargas
CC BY-SA 3.0
2014-12-29T23:15:35.323Z
@barrycarter I was looking for something similar to the answer. A mathematical explanation of why the meridian is the line where all stars have their maximum elevation. I'm reading also a great book called Elements of Positional Astronomy by José Portilla (don't know if there is a version in english, I read it in spanish).
Vladimir Vargas
<positional-astronomy>
Why is a star at its highest point when local sidereal time = RA (right ascension)?
Vladimir Vargas
# Why is a star at its highest point when local sidereal time = RA (right ascension)?

(Asked by: username_0 on 2014-12-27)

I'm currently taking a course called Introduction to Astronomy at Coursera. It's not very formal so they didn't present a proof of the statement in the title. I understand why logically one could argue that when a star lies on the projection of the observer's local meridean onto the celestial sphere (I understand that this projection defines the observer's local sidereal time) then this star is at its highest point on the sky.

I haven't found any mathematical proof of this statement and also do not know how to approach the problem geometrically (either euclidean geometry or analytic geometry).

I appreciate any hint or direct answer on how to approach the problem. Thanks.

**Tags:** positional-astronomy

> **Comments:**
> **username_0**: @barrycarter I understood that my sidereal time is the right ascension of the star that is on the projection of my meridean onto the celestial sphere.  
> **username_0**: @barrycarter I was looking for something similar to the answer. A mathematical explanation of why the meridian is the line where all stars have their maximum elevation. I'm reading also a great book called Elements of Positional Astronomy by José Portilla (don't know if there is a version in english, I read it in spanish).  

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## 1 answers

### Answer by username_1 on 2014-12-28. Score: 2 | Accepted answer

At some geocentric latitude $\phi$ and longitude $\lambda$, the elevation or altitude $a$ of a star with right ascension $\alpha$ and declination $\delta$ is given by
$$\begin{aligned}
\sin a = \sin\phi \sin \delta + \cos \phi \cos\delta \cos h&&&&&&(1)
\end{aligned}$$
where
$$\begin{aligned}
h=\theta\_\phi - \alpha\qquad&&\qquad&&\qquad&&\quad\,&&(2)
\end{aligned}$$
is the [hour angle](http://en.wikipedia.org/wiki/Hour_angle) to the star, expressed in terms of the local sidereal time $\theta\_{\phi}$ and the right ascension $\alpha$.

Differentiating (1) with respect to time yields
$$\begin{aligned}
\cos a \frac{da}{dt} = &
\phantom{+}\,(\cos\phi\sin\delta - \sin\phi \cos\delta \cos h)\,\frac{d\phi}{dt} \\
& + \,(\sin\phi \cos\delta - \cos\phi \sin\delta \cos h)\,\frac{d\delta}{dt} \\
& - \cos\phi \cos\delta \sin h\,\frac{dh}{dt} &\!\!\!(3)
\end{aligned}
$$
While not quite zero, the time derivatives of latitude $\phi$ and declination $\delta$ are negligibly small compared to the $2\pi$ radians per sidereal day time derivative of hour angle $h$. For all practical purposes, the above thus reduces to
$$\begin{aligned}
\cos a \frac{da}{dt} = - \cos\phi \cos\delta \sin h\,\frac{dh}{dt} \quad\!&&&&&&(4)
\end{aligned}
$$
The left hand side of (4) is zero at extrema of elevation angle. Thus we're looking for conditions that make the right hand side of (4) equal zero. Since $dh/dt$ is nonzero, the right hand side is zero only if one or more of $\cos \phi$, $\cos \delta$, or $\sin h$ is zero. The first two cases ($\cos \phi = 0$ and $\cos \delta = 0$) represent conditions where elevation is constant. The only condition of interest is $\sin h = 0$, which means an hour angle of 0° or 180° (hour angle is constrained to lie between 0° (inclusive) and 360° (exclusive)).

Since latitude $\phi$ and declination $\delta$ both lie between -90° and +90°, the condition $h=180^{\circ}$ represents the minimum possible elevation, while $h=0$ represents the maximum possible elevation. From (2), $h=0$ means $\theta=\alpha$, or local sidereal time being equal to right ascension.

> **Comments:**
> **username_0**: username_1, could you recommend me a book that introduces positional astronomy and develops all these concepts?  

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