Geometry of Earth and the Sun

Recently I've been thinking about the things that we take for granted. Our time system, the number of days in a month, the names of the week days...¹

The goal of this post is to find the curve that the sun takes in the sky. First we can say this depends on the day of the year. Thinking in spherical coordinates we can classify the Sun's position at any one time in the sky as the angle to the horizon, and then which compass direction it is. By observation, we know that the Sun rises in the east and sets in the west², let's call the Sun's angle to the horizon $\alpha(t)$, we know $\alpha(0)=\alpha(T)=0$ if $t\in[0, T]$ is the time where soon is not below the horizon, little less obvious, but nontheless true is that the sun's highest point is achieved at $arg \max \alpha(t)=T/2$. These are some basic properties that we'll keep in mind for or full formula later.

To calculate the Sun's height we need the day of year, time of day, and location on earth. But really the day of the year is not a fundamental quanty, rather we can see time of year as a point in $[0, 2\pi)$, but really this does not matter either because the Sun shines equally in any direction. What matters is Earth's tilt towards the sun, the declination $\delta\in[-23^\circ, 23^\circ]$. Then we have time of day, one day is one revolution around Earth's rotational axis. Hence, an angle think in terms of hours and minutes, but more natural is just a number $h\in[0, 2\pi)$ that says how much Earth has spun around its axis.

Then for the position, on the surface of a ball (Earth), every position can be given by a pair $(\phi, \lambda)$ but since the longitude $\lambda$ simply shifts the start of the day, we don't care about it and we can assume $\lambda=0$, i.e on the prime meridian. In the animation below I've set the latitude to $\phi\approx 60^\circ$, the same latitude as Sandviken, Sweden.

Now we have the goal, to find $$\alpha(\delta, t, \phi).$$

Sun                             Earth
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Image laying on your back looking up

Fact: first writen down 1100 years ago in text, and then in it's clean spherical trigonometry formula by Euler in 1748, after a decade of inventing half of physics and maths, and solving the Basel problem...

Laying on your back and looking up in the sky along the direction of unit normal vector $\mathbf{v}$,⁶

Let's assume we have a coordinate system with center in the center of Earth. We assume earth is a unit ball.

Let's assume, for now, that the earth spins on the orbital plane. We will define two unit vectors, one pointing towards the sun $\mathbf{s}$, and one normal vector, point straight up from our position on Earth, also known as the zenith. Taking the dot product of these two $\mathbf{s}\cdot \mathbf{v}=\cos \theta$, where $\theta$ is the angle of sun in relation to the zenith, but since we want the angle of the sun in relation to the horizion, we have $\theta+ \alpha=90^\circ$, so $$\mathbf{s}\cdot \mathbf{v}=\cos\left( 90^\circ - \alpha\right)=\sin \alpha.$$

This is the fundamental relation for $\alpha=\alpha(\phi, t, \delta)$

In this simple example it's just $$\mathbf{s} =\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}.$$ Then for our normal vector it's position is, assuming standing at the equator ($\phi=0$) $$\mathbf{v} =\begin{bmatrix}\cos t\\ \sin t\\ 0\end{bmatrix},$$ but if we assume we stand at an arbitrary latitude $\phi$

$$\mathbf{v} =\begin{bmatrix}\cos \phi \cos t\\ \cos \phi \sin t\\ \sin \phi \end{bmatrix}$$

TODO: Final step, add seasonality for sun vector and do the dot product...

Note: really both vectors should originate in the center of the Earth, but since it's only the angle that matters between them, the position does not matter. And hence we can visualize it like this.

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Now add in the seasons again

Assume we are in the coordinate system where the z-axis is Earth's spinning axis, and x axis. Because the coordinate system is based on Earth it, Earth will look fixed, while the sun looks like it moves up and down. Let's take the March equinox, this is when Earth's tilt is not towards or away from the Sun, but rather orthogonal to the line between Sun and Earth. At this point you see how half of Earth is lit up (as always), but now if you think about one revolution of Earth (one day), the time spent inside the lit half, and outside the lit half is equal. Hence, it's a special time of the year since at any point on earth the day is just as long as the night. We also see how it's the day when the Sun rises due (exactly) east, and sets due west (something that was easy to measure 1000 years ago.).

To not confuse things, Earth is always tilting $\epsilon\approx 23^\circ$, but this tilt is fixed in direction while orbiting the Sun, and hence it's only the tilt's component in the x-axis, the declination $\delta\in[-\epsilon, \epsilon]$ that affects the season. There is precesion so it's not entirely true but in our blog's geometric world it is.

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What about absolute time?

So far we derived everything in natural coordinates, solar time etc. Something relevant now adays is a mapping From $$Time \times Date \times Location \rightarrow Solar\;angle$$ So for example $\alpha(\texttt{Oct 6},\; \texttt{17:00 UTC},\; \texttt{Zurich})$

Gnomon

well, it's hard to have a bird's eye view of the solar system. So thinking locally, from our locally flat looking plane?

Facts:

• Sun is 1 austronimal unit or 8 lightminutes away from earth. When you look up it looks like 0.5 degrees in the sky. Formula for is radial size is. Sun is 400x times size of moon, but 400x times further away, so moon is also 0.5 degrees. Meaning there can be perfect solar eclipses. Since
• Since the sun is not a point source, when it shines on moon it creates a large penumbra, while the eclipse on jupiter has almost no penumbra, since the sun in that case can almost be seen as point source.
• euler angles. Precession so our north star changes a bit.
• to fully get suns path we also need its compass direction in relation to North, so we can project the sun vector onto north and east for example, and we get that.