# Sunrises on the compass

1. Apr 18, 2005

### chroot

Staff Emeritus
A recent thread (https://www.physicsforums.com/showthread.php?t=69970) by DaveC426913 got me thinking about differential geometry. The compass angle at which the sun rises each morning varies with latitude, but not linearly; it actually seems to be a rather complicated relationship. Let's look at a diagram:

This diagram is drawn above the north pole. It is summer in the northern hemisphere, and the north pole experiences continual sunlight.

Consider two points at dawn, along the terminator. Point A is on the equator. The due-east and normal (altitude) vectors are drawn in black, and the due-north vector points directly out of the page.

Point B is at about 60 degrees north latitude. The due-east, due-north, and normal vectors are also drawn in black. At each point, a vector pointing directly to the sun (assumed to be at infinite distance) is drawn in red.

It can be easily seen that the sun rises more northerly at higher latitudes. That fact that can be understood easily by considering parallel transport. If the vector triple at point A were slid (parallel-transported) from point A to the north pole via a line of longitude, the sun's compass position would vary directly with the latitude -- it'd be an easy problem.

On the other hand, if the vector triple at point A is parallel-transported along the terminator, its rotation is complicated. This is because the terminator is not a geodesic (great circle), while the line of longitude is.

I'd like to figure out how to understand this more deeply. I'd like to go from drawing a pretty picture to actually understanding the mathematics -- how to calculate the sunrises's compass angle on the horizon for any latitude. The summer solstice can be assumed.

Can anyone help?

- Warren

2. Apr 19, 2005

### Berislav

I'm basically a novice when it comes to differential geometry, so I don't think that I can help, but I will try.

The metric of a 2-sphere is given by:

$$ds^2=R^2(d\theta^2+\sin^2 \theta d\phi^2)$$

Denote the tangent space at point A with $V_a$ and the tangent vector of the equator curve with $A^\mu$. Similary denote the tangent space at B with $V_b$ and the parallely transported $A^\mu$ with $B^\mu$.

There exists a scalar field covering the entire area of the sphere through which we can define functions (with two variables - $\theta$ and $\phi$). These will be important as through them we can parametarise curves. The equator curve can parametarized by the function $a(\theta,\phi)$ by setting $\theta=0$. The terminator curve can also be parametarised this way. Since I don't know what that is I don't know how to parametarise it. Anyway, let's denote it's parametar by $c(\theta,\phi)$.

$A^\mu$ can be defined as

$$A^\mu=\frac{dx^\mu}{da}$$

Now we wish to parallely transport $A$ to point B. Introduce it's dual $A_\mu$.

$$B^\mu B_{\mu} = \int_0^{\theta_B}\int_{\phi_A}^{\phi_B} \frac{d(A^\mu g_{\mu\mu} A^\mu)}{dc} d\theta d\phi$$

This defines $B^\mu$.

The vector pointing to the sun $S$, expressed in the coordinate basis of A, $X^A_\mu$ will be equivalent to $A$.

$$S =\sum_{\mu=1}^2 A^\mu X^A_\mu$$

Since it doesn't change direction in point B, we don't parallely transport it, we only express it in the coordinate basis of B using the vector transformation law.

$$S^{B \nu} =\sum_{\nu=1}^2 A^\mu \frac{\partial x^{B \nu}}{\partial x^{A \mu}}$$

Since we know that this particular 2-sphere is embedded in Euclidian three-dimensional space we can find the coordinates of the basis of any point, in three-dimensional space, by imposing that they be orthonomal to the position vector, $\vec{r}$, which has it's origin at the center of the Earth.

The angle between $S$ and $B$ at $V_b$ can be found by simple methods of vector algebra once we define it's coordinate basis through the above procedure.

Hope I helped.

Last edited by a moderator: Apr 19, 2005
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