Summer solstice

A location on the earth's surface is defined by latitude $\phi$ and longitude $\lambda$ , so that in the earth-fixed frame of reference this point can be described in spherical coordinates by the vector

$\vec r = r_E \left( \begin{array}{c} \cos \phi \cos \lambda \\ \cos \phi \sin \lambda \\ \sin \phi \end{array} \right)$

where $r_E$ is the earth radius. Since the longitude is not relevant here, we set $\lambda = 0$ . In fact, the earth rotates counterclockwise from east to west with the angular velocity $\omega = \frac{2 \pi} {24 \,\text{h}}$ and is tilted by the angle $\delta = 23.5^\circ$ towards the ecliptic. In order to get from the earth-fixed vector $\vec r$ to the space-fixed coordinates $\vec R$ , we have to multiply the vector by two rotation matrices $R_t$ and $D_\delta$ which correspond respectively to the rotation and the tilt of the earth:

$\begin{aligned} \vec R &= D_\delta \cdot R_t \cdot \vec r \\ &= \left( \begin{array}{ccc} \cos \delta & 0 & -\sin \delta \\ 0 & 1 & 0 \\ \sin \delta & 0 & \cos \delta \end{array} \right) \cdot \left( \begin{array}{ccc} \cos \Omega t & -\sin \Omega t & 0\\ \sin \Omega t & \cos \Omega t & 0 \\ 0 & 0 & 1 \end{array} \right) \cdot r_E \left( \begin{array}{c} \cos \phi \\ 0 \\ \sin \phi \end{array} \right) \\ &= \left( \begin{array}{ccc} \cos \delta & 0 & -\sin \delta \\ 0 & 1 & 0 \\ \sin \delta & 0 & \cos \delta \end{array} \right) \cdot r_E \left( \begin{array}{c} \cos \phi \cos \Omega t \\ \cos \phi \sin \Omega t \\ \sin \phi \end{array} \right)\\ &= r_E \left( \begin{array}{c} \cos \delta \cos \phi \cos \Omega t - \sin \delta \sin \phi \\ \cos \phi \sin \Omega t \\ \sin \delta \cos \phi \cos \Omega t + \cos \delta \sin \phi \end{array} \right) \end{aligned}$

As connection vector between earth and sun we choose $\vec S = - a \vec e_X$ in negative X direction, so that the elevation of the sun is highest at midday $t = 12 \,\text{h}$ . Since the position vector $\vec R$ is perpendicular to the Earth's surface, the angle between the vectors $\vec S$ and $\vec R$ is just the angle of incidence of the Sun's rays:

$\begin{aligned} & & \angle(\vec S, \vec R)&= \tfrac{\pi}{2} - \theta \\ \Rightarrow & & \frac{\vec S \cdot \vec R}{|\vec S||\vec R| } &= \sin \delta \sin \phi - \cos \delta \cos \phi \cos \Omega t = \cos(\tfrac{\pi}{2} - \theta) = \sin \theta \\ \Rightarrow & & \theta &= \arcsin(\sin \delta \sin \phi - \cos \delta \cos \phi \cos \Omega t) \end{aligned}$

Negative elevation angles $\theta < 0$ occur at night while during the day is $\theta > 0$ . The times for sunrise and sunset are then given by the condition $\theta = 0$ :

$\begin{aligned} t_\text{sunrise} &= \begin{cases} \frac{1}{\Omega} \arccos(\tan \delta \tan \phi) & \phi < \frac{\pi}{2} - \delta \\ 0 & \phi \geq \frac{\pi}{2} - \delta \end{cases} \\ t_\text{sunfall} &= \frac{2 \pi}{\Omega} - t_\text{sunrise} \end{aligned}$

Here, the latitude $\phi = \frac{\pi}{2} - \delta = 66.5^\circ$ corresponds to the Arctic Circle, so that the sun rises at midnight and there is no night. The temporal course of the sun's position $\theta (t)$ is sketched in the diagram below. At the tropic ( $\phi = \delta)$ the curve $\theta (t)$ corresponds to a triangular function with a maximum at $\theta = \frac{\pi}{2} = 90^\circ$ . At the North Pole, on the other hand, the position of the sun is constant with $\theta = \delta = 23.5^\circ$ .

We can already exclude the answer option $\phi = 0$ (equator), because according to the diagram the elevation of the sun is always larger at the tropic than at the equator ( $\theta_{\phi = \delta}> \theta_{\phi = 0}$ ). For all other cases, the total incidence of light will have to be calculated by integration:

$\begin{aligned} I_\text{tot} &= I_0 \int_{t_\text{sunrise}}^{t_\text{sunfall}} \sin \theta(t) dt \\ &= I_0 \int_{t_\text{sunrise}}^{t_\text{sunfall}} ( \sin \delta \sin \phi - \cos \delta \cos \phi \cos \Omega t) dt \\ &= I_0 \left(\sin \delta \sin \phi (t_\text{sunfall} - t_\text{sunrise}) - \frac{\cos \delta \cos \phi}{\Omega} (\sin t_\text{sunfall} - \sin t_\text{sunrise} ) \right)\\ I_\text{tot} &= I_0 \cdot 24\,\text{h} \times \begin{cases} \approx 0.352 & \phi = 23.5^\circ \,(\text{case } B)\\ \cos (\delta)\sin(\delta) \approx 0.366 & \phi = 66.5^\circ\, (\text{case } C)\\ \sin(\delta) \approx 0.399 & \phi = 90^\circ\, (\text{case } D) \end{cases} \end{aligned}$

In fact, the solar irradiation at the North Pole reaches the highest value. Plotting the whole function $I_\text{tot}(\phi)$ also shows that this is the global maximum of the curve. Nevertheless, it is not really warm at the North Pole even at the summer solstice. Firstly, there is a high sun intensity only for a very short time period at the North Pole, so that the place has no time to warm up. In addition, the albedo of snow and ice is particularly high, so that the majority of the incident sunlight is directly reflected and does not contribute to the increase in temperature. Nevertheless, one should not forget the sunscreen at a North Pole expedition in summer.