Is zero shadow day the longest day Wherever I read, I read that June 21st is the longest day for Northern Hemisphere, full stop. However the zero shadow day is different for different places. It is that day where the sun reaches the vertical point above. So shouldn't that be the longest day? In other words, up to the Tropic of Cancer, the longest day should be the Zero shadow days and above that it should be 21st June and that would have made sense but I cannot understand why it would be otherwise.
Any thoughts regarding this would be helpful.
P.S. I am not sure if this is the correct place to ask (since I couldn't find required tags) Please direct me to the right place if I am mistaken.
 A: 
In other words, up to the Tropic of Cancer, the longest day should be the Zero shadow days and above that it should be 21st June and that would have made sense but I cannot understand why it would be otherwise.

There are a number of websites that calculate sunrise and sunset times for locations around the world. One such site is https://www.timeanddate.com/sun/. Whether you choose Singapore, Honolulu, or Edmonton, the longest day of 2018 for those three cities is June 21. All three cities are in the Northern Hemisphere. Singapore and Honolulu are in the tropics; Edmonton is well outside the tropics. The longest day of the year for any location north of the equator is June 21, plus or minus a day or so. Full stop.
Ignoring atmospheric effects and the Sun's nonzero size, sunrise and sunset occur when the center of the Sun is at 0° elevation, or
$$\cos h = \cos(\Theta_l - \alpha) = - \tan(\delta)\tan(\phi) \tag{1}$$
where


*

*$h$ is the hour angle,

*$\Theta_l$ is the observer's local sidereal time,

*$\alpha$ is the Sun's right ascension,

*$\delta$ is the Sun's declination,

*$\phi$ is the observer's latitude, and

*All quantities are expressed in radians.


Ignoring the small changes in the Sun's right ascension and declination over the course of a day, the length of a day in hours is
$$\frac{24}{\pi} \cos^{-1}(- \tan(\delta)\tan(\phi)) \tag{2}$$
Note that the above equations have no solution when $|\tan(\delta)\tan(\phi)| > 1$, e.g., observers above the Arctic Circle on the June Solstice / observers below the Antarctic Circle on the December Solstice.
A: In the Northern Hemisphere, the longest day always occurs at the summer solstice (June 20 or June 21). But for the days of shortest shadow, one must make a distinction between locations the north of the Tropic of Cancer and locations between the Tropic of Cancer and the equator.
For places to the north of the Tropic of Cancer, the summer solstice is also the day when the Sun casts the shortest shadow around noon. Indeed, every day around noon, the sun passes to the south of the local zenith. At the summer solstice, the position of the Sun at noon gets the closest to the local zenith, so then it casts the shortest shadow.
For places at of the Tropic of Cancer, the Sun passes (almost) directly over the local zenith at the summer solstice, so it casts no shadow at noon. So for these locations, the summer solstice is also a zero shadow day.
But for locations between the equator and the Tropic of Cancer, the sun can pass the zenith on the north side during summer. This means that for these locations there are two zero shadow days, which do not coincide with the summer solstice.
The image below illustrates this: it shows the daily motion of the Sun at several times of the year, for a location with latitude $\varphi = 11.72^\circ$, which is halfway between the equator and the Tropic of Cancer.

The declination of the Sun is denoted as $\delta_\odot$. Since the axial tilt of the Earth is $23.44^\circ$, the summer solstice corresponds with $\delta_\odot = 23.44^\circ$ and the winter solstice corresponds with $\delta_\odot = -23.44^\circ$. The equinoxes are the dates when $\delta_\odot = 0^\circ$. Note that at the summer solstice the Sun passes the local zenith $Z$ on the north side at noon.
The zero shadow days are the days when the Sun passes through the local zenith. This occurs twice a year, when $\delta_\odot = \varphi$ (so in this case when $\delta_\odot = 11.72^\circ$). It's straightforward to see why, because the Sun lies on the celestial equator when $\delta_\odot = 0^\circ$, which is perpendicular to the celestial north pole $P$, and the angle between $P$ and $Z$ is $90^\circ - \varphi$. A quick look at an ephemeris table tells me that the corresponding zero shadow dates for this location are around April 21 and August 22.
A: Take a star like Sirius which is to the North of the equator. Each year, overall, this star is visible for lesser duration of time for the people of the Northern Hemisphere than for the Southern, and this decreases as you go to higher Northern latitudes. As the tilt of the earth changes, which gives rise to the Northern and Southern transits of the Sun, the duration of visibility changes. It is visible for longest duration at the time of Summer solstice for people in the Northern hemisphere. During Winter solstice, in a way, it is blocked by the equatorial bulge and becomes invisible.
Something similar happens to the Sun too with duration of visibility.
