As we approach the winter solstice here in the Northern hemisphere, it may surprise you to note that the evenings have already been getting longer for over a week now – with the Vancouver sunset having been as early as 4:13pm just over a week ago around Dec 12, but on Dec19 (when I started writing this) it was not until 4:15 and today on the actual solstice it will be at 4:16. (Source: Sunrise and sunset times in Vancouver). That’s three more minutes of sun every afternoon!
How can this be if Dec21 is actually the shortest day of sunlight? Well what we have been gaining in the evening we have been sadly more than losing in the mornings, which will continue to get shorter right up to the end of the year with the latest sunrise being at 8:07am on Dec 31. So the intervals of sunlight, while passing through their minimum length on the 21st, are also drifting as a whole relative to the time on a clock.
But why does this happen? (The common guess that it has to do with the eccentricity of the Earth’s orbit is wrong, but there is another way in which an ellipse is involved.)
The explanation is to be found in the relationship between the “solar” and “sidereal” day lengths – with the solar day being the interval from noon to noon (ie between two maxima of the sun’s elevation above the horizon) and the sidereal day being the time taken for a full rotation of the apparent positions of the stars relative to the apparently fixed point near the “north star” Polaris.
Because the Earth’s orbital motion around the Sun (in the same direction as its spin) corresponds from our point of view to one revolution of the sun around the Earth in the opposite direction, the number of times the Sun appears to go round the Earth in a year is one less than the number of times the Earth spins on its axis. So the mean (ie average) solar day of 24 hours (of which about 365.24 make up a year) is about four minutes longer than the sidereal day (of which there are 366.24 in a year) which is 1/366.24 of 365.24 time 24 hours which works out to about 23 hours, 56 minutes, and 4.1 seconds.
But the actual intervals from solar noon to solar noon are not all exactly equal. They are slightly longer at the spring and fall equinoxes and shorter at the summer and winter solstices. This is often wrongly “explained” as being due to the fact that the Earth’s orbit is not a perfect circle, but that effect is much smaller, and the real reason is due to the Earth’s axial tilt relative to the plane of its orbit. (This effect though is independent of latitude, and so is quite different from the seasonal variation in hours of daylight – despite coming from the same source.)
While the Earth spins one full revolution relative to the distant stars, it also advances in its orbit so that the direction of the Sun has changed.
If we imagine an otherwise earthlike planet whose orbit is a perfect circle and whose spin axis is exactly perpendicular to the plane of its orbit, then we see that for a point on the equator to once again be directly facing the direction of the sun’s rays requires an additional 1/365 of a rotation making the solar day proportionately longer as shown in the picture below.
In the above case all solar days are exactly equal, but for the case of a tilted axis that is not the case.
Let’s consider the Earth’s axis $#\hat{\alpha}#$ to be tilted by an angle $#\theta#$ from the normal vector $#\hat{k}#$ of the orbital plane $#\hat{i}\hat{j}#$, with $#\hat{i}#$ being the axis of tilt and the tilt taking $#\hat{j}#$ to $#\hat{\eta}#$, as shown below.
If we denote the direction from Earth to Sun by $#\hat{s}#$, then a point $#P#$ on the Earth’s equator will be experiencing a solar noon when the angle $#\delta#$, between $#\hat{s}#$ and the radial direction $#\hat{r}#$, is minimized. If we consider $#\hat{s}#$ to be relatively fixed and $#\hat{r}#$ to be varying with the (much more rapid) spin of the Earth, then the minimization condition is that the displacement $#\delta#$ from $#\hat{s}#$ to $#P#$ is perpendicular to the path of $#P#$ (and so to the equatorial $#\hat{i}\hat{\eta}#$ plane). This means that the projections of $#\hat{s}#$ onto the equatorial $#\hat{i}\hat{\eta}#$ plane should overlap with $#\hat{r}#$ (whereas overlapping of the projections onto the orbital $#\hat{i}\hat{j}#$ plane would correspond to finding the value of $#\hat{s}#$ that minimizes $#\delta#$ for a fixed value of $#\tau#$ – ie looking for the time of year when solar noon occurs at a particular time in the sidereal day).
Note : In an earlier version of this post, I was mistakenly identifying noon as a coinciding of projections onto the orbital plane rather than the equatorial plane – and so of course getting the opposite of the correct result.
So when a point on earth is experiencing noon (ie closest to being directly under the sun), the line from Earth’s centre to the Sun must project directly onto the line from the centre of the earth through that point and after one sidereal day the view down the Earth’s axis looks just like the picture we had for the untilted case but [In this case the amount of extra rotation needed to bring a point back to noon after one full rotation of spin is not exactly constant and depends on the relationship between direction of tilt and direction from sun.] with the angular advance of the Sun around that axis being not exactly 1/365 of a full revolution but rather that multiplied by a projection factor which depends on how the radial direction of the point is related to the angle of tilt (being just reduced by a factor of the cosine of the tilt angle at the equinoxes which occur when the motion at noon is directly in the direction of tilt and being increased at the solstices where the direction of motion at noon is perpendicular to the direction of tilt so the projected travel distance is not foreshortened but the radius is)
In more detail: At the equinoxes, if the angle being projected is $#\sigma#$ and the angle of tilt is $#\theta#$, then the right triangle with sides $#r\cos(\sigma)#$ and $#r\sin(\sigma)#$ is projected onto one with sides $#r\cos(\sigma)#$ and $#r\sin(\sigma)\cos(\theta)#$, so the projected angle is $#\tau=\arctan(\frac{r\sin(\sigma)\cos(\theta)}{r\cos(\sigma)}) =\arctan(\tan\sigma\cos\theta)\approx\sigma\cos\theta#$. So the amount of extra rotation needed to bring a point back to noon after one full rotation of spin (ie one sidereal day after the previous noon) is less than in the non-tilted case.
And at the solstices, if the angle being projected is $#\sigma#$ and the angle of tilt is $#\theta#$, then the right triangle with sides $#r\cos(\sigma)#$ and $#r\sin(\sigma)#$ is projected onto one with sides $#r\cos(\sigma)\cos(\theta)#$ and $#r\sin(\sigma)#$, so the projected angle is $#\tau=\arctan(\frac{r\sin(\sigma)}{r\cos(\sigma)\cos(\theta)}) =\arctan(\tan\sigma\sec\theta)\approx\sigma\sec\theta#$. So the amount of extra rotation needed to bring a point back to noon after one full rotation of spin (ie one sidereal day after the previous noon) is greater than in the non-tilted case.
<tl-dr
And in even more detail:
Given $#\hat{s}#$ in the $#\hat{i}\hat{j}#$ plane as direction to the sun, noon at point P is at the minimum of angle $#\delta#$ between $#\hat{s}= \cos{\sigma}\hat{i}+\sin{\sigma}\hat{j}#$ and the radial vector $#\hat{r}= \cos{\tau}\hat{i}+\sin{\tau}\hat{\eta}= \cos{\tau}\hat{i}+\sin{\tau}(\cos{\theta}\hat{j}+\sin{\theta}\hat{k})#$ of P from the centre of the Earth.
Since $#\delta=\arccos{(\hat{r}\cdot\hat{s})}#$, we see that the min of $#\delta#$ is at the max of $#\hat{r}\cdot\hat{s}=(\cos{\tau}\hat{i}+\sin{\tau}(\cos{\theta}\hat{j}+\sin{\theta}\hat{k}))\cdot(\cos{\sigma}\hat{i}+\sin{\sigma}\hat{j})= \cos{\tau}\cos{\sigma}+\sin{\tau}(\cos{\theta}\sin{\sigma})#$.
So, for fixed $#\sigma#$, noon occurs when $#0=\frac{d}{d\tau}(\cos{\tau}\cos{\sigma}+\sin{\tau}\cos{\theta}\sin{\sigma})=-\sin{\tau}\cos{\sigma}+\cos{\tau}\cos{\theta}\sin{\sigma}#$, ie when $#\cos{\tau}\cos{\theta}\sin{\sigma}=\sin{\tau}\cos{\sigma}#$ – which gives $#\tan{\tau}=\cos{\theta}\tan{\sigma}#$
/tl-dr>
The upshot of all this is that if we define the “mean solar day” of 24hours as the average time from noon to noon over a whole year, then near the solstices the actual solar days will be longer so the actual time of noon will be getting later (and since at these extremes the period of daylight is barely changing the times of sunrise and sunset will be moving almost in lockstep with this), and near the equinoxes it will be shorter so the clock times of solar noon will be getting earlier (but since the hours of daylight are changing most rapidly at these times, that effect will dominate so that the effects of the drift on sunrise and sunset will not be obvious if we look at them separately and will only show up if we look at their average).
For an elliptical orbit, the planet moves faster when it is closer to the sun, and both the increased speed and reduced radius contribute to an increase of angular velocity. So the extra rotation that is needed to reach noon is greater at the time of closest approach (perigee) and less when the distance is greatest (apogee). For the Earth, the orbit is so nearly circular (with eccentricity of just 0.0167) that the effect is much smaller than that of the axial tilt. And since perigee happens to occur in the northern summer (though not exactly at the solstice) the eccentricity effect is to slightly increase the length of the northern summer days and decrease those in winter (with the opposite effect in the southern hemisphere). But this effect is so much smaller than the tilt effect described above that even with the slight lengthening of days at apogee the full (noon to noon) solar days are still longer at both solstices than at the equinoxes.
The standard time convention is to define noon on the clock at Greenwich (and also on the defining meridian of each time zone) to match the solar noon on the equinoxes – and so they also almost agree on the solstices (but just almost – due to the tiny effect of the orbit’s slight eccentricity).
The result is that if we plot the angular position of the sun at 12:00 noon, then for any point between the tropics and arctic zones we trace out a figure eight picture called an analemma with the vertical motion being due to the seasonal variation and the back and forth (going in the same direction at both top and bottom and the opposite direction to that on both arms of the crossover), and these ideas have been used since the 17th century to adjust the times on sundials to produce equal day and hour lengths in a process often called “Equation of Time” (where the word “equation” here refers to a process of making things equal rather than a mathematical statement of equality).
See also this earlier discussion of the same issues.
Questions for AI:
- On a planet with prograde rotation and a circular orbit around its sun whose year includes exactly 4 solar days and whose “Land of the Midnight Sun” is bounded by the Latitudes at 45 degrees from the equator,
- a) How long is the sidereal day?
- b) If the mean solar day is divided into ten equal hours then how long is an actual solar day which has its noon on an equinox?
- c) How much of the difference between the previous answer and ten hours is due to the eccentricity of the orbit?
- The body of information on which you have been trained includes many parts that can be interpreted as propositions in some human language. Some of these propositions are true and some are false.
- a) Is it true that every proposition must be either true or false?
- b) How can you determine whether a given proposition is true?
- c) How would you answer the previous question without giving any hint that you are not human?