The Sun

According to our model, the ecliptic longitude of the Sun, $\lambda$, satisfies

$$\tan(\lambda-\pi) = \frac{\sin(x+e_c\,\sin x-e_s\,\cos x)-e_s}{\cos(x+e_c\,\sin x-e_s\,\cos x)-e_c},$$ (46)

where $x=n\,(t-t_0)$, $n\,t_0=-(\omega+\phi_0)$, $e_s=e\,\sin\omega$, and $e_c=e\,\cos\omega$. (The ecliptic latitude of the Sun is, by definition, zero.) Here, $n=2\pi/T$ is the mean orbital angular velocity, $T$ the mean orbital period, $e$ the ellipticity, $\omega$ the argument of the perihelion, and $\phi _0$ the mean anomaly on day 0. Of course, these orbital elements are actually those of the Earth, rather than the Sun.

Figure 14: The apparent orbit of the Sun, $S$, around the Earth, $E$, showing the Spring Equinox, SE, Summer Solstice, SS, Autumn Equinox, AE, and Winter Solstice, WS.

Figure 14 shown a rough sketch of the Sun's apparent orbit around the Earth. The Spring Equinox (at which day and night are equally long) corresponds to $\lambda = 0^\circ$, the Summer Solstice (at which days are longest in the northern hemisphere) to $\lambda=90^\circ$, the Autumn Equinox (at which day and night are again equally long) to $\lambda = 180^\circ$, and the Winter Solstice (at which days are shortest in the northern hemisphere) to $\lambda=270^\circ$.

Figure 15 shows ecliptic latitude data for the Sun, covering the years 1995-1996. In order to fit our model to the data, we need four pieces of information: i.e., the times of the 1st Spring Equinox, the 1st Summer Solstice, the 1st Autumn Equinox, and the 2nd Spring Equinox. The times in question are $t_1=79.094$, $t_2=171.858$, $t_3=265.510$, and $t_4=444.336$, respectively. These data-points are indicated in the figure.

Figure 15: The ecliptic longitude of the Sun versus time. The vertical green lines indicate the times of the 1st Spring Equinox, 1st Summer Solstice, 1st Autumn Equinox, and 2nd Spring Equinox, respectively.

Now, it is clear, from Eq. (46), that $\lambda(t+T)=\lambda (t)$ for all $t$. Hence, the time difference between successive Spring Equinoxes must correspond to the Earth's orbital period, $T$. Thus, we obtain $T=t_4-t_1=365.24$ days.²

Figure 16: The ecliptic longitude of the Sun minus that of the mean Sun versus time. The blue and red curves indicate the prediction of the updated Almagest model, and the original data, respectively.

Now, Eq. (46) can also be written in the form

$$F(t,T,\lambda; e_s, e_c,t_0) = 0,$$ (47)

where

$$F = \left[\sin(x+e_c\,\sin x-e_s\,\cos x)-e_s\right]\cos\lambda$$

$$-\left[\cos(x+e_c\,\sin x-e_s\,\cos x)-e_c\right]\sin\lambda.$$ (48)

Hence, our first three data points yield

$$F(t_1,T,0;e_s,e_c,t_0) = 0,$$ (49)

$$F(t_2,T,\pi/2;e_s,e_c,t_0) = 0,$$ (50)

$$F(t_3,T,\pi;e_s,e_c,t_0) = 0,$$ (51)

which can be written more succinctly as

$${\bf F}({\bf x}) = {\bf0},$$ (52)

where ${\bf x} = (e_s,\,e_c,\,t_0)$. We can solve this problem via a standard Newton iteration procedure: i.e.,

$${\bf x}^{n+1} = {\bf x}^{n}- {\bf J}^{-1}({\bf x}^n)\cdot {\bf F}({\bf x}^n).$$ (53)

Here, ${\bf J} = \partial {\bf F}/\partial {\bf x}$ is the Jacobian matrix. Our initial guess is $e_s=e_c=0$, and $t_0=t_3$.

Figure 17: The residual in the ecliptic longitude of the Sun versus time.

Table 1 shows the orbital elements for the Earth obtained using the iteration scheme outlined above. The true orbital elements are given in Tab. 2. It can be seen that the orbital elements inferred from our updated Almagest model are remarkably accurate.

The mean Sun is a fictitious body which follows the apparent orbit of the true Sun, but rotates uniformly. The mean Sun is defined in such a manner that it coincides with the true Sun at the two equinoxes. It follows that the ecliptic longitude of the mean Sun is given by $\bar{\lambda} = \pi + n\,(t-t_0)$. Figure 16 shows the difference between the ecliptic longitudes of the true and mean Suns. The non-uniform rotation of the true Sun is clearly apparent from this figure. The figure also compares the predictions of our updated Almagest model, made using the orbital elements given in Tab. 1, to the original data. It can be seen that the two are essentially indistinguishable. Finally, Fig. 17 shows the residuals in the solar ecliptic longitude (i.e., the difference between the model prediction and the data). We can see that the residuals are less than 1 arc minute--i.e., less than 1/30th of the apparent size of the solar disk. Clearly, our updated version of Ptolemy's model does an excellent job of accounting for the Sun's apparent motion.