The Fixed Hebrew Calendar

Transcription

1 The Fixed Hebrew Calendar Moshe Lerman June, 2017 קול גלגל המתגלגל ממטה למעלה 0. Introduction The present paper is an extension of a paper entitled Gauss Formula for the Julian Date of Passover by Zvi Har'El [1]. For reasons of completeness, a lot of material is copied from this paper. The generational parameter n, introduced in Section 2, is new, and reveals a freedom in the Gauss algorithm. Section 7 deals with this freedom, and provides a way to determine the integer n and so to "fix" the calendar. Thus, a way exists to increase the validity of the calendar, using the same data for the "first molad," the first new moon, and the same length of the month, known from the Talmud and letters from the Geonic period, and the same rules for exceptions. The calendar takes the form of a Java class named Gauss. Finally, we introduce the integer parameter z, initially equal to zero, which increments every 19 values of n, and changes the molad by z days. This is to 2 set an "eternal" calendar, based on the same old Gauss class. 1. Definitions Nisan origin is the instant occurring before the beginning of 1 Nisan, in a time difference which is equal to the difference between the new moon, or molad, of Tishri and the beginning of Tishri in the following Hebrew year. Nissan origin of the year 1 AM is 29 Adar, 14 hours, because molad Tishri of the year 2 AM was on Friday, 14 hours, and 1 Tishri was a Saturday. March origin is the instant occurring one day before the beginning of 1 March. Thus, the March date of a given instant will be in fact the time difference between that instant and the March origin. Note that the Nisan origin of the year 1 AM is 33 March, 14 hours, because the Julian date of 1 Nissan 1 AM was 3 April, i.e. 34 March. Here we assume that the Julian day begins in the previous evening, as the Hebrew day does, to simplify discussion.

2 Let H(A) denote the Nisan origin and J(A) the March origin of the Hebrew year A. Let T be the Nisan origin of year 1, in March days: (1) T = H(1) J(1) = 33:14:0 = Time intervals are given in the Talmudic units. An hour is divided in 1080 parts, so that d days, h hours and p parts are written as d:h:p = (p/ h)/24 + d. The unit is days, which will often be left understood. Let K be 1/19 of the length of the lunar month: K = 29:12: = The "Metonic" approximation S M of the solar year is defined as S M = 235K (= ) The Julian approximation S J of the solar year is defined as Let L be the average excess value S J = 365:6:0 = L = S J S M = 0:1: The "true" (see section 7) length of the year = S E = = S J L 3K/1000, a very close approximation of length of the mean tropical year. Finally, the Gregorian approximation S G is S G = = S J and can also be treated as the true value (see section 11) of the solar year. 2. Nisan origin A cycle of 19 consecutive lunar years containing 235 months, arranged in 12 common years containing 12 months each, and 7 leap years, containing 13 months each. The length of a leap year is 13 19K = 247K, while the length of a common year is 12 19K = 228K. Subtracting these quantities from L + 235K, the length of the Julian solar year, we get the common excess value, L + 7K, and the (negative) leap excess value, L 12K.

3 Considering this, we get (2) H(A) H(1) = (A 1)S J Δ(A), where Δ(A) is the cumulative excess value which should satisfy Δ(1)= 0. Each year we add L + 7K to Δ(A), unless the year is leap, when we add L 12K. In this way, the coefficient of L is incremented continuously, while the coefficient of K is increased by 7 each time, until when it becomes 11+n or higher, when it is decreased by 19, where n is a free integer parameter with bounds, see below. According to convention, n equals 0. It follows from this that the leap years are conventionally the years A such that A 19 equals the values 0, 3, 6, 8, 11, 14, or 17, where x k = x k[x/k], where [x] is the integral part of rational x. In general, though, Δ(A) is given by (3) where Δ(A) =(10 a)k + (A - 1)L + nk, a = (12A n) 19, such that Δ(1)=0, for any integer value of -10 n 8. The expectation value at A, which is the 19-year average around A, is < Δ(A) > = (n + 1) K + (A - 1)L, which equation can be used to fix the free parameter n. For this, see Section 7. Finally, irrespective of the value of n, we can divide the cycle years into 4 categories, according to this table: a A 1 A A Common Leap Common Leap Common Common 5..6 Leap Common Leap 0..4 Common Common Leap 3. March origin Set J(A) J(1) = (A - 1)365 + δ(a) Where δ(a) is the number of intercalary days (29 February) between 1 March 1 AM and 1 March of the year A. Since the Hebrew year 1 AM corresponds the

4 year 3760 BCE, or CE which gives a remainder of 1 when divided by 4, we obtain that the Julian year A will contain an intercalary day if and only if A = 0 (mod 4). Thus δ(a) = [A/4], or, denoting we get Therefore or, finally (4) 4. March date of Passover b = A 4, δ(a) = A/4 b/4. J(A) J(1) = (A 1)365 + A/4 b/4, J(A) J(1) = (A 1) S J b/4 + 0:6:0. Subtracting (4) from (2), using (3) and (1), we get, for -10 n 8, or H(A) J(A) = T + (a 10 n)k (A 1)L + b/4 0:6:0, H(A) J(A) = T (10 + n)k + L + ak AL + b/4 0:6:0. This is the Julian March date of the Nisan origin of the Hebrew year A. We add 6 hours to implement the rule that if molad Tishri is at noon or later, 1 Tishri is postponed to the following day. Finally, we add 14 days to get the Julian March date of Nissan 15. Setting we get m 0 = T + (10 + n)k + L + 14 M + m = m 0 + ak AL + b/4, where M is the integral part (rounded to the integer below), and m is the positive fractional part: m = m 0 + ak AL + b/4 - M. Unless further exceptions apply (see below), M is the Julian March date of the first day of Passover of the Hebrew year A, and the Gregorian March date is 5. Week day of Passover M + [(A 3760)/100] [(A 3760)/400] 2.

5 Calculating modulo 7, we obtain for J(A) J(1): (A 1)365:6:0 b/4 + 0:6:0 = (A 1)1:6:0 b/4 + 0:6:0 = 5A/4 b/4 1 = A 1 + (A-b)/4 = A 1 + 8(A-b)/4 = A 1 + 2(A-b) = 3A 2b 1 = 3A + 5b 1 (mod 7) Since the March origin of 1 AM was on Friday, we get for the Julian March date M of the Hebrew year A, that c is the day of the week with c = 0 for Saturday: 6. Exceptions c = (M + 3A + 5b + 5) 7 In the discussion above, we assumed that 1 Tishri is the day on which molad Tishri has taken place, and established that the date of 15 Nissan is M March. We already mentioned one exception. If molad Tishri is at noon or later, 1 Tishri is postponed to the following day. We implemented this exception by adding 6 hours to the Nissan origin. However, there are three more exceptions. The second exception is the rule that 1 Tishri is excluded from being a Sunday, Wednesday, or Friday, and is postponed to the following day. To implement this rule, we notice that 15 Nisan and 1 Tishri are 163 days apart, i.e., 23 weeks plus 2 days. Thus, 15 Nisan is excluded from being a Friday, Monday or Wednesday, respectively. The last two exceptions are in sense the consequence of the previous one, and from the wish to restrict the length of the Hebrew year. As we have seen, the length of the common lunar year is 12 19K = 354:8:876 days, and the length of the leap lunar year is 13 19K =383:21:589 days. Of course, a calendar year must have an integral number of days. Thus, a common Hebrew year has 353, 354, or 355 days, while a leap Hebrew year has 383, 384, or 385 days. The third exception follows from restricting the common year to have at most 355 days. Molad Tishri of a common year A+1 and its successor are 354:8:876

6 days apart, i.e., 51 full weeks minus 2:15:204 days. Thus, if molad Tishri of A+1, after being moved 6 hours ahead, is on Tuesday, 15 hours and 204 parts or later, its successor is on Sunday. Then, 1 Tishri A+2 is on a Monday and if 1 Tishri is not postponed from Tuesday (to Thursday, as Wednesday is excluded), the year A+1 will have 356 days. Similarly, the fourth exception follows from restricting the leap year to have at least 383 days. Molad Tishri of a leap year A and its successor are 383:21:589 days apart, i.e., 54 full weeks plus 5:21:589 days. Thus if molad Tishri of A+1, after being moved 6 hours ahead, is on Monday, 21 hours and 589 parts or later, its predecessor is on Wednesday. Then, 1 Tishri of A is on a Thursday, and if 1 Tishri A+1 is not postponed from Monday (to Tuesday), the year will have 382 days. To implement the last two exceptions, we notice that 1 Tishri A+1 being a Monday or Tuesday implies that 15 Nissan A is a Saturday or Sunday, respectively. Also, if we consider the table in Section 2, we notice that A is leap if a 12 and A+1 common if a 7. Thus, setting m 1 = (13 19K) 1 = 0:21:589 m 2 = 1 - (12 19K) 1 = 0:15:204 we find that the following exceptions apply to the March date M of the first day of Passover (Section 4): M+1 if c=0, a 12, and m m 1 M+2, if c=1, a 7 and m m 2 M+1, if c=2, 4, or 6 7. Proposal to Fix the Parameter n The expectation value of (2) is <H(A) H(1)> = (A 1)S J (n + 1) K (A 1)L. Here n is a free parameter, which has to be determined in order to get a Hebrew calendar. The traditional choice is n = 0, which keeps the calendar such that in the fifth century CE, the earliest Passover would be the day before the Spring equinox. However simple the choice n = 0 may seem, it cannot be maintained if we are to keep Passover close to the Spring equinox.

7 The equation (2) has limited validity. In truth, the equation should be H(A) H(1) = (A 1)S T Δ'(A), which defines S T, the true length of the year, so that Δ'(A) can be a bounded to be fluctuating function, for all A. That is, <Δ'(A)> is a constant: <H(A) H(1)> + constant = (A 1)S T. Therefore, we obtain (for now, we consider only non-negative integer n), And because n = [(S J ST L)A/K + constant]. S E = = S J L 3K/1000 we get, assuming that S E is the same as S T, n = 3 [ A constant ] This can be used to fix the calendar. If we take the last constant to be 4333, and we currently live in generation n=4, and generation n=5 starts in the year That is easy to remember. So we propose, for all integer n: n = [ 3A ], for 3A and A< n = [ 3A ] 1, for 3A<13000 and A In order to create a version of the Gauss algorithm that is not bounded by A 1000 and A<7334, by 7334 the Talmudic data need to be resynchronized with the Moon data as in the next section. 8. Fixing the Molad The condition -10 n 8 connects us to the "first molad," through the function Δ(A,n), seen as a function with two independent parameters A and n, yielding 0 for A=1 as long as -10 n 8. But if we take n to be a function of A, as proposed above, we can define Δ(A) = Δ(A,n(A)), for all values of A and therefore for all n, and we have Δ(1) 0 in any case. Hence, the first molad ceases to be, in a sense, though T remains in the algorithm as a remembrance

8 of what once was thought to be real. Moreover, we can take the opportunity to adapt the molad according to the Moon data, considering the following. The length of the month is, according to the Talmud, 29:12:793 days. The real length of the month is days. The difference of the two, multiplied by the number of months in years,, is about half a day: )29:12: ( = 0.48 The best fix is to change the value m 0 every years. We make the natural choice that the "first time" the fix will work, will be when n = 9: (5) 3 with z being a whole number given by m 0 = T + (10 + n)k + L + 14 z 2 z = [ (10+n (10+n) 19) 19 z = [ (10+n (10+n) 19) 19 ], for 10 + n 0 ] - 1, for 10 + n < 0 This change gives a long validity to the calendar that comes out of the algorithm given below, such that the length of the month remains the value of the Talmud: 29:12:793 days. The first molad will change when n = 9, i.e., according the above proposal, in Summarizing, we propose, assuming the change of (5): n = [ 3A ], for 3A n = [ 3A ] 1, for 3A< We would like call the term that fixes the Molad after Ben Meir, who sought, in a dispute [2] with Rav Saadya Gaon, to subtract 0:0:642 from the first molad. Instead of the 0:0:642, we subtract the Ben Meir term, z A Java Implementation public class Gauss { private static final double T = / 24.;

9 private static final double L = ( / 1080.) / 24. / 19.; private static final double K = (29. + ( / 1080.) / 24.) / 19.; private static final double m1 = ( / 1080.) / 24.; private static final double m2 = ( / 1080.) / 24.; public int day, jul, date; public Gauss (int year, int n, int z) { int a, b, c, M; double m, m0; m0 = T - (10. + n) * K + L z/2.; a = (12 * year n) % 19; b = year % 4; m = m0 + K * a + b / 4. - L * year; if (m < 0) m--; M = (int)m; if (m < 0) m++; m -= M; switch (c = (M + 3 * year + 5 * b + 5) % 7) { case 0: if (a <= 11 m < m1) break; c = 1; M++; break; case 1: if (a <= 6 m < m2) break; c = 3; M += 2; break; case 2: c = 3; M++; break; case 4: c = 5; M++; break; case 6: c = 0; M++; break; } day = c; jul = M; M += (int)(year ) / 100 (int)(year ) / 400-2; date = M; } public Gauss(int year) { //traditional Gauss object this(year,0, 0); }

10 public static Gauss gauss(int year, int n) { return new Gauss (year, n, (int) ((10 + n - (10 + n) % 19)/19) - ((10 + n >=0)?0:1) ); } public static Gauss EternalGauss(int year){ //"Eternal" Gauss object, proposal return gauss(year, (int)((3*year )/1000) - ((3*year>=13000)?0:1) ); } This is included in the app and in the app Data I follow Shneer's notation [3] for the Qeviyot. This notation is (C L)(D R A)( ) where C stands for Common year and L stands for a leap year. D stands for a Deficient (353 or 383 days) year, R stands for Regular (354 or 384 days) year, A stands for Abundant (355 or 385 days) year. Finally, the number at the end denotes the day in the week Rosh HaShana falls (7 is Saturday). For instance, LR3 is a regular leap year (384 days) and it starts on a Tuesday. For instance, CA2 is abundant common year (355 days) and it starts on a Monday. The following is from 19*360 years, from the year 600 until year 7439, according to the calendar that we have proposed. Visit the link in order to see the complete listing. To see the complete Java program, visit the link

11 For every date, we first mention the year, ended by a colon, then the Qeviyah according to Shneer, then the Gregorian date of the first day of Passover. Two fragments are especially interesting. As 5777 is a multiple of 19 plus 1, the current cycle, the years : 5777: LD : CR : CA : LA : CD : LR : CA : CA : LD : CR : CA : LA : CR : LD : CA : CR : LD : CA : CR We see that generation n=4 has leap years in the years 1, 4, 6, 9, 12, 14, 17 of the cycle. Generation n=5, which will start according to our proposal in the year 6000, has leap years in the years 1, 4, 6, 9, 12, 15, 17 as we can see in the fragment : 6005: LR : CA : CA : LD : CR : LA : CA : CR : LD : CA : CR : LD

12 6017: CA : CR : LA : CD : LA : CR : CA The distribution of the possible days of first day of Passover is as follows. This serves as a summary of the 19*360 dates: 19 March: 6 times 20 March: 47 times 21 March: 110 times 22 March: 190 times 23 March: 232 times 24 March - 16 April: about 230 times 17 April: 224 times 18 April: 216 times 19 April: 177 times 20 April: 82 times 21 April: 30 times 22 April: 2 times Passover is on 19 March in the years 5016, 7001, 7039, 7354, 7373, And it is on 22 April only twice, in the years 875 and 970. As expected, we see a small shift of the Gregorian date, we see S E versus S G. If S E is the true length, then indeed the Gregorian month of March moves towards the Summer. Compare this to the current calendar, the output of which, for same years 600 until 7439 can be seen here This shows a very wide distribution of dates, ranging from 4 March (in the years 605, 624, 643, 757) until 2 May (in the years 7285, 7304, 7384). Passover has shifted a whole month. 11. Completeness Every 1000/3 years, the proposed calendar changes generation. The year that the calendar changes, two things might happen:

13 1. This year is a common year, but in the previous generation it would have been a leap year. If the fractional part of the date of Pesach of the previous year (m) is greater than 0:15:204, and if the previous Pesach falls on a Sunday, the exception rule did not work, as it assumed that this year would be a leap year. The result is that this year has 356 days. 2. In this year, the first Molad changes, i.e. the value of z. Half a day is subtracted from M + m. This means that if, with the previous value of z, the day was deficient, 353 or 383 days, now there is a chance that it will become a year with 352 or 382 days. The first scenario happens about once in 400 calendar generation switches, one time in every years. The second scenario happens slightly more often, about once 12 Molad switches, which amounts to once in every years. All in all, there is a chance of one in years of something happening. The first time that actually something happens is in the year 35334, and it is the first scenario that is happening. In the first years, the first scenario happens four times, whereas the second scenario plays out six times. The following two rules make the calendar mathematically complete: 1. If scenario 1 happens, the previous year (is a deficient year and) should have two extra days, making the previous year into an abundant year and this year into a regular year of 354 days. 2. If scenario 2 happens, the previous year (is an abundant year and) should have two days less, making the previous year into a deficient year and this year into a regular year of 354 or 384 days. I have validated the sufficiency of these rules for the first years, though it should be said that changes in the three motions regarding the Earth that underlie the calendar - the spinning of Earth, the rotation of Moon around the Earth, the rotation of the Earth around the Sun - will likely undermine the accurateness of the calendar. 12. An Alternative Proposal It is possible to define the Gregorian year S G = S T, the true year: H(A) H(1) = (A 1)S G Δ'(A)

14 where <Δ'(A)> is constant. The consequence is that n = [( L)A/K + constant], n [ A constant ]. 360 If we take the last constant to be 4200, and we currently live in generation n=4, and generation n=5 starts in the year This is exactly as we had it in section 7. Therefore, it is possible to not decide, or it is possible to change a decision as until year 6333 there is no material difference. The calendar is implemented by: public static Gauss EternalGauss(int year){ return gauss(year, (int)((year )/360) - ((year>=4200)?0:1) ); } With this calendar, the distribution of the possible days of first day of Passover is as follows., as a summary of the 19*360 dates, from year 600 until year 7439 (for details, 19 March: 6 times 20 March: 93 times 21 March: 179 times 22 March: 231 times 23 March 16 April: about 230 times 17 April: 228 times 18 April: 208 times 19 April: 118 times 20 April: 13 times 21 April: 1 time Passover is on 19 March in the years 1057, 3156, 3528, 4253, 4644, It is on 21 April only once, in the year As expected, this shows no shift with respect to the Gregorian calendar. 13. Conclusion We have shown how simple it is to change the Hebrew calendar, with the Gauss algorithm. There are now three options before us.

15 The first option is summarized by S E = S M 3K/1000 = telling us that having the calendar change itself every 1000/3 years gives us the tropical year ( The fixing of the molad (section 8) reduces, at the long range, the year to days. The second option, is to have the calendar change itself every 360 years, which approximately gives the Gregorian year: S G S M 3K/1080 = The fixing of the molad reduces the year to, approximately, days. Of course, there is always the option of doing nothing, or S M = What year do we want, S M, S G, or S E? Surely S E, and certainly the algorithm of section 9. It is the most natural calendar, with the deepest of roots, the most precise, and this calendar is waiting for its natural adoptor. 14. References אשריהם ההולכים בדרך אמת בעוה"ז 1. Har'El, Zvi, 2005, Gauss Formula for the Julian Date of Passover, 2. Sacha Stern, 2001, Calendar and Community. 3. James Shneer, 2013, The Jewish Calendar and the Torah - 2nd Edition.