When the calendar changes a page, how do astronomers change their time numbers?
Drag the clock through the day and see how the calendar and consecutive day numbers describe the same moment.
If you want to calculate the time between two observations, you can use a calendar to count the months and days. Astronomy often expands dates into consecutive days: a number is given to the time, and the interval is obtained by subtracting the two numbers. Let’s use a simple day first and see how this number progresses little by little.
Guess first
The calendar changes dates at midnight, so does the integer of JD also change at midnight?
Show answer and explanation
Not necessarily, JD replaces the integer at noon — JD's day-counting integer boundary is at noon, so midnight is usually .5 decimal. It still counts continuously.
Then connect experience and science
Put the date on the timeline
First, fix January 1, 2000 in the Gregorian calendar, drag the time and observe how the time in the date changes to decimals.
Fractional day = UT hour ÷ 24
Find the midnight number
The JD at 0:00 on this day is 2451544.5. The integer junction falls at noon, so the decimal fraction of midnight is not zero.
JD₀ = 2451544.5
Add the decimals of the day
12 hours is half a day. Add 0.5 and you get exactly 2451545.0.
JD = JD₀ + UT hours / 24
The difference between two numbers is the interval
Turn the slider to 24 o'clock, which is midnight the next day. A day is incremented by exactly 1, eliminating the need to calculate the month length.
Δt (day) = JD₂ − JD₁
Expand the calculation
JD = JD₀ + UTHour/24 Δtday = JD₂ − JD₁ ΔtSecond = 86400 × Δtday
JD₀ is the Julian day at 0:00 on that day. This lesson has fixed calendar dates; see the same chapter for complete calendar conversions. The integer junction of JD is at noon.
give it a try
Click midnight, noon and then midnight. NOTE: Calendar page flips and JDs span integers and occur at different times.
No need to memorize symbols
- JD
- Continuous counting of Julian days, unit: days
- UT
- Universal time; this lesson uses UT1 to understand the moment when the earth rotates
Algorithm, accuracy and scope of application
This lesson fixes the Gregorian calendar date and does not involve calendar switching. The calculation library supports specifying the Gregorian or Julian calendar; historical dates cannot be directly guessed based on today's administrative region. The definition of J2000.0 is TT 2000-01-01 12:00; what is demonstrated here is the JD corresponding to the UT date, and the two cannot be regarded as the same physical moment. Book Equation (7.1), printed pages 61–62. Scenario description: UT1 example with fixed Gregorian calendar 2000-01-01, 24 o'clock displays 00:00 the next day; not local time zone conversion. J2000.0 is defined using TT. Just because the JD numbers are the same as 2451545.0, we cannot regard UT1 noon in this lesson as the same physical instant.
my understanding check
How much does JD increase from midnight to noon on this day?
0.5 — 12 hours equals half a day, so the number of consecutive days increases by 0.5.
How do the calendar and JD change from midnight of the current day to midnight of the next day?
Change the calendar by one day and add 1 to JD — Both represent the same 24 hours, just expressed in different ways.
The JD of the two observations differs by 0.25. How long has passed?
6 hours — 0.25 days is one quarter of a day, 24 × 0.25 = 6 hours.
The calendar divides time into years, months, and days, and JD writes it as consecutive days. The two writing methods describe the same moment, but the integer boundaries are not at the same time.
Jean Meeus · Astronomical Algorithms, 2nd ed. · No. 7 chapter