# How many days has it been since the Cambrian explosion?

I want to know how to calculate the number of days since an event in Earth's past, and am using the Cambrian explosion as an example.

I define a day to be one rotation of Earth with respect to the sun. The key point is that Earth's rotation has been slowing down over millions of years, so it will not be as simple as multiplying the time in years by 365.

Is there a formula for the number of days x years ago that I can simply integrate to find the answer? Or some other way to approximate?

• @MichaelWalsby I disagree. There are estimates for a day's length that go back in time (which I don't remember now, but they exist). You can also assume that the Cambrian explosion was the day exactly 541 million years ago. You are missing the point of the question. This is a good question. – Gimelist Mar 10 '20 at 3:50
• @Perry Ainsworth You will be lucky to get an answer accurate to within 50 million years. There are a lot of days in 50 million years, especially if they are short days as they would surely be. Any answer you get will be little more than a wild guess. – Michael Walsby Mar 10 '20 at 8:29
• I think the point about the Cambrian explosion muddies the waters here - if we aren't sure to the exact year (and we aren't) then errors in the number of days will be small in comparison. You'll be saying something like 987,654,321,123,456 days ago give or take 500,000 days (numbers completely made up for illustrative purposes) - there is no point having the precision of the last five digits if we've got that large an uncertainty. What you actually seem to be asking, however, is if there is an equation/estimate of day length going back in time. – Lio Elbammalf Mar 10 '20 at 9:27
• There are surelly matlab scripts for that – user18590 Mar 10 '20 at 9:34
• @trondhansen: if there are reasonable estmates for how long the length of the day is as a function of time (and I'd expect people interested in the history of the solar system to have such things) then yes, the question is answerable (not, obviously to the day, but as $n \pm 0.1n$ or something. – tfb Mar 10 '20 at 11:15

## 2 Answers

As a very rough approximation, one could start with equation (9) from Arbab (2009), https://arxiv.org/abs/physics/0304093 to get the effective number of days per year:

$$T_{\text{eff.}} = T_0 \left(\frac{t_0-t}{t_0} \right)^{-2.6}$$

with

• $$t$$ the time difference between now and then
• $$T_0 = 365.25$$ the current number of days per year
• $$t_0=(13.799 \pm 0.021) \cdot 10^9$$ the present age of the universe (from Planck Collaboration et al. (2015), https://arxiv.org/abs/1502.01589 )

Assuming the Cambrian explosion happened about $$t_c =(541 \pm 0.13) \cdot 10^6$$ years ago (from Bowring et al. (2007), https://core.ac.uk/download/pdf/62875.pdf), then the number of days since can be approximated as

$$n = \int\limits_0^{t_c} T_0 \left(1 - \frac{t}{t_0} \right)^{-2.6} \mathrm{d}t = \left[0.625 \cdot T_0 t_0 \left(1-\frac{t}{t_0}\right)^{-\frac{8}{5}} \right]_0^{t_c} \approx 2.1 \cdot 10^{11}$$

(of course one has to keep in mind that the error on this value is rather large, but to have some ballpark figure it should be good enough)

For comparison $$t_c \cdot 365.25 = 1.98 \cdot 10^{11}$$

• Nice exercise, but as Lio already pointed to in a comment, with these uncertainties it is irrelevant to distinguish between 1.98 and 2.08. If you do the calculations again and propagate the uncertainties mentioned (not to mention the assumptions) one significant digit is enough (not three) and there is no significant difference. – Jan Doggen Mar 10 '20 at 14:09
• There is not complete agreement on the length of the Cambrian period. Dating methods are always subject to error and uncertainty, and the further you go back, the greater the error. That's before you take into account the shorter days. To imagine that the Earth's rotation has always slowed to the same extent as it slows today is another source of error. In Cambrian times the moon was closer and its influence and braking power therefore greater than today. – Michael Walsby Mar 10 '20 at 16:22
• I wouldn't give any credence to the Arbab paper; it incorrectly claims that the Earth's rotation has been slowing down due to the expansion of the Universe; not on tidal drag from the Moon. That's why the formula you used is dubious; it contains a term for the age of the Universe, which is unrelated to Earth-Moon dynamics. – Spencer Mar 11 '20 at 23:01

From this answer a modern day is 1,7ms longer than a century ago. The data is taken from wikipedia, but it is sourced to this book.

Asuming the slowing down on earth's rotation has been constant since Cambrian, and that the sideral year duration has been constant too, it can be aproximated with a script:

#Constants
yearsSinceCambrian = 541000000
slowDownSecondsDayEachYear = 0.000017
daysPerYearPresent = 365.25
secondsDayPresent = 86400
#Initiallizate the variable
daysSinceCambrianExplosion = 0

#Calculate the number of days of each year
def sumYearDays(year):
secondsDay = secondsDayPresent - slowDownSecondsDayEachYear*year
hoursDay = secondsDay/3600
daysYear = daysPerYearPresent*24/hoursDay
return daysYear
#Sum all days from Cambrian Explosion to year 0
for i in range (0,yearsSinceCambrian):
daysSinceCambrianExplosion += sumYearDays(i)

#Sum days since year zero
daysSinceYearZero = 737875
daysSinceCambrianExplosion += daysSinceYearZero

#Print the result
print (str(daysSinceCambrianExplosion) + " days approximately since life explosion.")


Output

208929424039.7131 days approximately since life explosion.


Said $$2.09 \cdot 10^{11}$$ days.

I know the moon has moved away since Cambrian, so I assume as a constant wich is not. There are some extra days not summed.

I calculated the days the year had 510 my ago on my script and I got 406 days. In this publication from NASA 510 million years ago the year had 424 days, apparently; 18 more days. I think they know it because of fossils. I must say if paleontologists are rigth, if you examine the table published the slowing down ratio has not been constant. Ocean basins and tectonic should influence. So I can't be sure to use 1,7ms/century is accurated for the hole serie. In the past it migth have been 1,8 or 1,9 ms/century who knows.

To estimate the epsylon of the script, you can multiply the half of the 18 extra days on Cambrian for the total years of the serie.

$$9 * 541000000 = 4.9*10^9$$.

That would give a total of $$2.14 \cdot 10^{11}$$ days. This is not very far from what the script calculates, so you can use it to calculate from other events, knowing it is an aproximation.

The exact number of days can't be calculated, but there is not neither a exact known day when Phanerozoic started. Nobody can say the Phanerozoic started a 1st january 541002020 years ago, as assumed in the script.

I will say so @samcarter_is_at_topanswers.xyz calcs should be correct and approximately $$2.1 \cdot 10^{11}$$ days ago.

• Good answer if we can assume the slowing of rotation is linear - have you got any sources to suggest it is? Given that the moon isn't the same distance away and the oceans haven't stayed a constant size I suspect the approximation may not be valid but don't have a source to support this. – Lio Elbammalf Mar 10 '20 at 15:39
• I think your answer implies a constant absolute duration of year (hence shorter days mean "longer" years, in terms of number of days). However, sidereal year has also changed through time, as Earth is slowly going away from the Sun (15 cm per year according to Krasinsky & Brumberg 2004: doi.org/10.1007/s10569-004-0633-z). So you'd have to take this into account... See also: astronomy.stackexchange.com/questions/13041/… – Jean-Marie Prival Mar 10 '20 at 15:49
• Relevant new study via AGU: "Earth turned faster at the end of the time of the dinosaurs than it does today, rotating 372 times a year..." news.agu.org/press-release/… – jeffronicus Mar 10 '20 at 16:01
• @Universal_learner FYI: running your script with 541 I get 2.09e11. Only 0.01e11 difference from the value I get – samcarter_is_at_topanswers.xyz Mar 10 '20 at 17:29
• @LioElbammalf Universal_learner's use of linear interpolation is perfectly valid because of the low slope; it's similar to the approximation of $sin x=x$ for very small angles. – Spencer Mar 11 '20 at 23:12