Raindrops keep
fallin’ on my head…
During the rainy season, sometimes one observes a gradual
onset of rain. A drop falls at one spot on the ground, then at another nearby
point, and so on. In one of Satyajit Ray’s films a raindrop falls on the head
of a bald man (I can empathise with that!), and then many raindrops fall in the
adjacent pond, each creating a set of ripples (as in Fig1 below).
Fig.1
Fig.2
So how long does it take before each point on the ground
is covered by raindrops? So we can try to estimate this time, but making
all sorts of assumptions. But… not for the case of the bald head, which would
be problematic , since the drops will not stay in place!
Assuming 1 mm dia raindrops, you need 1 mm of rain on the
ground to cover it (assuming raindrops as spherical – which is not the case).
If the rainfall rate is 1 mm/hr it would need 60 mins!
The problem is an exact analog of getting one monolayer of
atoms on a surface.
Assuming a 10 mm x 10 mm grid with 100 pixels (each of 1 mm
x 1 mm size) and 1 mm dia raindrops, with 1 drop per sec, it would take a
minimum of 100 secs.
If we take random falling of drops at the same rate, it
takes more time (see below).
The difficulty with raindrops is that they are not of the
same diameter (nor even spherical, due to the effect of drag). For atoms,
assuming only one species of atoms and with the same isotope, there is no size
variation.
a a) assume all raindrops are the same size, 1 mm diameter (unlike Fig.2 above) and the pixel size equals the diameter of the raindrop, assumed to be spherical.
b b) The rate at which raindrops fall is assumed to
be constant (assumed as 1 drop/sec for convenience).
c c) The location at which the raindrop falls is
assumed to be perfectly random.
d d) Implicitly it is assumed that the raindrops fall
vertically (no wind) on a flat surface.
e) The bouncing off and splintering of raindrops
after hitting the ground is also neglected.
f f) If more than one raindrop lands at the same
site, there is no spillover (see below). This problem does not occur in
molecular beam epitaxy, because one atom that lands on top of another will just
bond with it (or get reflected off).
There is some scaling, obviously, here,
both in time and in space. If we chose a rainfall rate of 0.2 drops/sec, the time
required for each process goes up by 5X. Also, one could just as well have
chosen raindrops of 2 mm diameter, and proportionately increased the total area
to 20 x 20 mm2, without affecting the end result.
Thus, initially, no two raindrops are
likely to land on the same mesh point. But, as coverage increases, it becomes
increasingly unlikely that a fresh raindrop will find an unoccupied site. We
assume no spillover: that is, even if multiple raindrops land at the same site,
they do not spill over on to adjacent sites: this is clearly an unrealistic
approximation, adopted for simplicity. (If the drop spills over, in which
direction does it fall?).
The calculation of the total time taken to
cover the surface assumes that each pixel will have at least one raindrop that
has fallen on it. As coverage increases, the total time a raindrop has to hunt
to find an unoccupied site increases proportionately to the fraction of
unoccupied sites. A complete calculation would also show how the number
distribution n(i,j), i.e. the number of raindrops that fell on each point
(i,j), where the calculation stops when n ³
1 for all i,j.Sticking to a 10x10 grid of pixels, the very
first raindrop to strike takes just 1 sec to find its spot, while the very last
raindrop has to make multiple attempts to find the very last unoccupied site,
so it takes 100 secs to find it. Intermediate drops find their time in between
these two limits, as calculated proportionately by:
Ti = 100/(101 – i), where i goes
from 1 to 100.
And the total time is: T = STi, summing
over all i.
It may be necessary to sum over a smaller
number of sites: say i = 1 to 99 for 99% coverage (see below).
I did the summation in Excel, but it could be implemented in other ways as well.
The next question that occurs is: what
happens if we increase the grid size from 10x10 to 20x20, keeping all else the
same? The above formula gets appropriately modified, which is straightforward. But
the total time to cover the surface increases. This is what one would expect.
So, something should converge – but what? The total time per site may converge
(divide the total coverage time by the total number of sites). This is plotted
in the Figure 3 below, as the topmost curve. If the curve is converging, it’s going to
take a while! But a little reflection will show that it will never completely
flatten: as you keep increasing the grid area, there will be some pixel
somewhere that will not get covered – but clumps and blobs will get more and
more drops.
Fig.3
With a 10x10 grid, the if we leave out the last raindrop, that corresponds
to 99% coverage. For a 50x50 grid, missing the last drop works out as 99.96%
coverage.
So actually, it makes more sense to look at
99% and 90% coverage, which is what is shown in the two lower curves of the
graph. And these do saturate.
For a 10x10 grid, and 90% coverage it takes 226
secs (instead of 100 secs if everything was uniform), and for 99% coverage, 419
secs. For a 50x50 grid (a better approximation), it takes 230 secs for 90%
coverage, and 459 sec for 99% coverage.
But one thing is clear: there is no one
answer to the question,” How long does it take for rain to cover the surface?” –
unless you specify the coverage.
As mentioned above, this problem is
analogous to the case of molecular beam epitaxy (MBE), which also has a
layer-by-layer growth mode. MBE also has two other growth modes, but those are applicable
only for atoms that can bond to make solid layers. Another difference from MBE
is that rainfall takes place from fairly large clouds, whereas in MBE the effusion
cells are quite small, and close to the substrate, so geometry effects dictate
a cosine dependence on angle.
Of course, the above analysis has made a
number of assumptions, listed above. The requirement of perfect randomness may
not be met. The assumption that all raindrops fall vertically is also
unrealistic: a slight, localized breeze will make nonsense of that.
Finally, the method of calculation used
above may not be rigorous: it will apply only on average.
The right and proper way is to do Monte Carlo
– which I have no intention of doing. I’m sticking to my own pay grade.
