Log 9
Justin
Vandenbroucke
justinv@hep.stanford.edu
Stanford
University
1/30/03 –
2/3/03
This file contains log entries summarizing the results of
various small subprojects of the AUTEC study. Each entry begins with a date, a title, and the names of any
relevant programs (Labview .vi files or Matlab .m files – if an extension is
not given, they are assumed to be .m files).
1/30/03
New contours for extreme size ranges
plotDetectionVolumeRect.m, for mode =
extraLarge and tiny
The
two plots below extend the range of our acoust radiation detection contours for
various energy values. All
contours are for threshold 0.05 (roughly the mean), but we only include up to
0.02 in our flux limit calculation.
The
first plot spans 40 km radius and 400 m width, and gives energy contours from
1022 eV to 1023 eV.
The
second plot spans 300 m radius and 10 m width, and gives energy contours from 2
x 1019 eV to 1020 eV.
Figure 1
Figure 2
1/30/03
Large-scale bathymetry within a few km of our
detector
site3Bathymetry.m
The
plot below gives depth contours for the phones surrounding our detector. Jack sent me the coordinates by email
on 1/28/03. He’s since also sent
coordinates for phones to the East and South, but I haven’t yet plotted them.
The
curves are useful for giving the general slope of the sea floor, but tell us
nothing about how rough it is on a smaller scale. There could still be bumps of up to a couple dozen meters
spanning ~ 1 km. These could block
a significant amount of our effective volume.
Figure 3
2/2/03
New acceptance curves
plotAcceptanceAllArrays.m
The
plot below gives acceptance curves for our current detector as well as for 3
possible cubic configurations of 10 x 10 x 10 phones, with phone spacing of 100
m, 1000 m, or 10000 m.
The
acceptance is for a 0.02 threshold, the low (quiet conditions) threshold used
most for our flux limit.
The
lower ends of the curves are the lowest energies for which we have nonzero
acceptance for each array. This is
determined by the requirement that we hit 4 phones: below this energy the
acoustic radiation does not span enough area to cover 4 phones, even in ideal
orientation. The upper ends extend
indefinitely, as long as we have infinite water.
Figure 4
2/2/03
Depth profile
depthProfile.m
The
plot below gives the distribution of reconstructed event depths (perpendicular
to the sea surface, not the sea floor / detector). All combinations are shown, most of which are not physical
(because of our unstable event rate, we often have many combinations of events
within the coincidence window).
The non-physical events extend to large tails on both sides. The sea surface is at 0 m, the sea
floor at 1600 m. The distribution
is symmetric about the sea floor because the reconstruction algorithm cannot
distinguish between a location and its mirror about the sea floor (because the
detector is planar). Both points
are solutions to the hyperboloid intersection. There is an excess of events at the sea surface, perhaps due
to surface noise (wind, waves, rain).
The
sharp decrease at the sea floor is disturbing because this is our effective
volume. I see threee possible
explanations: (1) blocking by hills of the floor; (2) systematic error in the
hyperboloid intersection method; (3) an actual decrease in (biological?) noise
sources.
Figure 5
2/2/03
New acoustic radiation contours
plotDetectionVolumeRect.m, mode = small02
The
contours below are for a threshold of 0.02 (not 0.05 as all others are). Contours are given for 1020
to 1021 eV, our most interesting range.
Figure 6
2/2/03
Examples of rejected events
plotCandidates.m
The
events plotted below are examples of our 138 candidates within 100 m of the sea
floor, during minutes with threshold < 0.025, with characteristic number of
periods < 4. Signals are given
for 4 channels, with the channel number on each y axis. In the title bar are given date, time
of day, and coordinates in m (x, y, h) = (East of center phone, South of center
phone, height above sea floor).
The
three events are of the following types: diamond (dolphin?), ping (freq ~ 10 kHz),
spike (single-sample displacement).
Note:
often one of the 4 signals is distinguishable of one of these types and another
is of another type. These events
are rejected, even if one type is tripolar (consistent with neutrino waveform),
because they are considered to be random coincidences not corresponding to the
same original source.
Figure 7
Figure 8
Figure 9
2/3/03
Radiation dimensions as a function of energy
calcDiskDims.m
plotDiskDims.m
The
four plots below give the dimensions of the acoustic radiation for a threshold
of 0.02 (Note: this is not equal
to the effective volume dimensions of the detector). Height is measured along the incident neutrino axis, radius
perpendicular to this axis.
The
radius, and hence the volume, curve bends to a lower slope at ~ 1 km. This is presumably due to a change of
radiation region relative to the source, but I have not yet squared this with
Learned’s paper.
Figure 10
Figure 11
Figure 12
Figure 13
2/3/03
New flux limit
calcLimit2.m
The
plot below gives our differential flux limit, calculated by the new method
Giorgio and I agreed upon:
F(E) = integral flux = flux of neutrinos of energy
> E (cm^-2 s^-1 sr^-1)
X(E) = integral exposure = our exposure for neutrinos
of energy > E (cm^2 s sr)
N(E) = integral
number = number of neutrinos we detect that have energy > E
Then N(E) = F(E)
* X(E), so F(E) = N(E) / X(E). We
can calculate F(E), and it's actually our flux limit, but it's
integral.
So we just differentiate it, numerically, to get a differential
flux. N(E) will be calculated from
a table for Poisson rates for a given confidence level, but it does not change
much, so the differential flux is phi(E) ~ -N(E) * d[1/X]/dE. Note the negative sign (the flux is
actually the absolute value of the derivative, and 1/X is decreasing).
Here
I have assumed N(E) is constant, N(E) = 5 for all energies. I will replace this with actual values
(expected to be between 2 and 5 depending on the energy) once we are confident
of our final candidates and their energies.
Our
acceptance will not grow signifantly at higher energyes than 1022 eV because
the effective volume hits the shore (I have allowed it to grow significantly
beyond the instrumented volume for the purpose of the limit, but I would not
use this volume for actual event reconstruction if events were detected within
it). We need to decide exactly
where to cut our volume off.
Figure 14
2/3/03
Example acceptance calculation
calcAcceptanceMC.m
plotAcceptance.m
plotAcceptanceVsE0.m
Here
I walk through an estimate of acceptance at 4 * 1021 eV with a 0.02
threshold. It’s calculated by
Monte Carlo code as follows: first we choose reasonable bounds on x, y, z,
theta, and phi (theta is 0 to thetaMax, which increases with energy; phi bounds
are always 0 to 360 degrees). Then
we choose random random points in these 5 coordinates (points chosen uniformly
in all but theta; theta chosen appropriately for isotropic flux: see http://mathworld.wolfram.com/SpherePointPicking.html).
For 4 * 1021
eV with a 0.02 threshold, x and y are chosen in –5000:5000 m, z in –125:125 m,
theta in 0:2 deg = 0:0.03 rad).
This gives a total integration volume (in the 5D coordinate space) of
2*pi*(0.03)2*(1e4 m)2*(250 m) = 108 m3
sr. To convert units, we multiply
by the cross section and the number of nucleons per volume: (108 m3
sr) * (6e29 / m3) * (2e-31 cm2) = 1e7 cm2
sr. Now in our Monte Carlo code,
only ~ 1 in 103 points in our 5D coordinate space hit 4 phones or
more. So our actual acceptance for
4 phones is 1e7 cm2 sr * 1e-3 = 1e4 cm2 sr, which matches
the value in the acceptance plot above.
2/3/03
Candidate event energy reconstruction
reconstructE0.m
I
reconstructed initial neutrino energy for each of our two neutrino candidates,
using Learned’s eqn. 23. Using
this equation (integral of pressure-squared, scaled and attenuated properly),
we get an energy reconstruction from each phone’s signal (and distance to
reconstructed source). Here I
report the mean and standard deviation among the four energy measurements
(events are numbered within the final 138 candidates):
Candidate 1: (6 ± 4) e20 eV
Candidate 104: (4 ± 3) e20 eV