1) This problem deals with supernovas in
general and Supernova 1987A in particular.
Reading the appropriate Zeilik sections
about supernovas and neutron stars (scattered
over chapters 17 and 18) will be helpful.
Most answers will be found in the "Supernova
1987A!" article, from the 6 May 1988 Science.
So that everybody starts with the same
data, let's assume that the resulting
neutron star remnant left after the supernova
explosion has the properties R = 15 km
and M = 1.4 Msun (see bottom of 1st column,
page 752 or first line on page 754).
The answers that you calculate may not match the
article's number perfectly, but they should
all be close.
a) Calculate the time for the free-fall
collapse of the pre-supernova core (i.e., the portion that
becomes the neutron star). Presumably
you'll know why the calculated time is different than
that which the authors give (1st paragraph,
p. 753).
b) The gravitational energy released in
the core implosion (calculated by Zeilik) virtually all
goes into neutrinos; calculate the neutrino
luminosity by using Zeilik's result and the result
of the previous part. (p. 754 for comparison).
c) Neutrinos are produced during the complete
neutronization of the core (which became the
neutron star), during its collapse, i.e.,
e- + p --> n + ne
Calculate the total number of neutrinos
produced during the collapse.
d) Calculate the resulting neutrino flux (in neutrinos per m2) on earth.
e) Your answers to parts (c/d) are actually
too small because additional neutrinos are
produced in the following process:
_
e+ + e- <---> g
+ g
<---> n
+ n
These neutrinos can be any kind (e, µ,
or t). About 10 times more neutrinos and antineutrinos
are produced in the above process than
in the neutronization of the core; correct your
answers to parts d,e appropriately. You
should come close to the article's number; see
bottom of column 1, page 754.
f) As the article says (p. 753, 2nd column),
virtually all of this gravitational energy comes out in
neutrinos. Use your answers to previous
parts (corrected) to calculate the average energy per
neutrino. (bottom of column 1, page 754
for comparison)
g) The cross-sectional area for neutrino
interactions with protons in water molecules is
dependent on the neutrino energy
En and is calculable from weak interaction theory:
n-H2O cross section = 10-48 (En/mec2)2 m2 per H2O molecule
where me is the electron rest
mass. (The magnitude of the numbers for cross section should
be reminiscent of those from the solar
neutrino discussion.) Use this expression, the average
neutrino energy from part (f), and the
(corrected) neutrino flux [part (e)] to determine the
number of neutrino interactions per human
being (which is just the probability of a neutrino
interaction with a human being). Assume,
for convenience, that humans are walking
collections of pure water molecules. The
method for this is exactly the same as that used
when we were calculating the number of
argons produced from chlorine by the solar neutrinos.
h) Use the results of the previous part
to calculate the number of human beings on earth who
experienced an interaction, up close and
personal, with Supernova 1987A (via its neutrinos)
on February 23, 1987. Your answer should
compare fairly well with that stated in the article.
the following 2 parts are from test of
3 years ago:
i) As Zeilik points out on page 373, the
luminosity of a supernova drops exponentially due to
the radioactive decay of the cobalt-56
produced in the explosion. Using the V (apparent
magnitude) curve in Fig 18-22, extrapolate
the linear portion (200 d < time < 400 d)
backwards to find the luminosity of the
supernova near t = 0.
j) The radioactive decay law gives the
rate at which nuclei decay: dN/dt = - .693 N/T
where T = the half-life
of the isotope (given in Zeilik) and N = the number of nuclei
present at
any time t . If 3.72
Mev is released in each cobalt decay; determine the amount of radioactive
cobalt (in solar masses) present at/near
t = 0. Your result should compare well with the
"accepted" answer (which is actually given
for nickel-56) in the article.
2) Once pulsars were identified with rotating
neutron stars, it was not long before people also
realized that the tremendous magnetic
fields generated (Zeilik p. 341) during the
pre-supernova collapse to neutron star
might also be causing intense synchrotron radiation
to be emitted by the collapsed pulsar.
Synchrotron emission (by charged particles spiraling
in a magnetic field) is discussed in Zeilik
sections 4-6(C) and 6-1(C) and Prelude 6-4, 5
(although there are mistakes in the latter
which will be corrected in class).
Synchrotron radiation is applied (sort
of) to pulsars on p. 341 and supernova remnants on
p. 369. On page 341, Zeilik arrives
at a rough value of the magnetic field of a neutron star
with a heuristic (handwaving) conservation
law. We'll do much better. Synchrotron radiation
shows up as continuous spectral emission
(like blackbody radiation, unlike fluorescence). We
might derive in class the expression for
the luminosity produced by a rotating (angular
frequency w
) star of radius R andmagnetic field B in class; in any event it is
Lsynchrotron = 8 p B2 R6 w4 sin2 q / ( 3 µo c3 )
(µo is that magnetic permeability
constant you should remember from Phys 52; q
is the angle
between the magnetic field axis and the
rotation axis of the pulsar: see Fig 17-6)
a) By setting the rate of loss of rotational
kinetic energy (Zeilik, p. 344-345) to the synchrotron
emission rate, find an expression for
the magnetic field of the pulsar in terms of M, R, P, q
,
and dP/dt.
b) Calculate the magnetic field of the
Crab
pulsar using the known P and dP/dt values (the
value of dP/dt in Table 17-1 is correct;
the value is on page 340 may be wrong in your
printing); use reasonable stuff for M,
R, and q . This magnetic field value is far superior to
that estimated via the method of p. 341;
this method should be used for all future calculations.