Exercise: Using the experimental value of 0.25 and the 48 kg/m2 value of the traversed
matter. Calculate the total spallation cross section of the light nuclei.
The cross section for spallation of CNO is from the table (adding all channels and
averaging over the three channels) 75 mbarn. That means that the number of g/cm2 is given
by the atomic weight is 1 g/mol:
1
g
cm2
610
10
7510
kg
m2
220
kg
m2
Now unfortunately there is a mistake in the formula for the ratio (could have seen that
because
the
dimensions
are
not
correct).
The
correct
formula
is:
1
So this gives
0.25
220
1
This is of course a pretty difficult equation to solve. The answer is in fact 215 kg/m2. And
from this one gets 80 mbarn.
Exercise: Use the values of the ratios of nuclei and the leaky box model to determine the
escape traversed matter or escape time.
1
The leaky box model gives a weight of
gived the final ratio:
for each ratio so the weighted average then
1
〉
〈
1
So
1
1
0.5
Exercise: Verify that the spallation timescale is indeed larger than the escape timescale.
Calculate the escape timescale from the above. Given the escape timescale and the
required mean traversed matter before escape (previous exercise), determine the mean
matter density.
1
1
1
1
0.028
0.1
1
1
310
0.28
1
1
310
1
310
1
7.510 The traversed matter is 7 kg/m2 divided by the length travelled in , i.e.
7.510
3
310 310
710 so the density is
10 kg/m which is 10 protons per m3.
Exercise: Given the flux of cosmic rays in Error! Reference source not found. calculate
the energy density of cosmic rays and verify the value given above.
The spectrum looks like:
.
100
10GeV
m‐2 sr ‐1 GeV ‐1
We need to multiply by E integrate over E over the solid angle and then divide by the
velocity of light so:
4 100
310
10GeV
0.6MeV/m3
.
GeVm‐3
4 100
310 10
.
1
10
0.7
.
610
Exercise: From the Lorentz force derive a formula that gives the radius of curvature
of the path that a high energy particle of charge and momentum will follow while in
a homogeneous magnetic field . Given the approximate galactic magnetic field
T give the radius for a particle of
eV. Compare
strength of
μGaussor
this to the height and radius of the Galaxy.
Thee magnetic field
f
is given in T and tthis is
so
o
eVs
m
0.3GeV
/
0.3
For 10-9 T andd 1015 eV this
t
gives R
R=3 1015 m = 0.1 pc so significcantly less than
t
the
dimensiions of the galaxy.
g
Exercisse: For a given
g
accellerator sitee calculatee the ratio of the raddius that a 1 PeV
proton enscribes and that off an Fe ion . If the accceleration iss limited too 1 PeV giv
ve a plot
showin
ng the size of
o the acceleration sou
urce versuss its associa
ated magneetic field.
Thiss is obvioussly a ratio of 28. The pllot is like th
he Hilas plot
Exercisse: Given a total inelastic prroton-proton cross section off
mbarn
calculaate the diffeerence in depth betweeen Fe and p interactiions with nnitrogen. Compare
this wiith Figure. Explain the fact th
hat both for
f Fe and p the deppth of inteeraction
increasses with eneergy, even though
t
thee total crosss section acctually risess with enerrgy.
/
/
So tthe interactiion length iss inversely pproportionaal to the
/
so one exxpects a ratio of 14
This is much larger difference than in the figure. In fact the scaling does not really scale
like this but rather (AN+Ap)2/3 vs (AN+AFe)2/3 so the ratio would be 2.8 still too large. But the
position is of the maximum of the shower. As many more particles are produced in the Fe
collisions they will extend to larger depths.
This is of course also the answer to the second question. The depth of maximum of the
shower grows as ln(E).