PHYS 645
Winter 2012
Problem Set 6 Solutions
1. RB Question 6.1
(a) Show that the angular momemtum is conserved for a central force
eld (F = F (r) er ).
Considering that p = mr,
L=
=
dL
=
dt
=
rp
mr r
d
mr r
dt
m rr+rr
Since
rr 0
and
mr r = r Fr
(Newton)
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Problem Set 3 Solution
1. From Equation 3.162, the energy loss rate dE/dt 2 , therefore, we can write
dE
= KE 2 ,
dt
where K is a constant depending on the B -eld and particle mass.
dE
= Kdt
E2
E dE
t
= K dt
E0 E 2
0
1
1
= Kt
E E0
E0
E=
E0 Kt + 1
To deter
Problem Set 4 Solution
1. From Equation 2.22, the Larmor frequency
L =
eB
.
mc
Put e = 4.8 1010 esu, m=9.1 1028 g, B = 104 G, we obtain L = 1758 s1 and hence
f = 280 Hz 300 Hz.
From Equation 3.169,
E2
c = L =
L .
mc2
Put mc2 = 511 keV, E = 109 eV gives f
PHYS 645
Winter 2012
Problem Set 5 Solutions
1. Order of Magnitude Estimates (from Rosswog & Brggen, p. 201)
u
(a) Conservation of angular momentum implies I1 1 = I2 2 . Therefore
2 = 1
I1
.
I2
For a solid homogeneous sphere, I = 2 mr2 . The above equatio
PHYS 645
Winter 2012
Problem Set 6
Solutions
1. Disk Formation (Carroll & Ostlie problem 12.13)
We begin with the formula for gravitational acceleration:
d2 r
GMr
= 2
2
dt
r
(1)
where Mr is the mass enclosed within radius r. Adding in the centripetal acce
PHYS 645
Winter 2012
Problem Set 8 Solutions
1. RB Question 8.1
Considering that:
L
dM
=2
dt
c
and that
LE =
4 G M e c
T
which can be rewritten as
M=
T LE
4 G e c
The growth time:
tg = M/M
M c2
=
L
T c2 LE
=
4 G e c L
c T LE
=
4 G e L
Note that e is th
PHYS 645
Winter 2012
Problem Set 9 Solutions
1. RB Question 8.4
P ( < 0 ) =
=
0
sin d
0
/2
sin d
0
cos |0
0
/2
cos |0
cos 0 cos 0
cos /2 cos 0
cos 0 1
=
01
= 1 cos 0
=
1 v 2 /c2 = 1/ ,
For large Lorentz factor sin =
P ( < 0 ) = 1 cos 0
= 1 1 sin2 0
= 1 1
PHYS 645
Winter 2012
Problem Set 10
Solutions
1. Order of magnitude estimate (R&B 7.1)
Solid angle for each cone
2
j
=
0
0
sin d = 2 (1 cos j ) .
d
d =
0
j
2
0
Solid angle of 2 cones compared to the total sky
2/4 = 2.4 103 .
Therefore, the true rate is 2/