Astronomy 253 — Problem Set #1 Solutions
Problem #1: Geomagnetic Fields
In regions not too far from the Earth (r . 10RE , where RE is the Earth radius), the geomagnetic
field can be approximated as a dipole field with magnetic moment ME ≈ 8.0 × 1025 erg G−1 (in
cgs units). In spherical polar coordinates (anti-)aligned with the Earth’s magnetic dipole moment,
the three field components are
B=−
ME
(2 sin λ, cos λ, 0) ,
r3
(1)
where λ ≡ π/2 − θ is the latitude.
(1). Show that the equation for a field line is r = r0 cos2 λ, and compute how field strength varies
with λ along a field line. (5 pts).
Solution: At location (r, θ), field line shape should satisfy
dr
Br
dr
d sin θ
=
= 2 cot θ ⇒
=2
⇒ r = r0 sin2 θ = r0 cos2 λ .
rdθ
Bθ
r
sin θ
(2)
We can now use r0 to label individual field lines. The three field components along a field line is
now given by
ME
B=− 3
(2 sin λ, cos λ, 0) .
(3)
r0 cos6 λ
Thus the field strength is
B(r0 , λ) =
ME p
1 + 3 sin2 λ .
r03 cos6 λ
(4)
(2). The Earth’s outer Van Allen Radiation Belt is filled with energetic particles (both electrons
and ions, 0.1 − 10 MeV) that are trapped, bouncing back and forth between mirror points. The
belt extends from about 3RE to 9RE , reaching its greatest intensity at about 4 − 5RE . For this
problem, we consider 3.0 × 104 km (∼ 4.7RE ) as a characteristic radius of the belt at equatorial
plane.
[2a]. Provide a rough estimate of maximum proton/electron energy (in units of eV) above which
guiding-center motion is no longer a good approximation. (5 pts)
Solution: Particle gyroradius must be much smaller than the scale at which magnetic field varies
(∼ RE ). Magnetic field strength at the equatorial plane is the weakest, corresponding to largest
gyro-radius. At R0 = 3 × 104 km, field strength is
B(R0 ) ≈
ME
≈ 3 × 10−3 G .
R03
(5)
Electrons are trans-relativistic, with gyro-radius
rL,e = 5.2 × 10−3 RE
1
E
MeV
B
mG
−1
.
(6)
Protons are non-relativistic, with gyro-radius
r
rL,i = 0.23RE
B −1
E
.
MeV mG
(7)
Thus, electron gyroradius is much smaller than RE unless their energy reaches GeV level. Proton
gyroradius becomes comparable to RE at a few tens of MeV.
[2b]. Particles would be lost due to collisions with neutrals if they manage to enter the Earth’s
atmosphere (thickness RE ). Compute the equatorial loss cone αleq : the critical particle pitch
angle in the equatorial plane below which particles would be lost. (5 pts)
2 /B = const. Together with energy conSolution: Conservation of magnetic moment demands v⊥
2
servation, we have sin α/B = const. Given equatorial pitch angle αeq , mirror reflection occurs at
latitude λm , which satisifies
sin2 αeq
1
=
.
(8)
B(R0 , λ = 0)
B(R0 , λm )
For critical pitch angle (loss cone), we further require
RE = R0 cos2 λm .
(9)
Altogether, and using the expression of B from problem (1), we obtain
cos6 λm
(RE /R0 )3
=p
sin2 αleq = p
≈ 5.2 × 10−3 .
2
4 − 3(RE /R0 )
1 + 3 sin λm
(10)
This pitch angle translates to about 4.1◦ , independent of particle energy.
[2c]. For trapped particles, provide a rough estimate of the bounce period τb between mirror
reflection points. (5 pts)
Solution: The bounce period can be expressed as
Z λm
dλ ds
τb = 4
,
vk dλ
0
where ds is the arc length along the field line, and for dipole field we have
p
p
ds/dλ = r 1 + 4 tan2 λ = R0 cos2 λ 1 + 4 tan2 λ .
(11)
(12)
Parallel velocity is given by
vk = v
p
1 − (v⊥ /v)2 = v(1 − B/Bm )1/2 ,
(13)
where Bm = B(R0 , λm ) if magnetic field strength at reflection point. Substituting it back, we
obtain
√
Z
4R0 λm cos2 λ 1 + 4 tan2 λdλ
(14)
τb =
v 0
(1 − B/Bm )1/2
2
While this integral can not be evaluated analytically, numerical integrations show that it is of order
unity and varies by within a factor of 2 for λm between 0 and 90◦ (from ∼ 0.7 to ∼ 1.4). Therefore,
a reasonable approximation for τb is simply τb ≈ 4R0 /v. For non-relativistic ions, we have
τb,i
4R0
∼
∼ 8.7 s
vi
E
MeV
−1/2
.
(15)
For trans-relativistic electrons, we have
τb,e ∼
4R0
∼ 0.4 s .
c
(16)
(3). Trapped particles will also undergo curvature and grad-B drifts (ignore the E × B drift due
to the Earth’s rotation since it is negligible). Derive the combined curvature and grad-B drift
velocities for electrons and ions (assuming they are non-relativistic). What is the direction of the
drift? How does drift velocity depend on latitude? What are the drift timescales at the equatorial
plane for electrons and ions? (10 pts)
Hint: the curvature drift term in Equation (2.24) in Boyd & Sanderson is not rigorous, but is
applicable when field configuration is current-free (∇ × B ≈ 0) as in a vacuum dipole field. Start
from the standard expression for curvature drift given in the lecture, v c = b × vk2 (b · ∇)b/Ω, and
use the following identities for unit vector b:
(b · ∇)b = −b × (∇ × b) ,
∇ × B = (∇B) × b + B(∇ × b) ,
(17)
to show (b · ∇)b = ∇⊥ B/B in the current-free limit, and then combine it with the grad-B drift
term.
Solution: The combined grad-B and curvature drift velocities are given by
2
v⊥
µ
1
1
2
2
∇B + vk (b · ∇)b = b ×
∇B + vk (b · ∇)b .
Vd = b×
Ω
m
Ω
2B
(18)
We have the following identities for unit vector b:
(b · ∇)b = −b × (∇ × b) ,
∇ × B = (∇B) × b + B(∇ × b) .
(19)
Since the (vacuum) dipole field is current-free (we assume the field is unaffected by the ring current,
which is generally a good approximation), or ∇ × B ≈ 0, thus
(b · ∇)b ≈
b
1
∇⊥ B
× [(∇B) × b] = [∇B − b(b · ∇B)] =
.
B
B
B
(20)
Substituting back to (18), we obtain
Vd=
2 /2 + v 2
v⊥
k
(B × ∇B) =
mv 2 c
(1 + cos2 α)(B × ∇B)
2qB 3
ΩB 2
2Ek c
=
(1 − B/2Bm )(B × ∇B)
qB 3
3
(21)
Note that because of different sign of charge, electrons and ions drift towards opposite directions.
Computing total field strength and its gradient is straightforward
ME p
3M
E
2
B= 3
1 + 3 cos θ, sin θ cos θ, 0 .
(22)
1 + 3 cos2 θ , ∇B = √
r
r4 1 + 3 cos2 θ
The resulting drift velocity is
vd = −
6Ek cr2
B sin θ(1 + cos2 θ)
6Ek cr02
B cos λ5 (1 + sin2 λ)
(1 −
)
eφ = −
(1 −
)
eφ .
2
2
qME
2Bm (1 + 3 cos θ)
qME
2Bm (1 + 3 sin2 λ)2
(23)
Therefore, particles drift in the azimuthal direction. The ions drift westward (negative drift velocity), while electrons drift eastward. Such drift motion creates a net current known as the ring
current. The drift velocity depends sensitively on latitude (due to the cos5 λ factor) and hence the
ring current is concentrated in the equator. At the equator, we have
v eq
d
6Ek cr02
B
6Ek c
r0 2 Ek
2
−1
=
(1 −
)≈
≈ 4.9 × 10 km s
qME
2Bm
qBr
4RE MeV
(24)
For both electrons and ions, the drift timescale is
teq
d
∼ 0.36 hours
4RE
r0
Ek
MeV
−1
(25)
In reality, particles only spend a fraction of time near the equatorial plane where the drift is the
fastest, and (25) underestimates the drift timescale by a factor of ∼ 3. Therefore, it typically
takes an hour for these energetic particles to complete one cycle of drift around the Earth. While
relativistic correction is needed for electrons, the derivations above suffices for our purposes.
Problem #2: Conservation of Energy
The goal of this problem is to show that ideal MHD conserves energy. We will do this by finding
equations governing the evolution of magnetic, internal, and kinetic energy in conservative form:
∂
(stuff) + ∇ · (flux of stuff) = (sources − sinks)
∂t
(26)
For reference, the equations of MHD that you may need are:
∂B
= −c∇ × E
∂t
c
J =
∇×B
4π
∂V
J×B
ρ
+ ρ (V · ∇) V =
− ∇p
∂t
c
V×B
E+
= 0
c
∇·B = 0
∂p
+ V · ∇p + γp∇ · V = 0
∂t
4
(27)
(28)
(29)
(30)
(31)
(32)
2
(a) The magnetic energy density is given by B
8π . Start from Faraday’s law and derive an equation
for the time evolution of magnetic energy
2 density in terms of the Poynting flux, electric field, and
∂B
∂
B
current density. Hint: 4π · ∂t = ∂t B
8π . You will need vector identities and Ampere’s law. (8
pts)
Start from Faraday’s law and use the hint:
∂B
= −c∇ × E
∂t
B ∂B
B
·
=
· (−c∇ × E)
4π ∂t
4π
∂ B2
c
= − B · (∇ × E)
∂t 8π
4π
Now use ∇ · (A × B) ≡ B · (∇ × A) − A · (∇ × B) to get
c
∂ B2
= − [∇ · (E × B) + E · (∇ × B)]
∂t 8π
4π
Then use Ampere’s law for the last term on the RHS and rearrange to get
∂ B2
cE × B
+∇·
= −E · J
∂t 8π
4π
p
(b) The internal energy density is given by γ−1
. Start from the energy equation and derive an
equation for the time evolution of the thermal energy density. You will need the vector identity
f ∇ · A ≡ −A · ∇f + ∇ · (f A), but no other equations from MHD. In the conservation equation, the
divergence should be taken of not just the internal energy flux but also contain the contribution
from work done on/by the plasma through pressure gradients forces. (8 pts)
The energy equation is
∂p
+ V · ∇p + γp∇ · V = 0
∂t
Use the vector identity from the problem
∂p
+ V · ∇p − γV · ∇p + γ∇ · (pV) = 0
∂t
∂p
+ γ∇ · (pV) − (γ − 1)V · ∇p = 0
∂t
Divide by (γ − 1) and rearrange:
∂
p
p
+∇·
V + pV = V · ∇p
∂t γ − 1
γ−1
2
(c) The kinetic energy density is given by ρV2 . Start from the momentum equation and derive
an equation for the time evolution of kinetic energy density (recalling the hint from part a). You
will need the continuity equation (multiplied by some factor) to get the additional terms from
5
evaluating
∂
∂t
ρV 2
2
with the product rule. You will need the ideal Ohm’s law to put it terms of a
2
Poynting flux, that A · (A · ∇A) = A · ∇ A2 , and the vector identity A · (B × C) = −(B × A) · C.
(8 pts)
The momentum equation is
ρ
∂V
+ ρV · ∇V =
∂t
J×B
− ∇p
c
Take V· this equation.
∂V
J×B
V· ρ
+ ρV · ∇V = V ·
− ∇p
∂t
c
2
V
(−V × B) · J
∂ V2
+ ρV · ∇
=
− V · ∇p
ρ
∂t 2
2
c
2
V
∂ V2
+ ρV · ∇
= E · J − V · ∇p
ρ
∂t 2
2
Let’s take the time derivative of ρV 2 /2 to see what cross terms we need:
V 2 ∂ρ
∂ V2
∂ ρV 2
=
+ρ
∂t
2
2 ∂t
∂t 2
This suggests that we will need to multiply the continuity equation by V 2 /2. Add the continuity
equation times V 2 /2:
2
V
V 2 ∂ρ
∂ V2
ρ
+ ρV · ∇
+
+ ∇ · (ρV) = E · J − V · ∇p
∂t 2
2
2 ∂t
2
V
∂ V2
V 2 ∂ρ
V2
+ ρV · ∇
ρ
+
+
∇ · (ρV) = E · J − V · ∇p
∂t 2
2 ∂t
2
2
{z
}
{z
}
|
|
∂
∂t
ρV 2
2
∇·
ρV 2
V
2
Combine the first and second terms, and the third and fourth terms on the LHS with the product
rule:
2 ∂ ρV 2
ρV
+∇·
V
= E · J − V · ∇p
∂t
2
2
(d) From parts (a)–(c), you should have three equations which add up to
2
∂ B 2 ρV 2
p
ρV
p
cE × B
+
+
+∇·
V+
V + pV +
= 0,
∂t 8π
2
γ−1
2
γ−1
4π
(33)
indicating that total energy is conserved. For each of the equations derived in parts (a)–(c), there
should be source and sink terms that show up on the right hand side that cancel when you add the
equations. What do each of these source and sink terms physically represent? (6 pts)
6
• The term E · J represents the transfer between magnetic energy and kinetic energy. It is the
work done by electromagnetic forces on the plasma. (Note that this is true in ideal MHD.
When resistivity is considered, the additional E · J leads to Joule heating where E = J /σ,
and σ is electric conductivity.)
• The term V · ∇p represents the transfer between kinetic energy and internal energy. It is the
work done by pressure gradients to accelerate or decelerate the plasma.
7