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Examples from 3rd edition of ‘Space Physics’
1
1.1
Charged Particles in Electromagnetic Fields
1. Example Sect. 2.2.1: A hot plasma with T = 1 keV is confined
in a homogenous
p
magnetic field B = 1 T. The thermal speeds are given by v = 2Wkin /m. Thus we get
ve = 18.7 · 106 km/s for the electrons and vp = 4.37 · 105 m/s for the protons. Gyro-radii
then are determined from
v⊥
mv⊥
rL =
=
ωc
|q|B
to be re = 0.1 mm and rp = 4.6 mm. From
ωc =
|q|B
m
we get for the cyclotron frequencies ωc,e = 1.8 · 1011 s−1 and ωc,p = 108 s−1 .
2. Example Sect. 2.2.2: So far, the highest proton energies observed in the galactic cosmic
radiation are 1020 eV. These protons gyrate in an interstellar magnetic field of about
B = 3 · 10−10 T. Since for such high energies the kinetic energy Wkin is approximately
equal to the total energy Etotal , the second term on the right hand in
p~ = m(v) ~v = γm0~v ,
and
p2 =
2
Wkin
− m0 c2
2
c
can be ignored and the particle momentum is p = Wkin /c. With
v⊥
p⊥
=
ωc,rel
|q| B
rL,rel =
we obtain a maximum Larmor radius of
rL =
p
Wkin
=
= 1021 m .
eB
ceB
This is approximately the size of the Milky Way.
3. Example Sect. 2.3.2: Wien filter: A magnetic field B = 5 · 10−4 T is perpendicular to
~ ×B
~
an electric field E = 1000 V/m. This configuration is a simple example for the E
drift. Since both fields are perpendicular, from
~vE×
~ B
~ =
~ ×B
~
E
2
B
2
6
we get a drift velocity vE×
~ B
~ = (EB)/B = E/B = 2 · 10 m/s. The drift velocity is
perpendicular to both the electric and magnetic field. An electron approaches perpendicular to both fields. On hitting the field combination, the electron will start to gyrate
around the magnetic field and drift along its original direction of motion. The size of
the gyro-orbit, of course, depends on the electron speed ve . A special case results for
ve = vE×
~ B
~ : in this case, the electron moves along a straight line. This can be understood as follows: if the electron moves along a straight line at constant speed, no forces
act on the electron. Thus the forces exerted by the electric and the magnetic field must
be equal: eE = eve B or ve = E/B which is the drift velocity derived above.
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4. Example Sect. 2.3.3: A proton with a kinetic energy of 1 keV (a 10 keV electron)
gyrates in the equatorial plane of the terrestrial magnetic field at a radial distance of
five Earth radii from the center of the Earth. All its kinetic energy is in the gyration.
The equatorial magnetic field at the surface of the Earth is 3.11·10−5 T, it falls off with
radial distance as r−3 .p
Thus the local field at the proton orbit is B = 2.5 · 10−7 T. The
proton speed is vp = 2Wkin /m = 4.4 · 105 m/s (ve = 5.9 · 107 m/s), its gyro-radius
according to
v⊥
mv⊥
rL =
=
ωc
|q|B
is rL,p = 1.8 · 104 m (rL,e = 1.4 · 103 m/s), which is still small compared to the
scales of the system, for instance, the drift path around the Earth has a length of
ldrift = 2πr = 50πrE = 109 m. The gravitational acceleration scales with r2 , thus it is
only g/25 with g being the gravitational acceleration at the surface of Earth. Since at
and above the equator the magnetic field is parallel to the surface, the gravitational
field is perpendicular to the magnetic field. The drift speed from
~vg =
~
m ~g × B
q B2
then is ~vg = 1.6 cm/s perpendicular to both fields: the particle drifts along a circle in
~ drift thus would give the particle such a small speed
the equatorial plane. The ~g × B
10
that it would take 6 × 10 s or almost 2000 years to drift around the entire Earth. Note
that the drift speed depends only on particle mass. Thus all protons drift with the same
speed, in dependent of their energy. Almost all protons: if the energy becomes too large,
the particles gyration radius may become that large that it either hits the atmosphere
and is absorbed during interaction or suddenly finds itself in interplanetary space and
escapes. The same argument holds also for all other particle species only their drift
speeds must be scaled by the ratio of their mass to the proton mass. The drift speed
of an electron therefore is smaller by a factor of 1836 compared to that of the proton.
~
5. Example Sect. 2.3.4: Let us briefly return to example 4 where we saw that the ~g × B
−3
drift is extremely slow. Since the magnetic field decreases with r , during its gyration
the particle scans parts of different magnetic field strength, thus a gradient drift results.
The gradient of the magnetic field has the same direction as the gravitational accele~ drift. Although
ration, thus the gradient drift is into the same direction as the ~g × B
the magnetic field is roughly a dipole field, for this special case it can be treated as
spherical symmetric since the particle gyrates in the equatorial plane and therefore is
not influenced by the latitudinal variation of B. From
∂φ 1 ∂φ ∂φ
,
,
∇φ =
∂r r ∂θ ∂z
we get the gradient of the magnetic field B(r) = Bo (ro /r)3 as ∇B = (−3B/r, 0, 0).
The cross-product in
~v∇B =
µ ~
v⊥ rL ~
B × ∇B = ±
B × ∇B
qB 2
2B 2
~ ×∇B| = 3B 2 /r and the drift speed becomes vD = 3mv 2 /(2rqB) = 26.4 m
than gives |B
for the protons. This is three orders of magnitude larger than the drift due to the crossed
magnetic and gravitational fields. Again, the drift speed of the electron is smaller by a
factor of 1836.
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6. Example Sect. 2.3.5:
p A proton plasma with a temperature of 10 MeV (corresponding
to a speed of v = 2Wkin /m = 4.3 · 107 m/s is confined by a homogenous magnetic
field inside a torus of diameter 1 m. The dimension of the torus also gives the diameter
of the gyro orbit. Thus a magnetic field B = mv⊥ /(qrL ) = 0.9 T. According to
~vR =
~
mvk2 ~rc × B
2
2
qB
rc
the curvature drift then is vr = mv 2 rL /(qB) = 0.24 m/s tangential to the particle path.
7. Example Sect. 2.3.7: In example 5 we have calculated the drift speeds of 1 keV protons
and 10 keV electrons. In the radiation belts both have a number density n = 107 m−3 .
With the drift speeds from example 5 we can determine the ring current densities
according to
X nk v 2 k⊥
~ × ∇B .
~jD∇B =
B
2B 3
k
For the protons we then get j = 4.3 × 10−11 A/m2 .
8. Example Sect. 2.4.2: Assume an isotropic distribution, that is the pitch angles are
distributed equally, of 10 keV electrons injected on a field line at the equator, Lo = 5
Earth radii from Earth’s center. The magnetic field varies as
p
BE 1 + 3 sin2 Φ
B(Lo , Φ) = 3
(1)
Lo
cos6 Φ
with Φ being the geomagnetic latitude and BE = 3.11 · 10−5 T being the equatorial
magnetic field strength at the surface. The equation of the magnetic field line is L =
Lo cos2 Φ. From the conservation of the magnetic moment we can determine the number
of particles reflected at geomagnetic latitudes of 30◦ and 60◦ . We only need the ratio
between the magnetic field strength
at the injection site and at the reflection point.
p
From (1) we get Brefl /Beq = 1 + sin2 Φrefl / cos6 Φrefl , which gives a magnetic field
ratio of 2.65 for Φ = 30◦ and 84.7 for Φ = 60◦ . From
s
s
B
B1
1
1
1
2 r
r
sin α1 =
=
or α1 = arcsin
= arcsin
Bmp
Rmp
Bmp
Rmp
we then find that all particles with pitch angle larger than 38◦ (6.3◦ ) are deflected at
geomagnetic latitudes below 30◦ (60◦ ) (that is 48% (83%) of the initial population).
From the equation of the field line we find that it intersects the surface of Earth (L = 1)
at a geomagnetic latitude of 63◦ .
9. Example Sect. 2.4.2: Magnetic pumping is a process in which the violation of an
adiabatic invariant is used to accelerate particles. To illustrate the idea, let us start
from an isotropic plasma gyrating in a homogenous magnetic field. Plasma temperatures and thus particle energies parallel and perpendicular to the field are the same:
Wkin,k,0 = Wkin,⊥,0 = Wo . Let us no increase the magnetic field slowly by a factor of two
such that the concept of adiabatic invariants can be applied but fast enough to prevent
an exchange between parallel and perpendicular energy. Then the energy parallel to
the field remains unchanged but since the magnetic moment is conserved, the perpendicular kinetic energy increases as B increases: Wkin,⊥ ∼ B. If we wait for a sufficiently
long time to allow temperature exchange between the parallel and perpendicular motions, we again have an isotropic plasma, now with Wkin,k,1 = Wkin,⊥,1 = 1.5 Wo and
3
Wkin,k,1 = 1.5Wkin,k,0 and Wkin,⊥,1 = 1.5Wkin,⊥,0 . We can now allow the magnetic field
to relax to its original value with the same speed. Again, the parallel kinetic energy
remains constant while that in the motion perpendicular to the field is reduced by a
factor of two: Wkin,k,1 = 0.75Wkin,k,0 and Wkin,⊥,1 = 1.5Wkin,⊥,0 . The total energy then
is W = 1.25Wo , that is the plasma has gained energy during this process which can be
repeated for further energy gain.
1.2
Magnetohydrodynamics
10. Example Sect. 3.3.1: A homogenous magnetic field of 5 T according to
pM =
B2
2µ0
exerts a magnetic pressure p = B 2 /2µo = 9.95 · 106 N/m2 = 995 hPa · 100, which is a
hundred times the atmospheric pressure at sea level.
11. Example Sect. 3.3.2: Again, consider a 5 T magnetic field. It is disturbed by a sinusoidal
velocity field v = vo sin kx with vo = 1 m/s and k = 5 m−1 acting for δt = 1 µs.
To determine the force density, we need the second derivative of the velocity: v 00 =
−vo k 2 sin kx. With
1 ∂ 2 uz 2
fz =
B dt
µ0 ∂x2 0
we then get f = 20 N/m2 sin(5/m x) and thus for x = 0 m f = 0 m because here the
magnetic field is not replaced from its original position and no restoring force acts on it;
for x = 2.5 m we get 14 N/m2 and for x = 5 m (maximum displacement) f = 20 N/m2 .
12. Example Sect. 3.4.2: A sunspot with a radius of about 20 000 km from
τ ≈ µ0 σL2 = L2 /DM
(2)
has, with the conductivity given above, a life time of about 1000 years. If we are looking
closer at the sunspot, in particular the granules around the spot, we find a spatial scale
of about 1000 km, that is 1/20 of the sunspots scale. Since the dissipation time depends
on the square of the length scale, it is only 1/400 in granules, that is 2.5 years – which
comes closer to the observed life time of a sunspot.
1.3
Plasma Waves
13. Example Sect. 4.2.1: Average parameters of the solar wind are a number density of
8 cm−3 , a magnetic field B = 7 nT and a temperature of about 2 × 105 K. The density
f the solar wind then is % = nmp and the Alfvén speed is
s
(7 nT)2
kg
km
vA =
8 × 1.673 × 10−27 −3 = 54
.
Vs
−7
m
s
4π × 10
Am
Here we used the fact that the unit T can be expressed as Vs/m2 and that the product
of the electrical units volt and ampere gives watt which easily can be expressed as a
mechanical unit.
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14. Example Sect. 4.2.2: Let us briefly return to the previous example where we have
already determined the Alfvén speed in the solar wind. To determine the speed of the
magneto-sonic wave we also need the sound speed. From
r
γa p0
vs =
%0
p
p
we get vs =
γa nkB T /(nmp ) =
γa kB T /mp = 70 km/s. For the speed of the
magneto-sonic wave we get with
2
vms
=
ω2
2
= vs2 + vA
k2
vms = 88.7 km/s.
15. Example Sect. 4.3.1: Solar radio bursts are an important tool for the diagnostic of
coronal disturbances. For the lower corona we can use a simplified density model n ∼
no exp −r/ro with a scale height of ro = 0.1r and a density no = 101 5 m3 at r = 1r
(ok, more correctly at the base of the corona 2000 km above the solar surface, but for
the numerical exercise these 2000 km can be ignored). Radio emission observed from
the ground is in the frequency range 10 MHz to about 200 MHz. According to
s
ne e2
ωpe =
ε0 me
this corresponds to coronal heights between 2 ro and 0.8 ro above the surface. The
main mechanisms for the generation of solar radio bursts are electron beams with
speeds of about c/3 (type III burst) and coronal shock waves with speeds of order of
1000 km/s (type II bursts). The height range over which these exciters propagate is
1.2ro = 0.12r = 8.4 × 104 km. The electrons with c/3 travel this distance within
about 0.3 s, thus the type III burst shows an extremely fast frequency drift of order
of 750 MHz/s. The shock wave, on the other hand, needs almost 9 s for this distance,
corresponding to an average frequency drift of about 20 MHz/s. Thus both bursts can
be easily identified by their frequency drift in radio spectrograms.
16. Example Sect. 4.3.3: The proton density upstream of a shock is ni = 8 cm−3 , for the
proton Zi = 1. Ion-acoustic waves (upstream waves) excited by particles streaming
away from the shock than have a ion–acoustic frequency ωpi = 3.7 × 103 Hz.
17. Example Sect. 4.4.3: The electron density in the quiet ionosphere at 120 km height is
about 2×105 cm−3 . Reflection occurs for n = 0, that is when the local plasma frequency
ωpe equals the wave frequency. From
s
ne e2
ωpe =
ε0 me
we find fpe = 2.4 MHz. During a sudden ionospheric disturbance, the electron density
is increased by a factor of 3. Now the critical frequency for reflection at 120 km height
becomes 7 MHz. The 2.4 MHz wave reflected originally at this height now is been
reflected at a lower height and consequently their range of propagation is reduced and
the signal is not received where it was supposed to be received: communication is
inhibited.
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1.4
Sun and Solar Wind
18. Example Sect. 6.8.4: For the shocks in Fig. 6.41, the following plasma parameter are
determined from the figure: day 217 (268), upstream density 30 cm−3 (14 cm−3 ), downstream density 60 cm−3 (105 cm−3 ), upstream speed 360 km/s (602 km/s), and downstream speed 420 km/s (1101 km/s). From
vs =
%d un,d − %u un,u
.
%d − %u
(3)
we then get local shock speeds of 480 km/s (1204 km/s), the compression ratios rn are
2 (7.3). Both shock speeds are a good approximations on the more accurate ones given
with the figure.
19. Example Sect. 6.8.9: Let us briefly return to example 18. From the details of the magnetic field a colleague has inferred the shock normal and the direction of the upstream
and downstream plasma flows relative to it. For simplicity, the shock normal is given
as (1, 0) with the x component in the direction of shock propagation. For the shock
of doy 217 the direction of the upstream flow is (0.94,0.34), that of the downstream
flow (0.98,0.17), the shock therefore is almost quasi-parallel as expected from the weak
compression in the magnetic field. The shock speed than is
0.98
0.94
60 ×
× 420 − 30 ×
× 360 0.17
0.34
km
1 km
·
= 485
vs =
0
60 − 30
s
s
which is slightly above the values of 480 km determined in example 18.
1.5
Energetic Particles
20. Example Sect. 7.5.2: Let us now assume a shock with an upstream speed of 800 km/s
in the shock rest frame and a ratio of flow speeds r = 3. With a typical upstream mean
free path λu = 0.1 AU for 10 MeV protons we get an upstream diffusion coefficient
Du = vλ/2 = 3.26 × 1017 m2 /s. With
p
3
Du
Dd
τa =
=
+
dp/dt
uu − ud uu
ud
we then get a characteristic acceleration time τ = 2.3 × 106 s, that is almost 27 days or
one solar rotation. Even if we increase scattering by a factor of 10, the characteristic
acceleration time would be almost 3 days, which is longer than the average travel
time of a shock between Sun and Earth. However, under these conditions, the mean
free path would be rather small and the acceleration still would be inefficient because
during the characteristic acceleration time the momentum is increased by a factor e
only. The situation is different at a lower particle speed, say 100 keV. In that case, the
diffusion coefficient is an order of magnitude smaller and consequently the characteristic
acceleration time is only a tenth of that of 10 MeV protons, that is roughly 7 hours
(for the optimistic case of strong scattering with λ = 0.01 AU). During a travel time
of 2 days, particle momentum therefore increases by about three orders of magnitude
(under the simplifying assumption of a diffusion coefficient independent of particle
energy/momentum).
21. Example Sect. 7.5.2: Typical values for the shock compression ratio r between slightly
above 1 and about 8, the average value is close to 2. With this value from J(E) = J0 E −γ
6
we get a spectral index γ = 2 which is rather flat compared to the observed spectral
indices which cluster around 3.
22. Example Sect. 7.5.2: With the parameters given in example 20 the scale length of the
upstream increase is 2.5 × 10−12 m−1 or 0.35 AU−1 . Let us assume that the shock
has reached steady-state. Then this upstream increase would be convected across the
observer with the shock speed. If this is 1200 km/s, a scale length of 0.35 AU takes
about 54000 s or 15 h to pass the observer. Thus in the intensity time profile upstream
of the shock we would expect an increase by a factor of e over a time of 15 h. Again the
situation is different for 100 keV protons. Here the diffusion coefficient and thus the
scale length are an order of magnitude smaller and consequently the rise in intensity
upstream of the shock will be much steeper: a factor of e in 1.5 h.
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