ionosphere - kth · time. room: subject. litterature: l1. 29/8. 13-15. e52. course description,...

Last lecture (5) • Ionosphere

- index of refraction - reflection of radio waves

- particle drift motion in magnetized plasma

Today’s lecture (6) • Ionosphere

-electrical conductivity in ionosphere

• Magnetosphere, introduction

• Magnetospheric size (standoff distance)

• Particle motion in the magnetosphere

Today

EF2240 Space Physics 2016

Activity Date Time Room Subject Litterature

L1 29/8 13-15 E52 Course description, Introduction, The Sun 1, Plasma physics 1

CGF Ch 1, 5, (p 110-113)

L2 1/9 15-17 L52 The Sun 2, Plasma physics 2 CGF Ch 5 (p 114-121), 6.3

L3 5/9 13-15 E51 Solar wind, The ionosphere and atmosphere 1, Plasma physics 3

CGF Ch 6.1, 2.1-2.6, 3.1-3.2, 3.5, LL Ch III, Extra material

T1 8/9 15-17 D41 Mini-group work 1 L4 12/9 13-15 E35 The ionosphere 2, Plasma physics 4 CGF Ch 3.4, 3.7, 3.8 L5 14/9 10-12 V32 The Earth’s magnetosphere 1, Plasma

physics 5 CGF 4.1-4.3, LL Ch I, II, IV.A

T2 15/9 15-17 E51 Mini-group work 2 L6 19/9 13-15 M33 The Earth’s magnetosphere 2, Other

magnetospheres CGF Ch 4.6-4.9, LL Ch V.

T3 22/9 15-17 E51 Mini-group work 3 L7 26/9 13-15 E31 Aurora, Measurement methods in space

plasmas and data analysis 1 CGF Ch 4.5, 10, LL Ch VI, Extra material

L8 28/9 10-12 L52 Space weather and geomagnetic storms CGF Ch 4.4, LL Ch IV.B-C, VII.A-C

T4 29/9 15-17 M31 Mini-group work 4 L9 3/10 13-15 E52 Interstellar and intergalactic plasma,

Cosmic radiation, CGF Ch 7-9

T5 6/10 15-17 E31 Mini-group work 5 L10 10/10 13-15 E52 Swedish and international space physics

research.

T6 13/10 15-17 E31 Round-up, old exams. Written examination

26/10 8-13 F2

EF22445 Space Physics II 7.5 ECTS credits, P2

• shocks and boundaries in space

• solar wind interaction with magnetized and unmagnetized bodies

• reconnection

• sources of magnetospheric plasma

• magnetospheric and ionospheric convection

• auroral physics

• storms and substorms

• global oscillations of the magnetosphere

Courses at the Alfvén Laboratory

EF2260 SPACE ENVIRONMENT AND SPACECRAFT ENGINEERING , 6 ECTS credits, period 2

• environments spacecraft may encounter in various orbits around the Earth, and the constraints this places on spacecraft design

• basic operation principles underlying the thermal control system and the power systems in spacecraft

• measurements principles in space

Projects: • Design power supply for

spacecraft

• Study of radiation effects on electronics

The Astrid-2 satellite Radiation environment in near-

earth space

Mini-groupwork 2

arctan sunsw

ru

ωψ =tan

sunsw

ru ωψ

=

a)

ωsun = 2π/T = 2.9·10-6 s-1 (T = 25 days at equator) r = 1 A.U. tan ψ = |By/Bx| ≈ 3.6/2.6 (from figure) (ψ = 41°) With these figures I get usw = 313 km/s

By

-Bx


http://www.google.co.uk/imgres?imgurl=http://humbabe.arc.nasa.gov/MarsDustWorkshop/NASA_Logo.gif&imgrefurl=http://humbabe.arc.nasa.gov/MarsDustWorkshop/DustHome.html&h=857&w=1005&sz=44&tbnid=m0YhM3vHjtBOYM:&tbnh=127&tbnw=149&prev=/images?q=nasa+logo&hl=en&usg=__0kPVdOvODWC3RSZO6MPmj4_eV48=&ei=jkWpSuv_Cs7S-QavzsjXBg&sa=X&oi=image_result&resnum=1&ct=image

Mini-groupwork 2

l = vt

b)


The magnetic Reynolds number is calculated by using typical plasma flow velocities vc and typical length scales of magnetic field variations lc Use solar wind velocity obtained in a) for typical flow velocity. To obtain lc, multiply the time t it takes the magnetic field structure (indicated in the figure), to pass over the satellite and use lc, = vt. I get lc = 2.8·108 m. Using a temperature of 5·104 K, we can evaluate the conductivity, remembering that the temperature should be given in eV. We get the conversion from

BW k T= which gives the result that 1 eV corresponds to a temperature of 7729 K. We then get T = 6.5 eV, and σ = 3.1·104 S/m Putting in the numbers I get Rm = µ0 σ vc lc ≈ 9.8·1014 >> 1 So the solar wind magnetic field is frozen into the plasma to a very good approximation.
The magnetic Reynolds number is calculated by using typical plasma flow velocities vc and typical length scales of magnetic field variations lc

Use solar wind velocity obtained in a) for typical flow velocity. To obtain lc, multiply the time t it takes the magnetic field structure (indicated in the figure), to pass over the satellite and use lc, = vt. I get lc = 2.8·108 m.

Using a temperature of 5·104 K, we can evaluate the conductivity, remembering that the temperature should be given in eV. We get the conversion from

B

WkT

=

which gives the result that 1 eV corresponds to a temperature of 7729 K. We then get T = 6.5 eV, and

= 3.1·104 S/m

Putting in the numbers I get

Rm = 0 vc lc ≈ 9.8·1014 >> 1

So the solar wind magnetic field is frozen into the plasma to a very good approximation.

_1535524061.unknown

Frozen in magnetic flux PROOF II

( ) 20

1t µ σ

∂= ∇× × + ∇

∂B v B B

A B Order of magnitude estimate:

( )0

22

0 0

1 m

v BA L vL RBB

L

µ σ

µ σ µ σ

∆∇× ×

= ≈ = ≡∆

∇

v B

B

Magnetic Reynolds number Rm:

Rm >> 1 ⇒ ( )t∂

= ∇× ×∂B v B

2

0

1t µ σ

∂= ∇

∂B BRm

Typical length scale L


B

L dB Bdx L

x

https://www.google.se/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0CAcQjRxqFQoTCNv_t6qzxMcCFUGGLAods8cDGw&url=https://commons.wikimedia.org/wiki/File:Simple_sine_wave.svg&ei=2nXcVduqIcGMsgGzj4_YAQ&psig=AFQjCNF-2gocZWIngNnu1M53FHHK37RC4Q&ust=1440597849251002

Energy - temperature

3223

B

B

E k T

ETk

= ⇒

=

1 eV = 1.6·10-19 J ⇒

19

23

2 2 1.6 10 J 7729KJ3 3 1.38 10K

B

ETk

−

−

⋅ ⋅= = =

⋅ ⋅

Average energy of molecule/atom:


But beware! In plasma physics, usually:

32 B

B

E k T

ETk

= ⇒

=

1 eV = 1.6·10-19 J ⇒

19

23

1.6 10 J 11594KJ1.38 10K

BE k T−

−

⋅= = =

⋅


Does the plasma follow the magnetic field (a) or the other way around (b)?

Depends on relative energy density (pressure)

pl Bp nk T=2

02B

Bpµ

=

pl

B

pp

β =

1β >>1β

Mini-groupwork 2

c)

e pn mρ = = 6.1·106·1.67·10-27 = 1.02·10-20

Then the kinetic energy density is (v = 313 km/s): ρv2/2 = 5.0·10-10 Jm-3 The magnetic energy density is (using values of figure)

2

02Bµ

= ( )22 2 2

02x y zB B B

µ

+ +=(2.62 + 3.62 + 1.72)·(10-9)2 /2µ0 = 9·10-12 Jm-3

The ratio between the kinetic and magnetic energy densities is approximately 50, thus the plasma motion determines the magnetic field configuration, and not the other way around.

ep

nm

r

=

= 6.1·106·1.67·10-27 = 1.02·10-20

Then the kinetic energy density is (v = 313 km/s):

v2/2 = 5.0·10-10 Jm-3

The magnetic energy density is (using values of figure)

2

0

2

B

m

=

(

)

2

222

0

2

xyz

BBB

m

++

=(2.62 + 3.62 + 1.72)·(10-9)2 /20 = 9·10-12 Jm-3

The ratio between the kinetic and magnetic energy densities is approximately 50, thus the plasma motion determines the magnetic field configuration, and not the other way around.

_1220724889.unknown

_1220724972.unknown

_1220721865.unknown

Reflection of radio waves

n1

n2

Total reflection at a sharp boundary (or

large gradient) if

2 1n n<


Plasma oscillations parallel to B


eF m a=

F eE= −0

E σε

=

een xσ =

sin( )pex tω=

2

0

epe

e

n em

ωε

≡

2 2

20

e

e

n e x d xm dtε

− =

⊗ L

+

+ +

+ +

+ +

+

+

+

+

+ +

+

+

+

+ +

+

+ +

- -

-

- -

-

-

-

-

-

-

-

-

-

-

-

-

-

L

x

d


Index of refraction for electromagnetic waves in a plasma


0 0 0 tµ µ ε ∂∇× = +

∂EB j

t∂

∇× = −∂BE(1)

(2)

(3) e een= −j v

(4) eem et∂

= −∂v E

Assume all quantities vary sinusoidally, with frequency ω, e.g.:

( )0

i te ω⋅ −= k rE E

( )t

∂∇× ∇× = −∇×

∂BE(1)

2

0 2 2

1t t c t

µ∂ ∂ ∂∇× = +∂ ∂ ∂B j E

(2)

∴ ( ) 22

201

tct ∂∂

−∂∂

−=×∇×∇EjE µ

( ) 22

202 1

tcten ee ∂

∂−

∂∂

=∇−⋅∇∇EvEE µ

Index of refraction for electromagnetic waves in a plasma


Does not represent E.M. wave

(4)

2 22 2 2 2 20 0

0 0

1e ee e

n e n ec k cm m

µ µω ωµ ε−

= − + = +

∴ 2 2 2 2pc kω ω= +

2 2 22 2 22

2 2 2 21p p

ph

c c knv

ω ω ωω ω ω

−= = = = −

∴

2 2

2 21 1p pfn

fωω

= − = −

( ) ( ) ( )EvEEkk 2202 1 ωωµ −−−=+⋅− cienk ee

( )EEE 2202 1 ωωωµ −−

−−=

cmieenik

ee

Large gradient when

pef f≈

h

ne

n2 n1

2

2ph

2 1

c 1v

pefnf

n n

= = −

⇒<

Higher frequencies → higher fpe (ne)

Where does the total reflection

take place?


Ionosonde

The pulse will be reflected where

f = fpe The altitude will be determined by

2h = ct

Where t is the time between when the pulse is sent out and the registered again.


Reflection of radio waves, oblique incidence

EI2433 - 2015

Snell’s law: βα sinsin0 nn =

n0

n

α

β

Refraction index

for plasma

2

2ph

c 1v

pefnf

= = −

n

f fpe

Imaginary

1

Constant density

n

fpe f

Imaginary

1

Constant frequency

2

0

epe

e

n em

ωε

≡

EI2433 - 2015

Reflection of radio waves, oblique incidence

EI2433 - 2015

Snell’s law + condition for reflection:

α

α

αα

α

β

βα

cos

cos

1cos1sin

sin

1sin,1

sinsin

22

22

0

0

0

p

p

p

ff

ff

ff

n

nnn

nnn

<

⇒−>−

⇒−=>−=

⇒=>

⇒

>=

=

n0

n

α

β

Reflection of radio waves

F2-layer during night:

11 -3

7

5 10 m

10 Hz = 10 MHz

= HF/short wave

e

pe

nf

= ⋅ ⇒

=

Ground wave

Sky wave


E B


Drift motion


( ),0,x zE E=E

Consider a charged particle in a magnetic field.

Assume an electric field in the x-z plane:

( )dm qdt

= × +v v B E

xy x

yx

zz

dvm qv B qEdtdv

m qv Bdt

dvm qEdt

= + = − =

Constant acceleration along z

22

2

2 22

2 2

y yxg g x

y x xg g y x

dv dvd v qB vdt m dt dt

d v dv dvqB q Bv Edt m dt dt m

ω ω

ω ω

= = = −

= − = − = − −

y

x B = B z

+

Drift motion


22

2

2 22

2 2

y yxg g x

y x xg g y x

dv dvd v qB vdt m dt dt

d v dv dvqB q Bv Edt m dt dt m

ω ω

ω ω

= = = −

= − = − = − −

∴ 2

22

2

22

xg x

xy

xg y

d v vdt

Ed vEB v

dt B

ω

ω

−

+ = − +

g x

g y

i tx

i txy

v v eEv v eB

ω δ

ω δ

+⊥

+⊥

=

= − +

Average over a gyro period:

( ), 2 2

yx x zdrift y

E E BvB B B

×= − = − =

E B

In general:

2 2 2driftq

B qB qB× × ×

= = =E B E B F Bv

Drift motion

F = 0

F = qE

F = mg

F = -µ grad B

2drift qB×

=F Bu


Suppose you apply an electric field E in the direction showed in the figure, and that one electron and one ion (charge –e and e) is present. What will the resulting current be?

Green

F = 0

F = qE

F = mg

F = -µ grad B

2qB×

=F Bu

E

Yellow ˆEeB

= −I x

Red 1 1ˆ ˆ2 2

E Ee eB B

= −I x y

Blue 0=I

ˆEeB

=I y

x

y E e e≡ −i eI u u


Blue

e e≡ −i eI u u

2 2 2ˆ ˆe EB E

qB eB B B× ×

= = = − = −iF B E Bu x x

2 2 2ˆ ˆe EB E

qB eB B B× − ×

= = = − = −−e

F B E Bu x x

( ) 0e e e≡ − = − =i e i eI u u u u


So, if there is no current when you apply an electric field, is the conductivity of the ionospheric plasma zero ?


What is the electron density at 100 km?

What is the neutral density at 100 km?


Gyro motion


ExB-drift

With collisions

i+

e-

E B

e-

i+

Without collisions


Electric conductivity in a magnetized plasma

• i// = parallel current • iP = Pedersen current • iH = Hall current


Birkeland, Hall, Pedersen


Kristian Birkeland 1867-1917

Norwegian scientist

Edwin Hall 1855-1938

American physicist

Peder Oluf Pedersen 1874-1941

Danish engineer and physicist

S

I d= ⋅∫ j S

i

Current density

The current density j is a vector field with dimension [i] = Am-2.

The total current I through the surface S is


2222 11

11

igii

egeeP τω

στω

σσ+

++

=

2222 11 igiigi

iege

egeeH τω

τωσ

τωτω

σσ+

−+

=

// e iσ σ σ= +

eee mne /2 τσ = iii mne /

2 τσ =

////// Ei σ=

⊥= Ei PP σ

⊥= Ei HH σ

Electric conductivity in a magnetized plasma II

or P H Bσ σ ⊥⊥ ⊥

×= +

B Ei E


May be formulated as a tensor equation

Eσi ⋅=

−=

//0000

σσσσσ

PH

HP

σ

Electric conductivity in a magnetized plasma II

conductivity tensor


Collisional frequency


Ionospheric conductivities


Consequence: Birkeland currents

When the conductivity out in the magnetosphere is low, it is easier for the current to close through the ionosphere via currents parallel to the geomagnetic field. Such currents are called Birkeland currents.

Region of low conductivity


Exemple: Electric field 700 km above the aurora.

-1

-1

ˆ ˆ

1Vm

1 μVm

x y

x

z

E E

EE

= +

=

=

E x y

jP = jx = 0.01 µAm-2

j// = jz = 40 µAm-2

jP = jx = 10.0 µAm-2

j// = jz = 4.0 µAm-2

B x

z

Nightside, solar maximum.

Yellow

Red


jP = jx = 1.0 µAm-2

j// = jz = 40 mAm-2 Blue

-8 -1

-1//

8 -2 -2

6 -2 -2// //

1 10 Sm

40Sm

1 10 Am 0.01μAm

40 10 Am 40μAm

P

P x P x

z z

j j Ej j E

σ

σ

σ

σ

−

−

≈ ⋅

≈

= = = ⋅ =

= = = ⋅ =

Yellow


How do we define ”the magnetosphere”?

The region in space where the magnetic field is dominated by the geomagnetic field.


Polar (spherical) coordinates


http://en.citizendium.org/images/e/e7/Spherical_unit_vectors.png

Geomagnetic field

Approximated by a dipole close to Earth.

3( ) cosEr pRB Br

θ=

3( ) sin2

p EB RBrθ

θ=

3

0

2 E pR Baπ

µ=

magnetic dipole moment Magnetic field at the

“north pole”

ˆrB r

ˆBθθθ


B

Geomagnetic field

Alternative formulation of dipole field

3( ) cosEr pRB Br

θ=

3( ) sin2

p EB RBrθ

θ=

3

0

2 E pR Baπ

µ=

magnetic dipole moment

03

1 cos2r

aBr

µ θπ

=

03

1 1 sin2 2

aBrθ

µ θπ

= ⋅ ⋅


Geomagnetic field

• Angle between dipole axis and spin axis: ≈ 11°

• The geographic north pole is a magnetic south pole, and vice versa.

• Bequator = 31 µT, Bpole = 62 µT


Geomagnetic field Modified by solar wind into tail-like configuration


Stand-off distance from pressure balance

2SWSWd vp ρ=

Dynamic pressure: 2

02B

Bpµ

=

Magnetic pressure:

solar wind

- dense plasma with high conductivity

- weak magnetic field • thin plasma • large magnetic field

magnetosphere


Meissner effect in super-conductors


Dynamic (kinetic) pressure

ρ,v A

v dt

( ) ( ) 21 1d

d mv mvF Av t vp vA dt A t A tA

ρ ρ∆ ⋅ ∆ ⋅

= = ≈ = =∆ ∆


⇒= Bd pp⇒

= 0

2

302 2/1

4µ

πµρ

ravSWSW

( ) 6/1203/1

0 24

−

= SWSW v

ar ρµπ

µ

2SWSWd vp ρ=

2

0

12B

p Bµ

=

30 1

4 ra

Bπ

µ=

Dynamic pressure:

Magnetic pressure:

Dipole field strength (in equatorial plane):

a = 8x1022 Am2, v=500 km/s, ρSW=107x1.7x10-27 kg/m3:

r = 7 Re (1 Re = 6378 km)

Magnetopause ”stand-off distance”

r Solar wind Magnetosphere


Standoff distance

Green

Yellow

Red

Blue r = 1.8 Re

( ) 6/1203/1

0 24

−

= SWSW v

ar ρµπ

µ

v=500 km/s, ρSW=107x1.7x10-27 kg/m3: r = 7 Re

r

How will the standoff distance change if the magnetosphere is hit by a coronal mass ejection (CME)? (ρ = 10ρSW , v = 1000 km/s)

r = 3.8 Re

r = 5.8 Re

r = 9.8 Re


( )( ) ( )1/3 1/31/ 6 1/ 62 2 1/ 60 0

0 02 20 2 41 40

4 SW SW SWSWa ar v vµ µµ ρ µ ρ

π π

− − − = =

40-1/6·7 = 0.54 ·7 = 3.8

Green r = 3.8 Re

Standoff distance


Particle motion in magnetic field gyro radius

mvqB

ρ ⊥=

gyro frequency

gqBm

ω =

magnetic moment µ = IA = q fgπ ρ2 = mv⊥2/2B

Adiabatic invariant

DEFINITION:

An adiabatic invariant is a property of a physical system which stays constant when changes are made slowly.

By ’slowly’ in the context of charged particle motion in magnetic fields, we mean much slower than the gyroperiod.

’First adiabatic invariant’ of particle drift:

2

2mv

Bµ ⊥=


Slide Number 1Slide Number 2EF22445 Space Physics II�7.5 ECTS credits, P2Slide Number 4Slide Number 5Slide Number 6Frozen in magnetic flux PROOF IITypical length scale LSlide Number 9Slide Number 10Slide Number 11Slide Number 12Reflection of radio wavesSlide Number 14Slide Number 15Slide Number 16Slide Number 17Where does the total reflection take place?IonosondeReflection of radio waves,�oblique incidenceRefraction index �for plasmaReflection of radio waves,�oblique incidenceReflection of radio wavesSlide Number 24Slide Number 25Slide Number 26Drift �motion Slide Number 28Slide Number 29Slide Number 30Slide Number 31Slide Number 32ExB-driftElectric conductivity in a magnetized plasmaBirkeland, Hall, Pedersen Slide Number 36Electric conductivity in a magnetized plasma IIElectric conductivity in a magnetized plasma IICollisional frequencySlide Number 40Consequence: �Birkeland currentsExemple: Electric field 700 km above the aurora. Slide Number 43Slide Number 44Slide Number 45Slide Number 46Slide Number 47Slide Number 48Geomagnetic fieldSlide Number 50Slide Number 51Dynamic (kinetic) pressureMagnetopause ”stand-off distance”Slide Number 54Slide Number 55Particle motion in magnetic fieldSlide Number 57Magnetic mirrorMagnetic mirrorMagnetic mirrorSlide Number 61Magnetic mirrorSlide Number 63Slide Number 64Particle motion in geomagnetic field Drift �motion Slide Number 67Ring current and particle motionRadiation beltsRadiation beltsCRAND (Cosmic Ray Albedo Neutron DecayRadiation beltsRadiation beltsParticle motion in geomagnetic field Structure of magnetosphereMagnetospheric structureOutflow from the ionosphereSlide Number 78Slide Number 79Frozen in magnetic field linesXReconnectionSlide Number 83Slide Number 84Slide Number 85Magnetospheric dynamicsSlide Number 87Slide Number 88Slide Number 89Slide Number 90

ionosphere - kth · time. room: subject. litterature: l1. 29/8. 13-15. e52. course description,...

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