Lecture 4 Waves

A simple wave: Sound wave in the air.—restoring pressure balanceIn the Sun, possible waves:

Alfven wave: magnetic tensionSound wave: pressureGravity wave: gravity forceor combination of above

Basic Equations

0Vtρ ρ∂+ ∇• =

∂

uv

ˆ( ) 2V BB gz Vt

ρ ρ ρ ρµ

∂= −∇ + ∇× × − − Ω×

∂

uv uvuv uv uv

( ) 0Pt rρ∂

=∂

( )B Bt

∂= ∇× ∇×

∂

uvuuv

adiabatic law

0B∇• =uv 1j B

µ= ∇×

uv mPTkρ

=

If 0B↑↑ z↑ 0 0( )z

z eρ ρ−Λ=

0

z

P P e−Λ=

gP

0

0

ρ=Λ Scale height 150Km in photosphere, 108m in corona

Let’s consider small departure from equilibrium

1010110 BB , PP ,v v, BPvvv

+=+==+= ρρρLinearize equation

0

)(

0))(()(

2)(

0)()(

1

011

0112

011

1010

111

0

10011

=•∇

××∇=∂∂

=∇•+∂∂

−∇•+∂∂

×Ω−−××∇+−∇=∂∂

=•∇+∇•+∂∂

B

BvtB

vt

CPvt

vzgnlBBP

tv

vvt

s

v

vv

vv

vv)vv

v

ρρρ

ρρρ

ρρρ

mkTCsγ

ργρ

==0

02 Where

Remove everything except 1vv

0

001

1111

221

2

))](([2)()1()(µρ

γ BBvtvvgvzgvC

tv

zs

vv)v

v×××∇×∇+

∂∂

×Ω−∇−•∇−−•∇∇=∂∂

Seek plane-wave solution

k2 2T frequency

tornumber vec wave

),( )(11

πλωπω

ω

==

= −•

k

evtrv trki

v

vv vv

If assume is a locally constant (WKB approx)0ρ

16) (4. ))](([2)()1()(0

11112

12

µρωγω BBvkkviigkvvkzgivkkCv zs

vvvvvvvv)vvvv ××××+×Ω−+•−+•=

Main objective: drive dispersion relation)(kv

ωω =

velocitygroup v, v,v

velocityphase

gzgygxzyx

p

kkk

kk

v

∂∂

=∂∂

=∂∂

=

=

ωωω

ω vv

Sound wave

corona Km/s 200 ephotospher Km/s 10

/ 166 5.0 35

)( 0

21

2221

21

20

=

===

====•==Ω==

s

sp

sgpsss

C

smTCmm

CvvkCCkvkkCvBg

γ

ωωω vvvv

Sound wave is longitudinal wave

Magnetic wave

Alfven speed tA nBBv

0

0162/1

0

0 108.2)(

×==µρ

In corona above sunspots n0=1016m-3, B0=10G, vA=300km/sIn photospheric network n0=1023m-3, B0=103G, vA=10km/s

Let’s ignore to derive Alfven waveGP & ,Ω

Look at fig. 4.1 Alfven wave could be longitudinal or transverse, or can be propogateObliquely. (4.16) becomes 2

0012 ))](([ AvBBvkkv

))vv××××=ω

expand kvBBkvkBBkvkvBkvv A

vvvvvvvvvvvvvvv)])(()[())(()(/ 10010011

20

21

2 ••−•+••−•=ω

Assume angle between kBvv

& 0 is Bθ

00

)](cos)[(cos)(cos/

1

10101222

12

=•→=•∇

•−•+•−=

BkB

kvBkvkBkvkvkvv BBBAvvvv

vvvvvvvvvv θθθωnormal to

1Bv

kv

0 also 10 =• vB vvPerturbed velocity normal to

0Bv

Eq. 4.21 becomes (4.23)

Two solutionsshear Alfven wave

Wave propagate fastest in direction, not at all in a direction normal to

Other propertiesenergy is carried at in direction

and in same direction lying in a plane parallel to wave frommagnetic field perturbation is normal to

Second solution to (4.23)compression Alfven waveenergy is propagate isotropicallylongitudinal wave,----pressuretransverse wave, ----tension

)( kv

•0)( 1

222 =•− vkvk Avv

ω

BAkvvk θω cos 01 ==• vv

BAp vv θcos=

0Bv

0Bv

0Bvv Ag

vv = 0Bv

Av

2/10

11 )(µρ

Bvv

v −=1Bv

1vv

010 =• BBvv

Akv=ωkvv Ag

vv =

0Bv

kv || , 90 1o=θ

kv ⊥= 1 , 0oθ

Internal Gravity waveconsider displacement of a blob of plasma, displace vertically from equilibrium. It remains in

Pressure equilibrium with surrounding. Density change inside the blob are adiabatic At equilibrium position

Outside the blob at z+dz

Adiabatic

New density inside the blob differs from ambient density at new height, the blob experiences aBouyancy force:

N: Brunt---Vaisala frequency

In the presence of horizontal magnetic fields

If T is constant

If the only force is due to buoyancy

This is a simple harmonic motion W=N, this is true only if N>0

gdzd

00 ρρ

−=

dzdzdzg 0

00 ρδρδρδρ −=−=

δρδρρ

2

0

02 SSr CPmrkTrPCconstP

====

])([)1(

)(

0

0

02

02

0

adS dz

dTdzdT

Tg

Cg

dzdgN

dzNg

−=+−=

−=−ρ

ρ

ρδρδρ

constant) (T )1( )1()( 02

22

20

SSad C

grNCgTr

dzdT −

=−−=

)1( 220

0

2

AS vCg

dzdgN

++−=

ρρ

)( 22

2

22

AS

S

S vCCr

CgN

+−=

zNdtzd

dtzdmF δρδρ 0

22

2

02

2

−==

If N<0

Go back to equation 4.15, assuming derive dispersion

This is called internal gravity wave ---- not surfaceN-1~50s only propagate horizontally

Intertial wavesConsider coriolis force alone

Effect: cause a small frequency splitting of Alfven wave

Magneto accoustic wavesLet’s consider P&B, ignore g&Ω. 4.21 becomes

Take scalar product with in turn

solution growinglly exponentiay instabilit convective)()( 0 →−>− addzdT

dzdT

SkC<<ω )1()1( 2222

kkgrC z

S−

−=ω

downwordenergy carry

)1(

2

21

2

2

kk

kv

kkN

z

zgz

z

ωω

ω

−=∂∂

=

−=

104 )(2 2 11 −

Ω•±=×Ω−=

∂∂

kkv

tv

vvvv

ω

)1(22

A

zA ω

ωωω ±=

kvBkvkvCBkvkvkvv BA

SBBA

v))vv)vvvv )](cos))(1[(cos)(cos/ 012

2

011222

12 •−•++•−= θθθω

0 & Bkvv

)()(cos

)(cos))((

1022

1023

122222

vBvkCk

vBvkvkvkCk

SB

AASv)vv

v)vv

•=•

•=•++−

ωθ

θω

Dispersion relation for magnetoacoustic waves

Solution

+: higher frequency fast magnetoacoustic wave-: lower frequency slow magnetoacoustic wave

If slow wave disappearsfast wave becomes sound wave

If slow wave disappearsfast wave becomes compression Alfven wave

If

fast: slow:

Acoustic gravity wavecompressibility and buoyancy force together, both acoustic and gravity modes occur

Dispersion relation

0cos)( 242222224 =++− BASAS kvCvCk θωω

21

2224422 ]cos221)(

21[ BASASAS vCvCvC

kθω

−+±+=

0=Av

0=SC

122

0

0

>>==BP

PP µβ

SCk≈

ωBAvk

θω cos≈

zCrgikk

kk

CgrNC

CrgN

CkNN

S

zg

S

s

SS

SgS

)vv22

22

2/1

2222222

2 1sin

frequencyVaisala -nt Bru)1( 22

)sin()(

+=′′

−=

−=

Λ==

′′−=−

θ

θωωω

Magnetoacoustic-gravity wave

If N=0 magnetoacoustic wave

If acoustic wave

magneto-gravity wave

Five minute oscillation (Helioseismology)Observations: periods: ~5 min

duration: 23~50 min Dopplergramsvelocity: 0.1~0.3 km/s orwavelength: 5000 to 10000 km Intensity mapsk-ω diagram: ridges

Theory: trapped acoustic standing wave surface reflecting point

interioreflectingpoint

0cossin)( 222422222224 =+++− BASgSAS vCkNCkvCk θθωω

0cos)( 222422224 =++− BASAS vCkvCk θωω22AS vC >> 222

SCk=ω

BAg vkN θθω 222222 cossin +=

Application:probe sound speed inside the Sun

get ρ, P, T as a function of depth.probe rotation speed inside the Sunprobe magnetic fields before emergence

to surface

Shock waves

Uniformly propagating wave: wave profile maintains a fixed shape each Part of the wave travels at the same speed.If crest of a sound wave moves fast than its leading edge – progressive steepering of wave front.If a steady wave shape is attained – shock wave is formed.

shock wave speed > CS strong shockshock wave speed < CS weak shock

Shocks can dissipate – convert flow energy to heatShock front is a very thin transition region -- a few mean free path wide.

2/1)(ρrpC =

11

221

1

2

)(

)(

vx

Evvvx

vxt

xv

tE

γδδ

ργδ

δδδδργ

δδ

=−

=

≈=

Reynold number 11 ≈γδ xv

Two reference frames: Rest frameshock frame

Effect of magnetic fields

3 wavesSlow slow magnetoacoustic shockIntermediate Alfven wave (no shock)Fast fast magnetoacoustic shock

Only fast shock can propagate directly across magnetic fileds

fast mode:

For conducting gas (ionized gas, there is an extra dissipative mechanism: ohmic heating due to finiteElectric conductivity. It does not affect gas before & after shock, but affects the thickness of shock.

22As vCv +=

σδδ 2jtE

≈

xBB

j yy

µδ21 −

=

Energy dissipation:

Ampere’s law:

EvBB

x yy

σσµδ

12

221 )( −

=

Hydrodynamic shock

Jump relationsConservation of mas

121111112

2222222

111222

1122

)21()

21( vvevpvvevp

vpvpvv

ρρρρ

ρρρρ

++=++

+=+=

Conservation of momentum

Conservation of energy

gasperfect for )1(

energy internal :ρ−

=rpe

21

1

122

2

2

21

)1(21

)1(v

rrpv

rrp

+−

=+− ρρThen:

Solution (prove as home work)

1)1(2

)( )1(

)1(2

)1(2

)1(

21

1

2

2/1

1

112

1

21

1

2

1

112

1

21

1

2

+−−

=

=+−+

=

=−+

+=

rrrM

pp

rpCMrMr

vv

CvM

MrMr

S

S

ρ

ρρ

Entropy S2>>S1 entropy increaserpCS v ρ

log=

(heating) TT pp vahead speed sound specshock 1 Mfind We

12121222

1

≥≥≥≤≥≥

ρρSC

Perpendicular shock (magnetic shock)

)1()1(

22 M solution

)22

1()2

()22

1()2

(

22

,

221

121

1

2

1

21

1

2

21

21

21

11

11

1

2

1

212

11111

21

12

222

22222

22

2

211

21

1222

22

2

11221122

xxrMppx

BBx

vv

rvC

Bp

Cvx

vBvevBpvBvevBp

vBpvBp

vBvBvv

A

S

s

−+−===

====

++++=++++

++=++

==

−− β

µβρρ

µρρ

µµρρ

µ

ρµ

ρµ

ρρ

0)1()2)1(2()2(2)( 211

2111

2 =+−+−++−= MrrrxMrxrxf βββSolution for x

41 :so 111 , 1

gas ideal 35~ 21

1

2

1

2211

211 <<

−+

<<+>>>>

<<

BB

rr

BBvCv

r

ASβ

Oblique shock

B field is in x-y planeSolutions

)1(2

)1( , )( , 1 , , 21

22

21

21

1

22

121

21

21

1

2

1

22

121

21

21

1

21

1

2

vv

Cxvrx

pp

xvvxvv

BB

BB

xvvvv

vv

xvv

SA

A

y

y

x

x

A

A

y

y

x

x −−

+=−−

==−−

== −

0)]1()1[()2(sin21)]1()1([cos

21)( 2

121

21

21

221

21

21

21 =−−+−−++

+−−+− rXrXvvzXrXvvrrXvXCXvv AASA θθ

There are 3 solutions: fast, Alfven waves, slowTwo special cases:

switch-off

switch-offshocks

2

21

11 A

A vvxvv ==

1/21

21

21

11 −−+

=>r

vCrxCv ASSA

Intermediate wave: wave front propagate at Alfven speed

yyxxyyyyy

y

y

y vvBBBBBBppBB

vv

12121221

2212

1

2

1

2 −==−====

It is actually not a shock, as no change in pressure or density

Homework

Use a sequence of CaK images obtained at Big Bear Solar Observatory on 4th of July 2004, derive 2-dimensional κ-ω diagram to show oscillation power spectra. The 3-D IDL save set can be found in a link in the course web page. Array fn1 contains time information. Image pixel size is 1”.

(1) What is the peak frequency?(2) Interpret the diagram.