Magnetic structure over plage region May 5 2020 Tetsu Anan

Photosphere

Chromosphere

HAZEL result Really?

Contents

• Observed plage region• IRIS inversion• GREGOR/GRIS data– Stray light– IQUV crosstalks– Expected choromospheric B field from Hanle diagram

• HAZEL inversion– Photosphere– Chromosphere

DefinitionΘ! : Inclination of B in LOS frame)! : Inclination of B in solar frameΦ! : Azimuth of B in LOS frame+! : Azimuth of B in solar frame

Where is the observed active region• Heliocentric coordinates (693”, −237”)• Solar radius 958”

Where is the observed active region

Plage has single polarityReply to a comment

IRIS inversion ’IRIS2’Reply to a comment

Stray light of GRIS �4 %Reply to a comment

Stray light derived by comparing intensitiesat the line center is 13%Wing is not consistent with the atlas

Stray light derived by comparing equivalent widths in a λ range between dashed lines is 4%

Atlas: Sac Peak Solar Flux Atlas by Kurucz et al.(1984)

GRIS Disk centerATLASCalibrated

Definition of Stokes parametersComparison with HMI SHARP

• GRIS (email from Manolo)– +Q is the solar N-S direction– +U rotate counter-clockwise from +Q on the solar surface– +V is right circular polarization on the solar surface

Not plage data

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HMI: a Lambert Cylindrical Equal-Area projection, Not heliocentric coordinates

Cross talk IóQ,U,V �0.01-0.02%Plage data

Polarization degree in continuum

Solid: mean polarization degreeDotted: standard deviation of

polarization degree

The crosstalk is too small to significantly change Stokes profiles

Cross talk VóQ, VóUPlage data

Kuhn et al. 1994Assume small Q <=> U cross talk,

because of the comparison with Hinode!!"# = !$%! − $%$%! − &'$%!%!"# = %$%! − (!!"#'!"# = '$%! − )!!"#

a, b, c, d are not homogeneous spatially => Small cross talksa b

c d

Time (# of data)

Crosstalk• Collados 2003 SPIE– Apply to the umbra of a spot, where the magnetic

field ensures (together with the large magnetic sensitivity of near infrared spectral limes) that the sigma and pi components are completely split apart

– Results of umbra data taken by GREGOR/GRIS(Si�10827 Å can not be split completely to σ and π on the umbrae)

• !!→# = #! $ #"!! % !"

= 0.009

• !!→& = &! $ &"!! % !"

= −0.004• (' − (( = !#→! )' + )( + !&→! +' + +(

– !!→#=0.01– !$→# = −0.09

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Linear polarization

• Method 20190828_{2,3,4}.pro1. Fit Stokes-I of He 10830 with a gaussian to get

line center wavelength and line width2. Fixing the line center wavelength and the line

width, Stokes-Q/Ic & -U/Ic profiles are fitted with a gaussian=> Stokes-Q&U amplitude

• Noise: standard deviation of Q/Ic & U/Ic at continuum

New slide

Thin: noiseThick: Pl= !! + #!

Noise improve withspatial binning

Signal in-depends on spatial binning

Data with bin 2will be inverted

Oct. 3 -005plage between sunspots, AR 12723

Polarization angleis homogeneous

New slide

Linear polarization in He�1083 nm

1. Fit Stokes-I of He 10830 with a gaussian to get line center wavelength and line width

2. Fixing the line center wavelength and the line width, Stokes-Q/Ic & -U/Ic profiles are fitted with a gaussian=> Stokes-Q&U amplitude Linear polarization aligns

along radial direction

Plage data

Hanle diagram

• ! = 0°• Heliocentric coordinates (693”, −237”)

Solar radius 958”=> % = 49.8°

• +Q = NS direction=> * = 108°

!!0°, 180°10°, 170°20°, 160°30°, 150°40°, 140°50°, 130°60°, 120°70°, 110°80°, 100°90°

ObservationNo 90° ambiguity!! ≈ 90°

& = 0.7∆+ = 11 km/s1 = 200 G

HAZEL

HAZEL inversion

HAZELLine of sight angles

• Line of sight angles– ! = 49.8°– ( = 0°– * = 90° (+Q is parallel to limb)

• Rotation of coordinate

XY

θQ

θQ

!! = tan"# &' = −18.9°QQ’

.′0′ = cos 2!! sin 2!!

−sin 2!! cos 2!!.0

Heliocentric coordinates (X,Y)=(693”, −237”)Solar radius 958”

HAZEL parameters

PhotosphereFunctions of optical depth

• BX in line of sight reference frame(LOS frame)

• BY in LOS frame

• BZ in LOS frame

• Temperature

• Electron pressure

• Microturbulent velocity

• Macroscopic velocity

• Filling factor = 1 (fixed)

Chromosphere• B

• Θ! inclination angle in LOS frame

• Φ! azimuth angle in LOS frame

• Optical depth

• Doppler velocity

• Doppler width

• Damping parameter

• Height of slab = 3.0” (fixed)

• Enhancement factor = 1 (fixed)

• Filling factor = 1 (fixed)

HAZELThe axis of coordinates for B field

This figure is for chromospheric B fieldX axis for photospheric B is parallel to e1

Test #1 #2γ 90� 0�BX (Si) −69 G −318 GBY (SI) 318 G −69 GΘ! (He) 51� 51�Φ! (He) 349� −10�

HAZEL, B in chromosphereambiguities

Sequence of HAZEL1. HAZEL inversion => result !!, Θ"! , Φ"! , ⋯2. HAZEL inversion with fixing Φ" = Φ"! + 180°3. HAZEL inversion with fixing Φ" = Φ"! + 90°4. HAZEL inversion with fixing Φ" = Φ"! − 90°

5. Calculate mean square of residual in Q and U within a wavelength range of line center � line width

MS = 10#1$

2$ − 3it$noise$

%

6. Best solution has minimum of MS& +MS'7. Select solution that has

MS& −MS&"()* < 1 & MS' −MS'"()* < 1

Fitting

Stokes I, Si Stokes Q, Si

Stokes U, Si Stokes V, Si

Stokes I, HeStokes Q, He

Stokes U, He Stokes V, He

line of sight component of B

Photo BLOS Chromo BLOS

B in photosphere

Θ! Φ!

!"#° ambiguityX for photospheric BX is parallel to limb

Θ! ≈ 45°90�

Correct the result

B in photosphere, 180° ambiguity 2 solutions

!!

!! "!

"!

LOS

! = 49.8°

Θ! ≈ 45°

Sun center

Correct the result

Distributions of the 2 solutionsGRIS + HAZEL2

New slide

Comparison with HMI SHARP180° ambiguity is resolved using a minimum energy method (Metcalf 1994; Leka et al. 2009)

The vertical solution of GRIS inversion is consistent with HMI

New slide

HMI: a Lambert Cylindrical Equal-Area projection

Comparison with HMI SHARP180° ambiguity is resolved using a minimum energy method (Metcalf 1994; Leka et al. 2009)

Spatial sampling: 0.2”, 0.1” for GRIS (diffraction limit 0.2”), 0.5” for HMI

New slide

Small horizontal B field over plage• Hinode/SOT SP reveals there is horizontal B fields over plage region

– Lifetime 1 – 10 min– Randomly located spatially and temporally– Not associated with existing long lived B field and appear in a region of

insignificant vertical fields– First appears inside the granule, subsequently moves to the inter-

granule lane, and finally disappears in the inter-granular lane

Ishikawa & Tsuneta 2009

New slide

Short summary of the 180�ambiguity

• Two solutions of the 180° ambiguity is classified under vertical and horizontal B fields– Vertical solution is consistent with HMI SHARP CEA– For magnetic elements, which is larger than 1”, the

vertical solution is good– There is a possibility that small horizontal B field exists

• Because it is randomly located spatially and temporally (and it is anisotropic), it is impossible to solve 180�ambiguity for the small B field in a region near the solar limb

• I will “discuss” results in terms of spatial scale

New slide

B in photosphere

0° radial direction

#! $!B

B in chromosphereInclination & azimuth angle

Dotted: The solutions of the 4 inversionSolid : The selected solution No !"° ambiguity (see Hanle diagram)

No 90° ambiguity

van Vleck angle

LOS

& = 49.8°

&! ≈ 90°

Sun center

Θ! Φ!

&! .!

No /0"° ambiguity• 3 cos &!" − 1 sinΘ!" =

3 cos &′!" − 1 sinΘ′!"• &! ≈ 90°, Θ! ≈ 45°Þ −0.5 ≈ 3 cos &′!" − 1 sinΘ′!" (1)• Φ! should be ≈ 180°

=> Θ′! ≈ &#! + & (2)• Substitute (2) into (1) => &#! ≈ 80°Stokes V rejects this solution

4 candidates

B in chromosphere

0° radial direction

#! $!B

New slideContinuum images taken by HMI, IRIS spectra, and GRIS are spatially aligned carefully

Discussion:Is it true that chromospheric B is horizontal?• Orientation of liner polarization is consistent with Hinode for Sunspot data

– Axis of coordinates can be OK– Cross talk between Stokes Q & U can be OK

• Cross talks among IQUVs are small– 2 methods: Kuhn et al. 1994 & Collados 2003 SPIE

• Horizontal B field is expected from Hanle diagram– Hanle diagram seems to be appropriate

• If it was not true, assumptions of the HAZEL would be wrong– Radiation field?

• The assumption can be OK, because it is granule pattern, which is spatially homogeneous in continuum radiation

– 1 slab atmospheric model?

Discussion: radiation field

• If B is normal to the solar surface in the chromosphere, the Hanle effect is small and linear polarization is produced by anisotropy of radiation field– Radiation irradiating from side should be stronger

than that from vertical– The anisotropy should be homogeneous in the

field of view and oriented to the radial direction on the solar surface

New slide

Comment

• Solarsoft: rot_xy enables us to track plage– JSOC��

• GRIS�HMI�"������–������contrast

• Ishikawa et al.#�����$���)%���(���!&(� �(�� �

• Radiative flux�He intensity����'��