IMF derivation from Pickup Ions observed by ASPERA 3

2005-4-27 1331-1344 UT

2005-7-12 1133-1146 UT

2005-6-3 0602-0615 UT

M. Yamauchi

B

ion

ion motion in velocity space is a ring! (spiral motion in real space)

In-coming directions of ions

Azimuth scan Elevation scan

Electric scan

Electric scanIM

A

0 3 6 9 12 15

0 3 6 9 12 1501

2

3

4

5

6

1514

Ion motion with V0=0Viewing from +B

Viewing from +E

Initial velocity to B is V0cos

velocity spacetrajectory

V*X

VY

VX

V*Y

V*ZVZ

V (SW frame) = V* (Martian frame) + VSW

B 3-D view

Ring 2D minimum variance direction (N) is parallel to B!

V0=0 in Martian frame means V0=

-VSW in SW rest frame where E=0. simple spiral motion, ring trajectory in velocity space.

2005-4-27 , ELS (ch8) + IMA

= 5~15

= 4

= 3

= 2

= 1

SW

ring

MEX orbit (light blue) during 1329-1357 UT in the cylindrical MSO coordinates. R2 = Y2+Z2.

average IMB

average BS

Mars

Mars

Distribution in velocity spaceM

SO

(X

YZ

)M

VM

(L

MN

)

Manual method vs. Automatic methodM

anu

alA

uto

mat

ic

However, IMA orientation is not always ideal

2005-6-30546-0640 UT

All

c

h8

Aligned in the azi-direction (points 2, 3, 4, 5)

Linear alignment (small data range in M direction)

IMA 2005-6-3

IMA light ion (H+ and He++), 3~8 keV =

5~

14

=

15

=

1

=

2

=3

=

4

Viewing from +B

Viewing from +E

general V0≠0 case

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

MAX

Minimum variance method gives clear L direction (// ±E), but not M & N directions.

We may not use ±N as estimate of B direction, but(1) E x VSW gives BY/BZ orientation ! (2) BX/B (or ) can be obtained intuitively

Check item(1) Alignment direction of the ring data by IMA

≈ Maximum variance direction (LX, LY, LZ) where LX << 1

BT=(0, BY, BZ) // VSW x E // (0, LZ, -LY)

(2) Sign of B : If L = evolution direction, then sign(BZ) = - sign(LY)

(3) Tilt toward X (=)

* ' = angle between max energy (MAX) direction and SW direction

* Ratio k = VMAX/VSW= (MAX/SW)1/2

If V0 = 0, two angles must be the same, i.e., ' = , and k = cos()

If cos(') = k , most likely = '

Result

Using (3)~(6) Changing IMF for 2005-6-3 event

0608 UT, northward IMF, X tilt ~ +35~40°, Y tilt ~ -10°

0613 UT, northward IMF, X tilt ~ +35°, Y tilt ~ -45°

0623 UT, northward IMF, X tilt ~ +20°, Y tilt ~ -30°

0633 UT, X tilt > +40°

Using (1)~(2) Constant IMF for 2005-4-27 event

1336-1357 UT, dawn-dusk oriented IMF, X tilt ~ 20°

Procedure(1) Manually select the ring distribution. (2) Apply the minimum variance method to determine L, M, and N. If the ring data is well arranged into a partial circle, ±N // B

If not, (3) Examine |LX| < 0.3. If yes, L // -ESW. If not, remove the direction that corresponds to the lowest energy from the selected set of data and re-calculate a new L. (4) Manually obtain MAX and '. (5) Check if MAX /4SW ≈ cos2(') is satisfied. If yes, we have ≈ ', where BX/|B| = sin().

(6) If possible, identify the evolution direction and determine the sign of L and ESW. BT is parallel to VSW x ESW.

Summary

Approximate IMF orientation can be derived from ring-like distributed protons as measured by the IMA.

The actual derivation of the ring plane is complicated (see procedure) due to the very limited viewing directions and angular resolution of the instrument.

New work (1) Beam ? 2005-6-1 event

New work (2) Use magnetosheath

2005-4-27 event (again)Can we obtain "sign" of B direction?

2005-7-10

2005-6-1

H+ only (no He++) !

= 3

= 2

= 4 = 3 = 2SW He++

Ring H+

SW H+

Motion of ring ions

Ion motion with V0=0

velocity spacetrajectory

V*X

VY

VX

V*Y

V*ZVZ

V (SW frame) = V* (Martian frame) + VSW

B 3-D view

2005-4-27 (easy case)

Beam-origin (V0 ≠ 0) ring distribution

Selected ring data : ~ 1344~1347 UTEl Az sensor direction (X, Y, Z) E (keV) VX (km/s) VY (km/s) VZ (km/s)02 03 (0.81, -0.29, 0.51) 1.37 -414 148 -26003 03 (0.85, -0.30, 0.42) 1.86 -508 181 -25104 03 (0.89, -0.32, 0.33) 2.27 -584 208 -21805 03 (0.91, -0.33, 0.24) 2.49 -630 225 -16408 03 (0.94, -0.34, -0.05) 2.74 -680 243 3909 03 (0.93, -0.33, -0.15) 2.74 -673 241 11010 03 (0.91, -0.33, -0.25) 2.74 -659 236 18012 03 (0.85, -0.30, -0.43) 2.49 -585 210 29813 03 (0.80, -0.29, -0.52) 2.27 -529 190 34114 03 (0.75, -0.27, -0.60) 2.00 -464 167 37015 03 (0.69, -0.25, -0.68) 1.68 -392 141 382

*1) L = (0.05, -0.02, 1.00) in MSO *2) M = (0.94, -0.34, -0.06) in MSO *3) N = (0.34, 0.94, -0.003) in MSO

Aligned in azimuthal direction

Almost 1-D alignment

Difficult to determine ring "plane"

From ring data at ~0623 UT (2005-6-3)

-750

-500

-250

0

250

-500 -250 0 250 500

-750

-500

-250

0

250

-250 0 250 500 750

VL (km/s) VN (km/s)

VM

(km

/s)

Selected ring data : ~ 0608 UTEl Az energy

(keV)VX

(km/s)VY

(km/s)VZ

(km/s)count VL *1

(km/s)VM *2(km/s)

VN *3(km/s)

13 01 4771 -711 -410 485 153 -341 -401 37213 01 4352 -679 -391 464 296 -324 -362 37114 01 5712 -728 -422 616 15 -336 -495 46614 01 5225 -696 -404 590 14 -320 -453 46115 01 5712 -671 -392 696 7 -291 -497 55813 02 4771 -810 -106 492 13 -46 -470 27214 02 5225 -793 -106 596 36 -31 -521 36314 02 4771 -758 -102 569 176 -28 -476 36314 03 5225 -769 205 602 14 280 -492 33614 03 4771 -734 196 575 11 269 -449 33815 03 4771 -677 179 647 27 265 -448 43115 03 4352 -647 171 618 8 255 -406 42814 04 4352 -572 442 554 71 517 -297 383

*1) L = (0.06, 0.99, 0.14) in MSO *2) M = (0.79, 0.04, -0.61) in MSO *3) N = (0.61, -0.15, 0.78) in MSO

Beam-origin

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

Alignment direction of the ring data by IMA

≈ Maximum variance direction (LX, LY, LZ)

where LX << 1

BT=(0, BY, BZ) // VSW x E // (0, LZ, -LY)

Sign of B

If L = evolution direction, sign(BZ) = - sign(LY)

Tilt toward X (=)

* ' = angle between max energy (MAX)

direction and solar wind direction

* Ratio k = VMAX/VSW= (MAX/SW)1/2

If V0 = 0, two angles must be the same,

i.e., ' = , and k = cos()

If cos(') = k , most likely = '

MAX