=1.5 and m s = 2.8 = 2r 1 /d = 1.1 , = r 2 /r 1 =0.9
DESCRIPTION
CME 2005-01-15 06:30. C2 + FR 06:30. C2: Synthetic image. =1.5 and M s = 2.8 = 2R 1 /d = 1.1 , = R 2 /R 1 =0.9 R tip = 7.1 t 1 = 06:30 UT N16W01 b 0 = 0.9 = 0.59, R sh = 1.89 R 0 , R 0 = R tip /(2+1/ ). CME 2005-01-15 06:30. C3 + FR 07:42. - PowerPoint PPT PresentationTRANSCRIPT
=1.5 and Ms = 2.8
= 2R1/d = 1.1 , = R
2/R
1 =0.9
Rtip
= 7.1 t1= 06:30 UT N16W01
b0= 0.9 = 0.59, Rsh
= 1.89 R0 , R
0 = R
tip
/(2+1/ )
CME 2005-01-15 06:30
C2 + FR 06:30 C2: Synthetic image
=1.5 and Ms = 3.5
= 2R1/d = 1.1 = R
2/R
1 =0.9
Rtip
= 25.3 t1= 07:42 UT N20W00
b0= 0.7 = 1.76, Rsh
= 1.74 R0 , R
0 = R
tip
/(2+1/ )
CME 2005-01-15 06:30
C3 + FR 07:42 C3: Synthetic image
The magnetic field is given by:
Bp(r) = 3Bp(r/a)(1-(r/a)^2+r^4/(3a^4)
Geometry showing that a FR consists of the current core a and the Bp while Bex outlines the boundary of the FR [Chen et al. , 1996].
Eruptive Flux-Rope (EFR)(Chen 1996)
Geometry showing that a FR consists of the current core a and the Bp while Bex outlines the boundary of the FR [Chen et al. , 1996].
The magnetic field of EFR is given by:
Eruptive Flux-Rope (EFR)(Chen 1996, Krall et al., 2006)
Where Bp = Bp(a) is the poloidal field and Bt is the average of toroidal field Bt(r) across the minor radius, with the Relation Bt(0) = 3Bt. The toroidal current-carrying (a ) loop model yields Bp, Bt, and a, which are used to specify the solution.
1) Core: a central current-channel (r = a) and prominence.2) Cavity : a density depletion around a central current-channel (r = a) that produces the magnetic-cloud-like fields at 1 AU.3) Rim: Overlapped external loops and shock wave disturbance .
Relation to coronagraph CME images:
(a) (b) (c) (d)
From left to right: (a) C3 origin image on 2005/01/15 07:42 UT; (b), (c) and (d) C3 running difference image superimposed with the EFR model projected wireframe for core (red curves), cavity (green curves), and shock waves (yellow curves).
Initialization of a flux-rope CME based on solar observations:
1) We assume the flux-rope CME consists of a current channel with an apex
radius a and a cavity surrounding the core current with an apex minor diameter d.
d is assumed to 4a in Chen 1996. We propose here to determine a and d and
the flux rope orientations (latitude, longitude and tilt angel) using Krall and St.
Cyr's (2006) geometric EFR model fitting to the core and the cavity of the CME
image, respectively as shown in Figure (b) and (c).
Rtip
: radial distance from the flux rope origin to apex.
R1, R
2: semi-major and semi-minor of the flux rope elliptical axis.
d: minor diameter of the flux rope (width).
Flux rope model are defined by the follow two ratios:
Aspect ratio = 2R1/d and ratio of the elliptical semi-minor to
semi major = R2/R
1.
Elliptic flux rope fitting (Krall & St.Cyr,2006)
2) We specify the flux rope CME density using the density model from Krall
& Chen 2005, ApJ, 628, Figure 3, where the density model was developed
for a given magnetic field configuration based on the continuity equation
with the plasma density depleted around the flux rope axis.
Fig 2. Synthetic coronagraph image for the EFR model
Flux rope density profile is from Krall & Chen 2005, ApJ, 628, Figure 5.Shook density profile is given by the ad hoc function below :
sh/bg = 4(1 - 5||/!pi)) if <
22˚
sh/bg = .2 if >= 22˚
where is the solid angle of the
bow shock with respect to its
symmetric axis, sh/bg are density
compression ratio of the shock sheath.
3) We determine Bp and Bt based on the close
the relationship between magnetic fields in active regions and in their
associated MCs (Qiu et al, 2007).
First, the magnetic flux magnitudes within the cloud in the poloidal
component is given by:
phi_p = L/x01(B0*R0),
where R0 and B0 are the radius and the magnetic field strength along the axis
of the MC mx01 is the first zero of the Bessel function J0 (~2.4048), L is the
total length of the cloud, assumed to be 1AU(or 150Gm) (Qiu et al, 2007)
By assuming that the poloidal MC flux phi_p is equal to the reconnection flux
phi_r, we can determine (B0*R0) = phi-r(
Second, using the self-similarity of the flux-rope R0/1AU = a(z)/z, where a(z)
is the current-channel radius at the CME height z, we have R0 = a(z)/z*1AU.
Third, after determining R0, we can get B0. Then using the magnetic field
scaling law we have Bt (r = 0) ~ B0*(z/1AU)^2 and Bp ~ Bt (r = 0)/1.7,
assuming that at height z, the magnetic field of EFR is relaxed to Lundquist
solution (Chen 1996).
Amodification of Eruptive Flux-Rope (EFR) model (Krall et al, 2000, 2006):
1) Geometry: Elliptical flux rope 2) Density: observed electron density from coronagraph images3) Physics: underlying flux rope field
the determination of thedeflection was accomplished by trial and error.
the overall width of the model ICME is 4a = 0.76AU, where the width of the current channel within the flux rope is 2a = 0:38 AU. It is within the current channel thatthe field direction rotates smoothly as in a magnetic cloud.
Furthermore, as a result of our past studies, we have determined that the width of the apex of our model flux rope is 4a, so the leadingedge lies at height Z + 2a, where Z is the distance from the photospheric source to the magnetic axis of the flux rope at its apex and a isthe radius of the current channel within the flux rope.
The flux rope is described by apex height above the photosphere, Z, and the current-channel radius at the apex, a.
the model curves correspond to insitu measurements through the center of an undistorted fluxrope. To obtain a more exact correspondence between amodel flux rope and magnetic cloud data, a more detailednumerical model would be needed.
=1.5 and Ms = 3.0? (shock is too close to the edge of C2 FOV)
= 2R1/d = .75 = R
2/R
1 =0.9
Rtip
= 8.5 t1= 23:06 UT N17W08
b0= 0.5 = 0.57, Rsh
= 1.84 R0 , R
0 = R
tip
/(2+1/ )
CME 2005-01-15 23:06
C2 + FR 23:06 C2: Synthetic image
=1.5 and Ms = 4.0
= 2R1/d = .8 = R
2/R
1 =1.05
Rtip
= 21.3 t1= 23:42 UT N24W17
b0= 0.4 = 1.66, Rsh
= 1.685 R0 , R
0 = R
tip
/(2+1/ )
CME 2005-01-15 23:06
C3 + FR 23:42 C3: Synthetic image
=1.5 and Ms = 3.1
= 2R1/d = .65 = R
2/R
1 =0.9
Rtip
= 6.5 t1= 09:30 UT N15W15
b0= 0.7 = 0.41, Rsh
= 1.81 R0 , R
0 = R
tip
/(2+1/ )
CME 2005-01-17 09:30
C2 + FR 09:30 C2: Synthetic image
=1.5 and Ms = 3.1
= 2R1/d = .75 = R
2/R
1 =0.9
Rtip
= 19.3 t1= 10:20 UT N15W17
b0= 0.7 = 1.29, Rsh
= 1.81 R0 , R
0 = R
tip
/(2+1/ )
CME 2005-01-17 09:30
C3 + FR 10:20 C3: Synthetic image
=1.5 and Ms = 1.4
= 2R1/d = .7 = R
2/R
1 =0.75
Rtip
= 3.3 t1= 08:29 UT N46W51
b0= 0.7 = .58, Rsh
= 4.03 R0 , R
0 = R
tip
/(2+1/ )
CME 2005-01-19 08:29
C2 + FR 08:29 C2: Synthetic image
=1.5 and Ms = 1.4
= 2R1/d = .9 = R
2/R
1 =0.65
Rtip
= 21. t1= 10:10 UT N35W51
b0= 0.9 = 3.08, Rsh
= 4.03 R0 , R
0 = R
tip
/(2+1/ )
CME 2005-01-19 08:29
C3 + FR 10:10 C3: Synthetic image