Interaction Of Magnetic Clouds In The Inner Heliosphere
E. Romashets*, P. Cargill_ , and J. Schmidt_
*
Institute of Terrestrial Magnetism, Ionosphere, and Radio Wave Propagation of Russian Academy of Sciences
(IZMIRAN), Troitsk, Moscow Region, Email: [email protected]
_
Imperial College of Science, Technology, and Medicine, London, UK.
Abstract. A method of potentials has been used in the past for the calculation of the force acting on isolated magnetic
bodies in solar corona and inner heliosphere, where large gradients of magnetic pressure exist. Since recent observations
showed that coronal mass ejections (CME) can leave the Sun more frequently than was expected before 1995, it is clear
that interactions between CMEs can play important role in the formation of geo-effective structures near the Earth’s
orbit. We present here an evaluation of two interacting CMEs and the field distribution around them, using potential
solution in bi-cylindrical coordinates.
INTRODUCTION
Here a is a parameter of the coordinate system,
ε n = 1 for n = 1, ε n = 2 for n ≥ 2 . The following
relations hold between cartesian coordinates and new
coordinates µ ,η and Z:
In attempts to find the value of a force acting on an
isolated body in solar corona, Parker [1] proposed a
potential magnetic field, and found an expression for
the force: the so called “melon seed mechanism”. The
magnetic structure inside the cloud was assumed to be
force-free [2]. Later another form of this distribution in
cylindrical geometry [3] was used in the literature for
interpretation of magnetic clouds observations.
Romashets and Vandas [4] used the method for the
study of more complicated, toroidal-shape clouds
dynamics in solar wind. There were many studies of
non-potential flow around such kinds of cylindrical
and spherical bodies [5-7] using an MHD approach,
both analytically and numerically. But no explicit
expression for field and force were found.
x=
a sinh µ ,
cosh µ − cos η
(2)
y=
a sin η
,
cosh µ − cos η
(3)
z = Z,
(4)
and the radii of cylinders are equal to
r0
=
a , the distance between their centers
sinh µ0
is: D = 2 r 0 cosh µ0 = 2acth µ0 .
CALCULATION: POTENTIAL FIELD
µ=
Assume two cylinders aligned along the Z axis are
inserted into initially uniform X-directed field. The
cloud magnetic fields have no connection with the
ambient field. In bi-cylindrical coordinates the initial
field potential is
µ0
defines the cylinder’s surfaces in new
coordinates. On insertion of the two clouds,
the ambient field has been changed and
distorted in such a way that
Bµ
=0
µ =µ 0
nη . (1)of the Tenth International Solar Wind Conference,
cosProceedings
ε Wind
e Ten:
Y = B x =CP679,
B a∑
Solar
∞
0
0
0
n=0
The equation
− nµ
n
edited by M. Velli, R. Bruno, and F. Malara
© 2003 American Institute of Physics 0-7354-0148-9/03/$20.00
794
(5)
The new potential is then
near the Earth’s orbit more adequately. The
∞
Ψ = B a ∑ ε (e + e
0
1
− nµ
nµ − 2 n µ 0
n
n=0
) cos nη ,
initial external field is
(6)
chosen among harmonic functions satisfying (5)
and asymptotic to equ. (1) at infinity.
THE DISTURBED FIELD
+ B1
=
By
= − B0 − B1
B0
{
0
∞
nµ − 2 n µ 0
n =1
− nµ
+ sinh µ sin η ∑ n(e
nµ − 2 n µ 0
n =1
}
+ e ) sin nη
− nµ
Ψ = B a ∑ ε (e + e
(7)
− nµ
0
1
{
nµ − 2 n µ 0
n =1
n=0
∞
− B a ∑ 2( e + e
− nµ
mµ − 2 n µ 0
n=0
− e ) cos nη +
− nµ
∞
+(cosh µ cos nη − 1)∑ n(e
n =1
nµ − 2 n µ 0
}
+ e ) sin nη
− nµ
(8)
n=0
− nµ
) cos nη −
) sin nη +
+ B a ∑ n(e + e
1
∞
mµ − 2 n µ 0
n
0
∞
B = 2 B − sinh µ sin η ∑ n(e
0
(13)
component. In this case the disturbed potential is:
− e ) cos nη +
∞
∞
y
y
a
B1
measures the gradient of each
a
the 1 AU.
x
(12)
where B0 is the averaged value of x-component
near the location being considered, for example near
The components of the disturbed field are:
B = 2 B (1 − cosh µ cos η)∑ n(e
x
a
Bx
mµ − 2 n µ 0
) cos nη
(14)
In Figure 1 one can see contours of magnetic field
magnitude.
where
2
2
1 (x + a ) + y ,
µ = ln
2
2 (x − a ) + y 2
(9)
2
η=
i ( y − ia ) + x 2
.
ln
2 ( y + ia )2 + x 2
(10)
THE FORCE ON THE FLUX ROPES
AND EXAMPLE
The force acting on each cylinder is
B
2B
aL ⋅
S 8π
π
2 (11)
 ∞

⋅ ∫ (cosh µ 0 cos η − 1)  ∑ n e − n µ sin nη dη
n = 1

F =∫
a
2
η
2
e dS = −
x
0
z
π
0
−π
and is directed along X axis. This force
FIGURE 1. Sample contours of magnetic field magnitude
disturbed by insertion of two cylindrical magnetic clouds
into medium with initially non-uniform field.
moves the CMEs closer to each other. The
following example can describe the situation
795
CONCLUSIONS
The force acting on each of two cylinders has the
following components:
Modification of the IMF around two cylindrical
CMEs was found using a potential field formulation in
bi-cylindrical coordinates. The results can be used for
a calculation of the force acting on both CMEs during
their motion from the Sun. Plasma parameters
distribution for this geometry were not found but can
be if the entire system of MHD equations is solved for
V, n, and T and will be the subject of future work. The
maximum increase of B around both clouds is a factor
of 2-3, and one can expect that the velocity will
increase by a similar amount. Streamlines can be
slightly different from field lines but resemble them in
general.
2a L
(15)
∫ (cosh µ cos η − 1) ( B p( µ 0,η)) dη
π
π
z
F =−
x
2
0
−π
2a L
∫ sinh µ sin η ( B p( µ 0,η)) dη
π
z
F =−
y
π
2
(16)
0
−π
If the equations of motions are solved for both
cylinders using (15) and (16), it will be seen that two
bodies are rotating with respect to each other. There is
only a stable orientation for
ϕ = 450 .
For evaluation of (15) and (16), we can use an
approximation of interplanetary magnetic field (IMF):
Br
B e 2,
r
r 2 e Bϕ
=
Be ,
r re
=
ACKNOWLEDGEMENTS
This work was supported by EU/INTAS/ESA grant
99-00727.
(17)
where B e = 5 nT is the averaged value of IMF
components at 1 AU, r e = 1 AU , r is the distance
REFERENCES
from the center of the Sun. Now we can find gradient
of magnetic field.
2B
∂B
=−
r
∂r
r
∂B
B
=− r
∂r
r
r
e
3
(18)
2
e
(19)
e
ϕ
2
e
From (12), (13) and (18), (19) it can be seen that we
3B a
B0 = Be and B1 = − e for conditions
2 re
can take
at 1 AU. Using these values in (14) we have:
3na 
π
cosh(nµ − µ ) cos nη +  ≈


2r 
4
3na 
π
≈ 2 B e µ 2 +
cosh( µ − µ ) cos nη + 



2r
4
∞
B ( µ, η) = 2 B ∑ n e
p
−n
e
µ0
n =1
−
e
2+
0
0
(20)
and from (15), (16) we have the force:
2
e
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0
F =32 aL B e
z
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e
e
x
1.
−2
µ0 

1+
3a 
2r 
(21)
e
796