Solar wind direction in IMA frame

In this document, you may find the way to calculate the solar wind direction in the IMA frame.

The description is on-the-fly document during the feature branch of feature/swdir_in_imaframe.

The solar wind direction

The solar wind is flowing almost parallel to the Sun-Mars line, while 2-3 degrees of deviation is found both in \(\phi\) and \(\theta\) directions (Solar wind parameters). Here \(\phi\) is measured from -x direction with positive toward +y and \(\theta\) is the angle from x-y plane. The MSO frame is used, while no abberation is yet considered.

The above statement can be expressed by the following math form.

\[\begin{split}Vx &= -V\cos\theta\cos\phi \\ Vy &= V\cos\theta\sin\phi \\ Vz &= V\sin\theta\end{split}\]

In addition, the abberation should be considered. This is originated from the motion of the spacecraft relative to the inertia frame (more precisely speaking, the Sun fixed frame).

It is not very trivial which velocity in what frame with what correction is most simple to get this information.

Which velocity is most simple (I think). The spacecraft velocity with respective to the Sun. It is very close to the Mars velocity relative to the Sun.

In what frame? It should be in the “Spacecraft” frame (or IMA frame, where translation in between is rigidly fixed). Who is the object? Spacecraft.

Mmm. Very confused. Spacecraft velocity in the spacecraft frame should be zero, intuitively… (And, indeed spice also returns so)

Velocity in SPICE

Try spice.

>>> ms.get_posvel(datetime.datetime(2010, 1, 1),
... target='MEX', origin='SUN', frame='MEX_SPACECRAFT',
... correction='None')
1.18832037e+00,  -1.82601002e-03,   2.05399347e+00

Well, it is not very clear what velocity it is. I expect velocity something like 25 km/s.

Indeed, this is something like “apparent” velocity in a non-inertia frame. In the calculation, spice derives the veloicyt as

\[\lim_{dt\to 0}\frac{\vec{x}(t+dt) - \vec{x}(t)}{dt}\]

and \(\vec{x}\) is derived in the frame at each time. Therefore, a very “distant” non-inertia frame will derive very large velocity quantity.

>>> print ms.get_posvel(datetime.datetime(2010, 1, 1),
... target='MEX', origin='SUN', frame='IAU_SUN',
... correction='None')[3:]
511.71998832    -438.67762313   -3.43707870

This is because the IAU_SUN frame rotates as the Sun rotates, and the velocity of the “frame” at Mars orbit is something like \(2\pi \cdot 2.5\times 10^8 \mathrm{km}/ (30 \cdot 86400 \mathrm{s})~600 \mathrm{km/s}\).

Indeed, the correction is not effective. No very big difference is found. However, considering the principle, no correction on the light time is needed, and no correction on the abberation of frame is needed. So that in SPICE, you may use “None” for correction.

An exercise

>>> import datetime
>>> from irfpy.mexpvat import mexspice as ms
>>> ms.init()
>>> target='MEX'
>>> origin='MARS'
>>> frame='J2000'
>>> for correction in ('None', 'LT', 'LT+S'):
...     print ms.get_posvel(datetime.datetime(2010, 1, 1),
...          target=target, origin=origin, frame=frame,
...          correction=correction)[3:]

By changing target, origin and frame, one can search for the velocity vector.

The results is as below.

Target	Origin	Frame	Corr	Vx	Vy	Vz	Vtot	theta	phi
MEX	SUN	MEX_SPACECRAFT	None	1.18832037	-0.00182601	2.05399347	2.37297236	30.051	90.0880
			LT	1.10997952	-0.00147104	1.91886902	2.21678033	30.047	90.0759
			LT+S	1.10997990	-0.00147093	1.91886880	2.21678033	30.047	90.0759
MEX	Mars	MEX_SPACECRAFT	None	1.26637929	-1.16411868	0.35015275	1.75541897	78.494	132.590
			LT	1.26631973	-1.16414917	0.35017201	1.75540006	78.493	132.592
			LT+S	1.26634670	-1.16408683	0.35014719	1.75537323	78.493	132.590
MEX	MEX	MEX_SPACECRAFT	None	0.00000000	0.00000000	0.00000000	0.00000000	0	0
			LT	0.00000000	0.00000000	0.00000000	0.00000000	0	0
			LT+S	0.00000000	0.00000000	0.00000000	0.00000000	0	0
MEX	Earth	MEX_SPACECRAFT	None	-13.77548614	-5.87634770	5.37282297	15.91108756	70.264	-113.10
MEX	SUN	J2000	None	-21.59374840	-7.05161227	-4.20846585	23.10251913	100.49	-108.08
			LT	-21.62033610	-7.10624758	-4.52783773	23.20428845	101.25	-108.19
			LT+S	-21.62033620	-7.10624830	-4.52783809	23.20428883	101.25	-108.19
MEX	Mars	J2000	None	-0.85566535	1.15877170	-1.00270765	1.75508908	124.84	-36.443
MEX	MEX	J2000	None	0.00000000	0.00000000	0.00000000	0.00000000	0	0
MEX	SUN	IAU_SUN	None	511.71998832	-438.67762313	-3.43707870	674.02315761	90.292	130.605
MEX	SUN	J2000	None	-21.59374840	-7.05161227	-4.20846585	23.10251913	100.49	-108.08
MEX	SUN	ECLIPJ2000	None	-21.59374836	-8.14375934	-1.05622166	23.10251910	92.620	-110.66
MEX	SUN	GALACTIC	None	9.38031143	-10.67655867	18.21392987	23.10251910	37.964	138.697
MEX	SUN	MSO	None	-2.37291154	-0.44958875	-1.41509286	2.79916542	120.36	-100.72

Now, what is the solar wind direction?

How to calculate the solar wind direction? Ok, assume that the solar wind is “fixed” in any inertia frame of Sun center. This is derived from the fact that the solar wind propagate almost straightly from the Sun without feeling any forces (force-free).

On the other hand, Mars frame (MSO) or spacecraft frame (SC) is traveling across the frame. This is the ‘abberation’ having discussed in this section.

Ok, I assume you know the solar wind velocity in the J2000 frame (\(\vec{V_{sw}}\)). It is usually assumed “radial” velocity vector with a few degrees deviation (indicated by WIND data. See Solar wind parameters.

Then, you may know the spacecraft velocity (or Mars velocity) in J2000 (\(\vec{V_{sc}}\)). You can subtract it from the solar wind velocity in J2000, you may know the s/c abberrated solar wind velocity, \(\vec{v_{sw}}=\vec{V_{sw}}-\vec{V_{sc}}\). Note that this is yet J2000 reference frame.

You also know, independently, a conversion matrix from J2000 to Spacecraft coordinate system. Multiplying the matrix, \(\textbf{M}\), to the solar wind vector, you may get the abberated solar wind velocity vector in the SC frame.

\[\vec{w_{sw}} = \textbf{M}\cdot\vec{v_{sw}}\]

To me, this is the most simple way to calculate the solar wind velocity in the SC frame.

The key is that the spacecraft velocity should be calculated from the “solar wind” as origin.

An exercise

>>> import numpy as np
>>> import datetime
>>> from irfpy.mexpvat import mexspice as ms
>>> ms.init()
>>> t0 = datetime.datetime(2007, 12, 10)
>>> et0 = ms.spice.str2et(t0.strftime('%FT%T'))
>>> rvm = ms.get_posvel(t0, target='MaRS', origin='SUN', correction='None', frame='J2000')
>>> rm = rvm[:3]
>>> vm = rvm[3:]
>>> print rm, vm
[  1.95308260e+07   2.11300227e+08   9.63893224e+07] [-23.2270546    3.48866494   2.22763659]
>>> rvsc = ms.get_posvel(t0, target='MEX', origin='SUN', correction='None', frame='J2000')
>>> rsc = rvsc[:3]
>>> vsc = rvsc[3:]
>>> print rsc, vsc
[  1.95239861e+07   2.11297999e+08   9.63996136e+07] [-21.93529536   3.04962571   2.33319464]
>>> Vsw = 400 * rm / np.sqrt((rm ** 2).sum())
>>> print 'Vsw (J2000)', Vsw
Vsw (J2000) [  33.51970279  362.64317846  165.42779284]
>>> vsw = Vsw - vsc
>>> print 'vsw (J2000/abr)', vsw
vsw (J2000/abr) [  55.45499815  359.59355275  163.0945982 ]
>>> M=np.array(ms.spice.pxform('J2000', 'MSO', et0))
>>> print 'M(J2000->MSO)', M
M(J2000->MSO) [[-0.08379926 -0.90660795 -0.41356948]
 [ 0.99618024 -0.06599384 -0.05718175]
 [ 0.02454839 -0.41678153  0.90867515]]
>>> print 'wsw (MSO)', M.dot(vsw)
wsw (MSO) [ -3.98108408e+02   2.21861803e+01  -3.10612044e-01]
>>> M=np.array(ms.spice.pxform('J2000', 'MEX_SPACECRAFT', et0))
>>> print 'M(J2000->S/C)', M
M(J2000->S/C) [[-0.0534572   0.8955422   0.44175389]
 [ 0.00678712 -0.44205037  0.89696455]
 [ 0.99854708  0.05094745  0.01755263]]
>>> print 'wsw (S/C)', M.dot(vsw)
wsw (S/C) [ 391.11440447  -12.29201183   76.55754057]