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Astronomical Applications Department, U.S. Naval Observatory - curvpath (Page 3)

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Astronomical Applications Department, U.S. Naval Observatory - curvpath
residuals, which will be denoted X
2
:
X
2
=
N
i=1
e
2
i
=
N
i=1
[p
0
+ v
0
t
i
+
1
2
zt
2
i
+ e
i
- p
0
- v t
i
]
2
=
N
i=1
[(p
0
- p
0
) + (v
0
- v )t
i
+
1
2
zt
2
i
+ e
i
]
2
=
N
i=1
[p + vt
i
+
1
2
zt
2
i
+ e
i
]
2
=
N
i=1
[p
2
+ p · vt
i
+
1
2
p · zt
2
i
+ p · e
i
+ p · vt
i
+ v
2
t
2
i
+
1
2
v · zt
3
i
+ v · e
i
t
i
+
1
2
p · zt
2
i
+
1
2
v · zt
3
i
+
1
4
zt
4
i
+
1
2
z · e
i
t
2
i
+ p · e
i
+ v · e
i
t
i
+
1
2
z · e
i
t
2
i
+ e
2
i
]
=
N
i=1
[p
2
+ 2p · vt
i
+ p · zt
2
i
+ v
2
t
2
i
+ v · zt
3
i
+
1
4
z
2
t
4
i
+ 2p · e
i
+ 2v · e
i
t
i
+ z · e
i
t
2
i
+ e
2
i
]
(5)
The terms in the last row, except for e
2
i
, disappear in summation because the e
i
are randomly oriented. We
are left with
X
2
= N p
2
+ 2p · v
N
i=1
t
i
+ p · z
N
i=1
t
2
i
+ v
2
N
i=1
t
2
i
+ v · z
N
i=1
t
3
i
+
1
4
z
2
N
i=1
t
4
i
+
N
i=1
e
2
i
(6)
Suppose N is large and the observations are evenly distributed in time, at intervals t, from t = 0 to
t = T . Then N = T /t and in the limit N we have
N
i=1
t
i
T
2
2t
,
N
i=1
t
2
i
T
3
3t
,
N
i=1
t
3
i
T
4
4t
,
and
N
i=1
t
4
i
T
5
5t
Substituting the above in eq. (6), we obtain
X
2
=
1
t
(p
2
T + p · vT
2
+
1
3
p · zT
3
+
1
3
v
2
T
3
+
1
4
v · zT
4
+
1
20
z
2
T
5
) +
N
i=1
e
2
i
(7)
If we define a coordinate system (x, y) in the plane of the sky, then in that system p, v, and z are
2-vectors:
p =
p
x
p
y
v =
v
x
v
y
z =
a
x
a
y
The components will normally be expressed in angular units (see later discussion).
In eq. (7), the last term is a constant, as are the acceleration z and the time parameters t and T .
We seek values for the components of p and v that minimize X
2
. The range of values that X
2
takes
on can be thought of as a surface within the 4-space defined by p
x
, p
y
, v
x
, and v
y
. If we define
= (/p
x
, /p
y
, /v
x
, /v
y
), then the condition for minimizing X
2
is
X
2
= 0, leading to
the normal equation
X
2
p
x
=
p
x
1
t
(p
2
T + p · vT
2
+
1
3
p · zT
3
+
1
3
v
2
T
3
+
1
4
v · zT
4
+
1
20
z
2
T
5
) +
N
i=1
e
2
i
= 0 (8)
3

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