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d
2
E
τ
dE
τ
(
t
)
dt
(
t
)
=
0
with
<
0
dt
2
we derive the following equation in
t
o
:
2
sin2
E
x
2
sin2
E
y
|
E
x
|
+|
E
y
|
E
tan(
E
x
tan 2
t
o
=
with
tan
t
o
−
)
>
0
,
2
cos2
E
x
2
cos2
E
y
|
E
x
|
+|
E
y
|
whence
E
y
=
|
E
y
|
cos(
t
o
−
)
x
)
=
|
E
y
|
E
tan(
E
x
)]cos
E
tan
[1
+
tan
t
o
−
.
(2
.
12)
E
E
|
E
x
|
|
E
x
|
cos(
t
o
−
In accord with (2.9), (2.12) this equation becomes:
2Re
P
E
E
cos
E
tan 2
=
2
=
tan 2
,
(2
.
13)
E
1
−|
P
E
|
E
where
E
is taken within quadrant I (0
≤
E
≤
/
2), if cos
≥
0 or within quad-
E
<
0.
rant IV (0
>
E
≥−
/
2), if cos
E
between semiaxes of the polarization ellipse. This parame-
ter is termed the field
ellipticity.
Substitution of
t
o
and
t
o
+
/
Next find the ratio
2
into
E
τ
(
t
) yields
the major and minor semi-axes of the ellipse:
1
1
2
2
+ |
P
E
|
+
2Im
P
E
+
+ |
P
E
|
−
2Im
P
E
a
E
=|
E
x
|
,
2
1
1
2
2
+ |
P
E
|
+
2Im
P
E
−
+ |
P
E
|
−
2Im
P
E
b
E
=|
E
x
|
,
2
>
where the quantity
b
E
is defined with its sign: it is positive for Im
P
E
0, that is,
E
>
0, and it is negative for Im
P
E
<
0, that is, for sin
E
<
0. Thus,
for sin
1
1
2
2
b
E
a
E
=
+ |
P
E
|
+
2Im
P
E
−
+ |
P
E
|
−
2Im
P
E
E
=
1
1
2Im
P
E
=
tan
E
,
(2
.
14)
2
2
+ |
P
E
|
+
2Im
P
E
+
+ |
P
E
|
−
where
1
2
arcsin (sin 2
E
sin
E
)
E
=
−
/
≤
≤
/
4
4
E
and
−
≤
≤
.
1
1
E
