as in the case of Fig. 7.6. If δ takes a negative value, on the other hand, the field traces
the ellipse clockwise.
If δ ¼ Æ π/2, in (7.94) we have
E x ¼ E 1 cos Àωt
ð
Þand E y ¼ E 1 cos Àωt Æ
π
2
:
ð7:95Þ
We examine the case of δ ¼ π/2 first. In this case, when t ¼ 0, E x ¼ E 1 and E y ¼ 0
(Point C 1 in Fig. 7.8b). If ωt ¼ π/4, E x ¼ E y ¼ 1=
ffiffi ffi
2
p
(Point C 2 in Fig. 7.8b). In turn,
if ωt ¼ π/2, E x ¼ 0 and E y ¼ E 1 (Point C 3 ). Again the electric field traces the circle
counterclockwise. In this situation, we see the light from above the z-axis. In other
words, we are viewing the light against the direction of its propagation. The wave is
said to be left-circularly polarized and have positive helicity. In contrast, when
δ ¼ À π/2, starting from Point C 1 the electric field traces the circle clockwise.
That light is said to be right-circularly polarized and have negative helicity.
With the left-circularly polarized light, (7.69) can be rewritten as
E = E 1 þ E 2 = E 1 e 1 þ ie 2
ð
Þ e
i kzÀωt
ð
Þ
:
ð7:96Þ
Therefore, a complex vector (e 1 + ie 2 ) characterizes the left-circular polarization. On
the other hand, (e 1 À ie 2 ) characterizes the right-circular polarization. To normalize
them, it is convenient to use the following vectors as in the case of Sect. 4.3 [3].
e þ
1
ffiffi ffi
2
p e 1 þ ie 2
ð
Þand e À
1
ffiffi ffi
2
p e 1 À ie 2
ð
Þ ,
ð4:45Þ
In the case of δ ¼ 0, we have a linearly polarized light. For this, the points A 1 , A 2 ,
and A 3 coalesce to be a point on a straight line of E y ¼ E x .
References
1. Arfken GB, Weber HJ, Harris FE (2013) Mathematical methods for physicists, 7th edn. Academic Press, Waltham
2. Pain HJ (2005) The physics of vibrations and waves, 6th edn. Wiley, Chichester
3. Jackson JD (1999) Classical electrodynamics, 3rd edn. Wiley, New York
References
293
the ellipse clockwise.
If δ ¼ Æ π/2, in (7.94) we have
E x ¼ E 1 cos Àωt
ð
Þand E y ¼ E 1 cos Àωt Æ
π
2
:
ð7:95Þ
We examine the case of δ ¼ π/2 first. In this case, when t ¼ 0, E x ¼ E 1 and E y ¼ 0
(Point C 1 in Fig. 7.8b). If ωt ¼ π/4, E x ¼ E y ¼ 1=
ffiffi ffi
2
p
(Point C 2 in Fig. 7.8b). In turn,
if ωt ¼ π/2, E x ¼ 0 and E y ¼ E 1 (Point C 3 ). Again the electric field traces the circle
counterclockwise. In this situation, we see the light from above the z-axis. In other
words, we are viewing the light against the direction of its propagation. The wave is
said to be left-circularly polarized and have positive helicity. In contrast, when
δ ¼ À π/2, starting from Point C 1 the electric field traces the circle clockwise.
That light is said to be right-circularly polarized and have negative helicity.
With the left-circularly polarized light, (7.69) can be rewritten as
E = E 1 þ E 2 = E 1 e 1 þ ie 2
ð
Þ e
i kzÀωt
ð
Þ
:
ð7:96Þ
Therefore, a complex vector (e 1 + ie 2 ) characterizes the left-circular polarization. On
the other hand, (e 1 À ie 2 ) characterizes the right-circular polarization. To normalize
them, it is convenient to use the following vectors as in the case of Sect. 4.3 [3].
e þ
1
ffiffi ffi
2
p e 1 þ ie 2
ð
Þand e À
1
ffiffi ffi
2
p e 1 À ie 2
ð
Þ ,
ð4:45Þ
In the case of δ ¼ 0, we have a linearly polarized light. For this, the points A 1 , A 2 ,
and A 3 coalesce to be a point on a straight line of E y ¼ E x .
References
1. Arfken GB, Weber HJ, Harris FE (2013) Mathematical methods for physicists, 7th edn. Academic Press, Waltham
2. Pain HJ (2005) The physics of vibrations and waves, 6th edn. Wiley, Chichester
3. Jackson JD (1999) Classical electrodynamics, 3rd edn. Wiley, New York
References
293