The emphasis is on understanding where the equations come from rather than simply quoting the final result.
Superposition of Two Counter-Propagating Circularly Polarized Waves
Consider two equal-amplitude monochromatic plane waves propagating in opposite directions along the
$z$-axis. Let their common electric-field amplitude be $E_0$, wave number be $k$, and angular frequency be $\omega$.
For electromagnetic waves in vacuum,
$$
\omega = ck.
$$
The electric field of a plane wave can be written as
$$
\mathbf{E}(\mathbf r,t)
=
\operatorname{Re}
\left\{
\mathbf E_0
e^{i(\mathbf k\cdot\mathbf r-\omega t)}
\right\}.
$$
(a) $+z$ wave is LCP and $-z$ wave is RCP
Wave propagating along $+z$
For the wave propagating in the $+z$ direction,
$$
\mathbf{k}_1=k\hat{\mathbf z}.
$$
Take the left-circularly polarized electric field to be
$$
\mathbf E_1
=
E_0
\operatorname{Re}
\left[
(\hat{\mathbf x}-i\hat{\mathbf y})
e^{i(kz-\omega t)}
\right].
$$
Taking the real part gives
$$
\boxed{
\mathbf E_1
=
E_0
\left[
\hat{\mathbf x}\cos(kz-\omega t)
+
\hat{\mathbf y}\sin(kz-\omega t)
\right]
}.
$$
Wave propagating along $-z$
For the second wave,
$$
\mathbf{k}_2=-k\hat{\mathbf z}.
$$
For the right-circularly polarized wave propagating in the $-z$ direction,
$$
\mathbf E_2
=
E_0
\operatorname{Re}
\left[
(\hat{\mathbf x}-i\hat{\mathbf y})
e^{i(kz+\omega t)}
\right].
$$
Therefore,
$$
\boxed{
\mathbf E_2
=
E_0
\left[
\hat{\mathbf x}\cos(kz+\omega t)
+
\hat{\mathbf y}\sin(kz+\omega t)
\right]
}.
$$
Superposition
The total electric field is
$$
\mathbf E=\mathbf E_1+\mathbf E_2.
$$
Using the trigonometric identities
$$
\cos(A-B)+\cos(A+B)=2\cos A\cos B,
$$
and
$$
\sin(A-B)+\sin(A+B)=2\sin A\cos B,
$$
we obtain
$$
\boxed{
\mathbf E
=
2E_0\cos\omega t
\left(
\hat{\mathbf x}\cos kz
+
\hat{\mathbf y}\sin kz
\right)
}.
$$
Interpretation
At a fixed value of $z$, the vector
$$
\hat{\mathbf x}\cos kz+\hat{\mathbf y}\sin kz
$$
has a fixed direction. Therefore, at any particular position, the electric field oscillates back and forth along a fixed line.
Thus the resulting field is linearly polarized at every fixed position, although the direction of that linear polarization changes as we move along the $z$-axis.
For example:
At $z=0$:
$$
\mathbf E=2E_0\hat{\mathbf x}\cos\omega t.
$$
At $z=\lambda/4$:
$$
\mathbf E=2E_0\hat{\mathbf y}\cos\omega t.
$$
Thus the polarization direction rotates spatially as we move along the propagation axis.
Magnetic field
For a plane wave,
$$
c\mathbf B=\hat{\mathbf k}\times\mathbf E.
$$
For the two counter-propagating waves,
$$
c\mathbf B
=
\hat{\mathbf z}\times\mathbf E_1
-
\hat{\mathbf z}\times\mathbf E_2.
$$
After simplification,
$$
\boxed{
c\mathbf B
=
-2E_0\sin\omega t
\left(
\hat{\mathbf x}\cos kz
+
\hat{\mathbf y}\sin kz
\right)
}.
$$
Hence
$$
\boxed{
\mathbf B
=
-\frac{2E_0}{c}\sin\omega t
\left(
\hat{\mathbf x}\cos kz
+
\hat{\mathbf y}\sin kz
\right)
}.
$$
Notice that $\mathbf E$ and $\mathbf B$ are parallel rather than perpendicular. This does not contradict the properties of a single plane electromagnetic wave because the total field here is a superposition of two oppositely propagating waves.
(b) Both waves are LCP
Now suppose both counter-propagating waves have left-circular polarization.
For the wave propagating along $+z$,
$$
\mathbf E_1
=
E_0
\left[
\hat{\mathbf x}\cos(kz-\omega t)
+
\hat{\mathbf y}\sin(kz-\omega t)
\right].
$$
For the LCP wave propagating along $-z$,
$$
\mathbf E_2
=
E_0
\left[
\hat{\mathbf x}\cos(kz+\omega t)
-
\hat{\mathbf y}\sin(kz+\omega t)
\right].
$$
Adding them gives
$$
\mathbf E
=
E_0\hat{\mathbf x}
[
\cos(kz-\omega t)+\cos(kz+\omega t)
]
$$
$$
+
E_0\hat{\mathbf y}
[
\sin(kz-\omega t)-\sin(kz+\omega t)
].
$$
Using
$$
\cos(A-B)+\cos(A+B)=2\cos A\cos B
$$
and
$$
\sin(A-B)-\sin(A+B)=-2\cos A\sin B,
$$
we obtain
$$
\boxed{
\mathbf E
=
2E_0\cos kz
\left(
\hat{\mathbf x}\cos\omega t
-
\hat{\mathbf y}\sin\omega t
\right)
}.
$$
Interpretation
Here the spatial dependence appears as the amplitude factor
$$
2E_0\cos kz,
$$
while the direction of the electric field changes with time.
Therefore the electric-field vector rotates in the $xy$-plane.
The magnetic field is
$$
\boxed{
c\mathbf B
=
2E_0\sin kz
\left(
\hat{\mathbf x}\cos\omega t
-
\hat{\mathbf y}\sin\omega t
\right)
}.
$$
or
$$
\boxed{
\mathbf B
=
\frac{2E_0}{c}\sin kz
\left(
\hat{\mathbf x}\cos\omega t
-
\hat{\mathbf y}\sin\omega t
\right)
}.
$$
Comparison
|
(a) LCP + RCP |
(b) LCP + LCP |
| Electric field |
$2E_0\cos\omega t(\hat{x}\cos kz+\hat{y}\sin kz)$ |
$2E_0\cos kz(\hat{x}\cos\omega t-\hat{y}\sin\omega t)$ |
| Spatial dependence |
Direction |
Amplitude |
| Time dependence |
Amplitude |
Direction |
| Field at fixed position |
Linear oscillation |
Rotating vector |
| Character |
Standing-wave field |
Standing-wave field |
Interpretation of $R(z)$ as the Radius of Curvature
We now consider a Gaussian beam. The phase of the beam, neglecting the Gouy phase $\alpha(z)$, can be written as
$$
\Phi(\rho,z,t)
=
kz+
\frac{k\rho^2}{2R(z)}
-\omega t.
$$
The quantity $R(z)$ is given by
$$
\boxed{
R(z)
=
z\left[
1+\left(\frac{z_R}{z}\right)^2
\right]
}
$$
or equivalently,
$$
\boxed{
R(z)=z+\frac{z_R^2}{z}.
}
$$
We want to understand why $R(z)$ can be interpreted as the radius of curvature of a constant-phase surface.
Constant-phase surface
At a fixed time, a surface of constant phase satisfies
$$
kz+\frac{k\rho^2}{2R(z)}
=
\text{constant}.
$$
Dividing by $k$,
$$
z+\frac{\rho^2}{2R(z)}
=
\text{constant}.
$$
The important feature here is the quadratic dependence on $\rho$.
Geometry of a sphere
Consider a sphere of radius $R$ and a point near its axis. Let $\rho$ be the transverse distance from the axis and let $\Delta z$ be the longitudinal displacement of the surface from the reference point.
From the geometry of the sphere,
$$
R^2=(R-\Delta z)^2+\rho^2.
$$
Expanding the right-hand side gives
$$
R^2
=
R^2-2R\Delta z+(\Delta z)^2+\rho^2.
$$
Canceling $R^2$,
$$
0=-2R\Delta z+(\Delta z)^2+\rho^2.
$$
Paraxial approximation
We are interested in points close to the propagation axis, so $\rho\ll R$. Consequently $\Delta z\ll R$, and the term $(\Delta z)^2$ can be neglected.
Therefore,
$$
0\approx-2R\Delta z+\rho^2.
$$
Hence,
$$
\boxed{
\Delta z\approx\frac{\rho^2}{2R}.
}
$$
This is the key geometric result.
Comparison with the Gaussian beam phase
The phase accumulated because of the additional distance $\Delta z$ is approximately
$$
k\Delta z
=
\frac{k\rho^2}{2R}.
$$
Therefore the phase of a spherical wave near its axis has the form
$$
kz+\frac{k\rho^2}{2R}.
$$
But this is exactly the transverse phase dependence of the Gaussian beam:
$$
\Phi
=
kz+
\frac{k\rho^2}{2R(z)}
-\omega t.
$$
Thus the Gaussian beam has, near its propagation axis, the same phase structure as a spherical wave whose radius of curvature is $R(z)$.
Therefore,
$$
\boxed{
R(z)\text{ is the radius of curvature of the constant-phase surface.}
}
$$
Behavior at the beam waist
At the waist, $z=0$. Since
$$
R(z)=z+\frac{z_R^2}{z},
$$
we have
$$
\boxed{
R(0)\rightarrow\infty.
}
$$
A surface with infinite radius of curvature is locally a plane. Hence the wavefront at the beam waist is approximately planar.
Far from the waist
For $|z|\gg z_R$,
$$
\frac{z_R^2}{z}\ll z,
$$
so
$$
R(z)\approx z.
$$
Thus far from the waist, the Gaussian beam behaves increasingly like a spherical wave whose radius of curvature is approximately the distance from the waist.
Summary
Example 16.2
For oppositely propagating circularly polarized waves, the superposition produces a standing-wave electromagnetic field.
For LCP in $+z$ and RCP in $-z$,
$$
\boxed{
\mathbf E
=
2E_0\cos\omega t
(\hat{x}\cos kz+\hat{y}\sin kz)
}
$$
The field is linearly polarized at each fixed position, but the polarization direction changes with $z$.
When both waves are LCP,
$$
\boxed{
\mathbf E
=
2E_0\cos kz
(\hat{x}\cos\omega t-\hat{y}\sin\omega t)
}
$$
The amplitude varies with position while the electric-field direction rotates with time.
Example 16.4
The phase of a Gaussian beam contains the term
$$
\frac{k\rho^2}{2R(z)}.
$$
A spherical surface near its axis satisfies
$$
\Delta z\approx\frac{\rho^2}{2R}.
$$
Therefore the quadratic transverse phase of the Gaussian beam is precisely the phase variation expected from a spherical wavefront. This provides the physical interpretation
$$
\boxed{
R(z)=\text{radius of curvature of the constant-phase surface}.
}
$$
