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
$$ \boxed{ R(z)=\text{radius of curvature of the constant-phase surface}. } $$
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