Polarisation describes the geometric orientation of the oscillations of an electromagnetic wave. While the direction of propagation tells us where the wave is going, polarisation tells us how the electric field vector behaves as it travels. In this post, we will derive the general equation of polarisation by superposing two orthogonal components and classify the resulting waves into linear, circular, and elliptical states.
Polarisation of Electromagnetic Waves: Derivation from Superposition
To understand the nature of polarisation, we analyze the superposition of two independent plane waves traveling in the same direction but oscillating in mutually perpendicular directions.
1. Mathematical Setup
Consider a monochromatic electromagnetic wave propagating along the positive \(z\)-axis. The wave vector is \(\mathbf{k} = k\hat{z}\). Since electromagnetic waves are transverse, the electric field vector \(\mathbf{E}\) must lie in the \(xy\)-plane.
We can decompose this field into two orthogonal components:
- A component oscillating along the \(x\)-axis.
- A component oscillating along the \(y\)-axis.
Let the instantaneous electric field components be:
\[ \begin{aligned} E_x &= E_1 \cos(kz - \omega t - \theta_1) \\ E_y &= E_2 \cos(kz - \omega t - \theta_2) \end{aligned} \]Here, \(E_1\) and \(E_2\) are the amplitudes, and \(\theta_1\) and \(\theta_2\) are the initial phase angles. The total field is:
\[ \mathbf{E} = E_x \hat{x} + E_y \hat{y} \]The Relative Phase Difference
The crucial factor determining the polarisation state is the phase difference between these two components. We define:
2. Derivation of the Trajectory Equation
Polarisation is defined by the locus (path) traced by the tip of the electric field vector at a fixed point in space as time evolves. To find this path, we must eliminate the time-dependent term \((kz - \omega t)\) from our equations.
Step 1: Normalize the equations
\[ \frac{E_x}{E_1} = \cos(kz - \omega t - \theta_1) \] \[ \frac{E_y}{E_2} = \cos(kz - \omega t - \theta_2) \]Using the subtraction formula \(\cos(A - B) = \cos A \cos B + \sin A \sin B\), we expand the phase terms relative to \(\theta\). Through algebraic manipulation (multiplying by \(\sin\) and \(\cos\) of the phase angles), we derive the following identities:
Identity A:
\[ \frac{E_x}{E_1}\sin\theta_2 - \frac{E_y}{E_2}\sin\theta_1 = \cos(kz-\omega t)\sin(\theta_2-\theta_1) \]Identity B:
\[ \frac{E_x}{E_1}\cos\theta_2 - \frac{E_y}{E_2}\cos\theta_1 = -\sin(kz-\omega t)\sin(\theta_2-\theta_1) \]Step 2: Eliminate the time dependence
Squaring and adding Identity A and Identity B utilizes the identity \(\sin^2 \phi + \cos^2 \phi = 1\), removing the \((kz - \omega t)\) term entirely. After simplifying, we arrive at the General Equation of Polarisation:
Mathematically, this is the equation of a conic section (specifically, a rotated ellipse) centered at the origin.
3. Classification of Polarisation States
The shape of the curve depends entirely on the phase difference \(\theta\) and the relative amplitudes \(E_1\) and \(E_2\). Let us analyze the special cases assuming equal amplitudes \(E_1 = E_2 = E_0\).
The equation simplifies to:
\[ \frac{E_x^2}{E_0^2} + \frac{E_y^2}{E_0^2} - \frac{2 E_x E_y}{E_0^2}\cos\theta = \sin^2\theta \]Case 1: Linear Polarisation
Condition: \(\theta = 0\) or \(\theta = \pi\) (The waves are in phase or in anti-phase).
- If \(\theta = 0\): \(\cos\theta = 1, \sin\theta = 0\).
\[ \left(\frac{E_x}{E_0} - \frac{E_y}{E_0}\right)^2 = 0 \implies E_y = E_x \]
Result: A straight line with slope +1. - If \(\theta = \pi\): \(\cos\theta = -1, \sin\theta = 0\).
\[ \left(\frac{E_x}{E_0} + \frac{E_y}{E_0}\right)^2 = 0 \implies E_y = -E_x \]
Result: A straight line with slope -1.
Case 2: Circular Polarisation
Condition: \(\theta = \pi/2\) or \(3\pi/2\) AND Amplitudes must be equal (\(E_1 = E_2\)).
- If \(\theta = \pi/2\): \(\cos\theta = 0, \sin\theta = 1\).
Substituting into the general equation:
Result: The tip of the electric vector traces a Circle.
- \(\theta = \pi/2\): Left Circularly Polarised (LCP)
- \(\theta = 3\pi/2\): Right Circularly Polarised (RCP)
Case 3: Elliptical Polarisation
Condition: Any other value of \(\theta\), or if amplitudes are unequal \(E_1 \neq E_2\).
If \(\theta\) is not an integer multiple of \(\pi/2\), the term \(E_x E_y\) remains in the equation, representing a tilted ellipse. This is the most general form of polarisation.
4. Summary Table
| Phase Difference (\(\theta\)) | Conditions | Resulting Polarisation |
|---|---|---|
| \(0, 2\pi, \dots\) | Any Amplitudes | Linear |
| \(\pi, 3\pi, \dots\) | Any Amplitudes | Linear |
| \(\pi/2, 3\pi/2\) | \(E_1 = E_2\) | Circular |
| \(\pi/2, 3\pi/2\) | \(E_1 \neq E_2\) | Elliptical (Axes aligned with coordinates) |
| Arbitrary | Any Amplitudes | Elliptical (Tilted) |
5. Visualising the Field Vector
Below are descriptions of how to visualize these mathematical results:
- Linear: Imagine a pen moving back and forth along a single diagonal line. The field vector grows and shrinks but never rotates.
- Circular: Imagine the second hand of a clock. The length of the vector stays constant, but it rotates continuously around the propagation axis.
- Elliptical: A combination of both—the vector rotates, but its length grows and shrinks as it turns, tracing an egg shape (ellipse).