Q.Poynting vector $\mathbf{S}$ is defined as a vector whose magnitude is equal to the wave intensity and whose direction is along the direction of wave propagation. Mathematically, it is given by $\mathbf{S} = \dfrac{1}{\mu_0}\mathbf{E} \times \mathbf{B}$. Show the nature of $S$ vs $t$ graph.

The key idea is that for an electromagnetic wave, $\mathbf{E}$ and $\mathbf{B}$ oscillate sinusoidally in phase, so their cross product varies as $\sin^2(\omega t)$.

Reasoning:

1. For a plane wave propagating along $x$, let $E = E_0 \sin(\omega t - kx)$ and $B = B_0 \sin(\omega t - kx)$, with $E_0 = c B_0$.
2. The magnitude of the Poynting vector is:

3. Since $\sin^2$ is always non-negative and oscillates between $0$ and $1$, $S$ is a unidirectional pulse — it never goes negative, rising and falling from zero to a maximum twice per cycle. …

Why This Approach Works

The Poynting vector $\mathbf{S}$ represents the energy flux density of an electromagnetic wave — the rate at which energy flows through a unit area perpendicular to the direction of propagation. For a plane wave traveling along the $x$-axis, both $\mathbf{E}$ and $\mathbf{B}$ oscillate sinusoidally in time. Since $\mathbf{S}$ involves the product of these two fields, its time dependence is not simply sinusoidal but follows a $\sin^2$ pattern. This means the energy flow is always forward (positive direction) but pulsates — it never reverses direction, because both $\mathbf{E}$ and $\mathbf{B}$ reverse sign together, keeping their cross product direction constant.

The key insight: when two sine waves are multiplied, the result oscillates at twice the original frequency and is always non-negative (for aligned fields).

Step-by-Step Derivation

1. Set up the wave equations

Consider a plane electromagnetic wave propagating along the $+x$ direction. The electric field oscillates along the $y$-axis and the magnetic field along the $z$-axis:

Here $E_0$ and $B_0$ are the amplitudes, $k = 2\pi/\lambda$ is the wave number, and $\omega = 2\pi f$ is the angular frequency.

2. Recall the relation between $E_0$ and $B_0$

From Maxwell's equations, for an electromagnetic wave in vacuum:

where $c = 1/\sqrt{\mu_0 \varepsilon_0}$ is the speed of light. This is a fundamental relation — the electric and magnetic fields are in phase and their amplitudes are linked by $c$.

3. Compute the cross product

The Poynting vector is:

Substituting our fields:

Using $\hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}}$:

Therefore:

4. Express in terms of $E_0$ alone

Using $B_0 = E_0/c$ and $c = 1/\sqrt{\mu_0 \varepsilon_0}$:

The quantity $\sqrt{\varepsilon_0/\mu_0}$ is the reciprocal of the characteristic impedance of free space. So:

5. Analyze the time dependence at a fixed point

At a fixed position, say $x = 0$, the magnitude becomes:

$S(t) = S_0 \sin^2(\omega t)$ …

Method: Determining the Time-Dependence Shape of a Product of Two In-Phase Sinusoids

Use this whenever a quantity is built as the product (or cross product) of two fields that individually oscillate sinusoidally in phase — e.g. the Poynting vector, instantaneous power, or any $E \times B$-type quantity — and you need to describe its behaviour over time, not just compute a single value.

Steps

Step 1: Write both oscillating quantities with the same phase argument.

For an EM wave, $E$ and $B$ are in phase, so both can be written as $A\sin(\omega t - kx)$ with different amplitudes but an identical argument. This is what makes their product simplify to a single $\sin^2$ term rather than a general product-to-sum expansion.

Step 2: Multiply and recognise the $\sin^2$ pattern.

Since $\sin^2\theta$ is never negative, this alone tells you the resulting quantity never reverses sign — a conclusion you can state before doing any further calculation.

Step 3: Use the double-angle identity to see the oscillation frequency and shape. …