Light in classical electrodynamics

Question

I am starting to learn elementary electrodynamics with Griffiths.

In the book, he has shown the natural correspondence between light and electromagnetic plane waves.

The problem that has agitated me is that plane waves are "global solutions", i.e. they have non-trivial EM field almost everywhere; while light seems to be localized phenomenon, when considered as stuff generated in region A and traveled to region B.

How is this paradox resolved classically(without QM)?

From Fourier Theory, it seems viable to create localized solutions by adding plane waves of different frequencies. However, this does not solve the paradox when monochromatic light is considered.

Answer 1

For an EM wave to be considered a plane wave the source must have been "turned on" for an indefinetely long time plus the wave must travel in a single direction. Since all light sources emit light during a finite time interval, there's no true plane wave radiation in nature (this doesn't stop us to use it as an approximation in some cases). For instance, you may approximate the temporal part of an "almost monochromatic wave" as $$ E(t) = E_0 \text{rect}(t_1,t_2) e^{-i\omega_0 t} $$ where $\text{rect}$ is the rectangular function $$ \text{rect}(t_1,t_2) = \begin{cases} 1 &\text{ if } t_1 < t < t_2 \\ 0 &\text{ otherwise} \end{cases} $$ This wave is composed of many frequencies (take the Fourier Transform of it to find them!), but the nice thing is: when me make $t_1 \to -\infty$ and $t_2 \to \infty$ the transform of $E(t)$ approaches a delta function $\delta(\omega - \omega_0)$, which is a monochromatic wave in frequency domain.

But why do we study plane waves all the time? Well, the wave equation is linear, so a sum of solutions is also a solution (add two waves and you get another wave). And the simplest solution to the wave equation is a plane wave. Thus we may build any other wave from a superposition of plane waves (Fourier theory, as you pointed out). This simplifies a lot of work: imagine having to calculate transmission/reflection coefficients with a wave like $$ \mathbf E(\mathbf x,t) = \int \mathbf {E} (\mathbf k) e^{i(\mathbf k \cdot \mathbf x - \omega t)} d^3 k $$ where $|\mathbf k| = \omega / c$. Instead, we calculate for each Fourier-component $\mathbf E(\mathbf k) e^{i(\mathbf k \cdot \mathbf x - \omega t)}$ and call the problem "solved", since if we want the "entire" picture we "just add" each plane wave and build the localized wave.

In conclusion: there's no plane wave, we just study it because it's our "building block" for any wave and all the algebra gets simpler using them.

Answer 2

@ErickShock's answer is perfectly correct, but let's look at some numbers. A HeNe laser spectrum shown:

with a peak at

$$\nu = c/\lambda = 473755.4646016\,{\rm GHz}$$

If we send out a $\tau = 1$ millisecond pulse, we can call that monochromatic pulse that 182 miles long--physically it is a very long pulse. It is not infinite (per the OP's complaint), but it is much larger than the lab.

Mathematically, we model it with:

$$ \sin{(2\pi\nu t)} \times {\rm rect(0,\tau)} $$

The Fourier spectrum of that then has a width of:

$$\Delta\nu \approx 1/\tau = 1\, {\rm kHz}$$.

So the modeled pulse has a frequency range:

$$ \nu = 473755.4646016 \pm 0.000001\,{\rm GHz}$$

(and hence a coherence length of 182 miles).

Meanwhile, an actual HeNe laser has a bandwidth on the order of 1.5 GHz ( 1 foot coherence length).

So, the deviations from "truly monochromatic" (in space or time) caused by the finite size of the apparatus or duration of operation are tiny compared with the inherent deviations in the device itself.

Answer 3

The reason is that light in real life is better approximated as "spherical", although when you zoom in, it looks like planes.

Spherical waves expand as they propagate, and therefore decrease in intensity due to conservation of energy. So if a wave carries a certain amount of energy power P (Joules per second), the intensity, which is P/A where A is area decreases, since A increases as the wave propagates. This creates the effect where light is "stronger" near the source and therefore "localized".