Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Diffraction can be observed in physical waves very easily, however when it comes to the diffraction of electromagnetic waves in things like the single slit experiment, I become a little confused. In things like water, it seems like there is something "holding together" the wave, although I'm not sure what it is. In the case of electromagnetic waves, why does diffraction occur? I know it works out when viewed via Huygens' principle, but what is the mechanism behind electromagnetic diffraction?

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

You seem to be asking two questions: (1) what is the medium for the EM field is and (2) what is the mechanism for diffraction.

  1. There is no material medium - this is what the Michelson-Morley experiment was all about. You could say that the separate harmonic oscillators in the second quantized EM field and the photon field they represent have replaced the "aether" sought in the MM experiment. The difference is that the second quantized field can be made Lorentz invariant (respecting special relativity), well almost (see footnote). The Wikipedia entry on this topic uses a gauge that is not Lorentz invariant, but this problem can be overcome and the Wikipedia page gives you a good idea of what's (messily!) involved.

  2. The simplest answer to what is the mechanism of diffraction I can think of is this: consider a field on a plane, say $z = 0$ and split it up using Fourier decomposition into constituent plane waves. Each constituent plane wave has a different direction defined by the wavevector $\left(k_x, k_y, k_z\right)$ with $k^2 = k_x^2 + k_y^2 + k_z^2$, i.e. all the wavevectors have the same magnitude but different directions. So, when we ask what the field looks like at a different value of $z$, we build the field up from our plane wave constituents at this point (use an inverse Fourier transform). However, now, because the wavevectors are all in different directions, the plane waves have all undergone different phase delays in reaching the new value of $z$ (even though their phase advances by $k$ radians per unit length in the direction of the respective wave vector). Therefore, the field's configuration gets scrambled by all these different phase delays.

You can try this thinking with your slit - set the analysis up in two dimensions so that there are only 2D wave vectors for simplicity. The screen with the slit is in the $z = 0$ plane and the one orthogonal direction is the $x$ axis. All the Cartesian components of the fields fulfil the same (Helmholtz) equation, so we can discuss the principles by just looking at one scalar field $\psi$ (say, the electric field's $x$-component). Each plane wave has the form $\psi(k_x) = \exp\left(i (k_x x + k_z z)\right)$ The Fourier transform of the field output from the slit is then $\frac{\sin\left(\frac{w k_x}{2}\right)}{k_x}$ where $w$ is the slit width, and unless the slit is very wide, the Fourier transform has a wide spread of frequencies. This means that for $z > 0$ ("downstream of the the slit's output) the field is the superposition $\int\limits_{-\infty}^\infty \frac{\sin\left(\frac{w k_x}{2}\right)}{k_x} \exp\left(i (k_x x + k_z z)\right) d k_x$. When we plug $z = 0$ in, the integral is simply the inverse FT and we get our original slit field. But now put some nonzero value of $z$ in: because $k_x^2 + k_z^2 = k^2$, we have $k_z = \sqrt{k^2 - k_x^2}$ (assuming the field is running in the $+z$ direction), we get $\int\limits_{-\infty}^\infty \frac{\sin\left(\frac{w k_x}{2}\right)}{k_x} \exp\left(i (k_x x + \sqrt{k^2 - k_x^2} z)\right) d k_x$. You can see the "scrambling", $k_x$-dependent phase factor $\exp\left(i \sqrt{k^2 - k_x^2} z\right) = \exp\left(i k \cos\theta_x\right)$ (where $\theta_x$ is the angle that the plane wave with wavevector $(k_x, k_z)$ makes with the $z$-axis) will yield the complicated scrambling you see with the slit. Various approximations, notably Fraunhofer and Fresnel, are applied to this integral. Interestingly, the $thinner$ your slit, the wider the direction spread and the more noticeable your diffraction. Wide slits with coherent light produce nondiverging beams (like a laser beam). This is related to the uncertainty principle: the space of momentum eigenstates is found by Fourier transforming the space of position eigenstates, so that the more spread out in one domain a wave is, the more tightly packed in the other it is.

Footnote: A second quantized EM field comprises an infinite number of oscillators - countably so in a box, uncountably so in $\mathbb{R}^3$. The ground state therefore has an energy $\frac{h \nu}{2}$ for each oscillator, and so holds infinite energy. Depending on the approach, this may or may not be a problem - there are procedures for subtracting this infinity out of calculations. However, one may want to "regularize" calculations by assuming that the oscillators don't go up to infinite frequency, but instead are "cut off" at some high frequency (presumably well above the frequencies of cosmic ray $\gamma$ rays). So, although the second quantized EM field can be made LOrentz invariant, I don't believe there is any way one can make this cutoff process truly Lorentz invariant (the cutoff frequency, unless infinite, will depend on observers).

share|improve this answer
    
Why does the field have components in all directions, not just the direction of propagation? –  Anthony Jan 23 at 4:34
    
Because a plane wave in one direction makes a purely sinusoidal variation across a transverse plane as it cuts through it. If it is travelling precisely in the direction of propagation, the phase front aligns with the transverse plane and there is no transverse variation. So any field with a non sinusoidal variation across a tranverse plane must have more than one plane wave Fourier component. Paraxial fields are fields comprising plane waves running at shallow angles relative to some "mean" propagation direction. –  WetSavannaAnimal aka Rod Vance Jan 23 at 4:37
    
I don't follow, I'm afraid. The wave front is moving forward through the aperture, isn't this a single plane wave? –  Anthony Jan 23 at 4:46
    
Or are you looking at the surface perpendicular to propagation? And we choose to Fourier transform this because of Huygens principle? –  Anthony Jan 23 at 4:49
    
The key thing to understand is that the aperture is of finite width. So there are Fourier components with nonzero spatial frequency. If the aperture is very wide, these spatial frequencies get smaller and smaller: in the limit of an infinitely wide aperture, we truly do have only a plane wave and there is NO diffraction. This is why laser beams diverge. Why do we use the Fourier transform? Because plane waves are the eigenfunctions of freespace: these are the fields that propagate by taking on a phase but their form stays the same. So Fourier transforming - phase delay - inverse Fourier ... –  WetSavannaAnimal aka Rod Vance Jan 23 at 7:59
show 8 more comments

In a standard experiment of diffraction by some aperture (monochromatic wave, wavefront parallel to the aperture), you may consider that the electromagnetic field at the aperture is coherent, that is the phase is identical for all the points of the aperture. If you want to calculate the resulting field at some point $X$, (and simplifying in considering a scalar field) you have to sum the contribution for all direct paths (straight lines) from $X$ to any point $O$ of the aperture. The phase difference $\Delta \phi$ is not the same for each of these paths. When summing all these paths, you get the wave amplitude at the point $X$, that is $A(X) \sim \sum_{paths(O \rightarrow X)} e^{i\Delta \phi(path)}$, so you may have constructive or destructive phenomena, depending on the size and shape of the aperture, and of the position of the point $X$, you have a diffraction pattern, meaning that for some points, there are (local) minimum/maximum for the square of the absolute value of the amplitude $|A(X)|^2$.

share|improve this answer
    
@Kyle : Thanks for the correction –  Trimok Jul 8 '13 at 18:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.