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Consider the following observations:

  1. A superposition of two electromagnetic waves with different frequencies will never produce visible interference patterns. Such waveforms will produce spatiotemporal beat patterns, but they rapidly propagate through space such that a stationary detector (e.g. screen, camera, eyes) will never see any static speckles or fringes whatsoever.

  2. To my understanding, simply increasing the number of discrete Fourier components will not produce stationary intensity patterns either. (I could be wrong here, but please see the closing remarks for an intuitive argument.)

  3. However, real life experience shows that it's quite straightforward to generate visible interference from virtually any source of almost arbitrary spectral properties and coherence, such as black body radiators (burning candle, lightbulbs, glowing metal surfaces, sunlight), broadband LEDs etc.

Between these extreme cases, where exactly do visible speckles/fringes start emerging out of chaos, i.e. what is the precise mathematical minimal condition for the emergence of stationary intensity patterns?

Alternative, but equivalent formulations of the question could be:

  • What does a burning candle or white LED provide that a superposition of potentially billions or trillions of sine waves cannot?
  • What mathematically constitutes a minimally coherent source that will "just begin" to produce stable correlations? E.g. exactly how many frequencies and what kind of correlations are required?

To better visualize the problem, have a look at the following 2D wave fields generated with William Leu's WaveToy:


Overview

Five point emitters and two square barriers rotated 45° to emulate a single slit. In the subsequent snapshots, spatial coherence was reduced by increasing the slit width, and temporal coherence was reduced by adding a frequency step $\Delta f$ from one emitter to the next.

Overview


Case I: Full spatiotemporal coherence

Fully coherent wave field, both in front and behind the single slit.

On a screen placed anywhere in front or behind the slit, a stationary intensity pattern would be observed (a more complex pattern in front and a single, broad lobe behind the slit), independent of the time scale of observation.

Spatiotemporal coherence


Case II: Spatial incoherence, temporal coherence ($\Delta f = 0$)

Same setup as above, but larger slit width. The wave field gets broken up in smaller areas of spatial coherence.

Important: These coherent areas are stationary and would produce visible/static interference fringes on a screen.

Spatial incoherence


Case III: Spatial coherence, temporal incoherence ($\Delta f = 0.1$)

Spatially coherent wave fronts, but now the wave field is broken up into "wave packets" propagating with the speed of light (image intensity rescaled to increase visibility of diffracted fields).

Important: The wave fronts between subsequent "packets" are no longer in phase (not easy to see here)! For a larger number of emitters, the phase relation would be essentially random. On a screen, one would observe either random flashes or a homogeneous illumination, depending on the speed of the detector (camera sensor, eyes).

Temporal coherence


Case IV: Spatiotemporal incoherence ($\Delta f = 0.1$)

Wave field is both spatially and temporally incoherent. Random$^*$ patches of (transiently?) coherent areas rapidly propagate through space.

On a screen, one would observe either essentially random$^*$ fluctuations (speckle noise) or, over more realistic integration times, a homogenous illumination. No visible interference fringes. As far as I understand, it is no longer possible to generate stationary intensity patterns from such a source by any (passive) means.

...yet, nothing is easier than generating stationary interference patterns from virtually any real light source, such as sunlight or a piece of red-hot metal!

Spatiotemporal incoherence

$^*$Here: Only quasi-random, see 2nd remark below.


Remark 1: Note that the coherence characteristics of the wave field behind the slit are a reflection of the "source field" characteristics before it. In those cases, where visible interference patterns can (or potentially could) be observed - namely I & II, the source field itself has a stationary intensity distribution. Whereas in cases III & IV, the source field itself does not have a stationary intensity distribution. To my current understanding, passive elements such as a diffraction slit therefore only serves to select a subset of field constituents with have a higher correlation (coherence) than the underlying ensemble.

Remark 2: The wave fields in cases II-IV are, in fact, not 100% incoherent when measured in the respective quantities (coherence area for spatial coherence, coherence time/length for temporal coherence). Rather, they are still partially coherent due to the small number of emitters and frequencies used in the simulation, so the term "incoherent" here is used as a relative one to compare different scenarios.

The latter point also results in a periodicity of the wave fields. Without actually (continuously) randomizing the emitters, the generated fields will repeat as can be seen in the following updated example using 10 sources with 10 different frequencies and phases (A, B, C representing repeating "wave packets"):

Periodicity

In this sense, the whole problem/question could also be formulated backwards:

  • On one hand, we have to add more and more constituent waves in order to further and further randomize the wave field and thereby reduce correlation.
  • On the other hand, upon superposing sufficiently (infinitely?) many components, we "suddenly" get precisely what we've tried to eliminate, namely a strong correlation in the form of stationary intensity distributions.

UPDATE 1: Two more simulations, this time using only 2 emitters with different frequencies (image intensities rescaled to increase visibility of diffracted fields).

Case IV.b: Spatiotemporal incoherence, single slit ($\Delta f = 0.1$)

The wave field looks similar as in case IV above, but with larger coherent patches. Again, these patches rapidly propagate through space. As one increases the number of emitters, the field is broken up into smaller and smaller patches.

Single slit with 2 emitters, 2 frequencies, and 2 phases.


Case IV.c: Spatiotemporal incoherence, double slit ($\Delta f = 0.1$)

Important: The distinct "fringe patterns" rapidly propagate through space and do not represent stationary interference fringes.

As the number of emitters is increased, the resulting wave fields will transition more and more into chaos.

Double slit with 2 emitters, 2 frequencies, and 2 phases.


UPDATE 2: Proof for the 1st observation in the question, namely that a superposition of two waves with different frequencies never produces a stationary interference pattern.

Starting with the modulus squared of a complex number

$$|z|^2 = \sqrt{zz^*}^2$$

we insert the superposition $z = z_1 + z_2$ with

$$z_1 = \tfrac{1}{2} e^{i k_1 x}e^{-i \omega_1 t} \\ z_2 = \tfrac{1}{2} e^{i k_2 x}e^{-i \omega_2 t}$$

using half-amplitudes for convenience (just to normalize the resulting intensity) and get

$$\begin{align} |z|^2 &= \sqrt{ (z_1 + z_2)(z_1 + z_2)^* }^2 \\ &= \sqrt{ z_1 z_1^* + z_2 z_2^* + z_1 z_2^* + z_1^* z_2 }^2 \\ &= \sqrt{ 1 + z_1 z_2^* + z_1^* z_2 }^2 \\ &= \sqrt{ 1 + z_1 z_2^* + (z_1 z_2^*)^* }^2 \\ &= \sqrt{ 1 + 2 \ \Re (z_1 z_2^*) }^2 \\ &= \sqrt{ 1 + 2 \ \Re (\tfrac{1}{4} e^{i k_1 x} e^{-i \omega_1 t} e^{-i k_2 x} e^{i \omega_2 t}) }^2 \\ &= \sqrt{ 1 + \tfrac{1}{2} \ \cos (k_1 x) \ \cos (-k_2 x) \ \cos (-\omega_1 t) \ \cos (\omega_2 t) }^2 \\ &= \sqrt{ 1 + \tfrac{1}{8} \ [\cos ((k_1 - k_2)x) + \cos ((k_1 + k_2)x)] \times [ \cos ((-\omega_1 + \omega_2)t) + \cos ((-\omega_1 - \omega_2)t)] }^2 \\ &= \sqrt{ 1 + \tfrac{1}{8} \ [\cos (k_- x) + \cos (k_+ x)] \times [ \cos (\omega_- t) + \cos (\omega_+ t)] }^2 \\ &= \sqrt{ 1 + \tfrac{1}{8} \ \Psi_{\text{spat. beat}}(x) \ \Psi_{\text{temp. beat}}(t) }^2 \\ \end{align}$$

with $k_1 \pm k_2 =: k_\pm$, $\omega_1 \pm \omega_2 =: \omega_\pm$, and using the cosine's symmetry and a prosthaphaeresis relation in the final steps.

The result is not simply a stationary sum of the intensity distributions from individual sources, but a time-dependent/oscillatory sum of "crossed" products of waves, i.e. a spatiotemporal beat pattern that rapidly propagates through space. This is also visualized in the simulations above (scenarios IV, IV.b, and IV.c).

Also, both spatial and temporal beat components are multiplicative (they're in a product term), so averaging over either one returns zero spatiotemporal modulation and, therefore, zero fringe contrast:

$$\begin{align} \langle |z|^2 \rangle_T \ &= \ \langle \sqrt{ 1 + \tfrac{1}{8} \ [\cos (k_- x) + \cos (k_+ x)] \times [ \cos (\omega_- t) + \cos (\omega_+ t)] }^2 \rangle_T \\ \ &= \ \sqrt{ 1 + \tfrac{1}{8} \ [\cos (k_- x) + \cos (k_+ x)] \ \times \langle [ \cos (\omega_- t) + \cos (\omega_+ t)] \rangle_T }^2 \\ \ &= \ \sqrt{ 1 + \tfrac{1}{8} \ [\cos (k_- x) + \cos (k_+ x)] \cdot 0 }^2 \\ \ &= \ 1 \end{align}$$

For the plane waves considered here, the result is simply the sum of source intensities ($\tfrac{1}{2} + \tfrac{1}{2} = 1$), as one would expect due to energy conservation. In the case of e.g. point sources, one would get the sum of $\tfrac{1}{r^2}$ laws for the source intensities. Experimentally, temporal averaging corresponds to the signal integration (limited time resolution/finite exposure time) inherent to any realistic sensor or detector, and spatial averaging corresponds to a finite detection volume (limited spatial resolution).


...to round things off (and prove point 3 in the introduction), here is an actual Michelson interferogram taken from the flame of a lighter, so basically combusting butane gas with black body radiation from glowing soot particles:

Butane flame interferogram

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    $\begingroup$ I'm not sure what the specific question is, but I think you're reaching towards describing the van Cittert–Zernike theorem. $\endgroup$
    – knzhou
    Mar 11, 2023 at 3:30
  • $\begingroup$ @knzhou I'm asking about the mathematical "boundary" between coherent and incoherent fields/sources, i.e. the exact minimal conditions both necessary and sufficient for stationary intensity distributions. The question is motivated by the contradiction or "constructivist gap" between 1.) my assumption that a superposition of even a "large" number of Fourier components does not produce visible interference and 2.) the observation that an overwhelmingly large class of everyday sources (e.g. candles, sunlight) does produce visible interference. $\endgroup$
    – srhslvmn
    Mar 11, 2023 at 4:08
  • $\begingroup$ I've been looking both into the Wiener–Khinchin and van Cittert-Zernike theorem, but haven't found a satisfactory explanation yet. To my understanding, they're somewhat equivalent in that they relate the spectral and spatial distribution (i.e. spectral "size" and angular size, respectively) of a source to statistical correlations (i.e. coherence) in the resulting fields. I've been, however, more focused on the Wiener-Khinchin theorem as, experimentally, increasing the spatial coherence is usually the easier/trivial part (simply decreasing the spot size of and/or distance to the source). $\endgroup$
    – srhslvmn
    Mar 11, 2023 at 7:19
  • $\begingroup$ Statement (1) is wrong for sure. Consider the superposition of two frequencies. The stationary diffraction pattern of the superposition is just the sum of the intensities of the diffraction patterns of the single frequencies. There is no reason why this sum should not show any interference pattern. E.g. if the two frequencies are very close, you get the sum of two similar patterns. What follows (1) is thus incomprehensible. $\endgroup$ Mar 12, 2023 at 20:50
  • $\begingroup$ @DorianoBrogioli Hi, you can easily verify this for yourself with the linked wave simulator. Just place two point sources (right click inside the simulation box) and set different frequencies. You'll see interference patterns, but they're rapidly propagating through space - it's basically a simpler scenario of cases III & IV in above simulations, only using 2 instead of 5 emitters. And the more sources you add, the more chaotic the resulting wave field becomes. $\endgroup$
    – srhslvmn
    Mar 12, 2023 at 23:07

2 Answers 2

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I do not understand the question, because it starts with a wrong statement:

A superposition of two electromagnetic waves with different frequencies will never produce visible interference patterns.

This statement is clearly false. Take, for example, the diffraction pattern from a double slit with a single frequency:

Double slit diffraction pattern with a single wavelength

If you take a wavelength that is 10% longer you get a slightly wider spacing:

Double slit diffraction pattern with a longer wavelength

When you superpose the two wavelength, the intensity that you get is simply the sum of the two. The result still has a very clear diffraction pattern, showing interference:

Double slit diffraction pattern with two wavelengths

It is easy to prove that the intensities of the two waves with different wavelength are simply added. Let us say that the first wave is: $$ \varphi_1 = \psi_1(x) e^{i \omega_1 t} $$ and the second: $$ \varphi_2 = \psi_2(x) e^{i \omega_2 t} $$ The square modulus of the superposition is: $$ |\varphi|^2 = \left| \varphi_1 + \varphi_2 \right|^2 $$ Substituting: $$ |\varphi|^2 = \left| \psi_1(x) e^{i \omega_1 t} + \psi_2(x) e^{i \omega_2 t} \right|^2 $$ This is equal to: $$ |\varphi|^2 = \left( \psi_1(x) e^{i \omega_1 t} + \psi_2(x) e^{i \omega_2 t} \right) \left( \psi_1(x) e^{i \omega_1 t} + \psi_2(x) e^{i \omega_2 t} \right)^* $$ The four terms are: $$ |\varphi|^2 = \psi_1(x) e^{i \omega_1 t} \psi_1(x)^* e^{-i \omega_1 t} + \psi_1(x) e^{i \omega_1 t} \psi_2(x)^* e^{-i \omega_2 t} + \psi_2(x) e^{i \omega_2 t} \psi_1(x)^* e^{-i \omega_1 t} + \psi_2(x) e^{i \omega_2 t} \psi_2(x)^* e^{-i \omega_2 t} $$ Some of the exponential cancel out: $$ |\varphi|^2 = \psi_1(x) \psi_1(x)^* + \psi_2(x) \psi_2(x)^* + 2 Re \left( \psi_1(x) \psi_2(x)^* e^{i (\omega_1-\omega_2) t} \right) $$ Calculating the intensity, the last term on the right averages to 0: $$ I = \psi_1(x) \psi_1(x)^* + \psi_2(x) \psi_2(x)^* = I_1 + I_2 $$

So, from the superposition of two diffraction patterns at different wavelengths you simply get the sum of the intensities. Normally, the interference will still be visible.

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – SuperCiocia
    Mar 17, 2023 at 21:29
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Ok, let's consider two plane waves:

\begin{align} \psi_1(x, y, z, t) =& E_1 e^{-i(k_{1,x} x + k_{1,z} z)} e^{i\omega_1 t}\\ \psi_2(x, y, z, t) =& E_2 e^{-i(k_{2,x} x + k_{2,z} z)} e^{i\omega_2 t} \end{align} Suppose $k_{i,y}=0$ for $i=1, 2$. Suppose that $k_{1,z} \not= k_{2, z}$ and $k_{1,x} \not = k_{2,x}$ in general. This means the two waves may be travelling at an angle to each other and that the waves will have different wavelengths and frequencies. $$ \omega_i = c \sqrt{k_{i, x}^2 + k_{i, z}^2} $$

Suppose we put a CCD sensor at $z=0$. The total field on the sensor is

$$ \psi_{tot}(x, 0, 0, t) = \psi_1(x, 0, 0, t) + \psi_2(x, 0, 0, t) = E_1 e^{-i k_{1, x} x} + E_2 e^{-i k_{2, x} x} $$

The CCD sensor is sensitive to the squared field $I_{tot} = \psi^*_{tot}\psi_{tot}$ which we calculate to be

\begin{align} I_{tot} =& |E_1|^2 + |E_2|^2 + \text{Re}\left(E_1 E_2^* e^{-i(k_{1,x} - k_{2, x})x}e^{i(\omega_1 - \omega_2)t}\right)\\ =& |E_1|^2 + |E_2|^2 + |E_1||E_2|\cos\left((\omega_1 - \omega_2)t - (k_{1,x} - k_{2, x}) x + (\phi_1 - \phi_2)\right) \end{align}

Where I've defined $E_i = |E_i| e^{i \phi_i}$ and done some complex trig. We see that the intensity on each point on the screen is spatially- ($x$) and temporally- ($t$) dependent.

The OP claims that:

  1. A superposition of two electromagnetic waves with different frequencies will never produce visible interference patterns. Such waveforms will produce spatiotemporal beat patterns, but they rapidly propagate through space such that a stationary detector (e.g. screen, camera, eyes) will never see any static speckles or fringes whatsoever.

This is not true. It is true that if $\omega_1 \not = \omega_2$ then the spatial interference pattern (the variation of $I_{tot}$ with $x$) will not be fixed in space and it will be moving. But that doesn't mean we cannot see the interference pattern. Whether we can see the (spatially moving) interference pattern or not depends on how fast it is moving and the temporal resolution (aka the bandwidth) of the sensor we are using.

For just two sinusoids like this, in time, the sinusoid travels in either the $+x$ or $-x$ direction at a constant velocity. We can see that it moves one transverse wavelength ($\lambda_{\perp} = 2\pi / (k_{1,x} - k_{2, x})$) in a time $T = 2\pi /(\omega_1 - \omega_2)$.

Suppose we have a camera sensor that the beams have a large enough angle between them and our sensor is big enough that we see about 10 fringes across the sensor when the two beams are at the same frequency. Now detune one of the tones by 0.1 Hz. On the sensor we will start to see the interference pattern slowly drift one way or the other on the sensor. As long as our sensor has a bandwidth greater than 0.1 Hz (and CCD sensors typically have bandwidths of 10s of Hz at least) we will be able to record and watch a movie of the intensity pattern slowly drifting.

This is a refutations of the OPs claim that electromagnetic waves of different frequencies can't produce a visible interference pattern.

Now, if the detuning is turned up to 1 kHz, or 1 MHz, our slow 100 Hz bandwidth CCD is no longer going to be able to see the motion of the interference pattern. In this case the slow bandwidth of the sensor will result in averaging over durations of $1/BW$ where $BW$ is the detector bandwidth. For these high frequency detunings the signal would totally wash out and we would just see the whole sensor lit up with the sum of the intensities of the two individual beams. So what you see depends on the specific timescales involved.

Notably, the interference pattern is easier to see when the detunings are small. This is sort of related to the statement that: lasers with small linewidths have better coherence properties. This is because those lasers are composed of frequencies close to the central frequency. So even though these tones are at different frequencies, the motion of the interference pattern is slow enough that you can still see it. Whereas if a laser has a large linewidth then you would need a fast sensor (a sensor with a bandwidth larger than the linewidth of the laser) to see interference signals from that laser.

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  • $\begingroup$ This answer starts considering two plane waves with different wavelengths. They will not produce interference patterns, after time average! In general, two waves with different wavelength will not produce interference between them; the interference terms average to 0. However, if one of the two waves produces interference patterns (with itself), then the interference will still be visible superposing it to another wave at a different wavelength. Simply, because the intensities of the two sum up. $\endgroup$ Mar 16, 2023 at 8:31
  • $\begingroup$ @Jagerber48 Thanks for your answer, I'll work through it later today. Also, to whomever downvoted this answer - I'd rather have a fruitful discussion (if not in the comment section, then in the chat), so please consider withdrawing $\endgroup$
    – srhslvmn
    Mar 16, 2023 at 12:18
  • $\begingroup$ @DorianoBrogioli you’re correct that the interference washes out upon time averaging, but if you DONT time average (because your detector has faster bandwidth than the timescale in which the pattern is changing) you can indeed observe interference fringes. They will just be in motion. $\endgroup$
    – Jagerber48
    Mar 16, 2023 at 12:59
  • $\begingroup$ This is correct, but the question claims the opposite thing: that no interference is visible if there are two different frequencies. This is false and even more false if the frequency difference is small: in the latter case, you do not only see the spatial effect of interference, but also the temporal oscillations. That is why I downvoted. The answer is technically correct but does not clarifies the problem. However, it is not a problem of the answer but of the (very confusing) question. $\endgroup$ Mar 16, 2023 at 16:32
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    $\begingroup$ @DorianoBrogioli I dunno, I think it does clarify the problem. Interference of waves results in oscillations of fields and field amplitudes in space and time. The OP seems to think interference is only "observable" if the field amplitude does NOT vary in time but DOES vary in space AFTER temporal averaging. The OP things you can't get this phenomenon for waves with different frequencies. Your answer shows that this claim is directly wrong by demonstrating waves at different frequencies whos amplitudes still vary in space but not vary in time after time averaging. $\endgroup$
    – Jagerber48
    Mar 16, 2023 at 17:58

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