# Interference requires countable wavefields

1. Dec 3, 2015

### Anton Alice

Hello,

if have a question concerning the difference between " interference" and "diffraction".

Interference can be created, by superposition of countable many wavefields.
So if I have a countable amount of wave sources,
then the superposition can be called an interference.

In contrary, for diffraction, I need infinitely many such sources. These are usually considered
to be the famous secondary sources of huygens.

I dont quite understand, why this should be the case.
Why isnt it possible to create an interference by superposition of infinitely many sources?
And why is it not possible to have diffraction with only countable many sources?

2. Dec 3, 2015

### Andy Resnick

You raise some subtle issues. Clearly, there is no such thing as a point source or monochromatic source. Yet, interference is observed- the resolution to this is to carefully account for a continuous distribution of source points and wavelengths, leading to the concepts of coherence area and coherence length. The full beast is called the "mutual coherence function".

Crudely put, if the path difference in a Mach-Zender interferometer is less than the coherence length (set by the spectral width), interference fringes are obtained. If two slits in a Young's interferometer are separated by less than the transverse coherence length, interference fringes are observed.

3. Dec 6, 2015

### Anton Alice

Yes I know that thing. Its made up of an auto correlation, and a normalization. But I dont quite see the relation to my question. Maybe you can elaborate on this?

Meanwhile I figured out a substitute explanation, to put off the evil hour:
Lets arrange some point sources along the x-axis. countable now could mean:
at any integer there is a source, or also
at any rational number there is a source
(or at any irrational number?).
In other words, the positions of the sources can be described as a sequence, therefore each source can be assigned with an index i.
In other words, there is always a seperation (although it can be an infinitely small distance) between each source.
Therefore the distribution of these sources is discret.
I could have infinitely many of these sources, but I could weight them such that the sum of all intensities converges, i.e. the sum over the elements Ii of the sequence converges (whereby Ii is the weighted intensity of source i at some position)

In contrast to this, if I would allow for real numbers on the x-axis to be source positions, then the distribution of source points could also be continuous, therefore not discret, therefore there would be no seperation between source points. And the other thing is, that each infinitesimal source has an infinitely small (converging to zero) contribution to the overall intensity, and therefore can not be considered as a "stand-alone" point source.
One could also say, that in this case the source points are merged to one single spatially expanded source. Therefore you can not speak about an interference of different point sources, because you only have one single expanded source.

4. Dec 6, 2015

### Andy Resnick

That's it- a spatially and/or spectrally extended source is expressed as a integral of a density (spectral and/or spatial), such that the integral is the total radiated power/intensity. In radiometry this is extended to include the direction of emitted power (the spectral radiance).

5. Dec 8, 2015

### Anton Alice

Ok. But still it seems to me like a play of words. Because that integral over the density I could also call "superposition", therefore the result of the integral I felt free to call interference.

Either there is still a deeper reason for the distinction between interference and diffraction, which you explained and I did not capture, Or its just a play of words in the end.

Would it be possible to get a good approximation of the wave field of an extended source, by just a countable amount of weighted sources? Or in other words: Can I simulate all aspects of diffraction by using interference? Or in other words:What problems could appear, if I try to explain diffraction with only a countable amount of sources?

You referred to the aspect of coherence:
It is often said, that an extended source has bad coherence properties, so one could not use them for a doubleslit experiment for example, unless one uses a small aperture to approximate a point source. I assume that with "extended" they mean our definition of extended, i.e. uncountable collection of sources.
I this case, even if the uncountable collection of sources have each an infinite temporal coherence (e.g. because they emit continuous sines), the spatial coherence is vulnerable, because at each slit there would be a superposition of all various phase relations, which is bad for the experiment, because each slit can then be seen as a point source with very bad temporal coherence (bad auto-correlation), right?

But what would happen with the coherence properties, If we used a countable collection of point sources (again perfectly temporal coherent) instead of an extended source? There are still various phase relations arriving at the slits.

6. Dec 8, 2015

### Andy Resnick

I'm not sure where you are going here. Are you struggling with the discrete -> continuum transition?

7. Dec 11, 2015

### Anton Alice

I don't think so. But I am still wondering, if there is a deeper meaning to the distinction between "interference" and "diffraction". To me this seems just as a play of words.
.. which I could also call "interference", interference of a uncountable amount of sources.

8. Dec 12, 2015

### Staff: Mentor

Diffraction happens at an aperture. Interference happens in free space.

9. Dec 13, 2015

### blue_leaf77

Trying to elaborate what Dale said, diffraction is referred to as the propagation of a light field following a passage through a hindrance, e.g. aperture,edge, or lens. In terms of spatial frequency spectrum, this means diffraction takes place whenever there is a change in the spatial frequency spectrum. Imagine you have a black box whose content is unknown to you, and you measure the spatial spectra (e.g. using focusing lens) before and after the beam enters this box, if you notice that there is a change in the spatial spectrum, then the beam must have experienced diffraction inside the box, otherwise, there should be no diffraction. And the whole thing about diffraction is a result of a wave exhibiting the property of interference as you said in post #7.

10. Dec 17, 2015

### Anton Alice

Ok, is it correct to say, that there is no actual interference in the real world, because there is no ideal point source? All sources have a spatial extension. Therefore, what we call interference is just an approximation of what is really happening: Diffraction = Interference of uncountable many sources.

11. Dec 17, 2015

### Staff: Mentor

I wouldn't say that.

I don't know why you are focused on point sources and countability. As I mentioned above, if I felt that it was necessary to make a distinction between diffraction and interference, then I would classify diffraction as what happens when waves meet a material aperture and interference as what happen when waves meet each other in space.

Diffraction is an interaction between the field and matter. Interference is just the fact that the field equations obey superposition.

Last edited: Dec 17, 2015
12. Dec 17, 2015

### blue_leaf77

(Overly) strictly speaking, interference is a special case of superposition where the superposing waves have exactly the same frequency, which necessarily implies that the two waves must be perfectly coherent. Because in reality, there is no such a thing as 100% perfect coherent light source, the term interference in the sense mentioned before is not present around us. So, the reason has nothing to do with the shape of the source.
As said above, that no perfectly coherent light exists, this in turn suggests that defining interference as stated at the beginning is a dead-end statement as we won't observe it anyway. So, one need not sweat too much on such matter, and will be better to soften the definition to include the approximate conditions.

13. Dec 20, 2015

### Shreyas Samudra

Can somebody please explain to me what the 'starter' means by countable wave fields ?

14. Dec 21, 2015

### Anton Alice

Why not asking the "starter" himself?

With "countable wave fields" I mean a countable number of point sources, that are associated with that one resulting wavefield.

I think post #11 brings it to the point. But I don't feel it is necessary to make a distinction.