Diffraction condition and the Fourier transform

In summary, the author is trying to derive the scattering amplitude for elastic scattering of a wave off a collection of atoms located at R_1, R_2, R_3, etc. and is using an equation from scattering theory.
  • #1
ehrenfest
2,020
1

Homework Statement


My book uses the following equation to derive the diffraction condition for electromagnetic waves scattering in a crystal lattice:

[tex]F= \int dV n(\mathbf{r}) \exp \left[i\Delta\mathbf{k}\cdot \mathbf{r} \right][/tex]

F is the scattering amplitude and n is the electron density. I just don't understand where that comes from. How does the Fourier transform relate to the diffraction? I have studied Fourier expansions in calculus and I understand how (virtually) every periodic function can be represented as a Fourier series.

Homework Equations


The Attempt at a Solution

 
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  • #2
so, for example, does the following equation for elastic scattering of a wave off a collection of atoms located at R_1, R_2, R_3, etc, make more sense to you?
... the scattering amplitude is roughly
[tex]
F\sim \sum_{i} e^{i(\vec k_0 - \vec k_f)\cdot\vec R_i}
[/tex]

can you derive the above equation? can you see how it relates to the equation you have written down?
 
  • #3
olgranpappy said:
so, for example, does the following equation for elastic scattering of a wave off a collection of atoms located at R_1, R_2, R_3, etc, make more sense to you?
... the scattering amplitude is roughly
[tex]
F\sim \sum_{i} e^{i(\vec k_0 - \vec k_f)\cdot\vec R_i}
[/tex]

can you derive the above equation? can you see how it relates to the equation you have written down?

Well, I see how my equation is just the integral version of your equation, but no your equation does not make sense to me
 
  • #4
okay. so. let's see... so, if you have an incoming plane wave [tex]e^{i\vec k_0\cdot\vec r}[/tex]and it scatters off some point scatterer at a position [tex]R_1[/tex], the amplitude of the wave at the position of
the scatterer, but before it scatters is
[tex]
e^{i\vec k_0 \cdot \vec R_1}
[/tex]
the wave scatters off the point scatterer and is thus now given by the original amplitude times a spherical wave of the form [tex]f e^{ik|\vec r-\vec R_1|}/|\vec r - \vec R_1|[/tex], where for simplicity f is just some number indep of scattering angle (cf. something like Jackson Third Ed. Eq. 10.2, I am writing the same thing for a scalar theory with a much simpler scattering f, and also I'm keeping around the initial phase factor for latter, cf. also Jackson's section on scattering by a collection of scatterers, like Eq. 10.19).

So, the amplitude is
[tex]
e^{i\vec k_0\cdot \vec R_1} \frac{fe^{ik|\vec r - \vec R_1|}}{|\vec r-\vec R_1|}
[/tex]

So, that's for a scatter at R_1

Next, add the amplitude from another scatterer at R_2, and another at R_3, etc.

Next take the limit where r >> R for all the R's and you can use
[tex]
|\vec r-\vec R| \approx r - \vec R\cdot\frac{\vec r}{r}
[/tex]
('member this from scattering theory, maybe in quantum... you know what to do)

and get my previous result up to the factors I left off and compensated for with the [tex]\sim[/tex] symbol which is certainly the lazy man's good friend.
 

1. What is the diffraction condition?

The diffraction condition is a fundamental principle in optics that describes the relationship between the shape and size of an aperture or slit and the resulting diffraction pattern. It states that in order to produce a well-defined diffraction pattern, the size of the aperture or slit must be on the same order of magnitude as the wavelength of the incident light.

2. How is the diffraction condition related to the Fourier transform?

The diffraction condition is closely related to the Fourier transform, which is a mathematical tool used to analyze the spatial frequency content of a signal. The Fourier transform of a diffraction pattern can reveal information about the size and shape of the aperture or slit that produced it, and vice versa.

3. What is the significance of the diffraction condition in optics?

The diffraction condition is crucial in understanding how light interacts with different structures and materials in optics. It helps us predict the diffraction patterns that will be produced and design optical systems with specific properties, such as focusing or dispersing light.

4. How does the diffraction condition affect the resolution of an optical system?

The diffraction condition plays a critical role in determining the resolution of an optical system. If the aperture or slit is too large compared to the wavelength of light, the resulting diffraction pattern will be blurry and the system will have a lower resolution. By adhering to the diffraction condition, we can optimize the resolution of an optical system.

5. Can the diffraction condition be applied to other wave phenomena besides light?

Yes, the diffraction condition can be applied to other wave phenomena, such as sound, water waves, and radio waves. In these cases, the size of the aperture or slit must be on the same order of magnitude as the wavelength of the corresponding wave. This principle is also known as the Huygens-Fresnel principle.

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