6
$\begingroup$

If you have a system of independently radiating electrons/point-charges, the far field distribution of the EM waves can be approximated by the Fraunhoffer diffraction integral, or simply by the Fourier transform of the charge/electron density distribution.

When taking the Fourier transform of something, there is always a zero frequency value. What does this represent in terms of the EM wave example? Is it the average of the near field? Is it the average of the entire field summed in all space (this case it should never vary)?

$\endgroup$
1
$\begingroup$

I don't believe it means a great deal. The reason is that the the Fraunhofer approximation or other farfield approximations assume that the point where you're calculating the EM field is at a distance from the source that is (i) big compared with the extent of the source and (ii) big compared with the wavelength. These two assumptions are needed to make e.g. the stationary phase method approximation of the diffraction integral that one finds in Born and Wolf work.

Once you're down to momentums near $k=0$, these approximations are violated and you're either doing electrostatics or the Fourier transform methods are inadequate: you're now in the near field régime, where the evanescent field can be important.

$\endgroup$
0
$\begingroup$

I assume that by zero frequency, you mean zero momentum transfer. Zero momentum transfer corresponds to the $k=0$ value of the Fourier transform. The value of this part of the fourier transform is the integral of the scattering strength over all space. So you can think of this value as being the total amount of stuff that is there.

Another thing to keep in mind, though, is that if the thing doing the diffraction scatters light only very weakly, like if it is a single molecule, most of the light will go through undiffracted and so at zero momentum transfer you will see a spot whose intensity is just approximately the laser's intensity regardless of how much stuff is there (in the limit of weak scattering).

$\endgroup$
  • $\begingroup$ Hmmmm, so thats what I thought at first. BUT, when I think about relativistic processes that result in radiation (transition radiation for example), the radiation is zero on axis and hence zero for (kx=0,ky=0)...I'm not quite sure what to make of this? $\endgroup$ – user1886681 Mar 3 '14 at 6:48
0
$\begingroup$

If you have a system of independently oscillating point charges as radiators and they do not have a coherence among themselves. Then, If you take single dipole it radiate in a dumbbell shape. If you orient these radiators randomly oriented in space the radiation will propagate as a spherical wave. If you let this wave pass through a slit then you will see the Airy pattern of the slit. If these radiators have random phase relationship among them then the final intensity will be

I=$\left|\left\langle\Sigma_i E_i\right \rangle\right|^2=$

The time averaged cross terms $\left\langle E_{i1}.E_{i2}\cos(\phi)\right \rangle$ will average to zero and finally you will get

$I=\sum I_i$

Hence you will see sum of all the intensities not any diffraction pattern. If the radiators are coherent you will see the diffraction pattern on the screen.

Laser speckle pattern is one of such diffractions.

I hope this will help

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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