Is the Wave Vector of a Plane Wave Affected by Diffraction?

AI Thread Summary
The discussion centers on the relationship between the wave vector of a plane wave and its diffraction characteristics. It is established that the wave vector's direction is not dependent on its position in an ideal plane wave, as it remains perpendicular to the wavefront. However, in realistic scenarios, such as Gaussian beams, the wave vector can exhibit divergence away from the beam's center. The complexity of diffraction is highlighted, indicating that a diffracted beam diverges from being a plane wave, necessitating specific conditions for analysis. Overall, understanding the wave vector's behavior requires careful consideration of the wave's characteristics and the effects of diffraction.
ecastro
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Considering a plane wave propagating in one specific direction, does its wave vector dependent on a certain point in the plane wave or is it dependent on a certain point on a plane parallel to the wave?
 
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I think the wave's direction depends on the point you want to discuss. It may be the function of time ##t## and position ##x.##
 
tommyxu3 said:
I think the wave's direction depends on the point you want to discuss. It may be the function of time ##t## and position ##x.##
Dropping the time dependence for simplicity, how will I calculate the components of the wave vector?
 
If it is a still wave, then it doesn't move. I think (but I'm not sure that's correct) that we should discuss the situation of the wave between ##t## and ##t+dt.##
 
You mean like the change in amplitude? What if I considered an instantaneous wave form?
 
What I states is also considering an instantaneous, isn't it? Take the change of the direction of the point concerned into consideration, then the direction of the wave vector of the point may be got.
Or, I'm not sure if I have misunderstood what you meant...
 
So, a wave vector positioned at ##\left(x_0, y_0\right)## directed to a point ##\left(x'_0, y'_0\right)## in the plane of interest is different for a wave vector positioned at ##\left(x_1, y_1\right)## directed at the same point ##\left(x'_0, y'_0\right)## at the plane of interest?
 
Yes, I think it is, for the two vectors is different.
 
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The wave-vector is a 'vector' which contains both amplitude and direction. The amplitude is equals to '2pi/lambda' (lambda is the effective wavelength in the specific medium);

Its direction is perpendicular to the 'wavefront'-a virtue surface of equal phase. For an ideal plane-wave, the direction of the k-vector is always perpendicular to the 'plane', regardless of its location. So, the simple answer to your question is 'the k-vector of a plane-wave does not dependent on its position.'

However, It is noteworthy that the perfect plane wave (requiring infinite space and energy) is not existed in nature. The realistic 'plane-wave' like beam is usually a gaussian beam with very small divergence angle. Only in its waist the direction of k-vector is uniformly following its optical axis in the transverse plane; otherwise, the k-vector will have some divergence angle when away with its center.
 
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Yes, and I just believe in the real world almost any time there is something bothering the process of a wave. So to make the most appropriate answer we want to get, I think We should take as much as info into consideration. So to get the result, whether it has been changed from what we expected it to do, observing the instantaneous change os the time we want is one of a ways to reach the answer.
 
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What will happen to the wave vector of a plane wave if the wave is diffracted?
 
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ecastro said:
What will happen to the wave vector of a plane wave if the wave is diffracted?

This question is very ambiguous, as diffraction is a very complicated process, normally you need to specify your condition.

But in general, the diffracted beam from a plane-wave is no longer a plane-wave. You can consider the diffraction as the interference (superpositions) from a sub-light-sources on the object (or hole, where the diffraction is induced). To find the wave-vector, the wave-front (equal-phase-surface) must be calculated, and plot the normal line to this surface.
 
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