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I find plane waves are uncompatible with light cone.

Perhaps plane waves are "virtual" and can never be measured in that case, shouldn't we call plane waves as "virtual plane waves"? (other option could be that plane waves allows waves travel faster than c)

I would like to clarify this point through this question.

If plane waves would exist(as measurable), then higher than c speed could be reached like this:

A wave going from X to Y at a speed c, it will reach Z at higher than c speed, because it will reach at same time, traveling more distance.

                   (X).
                    |   
                    v   
                  ________________________________________________  
plane waves       ________________________________________________  
going X to Y      ________________________________________________
                   (Y).                                       (Z).

In a real situation the wave will be a circle (or a sphere in 3d) so it will get Z later then that's not a plane wave.

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4 Answers 4

up vote 11 down vote accepted

A "plane wave" generally refers to an infinite and perfectly flat wavefront, which cannot exist in reality, of course. However, there is nothing at all impossible about a plane wave of finite extent. Such a wave will experience diffraction at its edges, of course, but can still propagate over long distances before losing its planar nature.

The problem with your question about faster than light information transmission, is that if X were the only point emitting a wave, then it would be a spherical wave, not a plane wave. A plane wave can be thought of as being composed as a superposition of spherical waves emitted in phase from every point on an infinite plane. So in your example, information would not be arriving at Z from X, but instead from some other source point within the causal past of Z. In your diagram that point could be at the same height as X, but directly above Z.

Considering a wavefront (planar or otherwise) as a superposition of spherical waves is the central feature of the Huygens-Fresnel Principle, which would be a good reading on the topic.

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Just to further illustrate how real a plane wave can be, any well collimated laser beam can be considered a close approximation to a true plane wave. Good optics can produce planar wavefronts that are perfect to within a nanometer or so. –  Colin K Apr 5 '11 at 15:26
    
@Colin K, Understood, in part, I still perceive a contradiction, you said "A plane wave can be thought of as being composed as a superposition of spherical waves emitted in phase from every point on an infinite plane", then you give an example of "good optics" producing a planar wavefronts, only explanation I found is that the wavefront can't never be wider than the "colimator", isn't it? thanks –  HDE Apr 5 '11 at 15:52
1  
@HDE: Yes, in reality you cannot have a plane wave that is larger than the aperture of your optics. This wave can be considered as a superposition of spherical waves emitted from every point on a finite plane, in that case. You should check out the reference I'm going to edit into my answer in a moment. –  Colin K Apr 5 '11 at 15:55
    
@HDE: Also, by "good optics" I mean optics of high quality. Less expensive optics can collimate a beam and give roughly planar wavefronts, but there are likely to be some significant deviations from perfect flatness of the wavefront due do errors in the shape of the lenses. –  Colin K Apr 5 '11 at 16:01
    
@Colin K, good +1, then is clear from this answer that a single source can't emit planar waves, only a space distributed source can, thanks –  HDE Apr 5 '11 at 16:06
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A plane wave $e^{-i(kx~-~\omega t)}$ is a case where the Fourier transform of the wave is delta function of momentum. The physical reality of a plane wave is then somewhat more in line with a mathematical “fiction.” Usually there is some envelope with the wave, such as a Gaussian packet, which attenuates the wave off to “infinity.” However, the plane wave is useful for waves with well known momentum, or a small spread or uncertainty in $\Delta k$ or $\Delta p$. Further, one can write a plane wave as an infinite sum of Bessel functions. The wave as it interacts with a hole in a screen selects out some of these components by Kirchoff’s rule, and the wave at the other side of the screen is a spherical wave front.

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As soon as your source is X, there is no wave propagation from Y to Z, whatever wave front looks like (spherical, plane, etc.).

About "virtuality":

A wave is real if it is of sufficient intensity, i.e., it consists of many-many photons. Then your detector at Y will work as well as your detector at Z.

For a low-intensity wave it is not the case and the wave describes the probability amplitude. If you have only one photon in it, only one detector will register it.

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Hello, the question state that Y and Z will receive -same signal- from X, but delayed, in case of spherical front, and the point is that if were a plane wave front they would receive the signal at same time, but it's partially answered in Colin K answer, we were discussing now the case of a collimator producing a plane wave from a non plane source, read comments in Colin K answer, thanks –  HDE Apr 5 '11 at 18:53
    
For a spherical wave front all points in this front receive the signal simultaneously but the signal comes always from the source. –  Vladimir Kalitvianski Apr 5 '11 at 18:59
    
of course that the signal comes from the source, the question was about if a source "could" generate plane waves, then the time to travel X->Y would be the same than X->Z, so as distance are different it would lead to higher than c paradox, so what I understand till now is that plane waves exist but a point source can't create them –  HDE Apr 5 '11 at 19:19
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The questioner is considering point sources. If you consider a source that is planar, you will find that, at least near the source, it produces plane waves. Of course this is all just an approximation, but, since all experiment has accuracy limitations, everything in physics is an approximation.

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