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I Is light ACTUALLY a sinsusoidal wave?

  1. Oct 22, 2016 #1
    Any function of the form f(x+ct) or f(x-ct) can be a solution to a wave equation - including the electromagnetic wave equation generated by combining maxwell's equations. Light is an electromagnetic wave which obeys maxwell's equations.

    If you were to meausure the electric field of light, is it really a sinuoid? Or is it just represented that way for convenience, and because any function can be represented by sinusoids?

    I understand that any well behaved function can be represented by a series of sine waves (resulting from fourier series or fourier transform).

    I also understand that light different colors of light correspond to different frequencies (what's the proof that these are frequencies of sine waves?).

    However say you were to measure the electric field magnitude of a light soure at an instant of time over a certain distance. What would the resulting function look like? Would it be a pure sinusoid? Would the function look different if the source were white light, or a laser source? Is there an easy experiment which you can do which would show this (or which has been done)? Or can someone provide me with a name of a textbook which shows this?
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  3. Oct 22, 2016 #2


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    Light in the form of a plane wave would be perfectly sinusoidal. However in reality light is always composed of at least a small range of frequencies and would thus be made up of a waveform that would vary non-sinusoidal.

    Lower frequency EM waves in the microwave and radio bands can be directly measured. It is trivial to show that these EM waves are sinusoidal or near-sinusoidal. Experiments have shown that light is also an EM wave and would thus behave identically.
  4. Oct 22, 2016 #3


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    If it's an ideal coherent monochromatic source you will get a sine wave. With anything else, you'll get whatever sum of sines happens to be passing at that time. As you note, the existence of Fourier transforms suggests that the function could be pretty much anything.
    Last edited: Oct 22, 2016
  5. Oct 22, 2016 #4


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    Why should a plane wave be sinusoidal? The equations for the potentials in radiation gauge read
    $$\Box \vec{A}=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
    Now assume you make the ansatz for a plane wave propagating in ##z## direction. Then you get
    $$\vec{A}=\vec{A}_0 a(t,z),$$
    leading to
    $$(\partial_t^2-\partial_z^2) a(t,z)=0$$
    with the general solution
    where ##a_R## and ##a_L## are arbitrary functions (not necessarily sinusoidal!). In fact the pure sine or cosine solutions are never realized in nature. Their energy would be infinite.

    The constraint tells you simply that ##\vec{A}_0 \cdot \vec{e}_z=0##, i.e., that you have a transverse wave.
  6. Oct 22, 2016 #5


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    I have always been confused by that aspect of the wave/particle duality. Thinking of photons one at a time, they do not interfere with each other. If reflected or refracted, each is a discrete event. But a group of photons passing in nearly the same direction at nearly the same time, should be treated as the superposition of the group.

    Adding to my confusion, the size of a photon either doesn't exist, or is difficult to define. Therefore, how can I define a time window such that two photons arriving within the window should be superimposed, but arriving outside of that window should be treated as discrete events?
  7. Oct 22, 2016 #6


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    Photons are neither classical particles (they don't even allow to define a position observable in the strict sense) nor classical waves but they are defined as certain states of the quantized electromagnetic field, the one-"particle" Fock states. Consequently in "modern quantum theory" (i.e., the theory developed in 1925/26 by Heisenberg, Born, Jordan and Schrödinger and Dirac) there is no wave-particle duality and consequently no apparent paradoxes of such an idea.
  8. Oct 22, 2016 #7


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    My explanation should only be considered in the classical sense. I have little knowledge of how quantum physics describes such phenomena.
  9. Oct 25, 2016 #8
    * I may be incorrect in some of the things I'm saying here, but I'll try to offer my own explanation.

    This is from a purely classical (non quantum mechanical) point of view. In reality we will never see a perfect electromagnetic plane wave. However, because the differential equations governing the electromagnetic field are fundamentally linear, we can decompose an arbitrary electromagnetic field as being a linear superposition of plane wave solutions to the electromagnetic field equations.

    As an analog to this problem, we can consider the problem of the one dimensional wave equation. This approximately describes the motion of waves on a string given that the waves have a small amplitude. If we hold the ends of the string fixed then it turns out that there are certain solutions to the wave equation, called standing waves, which solve the wave equation and solve the boundary conditions (that the ends of the string are held still). These standing waves are sinusoidal. Because of the linearity of the wave equation, we can add two of these solutions together and still have something that solves the boundary conditions and the wave equation. In reality, it would be essentially impossible to form a "true" standing wave, and the motion of the string would almost never be described by a standing wave solution. The utility of the sinusoidal standing wave solutions is that they form what is known as an orthogonal basis ; essentially we can describe ANY motion of the string by superimposing (adding together) an infinite number of the standing wave solutions.

    In a similar manner we can represent any state of the electromagnetic field as being a linear combination of sinusoidal plane wave solutions (with the caveat that we need to be careful about boundary conditions, we need to choose our set of basis functions in such a way that boundary conditions can be satisfied).
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