3

Here is a typical representation of light waves using wavefronts. In the diagram it depicts the peak as the lines and the throughs as the space half way between consecutive lines.

Considering light as individual photons, we see that it travels not unlike sound wave, with regions of concentrated particles and regions of no particles. This is by definition a longitudinal wave.

Of course this issue disappears when we consider light as an EM wave. But that feels cheaty. Like saying A is B because we define it so.

I think the problem with my understanding is that I'm not seeing the whole picture of light waves. Any enlightenments would be greatly appreciated.

  • When you say light waves have "regions of concentrated particles and regions of no particles" that's a dramatic over-simplification. Further photons don't interact with each other, so forces between the photons are not the mechanism for the wave oscillation as forces between particles are the mechanism for sound wave oscillation. (Also the spacing between peaks or troughs in the photon density are half the distance that we usually call the wavelength in an EM wave) – The Photon May 10 '22 at 15:32
  • Never mind that you can't actually localize a photon to any particular "region of concentrated particles" in the wave. – The Photon May 10 '22 at 15:33
  • Ok it's kind of funny how @ThePhoton himself answered my question. Anyways, can you elaborate on what you mean in "Also the spacing between peaks or troughs in the photon density are half the distance that we usually call the wavelength in an EM wave"? Thank you. – joshua mason May 11 '22 at 06:01
  • the photon density is proportional to the energy density in the wave. So photon density is high both when the E field is strong in the "+" direction and when the E field is strong in the "-" direction. Therefore the distance between peaks in the photon density is half the distance between positive peaks in the E-field strength. But the distance between E-field peaks is what we usually call the wavelength of the wave. – The Photon May 11 '22 at 14:56

2 Answers2

2

The confusion here is about a direction of a wave versus its polarization. Even sound waves can be longitudinal and transversal, depending on whether the particles in a wave oscillate along the direction of the wave propagation or perpendicular to it. Such transversal waves cannot exist in liquids and gases, but necessarily present in solids. A string of a musical instrument is also a good example of sound wave that is not longitudinal.

Electromagnetic wave (in vacuum) has its magnetic and electric fields oscillating in directions perpendicular to the wave vector (i.e., the direction of the wave), hence it is a transversal wave. This is not always the case for electromagnetic waves in media, where longitudinal component may emerge.

The Wikipedia article on Polarization formulates this as:

Polarization (also polarisation) is a property applying to transverse waves that specifies the geometrical orientation of the oscillations.1[3][4][5] In a transverse wave, the direction of the oscillation is perpendicular to the direction of motion of the wave.[4] A simple example of a polarized transverse wave is vibrations traveling along a taut string (see image); for example, in a musical instrument like a guitar string. Depending on how the string is plucked, the vibrations can be in a vertical direction, horizontal direction, or at any angle perpendicular to the string. In contrast, in longitudinal waves, such as sound waves in a liquid or gas, the displacement of the particles in the oscillation is always in the direction of propagation, so these waves do not exhibit polarization. Transverse waves that exhibit polarization include electromagnetic waves such as light and radio waves, gravitational waves,[6] and transverse sound waves (shear waves) in solids.

An electromagnetic wave such as light consists of a coupled oscillating electric field and magnetic field which are always perpendicular to each other; by convention, the "polarization" of electromagnetic waves refers to the direction of the electric field. In linear polarization, the fields oscillate in a single direction. In circular or elliptical polarization, the fields rotate at a constant rate in a plane as the wave travels. The rotation can have two possible directions; if the fields rotate in a right hand sense with respect to the direction of wave travel, it is called right circular polarization, while if the fields rotate in a left hand sense, it is called left circular polarization.

Comment
Let us consider a wave of an arbitrary nature for a vector quantity: $$ \mathbf{A}(\mathbf{x},t)=\mathbf{A}\cos(\mathbf{k}\cdot\mathbf{x}-\omega t). $$ The wave front is the surface where the quantity is the same, i.e., where the value of the argument of the cosine function is the same: $$ \mathbf{k}\cdot\mathbf{x}-\omega t=c $$ It is easy to see that this surface propagates with velocity $$ \mathbf{v}_{ph}=\frac{\mathbf{k}\omega}{k^2}, |\mathbf{v}_{ph}|=\frac{\omega}{k}. $$ (this is a so-called phase velocity, not to be confused with the group velocity, see here) The direction of this velocity is the same as that of $\mathbf{k}$, but completely unrelated to that of $\mathbf{A}$, which depends on the nature of the wave. By analyzing the derivation of waves (or wave equations) in different media, one can see that in liquids or gases, where the layers of matter are moved by excess or lowering of pressure, the direction of the particle movement can be only in the same direction as $\mathbf{k}$. However in solids, the existence of a shear stress leads to the possibility of transversal waves. Same is true for EM field, as can be seen from the Maxwell equations.

To restate the point differently: the particles (or the fields in the wave) are not moving along with the wave - they are oscillating near the same equilibrium position. The propagation of a wave has to do with the propagation of the phase surface (the "wave front"), not with the actual direction of motion of the matter or fields.

Roger V.
  • 58,522
  • I'm not convinced about your statements concerning sound. The wave produced by a transversely vibrating string propogates longitudinally (though the string's vibration is a superposition of waves travelling along the string). Even in a solid, unless you can show an atom or molecule vibrating transversely passes energy longitudinally via inter-molecule bonding, the sound wave itself is still a longitudinal wave. – Carl Witthoft May 10 '22 at 12:50
  • 1
    @CarlWitthoft You can just take a book on continuum mechanics/elasticity and convince yourself that there are three polarization for waves in solids: two transversal and one longitudinal. Same is true on molecular/atomic level: phonons have three polarization. See also the quote from Wikipedia that I added. – Roger V. May 10 '22 at 13:02
0

Considering light as individual photons, we see that it travels not unlike sound wave, with regions of concentrated particles and regions of no particles. This is by definition a longitudinal wave.
Great observation. In fact, this is exactly how a radio wave works.

In an antenna rod, surface electrons are periodically moved back and forth. With each acceleration, the electrons resist and emit photons. So a periodically rising and falling stream of photons arrives at the receiver.
However, this is not a longitudinal wave in the classical sense, because the definition of a wave is a transport of energy through the periodic oscillation of particles, which in the end remain in the same place.

And now it gets interesting. The radio wave also shows us the properties of the individual photons, which are difficult to measure on the single photon. This is because the photons leave the antenna rod in a polarised state. For a radio wave in the near field, we measure an electric polarisation parallel to the antenna rod and a magnetic polarisation perpendicular to it. It looks like this:

enter image description here

There is nothing to stop us from deducing the individual properties of polarised photons from their macroscopic properties. Each photon is a quantum of energy moving forward in vacuum at the speed of light, periodically converting an electric dipole into a magnetic one and vice versa, both 90° to the direction of motion and 90° to each other.


Two more comments on this.

  1. the electron is not only an electric charge, it is also an intrinsic (invariant under all circumstances) magnetic dipole. The electron inherits both fields on the photon during photon emission.

  2. Phonons and their varieties of resonances in materials are now a widely ramified field of research. It is therefore surprising that the interaction of the surface electrons of an edge with the two field components of the photons grazing it is completely ignored in the generation of intensity fringes.

HolgerFiedler
  • 10,334
  • If we're talking about a photon "moving forward in vacuum", why do we compare it to a near field field pattern rather than far field? Because the near field components are out of phase, they don't "move forward in vacuum". – The Photon May 11 '22 at 17:40