# Photons as a wave traveling in a vacuum

1. Sep 8, 2007

### Mr. Gus

I'm extremely uber layman, so forgive me for any ignorance or stupidity and my excessive use of the word "so".

I'm failing to understand either what light is or how it can travel through space (or both). I read that light is photons, and photons are particles that have no mass, and are a wave. So my understanding of that is that photons are not so much physical substance as they are energy that effect some aspect of physical substance.

So if it was a wave as in sound, it causes vibrations in the air of our atmosphere. A literal wave or ripple comes from vibrations of the surrounding water. So if photons as a "wave" are directly analogous to sound or a ripple in a pond in how they work, then that means that there is some sort of medium or atmosphere that they are effecting, and it is this effect on whatever medium or atmosphere that exists that we are measuring and referring to as photons and light.

So if this above is more or less correct, and space is really an empty vacuum apart from the odd rocks and debris, how does light get transmitted by nothing for millions or billions of years? Is there really something there that is transmitting light waves? Or is light really a physical something that is just moving really fast, and my lack of a proper scientific physics-based understanding of what mass really is is messing me up?

2. Sep 8, 2007

### meopemuk

Photons are particles - tiny indivisible objects that can be counted. In this respect they are not much different from other known particles, like electrons, protons, etc. Photons have zero mass, but this is not that important. Rules of relativistic quantum mechanics permit existence of massless particles.

As all other particles, photons obey rules of quantum mechanics. These rules basically say that propagation of particles in space is (to a certain degree) random. Generally, it is not possible to predict at which point in space the photon will be found if we decided to measure its position. Such a prediction can be done only probabilistically. Quantum mechanics provides rules for calculating these probabilities. According to these rules, the probability distribution should be obtained as a square of photon's wavefunction, which obeys a wave equation. This is where all wave properties of photons (diffraction, interference, etc.) come from.

The wavefunction should not be imagined as some kind of "substance" propagating in space. Actually, wavefunctions are just mathematical abstractions, whose only purpose is in assisting our calculations of probabilities of observed physical events (like detection of photons at a certain position in space). Since wavefunctions are not material "substances", there is no point to invent any "medium" for their propagation.

Eugene.

3. Sep 8, 2007

### olgranpappy

everybody just loves to jump into the whole "photon" discussion, don't they.

you might rather start off by thinking about the old ho-hum description of light as an electromagnetic wave... since that's how light, as we normally experience it--I.e., the stuff that goes into your eye that you actually see--is most conveniently described.

a charged particle gives off an electric field which puts forces on other charged particles, it the charged particle shakes and shimmies about then the electric field shakes a bit too, and if the shaking is in such a way that a wave can be produced (like the way that electrons in an antenna shake) then the shaking of the electric field can propagate out like a wave--it is a wave--an electric field wave, and that's light.

I admit that the above is a little vague, and if you really want to understand light you have to understand Maxwell's four equations which I have cutely unified into one equation for ya (forgive me, I couldn't resist, I still a bit tipsy... gaussian units btw):

$$(\nabla\cdot\vec E-4\pi\rho)^2+(\nabla\cdot\vec B)^2+(\nabla\times\vec B-4\pi \frac{\vec j}{c}-\frac{\partial \vec E}{c\partial t})^2+(\nabla\times E-\frac{\partial B}{c\partial t})^2=0$$

cheerz

4. Sep 8, 2007

### meopemuk

This continuous electromagnetic wave representation does a good job at explaining properties of high intensity light, which we experience in our everyday life. What about extremely low intensity light consisting of one or two photons? How can you reconcile its apparently discrete nature with your continuous wave description?

Obviously, your classical electromagnetic wave description is not valid in the low-intensity regime anymore. It should be replaced by a quantum theory. However, we know that the two-slit interference picture has exactly the same shape whether it was created by a high-intensity lightbulb or by photons released one-by-one. If we accept your logic, we should conclude that the same effect (interference) is described by two completely different theories in the high-intensity and low-intensity regimes. In the high intensity regime, the interference is due to vector addition of vectors E and B from different parts of the wave. In the low intensity regime, the interference is due to addition of complex wavefunction amplitudes.

It is more economical to have a single theory for all intensity levels. The quantum approach is general enough to cover both (high and low) limits. Then the question is: how classical vectors E and B are expressed through the quantum photon wavefunction? I don't have a good answer for this question yet.

Eugene.

5. Sep 8, 2007

### olgranpappy

well, the OP asked about "light" which commonly refers to photons en masse and at eV scale energies...

Oh, I wouldn't call it "my" description. It belongs to men much greater than I.

nor I. Something to do with coherent states, etc, maybe...

certainly I can write down a free-field expression for an electric field operator E and then consider expectation values
$$<E>$$

Cheers.