Wondering about the origins of light?

[jeebs]

Atoms can absorb and emit light of specific wavelengths by electron transitions between their quantized energy levels. I believe Hertz or Lorentz first thought of the electrons in an atom as being bound to the nucleus by springs that allowed dipole oscillations of the electrons and nucleus about some position.
What I do not understand is how an electron and nucleus wobbling back and forth produces an electromagnetic wave (doesn't a wave need a medium to propagate, and the proposed "aether" medium has been dismissed?), and why, if the effect that a point charge has on an object falls off as 1/r^2, this EM wave can propagate unimaginable distances in a vacuum without dimishing at all?
I mean, if we picked some position in space near an oscillating electric dipole, we would observe the electric field fluctuating up and down at that point. How come we can go to a further distance away and notice the exact same fluctuation at the same strength just at a later time given the 1/r^2 dependence of the electric field around a point charge?

Oh also, apparently photons can also be produced when particles collide or decay. What is the explanation of this, seeing as I can't see where any electric dipole oscillation would be occurring in this siutation?

Thanks.

 

[Dr Lots-o'watts]

As Stephen Hawkins said in his famous book, physics explains "how", it's the philosopher's business to answer "why".

 

[jeebs]

well that's cleared that up...

 

[Dr Lots-o'watts]

Unlike the case with sound waves, there is no medium for lightwaves. You just have to accept it. See Michelson-Morley's experiment to understand why aether has to be dismissed.

From what I've studied and read, all I've seen are equations that describe how light propagates. Equations don't say "why". Maybe someone else can better answer your questions, though.

 

[The_Duck]

If you are far away from an oscillating dipole you will *not* see the same amplitude of field oscillations as if you were up close to it. The amplitude of the field oscillations from an oscillating dipole falls off as 1/r, and the energy density of the radiation falls off as 1/r^2, with r being the distance of the source. If there was indeed no fall-off each star in the night sky would appear as bright as the sun.

EM waves don't need a medium to propagate, unless you think of the electric and magnetic fields themselves as a sort of medium.

Oh also, apparently photons can also be produced when particles collide or decay. What is the explanation of this, seeing as I can't see where any electric dipole oscillation would be occurring in this siutation?

These are quantum mechanical processes and I think you will not find a satisfying answer from the perspective of classical electromagnetism.

 

[granpa]

http://en.wikipedia.org/wiki/Near_and_far_field

 

[haael]

I mean, if we picked some position in space near an oscillating electric dipole, we would observe the electric field fluctuating up and down at that point. How come we can go to a further distance away and notice the exact same fluctuation at the same strength just at a later time given the 1/r^2 dependence of the electric field around a point charge?

This is the very core of quantum mechanics. If you like de Broglie-Bohm, then this [tex]1/r^2[/tex] is the pilot wave and the photon is the actual particle. I don't like dBB, I'm rather into multiverse and I say that the concept of "particle" is not fundamental, it is rather an approximation of something else.

 

[calhoun137]

The question "why does an electron emit a photon" is very puzzling indeed! This is the mystery of quantum mechanics, it's the same thing as asking "why does an electron land on this or that part of the wall". It's just the probabilistic nature of quantum theory.

But I think you are asking a deeper, physical question about the nature of light. At the most basic level, our theory of quantum fields tells us that electrons and photons are fundamental particles. In fact, there are two types of fundamental particles as you surely know, according to integral (photon type) and half integral spin (electron type). The electron type particles interact with each other by exchanging photon type particles. That's just the rules of the theory to make everything match experiments; so the "why" has to do with matching the results of experiments, but also searching for mathematical beauty in the spirit of Dirac.

Furthermore, the electromagnetic field is just three space where we treat each point as a quantum oscillator. The vector potential A(x,y,z,t) IS the electromagnetic field, and this field originates from the symmetry that we can always change our wave functions by an arbitrary phase, even when this phase depends on a spatial coordinate! It's easy to plug in a phase f(x) which depends on the spatial dimension x, into Schrodinger equation for the motion of an electron and to see that in order to compensate for new phase, we must introduce the vector potential.

A gauge transformation (local phase transformation) is sort of like an abstract coordinate transformation, so instead of thinking about the motion of the electron what we do is think about all the possible local phase transformations, and the vector fields that gauge symmetry induces. This process is very deep, mathematically. It involves primarily Lie Groups/Algebra's and Differential Geometry.

 

[haael]

A gauge transformation (local phase transformation) is sort of like an abstract coordinate transformation, so instead of thinking about the motion of the electron what we do is think about all the possible local phase transformations, and the vector fields that gauge symmetry induces. This process is very deep, mathematically. It involves primarily Lie Groups/Algebra's and Differential Geometry.

It appears little less deep, when you realize that this is just accounting the fifth dimension.

 

[jeebs]

damn... I can't wait until i know enough physics to really understand/appreciate half of this stuff... I get the feeling that could be years away . I think the very first time I've even heard de Broglie-Bohm mentioned was in a lecture I had earlier today.

 

[Chopin]

The first part of your question can be answered using only classical electrodynamics, without involving any quantum theory. The "medium" through which an electromagnetic wave propagates is the electric and magnetic fields themselves. A disturbance created in the fields will spread out from its source like ripples on a pond. This occurs because changing the electric field causes a change in the magnetic field, and vice versa. These two effects interact with each other to form a disturbance which essentially pushes itself outward through space.

The initial disturbance which gets the wave started is due to the movement of the dipole. Any time charge moves from one point in space to another, the resulting current causes a disturbance in the field, which initiates the wave, much as plucking a string causes vibrations, which continue after the plucking is done. All of these effects are described mathematically by Maxwell's Equations , which (along with the Lorentz equation) contain everything there is to know about classical electrodynamics.

Solving these equations reveals that even though the strength of a static (non-moving) field falls off at 1/r^2, a moving disturbance only decreases as 1/r. This means it can propagate much farther away from the source. Imagine a rope that stretches out in front of you--if you pick up one end, it will dangle down, and touch the ground not far away from you. But if you shake it, the energy you put into the rope will cause a wave which can move much farther away from your hand before dissipating. In the same way, an electromagnetic wave diminishes much more slowly than a static Coloumb field. This is the "far field" mentioned by granpa above.

Your second question, regarding the emission of photons during collisions and decay, cannot be explained by classical electrodynamics. For that you need quantum electrodynamics (QED). I'm not very familiar with that area, so somebody else will have to answer that part.