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Not HW: Why are light waves in the form of the sine wave, instead of some other wave?

by physixer
Tags: form, light, sine, wave, waves
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Bob S
#19
Feb7-10, 07:25 PM
P: 4,663
Quote Quote by kcdodd View Post
If a sine wave is a solution, then any wave is a solution, due to Fourier decomposition. A photon can be any shape. There seems to be 'something' special about sine waves, and it'd be interesting if someone had a physical reason why the universe decided not to quantize in terms of square, triangle, or teddy-bear waves...
I thought all the atomic transition lines, like the hydrogen transition 4d->2p, was a sine wave. See (Enter H for spectrum and 4000 to 6000 A for range):

http://physics.nist.gov/PhysRefData/ASD/lines_form.html

Here is the Fourier decomposition for a square wave.

http://mathworld.wolfram.com/Fourier...quareWave.html

Note that all the components of the square wave are sine waves. If any atomic transition line were a square wave, there would have to be the frequency harmonics sin(nωt) for n= 2, 3, 4 etc. But n>1 would violate the requirement that the transition energy is fixed at a single value. Are there any visible lines in light i.e., (photons) that are square waves, and have the necessary frequency harmonics?

Bob S
kcdodd
#20
Feb7-10, 08:56 PM
P: 192
Thats what I meant by quantize in terms of sine wave frequency. An energy transition in an atom gives a single energy, and that corresponds to a single sine wave frequency. I am fairly sure all properties eventually derive from this idea, as in dispersive media, involving photons interacting with the atom. Of course, it is not perfect because photons do not immediately occupy the entire universe. Due to uncertainty principle all photons would have a distribution of energy/momentum (ie frequencies). The uncertainty probably comes from the final kinetic energy of the atom imparted when the photon kicked out (conservation of momentum).

Now, you could artificially make a 'square' photon. Imagine it travelling along and someone traps it in a box. Well, it's shape has to conform for it to be inside. However, since we always quantize energy by sine waves, for whatever reason, all the modes are sine waves. But you can still get something fairly square like given enough energy bandwidth. At least, I think so...
Born2bwire
#21
Feb7-10, 09:26 PM
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Quote Quote by Bob S View Post
If you keep on putting neutral density filters in a visible light beam, you will eventually get to a point where there are only a few photons per second. These individual light photons have to satisfy the two Maxwell curl equations, and if you substitute one into the other, you will get (without conduction)

curl curl E+ ε0μ0E = 0

or del2E - ε0μ0E = 0

A sine wave is one of the few (only?) function that satisfies this wave equation (without attenuation).


I would like to believe this, but then if we look at the curl equation

curl E = - μ0H/∂t

E and H appear to be 90 degrees out of phase (E=-μ0H in EE parlance), just like in the Faraday induction Law..

Bob S
The curl of the electric field would bring out a -jk factor on the left-hand side cancelling out the imaginary factor on the right-hand side.
Bob S
#22
Feb8-10, 11:16 AM
P: 4,663
Quote Quote by Born2bwire View Post
The curl of the electric field would bring out a -jk factor on the left-hand side cancelling out the imaginary factor on the right-hand side.
Thanks. You are absolutely correct.

Now we have a slight ambiguity.

Faraday's law is commonly written (in air)

[1] E·dl = -(d/dt)μ0H·n da

which is closely related to the EE equation

[2] V = L dI/dt

where, if E(t) = E0sin(ωt), clearly E (and V) are 90 degrees out of phase with H (and I).

Using Stokes Law for the vector E

[3] E·dl = ∫(curl·E)·n da

which, by comparing to [1], leads to the Maxwell equation

[4] curl E = -μ0H/∂t

It was established in an earlier post that in [4], E and H were in phase and curl·E and H (and hence E) were 90 degrees out of phase, while in [1] E and H are 90 degrees out of phase.

Furthermore, in [2] (Stokes Law) we have a vector E which clearly is in phase on both sides of the equation, implying that E and curl·E are also in phase. How can this be?

Bob S

[added] After [2], change equation to read E(x,t) = E0(x)·sin(ωt).
Before [4], add where E(x,t) = E0·sin(ωt - kx)
kcdodd
#23
Feb8-10, 01:50 PM
P: 192
I think perhaps when you integrate E under stokes theorem and you draw your loop, it is over differing spatial locations, and so differing phase of E.
bjacoby
#24
Feb9-10, 03:58 AM
P: 132
Quote Quote by Bob S View Post
If you keep on putting neutral density filters in a visible light beam, you will eventually get to a point where there are only a few photons per second. These individual light photons have to satisfy the two Maxwell curl equations, and if you substitute one into the other, you will get (without conduction)

curl curl E+ ε0μ0E = 0

or del2E - ε0μ0E = 0

A sine wave is one of the few (only?) function that satisfies this wave equation (without attenuation).


I would like to believe this, but then if we look at the curl equation

curl E = - μ0H/∂t

E and H appear to be 90 degrees out of phase (E=-μ0H in EE parlance), just like in the Faraday induction Law..

Bob S
1. Mathematics is not more real than reality!

2. A sine wave is a mathematical construct which does not exist in reality. All real waves start and stop at some point making them a pulsed function. Such "modulation" creates a broadened bandwith and a "pure" single frequency is hence impossible.

3. Also even in a pure single frequency laser ignoring the pulsed nature of the output, there is the additional problem of limited coherence. This essentially amounts to phase modulation which like amplitude modulation broadens the frequency spectrum of the light.

4. What you guys are doing is deriving that the uncertainty principle is at work here. It is related to Fourier transform relations that are mutually exclusive.

5 Hence, it follows that your sine wave solution to the wave equation is correct but totally abstract. It is a mythological construct with no basis in reality. Mathematics is NOT more real than reality!


Part II.

Sorry. Magnetic and Electric fields in EM radiation are in phase. And No, they DO NOT "create each other".

Nor is Faraday's law correct as usually expressed where a changing magnetic field creates an electric field. Bzzzt. Wrong. The truth is that Maxwell's equations as usually expressed are NOT causal relationships! Yes, one side of the equation EQUALS the other side, but one side does not CAUSE the other side! The cause of BOTH E and B fields is CHARGE and it's variations. Hence in Faraday induction a changing electric current (charge) creates BOTH the changing magnetic field AND the electric fields resulting in EMF. Likewise for EM radiation the charges simultaneously create BOTH Electric and Magnetic fields which are propagated out through space from the source together and as it happens, in phase.


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