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Not HW: Why are light waves in the form of the sine wave, instead of some other wave? 
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#19
Feb710, 07:25 PM

P: 4,663

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 


#20
Feb710, 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... 


#21
Feb710, 09:26 PM

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#22
Feb810, 11:16 AM

P: 4,663

Now we have a slight ambiguity. Faraday's law is commonly written (in air) [1] ∫E·dl = (d/dt)μ_{0}∫H·n da which is closely related to the EE equation [2] V = L dI/dt where, if E(t) = E_{0}sin(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 = μ_{0}∂H/∂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) = E_{0}(x)·sin(ωt). Before [4], add where E(x,t) = E_{0}·sin(ωt  kx) 


#23
Feb810, 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.



#24
Feb910, 03:58 AM

P: 132

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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