# Waves in spherical emission

1. Dec 7, 2007

### kcodon

"Waves" in spherical emission...

Hi all,

I just thought of this and wondered why it had never occurred to me before. In any source emitting EM radiation (a lightbulb for example), it does so all around it, so a sphere of EM radiation propagates out from the source. However how in the world do we know consider this EM radiation as sinusoidal waves? Wouldn't waves be interfering with eachother, causing a whole mess, and you'd never actually get a nice sinusoidal waveform would you? I hope you see what I'm saying here, cause its confusing the buggery out of me

Its almost as if one has to consider the EM field propagating outward in the spherical manner, and the EM wave traveling through it as compressions and rarefractions like sound. One can't really make EM waves analagous to water waves can they, as water wave occur in a 2D plane so to speak...the surface of the water. If you then have a 3D "plane", waves don't travel like this. In fact a 3D "plane" for water would be simply sound underwater, thus does not this seem to support EM radiation being like sound in a sphere expanding at c?

Whenever we are shown waves in books etc, its a sinusoidal wave, however how one gets a sinusoidal ray from a sphere of EM, I've got no idea. Photons looks oh so amazingly appealing right now!

If anyone could spread some light on this it'd be much appreciated,

Kcodon

PS just clarifying, do EM waves self propagate, i.e. they aren't ripples in an EM field...they make their own EM field so to speak. Correct or not?

2. Dec 8, 2007

### ZapperZ

Staff Emeritus
Er... you might want to look at how, for example, the schrodinger equation is solved in 3D (after all, that is a 3D wave equation). You'll see the radial part having a solution in the form of a 3D Bessel function.

Zz.

3. Dec 8, 2007

### kcodon

Thanks Zz,

Um maybe I'll have to worry about this one at a later date...not quite up to scratch with the schrodinger equation or 3D Bessel functions lol. Other than that I'm guessing there is no such physical explanation for it...just mathematical?

Kcodon

4. Dec 8, 2007

### ZapperZ

Staff Emeritus
I think when you deal with special functions and you come across 3D Bessel functions, you might see such solutions in the situation you asked. As a preview, if you pick up Mary Boas's text "Mathematical Methods...... ", she explicitly does the 2D wave equation (such as a drum membrane) using such functions. This should give you a taste of the kind of wave you're asking for.

Zz.

5. Dec 8, 2007

### kcodon

Okey dokey I'll look into that,

Thanks again,

Kcodon

6. Dec 8, 2007

### cbacba

It might help to think of the wave as a dent in space traveling through the medium like a peak and trough of an ocean wave. Energy is being transferred by that 'dent' but no material matter is moving from one place to another, rather the matter is shifting up and down (in the water wave). One thinks of the sine waves as that is going one when creating the wave - such as a radio transmitter and the sine wave represents a pure or single frequency. For the wave propagation, special relativity demands that time is stopped when moving at the speed of light so there is no oscillation happening in the wave itself. Rather the oscillations appear to happen at a fixed location as the wave ripples past due to the fact it is traveling at a velocity (c) and it has a wavelength which is a separation of these 'dents' which are the presence of an electric field perpendicular to a magnetic field.

Sound waves are minor pressure variations traveling through the medium. These occur as slight movements of the molecules in the direction the wave is traveling. EM waves travel in a vacuum through space itself, hence no aether (or ether). They are the electric and magnetic field which are at right angles to each other and to the direction of propagation.

The EM waves form from a rest state where there is an oscillating electric field perpendicular to an oscillating magnetic field - a radio antenna. Again, special relativity demands that these are identical in nature to light waves that consist of detectable photons as if one were to travel at a very high velocity towards the source of a shortwave broadcasting station, the transmission would shift to much higher frequencies/shorter wavelengths including to the point where it appeared as light if the velocity was sufficiently high.

Note too that in radio, the concept of isotropic antenna is often used yet there is no such thing. One cannot construct an antenna to radiate in all directions equally. It provides a baseline for referencing the gain of an antenna in a particular direction.

With light, such as an incandescent, you have electricity transferring power to a small object inside to heat it up. The light follows a blackbody curve of intensity versus wavelength based upon the temperature of that object. It is all frequencies (up to a point) meaning all colors, all polarizations and all phases. It is what is described as incoherent light. A laser creates a coherent beam of light where the phase is common as is the frequency (or wavelength). This is where the name comes from. Energy goes into the material (like helium neon gas) and raises the internal energy state of the electrons in the atoms. An effect, stimulated emission, occurs where a nearby passing photon causes a higher energy state atom to transition to a lower state and emitting another photon with the same wavelength, the same phase and the same polarization, traveling in the same direction. The result is a single frequency (or wavelength) light source that is coherent.

You'll note there is tremendous differences in the appearance of something like the usual pocket pointer laser and a typical incandescent flashlight. That's because there's allsorts of interferences and shifts of phase that make an incandescent light totally chaotic with no ability to distinguish things about it such as a laser permits. Assuming someone could figure out how to draw such an example, one would never manage to recognize anything of it by looking at the drawing and the situation would change nanosecond by nanosecond.

As for sinewaves (or cosines) those are our representation of a single frequency wave. If a wave is not a simple sine or cosine, it's not monochromatic. If a wave isn't close to this but is rather something more commonplace in the real world, you wouldn't be able to describe or comprehend it due to the complexity. However, waves are going to behave so that the complex waves can be formed from the superposition of many simple waves, like the sound of a flute or trumpet by the combination of pure one frequency tones or the sound of a symphony orchestra from the combination of 50 or 100 different musical instruments.

7. Dec 8, 2007

### Staff: Mentor

The EM from a lightbulb is incoherent. This means that you don't see nice sinusoidal waves and you get a "whole mess".

However, you should be aware that sinusoids are not the only solutions to a wave equation, just the simplest to understand. If you had an EM generator that made a nice sharp EM pulse that pulse would propagate outward, not as a sinusoid, but as a pulse.

8. Dec 8, 2007

### Staff: Mentor

It's simpler to consider the radiation from an oscillating dipole, e.g. a simple transmitter for radio waves or microwaves. This page has a derivation of the E and B fields for an oscillating dipole (scroll down to equations 1090 and 1091 to see the final result):

http://farside.ph.utexas.edu/teaching/em/lectures/node94.html

9. Dec 8, 2007

### billiards

The wavefront is by definition a "surface" of constant phase. Huygen's principle stipulates that each point in space along the wavefront, may be considered as a source of a new wave front, all these sub-wavefronts will destructively interfere except in the direction the larger wavefront is propagating. Therefore the larger wavefront can be considered as the sum of lots of smaller wavefronts, this can be helpful when considering the effects of scatterers, although it is only valid under the condition we are dealing with high frequencies.

10. Dec 8, 2007

### Staff: Mentor

You also have to consider the time domain. The individual "sources" in a lightbulb effectively each fire off short bursts of radiation, at random times relative to each other. This gets into the issue of "coherent" versus "incoherent" light, which is discussed extensively in optics textbooks.

However, I had the impression from kcodon's original post that he/she was having trouble visualizing even a spherical wave from a single coherent source. That's why I brought up the oscillating dipole. Note in particular that if you follow a straight line outward from the source in any direction, varying r while keeping $\theta$ and $\phi$ constant, you get a sinusoidal wave whose amplitude decreases as 1/r.

11. Dec 8, 2007

### Antenna Guy

We know to consider EM radiation as waves because it behaves as such. Consider Green's function:

$$\phi = e^{(jkr)}$$

At a sufficiently large constant radius from a source, the observed phase of the EM energy will be constant - and that phase will vary uniformly (and predictably) as the observation radius is varied. In essense, the sinusoid varies as a function of radius - not time.

The Fresnel (pronounced franel) region about a large EM emitter (many wavelengths in size) will appear in the seemingly chaotic manner that you describe. This occurs not because the composite spherical wave is chaotic, but because there are effectively a multitude of phase centers (each at a different radius from a particular reference point) that cause the fractional energy emitted from each to interact within this region (although those packets of energy are not actually propogating in the same direction).

The Fraunhoffer region ( $$r \ge \frac{2D^2}{\lambda}$$, where D is the aperture size in wavelengths, or $$\lambda$$ ) describes the that point, beyond which, the phase terms from the multitude of sources approach a common change in phase with change in radius from a common phase center. The radiation pattern (as a function of spherical angles) no longer changes significantly with radius in the region.

Well, if those water waves were to propogate at the speed of light, their amplitude at constant radius from the point of excitation would no longer vary with time. However, since water waves do not propogate at the speed of light, the wave's amplitude at a constant radius still varies as a function of time.

A plane in 3D is no different than a plane in 2D (i.e z=0). However, you bring up a common misconception: EM waves are not planar - they are sperical. Always. They exist on a sphere of constant radius/time.

Consider, if you will, an electron orbiting an atom. That electron exists on a shell of constant energy about the atom. If this electron were to move to a lower energy shell, a photon is released to conserve energy. This photon propogates outward equally in all directions at the speed of light - compressing the spatial dimension representing the shell's diameter to zero (radially) by Lorentz contraction. At any given point in space, that pseudo electron shell (representing the change in energy) can then be pojected on a sphere of constant time with respect to the emitting atom. The phase of this pseudo-electron orbit does not vary with time at our given point, but it will vary with radial distance from the emitter. The entire orbit of this pseudo-electron can then be construed to exist at a point (or "patch") on a spere - only by varying radius can you "observe" the sinusoidal amplitude of the wave as its' complex amplitude varies through 360 degrees (one wavelength).

I hope the above was at least somewhat "illuminating". :)

For more background, you might consider looking for info regarding how nearfield antenna ranges work. Spherical, in particular.

I'm not sure I understand your question, but I'm inclined to answer "yes". I'm of the opinion that the propogation of EM waves are the causal equivalent of the past of one reference frame finding its' way into the past/present/future of another.

Regards,

Bill

12. Dec 8, 2007

### billiards

The water waves you are talking about are surface waves, these travel at a group velocity that is much much less than the speed of sound in water. These surface waves are in no way analagous to the EM waves you are talking about. Surface waves require a free-surface, I'm not sure what would constitute a free-surface in EM.

"Sound underwater" is a much more reasonable analogy, this would propagate spherically much like the EM wave.

13. Dec 8, 2007

### cbacba

A short EM pulse would imply the presence of all frequencies in-phase initially. Such can be simulated by combining a virtually infinite series of single frequency sinewaves. Good luck with recognizing anything or comprehending it in much of any meaningful fashion.

14. Dec 8, 2007

### kcodon

Wow thanks everyone, though I fear it was somewhat ignorant of me to ask the question without wanting the maths

Firstly thanks billiards for the Huygens idea, and remember that from a while back, and didn't realise it applied (or could be applied to) wave propagation as well as reflection etc.

Thanks jtbell, I'm sure you're link would help though the maths is a bit over my head. However when you said:
You were correct, I only used the light bulb as an example and I should specify that I meant monochromatic light, all in phase etc.

And thanks Antenna Guy for the effort but most of that was way over my head. However I found this interesting...
So theres no such thing as an EM "wave" so to speak? This was kind of my original question. Does that mean that any "wave" is actual an increasing sphere so to speak? Then my question was how does one get a sinusoidal wave from a sphere or vice a versa? And then a photon I believe is not spherical...but I think you explained this later on, I just didn't understand. This I also found interesting...
Do you mind elaborating?

And thanks billiards again:
So this is what I'm trying to picture here, is the EM wave like the sound wave underwater, in that the compressions/rarefractions are the changing EM fields? Is it only when one measures the amplitude of the EM fields at a specific point does one get the sinusoidal wave? If this is correct it would answer my question. Look at the image I attached. Now is this how the coherent monochromatic source emits waves...i.e. each photon a wave, and then it turns into a mess as shown. Or does it do as I said before, and is the alternations in the EM field analogous to sound in water sort of?

Also just randomly, to do with wave-particle duality...could one treat light as always a particle, it just appears to be a wave, in that the "wave" shape as we've kind of been discussing before, is the probability amplitude of finding the photon there, due to alternating EM fields or the like? Or another version...hmmm I'll try write this understandably. For a photon, E=hf. However could the f represent the "rms" frequency so to speak, analogous to the rms voltage in ac circuit. So this would mean that the photon has energy varying with respect to time. Then as in a wave the amplitude squared is the energy, could not the varying energy of the photon with respect to time? I'm pretty sure this idea isn't correct, but if the photon energy varied with time, could it maybe explain to some degree the uncertainty principle, in that if we assume the energy is constant, when it actually isn't? I think I've confused myself here so feel free not to reply to my ravings

Thanks anyway everyone,

Kcodon

15. Dec 9, 2007

### kcodon

Oops sorry I just realised I'm away all next week - somewhat disorganized yes - so can't reply, however I look forward to hearing what everyone has to say!!

Kcodon

16. Dec 9, 2007

### p764rds

One Photon in a spherical wave

Yes, yes, good question. To make it even more difficult, lets say we start one photon off in a 3D deep space. For example a molecule floating around in deep space that is made to emit a single photon by a well aimed photon shot from our remote control iPhotonEmitter. We could then place a telescope on the spherical suface swept out by the wave at distance 100 miles from its center. I assume a telescope anywhere on that huge spherical surface would detect it. How did one photon manage to spread itself out over 30000 square miles? If it were a wave, then when one portion of the wave hits our telescope does all the rest of the wave over the sphere disappear suddenly? Does Noethers Theory or Gauge Symmetry even know?

17. Dec 9, 2007

### Antenna Guy

To my knowledge, average energy density falls of with $$\frac{1}{4\pi r^2}$$ , and, presumably, $$r=ct$$. If the photon's energy is distributed uniformly in all directions (m=0 mode), the fractional energy received by the telescope would equal the ratio of the telescope's aperture area divided by $$4\pi r^2$$.

Regardles what Noethers Theory or Guage Stmmetry might imply ( I honestly have no idea anyway), the wave would not "collapse". One might completely cover the sphere in question with identical telescopes, and each one would receive the same energy (for the m=0 mode at least). Only the integration of the energy received by all those telescopes would yield the total energy emitted at $$t=-\frac{r}{c}$$ (i.e. where aperature area=surface area at $$r=ct$$ ).

Regards,

Bill

18. Dec 9, 2007

### p764rds

Only one photon to receive

But there is only one photon which must be totally absorbed
by one telescope. As far as I know, it could not be 'fractionally
absorbed' over a spherical surface area of 100 square miles
containing 5 million telescopes. I think only one of the telescopes
would get it and the rest will go empty handed, again as far as
I know (I have not done it practically)

19. Dec 9, 2007

### Antenna Guy

If you agree that the energy density falls off with $$\frac{1}{4\pi r^2}$$ , where does the rest of the energy go?

Regards,

Bill

20. Dec 9, 2007

### p764rds

Hey, thats a very good question. You've got me. I don't see
how one photon can obey that law. It easy to see how millions
could though - they just 'thin out'. But just one photon? No chance.

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