How do EM waves propagate in spherical emission?

In summary: With EM waves, its similar in that there is an oscillating electric field perpendicular to an oscillating magnetic field. However, the electric field and magnetic field are at right angles to each other and the direction of propagation. So when an EM wave hits something, the electric and magnetic fields oscillate and create photons.
  • #1
kcodon
81
0
"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 each other, 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 :confused:

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?
 
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  • #2
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
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
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
Okey dokey I'll look into that,

Thanks again,

Kcodon
 
  • #6
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
kcodon said:
However how in the world do we know consider this EM radiation as sinusoidal waves? Wouldn't waves be interfering with each other, causing a whole mess, and you'd never actually get a nice sinusoidal waveform would you?
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
DaleSpam said:
The EM from a lightbulb is incoherent. This means that you don't see nice sinusoidal waves and you get a "whole mess".

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
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
DaleSpam said:
I suppose you could consider a lightbulb to be a superposition of a large number of dipoles of incoherently varying frequency and phase.

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 [itex]\theta[/itex] and [itex]\phi[/itex] constant, you get a sinusoidal wave whose amplitude decreases as 1/r.
 
  • #11
kcodon said:
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 each other, 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 :confused:

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

[tex] \phi = e^{(jkr)} [/tex]

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 ( [tex] r \ge \frac{2D^2}{\lambda} [/tex], where D is the aperture size in wavelengths, or [tex]\lambda [/tex] ) 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.

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.

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.

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?

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.

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!

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).

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

Kcodon

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.


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?

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
kcodon said:
Hi all,
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?

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
DaleSpam said:
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.

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

Firstly thanks billiards for the Huygens idea, and remember that from a while back, and didn't realize 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:
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.
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...
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.
So there's 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...
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.
Do you mind elaborating?

And thanks billiards again:
"Sound underwater" is a much more reasonable analogy, this would propagate spherically much like the EM wave.
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 :smile:

Thanks anyway everyone,

Kcodon
 
  • #15
Oops sorry I just realized 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
One Photon in a spherical wave

Yes, yes, good question. To make it even more difficult, let's 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
p764rds said:
Yes, yes, good question. To make it even more difficult, let's 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?

To my knowledge, average energy density falls of with [tex]\frac{1}{4\pi r^2} [/tex] , and, presumably, [tex] r=ct [/tex]. 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 [tex] 4\pi r^2 [/tex].

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 [tex] t=-\frac{r}{c} [/tex] (i.e. where aperature area=surface area at [tex] r=ct [/tex] ).

Regards,

Bill
 
  • #18
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:rolleyes:)
 
  • #19
p764rds said:
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:rolleyes:)

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

Regards,

Bill
 
  • #20
Hey, that's 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.
 
  • #21
p764rds said:
Yes, yes, good question. To make it even more difficult, let's 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?

Let's say you had a 'photon drive' spacecraft that emitted 1 photon, perhaps a billion years ago and you went zooming off towards your destination at some pace that makes a snail look like light speed travel. Now the fact that a telescope camera detected that photon on Earth rather than somewhere in a galaxy that is about 45 degrees away from where your rocket propulsion was aimed. Now does that mean you wound up going to the wrong destination? Or, all of a sudden, did you translate from close to where you were headed to a far distant location 45 degrees away from your intended course?

Although emission and absorption appear to happen in quantized chunks of energy, generally, you're going to experience wave properties otherwise and I'm referring to real EM wave properties rather than qm probability stuff. What may seem like acceptable concepts over short distances in the lab can tend to breakdown in a big universe.

Take emission A that occurred 6 billion years ago (6 billion light years away) and proceeded to cross the universe to hit a ccd camera photosite from a camera manufactured in 2006 by a design made in 2005 by a species that has only been around a few thousands of years on a planet formed maybe 4 billion years ago located in a galaxy which was nowhere close to the location where the even took place back 6 billion years when this even started? Or, was there a precursor wave that traveled into the future billions of years to 'negotiate' the transaction and then report back 6 billion years in the past to 'inform' the emitting object the direction of the momentum which the 'photon' had decided to take and then notify the rest of the wavefront that it had no more need to exist? It sounds much more plausible when it's just a flashlight generated photon in the laboratory.

There are some simple experiments that were run this year. They were a slight modification of the young's doubleslit where a small wire was placed strategically so that it would be in a wavefront minima while a sensor - perhaps a canon digital camera - was used to detect photons as to which slit the photon came through. Conventional interpretations would imply one can measure either wave or particle in one experiment but not both. The cleverness of the experiment was to have a null result experiment verify the wave nature due to the position of the wire while the camera captured photon direction information. If the particle nature was present without the wave nature, the wire would have blocked a measurable amount of the test photons. the fact that it didn't - the null result- indicates the wave nature was present in the path even as the ultimate detection was particle in nature.

The experiment is repeatable but the interpretations have been somewhat controversial. It seems to call into question the validity of the copenhagen conventions and some of what has become common interpretations of qm effects. There are a number of arxiv papers concerning this. If there are severe problems at the foundations, then the secondary and tertiary concepts are likely to really be fouled up.
 
  • #22
p764rds said:
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

Yes, that is correct. In this situation you are no longer in the domain of classical E&M, and the [itex]1/r^2[/itex] law no longer applies to the actual energy received by the telescopes. Instead, you have to interpret it as related to the probability that any particular telescope receives the photon. As r increases, it takes more telescopes (each with the same aperture size) to cover the surface of a sphere of radius r, and so the probability that any particular telescope receives the photon decreases accordingly.
 
  • #23
Ok just quickly before I leave...does this mean that with p764rds example, that if you look with the one telescope, there is only a probability of seeing the photon...you probably won't be lucky enough for the photon to head toward your telescope.

So this means even though the energy of the photon could be possibly anywhere within this massive spherical radius (wave nature)...it is only actually at one specific point (photon nature)?

Also if someone could give some details on the questions I asked last, it'd be much appreciated, but no rush...I have a week before I get back!

Thanks,

Kcodon
 
  • #24
p764rds said:
Hey, that's 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.

I think the chance is pretty good, but I'm just a dumb engineer that's been working with EM for close to 20 years. :smile:

From where I sit, the wave model of a photon is omni-directional (to some extent), but the particle (quantum?) model portrayed here is construed as uni-directional. If a photon has zero mass, is it unreasonable to assume that its' energy can be spread over an ever increasing large area as the radius from the emitter grows with time? I think not, but I'm open to evidence to the contrary...

Regards,

Bill
 
  • #25
jtbell said:
Yes, that is correct. In this situation you are no longer in the domain of classical E&M, and the [itex]1/r^2[/itex] law no longer applies to the actual energy received by the telescopes. Instead, you have to interpret it as related to the probability that any particular telescope receives the photon. As r increases, it takes more telescopes (each with the same aperture size) to cover the surface of a sphere of radius r, and so the probability that any particular telescope receives the photon decreases accordingly.

?

How might one construct a meaningful surface integral at radius r=ct to account for the apparent change in flux at time t due to the radiated energy at time t in this scenario?

As r approaches infinity, doesn't a fixed size aperture approach the equivalent of differential area?

I think the quantum guys need to figure out how big a photon is at radius r, and how many differential area "quanta" are required to capture all of its' energy.

Until then, the intergrated probability that a closed sphere of detectors will record the total energy of an emitted photon is still unity - regardless how small the probability that any given detector will record (a portion of?) it.

Regards,

Bill
 
  • #26
Does this mean that one telescope 10 miles away from a single photon release into spherical 3D space will with 99.9 percent probability not 'see' the photon because said photon missed the telescope? Can anyone say yes or no without obfuscating to this quaestion, admit that we don't know you da you de ya.
 
  • #27
the original emission of the photon can be one of two methods, spontaneous or stimulated. Stimulated requires the presence of another photon triggering a molecule or atom in a higher state to drop down to a lower state, emitting an identical photon, same wavelength, direction of travel, polarization, phase etc. Spontaneous means just that, at some point in time the atom decides to drop down from a higher state to a lower. There is no telling exactly when this will occur or what orientation the molecule/atom might be in when it happens. Hence, there is a very great liklihood that it will go off in a random direction other than that of the telescope. Also, with intervening molecules/atoms in the way, there is also a distinct possibility that it gets absorbed somewhere else on the way. Any attempt to determine the orientation of the molecule or when it will emit will of course modify the molecule and change the outcome.

While a photon may have 0 rest mass - and it's meaningless to talk of a photon at rest as it doesn't exist, it does have momentum and velocity (c) so it has mass when moving (or perhaps more accurately, there is a mass equivalent of energy contained within it).

There is also conservation of momentum with the excited atome that emitted it as well. It will recoil. If there were such a thing as the photon traveling in all directions until it 'decided' to be really at one location, say to be absorbed (and detected) by the 3rd element from the top and fourth element from the right of a ccd sensor in a camera on the telescope, then there would be no recoil for the atom emitting it until that 'decision' was made.

Seems like I recall something of a 'pilot wave' - at least a thought experiment - trying to convey this concept of all directions. In terms of the universe, such things lose out and rather appear indicative of serious problems in interpretation of the concepts(at least to me).

The question is if there is nothing between a single atom and a spherical shell 10 miles away with one small telescope peering through a port, why would there be more photons reaching it than reaching any other point on the sphere?
 
  • #28
kcodon said:
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.
So there's 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.

The r=ct sphere (in my opinion) represents the leading edge of the propogating sinusoid. However, by my interpretation, the amplitude of the sinusoid snakes its' way through space along a poynting vector (direction of propogation). This places the trailing edge of the sinusoid in what amounts to "imaginary space" - physically inside the r=ct sphere, and yet to pass through (not "visible" at time t, per se). In order to observe the sinusoid as such, you would have to freeze time as the sinusoid passes, and vary position along the poynting vector. The sinusoid is exactly one wavelength long within this time-frozen 3-space.

Plane waves are merely a convenient way of approximating spherical waves at very large radii, and are only valid were the included spherical surface area has negligible curvature (phase error).

This I also found interesting...
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.
Do you mind elaborating?

The infamous r=ct sphere about whatever we have constructed in order to radiate a single photon represents the past of the item at the sphere's center. When some portion of that radiated energy reaches another item constructed to detect it at the sphere's surface, that energy arrives in that second item's present.

The emission effectively takes place in the future of the detector, arrives (later) in the present of the detector, from a point in the past of the emitter.

Follow?

Given a Euclidian sense of 4-space, time would be orthogonal to 3-space. Just as z=3 can be true anywhere on the x-y plane, t=0 can be true for any x-y-z. By the same token, any t can be true for any x,y,z - beit the present of whatever is at that point, or the past/future of another item that either has passed, or will pass through it (in some way). "Local time" is relative within the framework of absoloute time - because of stipulations regarding "c", and some arbitrary "points of reference" affected by time's passage.

Since a photon allegedly does not experience time (by virtue of its' velocity), a transaction occurring over tens of millions of years is effectively instantaneous within the photon's frame. Hence the notion of the past (energy decrease) finding its' way into the future (energy increase). Energy is still conserved ... over time.

Regards,

Bill
 
  • #29
Hello again everyone,

I can see some people are getting into this thread, and that's cool; its just a little over my head lol. So I was wondering if someone could give me simple answers on the following please?

#1 OK cbacba raises a really good, logical point
If there were such a thing as the photon traveling in all directions until it 'decided' to be really at one location ... then there would be no recoil for the atom emitting it until that 'decision' was made.
So this implies that there is no wave nature of light at all in the case of a single photon being emitted, or spherical nature, and that light must be a particle, does it not?

#2 The sphere in this case would be of r=ct and would not really physically represent anything...except that at one point on its surface the photon would be located - at least this is what I think??. Mathematically, would the surface area represent the probability of receiving the photon, as jtbell mentioned?

#3 So sort of combining the above two, can one then think that the "wave" nature of a photon, is actually just some form of mathematical representation of the probability of finding the photon at a given point in space and time? I assume this can be related to what Bill said in the last post about sinusoidal wave forms etc? So really light is "corpuscles" as Newton put it?

And finally, thanks Bill,
The emission effectively takes place in the future of the detector, arrives (later) in the present of the detector, from a point in the past of the emitter.

Follow?

I think I see what you're saying...I've never thought of that before.

Anywho, thanks for any replies on these 3 points, and hopefully some light can be shone on my questions,

Kcodon
 
  • #30
kcodon said:
#1 OK cbacba raises a really good, logical point
If there were such a thing as the photon traveling in all directions until it 'decided' to be really at one location ... then there would be no recoil for the atom emitting it until that 'decision' was made.
So this implies that there is no wave nature of light at all in the case of a single photon being emitted, or spherical nature, and that light must be a particle, does it not?
While that seems to be a reasonable interpretation, I don't think that it would be accurate. Consider the source of the emission: change in energy level of an *orbiting* electron.

For the sake of a description, let us assume that an electron orbits an atom in a plane with period T. Sensors at constant radius in this plane might sense the changing electric field emanating from the electron as having a maxima when the electron is closest, and minima when the electron is furthest - vis: a sinusoid of period T.

If this electron were to drop to a lower energy level, all sensors at constant radius will record the same change in amplitude of the radiated field. The detection of this change (photon?) becomes fleeting as r increases due to the rate at which the event propogates around the r=ct arc (or shell, for the 3-D version).

I doubt there would be any "recoil" to speak of - aside from, perhaps, a very brief oscillation with no net change in position/direction.

#2 The sphere in this case would be of r=ct and would not really physically represent anything...except that at one point on its surface the photon would be located - at least this is what I think??. Mathematically, would the surface area represent the probability of receiving the photon, as jtbell mentioned?

I'd offer that the r=ct sphere physically represents a region of "constant time" with respect to the emitter, and the probability that the photon exists on that sphere at time t is 1. However, given the description I provided above, it may well be true that at any given radius, the photon might only present itself at a particular point on that sphere - as it moves to a slightly different point on a slightly larger/smaller sphere (r=c(t+dt)) while following the angular trajetory of the affected electron. Curiously, this interpretation satisfies both the particle (one point on a sphere of radius r=ct) and wave properties (all spherical angles) of photons simultaneously.

I think it would be interesting to arrange an array of detectors over a narrow angular range, and broad radial range to see if it might be possible to record the same photon wave in multiple locations/times. I'd imagine that a steep cone arrangement would suffice.

#3 So sort of combining the above two, can one then think that the "wave" nature of a photon, is actually just some form of mathematical representation of the probability of finding the photon at a given point in space and time? I assume this can be related to what Bill said in the last post about sinusoidal wave forms etc? So really light is "corpuscles" as Newton put it?

The above strikes me as having a little too much emphasis on the particle properties, and too little on the wave. Both aspects should carry equal weight since either could be observed.

Regards,

Bill
 
  • #31
cbacba said:
Seems like I recall something of a 'pilot wave' - at least a thought experiment - trying to convey this concept of all directions. In terms of the universe, such things lose out and rather appear indicative of serious problems in interpretation of the concepts(at least to me).

In order for your pilot wave to work, you would have to concentrate the energy radiated in one general direction. The sperical distribution of the wave energy still holds (thanks to such phenomena as diffraction), but the energy density on the shell can be modified to favor a particular direction above all others.

Imagine an omnidirectional emitter at the focus of a parabolic bowl that creates a hemispherical "shadow" (focus and rim in plane with the emitter). The size of the aperture in terms of wavelengths relates directly to the configuration's directivity along the focal axis. A simple raytrace will show that any ray from the emitter/focus, to the parabola, and then to the aperture plane (plane of rim) will follow a path of equal length. Thus, the energy emitted in a spherical fashion becomes "columnated" along the focal axis as a disk of constant phase energy (rather than a hemisphere of constant phase).

While the modes of propogation are defined over the entire sphere, there are any number of non-zero order modes (limited by the aperture's size in wavelengths) that can be used to characterize the resultant spherical energy distribution.

The question is if there is nothing between a single atom and a spherical shell 10 miles away with one small telescope peering through a port, why would there be more photons reaching it than reaching any other point on the sphere?

Perhaps because that is where the emitter was "pointed" at that point in time.

Regards,

Bill
 
  • #32
I think the real point of qm is that you cannot detect the condition without altering the condition. That and it's a probability thing.

When it comes to the emission of a single photon, there are two types. These are spontaneous and stimulated. There are probability factors for each. Spontaneous means it's going to go off on its own time schedule and in its own direction - ostensibly, completely random within the context of what is known about the nature of the atom or molecule and the state it is in. Stimulated is a forced emission caused by the nearby passing of a photon of the same energy and direction and phase of the one that is being emitted (laser stuff).

BTW, it's not my 'pilot wave'. I don't believe it and was attempting to debunk it. I don't know whose it is but I can't help remembering seeing feynman in movie clip explaining it somewhere.
 
Last edited:
  • #33
I doubt there would be any "recoil" to speak of - aside from, perhaps, a very brief oscillation with no net change in position/direction.

Hmmm I think I disagree...photons have momentum and there must be some change in momentum of the atom emitting it - instantly - to conserve momentum. This would of course be much to small to detect etc etc, but still occurs. However:
I think the real point of qm is that you cannot detect the condition without altering the condition. That and it's a probability thing.
cbacba is right here I think...and this is what so much annoys me about QM, and although I'm know very little of it, I'm guessing the QM interpretation would indicate something like the molecule would be in a superposition of possible states until the photon emitted was measured, so it nicely avoids this problem.

The question is if there is nothing between a single atom and a spherical shell 10 miles away with one small telescope peering through a port, why would there be more photons reaching it than reaching any other point on the sphere?
Perhaps because that is where the emitter was "pointed" at that point in time.

Its just the whole probability thing again then. The small telescope would maybe only detect a photon one in every gazillion emissions, but it would eventually. Asking then why it hit that certain telescope is kind of like asking when you toss a coin and get heads, why it was heads and not tails...its a somewhat meaningless question. Lol I think I've unfortunately asked some of these questions in my time and been baffled by them too.

I also think in a round about way we agree on #2, however for #3...
The above strikes me as having a little too much emphasis on the particle properties, and too little on the wave. Both aspects should carry equal weight since either could be observed.
What is the wave properties of the emitted photon in this exact case, without passing it through a double slit or anything? We've all been saying - I think - that all the energy from the emission is concentrated in one place (particle nature), however there is a possibility of it being anywhere in the sphere r=ct (wave nature to some degree).

Kcodon
 
  • #34
cbacba said:
I think the real point of qm is that you cannot detect the condition without altering the condition. That and it's a probability thing.

The non-qm version (as I understand it) would require that if some amount of energy is received at radius r, then there is that much less energy available on the r+dr sphere. How the conservation of energy might be probablistic is something that I don't quite follow.

Does a photon cease to propogate further (in all other directions) once detected? Or, is the condition of the photon wave simply altered to account for the loss of energy?

I find it hard to rationalize how both must necessarily be true if a single isolated detector records a "hit".
When it comes to the emission of a single photon, there are two types. These are spontaneous and stimulated. There are probability factors for each. Spontaneous means it's going to go off on its own time schedule and in its own direction - ostensibly, completely random within the context of what is known about the nature of the atom or molecule and the state it is in. Stimulated is a forced emission caused by the nearby passing of a photon of the same energy and direction and phase of the one that is being emitted (laser stuff).

I find it curious that "stimulated emissions" presuppose the existense of photons already traveling in the direction you desire. Where would such a stimuli originate?

BTW, it's not my 'pilot wave'. I don't believe it and was attempting to debunk it. I don't know whose it is but I can't help remembering seeing feynman in movie clip explaining it somewhere.

Sorry, I didn't know it came with baggage - just that you brought it up. :smile:

If photon emission is unidirectional (and I'm not saying I think it is), you might be able to place a "spontaneous emitter" at the center of a perfectly reflective sphere and wait to see if it moves - once, upon emitting the photon, and accelerated once again by absorbing the reflection (since the reflection would pass back through the center of the sphere in the direction of the recoil).

Another possibility that would allow for detection/confirmation of the emission would be to place the emitter at one focus of a perfectly reflective ellipsoid, and an omnidirectional detector at the other. While the detector might see a photon traveling directly from one focus to the other, it should certainly notice a spherical wave converging on it from all directions.

I doubt the emitter would have moved in either case, but that's just my opinion.

Regards,

Bill
 
  • #35
kcodon said:
photons have momentum and there must be some change in momentum of the atom emitting it - instantly - to conserve momentum.

Is that so?

Consider that in the omni-directional case I attempted to describe (poorly I guess), for any photon traveling in one direction, there may well be another (the same?) traveling in the opposite direction (and 180deg out of phase). The momentum of each cancels the other, leaving the atom without change in momentum - just change in energy.

I've read over and over again how it takes an infinite amount of energy to accelerate a finite mass (however small) to the speed of light - so what better way to counter that infinite force than with another of equal magnitude and opposite direction?

An atom that emits a photon does still lose mass in the process; so how else could that instantaneous change in mass convert itself to a photon (wave, or otherwise)?

Just because it does?

Please re-read the 2-D scenario I described, and see if you can imagine how the detection points of a given photon might traverse a spiral of 360degs in phi for every one wavelength change in r.

The above strikes me as having a little too much emphasis on the particle properties, and too little on the wave. Both aspects should carry equal weight since either could be observed.
What is the wave properties of the emitted photon in this exact case, without passing it through a double slit or anything?

Well - the way I see it is that the total energy of the wave exists in different directions at different times due to the nature of the source (an orbiting electron). The suggestion that one could capture all of the wave energy at a single point in space strikes me as absurd.

We've all been saying - I think - that all the energy from the emission is concentrated in one place (particle nature), however there is a possibility of it being anywhere in the sphere r=ct (wave nature to some degree).

The other possibility is that the wave exists everywhere on the r=ct sphere, but is only detectible where the leading edge of the propogating wave form is not imaginary (essentially equivalent to not "existing" there yet). Refer back to my orbiting electron for the rationale for phase to vary at constant radius.

As an aside, has anyone noticed that if you increase the radius of a circle by one wavelength ( [itex] 2\pi [/itex] radians), the circumference grows by the wavelength squared ( [itex] 4\pi^2 [/itex] radians)?

Regards,

Bill
 
<h2>1. How do electromagnetic waves propagate in a spherical emission?</h2><p>Electromagnetic waves propagate in a spherical emission by radiating outwards in all directions from the source. This is known as spherical spreading, where the energy of the wave is spread over an increasingly larger area as it travels away from the source.</p><h2>2. What factors affect the propagation of EM waves in spherical emission?</h2><p>The propagation of EM waves in spherical emission is affected by several factors, including the distance from the source, the frequency of the wave, and the medium through which it is traveling. These factors can impact the strength and speed of the wave as it propagates.</p><h2>3. How does the distance from the source impact the propagation of EM waves in spherical emission?</h2><p>The distance from the source has a direct impact on the propagation of EM waves in spherical emission. As the wave travels further away from the source, it spreads out and becomes weaker. This is due to the inverse square law, which states that the intensity of the wave decreases as the distance from the source increases.</p><h2>4. Can EM waves propagate in spherical emission in a vacuum?</h2><p>Yes, EM waves can propagate in spherical emission in a vacuum. In fact, they travel faster in a vacuum compared to other mediums, such as air or water. This is because there is no resistance or interference from particles in a vacuum, allowing the wave to travel unimpeded.</p><h2>5. How does the frequency of EM waves affect their propagation in spherical emission?</h2><p>The frequency of EM waves also plays a role in their propagation in spherical emission. Higher frequency waves have shorter wavelengths and can travel further before losing their energy, while lower frequency waves have longer wavelengths and tend to lose energy more quickly. This can impact the distance and strength of the wave as it propagates.</p>

1. How do electromagnetic waves propagate in a spherical emission?

Electromagnetic waves propagate in a spherical emission by radiating outwards in all directions from the source. This is known as spherical spreading, where the energy of the wave is spread over an increasingly larger area as it travels away from the source.

2. What factors affect the propagation of EM waves in spherical emission?

The propagation of EM waves in spherical emission is affected by several factors, including the distance from the source, the frequency of the wave, and the medium through which it is traveling. These factors can impact the strength and speed of the wave as it propagates.

3. How does the distance from the source impact the propagation of EM waves in spherical emission?

The distance from the source has a direct impact on the propagation of EM waves in spherical emission. As the wave travels further away from the source, it spreads out and becomes weaker. This is due to the inverse square law, which states that the intensity of the wave decreases as the distance from the source increases.

4. Can EM waves propagate in spherical emission in a vacuum?

Yes, EM waves can propagate in spherical emission in a vacuum. In fact, they travel faster in a vacuum compared to other mediums, such as air or water. This is because there is no resistance or interference from particles in a vacuum, allowing the wave to travel unimpeded.

5. How does the frequency of EM waves affect their propagation in spherical emission?

The frequency of EM waves also plays a role in their propagation in spherical emission. Higher frequency waves have shorter wavelengths and can travel further before losing their energy, while lower frequency waves have longer wavelengths and tend to lose energy more quickly. This can impact the distance and strength of the wave as it propagates.

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