# How cross-section area of single E-M wave looks like?

1. May 31, 2014

### gfgts250

This question has been asked probably many times, and is inspired by area that appears in denominator of energy flux unit [W/m^2]. From what i have read so far, i come to conclusion, that this post should, first of all, try to explain that this question makes some sense.

First thing that should be explained is FIELD AREA (e.g. electric field area). We can substitute into mathematical expression of Coulomb's law (F = k * q1q2/r^2) any distance (r), and we always get specific force value. That's the reason for which is often assumed that the field is INFINITE. But it doesn't mean that we can not talk about it's shape and size. By assuming some numerical value (E) of electric field, we can draw an outline/contour (bounding surface) by connecting all the points in space around charge having this value [Pic. 1]. So, we've already got shape of the field (spherical, oval, etc.) and it's location relative to electric charge. Additionally, if we repeat this procedure for stronger charge, we can start talking about size, at least in comparative context (in this case: cross-section area of 2 times stronger charge will be 2 times bigger). We can also verify that all of this applies to any boundary value (E). So even if we substitute infinity for distance (r) or 0 for force (F), these rules will be fulfilled - it just mathematiacal apparatus we use, can not cope with such types of values. In other words: infinity of interaction doesn't mean that we can not talk about size of electric/magnetic field.

Let's imagine that we've a simple antenna [Pic. 2a] - piece of straight wire in vertical position. In this antena there is an electron (for this moment just one) which can move up and down. Doing so, it generates an electromagnetic (E-M) wave radiating in all directions (for simplicity let's constraint this model to 2 dimentions, like top-view picture [Pic. 2b]). I'm using this model intentionally because it's the best one i've found so far (unfortunately i don't remember address of that thread), in an attempt to explain absurdity of question about E-M wave's surface. The best doesn't mean correct (more precisely: not complete). Anyway, i use it because it provides a good base for further analysis.

This model shows that in practice we never generate a single E-M wave, but rather we are dealing with radiation which, at greater distances from the source, turns into almost PLANE WAVE (curvature of wave front decreases, area covered by radiation increases). That's why answer to the question "How to convert energy flux [W/m^2] to energy [J]?" often is: Multiply flux by time (this is understandable) and by area. What area? The area we are interested in, the area our illuminated object has, the area of the slit through our plane wave has passed - in other words: ANY area we want. It's easy to understand in context presented before, but usually it's not the intention of those who are asking. The question concerns the spatial size(extension) of field (e.g. electric) of SINGLE E-M WAVE with some given amplitude.

A single E-M wave? What does it mean? Does something like that even exist? Look at the picture, there are two patches/areas [Pic. 2] A1,A2 (at different distances from source) by which a couple of rays (let's consider that they are our single E-M waves) coming through. According to energy conservation law, the same amount of energy has to pass through both of them. So, the only possibility is to lower E-M wave amplitude, and this, in turn, is only possible by moving apart our rays, relative to each other. This process can be seen as opposite to interference. In the latter, when two separate waves approach each other, they E-M fields become indistinguishable and result is treated as a single wave (with doubled amplitude, of course). When moving apart, there should be some minimal distance from which we would be able to see two separate waves again.

But we also know, that all E-M radiation is quantized, it can not divide indefinitely - it can do this just up to the photon level. We should perform a simple experiment: Let's move our electron a little bit (for example: one time up and down - or something like that), just to generate a smallest unit of E-M radiation we can (the shorter the better, in ideal case it will be the shortest unit of E-M wave which can travel freely in space, but it's not a must). We know how much work we have done and how much energy E-M radiation has. We have to divide this by energy of single photon and we've got number of photons in the system. Let's define some minimal distance on projection screen (our measurement equipment) we can distinguish one photon from another and multiply it by photon count. We get some length, this is circumference of a circle. We just have to calculate the radius - distance from radiation source where we should perform our measurements (place projection screen there).

Now, everything depends on what pattern we will see: (1) single photons distributed evenly along screen [Pic. 3a] - it means that photons are generated with uniform spatial distribution of direction - direction of momentum, or if you like: direction of wave vector. It may be partially the result of some hyphotetical process like, let's call it for example: quantum dispersion (photons can diverge sponaneously, with some probability), but in practice it should have different characteristic [Pic. 4a] (be recognizable) then geometrical (dominant) one [Pic. 4b]. Conclusion: indeed, there is no such a thing like single E-M wave/ray in this dimension. In case (2) photons concentrated around some specific points [Pic. 3b] - it means that there are some predefined radiation directions (for example, due to the fact that angles are quantized in some way).

In case #(2) we've got already our single E-M wave! In case #(1) we have to continue searching in other dimensions. Let's generate again some smallest unit of E-M radiation, if photons hit exactly the same places as before [Pic. 3c], then we have found even something better: a single E-M wave with the smallest possible amplitude (stream of photons following each other with properties of continuous E-M wave). This can be called not only single but rather elementary E-M wave. If this also fails, it's time for last resort: so far our example has operated in 2-D (two dimensions) - it was like one slice. We have to take into account third dimension. There are other electrons in vertical wire (below, above), and they move synchronously with the first one. In this dimension, for this type of antenna photons/rays shouldn't diverge from each other (at least they should do it with different charachteristic - slower then in horizontal plane). If such photons are horizontally aligned (with those generated at other altitudes), we should be able to see vertical stripes on the projection screen [Pic. 3d]. Even if we can't get continuous wave this way, we still get a piece of such single wave. This is something more then a photon. A single photon is to small to perform measurement on it, but this piece of wave can have measurable amplitude (this is still a bunch of photons but with limited horizontal distribution - a "thin" wave) and it's E-M field strength is enough to extend to measurable sizes (E-M field is continuous, it can not just drop instantenously in one place. The stronger the source is, the larger is the area that can be detected using standard methods).

How to test all of this in practice? Maybe we don't have to perform all those measurement hundreds,thousand or millions miles away from the source of radiation (Note: Please, treat the following examples only as a theoretical, they are just to show very general idea, not real techniques). First, we can filter tight slice from radiation emitted in all directions (we can use narrow slit) - this will be our initial ray. Then, we can use for example two opposite mirrors to filter it further [Pic. 5a] (simulate longer distance). In theory, light can travel between them almost indefinitely (attenuation if present should be uniform, and maybe even can be helpful in faster reaching the final conditions). Perfecty flat surface of mirror doesn't change angle of incoming ray, just direction (the angle of incidence equals to the angle of reflection), so all we have to do is to wait for a while. If we worry about the interference of rays (incidented and reflected), we can use different mirror's configuration [Pic. 5b] . We can take advantage of the fact that we are only testing small fragment of wave not continuous one, and this fragment has to be shorter then 4L (ray can travel inside box xN times, and later be released on demand). You should also notice that all those mirrors are small, it allows rays diffracted too much to escape from box and without producing too much mess (E-M noise) inside. As i said, it was just theoretical example. Maybe there are better ideas? Anyone knowns about any similar experiments or can recall any knowledge/facts i've missed?

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2. May 31, 2014

### Simon Bridge

Wow - you seem to be very confused - was there a question in all that?
I'll have a go with what you talk about, see if I cannot give you the right words at least, and show you where you seem to be tripping up.
But I cannot tell what you are asking here.

That would be the units of flux density ... flux per unit area.

Why-ever would you want to do that? What use does the concept have?
The area in the flux-density units is not the area of the wave/field itself.

This is correct, you can use surfaces of constant E to describe EM fields.

Since electric field is a vector that is not terribly useful in general; better to use equi-potential lines.

We would call that sort of concept the "geometry" of the field - the specific geometric properties you have described (spherical, oval etc) may be called it's "symmetry", though there are more precise terms.

Nope - that concept does not translate.
For a point charge, the "line of equal E" would actually be a surface - a sphere to be exact.
The sphere for the same magnitude of E, for a charge 2x, would have twice the radius and thus 4x the surface area. Similarly the crossection of the constant-E surface would be 4x.

A plane wave is a single wave.
A wave that turns into a plane wave at long distances is still a single wave.

We are able to represent a wave as a superpositon of other waves.
The usual way to do this is by a Fourier Transform.

That is correct - converting the flux density into energy received will not answer questions about the geometry of the wave. But it is not supposed to.

The person asking is using the wrong words to ask the question.
If they mean to ask about spacial extent of a single EM wave, then they should not ask, as in your example, "How to convert energy flux..." That's just nonsense.

The existence of EM waves is well established empirically so "yes" EM waves exist.
"a single e-m wave" refers to a particular solution to the EM wave equations.

You could but this would be an erroneous interpretation of the picture.
The lines in the picture represent rays, not waves.

Careful not to mix up models here.
Photons belong to the particle model of light, while EM waves belong to the, well, wave model.
Also - individual photons may have any frequency, and, thus, and energy.

You are just confusing yourself more and more after this.

3. Jun 1, 2014

### gfgts250

I have two simple requests: please, read entire post first, then answer. And don't say that somone is confused just because you don't understand what he is talking about. Who knows maybe i was wrong at some point, but rather not confused.

My question is very simple: For E-M wave with given frequency (v) and given amplitude (A), please draw me an outline (it would be surface in 3D, but we are talking about cross-section, so let's constraint it to 2D) of electric field surrounding E-M wave for some given value (E).

If i ask this question for electric charge, it would be perfectly understandable. The answer is exactly on the [Pic. 1]. We've got an electric charge center, concentric circle around it where electric field = (E) and even specific vaue of radius (r) calculated from Coulomb's law. Now, I'm asking for the same for E-M wave. Of course, the concept of single E-M wave is needed for this question to be valid. And this is what i've tried to prove (define), for the larger part of my original post, that something like single E-M wave could really exist.

Now let's proceed to main explanation. This is the most important part, we have to agree on that somehow before further discussion.

The last two sentences are correct. The first two not so much (everything depends of knowledge one has).

Let's imagine that in dark room there are two identical lasers. Beams generated by them are parallel. At the beginning lasers are standing at some distance (for example: 1m) from each other. You should be able to define the boundary of each beam - it will be an area of lowest intensity of light your eyes can detect (of course, it will be imaginary boundary depending on precision of measurement instrument, but for for what i want to show, that's ok). If you can define boundary it's obvious you can clearly see two separate rays. Now, let's bring our lasers close together. When they become collinear (i mean they rays become collinear) you will be able to see only one beam but 2 times brighter (higher amplitude). And this is exactly how interference works (or superposition of waves if you like). You can also reverse entire procedure (split rays) and again you will see 2 separated rays (you can say: 2 times dimmer then the original one).

These were two opposite cases. But there is yet another one - quite interesting for us. Move beams very close to each other but NOT too close, there should be some distance between them (let's assume half of visible beam width should be ok). Your eyes should see a WIDER ray (not bigger but wider in some specific direction). Look at the [Pic. 1], when you bring closer those charges their electric fields (where E is considered boundary value) overlap. An outline (line where electric field = E) will surround two charges and will become oval like.

If you repeat this procedure: add more rays in horizontal and vertical directions you should get larger area of nearly (true plane wave has infinite dimensions) uniform intensity. And this is exactly what we are calling PLANE WAVE - previously we've just seen very small part of it! This is even how our base laser's beams are build (you can think that there are many smaller rays inside beam, some perfectly and some nearly collinear). We would be able to split them too, but in our example we don't have technical means to do that.

There are physical processes where such splitting naturally occurs (at least in one dimension). One of them is radiation emitted by antenna [Pic. 2]. One more time, let's position our lasers such their beams become collinear. Now start rotating one of them by some angle relative to the other. You would be able to see, in the same time, the 3 cases described previously. Near the axis of rotation (stage 1) you will see one ray with higher intensity (amplitude), at grater distances (stage 2) this ray becomes wider but dimmer and finally you will see 2 separated rays (stage 3) each one with 2 times lower intensity then ray in stage 1 (by the way, looking at stages 1 and 2 you can see how energy is conserved - how higher amplitude changes to wider area. Energy flux has to be the same, and there are two independent variables in equation: amplitude and area).

And this is the KEY POINT why we can't understand each other! The example using laser rays was a little bit artificial (but useful). In practice (pure wave interpretation) stage 3 should't exist, it would be rather stage 2 with wider area and lower amplitude. In such interpretation initial ray can be seen as consisting of infinite number of smaller rays each one with infinitely small amplitude There is no way to distinguish one from the other, and probably that's why you are calling this: a SINGLE WAVE (which is changing it's geometry with the distance). Well but we are not living 100 years ago, and we do know from "Photoelectric effect" experiment performed circa 1900, that light (E-M radiation) is QUANTIZED. It can not divide indefinitely. So stage 3 has to naturally appear! That's a fact. We know that energy coming through unitary area has to diminish with distance from the source of radiation. We know precisely what energy single photon has (E = h*nu). We know that photon can not be divided. Conclusion is obvious. Wave-particle nature of E-M radiation doesn't mean that you can arbitraly neglect some properties (there are NOT two separate models between which you can freely choose. There is permanent dependence between them).

Let's assume we've got an area with size of 1 m^2, we can calculate distance from radiation source (with known power) where energy coming throuh such erea equals energy of single photon. And from the fact, that photon can not be divided, we know that all energy has to be concentrated at one small point on such area. This is the consequence of the quantization, otherwise the energy should be spreaded evenly. Probability of finding photon could be uniform for such area but photon itself has finite size (like size of electric field used previously). Also electric field around photon is not uniform (decreases with distance) like it's uniform for ideal plane wave.

Honestly, we don't really know how those photons are arranged relative to each other (i've shown a couple of simple patterns [Pic. 3]). Even if they spread perfectly uniformly ([Pic. 1a])), they are doing so only in one plane (for assumed type of antenna). So, it's possible that there are factors causing photons to concentrate around some centers (for example stripes, like on [Pic. 3d]). And concentration area of photons is nothing more then: E-M WAVE (a bunch of photons going in the same direction)! And this is what I mean by saying about SINGLE WAVE. Now, i hope my question becomes clear to you. Why do we need all of this? Because the best (and maybe only) way we can learn something more about nature of E-M radiation is to examine E-M field around it. We still don't known much about it.

At the end, let's deal with the rest:

http://en.wikipedia.org/wiki/Energy_flux
"This is SOMETIMES called energy flux density, to distinguish it from the second definition"
Unit i've given [W/m^2] speaks for itself.

It has nothing to do with flux density. I've only said that questions about cross section of E-M are OFTEN INSPIRED by [m^2] that appears in flux density unit (there are a couple of threads at internet forums where others are asking about cross section). Of course, I'm still interested in such areas but not due to flux density interpretation.

My fault, I should rather say: "First thing that should be explained is what FIELD AREA means to me". Why do I want to do that? Because, lack of explicit limit of interaction (like in Coulomb's expression) causes some pople to think that we can not talk about anything, just sit and cry. This is also what i've learned reading other threads.

Well, later when we will be taking about fields of E-M wave, those vectors can even be changing with time, so why not to use derivative of vectors, or something like that (sorry i'm not good at math)? Seriously, I'm not doing any calculations with it. In the last sentence of second paragraph there is the main and only reason i was doing so: "...infinity of interaction doesn't mean that we can not talk about size of electric/magnetic field..." All of this was just to show that we can talk about SIZE. How we can define it? At this moment it doesn't matter.

Look at previous question and replace word SIZE with SHAPE...

Look at the expression: F = k * q1q2/r^2

F has to stay constant as well as k and q1 (or q2 - one of them). So if you double q2 in numerator, you have to double denominator, and because r is squared you have to choose such value for r, which squared gives you two times bigger value then previous squared r.

Let's assume that previous value of r was 1, 1^2 = 1, now you need value that squared gives you 2. And this vaule is about: 1.4, (1.4)^2 = 2. So, radius of 2 times stronger charge will be 1.4 times longer.

From my original post: "cross-section area of 2 times stronger charge will be 2 times bigger". Cross section of sphere is circle, and area of circle is: Pi * r^2. Now, substitute values from previous calculations and you will get the answer.

By the way: it's not about cross section, area, radius, volume. It was just an example, just to show that those values stay in SOME relation, that we can compare them for different charges and talk about SIZE of charge.

No, no ,no. You can't say that. As i mention previously, there is only one physical process responsible for whole E-M phenomenon. Division on particle model of light and wave model of light is artificial. We people are still trying to figure out everything about it. Such separate models temporary exist because we can't combine today's partial knowledge into one coherent whole.

You can't choose between models. You are not choosing anything, rather you are IGNORING some aspects of real E-M radiation!

Well, there are some situations when you can do so. For example when doing engineering calculations (designing antennas, optical apparatus or so). But condition is: Energy of E-M waves has to be much higher then energy of single photons. When energy of E-M wave with given frequency starts to approach energy of single photon of such frequency, something very interesting should happen. And this is exactly what i'm trying to analyse.

Of course photons may have any frequency, but it seems to me that you don't know exactly what you are talking about. Monochromatic E-M wave with given frequency consists ONLY of photons with such frequency! You can easily see this performing famous "Photoelectric effect" experiment (the first one revealing particle nature of light - this is how concept of photon was born). The so called stopping voltage depends on energy of photons colliding with electrons. And you can control energy of photons by choosing frequency of incident light. There is different stoping voltage for red, green and blue light.

In our example with antenna we are generating E-M wave with specific frequency, so we know exactly what energy photons building our E-M wave(s) have and at what distance from the source we should see them.

4. Jun 1, 2014

### Simon Bridge

Please read more carefully - I did not say you were confused anywhere in my response above.
I said you seemed to be confused. Spot the difference.

You were indeed wrong on many points, which may explain why you seemed confused.
You also are having trouble with the terminology - perhaps English is not your first language?
That would also explain why you seem confused.

This is not a question, it is an instruction ... and a very unclear one.
What are you trying to acheive?

You have not asked a question yet. What is the question?

You are not making sense.

You are saying that a line of constant electric field magnitude about a point charge is a circle.

Do you understand that diagrams like pic1 are not normally of constant E, usually such diagrams show lines of constant potential. Switching to electric potentials may help you but I cannot really tell because you have not asked a question yet.

You want a line of constant E for an E-M wave?
That is drawn the same way - except now the geometry need not be so simple.
i.e. if you have a spherical wave spreading out from a point source, the magnitude may look like this:

$E(r,t) = A\sin k(r-ct)$ (in spherical coordinates)

The surfaces of equal E are concentric spheres - the 2D version will have concentric circles.
The circles/spheres travel outward at speed c.

Proving is different from defining - which do you intend?
This is the sort of statement - where you conflate two different things - that makes you sound confused.

Similarly: in optics, a ray and a beam, are different things.
You appear to be confusing them but it may be you are just trying to use a language you are not familiar with.

Anyway - the topic is too big for these forums. I could go through an point out each place where it looks like there is some sort of mistake, where you may have got it wrong in some way, but that would take too long. It would really really help if you would just ask a question.

5. Jun 1, 2014

### Staff: Mentor

I am also confused. You seem to be asking about the size/shape of an EM wave that contains the energy of a single photon. Is that correct?

6. Jun 2, 2014

### Cthugha

You have some misconceptions here. First, it is important to note that a single photon does not necessarily have a single frequency. This was the first idea from the 1930s, which considered a Fourier decomposition of the light field. A single photon just means you have a fixed particle number, but that state can be polychromatic.

Having said that: The em wave associated with a single photon can have any shape you like. It depends on the emitter. It can be spherically symmetric or directed. Its temporal duration will be connected to its spectral distribution using a Fourier transform. The power spectral density and first order coherence function form the Fourier transform pair. It is a well known textbook fact that the single photon character does NOT appear at all when looking at the field alone. In fact, there is no way to say whether a field is classical (like a laser) or non-classical (like a single photon) just by looking at the fields alone. You need an intensity-sensitive measurement in order to determine that. Therefore, contrary to your claim the fields themselves are not of particular interest in quantum optics. Their higher-order moments are more interesting.

Pretty much every major development in optics starting from the 1960s. It looks like some kind of introductory review might be beneficial. Roy Glauber's Nobel prize talk might be a good starting point: http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.78.1267. It is free to read. However, these forums cannot replace a thorough course in quantum optics - including the math.

7. Jun 2, 2014

### gfgts250

Not exactly, but close. Your statement could be very good starting point for explanation.

Let's suppose our antenna in [Pic. 2a] has radiated some finite amount of energy (it has broadcasted by short moment). This energy propagates from the source in all directions in some characteristic way (characteristic to spherical wave). The same amount energy has to pass through larger and larger areas with distance from the source (see [Pic. 2b]). It's even more convenient to formulate it this way: Energy passing through some unitary area (for example 1m^2) is smaller and smaller with distance from the source. So, as you can see, at some distance, we naturally come to the point where energy passing through our 1m^2 equal energy of single photon (photon with the same frequency like our E-M wave of course).

And now, we have to decide how to interpret such phenomenon. We already know that light (E-M radiation) is quantized, it can not divide indefinitely, so it's rather not possible to uniformly cover such area with this amount of energy (not possible to divide photon into smaller parts and spread it uniformly) - this is the direct consequence of light's quantization. But on the other hand, it's also not possible photon has such large size (in our example it's 1m but it could be 1km or more, we can assume any size of testing area we just have to choose proper distance from the source) to cover such area (it would not even be able to do this uniformly, since strength of it's E-M field decreases with the distance from it's center). So the obvious conclusion (at least for me) is that with the distance from the source, we should start to see inhomogenities in E-M field passing through our unitary surface, and finally with even greater distance we should be able to detect single photons.

I try to clarify it even more: There are always single photons in E-M radiation, but there plenty of them and such close one together that there is no way we can detect them near the source (photons are bosons - they can exist in the same place in the same time, it's hard to even speak about such a thing like minimal distance between them. We can even say that they are indistingushable from each other). We can only register superposition of their E-M fields and we are used to call it: E-M WAVE. But depending on the characteristic of emitter of E-M radiation, this distance doesn't have to stay fixed. And this is exactly what i'm trying to exploit (the so called half-dipole antenna [Pic. 2] is very useful for that. It generates spherical wave - spherical at least in one direction).

For this type of emitter, from the very beginning, photons are starting to diverge, that's why amplitude becomes smaller and smaller with distance (but of course the total area covered by radiation increases at the same time). But even at the distance 1000 km there is huge concentration of them per unitary area, so we still are seeing uniform radiation (and still can use, for approximated description, the so called: wave model - which could be seen as model describing behaviour of bunch of photons). But with much, much more greater distances (which we can estimate) we are coming to the point, when spaces between photons will be measurable to our equipment, so we just start to see them. It's worth to note, that we are still dealing with the same phenomenon (division on wave model and particle model is artificial, made by people). Accuracy of our measuring equipment (and some laws of physics) has established boundary at which we were able to switch to photon interpretation. Wave didn't change into photons, rather we started to see single photons forming the wave. For spherical wave distance acts like magnifying glass.

Now we've got our photons, but one question remains: how are they arranged relative to each other? Are they spread uniformly, or maybe there are some centers around which they accumulate? It's worth to remember, there was specific reason why photons moved away from each other: Their initial vectors of momentum were different (photon can be treated as a particle, it can have momentum) - their directions were different (this is characteristic feature of spherical wave. For E-M waves generated by laser this phenomenon is not as visible - laser ray can travel longer distances without energy dispersion). So, if during generation, there were photons with exactly the same vector of momentum, they shouldn't diverge from each other. And if they don't diverge, they still should form bunch of photons even at large distances. And a bunch of photons is what we are calling... an E-M WAVE! And this wave is what i've called in my first post: SINGLE WAVE. And i've given a few examples of reasons why something like that could theoretically exist [Pic. 3]. (it's worth some experiment).

Think about photons on [Pic.3] like electric charges. If you superimpose their electric fields , you will get different SHAPE of resultant electric field for each arrangement. Electric field is continuous, we can sample it in different places of space, and this is how we can measure shape. And from shape we can guess arrangement of photons in such wave.

8. Jun 2, 2014

### Staff: Mentor

It appears that you are imagining photons as little particles with vectors that somehow add up to make an EM wave. It is my understanding that the reverse is true. That is, the EM wave is always a wave, and it is just the energy in the wave that is quantized. This is why the emissions from single-photon sources still behave exactly like waves, such as the double-slit experiment. Someone correct me if I'm wrong.

9. Jun 3, 2014

### Born2bwire

That is true. The photon, in QED, is represented as a field that permeates space. Interactions with this field occurs at points and the energy is quantized. Hence, the photon is akin to a classical field that interacts like a classical point particle. The electromagnetic field is an observable. The electromagnetic field of a single photon though is not well defined because the fields are not eigenstates of the photon number state. Only with very large numbers of photons will the electromagnetic field become a consistent observable. So you might as well ask what is the position of an electron in an atom. The answer is that it varies. Each time you check it will be different but after a large number of independent measurements you will arrive at a probability density.

10. Jun 3, 2014

### gfgts250

I have started this thread in: Physics Forums > Physics > Classical Physics. I repeat: Classical Physics. And I've done this for reason. For me classical physics includes also : black body radiation, Photoelectric effect, generally everything until the year circa 1920. Quantum nature of light (it consists of some quantas) was then already known, but today's interpretation (generally: Quantum mechanics) was not born yet. It doesn't mean that i belive in "Old quantum theory" or something like that. No. The only information i need is: Light is not just a wave, it's build from some quantas. That's all.

Don't get me wrong, i don't blame you for anything. All of this is 100% ma fault, because i've used one specific word: PHOTON. And this is probably the reason, why you are here.

According to my knowledge, up to circa 1920, chunk of light was called: QUANTA OF LIGHT. Term PHOTON was invented for newer interpretation just to distinguish it from this QUANTA. So i can not blame you that you are taking about quantum physics. But, try to understand me, using term: QUANTA OF LIGHT will create even more mess here. So i will stay with the term: PHOTON. Hoping, that everyone now knows what this term means to me.

Quantum physics in not needed here, i haven't asked for this! Even if I'm wrong it would be easy to explain it without that! There is enough mess here even without quantum mechanics. If you think otherwise, please go away.

I didn't said anything like that. And even don't want to speculate how photon is build. I've only said that photon (quanta if you like) is some chunk of energy. I don't even know if itself has ANY frequency, if we can speak about frequency of photon. All i know is that E-M wave with some specific frequency carries chunks with some specific energies (E = h*nu - i assume nu is the frequency of the wave from which the photon comes from, the only known property of photon to me is it's energy).

We also know from Photoelectric effect that monochromatic E-M wave with some specific frequency carries ONLY photons with ENERGIES specific to that frequency. That's why when we illuminate some material with one light (for example green) current flows in circuit, and when i do the same with other wavelength (for exampe red) it's not. So it is not so monochromatic E-M wave can consists of photons with different energies only arranged in a way that gives us wave with our base frequency. This is exactly what Photoelectric effect tells us.

In some specific conditions photon may change it's energy (turns into different photon) - good example is Compton experiment, but: first we don't know if we can use word: change - it could be different photon generated after absorbing original one (we are talking about Compton experiment). And second and most important to us - it can not change spontaneously. There has to be some interaction - like collision in Compton experiment. If it wasn't true, we could find at least a small amount of photons of some specific energy in any type of E-M radiation (see Photoelectric effect). In my examples, we've already got some set of photons in beginning. There is no interaction with anything and photons are rather transparent for each other. So we should be sure that there will be foton change in our analysed system.

And the second thing i said about photons is that OUTSIDE E-M wave they can be localized. I repeat: OUTSIDE - this is very important (i'm not sure if we even can speak about photon inside wave). So outside wave they have some properties of particles - we can measure its position (with some limited accuracy of course) and even direction of travel.

All those 2 properties, i mentioned, were known before 1920 and also they are not even main topic of my posts. I'm not talking about shape of the photon here. I'm talking about shape of standard E-M wave. Quantum physics in not needed here.

11. Jun 3, 2014

### gfgts250

I didn't said i'm imagining photons as little particles. I see them as the chunks of energies. I've assumed (and that's a fact) - that outside E-M wave they have PROPERTIES of particles (size, position, etc. - we can localize them). But those properties apply also to those little waves (as you said you see photons). Such little wave (wave-packet?) also has to have finite length - so you have got already size. And also it can traval in some direction, as ray of light do - so here is the second property: direction of motion represented by vector. And this is probably the reason you think i'm imagining photons as little particles - i've used word: MOMENTUM which is generally associated with particles. Well, you can replace it with WAVE VECTOR if you like. I just wanted to indicate their direction of travel somehow.

If you really want to give photon some geometrical form then yes: a small-wave is of course preferable one (also for me). It maintains properties of particle and explains some other effects. But it was not my intention to assume some shape of photons.

12. Jun 3, 2014

### gfgts250

EXACTLY! Yes, this is exactly what i'm asking for! I mean general idea is correct.

Let's continue with your spherical wave spreading out from a point source. Surround this source with spherical, opaque (for this wavelength) shell. There will be no E-M radiation outside it. Now drill a little holl in the shell and some single ray/beam? should appear.

Now, for this ray/beam do exactly the same, what you've done for point source before. If i'm right, the result will be a circle (2D version) on a plane perpendicular to ray/beam direction. Of course, this circle will be periodically changing it's size, but currently it's not such important.

That's not all. Cut narrow slit somewhere else in the shell and second ray/beam? will escape. You know what to do. Result will be similar to the previous one. You should describe me this like, well maybe not exactly squere but something similar to an ellipse (a shape extended in one direction).

But why to do all of that?!

Well, because life is cruel and sometimes you are getting results without causes. Suppose that i'm just giving you these two rays/beams and you have to guess which is which. How will you do this? Hint: E-M wave is nothing more then electric/magnetic field fluctuations which propagate in space. So what you can do? The only thing you can do is to measure such field!

And this my post all about: I suppose that in E-M radiation generated by antenna [Pic.2] could (i don't know for sure) be something what i've called: SINGLE WAVE (if exists, we should be able to see it at larger distances). I've told why i think so. And now i'm asking if anyone heard about something like that. If anyone know about some results of experiments, showing heterogeneity in intesity of radiationat at larger distances. Such heterogeneities should form some shape and this will be the shape of our single wave. What other question then about shape of E-M field can i ask? The shape will tell me what kind (if any) of relation between photons exists in source radiation.

By the way, question about shape is not such abstract like you may think. It's like analysing ray/beam of laser (such things are performed nowdays - of course for different reasons then my). You are placing detector at the front of ray/beam and you are registering intensity (in other words: strength of the field). It could be said: you are taking picture of ray/beam. And then, you are connecting points with the same intensity on the picture. And the result will be some SHAPE. Field is continuous, so no matter what reference intensity (E) you choose, the shape generally will be the same (it could be just smaller or bigger).

Can we agree, at least on what i've said?

Of course english is not my native language (it's not hard to guess that) - but i'm afraid it has nothing to do with confusion.

I try to carefully analyse your statement You've said: "seemed to be confused" not "be confused". But why? If "confuse" is such neutral word, why to use "seem to"? I see that there is some difference. The question is, do you know what this difference means? This is the language politicians and lawyers are using. But they are not doing so to respect someone's dignity. It's just a way not to be prosecuted by law for defamation. It's not the way to say something politely. You can say that you are confused after reading what i've written, but the only person that has right to say i'm confused is me.

By the way, why to use such statements? They bring nothing.

You are right. But it was my first post. I try to remember about it next time. Sorry.

13. Jun 3, 2014

### Cthugha

Light being built from quanta is NOT classical physics by any means. This idea did not exist in classical physics.

Yes, quantum optics is needed as soon as you really want to include the idea of light "build from some quantas". No, it is not easy to explain that classically. It is not even possible. We have that discussion here on a regular basis. But as you do not seem to be interested in how stuff works, but just in discussing your personal theory, I will indeed go away.

Photons inside and outside em waves? That does not even mean anything. Please keep in mind that we stick to accepted physics here and do not discuss personal theories.

14. Jun 3, 2014

### Staff: Mentor

I don't know what "outside" an EM wave even means.

Not really. Size is a difficult thing to talk about at the atomic level. The wave packet simply represents where a particle or photon might be, not its size. All elementary particles are considered to be point-like with zero size.

Photons, being a quanta of energy, cannot have a size or shape.

15. Jun 3, 2014

### Staff: Mentor

Rays and beams are not real. They are merely ways of making things easier. For example, in optical design the optical system is typically analyzed by use of ray tracing. This gets you a very good approximation of how the system will work, but if you want to be completely accurate you have to calculate wave properties like diffraction, which are much more math intensive than ray tracing.

In your example an EM wave will emerge from the hole and spread out according to the principles of diffraction. The smaller the hole is with respect to the wavelength, the more the hole acts like a point source.

The circle will get larger as the wave propagates, if that's what you mean.

Initially the wavefront will be the shape of the slit, but as it travels further it becomes closer and closer to spherical.

I have no idea what a "single wave" is supposed to be unless you mean that it's a single fluctuation in the EM field. If so, then it still doesn't have any relation to the "shape" of the EM wavefront.

16. Jun 3, 2014

### Simon Bridge

I cannot tell because you have not said what you want to do it for.

This is incorrect. Classical EM has everything to do with electric/magnetic field fluctuations which propagate in space.

All light can be represented as a single wave.
A wave, by definition, is any solution to the wave equation.

I think you are using the term "single wave" in a way that is non-standard.

"Heterogeniety" is, by definition, the property of being diverse.

Indeed there are many experiments showing the diversity in the intensities of light at many different distances.

I cannot tell what other question you can ask because you have not asked the first question yet.

What question about the shape of EM waves are you trying to ask?

I don't know what to think - you have not asked a question yet.

The process you describe is called "mapping the crossectional intensity of the laser beam".
Nobody is disputing that it is possible to do this - we'd usually map a series of curves to get something that looks like a contour map.

... There are native English speakers who write much worse than you, so I don't just assume this sort of thing. In English there is a saying: Never "assume", or you'll make an "***" out of "u" and "me".

... there is certainly a problem with communication. You are using words with a non-standard meaning for example. That is certainly very confusing for the people trying to understand you and is the sort of thing that happens when the speaker/writer is, themselves, confused about something.

Please try to put your ideas in as simple a form as possible.
Please try to stick to standard uses of words.

1. "confused" is not always a neutral word;
2. I did not know if you were confused or not.

You are in a scientific forum talking to scientists, not a political or legal forum. Scientifically we have to be careful to distinguish between how things seem and how they are.

You are wrong about that - it is also possible for someone to look like they are confused - by their actions or words.

In English, "confusion" is the state of being bewildered or unclear in one's mind about something.
An example if the use is "Mary looked about herself in confusion."

Confusion in others is a state which can be determined from observation. It is also possible for someone to be confused without realizing it - another way to say this is that they are mistaken, or that they have taken some ideas out of context or something more specific.

You are writing like someone who is unclear in their mind about the properties of EM radiation.
I allow that this may not be the case.

It is supposed to tell you that you are giving the impression of being confused and offers you a chance to help correct the matter.

The way to go about convincing people you are not confused is to make clear statements ... you have been so busy trying so hard to explain your question that you keep forgetting to say what you question actually is.

It is best practice to ask your question first, then, if other people are confused, they will tell you, and you can explain for them and make your meaning clear.

Perhaps you feel you have asked a clear question?
Then please restate it for me by itself so I can see it.

17. Jun 5, 2014

### gfgts250

I didn't wanted to be rude. I stand by my request rather not to include topics from quantum physics (modern quantum physics if i can say so) here, but of course i should formulate my will in a different way. My fault, sorry if i offended you. I don't want you go away (i have nothing personally against you), i even hope you still will be using your knowledge to help me. But, if i may ask, let it NOT be the knowledge from an area of modern quantum physics, at least not for this thread. I truly think it's not needed here (at least not yet. And later?, Who knows? I will ask THEN).

Well, let me summarize how i see it:

I do not only oppose the fact light can be seen as build from some chunks/pieces, I also think it's crucial information for what i try to discuss. I fully accept and use experiment's results showing this quantum nature of light (blackbody radiation, photoelectric effect even Compton's experiment - which shows that quanta has some momentum - property of particle).​

as you can see: i'm not trying to speak about light using physical knowledge from XVI century. But... on the other hand i feel i don't need:

wave functions, probabilities, quantum entanglement, QEDs, QFTs, all of that.​

So it doesn't work like: all or nothing. The question is: Can i do that? I think it all depends on the scope of knowledge i need from QP (quantum physics) area (For example: I don't need to learn special relitivity and know Lorenz transformations just too say trivial sentence like: "light is not standing but travels in space with some speed". do i?). From all QP i just really need 2 simple informations:

1. E-M radiation is quantized - we can not take arbitrary amount of energy we want from some monochromatic E-M wave with frequency (nu) We have to take at least E = h*nu - this is a lower limit.
2. Those chunks of energy are called quantas/photons. They have of course many interesting properties, but i'm particulary interested in one of them: they can be localized (with some limited accuracy of course) in space. We can detect their position and their direction of motion (where they comes from). It could be said they are behaving like particles - they are generally following in the direction initially given to them.
And this is all i need from QP. What theory do i need for that? You could be suprised but in my specific case, i don't need any theory. All this informaion i can get only from experiments. I can learn about quantization from Photoelectric effect experiment. And i can go to the store, buy some laboratory stuff and play with single quantas/photons. In many laboratory experiments single photons are generated so i assume I can do it too. I can also give them some direction (maybe i have no control over where excited atom releases photon, but i can filter, from previously generated set, these going where i want - it's just like small output window in laser works). The same with detection - there are many detectors which can detect single photons. By proper arrangement of those detectors in space i can determine photon's position and what was the direction of motion (i'm assuming that i know starting position). It's rather you could have some problem, trying to tell me i'm wrong with that. Since you need some experiment to do that. I my opinion (i hope this law still applies) experiment stands beyound any theory. By the way: The whole Quantum physics was born by experiment, not by theory. I'm not quite sure i can say the same for what is today called: modern quantum physics (modern interpretation of QP).

As for your statement: "This idea did not exist in classical physics". Well, everything depends what QP means to you. The most obvious approach is to assume QP as everything that has something to do with QUANTIZATION (hence the name. We can include knowledge from 1900-1920 in that) And belive me, I also THOUGHT so, but i was forced by people like you to change my mind. Since this approach has some consequences: key concept of QP in this case is just QUANTA. And until you are talking about quanta everything is OK. Then your statement is correct: indeed, classical physics is not enough. But, if you feel you can't speak about ANY (even basic ones) concepts of QP without wave function and other such things ("these forums cannot replace a thorough course in quantum optics - including the math") then it just means WAVE FUNCTION is key concept of QP form you, not QUANTA (and for unknown to me reasons you do not want to admit it). Then problem arises: where to classify knowledge from years 1900 - 1920 (maybe even up to 1930)? How people living then were able to speak about quantum physics without knowledge you are assuming essential today? Now, i hope, you understand why i've classified some aspects of QP as classical physics (i can't agree that some people are trying to change theories inpired by experimental results into experiments inspired by theories). So, no. Quantum optics is not needed if i really want to include the idea of light "build from some quantas" (this is from your previous post). This is what already Photoelectric effect told us in circa 1905 (experiment told us so, not theory).

Eventually, if it makes you happy, i can introduce division on: classical physics, quantum physics and modern quantum physics. In this case i can say i'm not interested in the last one.

In my original post i'm starting from E-M waves (classical physics) and i'm ending with E-M waves. Just in between i'm using a little bit QP, but only those things described previously. I must admit, later i metioned about things that COULD SUGGEST i want to talk about some aspects of modern QP but i've made it unintentionally. I mean, since no one understood me i've used some wrong examples and got into unneeded discussions. I regret that.

Is this such abstract? It's just like free electron and electron that is bound to an atom. The same way we can have free photon (i've unnecessary used term: "photon outside") - photon distant from the other photons if you like (it reminds us particle). We can localize it - detect it's position/direction of motion. Or we can have E-M wave, which COULD BE (just could) interpreted as stream of photons. This is how LASER works - single photons are generated inside and released in bunches. So i think it could be seen like that. In this case we are not be able to localize such photons, we only would be able to speak about position of the whole wave (and only say it's build from photons). I don't consider it as my personal theory - it's obvious.

18. Jun 5, 2014

### Staff: Mentor

A few things:

1. Classical physics and quantum physics are two different theories. The beginnings of quantum physics has its roots in the 1800's with several experiments and observations, but prior to the formulation of Quantum Mechanics in the 1920's and 1930's there was no theory that adequately described the atomic and subatomic world. The ideas put forward from about 1900 until the formulation of QM is termed the "old quantum theory" and is essentially a "stepping stone" from classical to quantum physics. Attempting to talk about EM waves must either involve photons and use a quantum theory, or it must not. One cannot talk about photons and not delve into modern quantum theories, as they are exactly what describe what a photon is and how it works. This isn't to say that we need to get into the nitty gritty details, it is simply to point out that attempting to talk about photons within the realm of classical physics is doomed to misunderstandings and confusion, as this very thread has pointed out.

2. The beginning of the 20th century had many experiments which disagreed with accepted physics. It was a time when we had a great wealth of experimental data, but no theoretical models, no theories, in which to explain them. It was only by building a mathematical theory that explained these results that moved us forward. So your claim that modern Quantum Physics was born by experiments and not theory is false. ALL scientific advances require both experiments and a theory. The biggest difference today in theoretical physics is that we have many theories but lack sufficient data to determine which one of them is correct.

3. Your understanding of EM waves, lasers, and photons is incorrect. A laser emits EM waves. Period. No one can claim that it doesn't because that is what light is, an EM wave. However, EM waves are quantized in their energy. A single photon sources emits an EM wave with an amount of energy equal to exactly one photon. Because this is still an EM wave, it will still show all the properties of waves, such as interference and diffraction. A laser does not emit a bunch of photons which happen to turn into an EM wave simply because there are a lot of them. It emits EM waves which are quantized.

Note that quantum physics does NOT do away with the idea of an EM wave. It doesn't say that EM waves are not really waves. What it does is it adds to the idea of an EM wave by including quantization of the energy as photons. See the following quote from: http://en.wikipedia.org/wiki/Photon#Wave.E2.80.93particle_duality_and_uncertainty_principles

Also:

The key here is that the probability distribution of the photons follows Maxwell's Equations and that the theory that fully describes EM waves using a quantum theory is Quantum Electrodynamics. Per QED, you can't even talk about the path the photon takes between emission and absorption. In order to find the probability of a photon arriving at a location after emission, you must take into account ALL possible paths, not just one. So you can be sure of the source of the photon, but not the exact path it took through space to get from the source to the detector. You can't even localize the photon without destroying it, unlike other particles.

I'll say it one more time to be clear. Photons are NOT little particles. EM waves are NOT the result of many of these particles being in the same place. A single photon is just as much of an EM wave as a billion photons are.

As always, someone correct me if I'm wrong.

19. Jun 5, 2014

### UltrafastPED

Every EM wave can be described in terms of its modes; the photons are the quantization of the modes of the electromagnetic wave.

PS: In laser physics it is often necessary to consider the light as consisting of photons; and just as often it is necessary to ignore the photons and consider everything as electromagnetic waves. I've done experiments where both were important considerations, but at different locations within the same experiment!

20. Jun 5, 2014

### Staff: Mentor

Interesting. A quick search on the modes of an em wave linked me to the following article: http://en.wikibooks.org/wiki/Quantum_Mechanics/Waves_and_Modes