Can the amplitude of an EM wave affect its physical size?

[fluidistic]

In the case of plane waves, E is orthogonal to B and they're both orthogonal to the direction of propagation, call it k.
I'm not sure I'm picturing well what such an EM wave is. For instance I know that E and B oscillates with respect to time.
Without looking to quantum electrodynamics that I don't know at all, let's say I have such a wave with a huge amplitude. I.e. the E and B fields are really strong. Does that mean that the EM wave occupy more place than a similar EM wave but with a low amplitude?

In fact I'm not sure whether the EM wave can be pictured as a line or as a 2 orthogonal "sine functions" in the 3d world. If the latter case is correct then a bigger amplitude seems to imply that the EM wave indeed occupy more place than a EM with a lower amplitude. My problem arises if say I have an enormous amplitude of E and B and that the frequency of the wave is about $$10^{-6}Hz$$. It could be possible in this example that in a time of about $$1/10^{-6}s$$, the space perturbation (the EM wave traveling) can reach say a distance of about 300 000 km from the source which is about $$10^{6}$$ times faster than the speed of light which is of course impossible. I understand that in the direction k the speed of the wave is c, or the speed of light. But I'm not understanding if the amplitude of the wave is related to the wave's place in the 3d world.

 

[gabbagabbahey]

Without looking to quantum electrodynamics that I don't know at all, let's say I have such a wave with a huge amplitude. I.e. the E and B fields are really strong. Does that mean that the EM wave occupy more place than a similar EM wave but with a low amplitude?

No. For example, let's say $\textbf{E}_1=\textbf{E}_0\cos(\textbf{k}\cdot\textbf{r}-\omega t)$ and $\textbf{E}_2=2\textbf{E}_0\cos(\textbf{k}\cdot\textbf{r}-\omega t)$ ...the only difference between the two fields is their amplitude. A test charge placed at any given point in space will experience twice as much force (in the same direction) from $\textbf{E}_2$ as it would from $\textbf{E}_1$.

In fact I'm not sure whether the EM wave can be pictured as a line or as a 2 orthogonal "sine functions" in the 3d world. If the latter case is correct then a bigger amplitude seems to imply that the EM wave indeed occupy more place than a EM with a lower amplitude.

You can't really picture EM waves in this way. At any given point in time, $\textbf{E}_1$, for example,. and its corresponding magnetic field, will occupy every point in the plane of oscillation. At each point, the field will have some specific magnitude between zero and its amplitude, and will point in the direction of $\textbf{E}_0$.

 

[fluidistic]

Hey there, thanks for your reply.

No. For example, let's say $\textbf{E}_1=\textbf{E}_0\cos(\textbf{k}\cdot\textbf{r}-\omega t)$ and $\textbf{E}_2=2\textbf{E}_0\cos(\textbf{k}\cdot\textbf{r}-\omega t)$ ...the only difference between the two fields is their amplitude. A test charge placed at any given point in space will experience twice as much force (in the same direction) from $\textbf{E}_2$ as it would from $\textbf{E}_1$.

Ok I understand.

You can't really picture EM waves in this way. At any given point in time, $\textbf{E}_1$, for example,. and its corresponding magnetic field, will occupy every point in the plane of oscillation. At each point, the field will have some specific magnitude between zero and its amplitude, and will point in the direction of $\textbf{E}_0$.

When you say "will occupy every point in the plane of oscillation", do you mean that the E and B fields have an influence on charges that are on this plane? But the plane is infinitely large, right? So does it mean I can know an EM wave has passed close to me if I get this E and B field, without "looking" directly at the wave? I.e. without absorbing the photon or the wave.

 

[jtbell]

See the following post, which contains my attempt to make a "better" diagram of a plane electromagnetic wave than the ones that you find in many books:

https://www.physicsforums.com/showpost.php?p=533190&postcount=6

Note also that the amplitude of the wave is independent of the size of volume of space that it occupies. The units are completely different: electric field strength in volts/meter or magnetic field strength in tesla, versus beam diameter in meters. A large diameter beam can contain either a large-amplitude ("strong") or a low-amplitude ("weak") wave, and a narrow beam can contain either a strong or weak wave. The amplitude of the wave corresponds to the intensity or "brightness," not to its physical size.

 

[fluidistic]

See the following post, which contains my attempt to make a "better" diagram of a plane electromagnetic wave than the ones that you find in many books:

https://www.physicsforums.com/showpost.php?p=533190&postcount=6

Note also that the amplitude of the wave is independent of the size of volume of space that it occupies. The units are completely different: electric field strength in volts/meter or magnetic field strength in tesla, versus beam diameter in meters. A large diameter beam can contain either a large-amplitude ("strong") or a low-amplitude ("weak") wave, and a narrow beam can contain either a strong or weak wave. The amplitude of the wave corresponds to the intensity or "brightness," not to its physical size.

Hey, thanks a lot for this insight.

Since all the plane is filled with this EM field, I can know an EM wave has passed close to me if I am an electron, I mean I'd suffer a force. So light passing close to conductors could potentially induce an at least very small emf?

 

[diazona]

Since all the plane is filled with this EM field, I can know an EM wave has passed close to me if I am an electron, I mean I'd suffer a force. So light passing close to conductors could potentially induce an at least very small emf?

If you're an electron, you'll know if an EM wave passes through you because you'll feel a force. Just passing close to you isn't enough.

 

[fluidistic]

If you're an electron, you'll know if an EM wave passes through you because you'll feel a force. Just passing close to you isn't enough.

Oops you're right. I still imagined the EM wave as a line propagating through space as time goes on.
Now looking back to jtbell's sketch, the E field is defined along the whole y-axis and the B field along the whole z-axis. The wave is propagating to the x-axis. So the wave is defined over an infinitely large plane? In fact I should say 3 dimensional region in space.
Am I right imagining the EM wave as if I was in a plane and looking to the sea, the front wave going to the sand if this is infinitely large? Meaning that the water wave will eventually cover the whole coast.

I don't think so... or is the jtbell's sketch+ its description a sort of steady state EM wave, meaning that the wave constantly exists where it was previously and has expanded along the whole y-axis and z-axis although in reality it's not possible? Hmm don't think so either, the only direction of propagation is the x-axis so what I've just thought can't be write. I'm still extremely confused.

 

[jtbell]

Note that in my instructions, you have to complete the picture of the E field by making many copies of my picture and "stacking" them one one above another, about a cm apart, so as to extend the picture along the +z and -z directions (in front of and behind the plane of the "paper").

The best way to display this would be a block of clear plastic with arrows embedded all through its interior, that you could hold in your hands and turn to look at it from varioius directions. Or maybe a 3-D computer graphic that you could rotate interactively. Unfortunately, neither my craftsmanship skills nor my programming skills are up to that task.

Also note that I didn't attempt to represent the boundaries of the beam (maximum +y, -y, +z, -z). There must be "fringing" effects at the boundaries, but I don't know what they look like. The diagram is supposed to represent what the fields look like deep inside the beam, far from the boundaries.

 

[fluidistic]

Note that in my instructions, you have to complete the picture of the E field by making many copies of my picture and "stacking" them one one above another, about a cm apart, so as to extend the picture along the +z and -z directions (in front of and behind the plane of the "paper").

The best way to display this would be a block of clear plastic with arrows embedded all through its interior, that you could hold in your hands and turn to look at it from varioius directions. Or maybe a 3-D computer graphic that you could rotate interactively. Unfortunately, neither my craftsmanship skills nor my programming skills are up to that task.

Also note that I didn't attempt to represent the boundaries of the beam (maximum +y, -y, +z, -z). There must be "fringing" effects at the boundaries, but I don't know what they look like. The diagram is supposed to represent what the fields look like deep inside the beam, far from the boundaries.

Oh I see. This is really helpful, I thought that the beam was infinitely "large".
I'm precisely having a doubt about its length. I guess the solution to the wave equation for EM waves could solve this doubt, right? Say I fix a time, could I see how extended is the wave? Does its largeness changes with time?
Am I right imagining an EM wave as the trace a plane leaves in the sky? At first it's thin and then it expands with time and the older trace is greater than the newer trace. The EM wave would affect charges that were not that far from the source even if the charges are not in the straight line of propagation of the wave. If the wave enlarges with time, any charge close enough to the source will feel the wave going through them. Is this a right picture?

Edit: Also, shouldn't there be a tiny emf in transparent materials like common glasses for example? Light passes through it and thus the free charges of the glass should suffer forces and charges in motion implies an emf. I'm not sure glass has a lot of free charges. But metals do, so light passing through metal should create an emf. Is this right?

 

[fluidistic]

Ok I'm maybe not explaining well my doubt. Please tell me if all is unclear to you.
Say I have a laser pointing on a wall. I put a static electron close to the laser beam (around 1 cm apart from it). So the electron is not on the path of the light, at least initially. Now I want to know if the electron, soon or later, will suffer the force from the EM waves of the laser. If not, it means that the EM waves extent in space, or better saying "largeness" or I don't know in English (épaisseur in French) is not infinite and is well defined. I would like to know how to calculate it. I.e. I'd like to know how close should I put an electric charge of an EM wave in order for it to suffer a Lorentz force. The answer is pretty obvious, you'll say on the path of the light beam, but my problem is that this isn't a line, isn't? If it's not a line, I'd like to know how "large" is the EM wave. So that I can finally imagine how it is.

 

[gabbagabbahey]

Ok I'm maybe not explaining well my doubt. Please tell me if all is unclear to you.
Say I have a laser pointing on a wall. I put a static electron close to the laser beam (around 1 cm apart from it). So the electron is not on the path of the light, at least initially. Now I want to know if the electron, soon or later, will suffer the force from the EM waves of the laser. If not, it means that the EM waves extent in space, or better saying "largeness" or I don't know in English (épaisseur in French) is not infinite and is well defined. I would like to know how to calculate it. I.e. I'd like to know how close should I put an electric charge of an EM wave in order for it to suffer a Lorentz force. The answer is pretty obvious, you'll say on the path of the light beam, but my problem is that this isn't a line, isn't? If it's not a line, I'd like to know how "large" is the EM wave. So that I can finally imagine how it is.

A laser isn't a plane wave, but rather an infinite sum of plane waves that results in net fields that are mostly confined to a small area---the fields of a laser die out rapidly as distance from the beam increases. So no, an electron placed a distance away from a laser beam won't experience any net force.

 

[fluidistic]

A laser isn't a plane wave, but rather an infinite sum of plane waves that results in net fields that are mostly confined to a small area---the fields of a laser die out rapidly as distance from the beam increases. So no, an electron placed a distance away from a laser beam won't experience any net force.

I thank you very much for your reply, once again. Pay attention to the part I put in bold please. Is there a mathematical formula I could look at in order to see this phenomenon mathematically? Is it true for any other EM field (light "beams")?

 

[jtbell]

I'd like to know how close should I put an electric charge of an EM wave in order for it to suffer a Lorentz force. The answer is pretty obvious, you'll say on the path of the light beam, but my problem is that this isn't a line, isn't? If it's not a line, I'd like to know how "large" is the EM wave. So that I can finally imagine how it is.

Put a card or something in the path of the laser beam and you can see the size of the beam, i.e. the diameter of the bright spot formed on the card. That's the region occupied by the EM wave of the light beam. Where there is no light, there is no EM wave, or at least it's negligible.

If you want to know how to predict (calculate) the size of the beam, that depends on the construction of the laser, probably on the size and shape of the laser cavity where the light is produced. I've never studied the details of laser design, so I don't know where you would find such information.

 

[fluidistic]

Put a card or something in the path of the laser beam and you can see the size of the beam, i.e. the diameter of the bright spot formed on the card. That's the region occupied by the EM wave of the light beam. Where there is no light, there is no EM wave, or at least it's negligible.

If you want to know how to predict (calculate) the size of the beam, that depends on the construction of the laser, probably on the size and shape of the laser cavity where the light is produced. I've never studied the details of laser design, so I don't know where you would find such information.

Ok, thank you for this,it's helpful information. More interesting than the diameter of the beam of light coming from a laser, I'm interested in knowing it for a single EM wave. If I send one EM wave, how thick will it be? Is it of a constant thickness? (I think it must be, otherwise as stated in my first post, since its speed is the speed of light in the direction of the wave's propagation, if it moves along any other axis then its total speed would be greater than the speed of light).
p.s.: the word I was looking for was "thickness".

 

[Born2bwire]

Ok, thank you for this,it's helpful information. More interesting than the diameter of the beam of light coming from a laser, I'm interested in knowing it for a single EM wave. If I send one EM wave, how thick will it be? Is it of a constant thickness? (I think it must be, otherwise as stated in my first post, since its speed is the speed of light in the direction of the wave's propagation, if it moves along any other axis then its total speed would be greater than the speed of light).
p.s.: the word I was looking for was "thickness".

What do you mean thickness and what do you mean by a single wave? I think you should probably sit down with an antennas textbook and you could probably find out what you want to know. Antennas are generally designed to direct the transmitted and received radiation. You can send a beam that covers a large area of space or a small area. You can send a long pulse in time or a short pulse.

 

[jtbell]

If I send one EM wave, how thick will it be?

What exactly do you mean by "one EM wave"? How do you propose to produce it?

If I send one sound wave, how thick will it be?

 

[Dale]

A wave is a solution to the wave equation:
$$\frac{\partial^2 u}{\partial t^2 } = c^2 \nabla^2 u$$

There are solutions to this equation which are thick or thin, there are solutions which occupy a whole plane of infinite extent, and those which are narrowly focused. The spatial extent of the solution is not constrained.

 

[fluidistic]

What do you mean thickness and what do you mean by a single wave? I think you should probably sit down with an antennas textbook and you could probably find out what you want to know. Antennas are generally designed to direct the transmitted and received radiation. You can send a beam that covers a large area of space or a small area. You can send a long pulse in time or a short pulse.

By thickness I mean that if the EM wave occupy a cylinder with expanding height at the speed of light then the thickness would be the radius of the cylinder.
I agree, I should read some books. Until now I've always seen the picture of the EM field oscillating in time and space while it's a field and not the dimensions of the wave. I'm interested in the dimensions of the wave, more precisely the "thickness" of it.

What exactly do you mean by "one EM wave"? How do you propose to produce it?

If I send one sound wave, how thick will it be?

I think the thickness of the sound wave would be the radius of the spherical perturbation. It's increasing with time and space. You point a good example, I was wondering if the EM waves were similar. Maybe the spherical waves?
About producing a single EM wave, I have no idea. Oh I just have an idea. Take a laser and close very slowly the emitting part. There will be a moment where no light can pass and we would see no light. Reopen it a very very bit, so that only "1 ray" of light pass.

A wave is a solution to the wave equation:
$$\frac{\partial^2 u}{\partial t^2 } = c^2 \nabla^2 u$$

There are solutions to this equation which are thick or thin, there are solutions which occupy a whole plane of infinite extent, and those which are narrowly focused. The spatial extent of the solution is not constrained.

That's what I was guessing in post #10.
About the solutions that occupies a whole plane of infinite extent, of course it doesn't happen in the universe and I guess it's a simplification we do and it works well for most of our experiments. But say, nevertheless, that the thickness is greater than 300 000 km. Now I send one of those waves in the z-axis direction. After half a second the wave has made 150 000 km in the z-axis, nothing astonishing. However in the x-y plane (or at least along the x-axis) the wave is already 300 000 km large (and this distance could remain fix, I'm not really sure, I'd have to look at the solution to the wave equation).
So the wave is in this case faster than the speed of light. Not in the sense of propagation but in its extent in the perpendicular planes to the axis of propagation.
Obviously it cannot be so. What am I misunderstanding?

 

[Dale]

So the wave is in this case faster than the speed of light. Not in the sense of propagation but in its extent in the perpendicular planes to the axis of propagation.
Obviously it cannot be so. What am I misunderstanding?

I don't understand what you are saying. Can you write an equation to describe the wave you are considering?

 

[fluidistic]

I don't understand what you are saying. Can you write an equation to describe the wave you are considering?

I'm not sure how such an equation would be. If r(x,y,z) and z is the direction of propagation, E(r,t) would be defined at x=300 000 and t=0. I just realize that my example in the end of my previous post can be simplified. How can the EM field be different from 0 when the wave is emitted precisely at t=0?


I have scanned some sketches. Part (1) was a doubt I had. That is, if the electric charge will eventually suffer a Lorentz's force due to the laser beam. Now I know that the answer is no. I represented the thickness of the laser beam as d. My question is what is the thickness of a single EM wave?

In (2) I show the infinite plane in which is defined the EM field at a fixed time. My question is: are the E and B field different from 0 for all points in the plane? If not, why do we say that the plane is infinite?
(3) shows the simplified example of my last post. E and B are defined at say a not null distance from the source (I forgot to draw it, it should be over the z-axis at the intersection with the sketched plane) at t=0, the time where the EM wave is emitted.


I'm considering plane waves, not spherical waves.
I hope my doubt is clearer.

All in all I'm asking what is the thickness of an EM plane wave? If it is finite, why do we say the planes where E and B are defined are infinite? Does this mean that E and B can be different from 0 anywhere in the infinite plane? If so... then I can define E and B so far if I divide the distance by the time since the wave has been emitted, I enconter a velocity greater than the one of the speed of light in vacuum. So at least one reply of my questions is no.

 

[fluidistic]

Last attempt to explain my doubt

Hey there. I will make a last attempt to explain my doubt.
To start, take the image I uploaded in my previous post. See figure 2.
Since E and B are vector fields of infinite extension, it doesn't matter if we represent them by a single vector centered over the z-axis or anywhere else, as long as the direction and modulus of them doesn't change.
My doubt is whether the E and B fields are defined in all the x-y plane and in case they aren't, why? What equation allow me to determine their extension in space? Is there a discontinuity in the fields if both have a finite extension? What exactly determine their extension?

 

[Dale]

Let me see if I can explain this well.

Suppose at some instant in time t=0 the spatial shape of the E field is:
$$\vec{E}(0,x,y,z)=f(z)\hat x$$
This is your figure 2.

Then the plane wave
$$\vec{E}(t,x,y,z)=f(z-ct)\hat x$$
is a valid solution of Maxwell's equations in vacuum for an appropriate choice of B.

However, if the spatial shape at t=0 is given by:
$$\vec{E}(0,x,y,z)=u(x)u(y)f(z)\hat x$$
where $$u(x) = \left\{ \begin{array}{lr} 1 & : 0<x<300000 \\ 0 & : otherwise \end{array} \right.$$
This is your figure 3.

Then
$$\vec{E}(t,x,y,z)=u(x)u(y)f(z-ct)\hat x$$
is not a valid solution of Maxwell's equations in vacuum.

 

[fluidistic]

Let me see if I can explain this well.

Suppose at some instant in time t=0 the spatial shape of the E field is:
$$\vec{E}(0,x,y,z)=f(z)\hat x$$
This is your figure 2.

Then the plane wave
$$\vec{E}(t,x,y,z)=f(z-ct)\hat x$$
is a valid solution of Maxwell's equations in vacuum for an appropriate choice of B.

However, if the spatial shape at t=0 is given by:
$$\vec{E}(0,x,y,z)=u(x)u(y)f(z)\hat x$$
where $$u(x) = \left\{ \begin{array}{lr} 1 & : 0<x<300000 \\ 0 & : otherwise \end{array} \right.$$
This is your figure 3.

Then
$$\vec{E}(t,x,y,z)=u(x)u(y)f(z-ct)\hat x$$
is not a valid solution of Maxwell's equations in vacuum.

Amazing, really amazing. I'd have to check it out. A little question: why did you assume that $$\vec{E}(0,x,y,z)=u(x)u(y)f(z)\hat x$$ for the figure 3 while there was no u(y) dependence for figure 2?
Thanks so much! I really have to see why Maxwell's equations are not satisfied. Maybe this weekend.

 

[Born2bwire]

Amazing, really amazing. I'd have to check it out. A little question: why did you assume that $$\vec{E}(0,x,y,z)=u(x)u(y)f(z)\hat x$$ for the figure 3 while there was no u(y) dependence for figure 2?
Thanks so much! I really have to see why Maxwell's equations are not satisfied. Maybe this weekend.

The discontinuity of the electric field in a homogeneous medium is causing the problems. The fields can be discontinuous, but this arises due to boundary conditions where the permittivity and/or permeability are discontinuous across the boundary. So an easy way to look at this is purely from the boundary conditions. The tangential electric field is continuous across a boundary and the normal electric flux density is continuous. If we take say x=0 as an artificial boundary, then \epsilon E_x must be continuous but this cannot happen since you have a discontinous electric field and a homogeneous background.

 

[Dale]

A little question: why did you assume that $$\vec{E}(0,x,y,z)=u(x)u(y)f(z)\hat x$$ for the figure 3 while there was no u(y) dependence for figure 2?

I don't understand, I didn't think I made any assumption, I thought that is what you drew. I thought you said figure 2 was an infinite x-y plane and figure 3 was a 300000 km square in the x-y plane. Did I misunderstand your drawing?

 

[fluidistic]

I was going to reply here but I decided to think more. I'll post a new sketch, my previous one was not very precise.
Ok thank you Born2bwire, this is of great interest.
Hey Dalespam, actually you understood well my sketch from your description, but now to me the sketch doesn't make the sense I intended to put in.
In case your answer (to Dalespam) still holds, please let me know.
---------------------------------------------------------------------
Description of the new sketch: Since the EM waves is a EM field, I represent the field as several vectors situated randomly in the field, instead of just 1 vector as usually seen in most books. I also didn't drew the B field which is of course present.
(2)- I represent the x-y plane at a given time. The E field has a constant value different from 0 over the whole plane, i.e. it is infinite in extent. So there's no discontinuity and therefore I'm almost sure such a field satisfies Maxwell's equation.

Sidenote: But I've been told in the early posts of this thread if I remember well, that the E field is not different from 0 in all the plane. Hence my suggestion that it was defined over the whole plane and that there was a discontinuity, passing from a non zero value to 0, but Born2bwire showed me such a field doesn't satisfies Maxwell's equations, so I'm sure such a field cannot exist.
Now I come up with a new idea:
(1)- Same as before: t=0. I drew a circle inside the x-y plane. Inside the circle the E field is absolutely constant. Over the circle the E field decreases quite fast and reaches 0 outside the circle, say by a function of the type of a Gaussian or so. Does such a wave satisfies Maxwell's equations? I see no reason why not.
----------------------------------------------------------------

The circle of the figure (1) is what I call "thickness" of the EM wave. I have never had any comment in the course I'm taking about the spatial representation of an EM wave. I wanted and still want to be able to imagine how is a single EM wave. That's why I started this thread.


Other comment: I first thought that (2) was possible, moreover I though that any ray of light had an infinite extension in the x-y plane, in classical EM. All this because I've read that the planes perpendicular to the axis of propagation of the wave are infinitely large. But any plane is actually infinite, so I thought they mean that the E field had a non zero value in all the plane, which clearly isn't true.
I want to see a mathematical equation that describes how the "borders" of a plane EM wave behave. I drew a circle because it was the most natural to me, but I'm not sure it makes sense.

Also notice that all the equations I've been given in class which satisfy the wave equation give information about the spatial extension on the z-axis (axis of propagation) and never give any information about the spatial extension of the wave in the x-y plane. I realize now that in class we've always assumed that if the value of E over the z-axis at a given time was say 10V/m, then it was so in all the x-y plane perpendicular to the given z. Now I know it's totally wrong and impossible to say so.

Lastly: The modified doubt I still have is the following one: since the E field is defined in a certain non infinite region in the x-y plane, is it so since the EM wave (I'm talking about a single EM wave) was created? What exactly determines the spatial extension in the x-y plane of the EM wave? If it is for example defined as a circle, at what speed was it created? I have a "video" in my mind like this: At t=0, a spherical wave with no spatial extension is created. Moments later it's like a circle in a plane I see from above. Now we only consider say the right part of the wave front. It looks like a plane wave when t increases. In this case the spatial extension when the wave was created (t=0) was 0, unlike what I previously thought.

Sorry if the message is too long or if I'm still not expressing myself well. Please let me know anything you want me to precise. I'm extremely worried about these basics doubt: I can't picture what an EM wave looks like in Classical electromagnetism. When I'll take a QM course I'll understand I can't picture such a wave, but in Classical theory, I think I must understand how to represent such a wave and understand well how it behaves.
Thanks a lot for all.

 

[Dale]

I wanted and still want to be able to imagine how is a single EM wave. That's why I started this thread.

I guess the part that confuses me is the use of the word "single" here. There are functions which are solutions to Maxwell's equations, and functions which are not solutions to Maxwell's equations. There isn't really a fundamental unit of Maxwell's equations that you could call a "single EM wave", any solution is equally valid as any other solution.

However, Maxwell's equations are linear so it is possible to expand an arbitrary solution in terms of some set of basis functions. However, that basis set is not natural nor unique and its basis functions are not physically more privileged than other solutions. Usually you choose the basis functions based on the geometry of the problem and for convenience.

One such basis set is the Fourier basis (sin and cos), which is essentially decomposing the solution into an infinite series of plane waves. Then, as long as you understand how one plane wave behaves, and you understand how to add plane waves, then you can understand any wave.

Do you know the Fourier transform?

 

[fluidistic]

I guess the part that confuses me is the use of the word "single" here. There are functions which are solutions to Maxwell's equations, and functions which are not solutions to Maxwell's equations. There isn't really a fundamental unit of Maxwell's equations that you could call a "single EM wave", any solution is equally valid as any other solution.

However, Maxwell's equations are linear so it is possible to expand an arbitrary solution in terms of some set of basis functions. However, that basis set is not natural nor unique and its basis functions are not physically more privileged than other solutions. Usually you choose the basis functions based on the geometry of the problem and for convenience.

One such basis set is the Fourier basis (sin and cos), which is essentially decomposing the solution into an infinite series of plane waves. Then, as long as you understand how one plane wave behaves, and you understand how to add plane waves, then you can understand any wave.

Do you know the Fourier transform?

Ok I see. I'm having lots of problems to imagine what I'm dealing with. As a single wave I think that it's the wave that would pass if I drill a very very tiny hole in a material plane. Am I right saying that if the hole is not large enough, no EM wave can pass? (it might depends on the frequency, but for any frequency there exist a diameter of a hole where the EM wave can start to pass). If the answer is yes then what I have in mind as a single EM wave is the wave passing through the thiner hole possible for that EM wave to pass through the hole of the plane. Or can it be considered as 1 photon? I mean, a single wave=1 photon?
Ok I get that Maxwell's equations are linear and so any independent set of solutions can be added together to form another solutions and that the sum of any solutions is also a solution and there's no privileged solution.
About Fourier transform, I've seen it as a commentary in my course (so for about 5 minutes) and we didn't have any exercise associated to it, so I vaguely have an idea about what it is, but not that much. I'm interested in it tough.

 

[Dale]

Ok I see. I'm having lots of problems to imagine what I'm dealing with. As a single wave I think that it's the wave that would pass if I drill a very very tiny hole in a material plane.

You are talking about diffraction by a circular aperture.
http://en.wikipedia.org/wiki/Diffraction#Diffraction_by_a_circular_aperture

In the limit as the aperture goes to zero you get a spherical wave from a point source.
http://en.wikipedia.org/wiki/Wave_equation#Spherical_waves

Ok I get that Maxwell's equations are linear and so any independent set of solutions can be added together to form another solutions and that the sum of any solutions is also a solution and there's no privileged solution.
About Fourier transform, I've seen it as a commentary in my course (so for about 5 minutes) and we didn't have any exercise associated to it, so I vaguely have an idea about what it is, but not that much. I'm interested in it tough.

Well, basically the point is that it is possible to use the Fourier transform to decompose an arbitrary wave into a sum of plane waves. If you wanted to you could think of each plane wave as a "single wave" just as easily as you could think of a point source as a "single wave".

 

[fluidistic]

Ah yes, you're right Dalespam.

Last question, if it makes any sense: How behaves the EM field at the border of a Ne-He laser beam light? By borders I mean the circumference of the cylinder the beam light could be represented. I hope I'm making sense.

 

[Dale]

It diverges, but not much.