Basic doubt and EM waves

1. May 3, 2010

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.

2. May 3, 2010

gabbagabbahey

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 $\textbf{E}_1$.

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 $\textbf{E}_0$.

3. May 3, 2010

fluidistic

Ok I understand.

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.

4. May 3, 2010

Staff: Mentor

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.

Last edited: Aug 2, 2014
5. May 3, 2010

fluidistic

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?

Last edited by a moderator: Aug 2, 2014
6. May 3, 2010

diazona

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.

Last edited by a moderator: Aug 2, 2014
7. May 3, 2010

fluidistic

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.

Last edited by a moderator: Aug 2, 2014
8. May 3, 2010

Staff: Mentor

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.

Last edited: May 3, 2010
9. May 5, 2010

fluidistic

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?

Last edited: May 5, 2010
10. May 6, 2010

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.

11. May 6, 2010

gabbagabbahey

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.

12. May 6, 2010

fluidistic

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")?

13. May 6, 2010

Staff: Mentor

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.

Last edited: May 6, 2010
14. May 6, 2010

fluidistic

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".

15. May 7, 2010

Born2bwire

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.

16. May 7, 2010

Staff: Mentor

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?

17. May 7, 2010

Staff: Mentor

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.

18. May 7, 2010

fluidistic

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.

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.
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?

19. May 7, 2010

Staff: Mentor

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

20. May 7, 2010

fluidistic

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.

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21. May 12, 2010

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?

22. May 12, 2010

Staff: Mentor

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:
$$\vec{E}(0,x,y,z)=f(z)\hat x$$

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 $$u(x) = \left\{ \begin{array}{lr} 1 & : 0<x<300000 \\ 0 & : otherwise \end{array} \right.$$

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.

Last edited: May 12, 2010
23. May 12, 2010

fluidistic

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.

24. May 12, 2010

Born2bwire

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.

25. May 13, 2010

Staff: Mentor

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?