# E+M fields for a localized pulse

1. Nov 19, 2007

### JustinLevy

Hello, I am having trouble finding solutions for a localized electromagnetic pulse in 3-D. Ideally, I'd like the feilds to be zero outside of a region centered on the pulse, but a gaussian wavepacket is fine with me as well.

Let's have the pulse travel in the z direction, the plane wave solutions look like:
$$\vec{E}(x,y,z) = \vec{E}_0 \cos(kz - \omega t)$$
where $$k = \omega/c$$

The constraint from Maxwell's equations, $$0 = \nabla \cdot \vec{E} = -[E_{0}]_z \ k \sin(kz - \omega t)$$ shows us that the electric field needs to be transverse to the direction of motion (ie. that E_z = 0).

Using Fourier transforms I can easily build a gaussian pulse (or any other shape) in the z direction using these plane waves. Basically, I can always write a solution of the form:
$$\vec{E}(x,y,z) = \vec{E}_0 \ f(kz - \omega t)$$
for any function f. But the result is still an infinite plane in the x-y direction. How do I make this a localized pulse?

I still want the pulse to travel in the z direction, and let's just choose linear polarization in the x direction, and then generalizing this with dependence in the x-y direction it would be something like:
$$\vec{E}(x,y,z) = \hat{x} \ E_0 \ g(x,y) \ f(kz - \omega t)$$

Now seeing the constraint on g(x,y) we find:
$$0 = \nabla \cdot \vec{E} = E_{0} \ f(kz - \omega t) \frac{\partial}{\partial x}g(x,y)$$

Which seems to say there can't be any dependence on x! What!?

What am I doing wrong? What is the correct way to continue to get some localized electromagnetic pulse. Heck, even localized in two dimensions (a continuous laser beam for example) would be useful.

2. Nov 19, 2007

### Claude Bile

I might have missed something obvious, but didn't you assume a planar wave form to start with? You wouldn't expect anything to depend on x in that case.

Try using an initial form that has a spectrum of k values. Or, you could assume the spatial shape of the pulse is Gaussian to begin with and just show that it is an acceptable solution to the scalar wave equation.

Claude.

3. Nov 20, 2007

### DaTario

Try superposing several k valued waves and you start having localization of the wave packet.

DaTario

4. Nov 22, 2007

### JustinLevy

I'm sorry if I am misunderstanding what both of you are suggesting, but you say:
"Try using an initial form that has a spectrum of k values"
or
"Try superposing several k valued waves and you start having localization of the wave packet."

But I have already done that. I can superimpose a spectrum of k to acheive localization in one direction. This I have already done above. I can put in any function I want for f(). But I can't figure out how to localize it in any other directions.

Are you saying there is a more general form for an electromagnetic wave moving in the z direction and polarized in the x direction than this:
$$\vec{E}(x,y,z) = \hat{x} \ E_0 \ g(x,y) \ f(z - ct)$$

That looks pretty general to me. But with the math above, I seem to show that this cannot have any x dependence and therefore there can be no such pulse localized in the x direction. We need to either conclude maxwell's equations don't allow such a localized state, or that there is a more general way to write this.

I assume that there is a more general form for an electromagnetic wave moving in the z direction and polarized in the x direction. The question is: What is this more general form?

That as far as I know is the most general form for such a wave.

5. Nov 25, 2007

### DaTario

localization has implications (see HUP) in dispersion. The more localized is the pulse the more dispersion it presents (one must be careful, however, to consider some special cases where interference is such that it produces a non dispersive pulse - or solitary waves).

Best Wishes

DaTario

6. Nov 25, 2007

### Claude Bile

You have separated out the z dependence from the x and y dependence, you can't do this in the general case. By separating out the variables in this way, you have actually enforced the condition that there is only a single value for k present (assuming a monochromatic wave) - because you have enforced the condition that the wavefronts are planar and invariant with distance.

Note that it is impossible to get a wave of the form you describe unless g(x,y) = 1. As soon as you truncate any real wave, the wavefront will begin to bend and z will no longer be separable from x and y.

Claude.

7. Nov 25, 2007

### JustinLevy

I am not discussing solutions in a dispersive media (instead I am discussing solutions in vaccuum). And I am also considering classical electrodynamics (Maxwell's equations) so it is not clear how bringing up Heisenburg's uncertainty principle (a relation in quantum mechanics) is useful for solving this problem.

If there is something unclear about my question, please let me know and I can try to clarify it as I would really like to solve this problem.

8. Nov 25, 2007

### JustinLevy

Oh, you posted while I was typing and I almost missed it.

This is not true. Again, I can use any function f there. It need not be a plane wave. For instance, let $$f(z - ct) = \exp( -\frac{(z-ct)^2}{a^2})$$, and the equation can still satisfy Maxwell's equations. I have already succeeded in localizing it in one direction, and to do so I already have to add an infinite number of "plane waves" together to get it (so this is definitely not monochromatic).

The question is, how do I localize in the remaining directions?
Are you saying the general form I should be starting with for a pulse that is linear polarized in the x direction is:

$$\vec{E}(x,y,z) = \hat{x} \ g(x,y,z-ct)$$

Then we have:
$$0 = \nabla \cdot \vec{E}(x,y,z) = \frac{\partial}{\partial x} g(x,y,z-ct)$$

This does not solve the problem, as it still seems to indicate we can't have x dependance.

If you are suggesting something else and I am misunderstanding, I am sorry. Can you please try describing it again? I have a feeling I am just misunderstanding.

9. Nov 25, 2007

### DaTario

Sorry, I was just mentioning HUP to make an analogy with QM, as momentum and position are related to each other through a Fourier Transform as do the wave number k and radius vector r.

Best Wishes

DaTario

10. Nov 25, 2007

### pervect

Staff Emeritus
I can give a few pointers to this problem:

Gaussian beams are not exact solutions to Maxwell's equations, but use an approximation known as the "paraxial" approximation.

I haven't looked it over closely, but http://people.seas.harvard.edu/~jones/ap216/lectures/ls_1/ls1_u3/ls1_unit_3.html

appears to talk about this in some depth. (I had some other sources on Gaussian beams I read more thoroughly back when I was thinking about a very similar issue, but I don't recall exactly where they were - the above looks good, but I've only glanced at it).

Note that Gaussian beams have a "throat" where they are narrowest.

Another more practical paper to look at would be http://www.rp-photonics.com/gaussian_beams.html, note that it also mentions the paraxial approximation.

Bessel beams aka non-diffractive beams are exact solutions to Maxwell's equations, but they have characteristic fringes. There's a little bit on them in the Wikipedia:

http://en.wikipedia.org/w/index.php?title=Bessel_beam&oldid=106117765

You can try google for "nodiffractive beams" and "Bessel beams", there's been at least some PF posts on them. I think I found out about nondiffractive beams from another poster, in fact.

11. Nov 26, 2007

### JustinLevy

12. Nov 26, 2007

### Claude Bile

The wave you have described is not a propagating wave, it is an evanescent wave, with still only a single (albeit imaginary) k value. It doesn't matter what function you choose for f, the wavefronts (for any given frequency) will still be planar.
We can describe a pulse localised in time as a continuous spectrum of frequencies. The principle is the same for the space domain. We can describe a pulse localised in space as a continuous spectrum of spatial frequency components where k = (k_x, k_y, k_z) is the conjugate variable of r = (x, y, z), just as f is with t for the time domain.
While x polarised components cannot contribute to confining the beam in x (since EM waves are transverse), they can aid in confining the beam in the y, z direction.

EDIT: To clarify the above statement, consider the case of two plane waves interfering with one another. Two x polarised waves MUST have a propagation vector in the yz plane due to the transverse nature of the wave. Hence x polarised waves contribute to spatial variation in the y-z plane (which is one step below confinement).

Claude.

Last edited: Nov 26, 2007
13. Dec 1, 2007

### JustinLevy

I think there is either some serious miscommunication, or you have made some mistakes.

It most assuredly is propagating. (plug in several values for t and watch the gaussian peak move in the z direction with speed c) Secondly, it would take an infinite sum of waves with a single k (wavevector) value to make that gaussian wave shape. Third, this is not an evanescent wave either. Evanescent waves occur during total internal reflection (which requires matter, yet we are focussing on vaccuum solutions ... which the above is).

The wavefront may be planar, but this does not make it a plane wave. By definition a plane wave is a constant frequency wave (and thus has an infinite number of constant amplitude "wavefronts").

The only functional form for f that would make this a plane wave is f(z) = A cos (kz - wt + phi) where w/k = c and A, phi are arbitrary constants.

That is exactly what I have been trying to do, but was only able to figure out how to localize it in one direction with this method. So I can't figure out how to add up the plane waves to get a travelling "3-D localized" pulse solution.

Pervect gave me some good stuff to read, and I think that will help a lot. (I read through some of it, but haven't worked it out for myself yet.)
But if you can show me how to obtain a solution to Maxwell's equations in vaccuum for a travelling "spatially localized" pulse using fourier transforms, please please do show me the details. I would very much like to see this.

You seem to be saying that a beam of light with linear polarization is a proof that Maxwell's equations are wrong. These are routinely made in the lab, so your statement cannot be correct. We can have linear polarization AND confinement in both directions perpendicular to propagation.

I think the problem in your statement is your jump from stating EM waves are transverse (a local statment of field amplitude) to concluding the electric fields can't vary spatially in the direction of the electric field. I don't think this jump is justified. I'm not sure how to show it mathematically, but again the fact that we can make beams of light with linear polarization shows that this cannot be true.

Last edited: Dec 1, 2007
14. Dec 2, 2007

### Claude Bile

Yes I made a mistake, I missed the ^2 somehow, whoops.

Anyway, so what we have is a wave-packet which has a continuum of values for k, but the direction of k is constant, so k looks like (0, 0, k_z) where k_z = 2*pi/wavelength.
A plane wave is a wave that has single values for k and w. Waves with a single frequency only are monochromatic, but not necessarily planar, e.g. a spherical wave is monochromatic but not planar.
This is the only functional monochromatic form.
Essentially what you have done thus far is confine the pulse temporally, which allows you to indirectly confine the pulse spatially in the direction of propagation.
I am assuming you understand Fourier transforms as they relate to the frequency and time domains, in other words, that we can obtain an arbitrary pulse shape (in time) by adding a continuous spectrum of frequencies.

Exactly the same principle applies to the 3 spatial domains (x, y, z). We can define "spatial frequencies" (k_x, k_y, k_z) where, for example k_x = 2*pi/x. Generally with these types of diffraction problems, the wave is assumed to be monochromatic, which adds constraints that the various components of k can take.

You can confine a pulse spatially in the same way you confined the pulse temporally. Assume a cross section of the form e^(-(x/a)^2). This wave consists of a number of plane wave components, all with different values for k_x (just as the temporal case consisted of a number of frequency components with different values for w). For a monochromatic beam this means that each plane wave component is traveling in a different direction. The consequence of this is a bending of the wavefront with propagation, and why diffraction is a natural consequence of beam propagation. It is also why more narrowly confined beams (which thus have a wider "spectrum" of k_x) diffract into a wider angle.

One can also derive the far-field diffraction limit for image resolution using this formulation by considering that the widest angle a beam can diffract into is pi (or 2*pi steradians in 3D), which limits the width of the k spectrum and thus how well light can be spatially confined.
There is a difference between a beam that is ideally linearly polarised and one that is close to linearly polarised. I would contend that being able to create a linearly polarised beam in a lab does not disprove what I have said since they are never perfectly linearly polarised. I'm afraid I can't give a more convincing argument than that, vector analysis of beam propagation is immensely complex.
If you are interfering two plane waves, unless the polarisation is perpendicular to the plane defined by the two k-vectors, you can't perfectly match the polarisation of the two wave components (which is required for them to interfere) without introducing a longitudinal component to one of the waves, now how well this translates into something as complex as a Gaussian beam, I'm not sure since this is a fairly simple case, but I think it answers your question as to why a linearly polarised wave must be invariant in the direction of polarisation.

Many of our misunderstandings I have realised is due to the fact (I think) that I was of a mindset where everything is monochromatic, whereas you initially had a spectrum of frequencies (and thus a spectrum of k values) to begin with.

Claude.

Last edited: Dec 2, 2007