Sketch of the electric field of a laser beam

[Justintruth]

Consider a vector field E:
Ex=e^(-x^2)
Ey=0
Ez=0

What is the divergence at x=1?

Imagine a unit cube centered around the point x=1, y=0, z=0

Clearly their is no flux through the planes perpendicular to y or z.

There is flux through the planes perpendicular to x. Specifically there is flux through the plane x=1.5 and x=.5 for this unit cube.

Also the flux out of the cube is not equal to the flux into the cube as the vectors at x=1.5 are not the same magnitude as the ones at x=.5.

e^(-1.5^2) *Area of a side < e^(-.5^2)*Area of a side

Now imagine another cube around the point x=1, y=0, z=0 but instead of a unit cube it is a cube with a side length of .5

All of the facts stated for the first cube are true for the second. But the flux into this cube is less because the flux on the x=1.25 side of the cube got closer to the flux on the x=.75 side

e^(-1.25^2) *Area of a side < e^(-.75^2)*Area of a side

Now imagine another cube around the point x=1, y=0, z=0 but instead of a unit cube it is a cube with a side length of .25

All of the facts stated for the first cube are true for the third...

e^(-1.125^2) *Area of a side < e^(-.875^2)*Area of a side

see how the flux into the cube keeps going down?

Etc

If you look at the difference between the total flux entering each of these cubes and then take the limit as the process continues then the LIMIT of the flux into the cube will be zero. So the divergence is zero.

Now imagine the same electric field but with one difference. For x <=1, Ex=e^(-x^2) and for x > 1 Ex= -e^(-x^2) (note the minus sign is the difference)

Do the same process. Now the divergence is not zero at x=1 but zero everywhere else.

Just because the amplitude of the E field is a Gaussian and therefore that a cube surrounding any point on the x-axis other than zero has non-zero flux into the cube does not mean that the limit of fluxes from successively smaller cubes does not approach 0.

Does that do it? Did I answer your real question? How about this?

Let len =length of the cube / 2

Limit of e^(-(x-len)^2)-e^(-(x+len)^2) as len -> 0 =

e^(-x^2) - e^(-x^2) =0 so no divergence.

Or this:

Consider a coulomb field of a point particle. The divergence of the electric field is everywhere 0 except at the point where the charge is.

Or this:

The divergence of the gradient of a scalar Gaussian field is everywhere zero but at the peak of the Gaussian

-but-

the divergence of a vector field where the vector magnitudes are Gaussian is undefined. You need to specify the directions of the vectors to get the divergence.

 

[Spinnor]

I am trying to sketch the electric field E in snapshot of a linearly polarized laser beam. Is it correct that the E vectors bend from vertical to longitudinal near the cylindrical surface of the beam, and that all field lines within a half wave segment are closed loops?

I don't mean to resurrect an old thread, which I bumped into while Googling, but to complete the thread I thought I would point out that this is very similar to a problem from Jackson's Classical Electrodynamics, from the 1975 edition, problem 7.20, page 333,

With a little thought one should be able to convert the above solution of the fields of a circularly polarized wave into the fields of a linearly polarized wave. It is interesting to plot with Mathematica. If I still had a copy I would plot it again, anyone else willing and able? Knowing that the divergence of both the magnetic and electric fields is zero we know all field lines form closed loops.

 

[Spinnor]

If I still had a copy I would plot it again, anyone else willing and able?

Mathematica lent me a copy for 2 weeks, that was nice of them. After an embarrassingly long time I came up with a plot of the electric field of the equation above with y=0 and with the above equation converted into a linearly polarized beam. The magnetic field is the same, just rotated 90 degrees about the x axis.

Field lines end and I can not figure out how to force them to continue, seems there is some maximum density that I can not override for now. Remember the divergence of the electric field is zero with no sources, field lines should form continuous loops. Also note the overall vertical scale should be stretched so that the beam is many wavelengths wide as per problem 7.20 above. Any suggestions on how to improve the above plot? I tried way too long to use VectorPlot3D but could not get any good results.

 

[tech99]

The magnetic field is the same, just rotated 90 degrees about the x axis.

If we assume a line source as an example, I thought the magnetic lines would be concentric rings around it.

 

[Spinnor]

If we assume a line source as an example, I thought the magnetic lines would be concentric rings around it.

No sources, electric field is of a laser beam in vacuum, somewhat distorted vertically.

 

[sophiecentaur]

Can you explain in the diagram how the field arrows can sometimes point towards or away from the source when we have a transverse wave? I don't think these arrows would be detected by an E-field sensor, such as a short dipole, positioned radially from the source.Thank you.

Sorry for picking this up so late. I must have been on holiday. The crucial part of your question is "when we have a transverse wave?" The term Transverse only applies for a plane wave that's infinitely wide or in the centre of a beam. At the edges of a directed beam, the E field has to bend away from the normal to the propagation direction so it's no longer 'transverse' but has a longitudinal component.
You can show this with a dipole probe placed at various locations around an HP transmitting dipole. The theory can be seen to work.

 

[Cutter Ketch]

A laser beam is just the superposition of a very large number of dipole radiators all in phase and nearly on axis, but distributed along the axis (backwards in time from a particular point in the phase front) and also distributed transversely according to the size of the gain medium and the intensity distribution in the cavity. From the point of view of a particular point in the beam, the resonant cavity causes there to appear to be another gain medium's worth of emitters one round trip further away in time and another and another as far back as either the pulse length, or the coherence length whichever is shorter. Here are 3 dipole emitters lined up along an axis, and one wavelength out of phase with each other.

you can see that each has a dipole field where the electric field is curled in the longitudinal direction at the edges. You can also see how they add up on axis but begin to interfere off axis. In reality there isn't a single line of emitting atoms, so there is a transverse distribution, and the transverse distribution of emitters generally isn't uniform. Diffraction plus the apertures in the cavity favor modes so that the transverse intensity in the cavity isn't uniform and so the participation of the emitters is also not uniform in the transverse dimension. However, all of that can be thought about later. For right now just imagine having a bunch of dipole emitters emitting in phase stretching back in time. The field at a plane somewhere away from the laser is the superposition of the fields of many of these emitters stretching back in time for the length of the gain medium and transversely across the gain medium AND from multiple passes through the gain medium so that it looks like there was a train of emitting volumes.

The net effect is approximately a plane wave which falls off quickly in intensity from the axis as the dipole fields interfere. All of the fields within the high intensity part of the beam are approximately transverse to the beam axis and polarized vertically (assuming the emitters all were aligned, usually enforced by a polarizer in the cavity throwing away any emissions that weren't) However in detail the electric field retains some curvature and far away from the axis where interference has added up to zero amplitude, each dipole field has components of electric field parallel to the axis.

So, yes the electric field from each emitter must curl around like a dipole field, but superposition and interference allow that to be negligible in the intense part of the beam.

 

[Spinnor]

So, yes the electric field from each emitter must curl around like a dipole field, but superposition and interference allow that to be negligible in the intense part of the beam.

So what I plotted is correct or incorrect given the caveats?

 

[Cutter Ketch]

So what I plotted is correct or incorrect given the caveats?

Although you had the correct idea that Maxwell's equations won't allow an E field to be truly be 100% transverse polarized, I think I would have to say your drawing was incorrect.

To put it another way, forget about lasers. You believe dipole fields are allowed. There is no fundamental reason why a long line of dipoles all in phase shouldn't be allowed, and in that construct you can't avoid the fact that by interference the intensity is axial and the electric field is polarized and transverse. Whether or not you believe that construct represents a laser, you have to believe that it does not violate Maxwell's equations. The only possible conclusion is that the electric field can be well polarized and highly transverse everywhere within the intense region of a beam.

 

[Spinnor]

I think I would have to say your drawing was incorrect.

The function I tried to graph was taken from a problem from Jackson's Electrodynamics. He gave the result, the problem was to prove the result. I just used the result Jackson gave. Because the transverse electric field of a beam of finite width most go to zero away from the beam and using the fact that the divergence of the electric field is zero I don't think any other result works, I think the electric field must have the form that I plotted? Granted the field plot must be stretched vertically but that does not change the basic form.

 

[sophiecentaur]

There is no fundamental reason why a long line of dipoles all in phase shouldn't be allowed,

That needs to be re-stated, actually. Their phases should be progressive along the length of the array so that the passing EM wave is in step with the radiation from each element. It's the basis of many directional antenna arrays!

 

[Cutter Ketch]

That needs to be re-stated, actually. Their phases should be progressive along the length of the array so that the passing EM wave is in step with the radiation from each element. It's the basis of many directional antenna arrays!

Yes, that’s better.