# I Sketch of the electric field of a laser beam

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1. Aug 3, 2017

### spareine

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?

2. Aug 3, 2017

### Gordianus

The electric field lines do not bend at the "top" or the "bottom". The electric field is stronger at the center and weakens as you move from the center to the border, but it always points in the same direction.

3. Aug 3, 2017

### spareine

The electric field cannot weaken along a field line, if the electric field is vertical everywhere. $\nabla \cdot \mathbf{E} = 0$ in the absence of electric charges (Gauss' law).

4. Aug 3, 2017

### davenn

does this help ?

5. Aug 3, 2017

### sophiecentaur

I can't reconcile that with a narrow beam in free space and each E field line around a radiating dipole follows a curve in a plane. The field strength in any particular elevation will follow a Cos θ law for a short dipole. A collinear / coaxial array will produce a more vertically directional pattern but, being the sum of dipole patterns, why would the E vector direction change? The same thing should apply for a very narrow beam, wouldn't it?

BTW The diagram at the top looks more like the fields in a circular waveguide (conducting walls). That wouldn't be the same boundary conditions as for a narrow beam in free space.

6. Aug 3, 2017

### spareine

I don't think polarization is the essence of my question. For me, the advantage of specifying the laser beam is linear polarized is that it is slightly simpler to describe and draw than other options.

The essence of my question is that I am wondering about the profile of the electric field on cross sections of the laser beam, and what the field lines look like, given that somehow they should be closed loops because of Gauss' law.

7. Aug 3, 2017

### spareine

I am afraid I do not understand your answer. My question is about a laser beam in free space, outside the optical cavity. I don't see how or why radiating dipoles are present in that laser beam in free space.

8. Aug 3, 2017

### davenn

the confusion is because the drawing you supplied implies light in a waveguide ... optical fibre, NOT free space

9. Aug 4, 2017

### spareine

Ok, but I am actually more interested in a laser beam in free space. Gordianus suggested the laser beam in free space is identical to an infinitely wide plane wave, multiplied by a gaussian-like beam profile in the cross sectional plane. My objection against that suggestion is that it makes $\nabla\cdot\mathbf{E}$ nonzero. Do you think this objection is invalid?

10. Aug 4, 2017

### tech99

I think there is a distinction between a Field LIne, which is a contour of equal field strength, and a Line of Force, which is the path taken by a positive charge when placed in the field. Field LInes will run circularly around a charge, whereas lines of force will project radially from it.
Lines of Force become further apart as we depart from a charge, so they indicate a weakening field. The original diagram seems to show Lines of Force, but describes them as Field Lines. The field is not entirely contained within the main beam, but at the top of it the field lines wander away into the rest of the page, and finally met up at the bottom. When using Lines of Force, the strength of the field is often shown by using small arrows of varying length.
Regarding polarisation, there are geometrical distortions arising from any source, and it differs depending on the type of radiator. Most usually, the polarisation is "correct" on the principle axes and then bends away to some extent off axis in the 45 degree planes.
I think the diagram from Davenn depicts the fields using arrows proportional to field strength, rather than showing the shape of the field.
(My apologies for using capitalisation to emphasise the terminology).

11. Aug 4, 2017

### sophiecentaur

EM waves are EM waves, whatever wavelength. Why would the E vector of light waves be any different from the E vector of RF waves? They don't know what generated them so they must behave the same way. I was, of course, discussing the far field pattern.

12. Aug 4, 2017

### spareine

I meant field lines in the sense that is common in electricity and magnetism: lines that follow the direction of the vector field (wikipedia). They can only form closed loops if there are no sources and sinks.

13. Aug 4, 2017

### tech99

In such a case, there is no need for the lines to turn over at the edge of the beam, they just continue outside the main beam but weaker..

14. Aug 4, 2017

### spareine

Could you draw that field line pattern? Wouldn't it be radial with a source at r=0? But there is no source at r=0.

15. Aug 4, 2017

### Staff: Mentor

I think the confusion comes because you are trying to use too "big" of a probe to measure the E-field at various points in the beam. Cosnider using a probe that is much smaller than the diameter of the beam. As you move that linearly polarized probe from the center of the beam toward the edge, you measure the Gaussian drop-off of the amplitude of the linearly polarized E-field. The polarization remains the same, just the amplitude drops off.

Like if the beam were aimed vertically in the plot below of E-field squared amplitude versus radius...

http://www.wavelength-tech.com/images/Top Hat Profile.jpg

16. Aug 4, 2017

### spareine

The problem I was trying to address, as mentioned in post #9, is more elementary. It does not involve measurement probes. Multiply a linearly polarized plane wave with a gaussian beam profile. The result is that $\nabla\cdot\mathbf{E}$ is nonzero everywhere (except at the axis of the beam). That is incompatible with free space, where charges are absent.

17. Aug 4, 2017

### tech99

18. Aug 4, 2017

### spareine

I see, so all field lines are closed loops around several side lobes. Thanks!

19. Aug 4, 2017

### sophiecentaur

It's been a long time since I used those operators but I don't think your inference can be correct. If you were right, it would mean that every antenna / radiator would have to be isotropic, I think. And none are.
I found this link and it has a diagram which accounts for what happens to the E field at the edges of the beams. It gives a ∇.E = 0 because the field vectors follow a curve in the other sense from the one you have drawn in your diagram.

20. Aug 4, 2017

### tech99

The circles in my diagram just indicate the nominal position of the main beam and some suggested sidelobes.The sidelobes are 180 degrees out of phase with the main beam.