Sketch of the electric field of a laser beam

[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?

 

[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.

 

[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).

 

[davenn]

I am trying to sketch the electric field E in snapshot of a linearly polarized laser beam.

does this help ?

 

[sophiecentaur]

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).

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.

 

[spareine]

does this help ?

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.

 

[spareine]

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.

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.

 

[davenn]

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.

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

 

[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?

 

[tech99]

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).

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).

 

[sophiecentaur]

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.

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.

 

[spareine]

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.

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.

 

[tech99]

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.

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..

 

[spareine]

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..

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.

 

[berkeman]

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.

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%20Hat%20Profile.jpg

 

[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.

 

[tech99]

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.

This is my field diagram if it opens for you:-
https://www.flickr.com/photos/58337586@N08/35537703754/in/dateposted-public/

 

[spareine]

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

 

[sophiecentaur]

. My objection against that suggestion is that it makes ∇⋅E nonzero.

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.

 

[tech99]

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

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.

 

[sophiecentaur]

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

I would say that's not what is going on. If you look at the link on my last post you will see the 'loops' in the antenna fields are not in the plane of the advancing wave front but they double back at the edges of the beam and join to field lines of the previous and next wave front. It is easier to look at the coarser structure of an antenna pattern than to try to draw on the scale of light waves. That also allows ∇.E to be zero without disturbing the polarisation at the edges. The field lines spread out, signifying a lower field at the edges.

 

[anorlunda]

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

You can not apply classical equations to quantum phenomena. A single photon has E and B fields but zero charge. Lasers won't lase except for quantum mechanical behavior of bosons.

QED (quantum elctrodyanmics) is the real theory. With simplifying assumptions, QED reduces to Maxwell's Equations. With simplifying assumptions, Maxwell's equations yield to circuit analysis where we can apply Kirchoff's Laws. But you can't go backward applying Kirchoff's Laws or Maxwell's Equations to QED.

 

[Gordianus]

You can not apply classical equations to quantum phenomena. A single photon has E and B fields but zero charge. Lasers won't lase except for quantum mechanical behavior of bosons.

QED (quantum elctrodyanmics) is the real theory. With simplifying assumptions, QED reduces to Maxwell's Equations. With simplifying assumptions, Maxwell's equations yield to circuit analysis where we can apply Kirchoff's Laws. But you can't go backward applying Kirchoff's Laws or Maxwell's Equations to QED.

Laser modes are derived from classical electromagnetic theory without resorting to QED. Siegman, in "Lasers", derives the paraxial wave equation and shows the gaussian mode is the simplest solution to that equation. Everything within the framework of classical electromagnetism.

 

[berkeman]

It seems like we should focus on a flashlight beam for now to answer the OP's question, IMO. It's an interesting question, and my previous reply may be wrong based on the resonse @tech99 -- I'll post a follow-up in a bit...

 

[berkeman]

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.

Interesting, so each linear polarizer that a linearly polarized beam would pass through would add more loss than just the impedance discontinuity reflection? The 180 degree polarization shifted components are attenuated? Does passing a linearly polarized Gaussian beam through multiple linear polarizers change the Gaussian shape due to more attenuation at the outsides of the beam where you show a different E-field polarization characteristic?

Maybe I'm not understanding this phase shift that you are graphing...?

 

[tech99]

Interesting, so each linear polarizer that a linearly polarized beam would pass through would add more loss than just the impedance discontinuity reflection? The 180 degree polarization shifted components are attenuated? Does passing a linearly polarized Gaussian beam through multiple linear polarizers change the Gaussian shape due to more attenuation at the outsides of the beam where you show a different E-field polarization characteristic?

Maybe I'm not understanding this phase shift that you are graphing...?

I have shown a general case of a beam of EM radiation from an aperture. There are different sorts of illumination of the aperture, which may be dipole-like, loop-like or a mixture and they have different cross polarisation patterns. But in general we expect the field lines at the aperture to be curved due to the characteristics of the source, and this translates into the radiation far field. For linear polarisation, it creates side lobes lying in the 45 degree planes. There are also cross polarised field components in the main beam, but they cancel to zero on-axis.
If the beam is passed through a linear polariser, it will remove the cross polarised sidelobes and will also remove the cross polarised components in the main beam. So there is a small loss of energy. At large angles from the axis of the polariser, geometry appears to dictate that cross polarised components will again be observed in the 45 degree planes.
In all these studies, it is important to be careful to understand the type of axes used if large angles are involved.

 

[sophiecentaur]

You can not apply classical equations to quantum phenomena. A single photon has E and B fields but zero charge. Lasers won't lase except for quantum mechanical behavior of bosons.
QED (quantum elctrodyanmics) is the real theory. With simplifying assumptions, QED reduces to Maxwell's Equations. With simplifying assumptions, Maxwell's equations yield to circuit analysis where we can apply Kirchoff's Laws. But you can't go backward applying Kirchoff's Laws or Maxwell's Equations to QED.

Sounds a good argument on the face of it but you would have to explain where the 'limits' of your statement apply. When does a Laser Beam become indistinguishable from a tight, conventional monochromatic beam? Even a laser beam beam will have a certain coherence length and a wider bandwidth than the source at a distance.
Your claim that "Lasers won't lase" is to do with the source and not with what happens to the beam, once it has been launched. How much will the Boson behaviour dominate over the basic EM field behaviour?

linear polarizers change the Gaussian shape due to more attenuation at the outsides of the beam

Yes. The term "linear polarisation" implies a plane wave. If the wave is constrained then the edges are not 'truncated' in a simple fashion (hence the OP topic). It's a matter of geometry. I still hold that the field pattern at the edge will have the same form as the diagram in my above post (here it is)

I can see no reason why that picture is irrelevant to the situation for a very narrow beam. It shows how the fields are less at the edges without any charge being involved. So, off axis, you are seeing E fields that are not actually parallel with the field at the centre - that's cross polarisation and a linear polariser will remove some of the energy.

 

[tech99]

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.

 

[davenn]

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.

who / what are you referring to ?
quoting would have been good

 

[tech99]

who / what are you referring to ?
quoting would have been good

My apologies, not very scientific! I was referring to the diagram in Post 27 from sophiecentaur.