Right handed vs Left handed circular polarization

In summary: And yes, the "right hand rule" is what you know it to be.Regards,BillIn summary, the discussion involved determining the conventions for circular polarization in both optics and high energy physics. It was determined that in optics, circular polarization is defined based on the direction of light coming towards the observer, while in high energy physics it is defined based on the point of view of the photon. The discussion also involved clarifying the use of complex exponentials and their role in representing circular polarization. The final conclusion was that both right-handed and left-handed circular polarization can occur depending on the direction of propagation and the phase difference between the x and y components of the electric field.
  • #1
h0dgey84bc
160
0
Hey,

I just wanted to clear up some confusion I've been having regarded which is which of these.

If I have the wave [tex] \vec{E}= E_{0X} cos(kz-\omega t)+ E_{0Y} sin(kz-\omega t) [/tex] and [tex] E_{0X}=E_{0Y} [/tex]. Then at z=0, t=0 the field is pointing completely in the x direction. Staying at z=0 ( [tex] \vec{E}= E_{0X} cos(-\omega t)+ E_{0Y} sin(-\omega t)=E_{0X}cos(\omega t)-E_{0Y}sin(\omega t) [/tex]. So that the x comp begins to decrease with time at z=0, and the y comp begins to increase negativley. Thus if the wave was coming toward you down the z-axis you'd see it rotating clockwise. If you were behind the wave you'd see it rotating counter clockwise.

This is right handed circ polarization?


Now if I have If I have the wave [tex] \vec{E}= E_{0X} cos(kz-\omega t)- E_{0Y} sin(kz-\omega t) [/tex] and [tex] E_{0X}=E_{0Y} [/tex]. Then at z=0, t=0 the field is pointing completely in the x direction again. Staying at z=0 ( [tex] \vec{E}= E_{0X} cos(-\omega t)+ E_{0Y} sin(-\omega t)=E_{0X}cos(\omega t)+E_{0Y}sin(\omega t) [/tex]. So that the x comp begins to decrease with time at z=0, and the y comp begins to increase positivley. Thus if the wave was coming toward you down the z-axis you'd see it rotating counter-clockwise. If you were behind the wave you'd see it rotating clockwise.

This is left handed circ polarization?

Does all this sound correct, and are these the conventions?

Thanks
 
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  • #2
You have the correct conventions for optics.
In high energy physics, the point of view is that of the photon, so you do look behind the wave. This means that negative helicity (or left-handed helicity) corresponds to right handed polarization.
 
  • #3
h0dgey84bc said:
... Thus if the wave was coming toward you down the z-axis you'd see it rotating clockwise. If you were behind the wave you'd see it rotating counter clockwise.

This is right handed circ polarization?

Not per IEEE-STD-145.

[tex]E_{CP}=E_x \pm jE_y[/tex]

Do you understand complex exponentials?

Regards,

Bill
 
  • #4
Not per IEEE-STD-145.

[tex]E_{CP}=E_x \pm jE_y[tex]

Do you understand complex exponentials?

Sure. You're referring to the phasor representation?

As long as the convention I'm using with be good with the GRE I'm happy, just don't want to lose marks stupidly for the wrong convention.
 
  • #5
Antenna Guy said:
Not per IEEE-STD-145.

[tex]E_{CP}=E_x \pm jE_y[/tex]

Do you understand complex exponentials?

Regards,

Bill
The first post is just the trig representation of your exponentials.
The question was is the + or - for right handed circular polarization.
 
  • #6
clem said:
The first post is just the trig representation of your exponentials.
The question was is the + or - for right handed circular polarization.

Note from phasors that a factor of [itex]j[/itex] is equivalent to a +90deg phase shift.

Assume that the phase convention is such that I am looking in the direction of propogation.

For RHCP, [itex]E_y[/itex] "lags" [itex]E_x[/itex], and a factor of j would rotate the phasor of [itex]E_y[/itex] onto that of [itex]E_x[/itex]. Another way of looking at it is that after a quarter wavelength of propogation, [itex]E_y[/itex] would have the phase that [itex]E_x[/itex] started with.

Hence:

[tex]E_R=E_x+jE_y[/tex]

[tex]E_L=E_x-jE_y[/tex]

If [itex]E_x[/itex] and [itex]E_y[/itex] are co-phase, the total field is linear (i.e. non-rotating). In this case, [itex]E_R[/itex] and [itex]E_L[/itex] are simply conjugates of one-another (same magnitude).

If [itex]E_x[/itex] and [itex]E_y[/itex] differ by exactly 90deg of phase (and have the same magnitude), the total field is either [itex]E_R[/itex] or [itex]E_L[/itex].

In any other case the total field is elliptical, and has both [itex]E_R[/itex] and [itex]E_L[/itex] components.

Regards,

Bill
 
  • #7
Bill: You "Assume that the phase convention is such that I am looking in the direction of propagation." The usual convention in optics is that you are looking at the light coming toward you. This is opposite to the direction of propagation.
You are using the high energy convention for photons, which is fine, but not what the original questioner asked about.
 
  • #8
h0dgey84bc said:
Hey,

I just wanted to clear up some confusion I've been having regarded which is which of these.

If I have the wave [tex] \vec{E}= E_{0X} cos(kz-\omega t)+ E_{0Y} sin(kz-\omega t) [/tex] and [tex] E_{0X}=E_{0Y} [/tex]. Then at z=0, t=0 the field is pointing completely in the x direction. Staying at z=0 ( [tex] \vec{E}= E_{0X} cos(-\omega t)+ E_{0Y} sin(-\omega t)=E_{0X}cos(\omega t)-E_{0Y}sin(\omega t) [/tex]. So that the x comp begins to decrease with time at z=0, and the y comp begins to increase negativley. Thus if the wave was coming toward you down the z-axis you'd see it rotating clockwise. If you were behind the wave you'd see it rotating counter clockwise.

This is right handed circ polarization?

Yes, it's right-handed. To see this let t=0. As you go in the +z direction, E rotates from +x to +y, then -x, then -y. Letting the fingers of your right hand curl around in the direction of rotation, your right thumb points in the +z direction. The picture is the same as the threading on a standard right-handed screw.

For your other case, the same argument works out if you use your left hand, so it is left-handed circular polarization.

You could also imagine going in the -z direction, these arguments would still work.
 
  • #9
clem said:
Bill: You "Assume that the phase convention is such that I am looking in the direction of propagation." The usual convention in optics is that you are looking at the light coming toward you. This is opposite to the direction of propagation.
You are using the high energy convention for photons, which is fine, but not what the original questioner asked about.

The OP never said anything about "optics" (or high energy physics).

I cited an international standard, and (hopefully) followed it accurately.

Regards,

Bill
 

1. What is circular polarization?

Circular polarization is a type of polarization of electromagnetic radiation, such as light waves, in which the electric field vector rotates in a circular pattern as the wave propagates. It is different from linear polarization, in which the electric field vector oscillates in a single plane.

2. What is the difference between right handed and left handed circular polarization?

The difference between right handed and left handed circular polarization lies in the direction of the rotation of the electric field vector. In right handed circular polarization, the electric field vector rotates in a clockwise direction, while in left handed circular polarization, it rotates in a counterclockwise direction.

3. How is circular polarization used in technology?

Circular polarization is used in various technologies, such as optical communication systems, 3D glasses, and satellite communications. It is also used in radar systems and medical imaging devices.

4. How is circular polarization produced?

Circular polarization can be produced by passing linearly polarized light through a quarter-wave plate or a Faraday rotator. It can also be generated by using specially designed antennas or filters.

5. Can circular polarization be converted to linear polarization?

Yes, circular polarization can be converted to linear polarization by passing it through a polarizing filter or a quarter-wave plate. The resulting linear polarization will be at an angle of 90 degrees to the original circular polarization.

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