Right handed vs Left handed circular polarization

Hey,

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

If I have the wave $$\vec{E}= E_{0X} cos(kz-\omega t)+ E_{0Y} sin(kz-\omega t)$$ and $$E_{0X}=E_{0Y}$$. Then at z=0, t=0 the field is pointing completely in the x direction. Staying at z=0 ( $$\vec{E}= E_{0X} cos(-\omega t)+ E_{0Y} sin(-\omega t)=E_{0X}cos(\omega t)-E_{0Y}sin(\omega t)$$. 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 $$\vec{E}= E_{0X} cos(kz-\omega t)- E_{0Y} sin(kz-\omega t)$$ and $$E_{0X}=E_{0Y}$$. Then at z=0, t=0 the field is pointing completely in the x direction again. Staying at z=0 ( $$\vec{E}= E_{0X} cos(-\omega t)+ E_{0Y} sin(-\omega t)=E_{0X}cos(\omega t)+E_{0Y}sin(\omega t)$$. 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

 PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
 Recognitions: Science Advisor 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.

 Quote by h0dgey84bc ... 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.

$$E_{CP}=E_x \pm jE_y$$

Do you understand complex exponentials?

Regards,

Bill

Right handed vs Left handed circular polarization

 Not per IEEE-STD-145. $$E_{CP}=E_x \pm jE_y[tex] Do you understand complex exponentials? Sure. You're referring to the phasor representation? As long as the convention I'm using with be good with the GRE I'm happy, just don't want to lose marks stupidly for the wrong convention. Recognitions: Science Advisor  Quote by Antenna Guy Not per IEEE-STD-145. [tex]E_{CP}=E_x \pm jE_y$$ 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.

 Quote by clem 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 $j$ 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, $E_y$ "lags" $E_x$, and a factor of j would rotate the phasor of $E_y$ onto that of $E_x$. Another way of looking at it is that after a quarter wavelength of propogation, $E_y$ would have the phase that $E_x$ started with.

Hence:

$$E_R=E_x+jE_y$$

$$E_L=E_x-jE_y$$

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

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

In any other case the total field is elliptical, and has both $E_R$ and $E_L$ components.

Regards,

Bill

 Recognitions: Science Advisor 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.

Mentor
Blog Entries: 10
 Quote by h0dgey84bc Hey, I just wanted to clear up some confusion I've been having regarded which is which of these. If I have the wave $$\vec{E}= E_{0X} cos(kz-\omega t)+ E_{0Y} sin(kz-\omega t)$$ and $$E_{0X}=E_{0Y}$$. Then at z=0, t=0 the field is pointing completely in the x direction. Staying at z=0 ( $$\vec{E}= E_{0X} cos(-\omega t)+ E_{0Y} sin(-\omega t)=E_{0X}cos(\omega t)-E_{0Y}sin(\omega t)$$. 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.

 Quote by clem 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