- #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
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