Polarization state conventions

[leroyjenkens]

Ok, so I keep reading about these concepts and every source says something different. Even within the same textbook, it says conflicting things.
So you have a circularly polarized light that is coming TOWARDS you. That is considered left-circularly polarized or right-circularly polarized? I've seen both stated.
And if you have two equations, one with no $\epsilon$ given for the initial phase, and the other with an initial phase of +$\pi$/2, which one is lagging and which one is leading? I've seen sources that say the no initial phase is leading and some that say the no initial phase is lagging.
Thanks

 

[mfb]

So you have a circularly polarized light that is coming TOWARDS you. That is considered left-circularly polarized or right-circularly polarized? I've seen both stated.

It can be both, depending on its polarization ;). "Left" and "right" is independent of your viewing direction.

And if you have two equations, one with no $\epsilon$ given for the initial phase, and the other with an initial phase of +$\pi$/2, which one is lagging and which one is leading? I've seen sources that say the no initial phase is leading and some that say the no initial phase is lagging.
Thanks

Depending on the definition of that phase in your equation.

 

[leroyjenkens]

I left out part of my question on that first one. I meant to ask if the electric field vector is rotating counter clockwise, with you seeing it as it's coming towards you, is that right-circularly polarized or left-circularly polarized. My book says right, which seems backwards. It's rotating left, but it's called right?

The definition of the phase in the equation is where the wave starts. So if it has +π/2 as its initial phase, that means it starts further along the x-axis (one half wavelength further) than the other wave. Is that considered leading or lagging?

Thanks.

 

[mfb]

Still depends on the definition.

If your phase is increasing and one is leading by pi/2, then this is leading.