Well it’s wrong. I suppose you are thinking so because Maxwell’s equations shows B depending on (∆E/∆t)[i cant write \partial symbol], BUT remember LHS is not B but is curl(B) so derivative acts on both sides. So they are in phase and not 90° out of phase. So the peaks should align, atleast in free space
Edit: neither of us are right, EM waves in vacuum do travel in phase but in conductors, there is a phase difference between 0° (very high resistance) and 45° (superconductor). So yeah, EM waves can “tell” they’re in a dummy load.
If two waves of the same frequency align with each others’ valleys, they align with each others’ peaks. Do you mean they should be aligned peak to valley? I don’t know how you’re deciding which direction of the axes is positive and which is negative though.
Just to point out, because it’s bothering me way more than it should…
But the electric and magnetic peaks align with each other’s valleys, not with each other’s peaks.
Well it’s wrong. I suppose you are thinking so because Maxwell’s equations shows B depending on (∆E/∆t)[i cant write \partial symbol], BUT remember LHS is not B but is curl(B) so derivative acts on both sides. So they are in phase and not 90° out of phase. So the peaks should align, atleast in free space
No, they fucking shouldn’t
https://en.wikipedia.org/wiki/Maxwell's_equations#Vacuum_equations,_electromagnetic_waves_and_speed_of_light
Edit: neither of us are right, EM waves in vacuum do travel in phase but in conductors, there is a phase difference between 0° (very high resistance) and 45° (superconductor). So yeah, EM waves can “tell” they’re in a dummy load.
If two waves of the same frequency align with each others’ valleys, they align with each others’ peaks. Do you mean they should be aligned peak to valley? I don’t know how you’re deciding which direction of the axes is positive and which is negative though.
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Uh, no?
Yeah, because it’s not pulsing in intensity, it’s pulsing between which kind of energy it’s being
You’re describing circular polarization. It’s not the only way.
Maybe it’s showing polarization superpositions of the E-field?