Meir Achuz said:
The direction of the arrows for E in the picture are just a convenient sign convention.
E Can have either sign, so the direction of the arrow does not presuppose anything.
Thank you. To verify that, I derived Fresnel equations in the case where ##E_r## is taken pointing in the opposite direction of ##E_i##. The only consequence I found is that the sign of the (amplitude) reflection coefficient ##r## is reversed. In that case ##r## is positive if the reflected field is reversed, while in the conventionnal case ##r## is negative if the field is reversed. So indeed it seems to be simply a matter of sign convention !
For the sake of completness, or if someone else stumble on this question, and for myself as a reminder, here are the two derivations (for a s-polarized, or TE, wave), for each sign convention :
1 - Usual convention (##E_i## and ##E_r## assumed in same direction)
For E :
##E_i + E_r = E_t##
##E_i + rE_i = tE_i##
##1 + r = t \quad\quad## (1)
Here since we assumed both ##E_i## and ##E_r## point in the same direction, a positive ##r## means they do indeed point in the same direction (they are in phase). However if ##r## turns out to be negative, it means our assumption was wrong so that ##E_i## and ##E_r## actually point in opposite directions (they are ##\pi## out of phase, since ##-1 = e^{i\pi}##). The sign of ##r## will automatically adjust. We will derive the expression of ##r## below.
For H (taking magnetic permeability equal for both mediums, for simplicity) :
The directions of the H fields follow from our choice of direction for the E fields. In the present convention, we get :
##-H_i cos(\theta_i) + H_r cos(\theta_r) = -H_t cos(\theta_t)##
##-k_i cos(\theta_i) + r k_r cos(\theta_r) = -t k_t cos(\theta_t)## (recall ##H = \frac{1}{\mu} \frac{k}{w}E##)
And, since ##\theta_i = \theta_r## (law of reflection) and ##k_i = k_r## (same medium) :
##k_i cos(\theta_i) (1 - r) = t k_t cos(\theta_t) \quad\quad## (2)
Similary, the sign of ##r## will take care of the real direction of the reflected H field.
Now combining equations (1) and (2) we get the usual Fresnel equations for a s polarized wave :
##r = \frac{k_i cos(\theta_i) - k_t cos(\theta_t)}{k_i cos(\theta_i) + k_t cos(\theta_t)}## and ##t = \frac{2k_i cos(\theta_i)}{k_i cos(\theta_i) + k_t cos(\theta_t)}##
2 - Opposite convention (##E_i## and ##E_r## assumed in opposite directions)
For E :
##E_i - E_r = E_t##
##E_i - rE_i = tE_i##
##1 - r = t \quad\quad## (1')
Here since we assumed that ##E_i## and ##E_r## point in opposite directions, a positive ##r## means they do indeed point in opposite directions (they are ##\pi## out of phase). However if ##r## turns out to be negative, it means our assumption was wrong so that ##E_i## and ##E_r## actually point in the same direction (they are in phase). This is the exact opposite of the situation we had earlier. The sign of ##r## will automatically adjust. We will derive the expression of ##r## below.
For H (taking magnetic permeability equal for both mediums, for simplicity) :
##-H_i cos(\theta_i) - H_r cos(\theta_r) = -H_t cos(\theta_t)##
##-k_i cos(\theta_i) - r k_r cos(\theta_r) = -t k_t cos(\theta_t)## (recall ##H = \frac{1}{\mu} \frac{k}{w}E##)
And, since ##\theta_i = \theta_r## (law of reflection) and ##k_i = k_r## (same medium) :
##k_i cos(\theta_i) (1 + r) = t k_t cos(\theta_t) \quad\quad## (2')
Similary, the sign of ##r## will take care of the real direction of the reflected H field.
Now combining equations (1') and (2') we get the Fresnel equations for a s polarized wave (with our new convention of sign) :
##r = \frac{k_t cos(\theta_t) - k_i cos(\theta_i)}{k_i cos(\theta_i) + k_t cos(\theta_t)}## and ##t = \frac{2k_i cos(\theta_i)}{k_i cos(\theta_i) + k_t cos(\theta_t)}##
We can see here that the sign of ##r## is reversed compared to the usual convention. Makes sense. In that aspect, the usual convention is perhaps more intuitive as a negative ##r## implies a reversed direction of the reflected E field.
Interestingly, the fact that ##E_r## can be reflected in the same direction/phase as ##E_i## (positive ##r## in the usual sign convention), thus implying that ##E_t## is
larger than ##E_i## (and that ##t = 1 + r > 1##), does not violate any physics.
This suprised me. But a quick searched
showed that what really matters is that the energy transported isn't higher (actually it must be lower if ##E_r \neq 0##), and indeed it is not, even in that case. If for instance I take a wave propagating in glass (index of refraction ##n \approx 1.5##) and arriving into air (##n \approx 1##) at normal incidence, I get (recall ##k = n \frac{w}{c}##) ##r = \frac{1.5 - 1}{1.5 + 1}## which is positive. But the Poynting vector is ##\vec{E} \times \vec{H} = \frac{1}{\mu} \vec{E} \times \frac{\vec{E}}{v} = \frac{1}{\mu} \frac{E^2}{v}##. Since ##v## is larger in air, even if ##E_t## is larger than ##E_i##, energy is still conserved.
Finally, if ##r## and ##t## are complex, then the reflected and transmitted fields are out of phase compared to the incident field, but not ##\pi## out of phase. This happens in lossy media.
Source (slide 17).
Hope all of this is correct !
