Direction of reflected E field on a dielectric surface?

[DoobleD]

When studying Fresnel equations in EM, I very often came across the following type of picture for a s-polarized incident wave on a dielectric interface (http://iqst.ca/quantech/pubs/2013/fresnel-eoe.pdf) :

Many similar diagrams can be found on the internet, for instance http://ecee.colorado.edu/~ecen3400/Chapter%2022%20-%20Reflection%20and%20Refraction%20of%20Plane%20Waves.pdf. My questions is : why is E_r pointing in the same direction as E_i does (e.g. towards the positive y) ?

The boundary condition on the E field at the interface is :

E_i + E_r = E_t

Since E_t can't have a higher amplitude than E_i, shouldn't E_r points in the opposite direction of E_i (towards negative y) ? And then H_r should of course be reversed as well on the diagram. However the E_i and E_r having same direction can be found in the pictures of many different sources, so I suppose it is not a mistake. How can it be ?

Also I know that each E field switch direction as it propagates, but in the above diagram and others similar that I found, E_r is drawn at the same z position as E_i and in the same direction, as if both were in phase.

 

[Meir Achuz]

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.

 

[DoobleD]

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 !