reflection plane

Parallel projection of electromagnetic fields with Geometric Algebra

March 8, 2015 ece1229 No comments , , , ,

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When computing the components of a polarized reflecting ray that were parallel or not-parallel to the reflecting surface, it was found that the electric and magnetic fields could be written as

\begin{equation}\label{eqn:gaFieldProjection:280}
\BE = \lr{ \BE \cdot \pcap } \pcap + \lr{ \BE \cdot \qcap } \qcap = E_\parallel \pcap + E_\perp \qcap
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:300}
\BH = \lr{ \BH \cdot \pcap } \pcap + \lr{ \BH \cdot \qcap } \qcap = H_\parallel \pcap + H_\perp \qcap.
\end{equation}

where a unit vector \( \pcap \) that lies both in the reflecting plane and in the electromagnetic plane (tangential to the wave vector direction) was

\begin{equation}\label{eqn:gaFieldProjection:340}
\pcap = \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}}
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:360}
\qcap = \kcap \cross \pcap.
\end{equation}

Here \( \qcap \) is perpendicular to \( \pcap \) but lies in the electromagnetic plane. This logically subdivides the fields into two pairs, one with the electric field parallel to the reflection plane

\begin{equation}\label{eqn:gaFieldProjection:240}
\begin{aligned}
\BE_1 &= \lr{ \BE \cdot \pcap } \pcap = E_\parallel \pcap \\
\BH_1 &= \lr{ \BH \cdot \qcap } \qcap = H_\perp \qcap,
\end{aligned}
\end{equation}

and one with the magnetic field parallel to the reflection plane

\begin{equation}\label{eqn:gaFieldProjection:380}
\begin{aligned}
\BH_2 &= \lr{ \BH \cdot \pcap } \pcap = H_\parallel \pcap \\
\BE_2 &= \lr{ \BE \cdot \qcap } \qcap = E_\perp \qcap.
\end{aligned}
\end{equation}

Expressed in Geometric Algebra form, each of these pairs of fields should be thought of as components of a single multivector field. That is

\begin{equation}\label{eqn:gaFieldProjection:400}
F_1 = \BE_1 + c \mu_0 \BH_1 I
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:460}
F_2 = \BE_2 + c \mu_0 \BH_2 I
\end{equation}

where the original total field is

\begin{equation}\label{eqn:gaFieldProjection:420}
F = \BE + c \mu_0 \BH I.
\end{equation}

In \ref{eqn:gaFieldProjection:400} we have a composite projection operation, finding the portion of the electric field that lies in the reflection plane, and simultaneously finding the component of the magnetic field that lies perpendicular to that (while still lying in the tangential plane of the electromagnetic field). In \ref{eqn:gaFieldProjection:460} the magnetic field is projected onto the reflection plane and a component of the electric field that lies in the tangential (to the wave vector direction) plane is computed.

If we operate only on the complete multivector field, can we find these composite projection field components in a single operation, instead of working with the individual electric and magnetic fields?

Working towards this goal, it is worthwhile to point out consequences of the assumption that the fields are plane wave (or equivalently far field spherical waves). For such a wave we have

\begin{equation}\label{eqn:gaFieldProjection:480}
\begin{aligned}
\BH
&= \inv{\mu_0} \kcap \cross \BE \\
&= \inv{\mu_0} (-I)\lr{ \kcap \wedge \BE } \\
&= \inv{\mu_0} (-I)\lr{ \kcap \BE – \kcap \cdot \BE} \\
&= -\frac{I}{\mu_0} \kcap \BE,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:gaFieldProjection:520}
\mu_0 \BH I = \kcap \BE.
\end{equation}

This made use of the identity \( \Ba \wedge \Bb = I \lr{\Ba \cross \Bb} \), and the fact that the electric field is perpendicular to the wave vector direction. The total multivector field is

\begin{equation}\label{eqn:gaFieldProjection:500}
\begin{aligned}
F
&= \BE + c \mu_0 \BH I \\
&= \lr{ 1 + c \kcap } \BE.
\end{aligned}
\end{equation}

Expansion of magnetic field component that is perpendicular to the reflection plane gives

\begin{equation}\label{eqn:gaFieldProjection:540}
\begin{aligned}
\mu_0 H_\perp
&= \mu_0 \BH \cdot \qcap \\
&= \gpgradezero{ \lr{-\kcap \BE I} \qcap } \\
&= -\gpgradezero{ \kcap \BE I \lr{ \kcap \cross \pcap} } \\
&= \gpgradezero{ \kcap \BE I I \lr{ \kcap \wedge \pcap} } \\
&= -\gpgradezero{ \kcap \BE \kcap \pcap } \\
&= \gpgradezero{ \kcap \kcap \BE \pcap } \\
&= \BE \cdot \pcap,
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:gaFieldProjection:560}
F_1
= (\pcap + c I \qcap ) \BE \cdot \pcap.
\end{equation}

Since \( \qcap \kcap \pcap = I \), the component of the complete multivector field in the \( \pcap \) direction is

\begin{equation}\label{eqn:gaFieldProjection:580}
\begin{aligned}
F_1
&= (\pcap – c \pcap \kcap ) \BE \cdot \pcap \\
&= \pcap (1 – c \kcap ) \BE \cdot \pcap \\
&= (1 + c \kcap ) \pcap \BE \cdot \pcap.
\end{aligned}
\end{equation}

It is reasonable to expect that \( F_2 \) has a similar form, but with \( \pcap \rightarrow \qcap \). This is verified by expansion

\begin{equation}\label{eqn:gaFieldProjection:600}
\begin{aligned}
F_2
&= E_\perp \qcap + c \lr{ \mu_0 H_\parallel } \pcap I \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ – \kcap \BE I \kcap \qcap I } \lr{\kcap \qcap I} I \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ \kcap \BE \kcap \qcap } \kcap \qcap (-1) \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ \kcap \BE (-\qcap \kcap) } \kcap \qcap (-1) \\
&= \lr{\BE \cdot \qcap} \qcap + c \gpgradezero{ \kcap \kcap \BE \qcap } \kcap \qcap \\
&= \lr{ 1 + c \kcap } \qcap \lr{ \BE \cdot \qcap }
\end{aligned}
\end{equation}

This and \ref{eqn:gaFieldProjection:580} before that makes a lot of sense. The original field can be written

\begin{equation}\label{eqn:gaFieldProjection:620}
F = \lr{ \Ecap + c \lr{ \kcap \cross \Ecap } I } \BE \cdot \Ecap,
\end{equation}

where the leading multivector term contains all the directional dependence of the electric and magnetic field components, and the trailing scalar has the magnitude of the field with respect to the reference direction \( \Ecap \).

We have the same structure after projecting \( \BE \) onto either the \( \pcap \), or \( \qcap \) directions respectively

\begin{equation}\label{eqn:gaFieldProjection:660}
F_1 = \lr{ \pcap + c \lr{ \kcap \cross \pcap } I} \BE \cdot \pcap
\end{equation}
\begin{equation}\label{eqn:gaFieldProjection:680}
F_2 = \lr{ \qcap + c \lr{ \kcap \cross \qcap } I} \BE \cdot \qcap.
\end{equation}

The next question is how to achieve this projection operation directly in terms of \( F \) and \( \pcap, \qcap \), without resorting to expression of \( F \) in terms of \( \BE \), and \( \BB \). I’ve not yet been able to determine the structure of that operation.

Resolving fields into components parallel to the reflecting plane

March 6, 2015 ece1229 No comments , , , , ,

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In order to apply the Fresnel equations, the field components have to be resolved into components where either the electric field or the magnetic field is parallel to the plane of reflection. The geometry of this, with the wave vector direction \( \kcap \) and the electric and magnetic field phasors perpendicular to that direction is sketched in fig. 1.

resolvingFieldsIncidentOnObliquePlaneFig1

fig. 1. Field components relative to reflecting plane

 

If the incident wave is a plane wave, or equivalently a far field spherical wave, it will have the form

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:20}
\BH = \inv{\mu_0} \kcap \cross \BE,
\end{equation}

with the field directions and wave vector directions satisfying

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:60}
\Ecap \cross \Hcap = \kcap
\end{equation}
\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:80}
\Ecap \cdot \kcap = 0
\end{equation}
\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:100}
\Hcap \cdot \kcap = 0.
\end{equation}

The key to resolving the fields into components parallel to the plane of reflection lies in the observation that the cross product of the plane normal \( \ncap \) and the incident wave vector direction \( \kcap \) lies in that plane. With

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:140}
\pcap = \frac{\kcap \cross \ncap}{\Abs{\kcap \cross \ncap}}
\end{equation}
\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:160}
\qcap = \kcap \cross \pcap,
\end{equation}

the field directions can be resolved into components

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:200}
\BE = \lr{ \BE \cdot \pcap } \pcap + \lr{ \BE \cdot \qcap } \qcap = E_\parallel \pcap + E_\perp \qcap
\end{equation}
\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:220}
\BH = \lr{ \BH \cdot \pcap } \pcap + \lr{ \BH \cdot \qcap } \qcap = H_\parallel \pcap + H_\perp \qcap.
\end{equation}

This subdivides the fields into two pairs, one with the electric field parallel to the reflection plane

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:240}
\begin{aligned}
\BE_1 &= \lr{ \BE \cdot \pcap } \pcap = E_\parallel \pcap \\
\BH_1 &= \lr{ \BH \cdot \qcap } \qcap = H_\perp \qcap,
\end{aligned}
\end{equation}

and one with the magnetic field parallel to the reflection plane

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:260}
\begin{aligned}
\BH_2 &= \lr{ \BH \cdot \pcap } \pcap = H_\parallel \pcap \\
\BE_2 &= \lr{ \BE \cdot \qcap } \qcap = E_\perp \qcap.
\end{aligned}
\end{equation}

This is most of what we need to proceed with the reflection and transmission analysis. The only task remaining is to determine the reflection angle.

Using a pencil with the tip on the table I was able to convince myself by observation that there is always a normal plane of incidence regardless of any oblique angle that the ray hits the reflecting surface. This was, for some reason, not intuitively obvious to me. Having done that, the geometry must be reduced to what is sketched in fig. 2.

resolvingAngleOfIncidenceFig1

fig. 2. Angle of incidence determination

 

Once \( \pcap \) has been determined, regardless of it’s orientation in the reflection plane, the component of \( \kcap \) that is normal, directed towards, the plane of reflection is

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:280}
\kcap – \lr{ \kcap \cdot \pcap } \pcap,
\end{equation}

with (squared) length

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:300}
\begin{aligned}
\lr{ \kcap – \lr{ \kcap \cdot \pcap } \pcap }^2
&=
1 + \lr{ \kcap \cdot \pcap }^2 – 2 \lr{ \kcap \cdot \pcap }^2 \\
&=
1 – \lr{ \kcap \cdot \pcap }^2.
\end{aligned}
\end{equation}

The angle of incidence, relative to the normal to the reflection plane, follows from

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:320}
\begin{aligned}
\cos\theta
&= \kcap \cdot \frac{
\kcap – \lr{ \kcap \cdot \pcap } \pcap }{
\sqrt{
1 – \lr{ \kcap \cdot \pcap }^2
}
} \\
&=
\sqrt{
1 – \lr{ \kcap \cdot \pcap }^2
},
\end{aligned}
\end{equation}

Expanding the dot product above gives

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:360}
\begin{aligned}
\kcap \cdot \pcap’
&=
\kcap \cdot \lr{ \pcap \cross \ncap } \\
&=
\frac{1}{\Abs{\kcap \cross \ncap} } \kcap \cdot \lr{ \lr{\kcap \cross \ncap} \cross \ncap },
\end{aligned}
\end{equation}

where

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:380}
\begin{aligned}
\kcap \cdot \lr{ \lr{\kcap \cross \ncap} \cross \ncap }
&=
k_r \epsilon_{r s t} \lr{\kcap \cross \ncap}_s n_t \\
&=
k_r \epsilon_{r s t} \epsilon_{s a b} k_a n_b n_t \\
&=
-k_r \delta_{r t}^{[a b]} k_a n_b n_t \\
&=
-k_r n_t \lr{ k_r n_t – k_t n_r } \\
&=
-1 + \lr{ \kcap \cdot \ncap}^2.
\end{aligned}
\end{equation}

That gives

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:400}
\begin{aligned}
\kcap \cdot \pcap’
&=
\frac{-1 + \lr{ \kcap \cdot \ncap}^2}{\sqrt{1 – \lr{ \kcap \cdot \ncap}^2} } \\
&=
-\sqrt{1 – \lr{ \kcap \cdot \ncap}^2},
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:420}
\begin{aligned}
\cos\theta
&= \sqrt{ 1 – \lr{-\sqrt{1 – \lr{ \kcap \cdot \ncap}^2}}^2 } \\
&= \sqrt{ \lr{ \kcap \cdot \ncap}^2 } \\
&= \kcap \cdot \ncap.
\end{aligned}
\end{equation}

This surprisingly simple result makes so much sense, it is an awful admission of stupidity that I went through all the vector algebra to get it instead of just writing it down directly.

The end result is the reflection angle is given by

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:340}
\boxed{
\theta = \cos^{-1} \kcap \cdot \ncap,
}
\end{equation}

where the reflection plane normal should off the back surface to get the sign right. The only detail left is the vector direction of the reflected ray (as well as the direction for the transmitted ray if that is of interest). The reflected ray direction flips the sign of the normal component of the ray

\begin{equation}\label{eqn:resolvingFieldsIncidentOnPlane:440}
\begin{aligned}
\kcap’
&= -\lr{\kcap \cdot \ncap} \ncap + \lr{ \kcap \wedge \ncap} \ncap \\
&= -\lr{\kcap \cdot \ncap} \ncap + \kcap – \lr{ \ncap \kcap} \cdot \ncap \\
&= \kcap -2 \lr{\kcap \cdot \ncap} \ncap.
\end{aligned}
\end{equation}

Here the sign of the normal doesn’t matter since it only occurs quadratically.

This now supplies everything needed for the application of the Fresnel equations to determine the reflected ray characteristics of an arbitrarily polarized incident field.