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Posts Tagged ‘transverse waveguide’

(A POSSIBLY WRONG) Superposition of transverse electromagnetic field solutions.

Posted by peeterjoot on August 3, 2009

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Disclaimer

FIXME: I’M NOT CONFIDENT THAT i GOT THIS RIGHT. An attempt to verify the final result didn’t work. I’ve either messed up along the way (perhaps right at the beginning), or my verification itself was busted. Am posting these working notes for now, and will revisit later after thinking it through again (or trying the verification again).

Motivation

In ([1]), a Geometric Algebra solution for the transverse components of the electromagnetic field was found. Here we construct a superposition of these transverse fields, keeping the propagation direction fixed, and allowing for continuous variation of the wave number and angular velocity. Evaluation of this superposition integral, first utilizing a contour integral, then utilizing an inverse Fourier transform allows for the determination of the functional form of a general wave packet moving along a fixed direction in space. This wave packet will be seen to have two time dependencies, an advanced time term, and a retarded time term. The phasors will be eliminated from both the propagation and the transverse fields and will provide operators with which the transverse field can be calculated from a general propagation field wave packet.

Superposition of phasor solutions.

When the field is required to have explicit sinusoidal time and propagation direction, the field components in the transverse (to the propagation) direction were found to be

\begin{aligned}F_t &= \frac{1}{{i \left( \pm k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{\omega}{c}\right) }} \boldsymbol{\nabla}_t F_z \\ &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t F_z \end{aligned} \quad\quad\quad(1)

Here F_z = F_z(x,y) = \mathbf{E}_z(x,y) + I\mathbf{B}_z(x,y)/\sqrt{\mu\epsilon} is required by construction to commute with \hat{\mathbf{z}}, but is otherwise arbitrary, except perhaps for boundary value constraints not considered here.

Removing the restriction to fixed wave number and angular velocity we can integrate over both to express a more general transverse wave propagation

\begin{aligned}F_t &= \int d\omega e^{-i\omega t} \int dk e^{ikz} \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left(k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t {F_z}^{k,\omega} \end{aligned} \quad\quad\quad(3)

Inventing temporary notation for convenience we write for the frequency dependent weighting function {F_z}^{k,\omega} = F_z(x,y,k,\omega). Also note that explicit \pm k has been dropped after allowing wave number to range over both positive and negative values.

Observe that each of these integrals has the form of a Fourier integral (ignoring constant factors that can be incorporated into the weighting function). Also observe that the integral kernel has two poles on the real k-axis (or real \omega-axis). These can be utilized to evaluate one of the integrals using an upper half plane semicircular contour. FIXME: picture here.

Assuming that {F_z}^{k,\omega} is small enough at \infty (on the large contour) to be neglected, we have for the integral after integrating around the poles at k = \pm \sqrt{\mu\epsilon}{\left\lvert{\omega/c}\right\rvert}

\begin{aligned} F_t &= \pi i \int d\omega e^{-i\omega t} {\left.e^{ikz} \frac{i}{k - \sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c}} \left(k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t {F_z}^{k,\omega}\right\vert}_{k=-\sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c}} \\ &+ \pi i \int d\omega e^{-i\omega t} {\left.e^{ikz} \frac{i}{k + \sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c}} \left(k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t {F_z}^{k,\omega}\right\vert}_{k=\sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c}} \end{aligned}

Writing {F_z}^{\omega+} = F_z(x,y,\omega, k=\sqrt{\mu\epsilon}{\left\lvert{\omega}\right\rvert}/c), and {F_z}^{\omega-} = F_z(x,y,\omega, k=-\sqrt{\mu\epsilon}{\left\lvert{\omega}\right\rvert}/c), for the values of our field at the poles, we have

\begin{aligned} F_t &= \frac{\pi}{2}\int d\omega e^{-i\omega t} e^{-i \sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c} z}\frac{\omega -\hat{\mathbf{z}}{\left\lvert{\omega}\right\rvert}}{{\left\lvert{\omega}\right\rvert}}\boldsymbol{\nabla}_t {F_z}^{\omega-} \\ &- \frac{\pi}{2}\int d\omega e^{-i\omega t} e^{i \sqrt{\mu\epsilon}\frac{{\left\lvert{\omega}\right\rvert}}{c} z} \frac{\omega +\hat{\mathbf{z}}{\left\lvert{\omega}\right\rvert}}{{\left\lvert{\omega}\right\rvert}}\boldsymbol{\nabla}_t {F_z}^{\omega+} \end{aligned}

Can one ignore the singularity at \omega = 0 since it divides out? If so, then we have

\begin{aligned} F_t &= \frac{\pi}{2}\int d\omega e^{-i\omega t} e^{-i \sqrt{\mu\epsilon}\frac{\left\lvert{\omega}\right\rvert}{c} z}(\text{sgn}(\omega) - \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega-} \\ &- \frac{\pi}{2}\int d\omega e^{-i\omega t} e^{i \sqrt{\mu\epsilon}\frac{\left\lvert{\omega}\right\rvert}{c} z} (\text{sgn}(\omega) + \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega+} \end{aligned}

Writing this out separately for the regions greater and lesser than zero we have

\begin{aligned} F_t &= \frac{\pi}{2}\int_{0+}^\infty d\omega e^{-i\omega t} e^{-i \sqrt{\mu\epsilon}\frac{\omega}{c} z}(1 - \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega-} - \frac{\pi}{2}\int_{0+}^\infty d\omega e^{-i\omega t} e^{i \sqrt{\mu\epsilon}\frac{\omega}{c} z} (1 + \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega+} \\ &- \frac{\pi}{2}\int_{-\infty}^{0-} d\omega e^{-i\omega t} e^{i \sqrt{\mu\epsilon}\frac{\omega}{c} z}(1 + \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega-} - \frac{\pi}{2}\int_{-\infty}^{0-} d\omega e^{-i\omega t} e^{-i \sqrt{\mu\epsilon}\frac{\omega}{c} z} (-1 + \hat{\mathbf{z}})\boldsymbol{\nabla}_t {F_z}^{\omega+} \\ \end{aligned}

Grouping by exponentials of like sign, and integrating over a region that omits some neighborhood around the origin, we have eliminated the {\left\lvert{\omega}\right\rvert} factors

\begin{aligned}F_t &= (1 - \hat{\mathbf{z}}) \boldsymbol{\nabla}_t\int d\omega e^{-i\omega \left(t + \sqrt{\mu\epsilon}\frac{z}{c}\right)}\frac{\pi}{2}({F_z}^{\omega-}\theta(\omega) + {F_z}^{\omega+}\theta(-\omega)) \\ &-(1 + \hat{\mathbf{z}}) \boldsymbol{\nabla}_t\int d\omega e^{-i\omega \left(t - \sqrt{\mu\epsilon}\frac{z}{c}\right)}\frac{\pi}{2}({F_z}^{\omega-}\theta(\omega) + {F_z}^{\omega+}\theta(-\omega)) \end{aligned}

The Heaviside unit step function has been used to group things nicely over the doubled integration ranges, and what is left now can be observed to be Fourier transforms from the frequency to time domain. The transformed functions are to be evaluated at a shifted time in each case.

\begin{aligned}F_t &= (1 - \hat{\mathbf{z}}) \boldsymbol{\nabla}_t\frac{\pi}{2}{\left.\mathcal{F}\left( ({F_z}^{\omega-}\theta(\omega) + {F_z}^{\omega+}\theta(-\omega)) \right)\right\vert}_{t=t + \sqrt{\mu\epsilon}\frac{z}{c}} \\ &-(1 + \hat{\mathbf{z}}) \boldsymbol{\nabla}_t\frac{\pi}{2}{\left.\mathcal{F}\left( ({F_z}^{\omega-}\theta(\omega) + {F_z}^{\omega+}\theta(-\omega)) \right)\right\vert}_{t=t - \sqrt{\mu\epsilon}\frac{z}{c}} \end{aligned}

Attempting to actually evaluate this Fourier transform shouldn’t actually be necessary since we the original angular velocity and wave number domain function F(x,y,\omega,k) was arbitrary (except for its commutation with \hat{\mathbf{z}}). Instead we just suppose (once again overloading the symbol F_z) that we have a function F_z(x,y,u) that at time u takes the value

\begin{aligned}F_z(x,y,u) = \frac{\pi}{2}\mathcal{F}\left( ({F_z}^{\omega-}\theta(\omega) + {F_z}^{\omega+}\theta(-\omega)) \right) \end{aligned}

The transform will not change the grades of the propagation field F_z so this still commutes with \hat{\mathbf{z}}. We can now write

\begin{aligned}F_t &= (1 - \hat{\mathbf{z}}) \boldsymbol{\nabla}_t F_z(x,y, t + \sqrt{\mu\epsilon}{z}/{c})-(1 + \hat{\mathbf{z}}) \boldsymbol{\nabla}_t F_z(x,y, t - \sqrt{\mu\epsilon}{z}/{c}) \end{aligned} \quad\quad\quad(4)

So, after a bunch of manipulation, we find exactly how the transverse component of the field is related to the propagation direction field, and have eliminated the phasor description supplied our single frequency/wave-number field relations.

What we have left is a propagation field that has the form of an arbitrary unidirectional wave packet, traveling in either direction through the medium with speed \pm c/\sqrt{\mu\epsilon}. If all this math is right, the transverse field for an advancing wave is generated by application onto the propagation field of the operator

\begin{aligned}-(1 + \hat{\mathbf{z}}) \boldsymbol{\nabla}_t \end{aligned} \quad\quad\quad(5)

Similarly, the transverse field for a receding wave is given by application of the operator

\begin{aligned}(1 - \hat{\mathbf{z}}) \boldsymbol{\nabla}_t \end{aligned} \quad\quad\quad(6)

Verification.

Given the doubt about the \omega =0 point in the integral, and the possibility for sign error and other algebraic mistakes it seems worthwhile now to go back to Maxwell’s equation and verify the correctness of these results. Then, presuming everything worked out okay, it would also be good to relate things back to the electric and magnetic field components of the field.

Maxwell’s equation in these units is

\begin{aligned}\left(\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F(x,y) = 0 \end{aligned} \quad\quad\quad(7)

and we want to compute the transverse direction projection from the propagation direction term

\begin{aligned}F_t &= \mathbf{E}_t + I \mathbf{B}_t = \frac{1}{{2}} (F - \hat{\mathbf{z}} F \hat{\mathbf{z}}) \\ F_z &= \mathbf{E}_z + I \mathbf{B}_z = \frac{1}{{2}} (F + \hat{\mathbf{z}} F \hat{\mathbf{z}}) \end{aligned} \quad\quad\quad(8)

Picking one of the terms of (4), we have

\begin{aligned}F_t = (1 + \pm (1 \mp \hat{\mathbf{z}}) \nabla_t) F_z(t \pm \sqrt{\mu\epsilon} z/c) \end{aligned}

Substiting into the left hand operator of Maxwell’s equation (7) and reducing I get

\begin{aligned}((\hat{\mathbf{z}} \pm 1) {\nabla_t}^2 + \nabla_t) F_z + \frac{\mu\epsilon}{c}(1 \pm \hat{\mathbf{z}}) F_z' \end{aligned}

I don’t see how this would neccessarily equal zero?

References

[1] Peeter Joot. Transverse electric and magnetic fields [online]. http://sites.google.com/site/peeterjoot/math2009/transverseField.pdf.

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Sumarizing: Transverse electric and magnetic fields

Posted by peeterjoot on August 1, 2009

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There’s potentially a lot of new ideas in the previous transverse field post (some for me even with previous exposure to the Geometric Algebra formalism). There was no real attempt to teach GA here, but for completeness the GA form of Maxwell’s equation was developed from the traditional divergence and curl formulation of Maxwell’s equations. That was mainly due to use of CGS units which differ since this makes Maxwell’s equation take a different form from the usual (see [1]).

This time a less exploratory summary of the previous results above is assembled.

In these CGS units our field F, and Maxwell’s equation (in absence of charge and current), take the form

\begin{aligned}F &= \mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}} \\ 0 &= \left(\boldsymbol{\nabla} + \frac{\sqrt{\mu\epsilon}}{c}\partial_t\right) F \end{aligned} \quad\quad\quad(30)

The electric and magnetic fields can be picked off by selecting the grade one (vector) components

\begin{aligned}\mathbf{E} &= {\left\langle{{F}}\right\rangle}_{1} \\ \mathbf{B} &= {\left\langle{{-I F}}\right\rangle}_{1} \end{aligned} \quad\quad\quad(32)

With an explicit sinusoidal and z-axis time dependence for the field

\begin{aligned}F(x,y,z,t) &= F(x,y) e^{\pm i k z - i \omega t} \end{aligned} \quad\quad\quad(34)

and a split of the gradient into transverse and z-axis components \boldsymbol{\nabla} = \boldsymbol{\nabla}_t + \hat{\mathbf{z}} \partial_z, Maxwell’s equation takes the form

\begin{aligned}\left(\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F(x,y) = 0 \end{aligned} \quad\quad\quad(35)

Writing for short F = F(x,y), we can split the field into transverse and z-axis components with the commutator and anticommutator products respectively. For the z-axis components we have

\begin{aligned}F_z \hat{\mathbf{z}} \equiv E_z + I B_z = \frac{1}{{2}} (F \hat{\mathbf{z}} + \hat{\mathbf{z}} F) \end{aligned} \quad\quad\quad(36)

The projections onto the z-axis and and transverse directions are respectively

\begin{aligned}F_z &= \mathbf{E}_z + I \mathbf{B}_z = \frac{1}{{2}} (F + \hat{\mathbf{z}} F \hat{\mathbf{z}}) \\ F_t &= \mathbf{E}_t + I \mathbf{B}_t = \frac{1}{{2}} (F - \hat{\mathbf{z}} F \hat{\mathbf{z}} ) \end{aligned} \quad\quad\quad(37)

With an application of the transverse gradient to the z-axis field we easily found the relation between the two
field components

\begin{aligned}\boldsymbol{\nabla}_t F_z &= i \left( \pm k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{\omega}{c}\right) F_t \end{aligned} \quad\quad\quad(39)

A left division of the exponential factor gives the total transverse field

\begin{aligned}F_t &= \frac{1}{{i \left( \pm k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{\omega}{c}\right) }} \boldsymbol{\nabla}_t F_z \end{aligned} \quad\quad\quad(40)

Multiplication of both the numerator and denominator by the conjugate normalizes this

\begin{aligned}F_t &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t F_z \end{aligned} \quad\quad\quad(41)

From this the transverse electric and magnetic fields may be picked off using the projective grade selection operations of (32), and are

\begin{aligned}\mathbf{E}_t &= \frac{i}{\mu\epsilon\frac{\omega^2}{c^2} -k^2} \left( \pm k \boldsymbol{\nabla}_t E_z - \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \right) \\ \mathbf{B}_t &= \frac{i}{\mu\epsilon\frac{\omega^2}{c^2} -k^2} \left( {\mu\epsilon}\frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t E_z \pm k \boldsymbol{\nabla}_t B_z \right) \end{aligned} \quad\quad\quad(42)

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

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Transverse electric and magnetic fields

Posted by peeterjoot on July 31, 2009

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Motivation

In Eli’s Transverse Electric and Magnetic Fields in a Conducting Waveguide blog entry he works through the algebra calculating the transverse components, the perpendicular to the propagation direction components.

This should be possible using Geometric Algebra too, and trying this made for a good exercise.

Setup

The starting point can be the same, the source free Maxwell’s equations. Writing \partial_0 = (1/c) \partial/{\partial t}, we have

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{B} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{E} &= - \partial_0 \mathbf{B} \\ \boldsymbol{\nabla} \times \mathbf{B} &= \mu \epsilon \partial_0 \mathbf{E} \end{aligned} \quad\quad\quad(1)

Multiplication of the last two equations by the spatial pseudoscalar I, and using I \mathbf{a} \times \mathbf{b} = \mathbf{a} \wedge \mathbf{b}, the curl equations can be written in their dual bivector form

\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{E} &= - \partial_0 I \mathbf{B} \\ \boldsymbol{\nabla} \wedge \mathbf{B} &= \mu \epsilon \partial_0 I \mathbf{E} \end{aligned} \quad\quad\quad(5)

Now adding the dot and curl equations using \mathbf{a} \mathbf{b} = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \wedge \mathbf{b} eliminates the cross products

\begin{aligned}\boldsymbol{\nabla} \mathbf{E} &= - \partial_0 I \mathbf{B} \\ \boldsymbol{\nabla} \mathbf{B} &= \mu \epsilon \partial_0 I \mathbf{E} \end{aligned} \quad\quad\quad(7)

These can be further merged without any loss, into the GA first order equation

\begin{aligned}\left(\boldsymbol{\nabla} + \frac{\sqrt{\mu\epsilon}}{c}\partial_t\right) \left(\mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}} \right) = 0 \end{aligned} \quad\quad\quad(9)

We are really after solutions to the total multivector field F = \mathbf{E} + I \mathbf{B}/\sqrt{\mu\epsilon}. For this problem where separate electric and magnetic field components are desired, working from (7) is perhaps what we want?

Following Eli and Jackson, write \boldsymbol{\nabla} = \boldsymbol{\nabla}_t + \hat{\mathbf{z}} \partial_z, and

\begin{aligned}\mathbf{E}(x,y,z,t) &= \mathbf{E}(x,y) e^{\pm i k z - i \omega t} \\ \mathbf{B}(x,y,z,t) &= \mathbf{B}(x,y) e^{\pm i k z - i \omega t} \end{aligned} \quad\quad\quad(10)

Evaluating the z and t partials we have

\begin{aligned}(\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}}) \mathbf{E}(x,y) &= \frac{i\omega}{c} I \mathbf{B}(x,y) \\ (\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}}) \mathbf{B}(x,y) &= -\mu \epsilon \frac{i\omega}{c} I \mathbf{E}(x,y) \end{aligned} \quad\quad\quad(12)

For the remainder of these notes, the explicit (x,y) dependence will be assumed for \mathbf{E} and \mathbf{B}.

An obvious thing to try with these equations is just substitute one into the other. If that’s done we get the pair of second order harmonic equations

\begin{aligned}{\boldsymbol{\nabla}_t}^2\begin{pmatrix}\mathbf{E} \\ \mathbf{B} \end{pmatrix}= \left( k^2 - \mu \epsilon \frac{\omega^2}{c^2} \right)\begin{pmatrix}\mathbf{E} \\ \mathbf{B} \end{pmatrix} \end{aligned} \quad\quad\quad(14)

One could consider the problem solved here. Separately equating both sides of this equation to zero, we have the k^2 = \mu\epsilon \omega^2/c^2 constraint on the wave number and angular velocity, and the second order Laplacian on the left hand side is solved by the real or imaginary parts of any analytic function. Especially when one considers that we are after a multivector field that of intrinsic complex nature.

However, that is not really what we want as a solution. Doing the same on the unified Maxwell equation (9), we have

\begin{aligned}\left(\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) \left(\mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}} \right) = 0 \end{aligned} \quad\quad\quad(15)

Selecting scalar, vector, bivector and trivector grades of this equation produces the following respective relations between the various components

\begin{aligned}0 = \left\langle{{\cdots}}\right\rangle &= \boldsymbol{\nabla}_t \cdot \mathbf{E} \pm i k \hat{\mathbf{z}} \cdot \mathbf{E} \\ 0 = {\left\langle{{\cdots}}\right\rangle}_{1} &= I \boldsymbol{\nabla}_t \wedge \mathbf{B}/\sqrt{\mu\epsilon} \pm i I k \hat{\mathbf{z}} \wedge \mathbf{B}/\sqrt{\mu\epsilon} - i \sqrt{\mu\epsilon}\frac{\omega}{c} \mathbf{E} \\ 0 = {\left\langle{{\cdots}}\right\rangle}_{2} &= \boldsymbol{\nabla}_t \wedge \mathbf{E} \pm i k \hat{\mathbf{z}} \wedge \mathbf{E} - i \frac{\omega}{c} I \mathbf{B} \\ 0 = {\left\langle{{\cdots}}\right\rangle}_{3} &= I \boldsymbol{\nabla}_t \cdot \mathbf{B}/\sqrt{\mu\epsilon} \pm i I k \hat{\mathbf{z}} \cdot \mathbf{B}/\sqrt{\mu\epsilon} \end{aligned} \quad\quad\quad(16)

From the scalar and pseudoscalar grades we have the propagation components in terms of the transverse ones

\begin{aligned}E_z &= \frac{\pm i}{k} \boldsymbol{\nabla}_t \cdot \mathbf{E}_t \\ B_z &= \frac{\pm i}{k} \boldsymbol{\nabla}_t \cdot \mathbf{B}_t \end{aligned} \quad\quad\quad(20)

But this is the opposite of the relations that we are after. On the other hand from the vector and bivector grades we have

\begin{aligned}i \frac{\omega}{c} \mathbf{E} &= -\frac{1}{{\mu\epsilon}}\left(\boldsymbol{\nabla}_t \times \mathbf{B}_z \pm i k \hat{\mathbf{z}} \times \mathbf{B}_t\right) \\ i \frac{\omega}{c} \mathbf{B} &= \boldsymbol{\nabla}_t \times \mathbf{E}_z \pm i k \hat{\mathbf{z}} \times \mathbf{E}_t \end{aligned} \quad\quad\quad(22)

A clue from the final result.

From (22) and a lot of messy algebra we should be able to get the transverse equations. Is there a slicker way? The end result that Eli obtained suggests a path. That result was

\begin{aligned}\mathbf{E}_t = \frac{i}{\mu\epsilon \frac{\omega^2}{c^2} - k^2} \left( \pm k \boldsymbol{\nabla}_t E_z - \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \right) \end{aligned} \quad\quad\quad(24)

The numerator looks like it can be factored, and after a bit of playing around a suitable factorization can be obtained:

\begin{aligned}{\left\langle{{ \left( \pm k + \frac{\omega}{c} \hat{\mathbf{z}} \right) \boldsymbol{\nabla}_t \hat{\mathbf{z}} \left( \mathbf{E}_z + I \mathbf{B}_z \right) }}\right\rangle}_{1}&={\left\langle{{ \left( \pm k + \frac{\omega}{c} \hat{\mathbf{z}} \right) \boldsymbol{\nabla}_t \left( E_z + I B_z \right) }}\right\rangle}_{1} \\ &=\pm k \boldsymbol{\nabla} E_z + \frac{\omega}{c} {\left\langle{{ I \hat{\mathbf{z}} \boldsymbol{\nabla}_t B_z }}\right\rangle}_{1} \\ &=\pm k \boldsymbol{\nabla} E_z + \frac{\omega}{c} I \hat{\mathbf{z}} \wedge \boldsymbol{\nabla}_t B_z \\ &=\pm k \boldsymbol{\nabla} E_z - \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \\ \end{aligned}

Observe that the propagation components of the field \mathbf{E}_z + I\mathbf{E}_z can be written in terms of the symmetric product

\begin{aligned}\frac{1}{{2}} \left( \hat{\mathbf{z}} (\mathbf{E} + I\mathbf{B}) + (\mathbf{E} + I\mathbf{B}) \hat{\mathbf{z}} \right)&=\frac{1}{{2}} \left( \hat{\mathbf{z}} \mathbf{E} + \mathbf{E} \hat{\mathbf{z}} \right) + \frac{I}{2} \left( \hat{\mathbf{z}} \mathbf{B} + \mathbf{B} \hat{\mathbf{z}} + I \right) \\ &=\hat{\mathbf{z}} \cdot \mathbf{E} + I \hat{\mathbf{z}} \cdot \mathbf{B} \end{aligned}

Now the total field in CGS units was actually F = \mathbf{E} + I \mathbf{B}/\sqrt{\mu\epsilon}, not F = \mathbf{E} + I \mathbf{B}, so the factorization above isn’t exactly what we want. It does however, provide the required clue. We probably get the result we want by forming the symmetric product (a hybrid dot product selecting both the vector and bivector terms).

Symmetric product of the field with the direction vector.

Rearranging Maxwell’s equation (15) in terms of the transverse gradient and the total field F we have

\begin{aligned}\boldsymbol{\nabla}_t F = \left( \mp i k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F \end{aligned} \quad\quad\quad(25)

With this our symmetric product is

\begin{aligned}\boldsymbol{\nabla}_t ( F \hat{\mathbf{z}} + \hat{\mathbf{z}} F) &= (\boldsymbol{\nabla}_t F) \hat{\mathbf{z}} - \hat{\mathbf{z}} (\boldsymbol{\nabla}_t F) \\ &=\left( \mp i k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F \hat{\mathbf{z}}- \hat{\mathbf{z}} \left( \mp i k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F \\ &=i \left( \mp k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) (F \hat{\mathbf{z}} - \hat{\mathbf{z}} F) \\ \end{aligned}

The antisymmetric product on the right hand side should contain the desired transverse field components. To verify multiply it out

\begin{aligned}\frac{1}{{2}}(F \hat{\mathbf{z}} - \hat{\mathbf{z}} F) &=\frac{1}{{2}}\left( \left(\mathbf{E} + I \mathbf{B}/\sqrt{\mu\epsilon}\right) \hat{\mathbf{z}} - \hat{\mathbf{z}} \left(\mathbf{E} + I \mathbf{B}/\sqrt{\mu\epsilon}\right) \right) \\ &=\mathbf{E} \wedge \hat{\mathbf{z}} + I \mathbf{B}/\sqrt{\mu\epsilon} \wedge \hat{\mathbf{z}} \\ &=(\mathbf{E}_t + I \mathbf{B}_t/\sqrt{\mu\epsilon}) \hat{\mathbf{z}} \\ \end{aligned}

Now, with multiplication by the conjugate quantity -i(\pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\omega/c), we can extract these transverse components.

\begin{aligned}\left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \left( \mp k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) (F \hat{\mathbf{z}} - \hat{\mathbf{z}} F) &=\left( -k^2 + {\mu\epsilon}\frac{\omega^2}{c^2}\right) (F \hat{\mathbf{z}} - \hat{\mathbf{z}} F) \end{aligned}

Rearranging, we have the transverse components of the field

\begin{aligned}(\mathbf{E}_t + I \mathbf{B}_t/\sqrt{\mu\epsilon}) \hat{\mathbf{z}} &=\frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t \frac{1}{{2}}( F \hat{\mathbf{z}} + \hat{\mathbf{z}} F) \end{aligned} \quad\quad\quad(26)

With left multiplication by \hat{\mathbf{z}}, and writing F = F_t + F_z we have

\begin{aligned}F_t &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t F_z \end{aligned} \quad\quad\quad(27)

While this is a complete solution, we can additionally extract the electric and magnetic fields to compare results with Eli’s calculation. We take
vector grades to do so with \mathbf{E}_t = {\left\langle{{F_t}}\right\rangle}_{1}, and \mathbf{B}_t/\sqrt{\mu\epsilon} = {\left\langle{{-I F_t}}\right\rangle}_{1}. For the transverse electric field

\begin{aligned}{\left\langle{{ \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t (\mathbf{E}_z + I \mathbf{B}_z/\sqrt{/\mu\epsilon}) }}\right\rangle}_{1} &=\pm k \hat{\mathbf{z}} (-\hat{\mathbf{z}}) \boldsymbol{\nabla}_t E_z + \frac{\omega}{c} \underbrace{{\left\langle{{I \boldsymbol{\nabla}_t \hat{\mathbf{z}}}}\right\rangle}_{1}}_{-I^2 \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t } B_z \\ &=\mp k \boldsymbol{\nabla}_t E_z + \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \\ \end{aligned}

and for the transverse magnetic field

\begin{aligned}{\left\langle{{ -I \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t (\mathbf{E}_z + I \mathbf{B}_z/\sqrt{\mu\epsilon}) }}\right\rangle}_{1} &=-I \sqrt{\mu\epsilon}\frac{\omega}{c} \boldsymbol{\nabla}_t \mathbf{E}_z+{\left\langle{{ \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t \mathbf{B}_z/\sqrt{\mu\epsilon} }}\right\rangle}_{1} \\ &=- \sqrt{\mu\epsilon}\frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t E_z\mp k \boldsymbol{\nabla}_t B_z/\sqrt{\mu\epsilon} \\ \end{aligned}

Thus the split of transverse field into the electric and magnetic components yields

\begin{aligned}\mathbf{E}_t &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \mp k \boldsymbol{\nabla}_t E_z + \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \right) \\ \mathbf{B}_t &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( - {\mu\epsilon}\frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t E_z \mp k \boldsymbol{\nabla}_t B_z \right) \end{aligned} \quad\quad\quad(28)

Compared to Eli’s method using messy traditional vector algebra, this method also has a fair amount of messy tricky algebra, but of a different sort.

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