## Mathematica notebooks updated, and a bivector addition visualization.

February 10, 2019 math and physics play No comments , , ,

This blog now has a copy of all my Mathematica notebooks (as of Feb 10, 2019), complete with a chronological index.  I hadn’t updated that index since 2014, and it was quite stale.

I’ve also added an additional level of per-directory indexing.  For example, you can now look at just the notebooks for my book, Geometric Algebra for Electrical Engineers.  That was possible before, but you would have had to clone the entire git repository to be able to do so easily.

This update includes a new notebook written today, which has a Manipulate visualization of 3D bivector addition that is kind of fun.

Bivector addition, at least in 3D, can be done graphically almost like vector addition.  Instead of trying to add the planes (which can be done, as in the neat illustration in Geometric Algebra for Computer Science), you can do the task more simply by connecting the normals head to tail, where each of the normals are scaled by the area of the bivector (i.e. it’s absolute magnitude).  The resulting bivector has an area equal to the length of that sum of normals, and a “direction” perpendicular to that resulting normal.  This fun little Manipulate lets you interactively visualize this process, by changing the radius of a set of summed bivectors, each oriented in a different direction, and observing the effects of doing so.

Of course, you can interpret this visualization as nothing more than a representation of addition of cross products, if you were to interpret the vector representing a cross product as an oriented area with a normal equal to that cross product (where the normal’s magnitude equals the area, as in this bivector addition visualization.)  This works out nicely because of the duality relationship between the cross and wedge product, and the duality relationship between 3D bivectors and their normals.

## PHY2403H Quantum Field Theory. Lecture 9: Unbroken and spontaneously broken symmetries, Higgs Lagrangian, scale invariance, Lorentz invariance, angular momentum quantization. Taught by Prof. Erich Poppitz

[Click here for a PDF of this post with nicer formatting (and a Mathematica listing that I didn’t include in this blog post’s latex export)]

### DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

## Last time

We followed a sequence of operations

1. Noether’s theorem
2. $$\rightarrow$$ conserved currents
3. $$\rightarrow$$ charges (classical)
4. $$\rightarrow$$ “correspondence principle”
5. $$\rightarrow \hatQ$$
• Hermitian operators
• “generators of symmetry”
\label{eqn:qftLecture9:20}
\hatU(\alpha) = e^{i \alpha \hatQ}

We found
\label{eqn:qftLecture9:40}
\hatU(\alpha) \phihat \hatU^\dagger(\alpha) = \phihat + i \alpha \antisymmetric{\hatQ}{\phihat} + \cdots

### Example: internal symmetries:

(non-spacetime), such as $$O(N)$$ or $$U(1)$$.

In QFT internal symmetries can have different “\underline{modes of realization}”.

[I]

1. “Wigner mode”. These are also called “unbroken symmetries”.
\label{eqn:qftLecture9:60}
\hatQ \ket{0} = 0

i.e. $$\hatU(\alpha) \ket{0} = 0$$.
Ground state invariant. Formally $$:\hatQ:$$ annihilates $$\ket{0}$$.
$$\antisymmetric{\hatQ}{\hatH} = 0$$ implies that all eigenstates are eigenstates of $$\hatQ$$ in $$U(1)$$. Example from HW 1
\label{eqn:qftLecture9:80}
\hatQ = \text{“charge” under $$U(1)$$}.

All states have definite charge, just live in QU.
2. “Nambu-Goldstone mode” (Landau-ginsburg). This is also called a “spontaneously broken symmetry”\footnote{
First encounter example (HWII, $$SU(2) \times SU(2) \rightarrow SU(2)$$). Here a $$U(1)$$ spontaneous broken symmetry.}.
$$H$$ or $$L$$ is invariant under symmetry, but ground state is not.

fig. 1. Mexican hat potential.

fig. 2. Degenerate Mexican hat potential ( v = 0)

Example:
\label{eqn:qftLecture9:100}
\LL = \partial_\mu \phi^\conj \partial^\mu \phi – V(\Abs{\phi}),

where
\label{eqn:qftLecture9:120}
V(\Abs{\phi}) = m^2 \phi^\conj \phi + \frac{\lambda}{4} \lr{ \phi^\conj \phi }^2.

When $$m^2 > 0$$ we have a Wigner mode, but when $$m^2 < 0$$ we have an issue: $$\phi = 0$$ is not a minimum of potential.
When $$m^2 < 0$$ we write
\label{eqn:qftLecture9:140}
\begin{aligned}
V(\phi)
&= – m^2 \phi^\conj \phi + \frac{\lambda}{4} \lr{ \phi^\conj \phi}^2 \\
&= \frac{\lambda}{4} \lr{
\lr{ \phi^\conj \phi}^2 – \frac{4}{\lambda} m^2 } \\
&= \frac{\lambda}{4} \lr{
\phi^\conj \phi – \frac{2}{\lambda} m^2 }^2 – \frac{4 m^4}{\lambda^2},
\end{aligned}

or simply
\label{eqn:qftLecture9:780}
V(\phi)
=
\frac{\lambda}{4} \lr{ \phi^\conj \phi – v^2 }^2 + \text{const}.

The potential (called the Mexican hat potential) is illustrated in fig. 1 for non-zero $$v$$, and in
fig. 2 for $$v = 0$$.
We choose to expand around some point on the minimum ring (it doesn’t matter which one).
When there is no potential, we call the field massless (i.e. if we are in the minimum ring).
We expand as
\label{eqn:qftLecture9:160}
\phi(x) = v \lr{ 1 + \frac{\rho(x)}{v} } e^{i \alpha(x)/v },

so
\label{eqn:qftLecture9:180}
\begin{aligned}
\frac{\lambda}{4}
\lr{\phi^\conj \phi – v^2}^2
&=
\lr{
v^2 \lr{ 1 + \frac{\rho(x)}{v} }^2
– v^2
}^2 \\
&=
\frac{\lambda}{4}
v^4 \lr{ \lr{ 1 + \frac{\rho(x)}{v} }^2 – 1 } \\
&=
\frac{\lambda}{4}
v^4
\lr{
\frac{2 \rho}{v} + \frac{\rho^2}{v^2}
}^2.
\end{aligned}

\label{eqn:qftLecture9:200}
\partial_\mu \phi =
\lr{
v \lr{ 1 + \frac{\rho(x)}{v} } \frac{i}{v} \partial_\mu \alpha
+ \partial_\mu \rho
} e^{i \alpha}

so
\label{eqn:qftLecture9:220}
\begin{aligned}
\LL
&= \Abs{\partial \phi^\conj}^2 – \frac{\lambda}{4} \lr{ \Abs{\phi^\conj}^2 – v^2 }^2 \\
&=
\partial_\mu \rho \partial^\mu \rho + \partial_\mu \alpha \partial^\mu \alpha \lr{ 1 + \frac{\rho}{v} }

\frac{\lambda v^4}{4} \frac{ 4\rho^2}{v^2} + O(\rho^3) \\
&=
\partial_\mu \rho \partial^\mu \rho
– \lambda v^2\rho^2
+
\partial_\mu \alpha \partial^\mu \alpha \lr{ 1 + \frac{\rho}{v} }.
\end{aligned}

We have two fields, $$\rho$$ : a massive scalar field, the “Higgs”, and a massless field $$\alpha$$ (the Goldstone Boson).

$$U(1)$$ symmetry acts on $$\phi(x) \rightarrow e^{i \omega } \phi(x)$$ i.t.o $$\alpha(x) \rightarrow \alpha(x) + v \omega$$.
$$U(1)$$ global symmetry (broken) acts on the Goldstone field $$\alpha(x)$$ by a constant shift. ($$U(1)$$ is still a symmetry of the Lagrangian.)

The current of the $$U(1)$$ symmetry is:
\label{eqn:qftLecture9:240}
j_\mu = \partial_\mu \alpha \lr{ 1 + \text{higher dimensional $$\rho$$ terms} }.

When we quantize
\label{eqn:qftLecture9:260}
\alpha(x) =
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} e^{i \omega_p t – i \Bp \cdot \Bx} \hat{a}_\Bp^\dagger +
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} e^{-i \omega_p t + i \Bp \cdot \Bx} \hat{a}_\Bp

\label{eqn:qftLecture9:280}
j^\mu(x) = \partial^\mu \alpha(x) =
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} \lr{ i \omega_\Bp – i \Bp } e^{i \omega_p t – i \Bp \cdot \Bx} \hat{a}_\Bp^\dagger +
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} \lr{ -i \omega_\Bp + i \Bp } e^{-i \omega_p t + i \Bp \cdot \Bx} \hat{a}_\Bp.

\label{eqn:qftLecture9:300}
j^\mu(x) \ket{0} \ne 0,

instead it creates a single particle state.

## Examples of symmetries

In particle physics, examples of Wigner vs Nambu-Goldstone, ignoring gravity the only exact internal symmetry in the standard module is
$$(B\# – L\#)$$, believed to be a $$U(1)$$ symmetry in Wigner mode.

Here $$B\#$$ is the Baryon number, and $$L\#$$ is the Lepton number. Examples:

• $$B(p) = 1$$, proton.
• $$B(q) = 1/3$$, quark
• $$B(e) = 1$$, electron
• $$B(n) = 1$$, neutron.
• $$L(p) = 1$$, proton.
• $$L(q) = 0$$, quark.
• $$L(e) = 0$$, electron.

The major use of global internal symmetries in the standard model is as “approximate” ones. They become symmetries when one neglects some effect( “terms in $$\LL$$”).
There are other approximate symmetries (use of group theory to find the Balmer series).

### Example from HW2:

QCD in limit
\label{eqn:qftLecture9:320}
m_u = m_d = 0.

$$m_u m_d \ll m_p$$ (the products of the up-quark mass and the down-quark mass are much less than a composite one (name?)).
$$SU(2)_L \times SU(2)_R \rightarrow SU(2)_V$$

### EWSB (Electro-Weak-Symmetry-Breaking) sector

When the couplings $$g_2, g_1 = 0$$. ($$g_2 \in SU(2), g_1 \in U(1)$$).

## Scale invariance

\label{eqn:qftLecture9:340}
\begin{aligned}
x &\rightarrow e^{\lambda} x \\
\phi &\rightarrow e^{-\lambda} \phi \\
A_\mu &\rightarrow e^{-\lambda} A_\mu
\end{aligned}

Any unitary theory which is scale invariant is also \underline{conformal} invariant. Conformal invariance means that angles are preserved.
The point here is that there is more than scale invariance.

We have classical internal global continuous symmetries.
These can be either

1. “unbroken” (Wigner mode)
\label{eqn:qftLecture9:360}
\hatQ\ket{0} = 0.
2. “spontaneously broken”
\label{eqn:qftLecture9:380}
j^\mu(x) \ket{0} \ne 0

(creates Goldstone modes).
3. “anomalous”. Classical symmetries are not a symmetry of QFT.
Examples:

• Scale symmetry (to be studied in QFT II), although this is not truly internal.
• In QCD again when $$\omega_\Bq = 0$$, a $$U(1$$ symmetry (chiral symmetry) becomes exact, and cannot be preserved in QFT.
• In the standard model (E.W sector), the Baryon number and Lepton numbers are not symmetries, but their difference $$B\# – L\#$$ is a symmetry.

## Lorentz invariance.

We’d like to study the action of Lorentz symmetries on quantum states. We are going to “go by the book”, finding symmetries, currents, quantize, find generators, and so forth.

Under a Lorentz transformation
\label{eqn:qftLecture9:400}
x^\mu \rightarrow {x’}^\mu = {\Lambda^\mu}_\nu x^\nu,

We are going to consider infinitesimal Lorentz transformations
\label{eqn:qftLecture9:420}
{\Lambda^\mu}_\nu \approx
{\delta^\mu}_\nu + {\omega^\mu}_\nu
,

where $${\omega^\mu}_\nu$$ is small.
A Lorentz transformation $$\Lambda$$ must satisfy $$\Lambda^\T G \Lambda = G$$, or
\label{eqn:qftLecture9:800}
g_{\mu\nu} = {{\Lambda}^\alpha}_\mu g_{\alpha \beta} {{\Lambda}^\beta}_\nu,

into which we insert the infinitesimal transformation representation
\label{eqn:qftLecture9:820}
\begin{aligned}
0
&=
– g_{\mu\nu} +
\lr{ {\delta^\alpha}_\mu + {\omega^\alpha}_\mu }
g_{\alpha \beta}
\lr{ {\delta^\beta}_\nu + {\omega^\beta}_\nu } \\
&=
– g_{\mu\nu} +
\lr{
g_{\mu \beta}
+
\omega_{\beta\mu}
}
\lr{ {\delta^\beta}_\nu + {\omega^\beta}_\nu } \\
&=
– g_{\mu\nu} +
g_{\mu \nu}
+
\omega_{\nu\mu}
+
\omega_{\mu\nu}
+
\omega_{\beta\mu}
{\omega^\beta}_\nu.
\end{aligned}

The quadratic term can be ignored, leaving just
\label{eqn:qftLecture9:840}
0 =
\omega_{\nu\mu}
+
\omega_{\mu\nu},

or
\label{eqn:qftLecture9:860}
\omega_{\nu\mu} = – \omega_{\mu\nu}.

Note that $$\omega$$ is a completely antisymmetric tensor, and like $$F_{\mu\nu}$$ this has only 6 elements.
This means that the
infinitesimal transformation of the coordinates is
\label{eqn:qftLecture9:440}
x^\mu \rightarrow {x’}^\mu \approx x^\mu + \omega^{\mu\nu} x_\nu,

the field transforms as
\label{eqn:qftLecture9:460}
\phi(x) \rightarrow \phi'(x’) = \phi(x)

or
\label{eqn:qftLecture9:760}
\phi'(x^\mu + \omega^{\mu\nu} x_\nu) =
\phi'(x) + \omega^{\mu\nu} x_\nu \partial_\mu\phi(x) = \phi(x),

so
\label{eqn:qftLecture9:480}
\delta \phi = \phi'(x) – \phi(x) =
-\omega^{\mu\nu} x_\nu \partial_\mu \phi.

Since $$\LL$$ is a scalar
\label{eqn:qftLecture9:500}
\begin{aligned}
\delta \LL
&=
-\omega^{\mu\nu} x_\nu \partial_\mu \LL \\
&=

\partial_\mu \lr{
\omega^{\mu\nu} x_\nu \LL
}
+
(\partial_\mu x_\nu) \omega^{\mu\nu} \LL \\
&=
\partial_\mu \lr{

\omega^{\mu\nu} x_\nu \LL
},
\end{aligned}

since $$\partial_\nu x_\mu = g_{\nu\mu}$$ is symmetric, and $$\omega$$ is antisymmetric.
Our current is
\label{eqn:qftLecture9:520}
J^\mu_\omega
=

\omega^{\mu\nu} x_\mu \LL
.

Our Noether current is
\label{eqn:qftLecture9:540}
\begin{aligned}
j^\nu_{\omega^{\mu\rho}}
&= \PD{\phi_{,\nu}}{\LL} \delta \phi – J^\mu_\omega \\
&=
\partial^\nu \phi\lr{ – \omega^{\mu\rho} x_\rho \partial_\mu \phi } + \omega^{\nu \rho} x_\rho \LL \\
&=
\omega^{\mu\rho}
\lr{
\partial^\nu \phi\lr{ – x_\rho \partial_\mu \phi } + {\delta^{\nu}}_\mu x_\rho \LL
} \\
&=
\omega^{\mu\rho} x_\rho
\lr{
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL
}
\end{aligned}

We identify
\label{eqn:qftLecture9:560}

{T^\nu}_\mu =
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL,

so the current is
\label{eqn:qftLecture9:580}
\begin{aligned}
j^\nu_{\omega_{\mu\rho}}
&=
-\omega^{\mu\rho} x_\rho
{T^\nu}_\mu \\
&=
-\omega_{\mu\rho} x^\rho
T^{\nu\mu}
.
\end{aligned}

Define
\label{eqn:qftLecture9:600}
j^{\nu\mu\rho} = \inv{2} \lr{ x^\rho T^{\nu\mu} – x^{\mu} T^{\nu\rho} },

which retains the antisymmetry in $$\mu \rho$$ yet still drops the parameter $$\omega^{\mu\rho}$$.
To check that this makes sense, we can contract
$$j^{\nu\mu\rho}$$ with $$\omega_{\rho\mu}$$
\label{eqn:qftLecture9:880}
\begin{aligned}
j^{\nu\mu\rho} \omega_{\rho\mu}
&= -\inv{2} \lr{ x^\rho T^{\nu\mu} – x^{\mu} T^{\nu\rho} }
\omega_{\mu\rho} \\
&=
-\inv{2} x^\rho T^{\nu\mu}
\omega_{\mu\rho}
– \inv{2} x^{\mu} T^{\nu\rho}
\omega_{\rho\mu} \\
&=
-\inv{2} x^\rho T^{\nu\mu}
\omega_{\mu\rho}
– \inv{2} x^{\rho} T^{\nu\mu}
\omega_{\mu\rho} \\
&=
– x^{\rho} T^{\nu\mu}
\omega_{\mu\rho},
\end{aligned}

which matches \ref{eqn:qftLecture9:580} as desired.

### Example. Rotations $$\mu\rho = ij$$

\label{eqn:qftLecture9:620}
\begin{aligned}
J^{0 i j} \epsilon_{ijk}
&=
\inv{2} \lr{ x^i T^{0j} – x^{j} T^{0i} } \epsilon_{ijk} \\
&=
x^i T^{0j} \epsilon_{ijk}.
\end{aligned}

Observe that this has the structure of $$(\Bx \cross \Bp)_k$$, where $$\Bp$$ is the momentum density of the field.
Let
\label{eqn:qftLecture9:640}
L_k \equiv Q_k = \int d^3 x J^{0ij} \epsilon_{ijk}.

We can now quantize and build a generator
\label{eqn:qftLecture9:660}
\begin{aligned}
\hatU(\Balpha)
&= e^{i \Balpha \cdot \hat{\BL}} \\
&= \exp\lr{i \alpha_k
\int d^3 x x^i \hat{T}^{0j} \epsilon_{ijk}
}
\end{aligned}

From \ref{eqn:qftLecture9:560} we can quantize with $$T^{0j} = \partial^0 \phi \partial^j \phi \rightarrow \hat{\pi} \lr{\spacegrad \phihat}_j$$, or
\label{eqn:qftLecture9:900}
\begin{aligned}
\hatU(\Balpha)
&=
\exp\lr{i \alpha_k
\int d^3 x x^i \hat{\pi} (\spacegrad \phihat)_j \epsilon_{ijk}
} \\
&=
\exp\lr{i \Balpha \cdot
\int d^3 x \hat{\pi} \spacegrad \phihat \cross \Bx
}
\end{aligned}

\label{eqn:qftLecture9:680}
\begin{aligned}
\phihat(\By) \rightarrow \hatU(\alpha) \phihat(\By) \hatU^\dagger(\alpha)
&\approx
\phihat(\By) +
i \Balpha \cdot
\antisymmetric{
\int d^3 x \hat{\pi}(\Bx) \spacegrad \phihat(\Bx) \cross \Bx
}
{
\phihat(\By)
} \\
&=
\phihat(\By) +
i \Balpha \cdot
\int d^3 x
(-i) \delta^3(\Bx – \By)
&=
\phihat(\By) +
\Balpha \cdot \lr{ \spacegrad \phihat(\By ) \cross \By}
\end{aligned}

Explicitly, in coordinates, this is
\label{eqn:qftLecture9:700}
\begin{aligned}
\phihat(\By)
&\rightarrow
\phihat(\By) +
\alpha^i
\lr{
\partial^j \phihat(\By) y^k \epsilon_{jki}
} \\
&=
\phihat(\By) –
\epsilon_{ikj} \alpha^i y^k \partial^j \phihat \\
&=
\phihat( y^j – \epsilon^{ikj} \alpha^i y^k ).
\end{aligned}

This is a rotation. To illustrate, pick $$\Balpha = (0, 0, \alpha)$$, so $$y^j \rightarrow y^j – \epsilon^{ikj} \alpha y^k \delta_{i3} = y^j – \epsilon^{3kj} \alpha y^k$$, or
\label{eqn:qftLecture9:n}
\begin{aligned}
y^1 &\rightarrow y^1 – \epsilon^{3k1} \alpha y^k = y^1 + \alpha y^2 \\
y^2 &\rightarrow y^2 – \epsilon^{3k2} \alpha y^k = y^2 – \alpha y^1 \\
y^3 &\rightarrow y^3 – \epsilon^{3k3} \alpha y^k = y^3,
\end{aligned}

or in matrix form
\label{eqn:qftLecture9:720}
\begin{bmatrix}
y^1 \\
y^2 \\
y^3 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & \alpha & 0 \\
-\alpha & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
y^1 \\
y^2 \\
y^3 \\
\end{bmatrix}.

## Antenna array design with Chebychev polynomials

Prof. Eleftheriades desribed a Chebychev antenna array design method that looks different than the one of the text [1].

Portions of that procedure are like that of the text. For example, if a side lobe level of $$20 \log_{10} R$$ is desired, a scaling factor

\label{eqn:chebychevSecondMethod:20}
x_0 = \cosh\lr{ \inv{m} \cosh^{-1} R },

is used. Given $$N$$ elements in the array, a Chebychev polynomial of degree $$m = N – 1$$ is used. That is

\label{eqn:chebychevSecondMethod:40}
T_m(x) = \cos\lr{ m \cos^{-1} x }.

Observe that the roots $$x_n’$$ of this polynomial lie where

\label{eqn:chebychevSecondMethod:60}
m \cos^{-1} x_n’ = \frac{\pi}{2} \pm \pi n,

or

\label{eqn:chebychevSecondMethod:80}
x_n’ = \cos\lr{ \frac{\pi}{2 m} \lr{ 2 n \pm 1 } },

The class notes use the negative sign, and number $$n = 1,2, \cdots, m$$. It is noted that the roots are symmetric with $$x_1′ = – x_m’$$, which can be seen by direct expansion

\label{eqn:chebychevSecondMethod:100}
\begin{aligned}
x_{m-r}’
&= \cos\lr{ \frac{\pi}{2 m} \lr{ 2 (m – r) – 1 } } \\
&= \cos\lr{ \pi – \frac{\pi}{2 m} \lr{ 2 r + 1 } } \\
&= -\cos\lr{ \frac{\pi}{2 m} \lr{ 2 r + 1 } } \\
&= -\cos\lr{ \frac{\pi}{2 m} \lr{ 2 ( r + 1 ) – 1 } } \\
&= -x_{r+1}’.
\end{aligned}

The next step in the procedure is the identification

\label{eqn:chebychevSecondMethod:120}
\begin{aligned}
u_n’ &= 2 \cos^{-1} \lr{ \frac{x_n’}{x_0} } \\
z_n &= e^{j u_n’}.
\end{aligned}

This has a factor of two that does not appear in the Balanis design method. It seems plausible that this factor of two was introduced so that the roots of the array factor $$z_n$$ are conjugate pairs. Since $$\cos^{-1} (-z) = \pi – \cos^{-1} z$$, this choice leads to such conjugate pairs

\label{eqn:chebychevSecondMethod:140}
\begin{aligned}
\exp\lr{j u_{m-r}’}
&=
\exp\lr{j 2 \cos^{-1} \lr{ \frac{x_{m-r}’}{x_0} } } \\
&=
\exp\lr{j 2 \cos^{-1} \lr{ -\frac{x_{r+1}’}{x_0} } } \\
&=
\exp\lr{j 2 \lr{ \pi – \cos^{-1} \lr{ \frac{x_{r+1}’}{x_0} } } } \\
&=
\exp\lr{-j u_{r+1}}.
\end{aligned}

Because of this, the array factor can be written

\label{eqn:chebychevSecondMethod:180}
\begin{aligned}
\textrm{AF}
&= ( z – z_1 )( z – z_2 ) \cdots ( z – z_{m-1} ) ( z – z_m ) \\
&=
( z – z_1 )( z – z_1^\conj )
( z – z_2 )( z – z_2^\conj )
\cdots \\
&=
\lr{ z^2 – z ( z_1 + z_1^\conj ) + 1 }
\lr{ z^2 – z ( z_2 + z_2^\conj ) + 1 }
\cdots \\
&=
\lr{ z^2 – 2 z \cos\lr{ 2 \cos^{-1} \lr{ \frac{x_1′}{x_0} } } + 1 }
\lr{ z^2 – 2 z \cos\lr{ 2 \cos^{-1} \lr{ \frac{x_2′}{x_0} } } + 1 }
\cdots \\
&=
\lr{ z^2 – 2 z \lr{ 2 \lr{ \frac{x_1′}{x_0} }^2 – 1 } + 1 }
\lr{ z^2 – 2 z \lr{ 2 \lr{ \frac{x_2′}{x_0} }^2 – 1 } + 1 }
\cdots
\end{aligned}

When $$m$$ is even, there will only be such conjugate pairs of roots. When $$m$$ is odd, the remainding factor will be

\label{eqn:chebychevSecondMethod:160}
\begin{aligned}
z – e^{2 j \cos^{-1} \lr{ 0/x_0 } }
&=
z – e^{2 j \pi/2} \\
&=
z – e^{j \pi} \\
&=
z + 1.
\end{aligned}

However, with this factor of two included, the connection between the final array factor polynomial \ref{eqn:chebychevSecondMethod:180}, and the Chebychev polynomial $$T_m$$ is not clear to me. How does this scaling impact the roots?

### Example: Expand $$\textrm{AF}$$ for $$N = 4$$.

The roots of $$T_3(x)$$ are

\label{eqn:chebychevSecondMethod:200}
x_n’ \in \setlr{0, \pm \frac{\sqrt{3}}{2} },

so the array factor is

\label{eqn:chebychevSecondMethod:220}
\begin{aligned}
\textrm{AF}
&=
\lr{ z^2 + z \lr{ 2 – \frac{3}{x_0^2} } + 1 }\lr{ z + 1 } \\
&=
z^3
+ 3 z^2 \lr{ 1 – \frac{1}{x_0^2} }
+ 3 z \lr{ 1 – \frac{1}{x_0^2} }
+ 1.
\end{aligned}

With $$20 \log_{10} R = 30 \textrm{dB}$$, $$x_0 = 2.1$$, so this is

\label{eqn:chebychevSecondMethod:240}
\textrm{AF} = z^3 + 2.33089 z^2 + 2.33089 z + 1.

With

\label{eqn:chebychevSecondMethod:260}
z = e^{j (u + u_0) } = e^{j k d \cos\theta + j k u_0 },

the array factor takes the form

\label{eqn:chebychevSecondMethod:280}
\textrm{AF}
=
e^{j 3 k d \cos\theta + j 3 k u_0 }
+ 2.33089
e^{j 2 k d \cos\theta + j 2 k u_0 }
+ 2.33089
e^{j k d \cos\theta + j k u_0 }
+ 1.

This array function is highly phase dependent, plotted for $$u_0 = 0$$ in fig. 1, and fig. 2.

fig 1. Plot with u_0 = 0, d = lambda/4

fig 2. Spherical plot with u_0 = 0, d = lambda/4

This can be directed along a single direction (z-axis) with higher phase choices as illustrated in fig. 3, and fig. 4.

fig 3. Plot with u_0 = 3.5, d = 0.4 lambda

fig 4. Spherical plot with u_0 = 3.5, d = 0.4 lambda

These can be explored interactively in this Mathematica Manipulate panel.

# References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

## Chebychev antenna array design

In our text [1] is a design procedure that applies Chebychev polynomials to the selection of current magnitudes for an evenly spaced array of identical antennas placed along the z-axis.

For an even number $$2 M$$ of identical antennas placed at positions $$\Br_m = (d/2) \lr{2 m -1} \Be_3$$, the array factor is

\label{eqn:chebychevDesign:20}
\textrm{AF}
=
\sum_{m=-N}^N I_m e^{-j k \rcap \cdot \Br_m }.

Assuming the currents are symmetric $$I_{-m} = I_m$$, with $$\rcap = (\sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta )$$, and $$u = \frac{\pi d}{\lambda} \cos\theta$$, this is

\label{eqn:chebychevDesign:40}
\begin{aligned}
\textrm{AF}
&=
\sum_{m=-N}^N I_m e^{-j k (d/2) ( 2 m -1 )\cos\theta } \\
&=
2 \sum_{m=1}^N I_m \cos\lr{ k (d/2) ( 2 m -1)\cos\theta } \\
&=
2 \sum_{m=1}^N I_m \cos\lr{ (2 m -1) u }.
\end{aligned}

This is a sum of only odd cosines, and can be expanded as a sum that includes all the odd powers of $$\cos u$$. Suppose for example that this is a four element array with $$N = 2$$. In this case the array factor has the form

\label{eqn:chebychevDesign:60}
\begin{aligned}
\textrm{AF}
&=
2 \lr{ I_1 \cos u + I_2 \lr{ 4 \cos^3 u – 3 \cos u } } \\
&=
2 \lr{ \lr{ I_1 – 3 I_2 } \cos u + 4 I_2 \cos^3 u }.
\end{aligned}

The design procedure in the text sets $$\cos u = z/z_0$$, and then equates this to $$T_3(z) = 4 z^3 – 3 z$$ to determine the current amplitudes $$I_m$$. That is

\label{eqn:chebychevDesign:80}
\frac{ 2 I_1 – 6 I_2 }{z_0} z + \frac{8 I_2}{z_0^3} z^3 = -3 z + 4 z^3,

or

\label{eqn:chebychevDesign:100}
\begin{aligned}
\begin{bmatrix}
I_1 \\
I_2
\end{bmatrix}
&=
{\begin{bmatrix}
2/z_0 & -6/z_0 \\
0 & 8/z_0^3
\end{bmatrix}}^{-1}
\begin{bmatrix}
-3 \\
4
\end{bmatrix} \\
&=
\frac{z_0}{2}
\begin{bmatrix}
3 (z_0^2 -1) \\
z_0^2
\end{bmatrix}.
\end{aligned}

The currents in the array factor are fully determined up to a scale factor, reducing the array factor to

\label{eqn:chebychevDesign:140}
\textrm{AF} = 4 z_0^3 \cos^3 u – 3 z_0 \cos u.

The zeros of this array factor are located at the zeros of

\label{eqn:chebychevDesign:120}
T_3( z_0 \cos u ) = \cos( 3 \cos^{-1} \lr{ z_0 \cos u } ),

which are at $$3 \cos^{-1} \lr{ z_0 \cos u } = \pi/2 + m \pi = \pi \lr{ m + \inv{2} }$$

\label{eqn:chebychevDesign:160}
\cos u = \inv{z_0} \cos\lr{ \frac{\pi}{3} \lr{ m + \inv{2} } } = \setlr{ 0, \pm \frac{\sqrt{3}}{2 z_0} }.

showing that the scaling factor $$z_0$$ effects the locations of the zeros. It also allows the values at the extremes $$\cos u = \pm 1$$, to increase past the $$\pm 1$$ non-scaled limit values. These effects can be explored in this Mathematica notebook, but can also be seen in fig. 1.

fig 1. T_3( z_0 x) for a few different scale factors z_0.

The scale factor can be fixed for a desired maximum power gain. For $$R \textrm{dB}$$, that will be when

\label{eqn:chebychevDesign:180}
20 \log_{10} \cosh( 3 \cosh^{-1} z_0 ) = R \textrm{dB},

or

\label{eqn:chebychevDesign:200}
z_0 = \cosh \lr{ \inv{3} \cosh^{-1} \lr{ 10^{\frac{R}{20}} } }.

For $$R = 30$$ dB (say), we have $$z_0 = 2.1$$, and

\label{eqn:chebychevDesign:220}
\textrm{AF}
= 40 \cos^3 \lr{ \frac{\pi d}{\lambda} \cos\theta } – 6.4 \cos \lr{ \frac{\pi d}{\lambda} \cos\theta }.

These are plotted in fig. 2 (linear scale), and fig. 3 (dB scale) for a couple values of $$d/\lambda$$.

fig 2. T_3 fitting of 4 element array (linear scale).

fig 3. T_3 fitting of 4 element array (dB scale).

To explore the $$d/\lambda$$ dependence try this Mathematica notebook.

# References

[1] Constantine A Balanis. Antenna theory: analysis and design. John
Wiley & Sons, 3rd edition, 2005.

## Mathematica CDF notebooks for ece1229 (antenna theory)

February 8, 2015 ece1229 No comments , ,

I put together a couple of cool Manipulate notebooks for some radiation plots.

I am not able to share these directly as blog posts since the CDF plugin that I am using in my wordpress instance appears to have a new plugin or version incompatibility, and is no longer working. The link above has some plain html javascript wrappers for these notebooks, and works at least with chrome and firefox on windows 7.