Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.3.9 The Asymptotic Light Functions

In this sec­tion, math­e­mat­i­cal cur­va­ture is con­sid­ered and then ap­plied in the con­text of math­e­mat­i­cal physics. In par­tic­u­lar, the hy­per­bol­ic as­ymp­tote serves as the bound­ary for the para­met­ric curve γ(t)=cosh⁡(t),sinh⁡(t)\gamma(t) = \cosh(t),\sinh(t), which ap­proach­es the as­ymp­tote per the limit

limt→∞⁡γ(t)=asymptote.\begin{array}{r} \lim_{t \rightarrow \infty}{\gamma(t)} = asymptote. \end{array}
(1.3.256)

In physics, this as­ymp­tote is a bound­ary sec­tion of the rel­a­tivis­tic light cone, which acts to sep­a­rate time from space. The light cone is the basis for Ein­stein-Minkows­ki space­time, which di­vides causal­i­ty into a frame­work of the future and the past, re­lat­ing to a tem­po­ral present. The light cone bound­ary sec­tions the geom­e­try into space­like, time­like, and light­like spaces. Figure 1.3.22 shows the time axis run­ning from past to present and then to future. Space is de­pict­ed as the area out­side the light cone, the time-space as in the cone, and the present/now as the center of the cone. We note that the space­like region is sit­u­at­ed on the hor­i­zon­tal plane pass­ing by the origin, the light­like is on the cone, and the time-like is in the volume space inside the light­cone. In sec­tion 1.1.26, we de­fined the D3𝕔D_{3\mathbb{c}} di­he­dral group, whose basis re­volves around an ex­cep­tion­al number, that being i\ \sqrt{i}, which has mul­ti­ple ex­pres­sions, ev­i­denced by the equiv­a­lent forms

i=1+i2=(−1)14=eiπ4.\begin{array}{r} \sqrt{i} = \frac{1 + i}{\sqrt{2}} = ( - 1)^{\frac{1}{4}} = e^{\frac{i\pi}{4}}. \end{array}
(1.3.257)

It is also es­sen­tial to con­sid­er that the neg­a­tive forms also suf­fice for i\sqrt{i},

i=−(1+i)2=−(−1)14=e−3iπ4,\begin{array}{r} \sqrt{i} = - \frac{(1 + i)}{\sqrt{2}} = - ( - 1)^{\frac{1}{4}} = e^{- \frac{3i\pi}{4}}, \end{array}
(1.3.258)

and more pre­cise­ly, i\sqrt{i} is equal to

i=±1+i2=±(−1)14=±eiπ4.\begin{array}{r} \sqrt{i} = \pm \frac{1 + i}{\sqrt{2}} = \pm ( - 1)^{\frac{1}{4}} = \pm e^{\frac{i\pi}{4}}. \end{array}
(1.3.259)
Figure
Figure 1.3.22 – The causal struc­ture of the Ein­stein-Minkows­ki space­time,

where the light cone sep­a­rates past, present, future, and space.

Its coun­ter­part, −i\sqrt{- i}, can be stated in a sim­i­lar way,

−i=±(i−1)2=±(−1)34=±e3iπ4,\begin{array}{r} \sqrt{- i} = \pm \frac{(i - 1)}{\sqrt{2}} = \pm ( - 1)^{\frac{3}{4}} = \pm e^{\frac{3i\pi}{4}}, \end{array}
(1.3.260)

for which, all four of these num­bers are lo­cat­ed at points on the bound­ary of the com­plex unit circle through which the hy­per­bol­ic as­ymp­totes em­anate. Figure 1.3.23 rep­re­sents how these num­bers are ori­ent­ed on the com­plex unit circle, demon­strat­ed as the points where the red as­ymp­totes meet the circle's bound­ary. The values that are bound­ary spe­cif­ic are de­pict­ed inside the com­plex unit circle, and the ex­po­nen­tial map­pings for the hor­i­zon­tal­ly ori­ent­ed unit hy­per­bo­la and circle are shown as the f(t)f(t) func­tions.

Figure
Figure 1.3.23 – The as­ymp­tot­ic num­bers e(iπ4+iπn2)e^{\left( \frac{i\pi}{4} + \frac{i\pi n}{2} \right)}, and axis func­tions f(t)=ee(±iπ2)t\ f(t) = e^{e^{\left( \pm \frac{i\pi}{2} \right)}t}, f(t)=e(±eϵ)t.f(t) = e^{\left( \pm e^{\epsilon} \right)t}.

The as­ymp­totes rep­re­sent a bal­ance be­tween the sphere's pos­i­tive cur­va­ture and the hy­per­bo­la's neg­a­tive cur­va­ture. The ques­tion be­comes, what func­tions serve as the as­ymp­totes in com­plex space? We will use the im­plied com­plex geom­e­try and the geo­met­ric prod­uct to find out. We start by taking the com­plex unit circle Lie group iso­mor­phism,

exp⁡(eIπ2t)=eIt,\begin{array}{r} \exp\left( e^{\frac{I\pi}{2}}t \right) = e^{It}, \end{array}
(1.3.261)

and ex­press this in vector format,

u=(cos⁡(t)e1,(Isin(t))e2),\begin{array}{r} u = \left( \cos(t)e_{1},\left( Isin(t) \right)e_{2} \right), \end{array}
(1.3.262)

where the cos⁡(t)\cos(t) and sin⁡(t)\sin(t) func­tions rep­re­sent vari­able-de­pen­dent scalar por­tions of the vector. Next, we define a vector, v=(e12,e22)v = \left( \frac{e_{1}}{\sqrt{2}},\frac{e_{2}}{\sqrt{2}} \right), to rep­re­sent the as­ymp­tot­ic Lie al­ge­bra, e(iπ4)e^{\left( \frac{i\pi}{4} \right)}, then con­sid­er the geo­met­ric prod­uct be­tween the two vec­tors

uv=u∙v+u∧v.\begin{array}{r} uv = u \bullet v + u \land v. \end{array}
(1.3.263)

By cal­cu­lat­ing the geo­met­ric prod­uct, we find the dot prod­uct equals

u∙v=(cos⁡(t)e1,(Isin(t))e2)∙(e12,e22)=cos⁡(t)2e1∙e1+(Isin(t))2e2∙e2=cos⁡(t)2+Isin(t)2,\begin{array}{r} u \bullet v = \left( \cos(t)e_{1},\left( Isin(t) \right)e_{2} \right) \bullet \left( \frac{e_{1}}{\sqrt{2}},\frac{e_{2}}{\sqrt{2}} \right) = \frac{\cos(t)}{\sqrt{2}}e_{1} \bullet e_{1} + \frac{\left( Isin(t) \right)}{\sqrt{2}}e_{2} \bullet e_{2} = \frac{\cos(t)}{\sqrt{2}} + \frac{{Isin}(t)}{\sqrt{2}}, \end{array}
(1.3.264)

and the wedge prod­uct

u∧v=(cos⁡(t)e1+(Isin(t))e2)∧(e12+e22)u \land v = \left( \cos(t)e_{1} + \left( Isin(t) \right)e_{2} \right) \land \left( \frac{e_{1}}{\sqrt{2}} + \frac{e_{2}}{\sqrt{2}} \right)
=cos⁡(t)2e1∧e1+cos⁡(t)2e1∧e2+(Isin(t))2e2∧e1+(Isin(t))2e2∧e2=Icos(t)2+sin⁡(t)2.\begin{array}{r} = \frac{\cos(t)}{\sqrt{2}}e_{1} \land e_{1} + \frac{\cos(t)}{\sqrt{2}}e_{1} \land e_{2} + \frac{\left( Isin(t) \right)}{\sqrt{2}}e_{2} \land e_{1} + \frac{\left( Isin(t) \right)}{\sqrt{2}}e_{2} \land e_{2} = \frac{{Icos}(t)}{\sqrt{2}} + \frac{\sin(t)}{\sqrt{2}}. \end{array}
(1.3.265)
Figure
Figure 1.3.24 – The geo­met­ric prod­uct of the vec­tors uu and vv, u∙v+u∧v.u \bullet v + u \land v.

By the iso­mor­phism i=Ii = I, the sum of the prod­ucts rep­re­sent­ing uvuv is

uv=(1+i)2cos⁡(t)+(1+i)2sin⁡(t),\begin{array}{r} uv = \frac{(1 + i)}{\sqrt{2}}\cos(t) + \frac{(1 + i)}{\sqrt{2}}\sin(t), \end{array}
(1.3.266)

a result of equal in mag­ni­tude con­tri­bu­tions from the dot and wedge prod­ucts. In Figure 1.3.24, the vec­tors that form this geo­met­ric prod­uct appear par­al­lel. De­spite this par­al­lel ap­pear­ance, in­cor­po­rat­ed is the nested bivec­tor II into the uu-vector, which acts to main­tain a con­tri­bu­tion from the wedge prod­uct in the result. Like the cosine and sine func­tion ex­am­ple in the last sec­tion, we can dis­trib­ute the bases con­tain­ing the bivec­tor II, to rep­re­sent the vector uu

u′=(cos⁡(t)e1,sin(t)e1),\begin{array}{r} u^{'} = \left( \cos(t)e_{1},sin(t)e_{1} \right), \end{array}
(1.3.267)

which will result in the same geo­met­ric prod­uct once cal­cu­lat­ed. Next, we arrange the result into real and imag­i­nary com­po­nents, defin­ing a two-di­men­sion­al para­met­ric vector f(t)f(t) in the com­plex plane, as

f(t)=12(cos⁡(t)+sin⁡(t)), i2(cos⁡(t)+sin⁡(t)).\begin{array}{r} f(t) = \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right),\ \frac{i}{\sqrt{2}}\left( \cos(t) + \sin(t) \right). \end{array}
(1.3.268)
Figure
Figure 1.3.25 – The para­met­ric equa­tion f(t)=12(cos⁡(t)+sin⁡(t)), i2(cos⁡(t)+sin⁡(t))f(t) = \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right),\ \frac{i}{\sqrt{2}}\left( \cos(t) + \sin(t) \right) in the com­plex plane.

The result is seen by plot­ting the func­tion f(t)f(t) in Figure 1.3.25, re­veal­ing a flat and di­ag­o­nal­ly ori­ent­ed curve con­sis­tent with the unit hy­per­bo­la's as­ymp­tote. This as­ymp­tot­ic con­sis­ten­cy is ap­pre­ci­at­ed given the flat curve in­ter­sects the com­plex unit circle at the points e(iπ4)e^{\left( \frac{i\pi}{4} \right)} and e(−3iπ4).e^{\left( \frac{- 3i\pi}{4} \right)}.The geo­met­ric in­tu­ition behind this ex­am­ple can be shown when we con­sid­er the com­plex plane, for ±isin(t)\pm isin(t) func­tions ori­ent­ed about the ver­ti­cal axis, and for ±cos(t)\pm cos(t) func­tions on the hor­i­zon­tal axis, as seen in Figure 1.3.26.

Figure
Figure 1.3.26 – Ro­tat­ing the e−it2\frac{e^{- it}}{\sqrt{2}} vector by π2\frac{\pi}{2} ra­di­ans in the com­plex trigono­met­ric plane.

When ro­tat­ing the e−it2\frac{e^{- it}}{\sqrt{2}} vector in the com­plex plane by iπ2\frac{i\pi}{2} ra­di­ans, it will align the vector func­tions. This is the same thing as adding the sta­tion­ary vectoreit2\frac{e^{it}}{\sqrt{2}} , to the ro­tat­ed vector ie−it2\frac{ie^{- it}}{\sqrt{2}} , rep­re­sent­ed by the ex­pres­sion

eit2+ie−it2=cos⁡(t)+isin(t)+icos(t)+sin⁡(t)2=(1+i)2(cos⁡(t)+sin⁡(t)),\begin{array}{r} \frac{e^{it}}{\sqrt{2}} + \frac{ie^{- it}}{\sqrt{2}} = \frac{\cos(t) + isin(t) + icos(t) + \sin(t)}{\sqrt{2}} = \frac{(1 + i)}{\sqrt{2}}\left( \cos(t) + \sin(t) \right), \end{array}
(1.3.269)

which is the same result as the geo­met­ric prod­uct cal­cu­la­tion. For t∈ℝt\in \mathbb{R}, the com­plex con­ju­gate of the func­tion f(t)f(t), is

f(t)∗=12(cos⁡(t)+sin⁡(t)), −i2(cos⁡(t)+sin⁡(t)),\begin{array}{r} f(t)^{*} = \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right),\ - \frac{i}{\sqrt{2}}\left( \cos(t) + \sin(t) \right), \end{array}
(1.3.270)

and using the iso­mor­phism, we set I=iI = i, for the f(t)f(t), f(t)∗f(t)^{*} func­tions, and pro­ceed to eval­u­ate their geo­met­ric prod­uct

f(t)f(t)∗=f(t)∙f(t)∗+f(t)∧f(t)∗.\begin{array}{r} f(t)f(t)^{*} = f(t) \bullet f(t)^{*} + f(t) \land f(t)^{*}. \end{array}
(1.3.271)

As these vec­tors are or­thog­o­nal­ly re­lat­ed, we find the dot prod­uct equals zero

f(t)∙f(t)∗=(12(cos⁡(t)+sin⁡(t))e1+I2(cos⁡(t)+sin⁡(t))e2)∙(12(cos⁡(t)+sin⁡(t))e1−I2(cos⁡(t)+sin⁡(t))e2)\begin{array}{r} f(t) \bullet f(t)^{*} = \left( \begin{array}{r} \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{1} \\ + \frac{I}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{2} \end{array} \right) \bullet \left( \begin{array}{r} \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{1} \\ - \frac{I}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{2} \end{array} \right) \end{array}
=(12(cos⁡(t)+sin⁡(t))e1)(12(cos⁡(t)+sin⁡(t))e1)+(I2(cos⁡(t)+sin⁡(t))e2)(I2(cos⁡(t)+sin⁡(t))e2)= \left( \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{1} \right)\left( \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{1} \right) + \left( \frac{I}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{2} \right)\left( \frac{I}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{2} \right)
=12(cos⁡(t)+sin⁡(t))2e1∙e1−12(cos⁡(t)+sin⁡(t))2e2∙e2=0,\begin{array}{r} = \frac{1}{2}\left( \cos(t) + \sin(t) \right)^{2}e_{1} \bullet e_{1} - \frac{1}{2}\left( \cos(t) + \sin(t) \right)^{2}e_{2} \bullet e_{2} = 0, \end{array}
(1.3.272)

and the wedge prod­uct

(12(cos⁡(t)+sin⁡(t))e1+I2(cos⁡(t)+sin⁡(t))e2)∧(12(cos⁡(t)+sin⁡(t))e1−I2(cos⁡(t)+sin⁡(t))e2)\left( \begin{array}{r} \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{1} \\ + \frac{I}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{2} \end{array} \right) \land \left( \begin{array}{r} \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{1} \\ - \frac{I}{\sqrt{2}}\left( \cos(t) + \sin(t) \right)e_{2} \end{array} \right)
=−I2(cos⁡(t)+sin⁡(t))2e1∧e2+I2(cos⁡(t)+sin⁡(t))2e2∧e1=(cos⁡(t)+sin⁡(t))2,\begin{array}{r} = - \frac{I}{2}\left( \cos(t) + \sin(t) \right)^{2}e_{1} \land e_{2} + \frac{I}{2}\left( \cos(t) + \sin(t) \right)^{2}e_{2} \land e_{1} = \left( \cos(t) + \sin(t) \right)^{2}, \end{array}
(1.3.273)

where ul­ti­mate­ly, the geo­met­ric prod­uct is solely due to the con­tri­bu­tion from the wedge prod­uct

f(t)f(t)∗=(cos⁡(t)+sin⁡(t))2.\begin{array}{r} f(t)f(t)^{*} = \left( \cos(t) + \sin(t) \right)^{2}. \end{array}
(1.3.274)

By way of the iden­ti­ty

sin⁡(2t)+1=(cos⁡(t)+sin⁡(t))2,\begin{array}{r} \sin(2t) + 1 = \left( \cos(t) + \sin(t) \right)^{2}, \end{array}
(1.3.275)

we find that the geo­met­ric prod­uct f(t)f(t)∗f(t)f(t)^{*} pro­duces a real-valued sine wave that is always ≥0\geq 0, given that the +1+ 1 in the ex­pres­sion shifts the wave­form up by one on the ver­ti­cal axis. To find the other as­ymp­tot­ic curve func­tion, we re­verse the vector order in the geo­met­ric prod­uct that yield­ed the f(t)f(t) func­tion, to vuvu'

vu′=v ∙u′+v∧u′,\begin{array}{r} vu' = v\ \bullet u' + v \land u', \end{array}
(1.3.276)

where we dis­trib­uted the bivec­tor, II, in the bases of the orig­i­nal uu-vector, be­com­ing, u′u',

u′=(cos⁡(t)e1+sin⁡(t)e10).\begin{array}{r} u' = \left( \begin{array}{r} \cos(t)e_{1} + \sin{(t)e_{1}} \\ 0 \end{array} \right). \end{array}
(1.3.277)

The dot prod­uct cal­cu­lates to

v ∙u′=12(e1e2)∙(cos⁡(t)e1+sin⁡(t)e10)=cos⁡(t)+sin(t)2,\begin{array}{r} v\ \bullet u' = \frac{1}{\sqrt{2}}\left( \begin{array}{r} e_{1} \\ e_{2} \end{array} \right) \bullet \left( \begin{array}{r} \cos(t)e_{1} + \sin{(t)e_{1}} \\ 0 \end{array} \right) = \frac{\cos(t) + sin(t)}{\sqrt{2}}, \end{array}
(1.3.278)

and the wedge prod­uct

v∧u′=12(e1e2)∧(cos⁡(t)e1+sin⁡(t)e10)=(cos⁡(t)+sin(t))e2∧e12,\begin{array}{r} v \land u' = \frac{1}{\sqrt{2}}\left( \begin{array}{r} e_{1} \\ e_{2} \end{array} \right) \land \left( \begin{array}{r} \cos(t)e_{1} + \sin{(t)e_{1}} \\ 0 \end{array} \right) = \frac{\left( \cos(t) + sin(t) \right)e_{2} \land e_{1}}{\sqrt{2}}, \end{array}
(1.3.279)

lead­ing to the vu′vu'result

vu′=(1−I)2(cos⁡(t)+sin(t)).\begin{array}{r} vu^{'} = \frac{(1 - I)}{\sqrt{2}}\left( \cos(t) + sin(t) \right). \end{array}
(1.3.280)

After ar­rang­ing it into a vector, we find vu′vu^{'} is iso­mor­phic to the com­plex con­ju­gate form f(t)∗f(t)^{*}

f(t)∗=(12(cos⁡(t)+sin⁡(t)), −i2(cos⁡(t)+sin⁡(t))).\begin{array}{r} f(t)^{*} = \left( \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right),\ - \frac{i}{\sqrt{2}}\left( \cos(t) + \sin(t) \right) \right). \end{array}
(1.3.281)
Figure
Figure 1.3.27 – Equa­tions f(t)f(t) and f(t)∗f(t)^{*} rep­re­sent the as­ymp­totes of the unit hy­per­bo­la.

In Figure 1.3.27, we plot the f(t),f(t)∗f(t),f(t)^{*} func­tions in the com­plex plane with the unit hy­per­bo­la, where it can read­i­ly be seen these func­tions are the as­ymp­tote func­tions to the unit hy­per­bo­la. Given that

12(cos⁡(t)+sin⁡(t))=sin⁡(t+π4),\begin{array}{r} \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right) = \sin\left( t + \frac{\pi}{4} \right), \end{array}
(1.3.282)

we can take the f(t)f(t) func­tion to define the para­met­ric func­tion γ(t)\gamma(t), and its com­plex con­ju­gate γ(t)∗\gamma(t)^{*}, as fol­lows

γ(t)=sin⁡(t+π4),isin(t+π4),\begin{array}{r} \gamma(t) = \sin\left( t + \frac{\pi}{4} \right),{isin}\left( t + \frac{\pi}{4} \right), \end{array}
(1.3.283)
γ(t)∗=sin⁡(t+π4),−isin(t+π4).\begin{array}{r} \gamma(t)^{*} = \sin\left( t + \frac{\pi}{4} \right),{- isin}\left( t + \frac{\pi}{4} \right). \end{array}
(1.3.284)
Figure
Figure 1.3.28 – The para­met­ric plots γ(t)\gamma(t) and γ(t)∗\gamma(t)^{*}, extend to the circle of

radius, r=2r = \sqrt{2} (dashed), for γ(t)=12(cos⁡(t)+sin⁡(t))+i2(cos⁡(t)+sin⁡(t)).\gamma(t) = \frac{1}{\sqrt{2}}\left( \cos(t) + \sin(t) \right) + \frac{i}{\sqrt{2}}\left( \cos(t) + \sin(t) \right).

In Figure 1.3.28, the func­tions γ(t)\gamma(t) and γ(t)∗\gamma(t)^{*} are plot­ted, re­veal­ing their flat curves extend beyond the unit circle to a max­i­mum mod­u­lus equal to 2\sqrt{2}, for n∈ℕn\in \mathbb{N}, oc­cur­ring when γ(t)\gamma(t) equals

γ(π4±πn)=|1+i|.\begin{array}{r} \gamma\left( \frac{\pi}{4} \pm \pi n \right) = |1 + i|. \end{array}
(1.3.285)

This ex­ten­sion beyond the unit circle bound­ary is due to the per­pet­u­al pres­ence of equal mag­ni­tude real and imag­i­nary values. This should be clear since the same trigono­met­ric func­tion rep­re­sents the real and imag­i­nary di­men­sions of γ(t).\gamma(t).We now define the time-based func­tion f(τ)f(\tau), by using this com­pact form

f(τ)=sin⁡(t+π4)+isin⁡(t+π4),\begin{array}{r} f(\tau) = \sin\left( t + \frac{\pi}{4} \right) + i\sin\left( t + \frac{\pi}{4} \right), \end{array}
(1.3.286)

which moves through time tt in the com­plex plane. The real por­tion of the wave func­tion is the sin⁡(t+π4)\sin\left( t + \frac{\pi}{4} \right) com­po­nent, and the imag­i­nary com­po­nent isin(t+π4){isin}\left( t + \frac{\pi}{4} \right) is its or­thog­o­nal­ly ori­ent­ed coun­ter­part.

In Figure 1.3.29, we plot the func­tion f(τ)f(\tau), which can be seen to move to the right in time, having or­thog­o­nal­ly ori­ent­ed wave com­po­nents os­cil­lat­ing in the real and imag­i­nary di­men­sions. The f(τ)f(\tau) func­tion has pe­ri­od­ic roots in the time di­men­sion, oc­cur­ring when t=−π4±πnt = - \frac{\pi}{4} \pm \pi n ,,and wave com­po­nents demon­strat­ing peaks and troughs for t=π4±πnt = \frac{\pi}{4} \pm \pi n. For the reader fa­mil­iar with elec­tro­mag­net­ic waves, these or­thog­o­nal waves are per­fect­ly con­sis­tent with the char­ac­ter­is­tics of light. Also cap­tured in the di­a­gram is the in­tu­ition behind the wave-par­ti­cle du­al­i­ty of light, mean­ing the phe­nom­e­nol­o­gy of light ap­pears to be wave-like and par­ti­cle-like. The par­ti­cle-like char­ac­ter­is­tics of light man­i­fest at each pe­ri­od­ic root along the time di­men­sion for t=−π4±πnt = - \frac{\pi}{4} \pm \pi n, oc­cur­ring twice per wave period as each wave com­po­nent hits the time axis in unison. While the wave char­ac­ter­is­tic, in­her­ent to its form, is de­fined by two or­thog­o­nal­ly ori­ent­ed waves os­cil­lat­ing through time.

Figure
Figure 1.3.29 - The wave func­tion f(τ)=sin⁡(t+π4)+isin⁡(t+π4)f(\tau) = \sin\left( t + \frac{\pi}{4} \right) + i\sin\left( t + \frac{\pi}{4} \right) moving through time,

with or­thog­o­nal­ly ori­ent­ed waves os­cil­lat­ing in re­spec­tive real and imag­i­nary di­men­sions.

Elec­tro­mag­net­ic waves also have spa­tial forms that travel through spa­tial dis­tances in time, and we can define the com­plex plane wave with spa­tial and tem­po­ral com­po­nents as fol­lows

exp⁡(eiπ2(−kx+ωt))=ei(−kx+ωt).\begin{array}{r} \exp\left( e^{\frac{i\pi}{2}}( - kx + \omega t) \right) = e^{i( - kx + \omega t)}. \end{array}
(1.3.287)

We then ex­press ei(−kx+ωt)e^{i( - kx + \omega t)}, as the vector uu

u=(cos⁡(−kx+ωt)e1, (Isin(−kx+ωt))e2),\begin{array}{r} u = \left( \cos( - kx + \omega t)e_{1},\ \left( Isin( - kx + \omega t) \right)e_{2} \right), \end{array}
(1.3.288)

and as before, the vector vv, is de­fined

v=(12e1,12e2).\begin{array}{r} v = \left( \frac{1}{\sqrt{2}}e_{1},\frac{1}{\sqrt{2}}e_{2} \right). \end{array}
(1.3.289)

We pro­ceed to cal­cu­late the geo­met­ric prod­uct be­tween the two vec­tors

uv=u∙v+u∧v,\begin{array}{r} uv = u \bullet v + u \land v, \end{array}
(1.3.290)

find­ing the dot prod­uct por­tion equals

u∙v=cos⁡(−kx+ωt)2+Isin(−kx+ωt)2,\begin{array}{r} u \bullet v = \frac{\cos( - kx + \omega t)}{\sqrt{2}} + \frac{{Isin}( - kx + \omega t)}{\sqrt{2}}, \end{array}
(1.3.291)

and the wedge prod­uct por­tion equals

u∧v=Icos(−kx+ωt)2+sin⁡(−kx+ωt)2,\begin{array}{r} u \land v = \frac{{Icos}( - kx + \omega t)}{\sqrt{2}} + \frac{\sin( - kx + \omega t)}{\sqrt{2}}, \end{array}
(1.3.292)

lead­ing to the fol­low­ing result for uvuv

uv=12(cos⁡(−kx+ωt)+sin⁡(−kx+ωt))+i2(cos⁡(−kx+ωt)+sin⁡(−kx+ωt)),\begin{array}{r} uv = \frac{1}{\sqrt{2}}\left( \cos( - kx + \omega t) + \sin( - kx + \omega t) \right) + \frac{i}{\sqrt{2}}\left( \cos( - kx + \omega t) + \sin( - kx + \omega t) \right), \end{array}
(1.3.293)

which sim­pli­fies to the form

uv=(1+i)sin⁡(−kx+ωt+π4).\begin{array}{r} uv = (1 + i)\sin\left( - kx + \omega t + \frac{\pi}{4} \right). \end{array}
(1.3.294)

Using this result, we define the func­tion f(x,t)f(x,t) to rep­re­sent a wave trav­el­ing through the spa­tial and tem­po­ral di­men­sions, as

f(x,t)=(1+i)sin⁡(−kx+ωt+π4).\begin{array}{r} f(x,t) = (1 + i)\sin\left( - kx + \omega t + \frac{\pi}{4} \right). \end{array}
(1.3.295)

Maxwell's equa­tions de­scribe elec­tro­mag­net­ism, and are de­fined by the fol­low­ing four equa­tions in their dif­fer­en­tial forms:

Gauss′s Law: ∇∙E⇀=4πρ,\begin{array}{r} {Gauss}^{'}s\ Law:\ \nabla \bullet \overset{\rightharpoonup}{E} = 4\pi\rho, \end{array}
(1.3.296)
Magnetic law: ∇∙B⇀=0,\begin{array}{r} Magnetic\ law:\ \nabla \bullet \overset{\rightharpoonup}{B} = 0, \end{array}
(1.3.297)
 Faraday’s law: ∇×E⇀=−1c∂B⇀∂t,\begin{array}{r} \ Faraday’s\ law:\ \nabla \times \overset{\rightharpoonup}{E} = - \frac{1}{c}\frac{\partial\overset{\rightharpoonup}{B}}{\partial t}, \end{array}
(1.3.298)
Generalized Ampere’s law:∇×B⇀=4πcJ⇀+1c∂E⇀∂t,\begin{array}{r} Generalized\ Ampere’s\ law:\nabla \times \overset{\rightharpoonup}{B} = \frac{4\pi}{c}\overset{\rightharpoonup}{J} + \frac{1}{c}\frac{\partial\overset{\rightharpoonup}{E}}{\partial t}, \end{array}
(1.3.299)

where E⇀\overset{\rightharpoonup}{E} rep­re­sents the elec­tric field, and B⇀\overset{\rightharpoonup}{B} is the mag­net­ic field. The first two equa­tions are dot prod­ucts, while the second two are cross-prod­ucts. For the static equa­tions, Gauss's law states that the elec­tric flux pass­ing through a closed sur­face is pro­por­tion­al to the total charge on its sur­face, and the mag­net­ic flux pass­ing through a closed sur­face is always equal to zero. While for the dy­nam­ic equa­tions, Fara­day's law states that a chang­ing mag­net­ic field in time pro­duces a mag­net­ic field, and Ampere's law says a chang­ing elec­tric field in time cre­ates a mag­net­ic field. The equa­tions in free space with­out sources can be rep­re­sent­ed as

∇∙E⇀=0,\begin{array}{r} \nabla \bullet \overset{\rightharpoonup}{E} = 0, \end{array}
(1.3.300)
∇∙B⇀=0,\begin{array}{r} \nabla \bullet \overset{\rightharpoonup}{B} = 0, \end{array}
(1.3.301)
∇×E⇀=−1c∂B⇀∂t,\begin{array}{r} \nabla \times \overset{\rightharpoonup}{E} = - \frac{1}{c}\frac{\partial\overset{\rightharpoonup}{B}}{\partial t}, \end{array}
(1.3.302)
∇×B⇀=4πcJ⇀+1c∂E⇀∂t\begin{array}{r} \nabla \times \overset{\rightharpoonup}{B} = \frac{4\pi}{c}\overset{\rightharpoonup}{J} + \frac{1}{c}\frac{\partial\overset{\rightharpoonup}{E}}{\partial t} \end{array}
(1.3.303)

where ρ\rho and J⇀\overset{\rightharpoonup}{J} are now equal to zero.

Next, we define the E⇀(x,t)\overset{\rightharpoonup}{E}(x,t) and B⇀(x,t)\overset{\rightharpoonup}{B}(x,t) fields to be equal to the f(x,t)f(x,t)-wave func­tion,

E⇀(x,t),B⇀(x,t)=(1+i)sin⁡(−kx+ωt+π4),\begin{array}{r} \overset{\rightharpoonup}{E}(x,t),\overset{\rightharpoonup}{B}(x,t) = (1 + i)\sin\left( - kx + \omega t + \frac{\pi}{4} \right), \end{array}
(1.3.304)

and con­sid­er a lin­ear­ly po­lar­ized wave trav­el­ing at the speed of light in the xx-di­rec­tion, with the elec­tric field E⇀\overset{\rightharpoonup}{E} con­strained to the yy di­rec­tion, and the mag­net­ic field B⇀\overset{\rightharpoonup}{B} con­strained to the zz-di­rec­tion (Figure 1.3.30).

Figure
Figure 1.3.30 – A lin­ear­ly po­lar­ized plane wave is trav­el­ing at the speed of light along the xx-axis.

We set the fol­low­ing basis vec­tors for the three di­men­sions

e1=(100),   e2=(010),   e3=(001),\begin{array}{r} e_{1} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\ \ \ e_{2} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\ \ \ e_{3} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \end{array}
(1.3.305)

which lend them­selves to the fol­low­ing vector field rep­re­sen­ta­tions

E⇀(x,t)=E(x,t)e2,\begin{array}{r} \overset{\rightharpoonup}{E}(x,t) = E(x,t)e_{2}, \end{array}
(1.3.306)
B⇀(x,t)=B(x,t)e3.\begin{array}{r} \overset{\rightharpoonup}{B}(x,t) = B(x,t)e_{3}. \end{array}
(1.3.307)

By eval­u­at­ing the di­ver­gence for both fields

∇∙E⇀=(∂∂xe1+∂∂yE(x,t)e2+∂∂ze3)=0,\begin{array}{r} \nabla \bullet \overset{\rightharpoonup}{E} = \left( \frac{\partial}{\partial x}e_{1} + \frac{\partial}{\partial y}E(x,t)e_{2} + \frac{\partial}{\partial z}e_{3} \right) = 0, \end{array}
(1.3.308)
∇∙B⇀=(∂∂xe1+∂∂ye2+∂∂zB(x,t)e3)=0,\begin{array}{r} \nabla \bullet \overset{\rightharpoonup}{B} = \left( \frac{\partial}{\partial x}e_{1} + \frac{\partial}{\partial y}e_{2} + \frac{\partial}{\partial z}B(x,t)e_{3} \right) = 0, \end{array}
(1.3.309)

we find they equal zero, agree­ing with Gauss' law and the mag­net­ic law from the equa­tions in free space. Next, we eval­u­ate Fara­day's law for our field func­tions, to find

∇×E⇀=−1c∂B⇀∂t→∇×E(x,t)e2=|e1e2e3∂∂x∂∂y∂∂z0E(x,t)0|=∂∂xE(x,t)e3\begin{array}{r} \nabla \times \overset{\rightharpoonup}{E} = - \frac{1}{c}\frac{\partial\overset{\rightharpoonup}{B}}{\partial t} \rightarrow \nabla \times E(x,t)e_{2} = \left| \begin{matrix} e_{1} & e_{2} & e_{3} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & E(x,t) & 0 \end{matrix} \right| = \frac{\partial}{\partial x}E(x,t)e_{3} \end{array}
=−k(1+i)cos⁡(−kx+ωt+π4)e3,\begin{array}{r} = - k(1 + i)\cos{\left( - kx + \omega t + \frac{\pi}{4} \right)e_{3}}, \end{array}
(1.3.310)

noting the speed of light is equal to the fol­low­ing wavenum­ber-an­gu­lar fre­quen­cy re­la­tion­ship

c=ωk.\begin{array}{r} c = \frac{\omega}{k}. \end{array}
(1.3.311)

The time de­riv­a­tive of the mag­net­ic field is

∂B⇀∂t=ω(1+i)cos⁡(−kx+ωt+π4)e3,\begin{array}{r} \frac{\partial\overset{\rightharpoonup}{B}}{\partial t} = \omega(1 + i)\cos{\left( - kx + \omega t + \frac{\pi}{4} \right)e_{3},} \end{array}
(1.3.312)

and we equate the mag­ni­tudes of both sides of Fara­day's law, and sim­pli­fy

−k(1+i)cos⁡(−kx+ωt+π4)=−1c(ω(1+i)cos⁡(−kx+ωt+π4))- k(1 + i)\cos\left( - kx + \omega t + \frac{\pi}{4} \right) = - \frac{1}{c}\left( \omega(1 + i)\cos\left( - kx + \omega t + \frac{\pi}{4} \right) \right)
→k=ωc→k=ωωk→k=k,\begin{array}{r} \rightarrow k = \frac{\omega}{c} \rightarrow k = \frac{\omega}{\frac{\omega}{k}} \rightarrow k = k, \end{array}
(1.3.313)

for which the dif­fer­en­tial equa­tion holds. Next, we eval­u­ate Ampere's law for our trav­el­ing wave

∇×B⇀=1c∂E⇀∂t→∇×B(x,t)e3=|e1e2e3∂∂x∂∂y∂∂z00B(x,t)|=−∂∂xB(x,t)e2,\nabla \times \overset{\rightharpoonup}{B} = \frac{1}{c}\frac{\partial\overset{\rightharpoonup}{E}}{\partial t} \rightarrow \nabla \times B(x,t)e_{3} = \left| \begin{matrix} e_{1} & e_{2} & e_{3} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & 0 & B(x,t) \end{matrix} \right| = - \frac{\partial}{\partial x}B(x,t)e_{2},
=k(1+i)cos⁡(−kx+ωt+π4)e2,\begin{array}{r} = k(1 + i)\cos\left( - kx + \omega t + \frac{\pi}{4} \right)e_{2}, \end{array}
(1.3.314)

and the time de­riv­a­tive of the elec­tric field

∂E⇀∂t=ω(1+i)cos⁡(−kx+ωt+π4)e2.\begin{array}{r} \frac{\partial\overset{\rightharpoonup}{E}}{\partial t} = \omega(1 + i)\cos{\left( - kx + \omega t + \frac{\pi}{4} \right)e_{2}.} \end{array}
(1.3.315)

We equate the mag­ni­tudes of both sides of Ampere's law

k(1+i)cos⁡(−kx+ωt+π4)=1c(ω(1+i)cos⁡(−kx+ωt+π4))k(1 + i)\cos\left( - kx + \omega t + \frac{\pi}{4} \right) = \frac{1}{c}\left( \omega(1 + i)\cos\left( - kx + \omega t + \frac{\pi}{4} \right) \right)
→k=ωc→kc=ω→ω=ω,\begin{array}{r} \rightarrow k = \frac{\omega}{c} \rightarrow kc = \omega \rightarrow \omega = \omega, \end{array}
(1.3.316)

to find this dif­fer­en­tial equa­tion holds as well. The fact that the func­tion f(x,t)f(x,t) is con­sis­tent with Maxwell's equa­tions, shows that these as­ymp­tot­ic num­bers and func­tions are light-based, specif­i­cal­ly elec­tro­mag­net­ic waves. The treat­ment of Maxwell's equa­tions was cur­so­ry in this ex­am­ple, but we will return to this in more depth in the forth­com­ing volume con­cern­ing the dy­nam­ics of light.

The uni­fied light func­tions are the final and ar­guably most im­pres­sive as­ymp­tot­ic light func­tions we con­sid­er in this sec­tion. The term uni­fied, means that the spher­i­cal and hy­per­bol­ic geome­tries have been brought into align­ment for the func­tions.

The de­fined vector uu

u=(ete1+eIte2),\begin{array}{r} u = \left( e^{t}e_{1} + e^{It}e_{2} \right), \end{array}
(1.3.317)

rep­re­sents the e1,e2e_{1},e_{2}-plane, for the ete^{t}, eIt\ e^{It} ex­po­nen­tial func­tions ori­ent­ed by their bases. We can rotate to any point on the func­tion plane using the fol­low­ing geo­met­ric prod­uct,

u(ae1+be2)=u∙(ae1+be2)+u∧(ae1+be2),\begin{array}{r} u\left( ae_{1} + be_{2} \right) = u \bullet \left( ae_{1} + be_{2} \right) + u \land \left( ae_{1} + be_{2} \right), \end{array}
(1.3.318)

where ‖(ae1+be2)‖=1,\left\| \left( ae_{1} + be_{2} \right) \right\| = 1,rep­re­sents a point on the bound­ary of the unit circle in the e1,e2e_{1},e_{2}-plane.

Figure
Figure 1.3.31 – Bivec­tor Ro­ta­tions

In Figure 1.3.31, we depict the e1,e2e_{1},e_{2}-plane with hy­per­bol­ic and cir­cu­lar ex­po­nen­tial func­tions ori­ent­ed along the pos­i­tive and neg­a­tive e1e_{1},e2e_{2}-axes, adding red vec­tors to rep­re­sent the axes of ro­ta­tion for the e1,e2e_{1},e_{2}-plane seg­ments, using (±e2,±ie2)\left( \pm \frac{e}{\sqrt{2}}, \pm \frac{ie}{\sqrt{2}} \right), as the ro­ta­tion values. It turns out that ro­tat­ing the e1,e2e_{1},e_{2}-plane about these as­ymp­tot­ic axes, re­sults in the light func­tions we are con­sid­er­ing. By set­ting v1=(ae1+be2)=12(e1+e2),v_{1} = \left( ae_{1} + be_{2} \right) = \frac{1}{\sqrt{2}}\left( e_{1} + e_{2} \right),we trans­form the vector uu, as fol­lows

uv1=u(e12+e22)=(ete1+eIte2)∙(e12+e22)+(ete1+eIte2)∧(e12+e22)uv_{1} = u\left( \frac{e_{1}}{\sqrt{2}} + \frac{e_{2}}{\sqrt{2}} \right) = \left( e^{t}e_{1} + e^{It}e_{2} \right) \bullet \left( \frac{e_{1}}{\sqrt{2}} + \frac{e_{2}}{\sqrt{2}} \right) + \left( e^{t}e_{1} + e^{It}e_{2} \right) \land \left( \frac{e_{1}}{\sqrt{2}} + \frac{e_{2}}{\sqrt{2}} \right)
=(Iet2+Isin(t−π4),et2+sin⁡(t+π4)),\begin{array}{r} = \left( \frac{Ie^{t}}{\sqrt{2}} + {Isin}\left( t - \frac{\pi}{4} \right),\frac{e^{t}}{\sqrt{2}} + \sin\left( t + \frac{\pi}{4} \right) \right), \end{array}
(1.3.319)

for which, we have arranged the result by real and imag­i­nary valued com­po­nents. Real and imag­i­nary hy­per­bol­ic and wave-like el­e­ments are noted in this result, which is read­i­ly seen in the com­plex plane by the para­met­ric plot in Figure 1.3.32. The sym­met­ric bal­ance be­tween the real and imag­i­nary hy­per­bol­ic el­e­ments, re­sults in a de­fin­i­tive­ly flat geom­e­try γh(t),\gamma_{h}(t),seen in the plot, while the wave el­e­ments com­prise the unit circle γs(t)\gamma_{s}(t) geom­e­try.

Figure
Figure 1.3.32 – Para­met­ric plots of the hy­per­bol­ic and wave-like spher­i­cal el­e­ments of uv1uv_{1}.

The vector uv1uv_{1} is com­posed of scalar por­tions and bivec­tors por­tions, form­ing a func­tion that is a hy­bridiza­tion of the unit hy­per­bo­la with the unit circle. We con­tin­ue the process, ro­tat­ing around the plane in a clock­wise manner to all the as­ymp­tot­ic points and axes, by using the fol­low­ing prod­ucts

uv2=u(e1)=et−IeIt,\begin{array}{r} uv_{2} = u\left( e_{1} \right) = e^{t} - {Ie}^{It}, \end{array}
(1.3.320)
uv3=u(e12−e22)=(et2+sin⁡(t−π4)−Iet2−Isin(t+π4)),\begin{array}{r} uv_{3} = u\left( \frac{e_{1}}{\sqrt{2}} - \frac{e_{2}}{\sqrt{2}} \right) = \left( \frac{e^{t}}{\sqrt{2}} + \sin\left( t - \frac{\pi}{4} \right) - \frac{Ie^{t}}{\sqrt{2}} - {Isin}\left( t + \frac{\pi}{4} \right) \right), \end{array}
(1.3.321)
uv4=u(−e2)=−eIt−Iet,\begin{array}{r} uv_{4} = u\left( {- e}_{2} \right) = - e^{It} - Ie^{t}, \end{array}
(1.3.322)
uv5=u(−e12−e22)=(−et2−sin⁡(t+π4)−Iet2−Isin(t−π4)),\begin{array}{r} uv_{5} = u\left( - \frac{e_{1}}{\sqrt{2}} - \frac{e_{2}}{\sqrt{2}} \right) = \left( - \frac{e^{t}}{\sqrt{2}} - \sin\left( t + \frac{\pi}{4} \right) - \frac{Ie^{t}}{\sqrt{2}} - {Isin}\left( t - \frac{\pi}{4} \right) \right), \end{array}
(1.3.323)
uv6=u(−e1)=−et+IeIt,\begin{array}{r} uv_{6} = u\left( {- e}_{1} \right) = - e^{t} + Ie^{It}, \end{array}
(1.3.324)
uv7=u(−e12+e22)=(−et2−sin⁡(t−π4)+Iet2+Isin(t+π4)),\begin{array}{r} uv_{7} = u\left( - \frac{e_{1}}{\sqrt{2}} + \frac{e_{2}}{\sqrt{2}} \right) = \left( - \frac{e^{t}}{\sqrt{2}} - \sin\left( t - \frac{\pi}{4} \right) + \frac{Ie^{t}}{\sqrt{2}} + {Isin}\left( t + \frac{\pi}{4} \right) \right), \end{array}
(1.3.325)
uv8=u(e2)=eIt+Iet,\begin{array}{r} uv_{8} = u\left( e_{2} \right) = e^{It} + Ie^{t}, \end{array}
(1.3.326)

where we find re­sults that are axis ori­ent­ed for uvjuv_{j} with even jj-values, and func­tions that are as­ymp­tote ori­ent­ed for odd jj-values. The reader can orient them­selves to the re­sult­ing vector arrange­ments by view­ing Figure 1.3.33, from the per­spec­tive of the e1,e2e_{1},e_{2}-plane, where the odd as­ymp­tot­ic vector func­tions are de­pict­ed in red, and the axis vector func­tions in black.

The square plane seg­ments should stand out to the reader in Figure 1.3.33, whose sig­nif­i­cance is that the ro­ta­tion of light occurs by moving ori­ent­ed plane seg­ments, which are the bivec­tor forms of the geo­met­ric prod­uct. This dif­fers from ro­tat­ing a point or a vector in the com­plex two-di­men­sion­al plane, where the scalars and the e1,e2e_{1},e_{2} bases are suf­fi­cient to de­scribe the process. In the case of the as­ymp­tot­ic light func­tions, ro­ta­tions occur in a higher-di­men­sion­al man­i­fold, ev­i­denced by the dy­nam­ics of bivec­tor ro­ta­tions.

Figure
Figure 1.3.33 – Ori­ent­ed uvjuv_{j} vec­tors and plane seg­ments in the e1/e2e_{1}/e_{2}-plane.

Next, we take a closer look at the uv1uv_{1} light func­tion by plot­ting uv1uv_{1} in Figure 1.3.34, its imag­i­nary and real com­po­nents

Re(uv1)=et2+sin⁡(t+π4),\begin{array}{r} Re\left( uv_{1} \right) = \frac{e^{t}}{\sqrt{2}} + \sin\left( t + \frac{\pi}{4} \right), \end{array}
(1.3.327)
i×Im(uv1)=iet2+isin(t−π4),\begin{array}{r} i \times Im\left( uv_{1} \right) = \frac{ie^{t}}{\sqrt{2}} + {isin}\left( t - \frac{\pi}{4} \right), \end{array}
(1.3.328)

which con­veys that these light wave­forms are a hy­bridiza­tion of wave and radial el­e­ments.

Figure
Figure 1.3.34 – Wave­form plots: Re(uv)=et2+sin⁡(t+π4)Re(uv) = \frac{e^{t}}{\sqrt{2}} + \sin\left( t + \frac{\pi}{4} \right), i∙\ i \bullet Im(uv)=iet2+isin(t−π4)Im(uv) = \frac{ie^{t}}{\sqrt{2}} + {isin}\left( t - \frac{\pi}{4} \right),

and the para­met­ric plot, γ(t)=(Re(uv),i∙Im(uv))\gamma(t) = \left( Re(uv),i \bullet Im(uv) \right).

The cir­cu­lar trigono­met­ric func­tions impart wave char­ac­ter­is­tics, while the hy­per­bol­ic func­tions impart radial ex­ten­sion. For t<0t < 0, the ex­po­nen­tial wave por­tion of each func­tion dom­i­nates, while for t>0t > 0, the ex­po­nen­tial radius por­tion be­comes om­nipresent. This qual­i­ty of the light func­tions clar­i­fies what occurs when the macroscopy of clas­si­cal physics be­comes quan­tum physics at the mi­cro­scop­ic level. De­spite the dom­i­nance of the radial ex­ten­sion of the ex­po­nen­tial func­tion in the mor­phol­o­gy to the right of the origin, the wave-like char­ac­ter­is­tics remain present in the form. On the other hand, to the left of the origin, it can also be said that de­spite the mor­pho­log­i­cal wave dom­i­nance, the radial ex­ten­sion from the ex­po­nen­tial func­tion re­mains present, albeit in an at­ten­u­at­ed form. The func­tion plots show this as wave mor­phol­o­gy dom­i­nance for t<0t < 0, trans­form­ing to hy­per­bol­ic dom­i­nance for t>0t > 0, after pass­ing through 2\sqrt{2} in that process.

The vjv_{j}-ro­ta­tion vec­tors are not ar­bi­trary, being all number forms re­lat­ed to i\sqrt{i} , as men­tioned at the onset of the sec­tion. The number eiπ4=(12+i2 ),e^{\frac{i\pi}{4}} = \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\ \right),is close­ly re­lat­ed to the vector v1=(e12+e22).v_{1} = \left( \frac{e_{1}}{\sqrt{2}} + \frac{e_{2}}{\sqrt{2}} \right).What should not be over­looked is that the hand­ed­ness of the geo­met­ric prod­uct can result in ro­ta­tions in the op­po­site di­rec­tion. So, the ini­tial prod­uct eval­u­at­ed was uv1uv_{1}, with v1v_{1} on the right side, which ef­fec­tive­ly ro­tates uu in the in­verse di­rec­tion of the re­lat­ed ro­ta­tion e(iπ4)e^{\left( \frac{i\pi}{4} \right)}, which is then ef­fec­tive­ly a ro­ta­tion of uu by e(−iπ4).\ e^{\left( - \frac{i\pi}{4} \right)}.

The uvjuv_{j} func­tions form the as­ymp­tot­ic cones having the fol­low­ing matrix rep­re­sen­ta­tions:

Cone 1:(uv1∈e1uv3∈e2uv1∈e2uv3∈e1)=(et2e1−IeIt2e1−Iet2e2−eIt2e2Iet2e2+eIt2e2et2e1−IeIt2e1),\begin{array}{r} Cone\ 1:\begin{pmatrix} uv_{1} \in e_{1} & uv_{3} \in e_{2} \\ uv_{1} \in e_{2} & uv_{3} \in e_{1} \end{pmatrix} = \begin{pmatrix} \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} & - \frac{{Ie}^{t}}{\sqrt{2}}e_{2} - \frac{e^{It}}{\sqrt{2}}e_{2} \\ \frac{{Ie}^{t}}{\sqrt{2}}e_{2} + \frac{e^{It}}{\sqrt{2}}e_{2} & \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} \end{pmatrix}, \end{array}
(1.3.329)
Cone 2:(uv3∈e1uv5∈e2uv3∈e2uv5∈e1)=(et2e1−IeIt2e1−Iet2e2−eIt2e2−Iet2e2−eIt2e2−et2e1+IeIt2e1),\begin{array}{r} Cone\ 2:\begin{pmatrix} uv_{3} \in e_{1} & uv_{5} \in e_{2} \\ uv_{3} \in e_{2} & uv_{5} \in e_{1} \end{pmatrix} = \begin{pmatrix} \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} & - \frac{{Ie}^{t}}{\sqrt{2}}e_{2} - \frac{e^{It}}{\sqrt{2}}e_{2} \\ - \frac{{Ie}^{t}}{\sqrt{2}}e_{2} - \frac{e^{It}}{\sqrt{2}}e_{2} & \frac{{- e}^{t}}{\sqrt{2}}e_{1} + \frac{Ie^{It}}{\sqrt{2}}e_{1} \end{pmatrix}, \end{array}
(1.3.330)
                    Cone 3:(uv5∈e1uv7∈e2uv5∈e2uv7∈e1)=(−et2e1+IeIt2e1Iet2e2+eIt2e2−Iet2e2−eIt2e2−et2e1+IeIt2e1),\begin{array}{r} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Cone\ 3:\begin{pmatrix} uv_{5} \in e_{1} & uv_{7} \in e_{2} \\ uv_{5} \in e_{2} & uv_{7} \in e_{1} \end{pmatrix} = \begin{pmatrix} \frac{- e^{t}}{\sqrt{2}}e_{1} + \frac{Ie^{It}}{\sqrt{2}}e_{1} & \frac{Ie^{t}}{\sqrt{2}}e_{2} + \frac{e^{It}}{\sqrt{2}}e_{2} \\ - \frac{{Ie}^{t}}{\sqrt{2}}e_{2} - \frac{e^{It}}{\sqrt{2}}e_{2} & - \frac{e^{t}}{\sqrt{2}}e_{1} + \frac{Ie^{It}}{\sqrt{2}}e_{1} \end{pmatrix}, \end{array}
(1.3.331)
Cone 4:(uv7∈e1uv1∈e2uv7∈e2uv1∈e1)=(−et2e1+IeIt2e1Iet2e2+eIt2e2Iet2e2+eIt2e2et2e1−IeIt2e1).\begin{array}{r} Cone\ 4:\begin{pmatrix} uv_{7} \in e_{1} & uv_{1} \in e_{2} \\ uv_{7} \in e_{2} & uv_{1} \in e_{1} \end{pmatrix} = \begin{pmatrix} - \frac{e^{t}}{\sqrt{2}}e_{1} + \frac{Ie^{It}}{\sqrt{2}}e_{1} & \frac{{Ie}^{t}}{\sqrt{2}}e_{2} + \frac{e^{It}}{\sqrt{2}}e_{2} \\ \frac{Ie^{t}}{\sqrt{2}}e_{2} + \frac{e^{It}}{\sqrt{2}}e_{2} & \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} \end{pmatrix}. \end{array}
(1.3.332)

The top row of each cone matrix, de­not­ed cj,c_{j},can take the para­met­ric func­tion form

γc(t)=cj(a1,1 (t),a1,2(t)),\begin{array}{r} \gamma_{c}(t) = c_{j}\left( a_{1,1\ }(t),a_{1,2}(t) \right), \end{array}
(1.3.333)

and the bottom row, the form

γc(t)=cj(a2,1 (t),a2,2(t)).\begin{array}{r} \gamma_{c}(t) = c_{j}\left( a_{2,1\ }(t),a_{2,2}(t) \right). \end{array}
(1.3.334)

The ori­en­ta­tion of the cone seg­ments can be seen in Figure 1.3.35, where the axes e1,−e2,−e1,e2e_{1}, - e_{2}, - e_{1},e_{2} are the re­spec­tive cone axes for c1,c2,c3,c4.c_{1},c_{2},c_{3},c_{4}.The cone ma­tri­ces all have de­ter­mi­nants equal to zero, which occurs since the vector arrange­ment mimics the dot prod­uct. Using the first cone matrix as an ex­am­ple, we ex­press the uv1uv_{1},uv3uv_{3} vec­tors, as

uv1=(et2e1−IeIt2e1Iet2e2+eIt2e2),  uv3=(et2e1−IeIt2e1−Iet2e2−eIt2e2).\begin{array}{r} uv_{1} = \left( \begin{array}{r} \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} \\ \frac{{Ie}^{t}}{\sqrt{2}}e_{2} + \frac{e^{It}}{\sqrt{2}}e_{2} \end{array} \right),\ \ uv_{3} = \left( \begin{array}{r} \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} \\ - \frac{{Ie}^{t}}{\sqrt{2}}e_{2} - \frac{e^{It}}{\sqrt{2}}e_{2} \end{array} \right). \end{array}
(1.3.335)
Figure
Figure 1.3.35 – The as­ymp­tot­ic cone sec­tions in the e1/e2e_{1}/e_{2}-plane.

We now cal­cu­late the dot prod­uct of the uv1uv_{1},uv3uv_{3} vector com­po­nents

uv1∙uv3=(et2e1−IeIt2e1Iet2e2+eIt2e2)∙(et2e1−IeIt2e1−Iet2e2−eIt2e2)=(et2−IeIt2)(et2+Ie−It2)+(Iet2+eIt2)(Iet2−Ie−It2)=0,\begin{array}{r} uv_{1} \bullet uv_{3} = \left( \begin{array}{r} \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} \\ \frac{{Ie}^{t}}{\sqrt{2}}e_{2} + \frac{e^{It}}{\sqrt{2}}e_{2} \end{array} \right) \bullet \left( \begin{array}{r} \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} \\ - \frac{{Ie}^{t}}{\sqrt{2}}e_{2} - \frac{e^{It}}{\sqrt{2}}e_{2} \end{array} \right) = \left( \frac{e^{t}}{\sqrt{2}} - \frac{Ie^{It}}{\sqrt{2}} \right)\left( \frac{e^{t}}{\sqrt{2}} + \frac{Ie^{- It}}{\sqrt{2}} \right) + \left( \frac{Ie^{t}}{\sqrt{2}} + \frac{e^{It}}{\sqrt{2}} \right)\left( \frac{Ie^{t}}{\sqrt{2}} - \frac{Ie^{- It}}{\sqrt{2}} \right) = 0, \end{array}
(1.3.336)

where the result equals zero, and this is con­sis­tent with the two or­thog­o­nal vec­tors. The geo­met­ric prod­uct uv1uv3uv_{1}uv_{3} is there­fore, equal to the wedge prod­uct uv1∧uv3uv_{1} \land uv_{3}

uv1∧uv3=(et2e1−IeIt2e1Iet2e2+eIt2e2)∧(et2e1−IeIt2e1−Iet2e2−eIt2e2)=(−Ie2t2+Ie2It2−eItet)e1∧e2+(Ie2t2−Ie2It2+eItet)e2∧e1uv_{1} \land uv_{3} = \left( \begin{array}{r} \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} \\ \frac{{Ie}^{t}}{\sqrt{2}}e_{2} + \frac{e^{It}}{\sqrt{2}}e_{2} \end{array} \right) \land \left( \begin{array}{r} \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} \\ - \frac{{Ie}^{t}}{\sqrt{2}}e_{2} - \frac{e^{It}}{\sqrt{2}}e_{2} \end{array} \right) = \left( - \frac{{Ie}^{2t}}{2} + \frac{Ie^{2It}}{2} - e^{It}e^{t} \right)e_{1} \land e_{2} + \left( \frac{{Ie}^{2t}}{2} - \frac{Ie^{2It}}{2} + e^{It}e^{t} \right)e_{2} \land e_{1}
=e2t−e2It−2IeItet=(et−IeIt)2=(uv2)2,\begin{array}{r} = e^{2t} - e^{2It} - 2Ie^{It}e^{t} = \left( e^{t} - Ie^{It} \right)^{2} = \left( uv_{2} \right)^{2}, \end{array}
(1.3.337)

where the wedge prod­uct is equal to the uv2uv_{2} vector squared. Now, we con­sid­er the vec­tors C1C_{1} and C1′C_{1}^{'}, which rep­re­sent the columns of the Cone 1 group

C1=uv1∈e1+ uv1∈e2,\begin{array}{r} C_{1} = uv_{1} \in e_{1} + \ uv_{1} \in e_{2}, \end{array}
(1.3.338)
C1′=uv3∈e2+ uv3∈e1,\begin{array}{r} C_{1}^{'} = uv_{3} \in e_{2} + \ uv_{3} \in e_{1}, \end{array}
(1.3.339)

for which, the values for each vector rep­re­sent the column vec­tors from the Cone 1Cone\ 1 matrix. Next, we add the two vec­tors to­geth­er

C1+C1′=((et2e1−IeIt2e1+Iet2e2+eIt2e2)+(et2e1−IeIt2e1−Iet2e2−eIt2e2))=2(et−IeIt)e1,\begin{array}{r} C_{1} + C_{1}^{'} = \left( \begin{array}{r} \left( \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} + \frac{{Ie}^{t}}{\sqrt{2}}e_{2} + \frac{e^{It}}{\sqrt{2}}e_{2} \right) \\ + \left( \frac{e^{t}}{\sqrt{2}}e_{1} - \frac{Ie^{It}}{\sqrt{2}}e_{1} - \frac{{Ie}^{t}}{\sqrt{2}}e_{2} - \frac{e^{It}}{\sqrt{2}}e_{2} \right) \end{array} \right) = \sqrt{2}\left( e^{t} - Ie^{It} \right)e_{1}, \end{array}
(1.3.340)

and the e2e_{2} basis por­tions cancel, lead­ing to a vector sum equal to the uv2uv_{2} geo­met­ric prod­uct scaled by 2\sqrt{2}, ori­ent­ing with the e1e_{1} axis

C1+C1′=2(et−IeIt)e1=2uv2e1.\begin{array}{r} C_{1} + C_{1}^{'} = \sqrt{2}\left( e^{t} - Ie^{It} \right)e_{1} = \sqrt{2}uv_{2}e_{1}. \end{array}
(1.3.341)

It is easily ver­i­fied that sim­i­lar vector sums for the other cones hold. Ob­tained as fol­lows:

C2+C2′=2(−Iet−eIt)e2=2uv4e2,\begin{array}{r} C_{2} + C_{2}^{'} = \sqrt{2}\left( {- Ie}^{t} - e^{It} \right)e_{2} = \sqrt{2}uv_{4}e_{2}, \end{array}
(1.3.342)
C3+C3′=2(−et+IeIt)e1=2uv6e1,\begin{array}{r} C_{3} + C_{3}^{'} = \sqrt{2}\left( {- e}^{t} + Ie^{It} \right)e_{1} = \sqrt{2}uv_{6}e_{1}, \end{array}
(1.3.343)
C4+C4′=2(Iet+eIt)e1=2uv8e2.\begin{array}{r} C_{4} + C_{4}^{'} = \sqrt{2}\left( {Ie}^{t} + e^{It} \right)e_{1} = \sqrt{2}uv_{8}e_{2}. \end{array}
(1.3.344)

Geo­met­ri­cal­ly, these sums rep­re­sent the di­ag­o­nals of the ori­ent­ed par­al­lel­o­grams, formed by the pair of vec­tors. We demon­strate this for the case of the C1+C1′C_{1} + C_{1}^{'} vector sum, in the figure below.

Figure
Figure 1.3.36 – The vector sum C1+C1′=2(et−IeIt)e1C_{1} + C_{1}^{'} = \sqrt{2}\left( e^{t} - Ie^{It} \right)e_{1}, rep­re­sents

the di­ag­o­nal of the ori­ent­ed par­al­lel­o­gram formed by the vec­tors.

The im­por­tance of these vector ad­di­tion re­sults is that 2\sqrt{2} plays a cen­tral role in rel­a­tiv­i­ty and the per­cep­tion of the speed of light. It will be con­sid­ered in more detail toward the end of the book, but suf­fice it to say, 2\sqrt{2} rep­re­sents the di­ag­o­nal length of the com­plex unit square, which plays a sig­nif­i­cant role as the con­stant for the com­plex unit cube's shear trans­for­ma­tions and main­tains this in­vari­ant area rep­re­sent­ing the speed of light.

Figure
Figure 1.3.37 – Tan­gent vector map of the Light func­tions in the com­plex plane and angle space.

The light cones and light axes result from a dy­nam­ic process of cir­cu­lar and hy­per­bol­ic geome­tries moving past one an­oth­er. These geome­tries should be con­sid­ered out of phase in the com­plex angle space, which was al­lud­ed to ear­li­er in the chap­ter and elu­ci­dat­ed by the simple ex­am­ple eit=eBte^{it} = e^{Bt}, which holds for B=e(iπ2).B = e^{\left( \frac{i\pi}{2} \right)}.The light as­ymp­totes have or­thog­o­nal re­la­tion­ships to the re­spec­tive as­ymp­totes ori­ent­ed a dis­tance e(±iπ2)e^{\left( \pm \frac{i\pi}{2} \right)} away, as do the light axes. This phase dif­fer­ence, e(±iπ2)e^{\left( \pm \frac{i\pi}{2} \right)}, de­fines the or­thog­o­nal­i­ty in the com­plex plane for the light func­tions. The hy­per­bol­ic and cir­cu­lar planes can be viewed as ro­tat­ing past one an­oth­er in op­po­site di­rec­tions. This is shown in Figure 1.3.37, where the chosen rep­re­sen­ta­tion has the red tan­gent vec­tors of the eite^{it} cir­cu­lar geom­e­try with neg­a­tive/clock­wise ori­en­ta­tion, and the black tan­gent vec­tors of the ete^{t} hy­per­bol­ic geom­e­try with pos­i­tive/coun­ter­clock­wise ori­en­ta­tion. When the light func­tions are viewed in this manner, we can define the gen­er­al­ized uni­fied light func­tions:

De­f­i­n­i­tion (1.3.345) – The gen­er­al­ized spa­tial-tem­po­ral uni­fied light func­tions 𝔏(𝐭,−𝛕)\mathfrak{L}\left( \mathbf{t}, - \mathbf{\tau} \right), 𝔏(𝐭,𝛕)\mathfrak{L}\left( \mathbf{t},\mathbf{\tau} \right) are de­fined, as

𝔏(t,−τ)=eiωt−iθτ+eωt−iθτ+iπ2,\mathfrak{L}(t, - \tau) = e^{i\omega t - i\theta\tau} + e^{\omega t - i\theta\tau + \frac{i\pi}{2}},
𝔏(t,τ)=eiωt+iθτ+eωt+iθτ−iπ2,\mathfrak{L}(t,\tau) = e^{i\omega t + i\theta\tau} + e^{\omega t + i\theta\tau - \frac{i\pi}{2}},

where the time-phased, iωi\omega, will be a mul­ti­ple of the as­ymp­tot­ic angle, ±iπ4\pm \frac{i\pi}{4} ra­di­ans, the func­tions may also take on the spa­tial forms

𝔏(x,−τ)=eikx−iθτ+ekx−iθτ+iπ2,\mathfrak{L}(x, - \tau) = e^{ikx - i\theta\tau} + e^{kx - i\theta\tau + \frac{i\pi}{2}},
𝔏(x,τ)=eikx+iθτ+ekx+iθτ−iπ2.\mathfrak{L}(x,\tau) = e^{ikx + i\theta\tau} + e^{kx + i\theta\tau - \frac{i\pi}{2}}.

Using the 𝔏(t,−τ)\mathfrak{L}(t, - \tau) func­tion, we set θ=π4\theta = \frac{\pi}{4}, and con­sid­er in­te­ger values for τ\tau, start­ing at τ=0\tau = 0 , then watch how the light system evolves

𝔏(t,0)=eiωt+eωt+iπ2=eiωt+ieωt,\begin{array}{r} \mathfrak{L}(t,0) = e^{i\omega t} + e^{\omega t + \frac{i\pi}{2}} = e^{i\omega t} + ie^{\omega t}, \end{array}
(1.3.346)
𝔏(t,−1)=eiωt−iπ4+eωt−iπ4+iπ2=(12−i2)eiωt+(12+i2)eωt,\begin{array}{r} \mathfrak{L}(t, - 1) = e^{i\omega t - \frac{i\pi}{4}} + e^{\omega t - \frac{i\pi}{4} + \frac{i\pi}{2}} = \left( \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \right)e^{i\omega t} + \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \right)e^{\omega t}, \end{array}
(1.3.347)
𝔏(t,−2)=eiωt−iπ2+eωt−iπ2+iπ2=−ieiωt+eωt,\begin{array}{r} \mathfrak{L}(t, - 2) = e^{i\omega t - \frac{i\pi}{2}} + e^{\omega t - \frac{i\pi}{2} + \frac{i\pi}{2}} = - ie^{i\omega t} + e^{\omega t}, \end{array}
(1.3.348)
𝔏(t,−3)=eiωt−3iπ4+eωt−3iπ4+iπ2=(−12+i2)eiωt+(12−i2)eωt,\begin{array}{r} \mathfrak{L}(t, - 3) = e^{i\omega t - \frac{3i\pi}{4}} + e^{\omega t - \frac{3i\pi}{4} + \frac{i\pi}{2}} = \left( - \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \right)e^{i\omega t} + \left( \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \right)e^{\omega t}, \end{array}
(1.3.349)
𝔏(t,−4)=eiωt−iπ+eωt−iπ+iπ2=−eiωt−ieωt.\begin{array}{r} \mathfrak{L}(t, - 4) = e^{i\omega t - i\pi} + e^{\omega t - i\pi + \frac{i\pi}{2}} = {- e}^{i\omega t} - ie^{\omega t}. \end{array}
(1.3.350)

Ap­par­ent from the above, is the light system ro­tates in the clock­wise di­rec­tion with re­spect to the uni­fied light func­tion values for the as­ymp­totes and axes. The rep­re­sen­ta­tion of the light func­tions in Figure 1.3.37 is linked to the con­cept of torque in physics. Torque is typ­i­cal­ly de­fined as the mag­ni­tude of a tan­gen­tial force com­po­nent times the dis­tance, im­ply­ing the work done per unit angle. A torque vector 𝛕\mathbf{\tau}\mathbf{} is rep­re­sent­ed, as

𝛕=𝐫×𝐅,\begin{array}{r} \mathbf{\tau}= \mathbf{r} \times \mathbf{F}, \end{array}
(1.3.351)

for 𝐫×𝐅\mathbf{r} \times \mathbf{F}, the cross prod­uct be­tween a po­si­tion vector 𝐫\mathbf{r}, orig­i­nat­ing at a point where torque is being mea­sured to the point that a force is being ap­plied, and a vector 𝐅\mathbf{F}, rep­re­sent­ing the ap­plied force. Given the pres­ence of the cross prod­uct, it pro­duces a vector t­­hat is or­thog­o­nal to both 𝐫\mathbf{r} and 𝐅\mathbf{F}, with an ori­en­ta­tion that fol­lows the right-hand rule. With the torque vector com­po­nents de­scribed, torque can be de­fined, as

τ=‖𝐫‖‖𝐅‖sin⁡(θ),\begin{array}{r} \tau = \left\| \mathbf{r} \right\|\left\| \mathbf{F} \right\|\sin(\theta), \end{array}
(1.3.352)

for θ\theta, the angle be­tween the force vector 𝐅\mathbf{F}\mathbf{} and the po­si­tion vector 𝐫\mathbf{r}. We can con­sid­er torque in terms of geo­met­ric al­ge­bra by con­sid­er­ing the po­si­tion vector 𝐫\mathbf{r} in the para­met­ric form,

𝐫=r(cos⁡(θ)e1+sin⁡(θ)e2)=re1(cos⁡(θ)+sin⁡(θ)e1e2),\begin{array}{r} \mathbf{r} = r\left( \cos(\theta)e_{1} + \sin(\theta)e_{2} \right) = re_{1}\left( \cos(\theta) + \sin(\theta)e_{1}e_{2} \right), \end{array}
(1.3.353)

for which, the unit bivec­tor in the plane is the imag­i­nary number ii, and noting the iso­mor­phism i≅e1∧e2=e1e2=Ii \cong e_{1} \land e_{2} = e_{1}e_{2} = I, lead­ing to the ex­pres­sion

𝐫=re1eIθ.\begin{array}{r} \mathbf{r} = re_{1}e^{I\theta}. \end{array}
(1.3.354)

In physics, the de­f­i­n­i­tion of work is the amount of energy trans­fer as­so­ci­at­ed with the move­ment of an object over some dis­tance, which is ac­com­plished by an ex­ter­nal force. For a system com­posed of a ro­tat­ing body about a fixed axis, we can define the work done dWdW by a tan­gen­tial force Ftan⁡F_{\tan}

dW=Ftan⁡ds.\begin{array}{r} dW = F_{\tan}ds. \end{array}
(1.3.355)

The in­fin­i­tes­i­mal change in dis­tance can be ex­pressed in the radian mea­sure ds=𝐫dθds = \mathbf{r}d\theta, for 𝐫\mathbf{r}, the radius vector, and dθd\theta, the in­fin­i­tes­i­mal dis­place­ment angle. We can rewrite the ex­pres­sion for work, as

dW=Ftan⁡𝐫dθ,\begin{array}{r} dW = F_{\tan}\mathbf{r}d\theta, \end{array}
(1.3.356)

for Ftan⁡𝐫F_{\tan}\mathbf{r}, the torque τ\ \tau, re­sult­ing from the ex­ter­nal and tan­gen­tial force Ftan⁡F_{\tan}, lead­ing to the iden­ti­ty

dW=τdθ,\begin{array}{r} dW = \tau d\theta, \end{array}
(1.3.357)

which can be solved for torque

τ=dWdθ.\begin{array}{r} \tau = \frac{dW}{d\theta}. \end{array}
(1.3.358)

In vector no­ta­tion, the ex­ter­nal force is ex­pressed as 𝐅\mathbf{F}, which has a tan­gen­tial com­po­nent force acting on a ro­tat­ing body, as­sumed to be con­stant. We now state torque in terms of our vec­tors, as

τ=dWdθ=𝐅∙d𝐫dθ,\begin{array}{r} \tau = \frac{dW}{d\theta} = \mathbf{F} \bullet \frac{d\mathbf{r}}{d\theta}, \end{array}
(1.3.359)

where drdθ\frac{dr}{d\theta} is the change of the 𝐫\mathbf{r} vector over an in­fin­i­tes­i­mal dis­place­ment angle

d𝐫dθ=(re1eIθ)ddθ=re1eIθI=𝐫I.\begin{array}{r} \frac{d\mathbf{r}}{d\theta} = \left( re_{1}e^{I\theta} \right)\frac{d}{d\theta} = re_{1}e^{I\theta}I = \mathbf{r}I. \end{array}
(1.3.360)

This en­ables us to ex­press torque using geo­met­ric al­ge­bra, as

τ=𝐅∙d𝐫dθ=𝐅∙𝐫I,\begin{array}{r} \tau = \mathbf{F} \bullet \frac{d\mathbf{r}}{d\theta} = \mathbf{F} \bullet \mathbf{r}I, \end{array}
(1.3.361)

and by ex­pand­ing the po­si­tion vector, we find

𝐫I=(r1e1+r2e2)e1∧e2=r1e2−r2e1,\begin{array}{r} \mathbf{r}I = \left( r_{1}e_{1} + r_{2}e_{2} \right)e_{1} \land e_{2} = r_{1}e_{2} - r_{2}e_{1}, \end{array}
(1.3.362)

which we use to take the dot prod­uct with 𝐅\mathbf{F}

𝐅∙𝐫I=(F1e1+F2e2+F⊥e3)(r1e2−r2e1)=r1F2−r2F1.\begin{array}{r} \mathbf{F} \bullet \mathbf{r}I = \left( F_{1}e_{1} + F_{2}e_{2} + F_{\bot}e_{3} \right)\left( r_{1}e_{2} - r_{2}e_{1} \right) = {r_{1}F}_{2} - {r_{2}F}_{1}. \end{array}
(1.3.363)

In the above, we used the per­pen­dic­u­lar basis vector e3e_{3} in the force vector F⊥e3F_{\bot}e_{3}, to be con­sis­tent with the cross prod­uct's ori­ent­ed normal vector. No­tably, the result is con­sis­tent with the wedge prod­uct, given that

(r1e1+r2e2)∧(F1e1+F2e2)=r1F2−r2F1,\begin{array}{r} \left( r_{1}e_{1} + r_{2}e_{2} \right) \land \left( F_{1}e_{1} + F_{2}e_{2} \right) = {r_{1}F}_{2} - {r_{2}F}_{1}, \end{array}
(1.3.364)

and this wedge prod­uct 𝐫∧𝐅\mathbf{r} \land \mathbf{F} rep­re­sents the torque of a system with­out 𝐫\mathbf{r} and 𝐅\mathbf{F} ex­pressed as mag­ni­tudes per the tra­di­tion­al de­f­i­n­i­tion. This leads us to define the torque vector in geo­met­ric al­ge­bra no­ta­tion, as

T=𝐫∧𝐅.\begin{array}{r} \mathbf{\mathrm{T}}= \mathbf{r} \land \mathbf{F}. \end{array}
(1.3.365)

Using the bivec­tors native to geo­met­ric al­ge­bra, we do away with the need for a normal axis of ro­ta­tion, which re­sults from the cross-prod­uct. This turns out to be more con­sis­tent with the frame­work of the com­plex plane and its ro­ta­tions, given that a normal axis to the plane of ro­ta­tion is not math­e­mat­i­cal­ly re­quired in the for­mal­ism. With torque de­fined in the con­text of geo­met­ric al­ge­bra, we apply it to the light func­tions de­scribed in this sec­tion. We con­sid­er the light system com­posed of two torques, by first, defin­ing two po­si­tion vec­tors

𝐫1=(et−IeIt)e1,  𝐫2=(eIt+Iet)e2,\begin{array}{r} \mathbf{r}_{1} = \left( e^{t} - {Ie}^{It} \right)e_{1},\ \ \mathbf{r}_{2} = \left( e^{It} + {Ie}^{t} \right)e_{2}, \end{array}
(1.3.366)

where we ab­stract that a po­si­tion in the geo­met­ric space de­fines a time-de­pen­dent ori­en­ta­tion of the com­plex unit circle and unit hy­per­bo­la, co­in­cid­ing with the pre­vi­ous­ly de­rived uv2uv_{2}, uv8uv_{8} geo­met­ric prod­ucts. The ap­plied forces on each of the po­si­tion vec­tors are

𝐅1=e22,  𝐅2=e12,\begin{array}{r} \mathbf{F}_{1} = \frac{e_{2}}{\sqrt{2}},\ \ \mathbf{F}_{2} = \frac{e_{1}}{\sqrt{2}}, \end{array}
(1.3.367)

having both mag­ni­tude and ori­en­ta­tion. Next, we cal­cu­late the in­di­vid­ual torque com­po­nents

T1=𝐫1∧𝐅1=(et−IeIt)e1∧e22,\begin{array}{r} \mathbf{\mathrm{T}}_{1} = \mathbf{r}_{1} \land \mathbf{F}_{1}\mathbf{=}\left( e^{t} - {Ie}^{It} \right)e_{1} \land \frac{e_{2}}{\sqrt{2}}, \end{array}
(1.3.368)
T2=𝐫2∧𝐅2=(eIt+Iet)e2∧e12,\begin{array}{r} \mathbf{\mathrm{T}}_{2} = \mathbf{r}_{2} \land \mathbf{F}_{2}\mathbf{=}\left( e^{It} + {Ie}^{t} \right)e_{2} \land \frac{e_{1}}{\sqrt{2}}, \end{array}
(1.3.369)

which are de­pict­ed in Figure 1.3.38, as the ori­ent­ed bivec­tors T1↻\mathbf{\mathrm{T}}_{1}\circlearrowright , T2↺\mathbf{\mathrm{T}}_{2}\circlearrowleft. These torque bivec­tors can be seen as equal in area but op­po­site in ori­en­ta­tion.

T1+T2=(et−IeIt)e1∧e22+(eIt+Iet)e2∧e12\mathbf{\mathrm{T}}_{1} + \mathbf{\mathrm{T}}_{2} = \left( e^{t} - {Ie}^{It} \right)e_{1} \land \frac{e_{2}}{\sqrt{2}} + \left( e^{It} + {Ie}^{t} \right)e_{2} \land \frac{e_{1}}{\sqrt{2}}
Figure
Figure 1.3.38 – The geo­met­ric rep­re­sen­ta­tion of the torques T1=𝐫1∧𝐅1\mathbf{\mathrm{T}}_{1} = \mathbf{r}_{1} \land \mathbf{F}_{1} and T2=𝐫2∧𝐅2\mathbf{\mathrm{T}}_{2} = \mathbf{r}_{2} \land \mathbf{F}_{2}.

The net torque of the func­tion system is found by adding the torque com­po­nents, which eval­u­ate to

=(et+e2e1eIt)e1e22+(eIt+e1e2et)e2e12=Iet+eIt2+et+−IeIt2= \frac{\left( e^{t} + e_{2}e_{1}e^{It} \right)e_{1}e_{2}}{\sqrt{2}} + \frac{\left( e^{It} + {e_{1}e_{2}e}^{t} \right)e_{2}e_{1}}{\sqrt{2}} = \frac{Ie^{t} + e^{It}}{\sqrt{2}} + \frac{e^{t} + {- Ie}^{It}}{\sqrt{2}}
=(Iet2+Isin(t−π4)+et2+sin⁡(t+π4))=uv1,\begin{array}{r} = \left( \frac{Ie^{t}}{\sqrt{2}} + {Isin}\left( t - \frac{\pi}{4} \right) + \frac{e^{t}}{\sqrt{2}} + \sin\left( t + \frac{\pi}{4} \right) \right) = uv_{1}, \end{array}
(1.3.370)

noting a result which is con­sis­tent with our pre­vi­ous­ly de­rived, uv1uv_{1}, geo­met­ric prod­uct.

So, the ori­ent­ed torque bivec­tor result de­pends on the func­tion's dis­tance in com­plex space, which varies by the value of the time vari­able tt. In­tu­itive­ly, this should make sense, given that the 𝐫\mathbf{r}-vector rep­re­sents an ori­ent­ed dis­tance in the com­plex plane for which, a tan­gen­tial force acts on the form as­so­ci­at­ed with its vector head. Put in simple terms, T1+T2\mathbf{\mathrm{T}}_{1} + \mathbf{\mathrm{T}}_{2} is a torque func­tion with re­sults that orient in the com­plex plane. It should be noted, the 𝐫1\mathbf{r}_{1}, 𝐫2\mathbf{r}_{2}, dis­tance func­tion vec­tors have a de­fault torque that is par­tic­u­lar to their forms, which stems from the pres­ence of the com­plex unit circle in each. For in­stance, we take the −IeIt- {Ie}^{It} por­tion from the T1\mathbf{\mathrm{T}}_{1} torque cal­cu­la­tion, to con­sid­er the form

−IeIte1=eIte2,\begin{array}{r} - {Ie}^{It}e_{1} = e^{It}e_{2}, \end{array}
(1.3.371)

in which, the sim­pli­fi­ca­tion shows that e2e_{2} is the ori­en­ta­tion basis of the func­tion. The ex­po­nen­tial func­tion eIte^{It} does have an an­gu­lar fre­quen­cy, which is simply uni­tary ω=1\omega = 1, and set­ting ω=sin⁡(π2)\omega = \sin\left( \frac{\pi}{2} \right), the ori­ent­ed com­plex unit circle be­comes

ee1∧e2ωte2=ee1∧e2sin⁡(π2)te2.\begin{array}{r} e^{e_{1} \land e_{2}\omega t}e_{2} = e^{e_{1} \land e_{2}\sin\left( \frac{\pi}{2} \right)t}e_{2}. \end{array}
(1.3.372)

So, a torque occurs in the ex­po­nent of the func­tion with an or­thog­o­nal­ly ori­ent­ed force, 𝐅=e2sin⁡(π2),\mathbf{F} = {e_{2}\sin}\left( \frac{\pi}{2} \right),which is wedged with the unit radius basis vector e1e_{1}. The ex­po­nent por­tion of the com­plex unit circle can now be seen in terms of its torque re­la­tion­ship, as

T=e1∧(e2sin⁡(π2)t),\begin{array}{r} \mathbf{\mathrm{T}} = e_{1} \land \left( e_{2}\sin{\left( \frac{\pi}{2} \right)t} \right), \end{array}
(1.3.373)

which amounts to a nested torque lo­cat­ed in the ex­po­nent, rep­re­sent­ing work done over a dis­tance in time, man­i­fest­ed as the radian dis­tance in the tan­gent space. The ex­po­nen­tial func­tion ete^{t} has no an­gu­lar ro­ta­tion com­po­nent, and there­fore, does not have an innate torque like the com­plex unit circle. The reader should be aware that if we have 𝐫j\mathbf{r}_{j}-vec­tors, as fol­lows

𝐫1=Ae1,  𝐫2=Ae2,\begin{array}{r} \mathbf{r}_{1} = Ae_{1},\ \ \mathbf{r}_{2} = Ae_{2}, \end{array}
(1.3.374)

being the same ab­solute dis­tance from the origin, when the same-valued forces are ap­plied

𝐅1=e22,  𝐅2=e12,\begin{array}{r} \mathbf{F}_{1} = \frac{e_{2}}{\sqrt{2}},\ \ \mathbf{F}_{2} = \frac{e_{1}}{\sqrt{2}}, \end{array}
(1.3.375)

the in­di­vid­ual torque com­po­nents are found to be

T1=𝐫1∧𝐅1=Ae1∧e22,\begin{array}{r} \mathbf{\mathrm{T}}_{1} = \mathbf{r}_{1} \land \mathbf{F}_{1}\mathbf{=}Ae_{1} \land \frac{e_{2}}{\sqrt{2}}, \end{array}
(1.3.376)
T2=𝐫2∧𝐅2=Ae2∧e12.\begin{array}{r} \mathbf{\mathrm{T}}_{2} = \mathbf{r}_{2} \land \mathbf{F}_{2}\mathbf{=}Ae_{2} \land \frac{e_{1}}{\sqrt{2}}. \end{array}
(1.3.377)

This leads to a net-torque force that is equal to zero, seen by eval­u­at­ing the sum of torque com­po­nents

T1+T2=Ae1∧e22+Ae2∧e12\mathbf{\mathrm{T}}_{1} + \mathbf{\mathrm{T}}_{2} = Ae_{1} \land \frac{e_{2}}{\sqrt{2}} + Ae_{2} \land \frac{e_{1}}{\sqrt{2}}
=Ae1e22+Ae2e12=Ae1e22−Ae1e22=0.\begin{array}{r} = \frac{Ae_{1}e_{2}}{\sqrt{2}} + \frac{Ae_{2}e_{1}}{\sqrt{2}} = \frac{Ae_{1}e_{2}}{\sqrt{2}} - \frac{Ae_{1}e_{2}}{\sqrt{2}} = 0. \end{array}
(1.3.378)

So, we can think of a net-zero torque as the geo­met­ric prod­uct be­tween two func­tion vec­tors, ori­ent­ed in an or­thog­o­nal basis in re­la­tion to one an­oth­er. For two or­thog­o­nal func­tion vec­tors ue1⊥ve2{ue}_{1}\bot ve_{2}, with a chosen ori­en­ta­tion ue1∧e2=v{ue}_{1} \land e_{2} = v and ve2∧e1=u,{ve}_{2} \land e_{1} = u,the torque will be equiv­a­lent to the geo­met­ric prod­uct

T=(ue1+ve2)(𝐅1e2+𝐅2e1).\begin{array}{r} \mathbf{\mathrm{T}}\mathbf{=}\left( {ue}_{1} + ve_{2} \right)\left( \mathbf{F}_{1}e_{2}\mathbf{+}\mathbf{F}_{2}e_{1} \right). \end{array}
(1.3.379)

If the net-torque

𝐫∧𝐅=(ue1+ve2)∧(𝐅1e2+𝐅2e1),\begin{array}{r} \mathbf{r} \land \mathbf{F} =\left( ue_{1} + ve_{2} \right) \land \left( \mathbf{F}_{1}e_{2}\mathbf{+}\mathbf{F}_{2}e_{1} \right), \end{array}
(1.3.380)

has ro­ta­tion forces 𝐅𝟏\mathbf{F}_{\mathbf{1}} and 𝐅2\mathbf{F}_{2}, that are equal value, this leads to a wedge prod­uct equal to zero

(e1+e2)∧(𝐅1e1+𝐅2e2)=0,\begin{array}{r} \left( e_{1} + e_{2} \right) \land \left( \mathbf{F}_{1}e_{1}\mathbf{+}\mathbf{F}_{2}e_{2} \right) = 0, \end{array}
(1.3.381)

given we are wedg­ing a par­al­lel scaled vector along its basis vector. By set­ting, 𝐅1=𝐅2\mathbf{F}_{1}\mathbf{=}\mathbf{F}_{2}, by as­so­cia­tiv­i­ty, we have the iden­ti­ty

(e1+e2)∧(𝐅1e2+𝐅2e1)=𝐅1(e1+e2)∧(e1+e2)=0.\begin{array}{r} \left( e_{1} + e_{2} \right) \land \left( \mathbf{F}_{1}e_{2}\mathbf{+}\mathbf{F}_{2}e_{1} \right) = \mathbf{F}_{1}\left( e_{1} + e_{2} \right) \land \left( e_{1} + e_{2} \right) = 0. \end{array}
(1.3.382)

Eval­u­at­ing the geo­met­ric prod­uct (ue1+ve2)(𝐅1e2+𝐅2e1)\left( ue_{1} + ve_{2} \right)\left( \mathbf{F}_{1}e_{2}\mathbf{+}\mathbf{F}_{2}e_{1} \right), we find

(ue1+ve2)(𝐅1e2+𝐅2e1)=(ue1+ve2)∙(𝐅1e2+𝐅2e1)+(ue1+ve2)∧(𝐅1e2+𝐅2e1)\left( ue_{1} + ve_{2} \right)\left( \mathbf{F}_{1}e_{2}\mathbf{+}\mathbf{F}_{2}e_{1} \right) = \left( ue_{1} + ve_{2} \right)\mathbf{\bullet}\left( \mathbf{F}_{1}e_{2}\mathbf{+}\mathbf{F}_{2}e_{1} \right) + \left( ue_{1} + ve_{2} \right) \land \left( \mathbf{F}_{1}e_{2}\mathbf{+}\mathbf{F}_{2}e_{1} \right)
=0+u𝐅1e1∧e2+v𝐅2e2∧e1=uI𝐅1−vI𝐅2,\begin{array}{r} = 0 + u\mathbf{F}_{1}e_{1} \land e_{2} + v\mathbf{F}_{2}e_{2} \land e_{1} = uI\mathbf{F}_{1} - vI\mathbf{F}_{2}, \end{array}
(1.3.383)

where the dot prod­uct por­tion is zero, given that 𝐅1\mathbf{F}_{1} acts on ue1ue_{1} per­pen­dic­u­lar­ly, as does F2F_{2} on ve2.ve_{2}\mathbf{.} By eval­u­at­ing the net-torque T=𝐫∧𝐅\mathbf{\mathrm{T}}= \mathbf{r} \land \mathbf{F}, we find

T=𝐫∧𝐅=(ue1+ve2)∧(𝐅1e2+𝐅2e1)=uI𝐅1−vI𝐅2,\begin{array}{r} \mathbf{\mathrm{T}}= \mathbf{r} \land \mathbf{F} =\left( ue_{1} + ve_{2} \right) \land \left( \mathbf{F}_{1}e_{2}\mathbf{+}\mathbf{F}_{2}e_{1} \right) = uI\mathbf{F}_{1} - vI\mathbf{F}_{2}, \end{array}
(1.3.384)

which is equiv­a­lent to the geo­met­ric prod­uct of or­thog­o­nal vec­tors. We can fur­ther state by the de­fined or­thog­o­nal ori­en­ta­tion of the vec­tors u,vu,v, that

uI𝐅1=v𝐅1,  −vI𝐅2=u𝐅2,\begin{array}{r} uI\mathbf{F}_{1}\mathbf{=}v\mathbf{F}_{1}\mathbf{,\ \ } - vI\mathbf{F}_{2}\mathbf{=}u\mathbf{F}_{2}\mathbf{,} \end{array}
(1.3.385)

which im­plies the geo­met­ric prod­uct equals

(ue1+ve2)(𝐅1e2+𝐅2e1)=(ue1+ve2)∧(𝐅1e2+𝐅2e1)=u𝐅2+v𝐅1.\begin{array}{r} \left( ue_{1} + ve_{2} \right)\left( \mathbf{F}_{1}e_{2}\mathbf{+}\mathbf{F}_{2}e_{1} \right) = \left( ue_{1} + ve_{2} \right) \land \left( \mathbf{F}_{1}e_{2}\mathbf{+}\mathbf{F}_{2}e_{1} \right) = u\mathbf{F}_{2} + v\mathbf{F}_{1}. \end{array}
(1.3.386)

This primer on torque shows how it can be ap­plied to the frame­work of the geo­met­ric prod­uct. In­stead of being a scalar value, its mean­ing has been ex­tend­ed to that of a func­tion ori­ent­ed in the com­plex plane. We can math­e­mat­i­cal­ly con­cep­tu­al­ize the forces acting on the cir­cu­lar and hy­per­bol­ic ex­po­nen­tial func­tions in con­text to the uni­fied light func­tions. To show this, we take the tem­po­ral uni­fied light func­tion as the ex­am­ple

𝔏(t,−τ)=eiωt−𝐢𝛉𝛕+eωt−𝐢𝛉𝛕+iπ2,\begin{array}{r} \mathfrak{L}(t, - \tau) = e^{i\omega t- \mathbf{i\theta\tau}} + e^{\omega t- \mathbf{i\theta\tau} + \frac{i\pi}{2}}, \end{array}
(1.3.387)

where we have em­pha­sized in bold, the −iθτ- i\theta\tau por­tion in the ex­po­nents. This rep­re­sents the forces acting on the func­tion com­po­nents eiωte^{i\omega t}, eωte^{\omega t}, each serv­ing to create torque in the com­plex func­tion­al plane. This is ap­pre­ci­at­ed by defin­ing the func­tion­al dis­tance vector 𝐫\mathbf{r}, as

𝐫=((eωt−IeIωt)e1+(Ieωt+eIωt)e2),\begin{array}{r} \mathbf{r}\mathbf{=}\left( \left( e^{\omega t} - {Ie}^{I\omega t} \right)e_{1} + \left( {Ie}^{\omega t} + e^{I\omega t} \right)e_{2} \right), \end{array}
(1.3.388)

and the force vector 𝐅\mathbf{F}, is

𝐅=(𝐅2e1+𝐅1e2)=(Re(eIθτ)e1+Im(eIθτ)e2).\begin{array}{r} \mathbf{F} = \left( \mathbf{F}_{2}e_{1} + \mathbf{F}_{1}e_{2} \right) = \left( Re\left( e^{I\theta\tau} \right)e_{1} + {Im(e}^{I\theta\tau})e_{2} \right). \end{array}
(1.3.389)

Note, we have changed the sign of the ex­po­nent for e−iθτe^{- i\theta\tau}, given the torque equa­tion has the force on the right-hand side. Using the ex­pres­sion T=𝐫∧𝐅\mathbf{\mathrm{T}}= \mathbf{r} \land \mathbf{F}, we cal­cu­late the func­tion­al geo­met­ric torque

T=𝐫∧𝐅=((eωt−IeIωt)e1+(Ieωt+eIωt)e2)∧(Re(eIθτ)e1+Im(eIθτ)e2)\mathbf{\mathrm{T}}= \mathbf{r} \land \mathbf{F} =\left( \left( e^{\omega t} - {Ie}^{I\omega t} \right)e_{1} + \left( {Ie}^{\omega t} + e^{I\omega t} \right)e_{2} \right) \land \left( Re\left( e^{I\theta\tau} \right)e_{1} + {Im(e}^{I\theta\tau})e_{2} \right)
=(eωt−IeIωt)Im(eiθτ)e1∧e2−(Ieωt+eIωt)Re(eiθτ)e1∧e2,\begin{array}{r} = \left( e^{\omega t} - {Ie}^{I\omega t} \right){Im(e}^{i\theta\tau})e_{1} \land e_{2} - \left( {Ie}^{\omega t} + e^{I\omega t} \right)Re\left( e^{i\theta\tau} \right)e_{1} \land e_{2}, \end{array}
(1.3.390)

find­ing the 𝐅1=Im(eiθτ)\mathbf{F}_{1} = {Im(e}^{i\theta\tau}) force acts on the func­tion (eωt−IeIωt)\left( e^{\omega t} - {Ie}^{I\omega t} \right), and 𝐅2=Re(eiθτ)\mathbf{F}_{2} = Re\left( e^{i\theta\tau} \right) acts on the func­tion (Ieωt+eIωt)\left( {Ie}^{\omega t} + e^{I\omega t} \right). We dis­trib­ute the re­spec­tive bivec­tors into the func­tions, find­ing that the net geo­met­ric torque equals

T=𝐫∧𝐅=(Ieωt + eIωt)Im(eIθτ)+(eωt−IeIωt)Re(eIθτ),\begin{array}{r} \mathbf{\mathrm{T}}= \mathbf{r} \land \mathbf{F} = \left( {Ie}^{\omega t}{\ + \ e}^{I\omega t} \right){Im(e}^{I\theta\tau}) + \left( e^{\omega t} - Ie^{I\omega t} \right)Re\left( e^{I\theta\tau} \right), \end{array}
(1.3.391)

rep­re­sent­ing a sum of prod­ucts, with the real force com­po­nent matched to the e1e_{1}-axis func­tion­al and the imag­i­nary force com­po­nent matched to the e2e_{2}-axis func­tion­al. We set a pref­er­ence to the e2e_{2}-axis func­tion (Ieωt + eIωt)\left( {Ie}^{\omega t}{\ + \ e}^{I\omega t} \right), to denote the geo­met­ric torque as

T=𝐫∧𝐅=(Ieωt + eIωt)(Im(eIθτ)−IRe(eIθτ)),\begin{array}{r} \mathbf{\mathrm{T}}= \mathbf{r} \land \mathbf{F} =\left( {Ie}^{\omega t}{\ + \ e}^{I\omega t} \right)\left( {Im(e}^{I\theta\tau}) - IRe\left( e^{I\theta\tau} \right) \right), \end{array}
(1.3.392)

and given that

Im(eIθτ)=sin⁡(θτ), −IRe(eIθτ)=−Icos(θτ),\begin{array}{r} {Im(e}^{I\theta\tau}) = \sin(\theta\tau),\ - IRe\left( e^{I\theta\tau} \right) = - Icos(\theta\tau), \end{array}
(1.3.393)

T\mathbf{\mathrm{T}} is equiv­a­lent to

T=(Ieωt + eIωt)(−IeIθτ),\begin{array}{r} \mathbf{\mathrm{T}} =\left( {Ie}^{\omega t}{\ + \ e}^{I\omega t} \right)\left( - Ie^{I\theta\tau} \right), \end{array}
(1.3.394)

which rep­re­sents the geo­met­ric al­ge­bra form of the uni­fied light equa­tions.

This sec­tion has served as our in­tro­duc­tion to light's geo­met­ric and dy­nam­i­cal prop­er­ties in the com­plex plane. The meth­ods re­flect­ed here will soon be ap­plied when we con­sid­er the higher-di­men­sion­al rep­re­sen­ta­tions of form.