Observer field split and exact classical wave Doppler reading

(Clause (a): laboratory components of $e$ and $b$) By [F2], $n^0={\gamma }_w$ and $n^i={\gamma }_w{w}_i\mathrm{/}c$, so with the dictionary [F1] the covariant components are $e_0=F_{00}{n}^0+F_{0i}{n}^i=-({\gamma }_w\mathrm{/}{c}^2)E\cdot w$ and $e_i=F_{i0}{n}^0+F_{ij}{n}^j=\frac{{\gamma }_w}{c}{E}_i+\frac{{\gamma }_w}{c}{ϵ}_{ijk}{w}_j{B}_k=\frac{{\gamma }_w}{c}(E_i+(w\times B{)}_i)$, using $F_{i0}=-F_{0i}=E_i\mathrm{/}c$ and $(w\times B{)}_i={ϵ}_{ijk}{w}_j{B}_k$ [F1]. For the dual, the definition $(\ast F{)}_{ab}=\frac{1}{2}{\epsilon }_{ab}{}^{cd}{F}_{cd}$ and the components of [F1] give $(\ast F{)}_{0i}=B_i$ and $(\ast F{)}_{ij}={ϵ}_{ijk}{E}_k\mathrm{/}c$: indeed ${\epsilon }_{0i}{}^{jk}={\epsilon }_{0ijk}={ϵ}_{ijk}$ because raising the two spatial indices is free and $(0,i,j,k)$ is an even permutation of $(0,1,2,3)$ exactly when $(i,j,k)$ is even, so $(\ast F{)}_{0i}=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}=\frac{1}{2}{ϵ}_{ijk}{ϵ}_{jk\mathrm{\ell }}{B}_{\mathrm{\ell }}=B_i$ with ${ϵ}_{ijk}{ϵ}_{jk\mathrm{\ell }}=2{\delta }_{i\mathrm{\ell }}$ (direct enumeration); and ${\epsilon }_{ij}{}^{0k}={\eta }^{00}{\eta }^{kk}{\epsilon }_{ij0k}=-{\epsilon }_{ij0k}=-{ϵ}_{ijk}$, since $(0,i,j,k)\mapsto (i,j,0,k)$ is an even permutation, so $(\ast F{)}_{ij}={\epsilon }_{ij}{}^{0k}{F}_{0k}=(-{ϵ}_{ijk})(-E_k\mathrm{/}c)={ϵ}_{ijk}{E}_k\mathrm{/}c$. Hence $b_0=(\ast F{)}_{00}{n}^0+(\ast F{)}_{0i}{n}^i=\frac{{\gamma }_w}{c}B\cdot w$ and $b_i=(\ast F{)}_{i0}{n}^0+(\ast F{)}_{ij}{n}^j=-{\gamma }_w{B}_i+\frac{{\gamma }_w}{c^2}{ϵ}_{ijk}{w}_j{E}_k=-{\gamma }_w(B_i-\frac{(w\times E{)}_i}{c^2})$. These are the four displayed component formulas of clause (a); in particular the laboratory temporal components $e_0,b_0$ need not vanish, so the orthogonality of the decomposition cannot be read as the vanishing of these laboratory components.

(Clause (a): annihilation, raising, reconstruction) Since $F$ and $\ast F$ are alternating, $e(n)=F_{ab}{n}^a{n}^b=0$ and $b(n)=(\ast F{)}_{ab}{n}^a{n}^b=0$; equivalently $n^a{e}_a=n^a{b}_a=0$ [F1]. For the metric-dual vectors $e^{\mathrm{♯}},b^{\mathrm{♯}}$, defined by $\eta (e^{\mathrm{♯}},v)=e(v)$ and $\eta (b^{\mathrm{♯}},v)=b(v)$ for all $v$, this gives $\eta (e^{\mathrm{♯}},n)=\eta (b^{\mathrm{♯}},n)=0$, so $e^{\mathrm{♯}},b^{\mathrm{♯}}\in n^{\perp }$; since both covectors annihilate $n$, restricting them to $n^{\perp }$ and raising with the positive rest metric of [F2] gives the same vectors, and this is the raising used in the definitions $E_n=c{e}^{\mathrm{♯}}$, $B_n=-b^{\mathrm{♯}}$. By [F1] the reconstruction $F=n^{\mathrm{♭}}\wedge e-\ast (n^{\mathrm{♭}}\wedge b)$ holds, and because the dictionary formulas of [F1] recover the six components of $F$ from $(e,b)$, the six rest components determine $F$ at the event. The measured vectors lie in the three-dimensional rest space $n^{\perp }$; they are not the laboratory coordinate triples $(e_1,e_2,e_3)$ and $(b_1,b_2,b_3)$.

(Clause (a): the observer's inertial rest frame and the observed components) At $w=0$, $B(0)$ is the identity and $n=e_0$, so the rest frame is the laboratory frame and the displayed identity is immediate without a division by $\mathrm{\mid }w\mathrm{/}c{\mathrm{\mid }}^2$. For $w\mathrm{\ne }0$, applying the matrix of [F3] to the vector $n={\gamma }_w(1,w\mathrm{/}c)$ of [F2] gives $(B(w\mathrm{/}c)n{)}^0={\gamma }_w({\gamma }_w-\frac{w}{c}\cdot \frac{{\gamma }_{w}w}{c})={\gamma }_w^2(1-\mathrm{\mid }w{\mathrm{\mid }}^2\mathrm{/}{c}^2)=1$ and $(B(w\mathrm{/}c)n{)}_{\mathrm{v}\mathrm{e}\mathrm{c}}=\frac{{\gamma }_{w}w}{c}+\frac{{\gamma }_w-1}{\mathrm{\mid }w\mathrm{/}c{\mathrm{\mid }}^2}(\frac{w}{c}\cdot \frac{{\gamma }_{w}w}{c})\frac{w}{c}-{\gamma }_w\frac{w}{c}{\gamma }_w={\gamma }_w\frac{w}{c}[1+({\gamma }_w-1)-{\gamma }_w]=0$; hence $n$ has coordinates $(1,0,0,0)$ in the chart $x^{\mathrm{\prime }}=B(w\mathrm{/}c)x+b$, that is $n=e_0^{\mathrm{\prime }}$: the chart $x^{\mathrm{\prime }}$ is the observer's inertial rest frame, and it is an admissible inertial change of description because $B(w\mathrm{/}c)$ is proper orthochronous [F3]. By clause (a′) of [F3] the components of the same $F$ read in $x^{\mathrm{\prime }}$ are the boosted components with vector parameter $v=w$: $E_{\parallel }^{\mathrm{\prime }}=E_{\parallel }$, $B_{\parallel }^{\mathrm{\prime }}=B_{\parallel }$, $E_{\perp }^{\mathrm{\prime }}={\gamma }_w(E+w\times B{)}_{\perp }$ and $B_{\perp }^{\mathrm{\prime }}={\gamma }_w(B-w\times E\mathrm{/}{c}^2{)}_{\perp }$ along $\widehat{w}$, with the displayed closed formulas, and $E^{\mathrm{\prime }}=E$, $B^{\mathrm{\prime }}=B$ when $w=0$. Reading the dictionary [F1] in the rest frame with $n=e_0^{\mathrm{\prime }}$ gives $e_0^{\mathrm{\prime }}=b_0^{\mathrm{\prime }}=0$, $e_i^{\mathrm{\prime }}=F_{i0}^{\mathrm{\prime }}=-F_{0i}^{\mathrm{\prime }}=E_i^{\mathrm{\prime }}\mathrm{/}c$ and $b_i^{\mathrm{\prime }}=(\ast F^{\mathrm{\prime }}{)}_{i0}=-(\ast F^{\mathrm{\prime }}{)}_{0i}=-B_i^{\mathrm{\prime }}$; hence the vectors $E_n=c{e}^{\mathrm{♯}}$ and $B_n=-b^{\mathrm{♯}}$ have rest-frame components $(0,E^{\mathrm{\prime }})$ and $(0,B^{\mathrm{\prime }})$. Under the adopted model assumption [A1] the observed electric and magnetic fields in the rest frame are $E^{\mathrm{\prime }}$ and $B^{\mathrm{\prime }}$, that is: the components along $w$ are $E_{\parallel }$ and $B_{\parallel }$, the components perpendicular to $w$ are ${\gamma }_w(E+w\times B{)}_{\perp }$ and ${\gamma }_w(B-w\times E\mathrm{/}{c}^2{)}_{\perp }$, and at $w=0$ the observer is the laboratory rest observer reading $E,B$ themselves with no distinguished direction of motion. The identification of the mathematical vectors with apparatus readings is exactly [A1] and is assumed, not proved.

(Clause (b): the phase covector, nonzero, future and null) By [F4] the wavevector is $\mathbf{k}=Ks$ with $\mathrm{\mid }\mathbf{k}\mathrm{\mid }=K=\omega \mathrm{/}c$ and propagation direction $\widehat{\mathbf{k}}=s$, and the phase is $\theta (x,t)=\mathbf{k}\cdot x-\omega t$. Its coordinate derivative components are $k_0=\mathrm{\partial }\theta \mathrm{/}\mathrm{\partial }{x}^0=c^{-1}{\mathrm{\partial }}_t\theta =-\omega \mathrm{/}c$ and $k_i=\mathrm{\partial }\theta \mathrm{/}\mathrm{\partial }{x}^i=(\omega \mathrm{/}c)s_i$ (Directional derivatives and partial derivatives of a map $U\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^n$ with $x^0=ct$), so the phase covector is $k_a=(-\omega \mathrm{/}c,(\omega \mathrm{/}c)s_1,(\omega \mathrm{/}c)s_2,(\omega \mathrm{/}c)s_3)$ and $\theta (p)=k_a{x}^a(p)$ up to an additive constant. Raising with $\eta$ gives $k^0=-k_0=\omega \mathrm{/}c>0$ and $k_{\mathrm{v}\mathrm{e}\mathrm{c}}=(k_1,k_2,k_3)=(\omega \mathrm{/}c)s$, so $\eta (k^{\mathrm{♯}},k^{\mathrm{♯}})=-(\omega \mathrm{/}c{)}^2+(\omega \mathrm{/}c{)}^2\mathrm{\mid }s{\mathrm{\mid }}^2=0$ and $k^{\mathrm{♯}}\mathrm{\ne }0$: the phase covector is nonzero, future ($k^0>0$) and null. By the amplitude equations of [F4] the fields are a source-free smooth Maxwell field on the whole-space region.

(Clause (b): exact frequency reading) At the reception event the ideal observer of [F5] with four-velocity $U_o=cn$ reads ${\omega }_n=-ck(n)=-k(U_o)$. With $k(n)=k_a{n}^a=-\frac{\omega }{c}{\gamma }_w+\frac{\omega }{c}\frac{w\cdot s}{c}{\gamma }_w=-\frac{\omega }{c}{\gamma }_w(1-\frac{w\cdot s}{c})$ from [F2] and step 1.4, this gives ${\omega }_n={\gamma }_w(1-\frac{w\cdot s}{c})\omega >0$, where the positivity is clause (ii) of [F6], whose formula this is with the frame-$x$ data $(\omega ,s)$ of [F4] and the observer velocity $w$; its hypotheses hold by step 1.4 and the adopted model postulates [F5], [F6]. At $w=0$ one has ${\gamma }_w=1$ and ${\omega }_n=\omega$, and the reader is the laboratory rest observer of clause (a) reading $E,B$.

(Clause (b): aberration, longitudinal and transverse protocols) The data $(\omega ,s)$ of [F4] are exactly the frame-$x$ data of the Doppler theorem [F6] for its future null wave covector (step 1.4), and the ideal plane-wave, ideal-clock and covariance postulates it adopts are those of the framework [F5], [F8]. Hence its clauses (i)-(v) apply verbatim to this wave: (i) in a chart $x^{\mathrm{\prime }}=B(\beta )x+b$ the frame data $({\omega }^{\mathrm{\prime }},s^{\mathrm{\prime }})$ satisfy ${\omega }^{\mathrm{\prime }}=\gamma (1-\beta \cdot s)\omega >0$ with the displayed $s^{\mathrm{\prime }}$ satisfying $\mathrm{\mid }{s}^{\mathrm{\prime }}\mathrm{\mid }=1$ (the direction transformation, i.e. aberration); (iii) $s=\pm \widehat{\beta }$ give the longitudinal red- and blue-shift factors $\omega \sqrt{(1\mp \mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}(1\pm \mathrm{\mid }\beta \mathrm{\mid })}$ with the stated wavelength ratios; (iv) $\beta \cdot s=0$ gives ${\omega }^{\mathrm{\prime }}=\gamma \omega$ and $\beta \cdot s=\mathrm{\mid }\beta {\mathrm{\mid }}^2$ gives ${\omega }^{\mathrm{\prime }}=\omega \mathrm{/}\gamma$, the two transverse protocols, and the two factors differ when $\beta \mathrm{\ne }0$; (v) the aberration relation $\cos {\theta }_{x^{\mathrm{\prime }}}=(\cos {\theta }_x-\mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}(1-\mathrm{\mid }\beta \mathrm{\mid }\cos {\theta }_x)$ with the map a strictly increasing bijection of $[-1,1]$ onto itself and $\cos {\theta }_x=0\Rightarrow \cos {\theta }_{x^{\mathrm{\prime }}}=-\mathrm{\mid }\beta \mathrm{\mid }$. For $\beta =0$ clause (i) gives ${\omega }^{\mathrm{\prime }}=\omega$ and $s^{\mathrm{\prime }}=s$, so frequency and direction are unchanged and no aberration angle is defined; each formula carries its nonzero-motion qualification.

(Clause (b): composition of successive inertial changes) Let $x_1={\mathrm{\Lambda }}_1{x}_0+b_1$ and $x_2={\mathrm{\Lambda }}_2{x}_1+b_2$ be successive inertial changes with proper orthochronous ${\mathrm{\Lambda }}_1,{\mathrm{\Lambda }}_2$. The covector law [F7] gives $k_{1,a}=({\mathrm{\Lambda }}_1^{-1}{)}^b{}_a{k}_{0,b}$ and $k_{2,a}=({\mathrm{\Lambda }}_2^{-1}{)}^b{}_a{k}_{1,b}=(({\mathrm{\Lambda }}_2{\mathrm{\Lambda }}_1{)}^{-1}{)}^b{}_a{k}_{0,b}$, with the affine phase constants transformed along with the translations. To compute the wave data for a general stage $x_j={\mathrm{\Lambda }}_j{x}_{j-1}+b_j$, factor ${\mathrm{\Lambda }}_j=B(-{\alpha }_j){\widehat{R}}_j$ with $\mathrm{\mid }{\alpha }_j\mathrm{\mid }<1$ and ${\widehat{R}}_j=\mathrm{diag}(1,R_j)$ a spatial rotation, by [F9]. The intermediate rotation leaves $k_0=-{\omega }_{j-1}\mathrm{/}c$ unchanged and sends the spatial covector triple to $R_j{k}_{\mathrm{v}\mathrm{e}\mathrm{c}}$, since $R_j^{-1}=R_j^{\mathsf{T}}$; hence it leaves the frequency unchanged and sends $s_{j-1}$ to $R_j{s}_{j-1}$. Apply [F6] to the remaining boost with parameter ${\beta }_j=-{\alpha }_j$ and ${\gamma }_j:=(1-\mathrm{\mid }{\beta }_j{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}$: ${\omega }_j={\gamma }_j(1-{\beta }_j\cdot R_j{s}_{j-1}){\omega }_{j-1}>0$, and $s_j$ is the aberration formula of [F6] applied to $R_j{s}_{j-1}$. This includes zero boosts and pure rotations. Therefore ${\omega }_2\mathrm{/}{\omega }_0=({\omega }_2\mathrm{/}{\omega }_1)({\omega }_1\mathrm{/}{\omega }_0)$, with the direction updated at every stage. This proves the composition claim for general inertial changes.

(Clause (c): status, scope and hypotheses) Clause (a) is steps 1.1-1.3: the four component formulas, the annihilation $e(n)=b(n)=0$, the raising into $n^{\perp }$, the reconstruction determining $F$ from $(e,b)$ and the identification of the observed rest-frame fields are established there, the last only under the adopted model assumption [A1]. Clause (b) is steps 1.4, 2.1, 2.2 and 2.3: the phase covector is nonzero, future and null; the frequency reading is ${\omega }_n={\gamma }_w(1-w\cdot s\mathrm{/}c)\omega >0$; the aberration, longitudinal and transverse conclusions are clauses (i)-(v) of [F6]; and successive inertial changes compose as stated. Clause (c) records that the split and reconstruction are pointwise for every future unit observer $n$, whereas the reading and aberration formulas were derived for an observer whose coordinate velocity is that of an inertial rest frame — an accelerating observer requires its own analysis, since no single rest frame exists and differentiating the decomposition adds derivatives of $n$ — and that the identification of $(E_n,B_n)$ and ${\omega }_n$ with apparatus readings is the adopted assumption [A1] together with the ideal-plane-wave and ideal-clock and ideal-light models [F5], [F8]. No calibration, apparatus, statistical or empirical claim is made, no photon model and no identification of frequency with energy by a factor $\mathrm{\hslash }$ is used, and media, dispersion, modulation, boundaries and finite line widths are excluded. Only the dictionary [F1], the observer data [F2], the boost [F3], the plane-wave and Doppler data [F4], [F6], the reading rule [F5], the covector law [F7], the ideal clock and light assumptions [F8] and the adopted model assumption [A1] were used; no choice principle is used.