Doppler shift and aberration for an ideal plane wave between inertial descriptions

Setup, and the relative velocity of the frames. By [L1] the frame-$x$ data of $k$ is $(\varpi ,s)$ with $\varpi =\omega \mathrm{/}c>0$. By the frame relation of [L5], $e_0^{\mathrm{\prime }}={\sum }_b({\mathrm{\Lambda }}^{-1}{)}^b{}_0{e}_b$ is the first column of $B(-\beta )$, namely $(\gamma ,\gamma {\beta }_1,\gamma {\beta }_2,\gamma {\beta }_3)$ by [L3], so $e_0^{\mathrm{\prime }}=\gamma (e_0+{\sum }_i{\beta }_i{e}_i)$. Then $\eta (e_0,e_0^{\mathrm{\prime }})=\gamma \eta (e_0,e_0)=-\gamma$ and $h(e_0^{\mathrm{\prime }})=e_0^{\mathrm{\prime }}+\eta (e_0,e_0^{\mathrm{\prime }})e_0=\gamma e_0+\gamma {\sum }_i{\beta }_i{e}_i-\gamma e_0=\gamma {\sum }_i{\beta }_i{e}_i$; by [L6] with $n=e_0$ and $u=e_0^{\mathrm{\prime }}$, the velocity of the $x^{\mathrm{\prime }}$-rest observer relative to the $x$-rest observer is $ch(e_0^{\mathrm{\prime }})\mathrm{/}(-\eta (e_0,e_0^{\mathrm{\prime }}))=c\gamma {\sum }_i{\beta }_i{e}_i\mathrm{/}\gamma =c\beta$, of magnitude $c\mathrm{\mid }\beta \mathrm{\mid }$ in the direction ${\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}$ when $\beta \mathrm{\ne }0$.

Clause (i). By the covector component law [L5] and [L3], $k_a^{\mathrm{\prime }}=(B(-\beta ){)}^b{}_a{k}_b$; with $k_0=-\omega \mathrm{/}c$ and $k_i=(\omega \mathrm{/}c)s_i$ from [L1] and the matrix entries of $B(-\beta )$, this gives $k_0^{\mathrm{\prime }}=\gamma (k_0+\beta \cdot k_{\mathrm{v}\mathrm{e}\mathrm{c}})=-\gamma (1-\beta \cdot s)\omega \mathrm{/}c$, with $1-\beta \cdot s>0$ by the positivity clause of [L2], and $k_i^{\mathrm{\prime }}=k_i+\gamma {\beta }_i{k}_0+\frac{\gamma -1}{\mathrm{\mid }\beta {\mathrm{\mid }}^2}{\beta }_i(\beta \cdot k_{\mathrm{v}\mathrm{e}\mathrm{c}})=\frac{\omega }{c}[s_i-\gamma {\beta }_i+\frac{\gamma -1}{\mathrm{\mid }\beta {\mathrm{\mid }}^2}(\beta \cdot s){\beta }_i]$ for $\beta \mathrm{\ne }0$ (and $k^{\mathrm{\prime }}=k$ for $\beta =0$). By the frame-data lemma [L2] the frame-$x^{\mathrm{\prime }}$ data is $({\varpi }^{\mathrm{\prime }},s^{\mathrm{\prime }})$ with ${\varpi }^{\mathrm{\prime }}=\gamma (1-\beta \cdot s)\varpi =\gamma (1-\beta \cdot s)\omega \mathrm{/}c$, $s^{\mathrm{\prime }}=[s+\frac{\gamma -1}{\mathrm{\mid }\beta {\mathrm{\mid }}^2}(\beta \cdot s)\beta -\gamma \beta ]\mathrm{/}[\gamma (1-\beta \cdot s)]$, $\mathrm{\mid }{s}^{\mathrm{\prime }}\mathrm{\mid }=1$ and $k_a^{\mathrm{\prime }}=(-{\varpi }^{\mathrm{\prime }},{\varpi }^{\mathrm{\prime }}{s}^{\mathrm{\prime }})$; in wave data this is ${\omega }^{\mathrm{\prime }}=c{\varpi }^{\mathrm{\prime }}=\gamma (1-\beta \cdot s)\omega >0$ and the displayed $s^{\mathrm{\prime }}$, and the axis case is the axis clause of [L2].

Clause (ii). By [L1] the rest observer of $x$ has four-velocity $c{e}_0$ and reads ${\omega }_{e_0}=-ck(e_0)=c\varpi =\omega$, and the rest observer of $x^{\mathrm{\prime }}$ has four-velocity $c{e}_0^{\mathrm{\prime }}$ by [L4] and reads ${\omega }_{e_0^{\mathrm{\prime }}}=-ck(e_0^{\mathrm{\prime }})=c{\varpi }^{\mathrm{\prime }}={\omega }^{\mathrm{\prime }}=\gamma (1-\beta \cdot s)\omega$ by step 2.1, so ${\nu }^{\mathrm{\prime }}={\omega }^{\mathrm{\prime }}\mathrm{/}(2\pi )=\gamma (1-\beta \cdot s)\nu$ and ${\lambda }^{\mathrm{\prime }}=c\mathrm{/}{\nu }^{\mathrm{\prime }}=\lambda \mathrm{/}[\gamma (1-\beta \cdot s)]$ by the relations $\nu =\omega \mathrm{/}(2\pi )$, $\lambda =c\mathrm{/}\nu$ of [L1]. For the general observer, [L6] gives $U_o={\gamma }_w(c,w_{\mathrm{v}\mathrm{e}\mathrm{c}})$, so $n=U_o\mathrm{/}c={\gamma }_w(1,w\mathrm{/}c)$ is future unit timelike and, using linearity of $k$ and $k(e_0)=-\omega \mathrm{/}c$, $k(e_i)=(\omega \mathrm{/}c)s_i$ [L1], ${\omega }_o=-ck(n)=-c{\gamma }_w[k(e_0)+{\sum }_i(w_i\mathrm{/}c)k(e_i)]={\gamma }_w\omega (1-(w\cdot s)\mathrm{/}c)$; with ${\beta }_w=w\mathrm{/}c$ this is ${\gamma }_w(1-{\beta }_w\cdot s)\omega$.

Clause (iii). Assume $\beta \mathrm{\ne }0$. If $s={\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}$ then $\beta \cdot s=\mathrm{\mid }\beta \mathrm{\mid }$ by [L7], so step 2.1 together with $\omega =c\varpi$ and [L2] gives ${\omega }^{\mathrm{\prime }}=\gamma (1-\mathrm{\mid }\beta \mathrm{\mid })\omega$, and since $0\le \mathrm{\mid }\beta \mathrm{\mid }<1$, $\gamma (1-\mathrm{\mid }\beta \mathrm{\mid })=(1-\mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}\sqrt{1-\mathrm{\mid }\beta {\mathrm{\mid }}^2}=\sqrt{(1-\mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}(1+\mathrm{\mid }\beta \mathrm{\mid })}$, which is $<1$ for $\mathrm{\mid }\beta \mathrm{\mid }>0$ and $=1$ for $\beta =0$; substituting $s={\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}$ in the formula of [L2] gives $s^{\mathrm{\prime }}={\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}$, so the propagation direction relative to the boost axis is unchanged. The wavelength and frequency ratios are ${\nu }^{\mathrm{\prime }}={\omega }^{\mathrm{\prime }}\mathrm{/}(2\pi )=\gamma (1-\mathrm{\mid }\beta \mathrm{\mid })\nu$ and ${\lambda }^{\mathrm{\prime }}=c\mathrm{/}{\nu }^{\mathrm{\prime }}=\lambda \mathrm{/}[\gamma (1-\mathrm{\mid }\beta \mathrm{\mid })]=\lambda \sqrt{(1+\mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}(1-\mathrm{\mid }\beta \mathrm{\mid })}$ by $\nu =\omega \mathrm{/}(2\pi )$, $\lambda =c\mathrm{/}\nu$ [L1] and ${\omega }^{\mathrm{\prime }}=\gamma (1-\mathrm{\mid }\beta \mathrm{\mid })\omega$. If $s=-{\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}$ the same computation with $-{\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}$ gives ${\omega }^{\mathrm{\prime }}=\gamma (1+\mathrm{\mid }\beta \mathrm{\mid })\omega =\omega \sqrt{(1+\mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}(1-\mathrm{\mid }\beta \mathrm{\mid })}>\omega$ for $\mathrm{\mid }\beta \mathrm{\mid }>0$, $s^{\mathrm{\prime }}=-{\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}$ and ${\lambda }^{\mathrm{\prime }}=\lambda \sqrt{(1-\mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}(1+\mathrm{\mid }\beta \mathrm{\mid })}$; the source language in the statement is an interpretation of these factors, the model carrying no source.

Clause (iv). If $\beta \cdot s=0$, step 2.1 gives ${\omega }^{\mathrm{\prime }}=\gamma \omega$, which is greater than $\omega$ exactly when $\gamma >1$, that is exactly when $\beta \mathrm{\ne }0$, and equals $\omega$ when $\beta =0$, by [L3]. If $\beta \cdot s=\mathrm{\mid }\beta {\mathrm{\mid }}^2$, then by [L2] $\beta \cdot s^{\mathrm{\prime }}=0$, and for $\beta \mathrm{\ne }0$ this is equivalent to $s^{\mathrm{\prime }}\cdot {\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}=(\beta \cdot s^{\mathrm{\prime }})\mathrm{/}\mathrm{\mid }\beta \mathrm{\mid }=0$; conversely $s^{\mathrm{\prime }}\cdot {\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}=0$ with $\beta \mathrm{\ne }0$ gives $\beta \cdot s^{\mathrm{\prime }}=0$ and hence $\beta \cdot s=\mathrm{\mid }\beta {\mathrm{\mid }}^2$ by [L2]. With $\beta \cdot s=\mathrm{\mid }\beta {\mathrm{\mid }}^2$ step 2.1 gives ${\omega }^{\mathrm{\prime }}=\gamma (1-\mathrm{\mid }\beta {\mathrm{\mid }}^2)\omega =\omega \mathrm{/}\gamma$, which is less than $\omega$ when $\beta \mathrm{\ne }0$ and equals $\omega$ when $\beta =0$.

Clause (v). For $\beta \mathrm{\ne }0$ put $u:=\cos {\theta }_x={\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}\cdot s=(\beta \cdot s)\mathrm{/}\mathrm{\mid }\beta \mathrm{\mid }$ and $u^{\mathrm{\prime }}:=\cos {\theta }_{x^{\mathrm{\prime }}}={\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}\cdot s^{\mathrm{\prime }}=(\beta \cdot s^{\mathrm{\prime }})\mathrm{/}\mathrm{\mid }\beta \mathrm{\mid }$ by [L7]; by the formula for $s^{\mathrm{\prime }}$ of step 2.1, equivalently by [L2], $u^{\mathrm{\prime }}=(\beta \cdot s-\mathrm{\mid }\beta {\mathrm{\mid }}^2)\mathrm{/}[\mathrm{\mid }\beta \mathrm{\mid }(1-\beta \cdot s)]=(u-\mathrm{\mid }\beta \mathrm{\mid })\mathrm{/}(1-\mathrm{\mid }\beta \mathrm{\mid }u)$. Write $b:=\mathrm{\mid }\beta \mathrm{\mid }$ and $f(u):=(u-b)\mathrm{/}(1-bu)$; since $0\le b<1$ one has $1-bu\ge 1-b>0$ for every $u\in [-1,1]$, so $f$ is defined on $[-1,1]$. Moreover $1-bu-(u-b)=(1+b)(1-u)\ge 0$ and $1-bu+(u-b)=(1-b)(1+u)\ge 0$, hence $-1\le f(u)\le 1$. The map $g(y):=(y+b)\mathrm{/}(1+by)$ is likewise defined on $[-1,1]$, and the comparisons $1+by-(y+b)=(1-b)(1-y)\ge 0$ and $1+by+(y+b)=(1+b)(1+y)\ge 0$ show $-1\le g(y)\le 1$, so $g$ maps $[-1,1]$ into itself; direct expansion of the rational expressions gives $f(g(y))=[(y+b)-b(1+by)]\mathrm{/}[(1+by)-b(y+b)]=[y(1-b^2)]\mathrm{/}(1-b^2)=y$ and symmetrically $g(f(u))=u$ for all $u,y\in [-1,1]$, so $f$ is a bijection of $[-1,1]$ onto itself with inverse $g$. For $v>u$ in $[-1,1]$ one has $f(v)-f(u)=[(v-b)(1-bu)-(u-b)(1-bv)]\mathrm{/}[(1-bv)(1-bu)]=(v-u)(1-b^2)\mathrm{/}[(1-bv)(1-bu)]>0$, so $f$ is strictly increasing; in particular $f$ is the stated strictly increasing bijection of $[-1,1]$ onto itself, with $f(-1)=-1$ and $f(1)=1$, and $u=0$ gives $u^{\mathrm{\prime }}=f(0)=-b=-\mathrm{\mid }\beta \mathrm{\mid }$.

Clause (vi), $\beta =0$, and scope. If $\beta =0$ then $B(0)=\mathrm{id}$ by [L3], so $x^{\mathrm{\prime }}=x+b$ is a pure translation; by [L2] the frame data agree, ${\varpi }^{\mathrm{\prime }}=\varpi$ and $s^{\mathrm{\prime }}=s$, whence ${\omega }^{\mathrm{\prime }}=\omega$ and the x- and x'-inertial axes carry the same wave data, while the affine phase constant changes as follows: since $\beta =0$ gives $x^{\mathrm{\prime }a}=x^a+b^a$ and [L5] gives $k^{\mathrm{\prime }}=k$, comparing the two chart forms $\theta (p)=k_a{x}^a(p)+\theta (o)$ and $\theta (p)=k_a^{\mathrm{\prime }}{x}^{\mathrm{\prime }a}(p)+\theta (o^{\mathrm{\prime }})$ of the affine phase yields $\theta (o^{\mathrm{\prime }})=\theta (o)-k_a^{\mathrm{\prime }}{b}^a$. All clauses (i)-(v) were derived from the adopted postulates [A1], [A2], [A3] and the cited mathematical lemmas, are conditional on the idealizations recorded in [A1]-[A3] and the scope clause, and make no measurement, statistical or empirical claim; no choice principle was used.