Light and Doppler — Examples

Setup. Here ${\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}=e_1$, $\mathrm{\mid }\beta \mathrm{\mid }=3\mathrm{/}5$ and $\gamma =(1-9\mathrm{/}25{)}^{-1\mathrm{/}2}=(16\mathrm{/}25{)}^{-1\mathrm{/}2}=5\mathrm{/}4$ by [L3]; both witness directions $s=\pm e_1$ have $\mathrm{\mid }s\mathrm{\mid }=1$, so the wave is in the model [A1].

Case (a), $s=e_1$. Then $\beta \cdot s=3\mathrm{/}5$, so [L1] gives ${\omega }^{\mathrm{\prime }}=\gamma (1-3\mathrm{/}5)\omega =(5\mathrm{/}4)(2\mathrm{/}5)\omega =\omega \mathrm{/}2$ and hence ${\nu }^{\mathrm{\prime }}={\omega }^{\mathrm{\prime }}\mathrm{/}(2\pi )=\nu \mathrm{/}2$ and ${\lambda }^{\mathrm{\prime }}=c\mathrm{/}{\nu }^{\mathrm{\prime }}=2c\mathrm{/}\nu =2\lambda$; the axis formulas of [L1] with $b=3\mathrm{/}5$, $s_1=1$, $s_2=s_3=0$ give $s_1^{\mathrm{\prime }}=(1-3\mathrm{/}5)\mathrm{/}(1-3\mathrm{/}5)=1$ and $s_2^{\mathrm{\prime }}=s_3^{\mathrm{\prime }}=0$, so $s^{\mathrm{\prime }}=e_1$; the frequency factor agrees with the longitudinal clause [L2], ${\omega }^{\mathrm{\prime }}\mathrm{/}\omega =\sqrt{(1-3\mathrm{/}5)\mathrm{/}(1+3\mathrm{/}5)}=\sqrt{(2\mathrm{/}5)\mathrm{/}(8\mathrm{/}5)}=1\mathrm{/}2$, since $s={\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}$.

Case (b), $s=-e_1$. Then $\beta \cdot s=-3\mathrm{/}5$, so [L1] gives ${\omega }^{\mathrm{\prime }}=\gamma (1+3\mathrm{/}5)\omega =(5\mathrm{/}4)(8\mathrm{/}5)\omega =2\omega$, ${\nu }^{\mathrm{\prime }}=2\nu$ and ${\lambda }^{\mathrm{\prime }}=c\mathrm{/}{\nu }^{\mathrm{\prime }}=\lambda \mathrm{/}2$; the axis formulas of [L1] with $b=3\mathrm{/}5$, $s_1=-1$ give $s_1^{\mathrm{\prime }}=(-1-3\mathrm{/}5)\mathrm{/}(1+3\mathrm{/}5)=-1$ and $s_2^{\mathrm{\prime }}=s_3^{\mathrm{\prime }}=0$, so $s^{\mathrm{\prime }}=-e_1$; the frequency factor agrees with the longitudinal clause [L2], ${\omega }^{\mathrm{\prime }}\mathrm{/}\omega =\sqrt{(1+3\mathrm{/}5)\mathrm{/}(1-3\mathrm{/}5)}=\sqrt{(8\mathrm{/}5)\mathrm{/}(2\mathrm{/}5)}=2$, since $s=-{\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}$. In both cases ${\omega }^{\mathrm{\prime }}>0$ and $\mathrm{\mid }{s}^{\mathrm{\prime }}\mathrm{\mid }=1$ as the theorem asserts. This proves (a) and (b) by exact rational arithmetic; the computation is conditional on the ideal plane-wave model, no measurement is performed and no empirical claim is made; no choice principle is used.

Case (a). Here $s=e_2$ has $\mathrm{\mid }s\mathrm{\mid }=1$ and $\beta \cdot s=(3\mathrm{/}5)(0)=0$, so the propagation direction is perpendicular to the relative velocity in $x$, and [L2] gives ${\omega }^{\mathrm{\prime }}=\gamma \omega =(5\mathrm{/}4)\omega$, which is greater than $\omega$ because $\gamma =5\mathrm{/}4>1$ by [L3]; the axis formulas of [L1] with $b=3\mathrm{/}5$, $s_1=0$, $s_2=1$, $s_3=0$ give $s_1^{\mathrm{\prime }}=(0-3\mathrm{/}5)\mathrm{/}(1-0)=-3\mathrm{/}5$, $s_2^{\mathrm{\prime }}=1\mathrm{/}[(5\mathrm{/}4)(1-0)]=4\mathrm{/}5$ and $s_3^{\mathrm{\prime }}=0$, and $\mathrm{\mid }{s}^{\mathrm{\prime }}{\mathrm{\mid }}^2=9\mathrm{/}25+16\mathrm{/}25=1$.

Case (b). Here $s=(3\mathrm{/}5,4\mathrm{/}5,0)$ has $\mathrm{\mid }s{\mathrm{\mid }}^2=9\mathrm{/}25+16\mathrm{/}25=1$ and $\beta \cdot s=(3\mathrm{/}5)(3\mathrm{/}5)=9\mathrm{/}25=\mathrm{\mid }\beta {\mathrm{\mid }}^2$; by [L2] this is equivalent to $s^{\mathrm{\prime }}\cdot {\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}=0$ with ${\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}=e_1$, the propagation direction is perpendicular to the relative velocity in $x^{\mathrm{\prime }}$, and ${\omega }^{\mathrm{\prime }}=\gamma (1-9\mathrm{/}25)\omega =(5\mathrm{/}4)(16\mathrm{/}25)\omega =(4\mathrm{/}5)\omega =\omega \mathrm{/}\gamma$, which is less than $\omega$; the axis formulas of [L1] give $s_1^{\mathrm{\prime }}=(3\mathrm{/}5-3\mathrm{/}5)\mathrm{/}(1-9\mathrm{/}25)=0$, $s_2^{\mathrm{\prime }}=(4\mathrm{/}5)\mathrm{/}[(5\mathrm{/}4)(16\mathrm{/}25)]=(4\mathrm{/}5)\mathrm{/}(4\mathrm{/}5)=1$, $s_3^{\mathrm{\prime }}=0$, so $s^{\mathrm{\prime }}=e_2$.

Conclusion. In case (a) the shift is by $5\mathrm{/}4$, in case (b) by $4\mathrm{/}5$; both configurations are called transverse because the propagation direction is perpendicular to the relative velocity in one of the two frames, and the two factors are different and neither is $1$ for the stated nonzero $\beta$. Hence a transverse Doppler statement is incomplete until the frame in which transversality holds is specified. The computation is conditional on the ideal plane-wave model and makes no measurement or empirical claim; no choice principle is used.

The transformed direction. Here $s=(4\mathrm{/}5,3\mathrm{/}5,0)$ has $\mathrm{\mid }s{\mathrm{\mid }}^2=16\mathrm{/}25+9\mathrm{/}25=1$ and $s_1=4\mathrm{/}5$; the axis formulas of [L1] with $b=3\mathrm{/}5$ and $\gamma =5\mathrm{/}4$ [L3] give $s_1^{\mathrm{\prime }}=(4\mathrm{/}5-3\mathrm{/}5)\mathrm{/}(1-(3\mathrm{/}5)(4\mathrm{/}5))=(1\mathrm{/}5)\mathrm{/}(13\mathrm{/}25)=5\mathrm{/}13$, $s_2^{\mathrm{\prime }}=(3\mathrm{/}5)\mathrm{/}[(5\mathrm{/}4)(13\mathrm{/}25)]=(3\mathrm{/}5)\mathrm{/}(13\mathrm{/}20)=12\mathrm{/}13$ and $s_3^{\mathrm{\prime }}=0$, so $s^{\mathrm{\prime }}=(5\mathrm{/}13,12\mathrm{/}13,0)$ with $\mathrm{\mid }{s}^{\mathrm{\prime }}{\mathrm{\mid }}^2=25\mathrm{/}169+144\mathrm{/}169=1$; the same value follows from the aberration clause [L2], $\cos {\theta }^{\mathrm{\prime }}=(4\mathrm{/}5-3\mathrm{/}5)\mathrm{/}(1-(3\mathrm{/}5)(4\mathrm{/}5))=(1\mathrm{/}5)\mathrm{/}(13\mathrm{/}25)=5\mathrm{/}13$.

Frequency and wavelength. By [L1] with $\beta \cdot s=(3\mathrm{/}5)(4\mathrm{/}5)=12\mathrm{/}25$: ${\omega }^{\mathrm{\prime }}\mathrm{/}\omega =\gamma (1-12\mathrm{/}25)=(5\mathrm{/}4)(13\mathrm{/}25)=13\mathrm{/}20$, so ${\nu }^{\mathrm{\prime }}={\omega }^{\mathrm{\prime }}\mathrm{/}(2\pi )=(13\mathrm{/}20)\nu$ and ${\lambda }^{\mathrm{\prime }}=c\mathrm{/}{\nu }^{\mathrm{\prime }}=(20\mathrm{/}13)\lambda$ by [A1].

The apparent direction. By the incoming-direction convention in the Example, the apparent source direction is $-s$ in $x$ and $-s^{\mathrm{\prime }}$ in $x^{\mathrm{\prime }}$; its components along the boost axis ${\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}=e_1$ are $-s_1=-4\mathrm{/}5$ and $-s_1^{\mathrm{\prime }}=-5\mathrm{/}13$ by step 1.1. Since $5\mathrm{/}13<4\mathrm{/}5$ (as $25<52$), also $-5\mathrm{/}13>-4\mathrm{/}5$: the axis component of the apparent direction increases, so the apparent direction moves toward $+e_1$, the direction of motion of the frame $x^{\mathrm{\prime }}$ relative to $x$, which is the sign of stellar aberration in this configuration. The computation is conditional on the ideal plane-wave model, records the exact direction transformation, and makes no measurement or empirical claim; no choice principle is used.

Clause (a). By [L1], $k^{\mathrm{♯}}=\varpi (e_0+{\sum }_i{s}_i{e}_i)$ with $\varpi >0$ and $\mathrm{\mid }s\mathrm{\mid }=1$, so $\eta (k^{\mathrm{♯}},k^{\mathrm{♯}})={\varpi }^2(-1+{\sum }_i{s}_i^2)={\varpi }^2(\mathrm{\mid }s{\mathrm{\mid }}^2-1)=0$ by [L1, L3], and $k(k^{\mathrm{♯}})=\eta (k^{\mathrm{♯}},k^{\mathrm{♯}})$ by [L1]. For every $p\in A$ and $\lambda \in \mathbb{R}$ the defining identity of the phase gives $\theta (p+\lambda k^{\mathrm{♯}})=\theta (p)+\lambda k(k^{\mathrm{♯}})=\theta (p)$, so $\theta$ is constant on the lines $p+\mathbb{R}{k}^{\mathrm{♯}}$.

Clause (b), kernel and sign. By [L2] and $\varpi >0$, $v\in \ker k$ is equivalent to $v^0=s\cdot v_{\mathrm{v}\mathrm{e}\mathrm{c}}$, and $\ker k$ is a $3$-dimensional subspace [L2]. For $v\in \ker k$, substituting $v^0=s\cdot v_{\mathrm{v}\mathrm{e}\mathrm{c}}$ into [L3] gives $\eta (v,v)=-(s\cdot v_{\mathrm{v}\mathrm{e}\mathrm{c}}{)}^2+\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mathrm{\mid }}^2$; by [L4] with $\mathrm{\mid }s\mathrm{\mid }=1$ this is $\ge 0$, and it is $0$ exactly when $\mathrm{\mid }s\cdot v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }=\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }$, which by the equality clause of [L4] holds exactly when $v_{\mathrm{v}\mathrm{e}\mathrm{c}}=\lambda s$ for some real $\lambda$. For such a $v$, $v^0=s\cdot v_{\mathrm{v}\mathrm{e}\mathrm{c}}=\lambda$, so $v=\lambda (e_0+{\sum }_i{s}_i{e}_i)=\lambda {\varpi }^{-1}{k}^{\mathrm{♯}}$ lies in $\mathrm{span}(k^{\mathrm{♯}})$; conversely every multiple of $k^{\mathrm{♯}}$ lies in $\ker k$ because $k(k^{\mathrm{♯}})=0$ by step 1.1. For every $w\in \ker k$ one has $\eta (k^{\mathrm{♯}},w)=k(w)=0$, so this line lies in the radical; conversely a radical vector has zero self-pairing and thus lies in this line by the equality calculation. Hence $\eta {\mathrm{\mid }}_{\ker k}$ is positive semidefinite, its radical is $\mathrm{span}(k^{\mathrm{♯}})$, and its only null elements are the nonzero multiples of $k^{\mathrm{♯}}$.

Clause (b), null directions. By [L1] and [L6] the vector $k^{\mathrm{♯}}$ is future null and $-k^{\mathrm{♯}}$ is past null, so the nonzero multiples of $k^{\mathrm{♯}}$ are exactly the two rays ${\mathbb{R}}_{>0}{k}^{\mathrm{♯}}$ (future null) and ${\mathbb{R}}_{>0}(-k^{\mathrm{♯}})$ (past null) of the line $\mathrm{span}(k^{\mathrm{♯}})$; combined with step 2.1 and [L5], the line through $k^{\mathrm{♯}}$ is the only null direction contained in $\ker k$.

Clause (c). By [L2] and step 2.1 the kernel of the nonzero functional $k$ is the $3$-dimensional subspace $\ker k$, and the level set $\mathrm{\Sigma }=\{p:\theta (p)=\theta (p_0)\}$ is the affine hyperplane $p_0+\ker k$ through $p_0$; its direction subspace is $\ker k$, whose normal covector is $k$ (the linear functional defining the direction subspace), and $k$ is null by hypothesis. Since $k^{\mathrm{♯}}\in \ker k$ by step 1.1, both rays $p_0+{\mathbb{R}}_{\ge 0}{k}^{\mathrm{♯}}$ and $p_0+{\mathbb{R}}_{\le 0}{k}^{\mathrm{♯}}$ lie in $\mathrm{\Sigma }$; every other direction vector of $\mathrm{\Sigma }$ lies in $\ker k$ outside $\mathrm{span}(k^{\mathrm{♯}})$ and is spacelike by steps 2.1 and 3.1 and [L5]. This proves (a), (b) and (c). The computation is exact algebra and no choice principle was used.

First witness. With $\beta =(3\mathrm{/}5)e_1$ one has $\mathrm{\mid }\beta {\mathrm{\mid }}^2=9\mathrm{/}25$ and $\gamma =(1-9\mathrm{/}25{)}^{-1\mathrm{/}2}=(16\mathrm{/}25{)}^{-1\mathrm{/}2}=5\mathrm{/}4$. Take $s=e_2$; then $\beta \cdot s=(3\mathrm{/}5)(0)=0$, so the propagation direction is perpendicular to the relative velocity in the frame $x$, and [L2] gives ${\omega }^{\mathrm{\prime }}=\gamma \omega =(5\mathrm{/}4)\omega \mathrm{\ne }\omega$ because $\omega >0$. Thus the antecedent of the refuted universal claim holds in $x$ while its conclusion fails.

Second witness. Take $s=(3\mathrm{/}5,4\mathrm{/}5,0)$; then $\mathrm{\mid }s{\mathrm{\mid }}^2=9\mathrm{/}25+16\mathrm{/}25=1$ and $\beta \cdot s=(3\mathrm{/}5)(3\mathrm{/}5)=9\mathrm{/}25=\mathrm{\mid }\beta {\mathrm{\mid }}^2$, so by [L2] $s^{\mathrm{\prime }}\cdot {\beta }_{\mathrm{h}\mathrm{a}\mathrm{t}}=0$, the propagation direction is perpendicular to the relative velocity in the frame $x^{\mathrm{\prime }}$, and [L2] gives ${\omega }^{\mathrm{\prime }}=\gamma (1-9\mathrm{/}25)\omega =(5\mathrm{/}4)(16\mathrm{/}25)\omega =(4\mathrm{/}5)\omega \mathrm{\ne }\omega$. Explicitly, the axis formulas of [L1] give $s_1^{\mathrm{\prime }}=(3\mathrm{/}5-3\mathrm{/}5)\mathrm{/}(1-9\mathrm{/}25)=0$, $s_2^{\mathrm{\prime }}=(4\mathrm{/}5)\mathrm{/}[(5\mathrm{/}4)(16\mathrm{/}25)]=(4\mathrm{/}5)\mathrm{/}(4\mathrm{/}5)=1$ and $s_3^{\mathrm{\prime }}=0$, so $s^{\mathrm{\prime }}=e_2$.

Conclusion. Each witness satisfies the antecedent of the refuted claim — transversality in $x$ in step 1.1 and transversality in $x^{\mathrm{\prime }}$ in step 1.2 — and yields ${\omega }^{\mathrm{\prime }}\mathrm{\ne }\omega$; the two factors $(5\mathrm{/}4)$ and $(4\mathrm{/}5)$ differ and neither equals $1$ for the stated nonzero $\beta$, so the universal claim is false in the ideal model, and the direction and size of the shift depend on the protocol (the frame in which transversality is required). The counterexample is conditional on the adopted idealizations, performs no measurement and claims no empirical falsification; no choice principle is used.