Using robphy's answer to my initial question, I try to implement what I understand to be his suggested approach, and I get it to almost work, but I get a small error at the end that I hope someone can spot.

As a reminder, I am using the wave 4-vector definition $$\kappa^\mu = \frac{2\pi}\lambda (1,\mathbf n) $$

where $\pmb n$ is the unit 3-vector in the direction photon motion and $\lambda$ is the photon's wavelength. To be a spacetime 4-vector, my text book requires an object $\kappa^\mu$ to satisfy the Lorentz transformation equation $\kappa^{\mu^\prime}=\Lambda^{\mu^\prime}_\nu\kappa^\nu$ between inertial frames K and $K^\prime$.

I define the inertial frame $K^\prime$ to be a boost with speed v in the $x^1$ direction of frame K. A photon is fired from a source at the origin $O^\prime$ of frame $K^\prime$, travels in the $x^1x^2$ plane, and arrives at the origin $O$ of frame K along a path that makes an angle $\theta$ with the $x^1$-axis. I set the speed of frame $K^\prime$ to be $v = c\;cos\,\theta.$

Pic of Frame <span class=$K^\prime$ offset from Frame K" />

Because the speed v is less than c (in my original attempt, I tried to use speed v = c for the photon), we can now define $$\gamma\equiv\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-\frac{c^2 cos^2\theta}{c^2}}}=\frac{1}{sin\theta}.$$

The Lorentz equation is then the boost equation with offset, namely $$x^{\mu^\prime}=\Lambda^{\mu^\prime}_\nu x^\nu+a^\mu$$

where $a^\mu$ is the offset and the matrix form of the Jacobian is

$$\Lambda^{\mu^\prime}_\nu =\begin{pmatrix} \gamma & -\frac{\gamma v}{c} & 0 & 0 \\ -\frac{\gamma v}{c} & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\;.$$

An observer in frame $K^\prime$ sees the photon travel along the $x^{2^\prime}$-axis from $O^\prime$ to $O$, so that $\mathbf n^\prime$ is a unit vector along the $x^{2^\prime}$ axis as shown. An observer in frame K sees the photon travel diagonally and so its unit vector $\mathbf n$ is on the diagonal path as shown. So,

$$\mathbf n=(cos\theta, sin\theta, 0),\qquad \mathbf n^\prime=(0, 1, 0),$$

$$k^\mu=\frac{2\pi}{\lambda}(1,\mathbf n)=\frac{2\pi}{\lambda}(1,cos\theta,sin\theta,0),$$

$$k^{\mu^\prime}=\frac{2\pi}{\lambda^\prime}(1,\mathbf n^\prime)=\frac{2\pi}{\lambda^\prime}(1,0,1,0).$$


$$\frac{\gamma v}{c}=\frac{1}{c}\frac{1}{sin\theta}\,(c\;cos\theta)=\frac{cos\theta}{sin\theta.}$$ So,

$$\Lambda^{\mu^\prime}_\nu = \begin{pmatrix}\frac{1}{sin\theta} & -\frac{cos\theta}{sin\theta} & 0 & 0 \\ -\frac{cos\theta}{sin\theta} & \frac{1}{sin\theta} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix},$$


$$\Lambda^{\mu^\prime}_\nu k^\nu=\begin{pmatrix} sin\theta \\ 0 \\sin\theta \\ 0 \\ \end{pmatrix}$$

and, so, the Lorentz transformation for $\kappa$ is

$$\frac{2\pi}{\lambda^\prime}\;\begin{pmatrix} 1 \\0 \\1 \\0 \\ \end{pmatrix}=\frac{2\pi}{\lambda}\;\begin{pmatrix} sin\theta \\0 \\sin\theta \\0 \\ \end{pmatrix}.$$

Equations 2 and 4 are 0 = 0 and equations 1 and 3 are $\frac{2\pi}{\lambda^\prime}=\frac{2\pi}{\lambda}\,sin\theta, or,$


That is, $\kappa$ satisfies this Lorentz transformation if and only if this ratio is true. So, my last step is to confirm the ratio but I get that


so either my proof or my confirmation (or both) has a mistake that I hope someone can spot.

I get my confirmation result by using the boost equation with offset, given just under the figure. I use just the equation for $x^{0^\prime}$. To use the equation, we must identify the offset $a^0$, which represents the time instant when origin O lies on the (moving) $x^{2^\prime}$ axis. WLOG we define that to be time $t=a^0=0$. With that convention the $x^{0^\prime}$ equation is

$$c t^{\prime}=x^{0^\prime}=\gamma x^0 - \gamma \frac{v}{c}x^1 + a^0=\gamma(c t - \frac{c\: cos\theta}{c} x^1)= \gamma (c t - cos\theta\; x^1).$$

If we set $t=t_\lambda$, the time in frame K for the photon to travel 1 wavelength, the corresponding time for one wavelength in $K^\prime$ is $t^\prime=t_{\lambda^\prime}$, and the $x^1$ distance, which corresponds to one wavelength $\lambda$ along the vector n, is $x^1=\lambda\; cos\theta$. So,

$$\lambda^\prime=c\; t_{\lambda^\prime}=\gamma (c t_\lambda - cos\theta\; x^1)=\gamma(\lambda - \lambda cos^2\theta)= \gamma \lambda sin^2\theta$$

which implies that

$$\frac{\lambda^\prime}{\lambda}=\gamma sin^2\theta=\frac{sin^2\theta}{sin\theta}=sin\theta,$$

not quite what I was trying to show.


2 Answers 2


The Lorentz Transformations are used to map between inertial frames of timelike-observers. The $"v"$ in the boost transformation is the relative spatial-velocity between the frames. In some sense, the boost transforms to a frame in which one of the timelike inertial observers is at rest.

However, the photon does not have a timelike 4-momentum... it has a lightlike one. So, one cannot use the spatial-velocity of the photon as $"v"$ in the transformation (since one can't transform into a frame where the photon is at rest... no such timelike-frame exists.)

However, the Lorentz Transformation still applies... between two inertial timelike-observers. So, pick two such that their 4-velocities are co-planar [in spacetime] with the photon 4-momentum. (That is, choose inertial observers such that one observer can use this photon to signal the other.)

In fact, if one solves the eigenvalue problem for the boost transformation, one finds that your photon 4-momentum is an eigenvector [the principle of the speed of light], and the eigenvalue is associated with the relative-Doppler factor between frames. (One ends up with light-cone coordinates, which underlies the Bondi k-calculus.)

My answer is primarily about the (1+1)-case since that already addresses the essence of your question.

  • $\begingroup$ Thank you. I was able to do a proof using two inertial frames as you suggested. I could post the proof in the "Answer your question" section, but only if anyone is interested and if doing so is appropriate in this forum. $\endgroup$
    – matrixbud
    Commented Apr 16, 2022 at 15:31
  • $\begingroup$ @matrixbud I think it would useful for others. I would think it would be appropriate. $\endgroup$
    – robphy
    Commented Apr 16, 2022 at 15:46
  • $\begingroup$ I discovered a problem with my "solution", so I need to post what I have done and request further help.I was able to use your suggestion to express the Lorentz transformation but it implied that $\lambda^\prime/\lambda = \gamma$ . When I try to confirm this using the frames I have set up I get $\lambda^\prime/\lambda = 1/\gamma$. In this forum, should I post this as a new question, or post it as "Answer your question" even though it has a flaw? $\endgroup$
    – matrixbud
    Commented Apr 18, 2022 at 19:16
  • $\begingroup$ @matrixbud I think it’s appropriate to continue it as an “answer” to your original question… (indicating what you did, where you got stuck, and where you are trying to go). Then the community can suggest changes to improve your answer. $\endgroup$
    – robphy
    Commented Apr 18, 2022 at 20:50

$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mc}[1]{\mathcal {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\vlr}[1]{\left\vert#1\right\vert} \newcommand{\Vlr}[1]{\left\Vert#1\right\Vert} \newcommand{\lara}[1]{\left\langle#1\right\rangle} \newcommand{\lav}[1]{\left\langle#1\right|} \newcommand{\vra}[1]{\left|#1\right\rangle} \newcommand{\lavra}[2]{\left\langle#1\right|\left#2\right\rangle} \newcommand{\lavvra}[3]{\left\langle#1\right|#2\left|#3\right\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\Vp}[1]{\vphantom{#1}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\ox}{\bl\otimes} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad} \newcommand{\qqLraqq}{\qquad\boldsymbol{\e\!\e\!\e\!\e\!\Longrightarrow}\qquad} \newcommand{\tl}[1]{\tag{#1}\label{#1}} \newcommand{\hebl}{\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}}$

REFERENCE-01 : My answer here Transformation of 4− velocity.

REFERENCE-02 : My answer here Deriving relativistic Doppler shift in terms of wavelength.

The $\,3\p 1\m$Lorentz boost transformation with arbitrary velocity $\:\mb u$ expressed in differential form is

\begin{align} \rm d\mb x' & \e \rm d\mb x\p\dfrac{\gamma^2_{\rm u}}{c^2 \plr{\gamma_{\rm u}\p 1}} \plr{\mb u\bl\cdot\rm d\mb x}\mb u\m\gamma_{\rm u}\mb u\,\mathrm dt \tl{01a}\\ \rm dt' & \e \gamma_{\rm u}\plr{\rm dt\m \dfrac{\mb u\bl\cdot\rm d\mb x}{c^2}} \tl{01b}\\ \gamma_{\rm u} & \e \plr{1\m\dfrac{\rm u^2}{c^2}}^{\m\frac12} \tl{01c} \end{align} see equations (03a),(03b),(02c) and Figure in REFERENCE-01.

Dividing equations \eqref{01a}/\eqref{01b} side-by-side we have the Lorentz transformation of velocity 3-vectors \begin{equation} \mb w'\e\dfrac{\mb w\p\dfrac{\gamma^2_{\rm u}\plr{\mb u\bl\cdot\mb w}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mb u\m\gamma_{\rm u}\mb u}{\gamma_{\rm u}\plr{1\m\dfrac{\mb u\bl\cdot\mb w}{c^2}}\vphantom{\dfrac{a'}{b'}}} \tl{02} \end{equation} see equation (08) in REFERENCE-01.

If \begin{equation} \mb w\e c\,\mb n\quad \texttt{where}\quad \Vlr{\mb n}\e 1 \tl{03} \end{equation} is the velocity 3-vector of a photon, then inserting this expression in the rhs of \eqref{02} we have \begin{equation} \mb w'\e c\, \dfrac{\mb n\p\dfrac{\gamma^2_{\rm u}\plr{\mb u\bl\cdot\mb n}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mb u\m\dfrac{\gamma_{\rm u}}{c}\mb u}{\gamma_{\rm u}\plr{1\m\dfrac{\mb u\bl\cdot\mb n}{c}}\vphantom{\dfrac{a'}{b'}}}\e c\, \mb n' \tl{04} \end{equation} where \begin{equation} \mb n'\e \dfrac{\mb n\p\dfrac{\gamma^2_{\rm u}\plr{\mb u\bl\cdot\mb n}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mb u\m\dfrac{\gamma_{\rm u}}{c}\mb u}{\gamma_{\rm u}\plr{1\m\dfrac{\mb u\bl\cdot\mb n}{c}}\vphantom{\dfrac{a'}{b'}}} \tl{05} \end{equation} It could be proved easily that $\:\Vlr{\mb n'}\e 1 $.

Now, if we have a light signal propagating with velocity $\:\mb w\e c\,\mb n\:$ of frequency $\:\nu\:$ and wavelength $\:\lambda\:$ ($\lambda\nu\e c$) in the rest frame $\:\rm S$, then for its frequency $\:\nu'\:$ and wavelength $\:\lambda'\:$ in the moving frame $\:\rm S'\:$ we have from the relativistic Doppler shift \begin{equation} \dfrac{\nu'}{\nu}\e\dfrac{\lambda}{\lambda'}\e\gamma_{\rm u}\plr{1\m\dfrac{\mb u\bl\cdot\mb w}{c^2}\vphantom{\dfrac{a'}{b'}}}\e \gamma_{\rm u}\plr{1\m\dfrac{\mb u\bl\cdot\mb n}{c}\vphantom{\dfrac{a'}{b'}}} \tl{06} \end{equation} You could prove equation \eqref{06} following the steps of Solution 1 in REFERENCE-02.

We express equation \eqref{06} as equation \eqref{07b} and use it in \eqref{02} to produce equation \eqref{07a} so that \begin{align} \blr{\dfrac{\mb w'}{\lambda'}} & \e \blr{\dfrac{\mb w}{\lambda}\vphantom{\dfrac{a'}{b'}}}\p\dfrac{\gamma^2_{\rm u}}{c^2 \plr{\gamma_{\rm u}\p 1}} \plr{\mb u\bl\cdot\blr{\dfrac{\mb w}{\lambda}\vphantom{\dfrac{a'}{b'}}}}\mb u\m\gamma_{\rm u}\mb u\,\blr{\dfrac{1}{\lambda}\vphantom{\dfrac{a'}{b'}}} \tl{07a}\\ \blr{\dfrac{1}{\lambda'}\vphantom{\dfrac{a'}{b'}}} & \e \gamma_{\rm u}\plr{\blr{\dfrac{1}{\lambda}\vphantom{\dfrac{a'}{b'}}}\m \dfrac{\mb u}{c^2}\bl\cdot\blr{\dfrac{\mb w}{\lambda}\vphantom{\dfrac{a'}{b'}}}} \tl{07b} \end{align}

Comparing equations \eqref{07a},\eqref{07b} with \eqref{01a},\eqref{01b} respectively we conclude that the 4-vector \begin{equation} \plr{\dfrac{1}{\lambda}\vphantom{\dfrac{a'}{b'}},\dfrac{\mb w}{\lambda}\vphantom{\dfrac{a'}{b'}}} \tl{08} \end{equation} is transformed as the Lorentz 4-vector $\:\plr{\mr dt,\mr d\mb x}$.


ADDENDUM A : Proof that in equation \eqref{05} $\:\Vlr{\mb n'}\e 1$.

\begin{equation} \begin{split} & \Vlr{\mb n'}^2\e\mb n'\bl\cdot \mb n'\e \dfrac{\blr{\mb n\p\dfrac{\gamma^2_{\rm u}\plr{\mb u\bl\cdot\mb n}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mb u\m\dfrac{\gamma_{\rm u}}{c}\mb u}\bl\cdot\blr{\mb n\p\dfrac{\gamma^2_{\rm u}\plr{\mb u\bl\cdot\mb n}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mb u\m\dfrac{\gamma_{\rm u}}{c}\mb u}}{\blr{\gamma_{\rm u}\plr{1\m \dfrac{\mb u\bl\cdot\mb n}{c}}}^2}\e\dfrac{\mc A}{\mc B}\quad \texttt{where}\\ & \mc A \e \blr{\mb n\p\dfrac{\gamma^2_{\rm u}\plr{\mb u\bl\cdot\mb n}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mb u\m\dfrac{\gamma_{\rm u}}{c}\mb u}\bl\cdot\blr{\mb n\p\dfrac{\gamma^2_{\rm u}\plr{\mb u\bl\cdot\mb n}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mb u\m\dfrac{\gamma_{\rm u}}{c}\mb u}\quad\texttt{and}\quad \mc B \e \blr{\gamma_{\rm u}\plr{1\m \dfrac{\mb u\bl\cdot\mb n}{c}}}^2\\ \end{split} \tl{09} \end{equation} Now \begin{equation} \begin{split} &\mc A \e \blr{\mb n\p\dfrac{\gamma^2_{\rm u}\plr{\mb u\bl\cdot\mb n}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mb u\m\dfrac{\gamma_{\rm u}}{c}\mb u}\bl\cdot\blr{\mb n\p\dfrac{\gamma^2_{\rm u}\plr{\mb u\bl\cdot\mb n}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mb u\m\dfrac{\gamma_{\rm u}}{c}\mb u}\e\\ &\plr{\mb n\bl\cdot\mb n}\p\dfrac{\gamma^4_{\rm u}\plr{\mb u\bl\cdot\mb n}^2}{c^4 \plr{\gamma_{\rm u}\p 1}^2}\mr u^2\p\dfrac{\gamma^2_{\rm u}}{c^2}\mr u^2\p 2\dfrac{\gamma^2_{\rm u}\plr{\mb u\bl\cdot\mb n}^2}{c^2 \plr{\gamma_{\rm u}\p 1}}\m 2\dfrac{\gamma_{\rm u}}{c}\plr{\mb u\bl\cdot\mb n}\m 2\dfrac{\gamma^3_{\rm u}\plr{\mb u\bl\cdot\mb n}}{c^3 \plr{\gamma_{\rm u}\p 1}}\mr u^2\e\\ &\underbrace{\blr{\plr{\mb n\bl\cdot\mb n}\p\dfrac{\gamma^2_{\rm u}}{c^2}\mr u^2}}_{\gamma^2_{\rm u}}\m 2\dfrac{\gamma_{\rm u}}{c}\underbrace{\blr{1\p\dfrac{\gamma^2_{\rm u}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mr u^2}}_{\gamma_{\rm u}}\plr{\mb u\bl\cdot\mb n}\p\dfrac{\gamma^2_{\rm u}}{c^2 \plr{\gamma_{\rm u}\p 1}}\underbrace{\blr{2\p\dfrac{\gamma^2_{\rm u}}{c^2 \plr{\gamma_{\rm u}\p 1}}\mr u^2}}_{\gamma_{\rm u}\p 1}\plr{\mb u\bl\cdot\mb n}^2\e\\ &\gamma^2_{\rm u}\blr{1\m 2\dfrac{\mb u\bl\cdot\mb n}{c}\p\dfrac{\plr{\mb u\bl\cdot\mb n}^2}{c^2}}\e \gamma^2_{\rm u}\plr{1\m \dfrac{\mb u\bl\cdot\mb n}{c}}^2\e \blr{\gamma_{\rm u}\plr{1\m \dfrac{\mb u\bl\cdot\mb n}{c}}}^2\e\mc B\\ \end{split} \tl{10} \end{equation} that is in equation \eqref{09} we have $\:\mc A=\mc B\,$, so $\:\Vlr{\mb n'}\e 1$. Note that in equation \eqref{10} we make use of the relation \begin{equation} \dfrac{\gamma^2_{\rm u}}{c^2}\mr u^2\e \gamma^2_{\rm u}\m 1 \tl{11} \end{equation}


enter image description here

Irrelevant here but interesting : Equation \eqref{04} and Figures-01,-02 have close relation to the relativistic aberration of light.

enter image description here

  • $\begingroup$ Frobenius, I appreciate that you introduced me to the the 3+1 version. The equations are actually more compact and easier to follow than 1+1. I learned a lot. I eventually figured out everything you put down except the highlighted equation for the Doppler shift. Thank you for doing this. $\endgroup$
    – matrixbud
    Commented Jun 13, 2022 at 0:18
  • $\begingroup$ @matrixbud : Self-studing Special Relativity I realize from the first moments that in many cases it would be difficult or even impossible to have answers using the restricted 1+1 Lorentz transformation. That's why I succeded very soon to derive (prove) from the 1+1 Lorentz tranformation the 3+1 Lorentz one (with velocity along an arbritrary direction, known as boost). In case you are curious enough I give you two links for this derivation...(*) $\endgroup$
    – Frobenius
    Commented Jun 13, 2022 at 2:33
  • $\begingroup$ @matrixbud : (*)...One is an old answer as user82794 (former diracpaul) here Two sets of coordinates each in frames O and O′ - Lorentz transformation and the other new one here Deriving Λij components of the Lorentz transformation matrix. $\endgroup$
    – Frobenius
    Commented Jun 13, 2022 at 2:35
  • $\begingroup$ @matrixbud : By the way, which 'highlighted equation for the Doppler shift' you couldn't figure out? $\endgroup$
    – Frobenius
    Commented Jun 13, 2022 at 2:39
  • $\begingroup$ In Figure 02 of The Relativistic Dopppler Effect, I don't follow the step in the highlighted equation (C): $$ \frac{1-\beta cos\theta}{\sqrt{1 - \beta^2}}=\frac{\nu'}{\nu}$$ $\endgroup$
    – matrixbud
    Commented Jun 15, 2022 at 0:22

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