Longitudinal Relativistic Doppler Effect

An ambulance goes past and the siren drops in pitch. Everybody knows why: the crests bunch up ahead of it and stretch out behind. It is the most familiar wave phenomenon there is, and the formula is a one-liner. Except that the acoustic formula comes in two versions, and they disagree. If the source moves through still air at $\beta_s = v/c_s$ , the received frequency is $f_0/(1+\beta_s)$ . If instead the observer moves through still air at the same speed, it is $f_0(1-\beta_s)$ . Those are different numbers — they differ at order $\beta_s^2$ — and an acoustician can tell you which is which, because the air is there, and "moving relative to the air" is a fact you can check. Now do it with light. There is no air. Relativity forbids any experiment from distinguishing "the source moved" from "the observer moved" — so the two formulas must collapse into one, and there is only one honest place for the answer to sit: exactly between them. And it does, in the sharpest possible sense. The relativistic Doppler factor is the geometric mean of the two acoustic answers:

\mathcal{D} \;=\; \sqrt{(1-\beta)\cdot\frac{1}{1+\beta}} \;=\; \sqrt{\frac{1-\beta}{1+\beta}}.

At $\beta = 0.6$ the two classical answers are $0.400$ and $0.625$ ; the true answer is $\sqrt{0.400\times0.625} = 0.500$ , exactly. The factor that splits the difference is $1/\gamma$ — time dilation. The medium's fingerprint, and the moving clock, are the same missing term.

Key takeaways

Signature $(+,-,-,-)$ . Collinear geometry: the source recedes along the line of sight at $\beta = v/c$ , positive for recession.
Recession $\ \mathcal{D} = f_{\text{obs}}/f_{\text{src}} = \sqrt{(1-\beta)/(1+\beta)}$ ; approach $\ \mathcal{D} = \sqrt{(1+\beta)/(1-\beta)}$ . Only the relative velocity appears — there is no medium, hence no "source moving" versus "observer moving" distinction.
Angle convention (band-locked): for a general direction, $\mathcal{D} = 1/[\gamma(1-\beta\cos\theta)]$ with $\theta$ the lab-frame angle between the source's velocity and the photon's propagation direction. $\theta = 0$ is approach, $\theta = \pi$ is recession.
In rapidity, $\mathcal{D} = e^{\mp\varphi}$ with $\beta = \tanh\varphi$ . Collinear Doppler shifts therefore MULTIPLY while rapidities ADD — recall aq-sr-rapidity-hyperbolic-boosts and aq-sr-rapidity.
Owned result: $\mathcal{D}_{\text{rel}}$ is the exact geometric mean of the two acoustic answers — equivalently, $\varphi = \operatorname{artanh}\beta$ is the exact arithmetic mean of their logarithms. Relativity does not correct one classical formula; it splits the difference between two, and the splitting factor is $1/\gamma$ .
Key idea: strip away the medium and the two Doppler formulas of acoustics have nowhere to go but the same place. Their geometric mean is the relativistic answer, and the extra $1/\gamma$ that gets you there is the moving clock.

The core insight

Classical Doppler is geometry: crests from a moving point pile up or spread out, and the counting is elementary. Relativistic Doppler is that same geometry times one extra factor — the source's clock runs slow by $\gamma$ , so it emits fewer crests per unit of your time than it thinks it does. Two effects, one product: $\mathcal{D} = (1/\gamma)\times\left[\text{classical crest-counting}\right]$ . The rest of this article is bookkeeping on that sentence.

Where this lives today

Radial velocities, and the planets they find. Every exoplanet found by the radial-velocity method is a Doppler measurement: the star wobbles by metres per second about the barycentre and a spectrograph reads the periodic line shift. At $v \sim 1\ \mathrm{m\,s^{-1}}$ , $\beta \sim 3\times10^{-9}$ , so the relativistic correction is of order $\beta^2 \sim 10^{-17}$ — utterly irrelevant. Worth saying plainly: the relativistic Doppler effect matters only when $\beta^2$ is measurable, which for spectroscopy means ion beams, storage rings, and jets.

Ion beams, where it does matter. Accelerated ions reach $\beta$ of a few tenths, where the second-order term is a part in $10^{2}$ and easy to see. This is the regime of Ives–Stilwell and its modern storage-ring descendants, where the longitudinal formula stops being a convenience and becomes a test of relativity itself (aq-sr-transverse-doppler-effect).

Cosmology, where the formula is borrowed and handed back. Astronomers quote $1+z$ rather than a velocity precisely because of the multiplicativity derived above: $(1+z)$ factors cleanly across a chain of effects — expansion, peculiar velocity, gravitational shift — where velocities would not. For a peculiar velocity on top of the Hubble flow, equation (4) is exactly right. For the expansion itself it is not a Doppler shift at all.

And the twins. The Doppler count is the cleanest resolution of the twin paradox, because it uses no simultaneity convention whatever: each twin counts flashes, and integers are invariant. The traveller sees red for half her trip and blue for half; the Earth sees red for far longer, because news of the turnaround is delayed by the light-travel time. That asymmetry — of when the shift changes, not what it is — is the answer. See aq-sr-twin-paradox-doppler-resolution.

The physical picture

Fix the geometry once. The source recedes from you along your line of sight at speed $\beta c$ , emitting wave crests at proper frequency $f_{\text{src}}$ — that is, $f_{\text{src}}$ crests per second by its own clock. You sit still and count crests as they arrive. Two things happen, and they are logically independent.

The clock. The source's clock is time-dilated. In your frame it emits at only $f_{\text{src}}/\gamma$ crests per second. This has nothing to do with the source moving away; it would be true if the source were circling you at constant distance. It is a fact about clocks, and it is the entire content of the transverse Doppler effect (aq-sr-transverse-doppler-effect).

The geometry. Consecutive crests leave from points further and further away, so each has further to fly. If crests leave $\Delta t$ apart in your frame, the second is emitted $\beta c\,\Delta t$ more distant and arrives an extra $\beta\,\Delta t$ later, stretching the arrival interval to $\Delta t(1+\beta)$ . This has nothing to do with clocks; it is true for sound, for water waves, and for a runner throwing tennis balls over her shoulder. Multiply: the arrival interval is $(\gamma/f_{\text{src}})\times(1+\beta)$ , so

f_{\text{obs}} = \frac{f_{\text{src}}}{\gamma(1+\beta)} = f_{\text{src}}\sqrt{\frac{1-\beta}{1+\beta}}.

The square root is not a mysterious relativistic flourish. It is $\sqrt{1-\beta^2}/(1+\beta)$ with the difference of two squares factored — clock times geometry, tidied.

Figure

PUNCHLINE — the relativistic Doppler factor is the GEOMETRIC MEAN of the two classical answers. Acoustics has two formulas because the air is a real thing you can move relative to: source-moving gives 1/(1+β), observer-moving gives (1−β). Light has no air, so relativity must pick ONE — and it picks exactly the geometric mean of the two, at every β. At β = 0.6 the classical pair is 0.625 and 0.400, and √(0.625 × 0.400) = 0.500, exactly. The factor that carries you from either classical curve to the true one is 1/γ: the moving clock

The core equations

\boxed{\;\mathcal{D} \;\equiv\; \frac{f_{\text{obs}}}{f_{\text{src}}} \;=\; \sqrt{\frac{1-\beta}{1+\beta}} \quad (\text{recession}), \qquad \mathcal{D} = \sqrt{\frac{1+\beta}{1-\beta}} \quad (\text{approach})\;} \qquad (1)

where: $\beta = v/c$ is the relative speed of source and observer. Physical reading: the two formulas are the same formula with $\beta \to -\beta$ . There is no third case, because there is no medium: "the source moved" and "the observer moved" are not different experiments.

\mathcal{D} = \frac{1}{\gamma\,(1 - \beta\cos\theta)}, \qquad \theta = \text{LAB-frame angle between } \mathbf{v} \text{ and the photon's propagation direction} \qquad (2)

where: $\theta = 0$ is a photon flying along the source's velocity, straight at you — approach; $\theta = \pi$ is a photon flying backwards from a receding source — recession; $\theta = \pi/2$ gives $\mathcal{D} = 1/\gamma$ , pure time dilation. Physical reading: this is the band-locked convention, and it is a $(1 - \beta\cos\theta)$ , not a $(1+\beta\cos\theta)$ . The AllQFT base printed a plus sign (defect D6) by tying $\theta$ to the observer's line of sight to the source rather than to the ray's direction of travel — the two are anti-parallel, hence the flip. Both bookkeepings can be made consistent; mixing them cannot. The master formula is derived from the invariant $k_\mu u^\mu$ in aq-sr-doppler-four-momentum.

\mathcal{D} = e^{-\varphi} \ \ (\text{recession}), \qquad \beta = \tanh\varphi, \qquad \mathcal{D}_{\text{tot}} = \mathcal{D}_1\mathcal{D}_2 = e^{-(\varphi_1+\varphi_2)} \qquad (3)

where: $\varphi = \operatorname{artanh}\beta$ is the rapidity. Physical reading: Doppler factors multiply because rapidities add. Composing two collinear recessions is one multiplication, not an application of the velocity-addition law — the awkwardness of $\beta$ -composition simply evaporates. (Rapidity additivity is owned by aq-sr-rapidity-hyperbolic-boosts and aq-sr-rapidity; used here, not re-derived.)

1 + z \;\equiv\; \frac{\lambda_{\text{obs}}}{\lambda_{\text{src}}} \;=\; \frac{1}{\mathcal{D}} \;=\; \sqrt{\frac{1+\beta}{1-\beta}} \;=\; e^{\varphi} \qquad\Longleftrightarrow\qquad \beta = \frac{(1+z)^2 - 1}{(1+z)^2 + 1} \qquad (4)

where: $z$ is the redshift. Physical reading: the inversion matters. A naive reader sets $\beta = z$ ; that is the first term of an expansion and it fails fast. At $z = 0.158$ the true kinematic $\beta$ is $0.1457$ , not $0.158$ — an $8.5\%$ error already.

Where the formula comes from

Derivation ALongitudinal Doppler by counting crests

This is the article's own derivation, and it deliberately avoids four-vectors. The covariant route — package $(\omega/c, \mathbf{k})$ into a null four-vector and boost it — is faster, more general, and belongs to aq-sr-doppler-four-momentum, which owns it. What crest-counting buys instead is separability: it shows you the classical effect and the relativistic effect as two distinct factors that you can point at, and it makes it obvious which one Newton would have got right.

Step 1. Set up in the observer's frame. The source recedes along $+\hat{x}$ at $\beta c$ , emitting crests once per proper time $1/f_{\text{src}}$ . You sit at the origin counting arrivals. Why: everything is computed in one frame, yours. No transformation is needed — only careful bookkeeping about when light was emitted and when it arrives.

Step 2. Dilate the emission rate. The source's proper time runs slow by $\gamma$ in your frame, so successive crests are emitted $\Delta t_{\text{emit}} = \gamma/f_{\text{src}}$ apart by your clock. f_{\text{emit}} = \frac{f_{\text{src}}}{\gamma} Why: this is the whole of relativity's contribution. Delete it and everything below is Newtonian. It is a statement about the source's clock and would hold even if the source were moving transversely, at fixed distance from you.

Step 3. Add the light-travel geometry. In the interval $\Delta t_{\text{emit}}$ the source retreats by $\beta c\,\Delta t_{\text{emit}}$ , so crest $n{+}1$ flies that much further, taking an extra $\beta\,\Delta t_{\text{emit}}$ to reach you: \Delta t_{\text{obs}} = \Delta t_{\text{emit}}(1+\beta). Why: pure geometry, no relativity. Identical for sound — and the only factor Doppler himself had in 1842.

Step 4. Take the reciprocal and factor. f_{\text{obs}} = \frac{1}{\Delta t_{\text{obs}}} = \frac{f_{\text{src}}}{\gamma(1+\beta)} = f_{\text{src}}\,\frac{\sqrt{1-\beta^2}}{1+\beta} = \boxed{f_{\text{src}}\sqrt{\frac{1-\beta}{1+\beta}}} using $1-\beta^2 = (1-\beta)(1+\beta)$ . Why: the notorious square root is not a new physical input. It is $(1/\gamma)\times 1/(1+\beta)$ with a difference of squares cancelled — clock times geometry, tidied.

Step 5. Check the anchors. $\beta = 0.6$ : $\gamma = 5/4$ , so $\mathcal{D} = 1/[(5/4)(1.6)] = 1/2$ — exactly one half, and the wavelength doubles ( $z = 1$ ). $\beta = 0.8$ : $\gamma = 5/3$ , $\mathcal{D} = 1/[(5/3)(1.8)] = 1/3$ ( $z = 2$ ). Why: these two velocities are the ones every problem set uses, precisely because they give exact rationals. They also have rapidities $\ln2$ and $\ln3$ , which is the reason. *Verified in verify_aq-sr-longitudinal-doppler-effect.py: $1/[\gamma(1+\beta)] = \sqrt{(1-\beta)/(1+\beta)}$ symbolically; the anchors $\mathcal{D}(0.6) = 1/2$ and $\mathcal{D}(0.8) = 1/3$ exactly; the general-angle formula (2) reduces to (1) at $\theta = \pi$ and $\theta = 0$ ; and $\mathcal{D}(\pi/2) = 1/\gamma$ .*

Go deeper · Derivation B · 2 minThe relativistic factor is the geometric mean of the two acoustic answers

Derivation BThe relativistic factor is the geometric mean of the two acoustic answers

Here is the article's owned structural result, and it is an exact identity rather than an analogy. Acoustics has two Doppler formulas because a medium exists and you can move relative to it. Relativity must have one, because it forbids you from finding out. The question is where the single answer sits relative to the two it replaces, and the answer is not "somewhere between" — it is exactly the geometric mean, at every speed, with the deviation of each classical answer from the truth being precisely $\gamma^{\pm1}$ .

Step 1. Write the two classical answers. Source receding through a still medium at $\beta$ (the crests are stretched at emission): $\mathcal{D}_{\text{src}} = 1/(1+\beta)$ . Observer receding through a still medium at $\beta$ (the crests are outrun at reception): $\mathcal{D}_{\text{obs}} = 1-\beta$ . Why: these are genuinely different physical situations in acoustics, and an experiment with a wind gauge can tell them apart. They agree to first order and split at $O(\beta^2)$ : $1-\beta+\beta^2$ against $1-\beta$ .

Step 2. Take the geometric mean. \sqrt{\mathcal{D}_{\text{src}}\,\mathcal{D}_{\text{obs}}} = \sqrt{\frac{1-\beta}{1+\beta}} = \boxed{\mathcal{D}_{\text{rel}}} Why: an exact identity, not an approximation, not a limit — the product of the two classical answers is $(1-\beta)/(1+\beta)$ and its square root is the relativistic Doppler factor on the nose.

Step 3. Read it in logs, where it becomes an arithmetic mean. Taking $\ln$ : $\tfrac12\left[\ln(1-\beta) + \ln\tfrac{1}{1+\beta}\right] = \tfrac12\ln\frac{1-\beta}{1+\beta} = -\operatorname{artanh}\beta = -\varphi$ . Why: rapidity is literally the average of the two classical log-shifts. The variable in which relativity is linear is the variable in which "split the difference" is ordinary averaging.

Step 4. Identify the splitter as $\gamma$ , and see why the mean is geometric. Note that $\mathcal{D}_{\text{rel}} = 1/[\gamma(1+\beta)] = \gamma(1-\beta)$ — the two forms are equal, which is the same difference-of-squares as before. Therefore \frac{\mathcal{D}_{\text{src}}}{\mathcal{D}_{\text{rel}}} = \frac{1/(1+\beta)}{1/[\gamma(1+\beta)]} = \gamma, \qquad\qquad \frac{\mathcal{D}_{\text{rel}}}{\mathcal{D}_{\text{obs}}} = \frac{\gamma(1-\beta)}{1-\beta} = \gamma. The three numbers form a geometric progression with common ratio $\gamma$ — and the middle term of a three-term geometric progression is the geometric mean of the outer two. Why: this is the whole theorem in one line. The two classical answers sit one factor of $\gamma$ above and one factor of $\gamma$ below the truth, and time dilation is the entire discrepancy — applied once, to whichever clock you were implicitly privileging by assuming a medium.

Step 5. Verify numerically at $\beta = 0.6$ . $\mathcal{D}_{\text{src}} = 1/1.6 = 0.625$ . $\mathcal{D}_{\text{obs}} = 0.4$ . $\sqrt{0.625\times0.4} = \sqrt{0.25} = 0.5$ ✓. And $0.625\times(1/\gamma) = 0.625\times0.8 = 0.5$ ✓, $0.4\times\gamma = 0.4\times1.25 = 0.5$ ✓. Why: three routes, one number. The classical formulas are not "wrong by a small correction" — they are wrong by exactly one factor of the Lorentz factor, in opposite senses. *Verified in verify_aq-sr-longitudinal-doppler-effect.py: $\sqrt{(1-\beta)\cdot(1+\beta)^{-1}} = \mathcal{D}_{\text{rel}}$ symbolically; $\tfrac12[\ln(1-\beta) - \ln(1+\beta)] = -\operatorname{artanh}\beta$ symbolically; $\mathcal{D}_{\text{src}}/\mathcal{D}_{\text{rel}} = \gamma$ and $\mathcal{D}_{\text{rel}}/\mathcal{D}_{\text{obs}} = \gamma$ symbolically; and the $\beta = 0.6$ triple $(0.625,\ 0.500,\ 0.400)$ recomputed from the article text.*

Go deeper · Derivation C · 1 minDoppler factors multiply because rapidities add

Derivation CDoppler factors multiply because rapidities add

The velocity-addition law is a nuisance to iterate: compose $0.6c$ with $0.6c$ and you must remember $(\beta_1+\beta_2)/(1+\beta_1\beta_2)$ , carry an exact fraction, and then re-evaluate a square root. In rapidity there is nothing to remember. This card shows that the Doppler factor is literally an exponential of the rapidity, so that chained shifts compose by multiplication — and it checks the claim against the velocity-addition law it replaces.

Step 1. Rewrite $\mathcal{D}$ in rapidity. With $\beta = \tanh\varphi$ : \mathcal{D} = \sqrt{\frac{1-\tanh\varphi}{1+\tanh\varphi}} = \sqrt{\frac{\cosh\varphi - \sinh\varphi}{\cosh\varphi + \sinh\varphi}} = \sqrt{\frac{e^{-\varphi}}{e^{+\varphi}}} = \boxed{e^{-\varphi}} Why: multiply numerator and denominator by $\cosh\varphi$ and use $\cosh\varphi \pm \sinh\varphi = e^{\pm\varphi}$ . The square root has evaporated; the Doppler factor was an exponential all along.

Step 2. Recognise what this is. $e^{\pm\varphi}$ are the eigenvalues of the Lorentz boost acting on the two null directions — the light cone's generators are the boost's eigenvectors. So $\mathcal{D}$ is not merely like a boost eigenvalue; it is one. (Owned by aq-sr-lorentz-boosts-hyperbolic-rotations and aq-sr-the-light-cone; recalled here.) Why: this is the structural reason redshifts multiply. Eigenvalues of composed linear maps multiply, and rapidities — the logarithms of those eigenvalues — add.

Step 3. Compose. Two successive collinear recessions of rapidity $\varphi_1$ and $\varphi_2$ give \mathcal{D}_{\text{tot}} = e^{-(\varphi_1+\varphi_2)} = e^{-\varphi_1}e^{-\varphi_2} = \mathcal{D}_1\mathcal{D}_2. Why: the shifts multiply — exactly, with no correction term. This is why cosmologists write $(1+z_{\text{tot}}) = (1+z_1)(1+z_2)$ and why $z$ itself is not additive.

Step 4. Check against velocity addition. Compose $\beta = 0.6$ with itself: Einstein's law gives $(0.6+0.6)/(1+0.36) = 1.2/1.36 = 15/17 = 0.882353$ . Then \mathcal{D}(15/17) = \sqrt{\frac{1 - 15/17}{1 + 15/17}} = \sqrt{\frac{2/17}{32/17}} = \sqrt{\frac{1}{16}} = \frac14, and independently $\mathcal{D}(0.6)\times\mathcal{D}(0.6) = \tfrac12\times\tfrac12 = \tfrac14$ ✓. Why: identical, exactly, with the messy $15/17$ never needing to appear. In rapidity: $\ln2 + \ln2 = \ln4$ , and $e^{-\ln4} = 1/4$ .

Step 5. Note what did not add. $z$ is not additive ( $z_1 = z_2 = 1$ but $z_{\text{tot}} = 3$ , not $2$ ); $\beta$ is not additive; $\gamma$ is not additive. Only $\varphi$ adds — and $\ln(1+z) = \varphi$ exactly. Why: the search for the additive variable is the whole content of this subject, and $\ln(1+z)$ is it. Anyone quoting "velocity" $= cz$ at large $z$ is using a variable that neither adds nor bounds. *Verified in verify_aq-sr-longitudinal-doppler-effect.py: $\sqrt{(1-\tanh\varphi)/(1+\tanh\varphi)} = e^{-\varphi}$ symbolically; $\mathcal{D}(\beta_1\oplus\beta_2) = \mathcal{D}(\beta_1)\mathcal{D}(\beta_2)$ symbolically under Einstein composition; $0.6\oplus0.6 = 15/17$ with $\mathcal{D} = 1/4$ exactly; $\ln(1+z) = \operatorname{artanh}\beta$ symbolically.*

Figure

Crest counting, drawn. The source recedes at β = 0.6, emitting a crest once per unit of ITS OWN proper time. Two independent things stretch the arrival interval. (1) TIME DILATION: by the observer's clock the crests leave γ = 1.25 times further apart than the source thinks. (2) LIGHT-TRAVEL GEOMETRY: each crest is emitted from further away and has an extra β·Δt of flight, stretching arrivals by (1+β) = 1.6. Product: 1.25 × 1.6 = 2.0, so 𝒟 = 1/2. Delete the dilation factor and you have Doppler's 1842 formula — and you are wrong by exactly γ

Limiting cases

Case	Result	Meaning
$\beta \to 0$	$\mathcal{D} \to 1 - \beta + \tfrac12\beta^2$	first order is classical; the $\tfrac12$ (not $1$ ) is where relativity first bites
classical, source moves	$1/(1+\beta) = 1 - \beta + \beta^2 - \cdots$	differs from the truth by exactly $\gamma$
classical, observer moves	$1 - \beta$	differs from the truth by exactly $1/\gamma$
$\beta = 0.6$	$\mathcal{D} = 1/2$ , $z = 1$ , $\varphi = \ln 2$	the exact-rational anchor
$\beta = 0.8$	$\mathcal{D} = 1/3$ , $z = 2$ , $\varphi = \ln 3$	the second anchor
$\beta \to 1^-$	$\mathcal{D} \to 0$ , $z \to \infty$	infinite redshift; $\varphi \to \infty$
$\theta = \pi/2$	$\mathcal{D} = 1/\gamma$	pure time dilation — aq-sr-transverse-doppler-effect
two collinear shifts	$\mathcal{D}_1\mathcal{D}_2$ , $\varphi_1 + \varphi_2$	shifts multiply, rapidities add — $z$ does neither

Common trapThe sign in the angular formula — and what the base got wrong

The band-locked general formula is $\mathcal{D} = 1/[\gamma(1-\beta\cos\theta)]$ , with $\theta$ the lab-frame angle between the source's velocity and the photon's direction of propagation. The AllQFT base for this article printed $f_{\text{obs}} = f_{\text{src}}/[\gamma(1+\beta\cos\theta)]$ — a plus sign (defect D6) — which directly contradicts the sibling articles 1904, 1906 and 1907. The base is not stupid: it was implicitly measuring $\theta$ from the observer's line of sight towards the source, which is anti-parallel to the ray's direction of travel, so its plus sign is self-consistent within its own convention. But a corpus cannot carry two conventions, and a student cannot be expected to notice which one a given page is using. Tie $\theta$ to the ray, always. Then $\theta = 0$ is approach and blueshift, $\theta = \pi$ is recession and redshift, and no formula in the chapter ever needs a sign flip.

Common trapThe half in the low-speed expansion is not simply time dilation

Expanding, $\mathcal{D}_{\text{rel}} = 1 - \beta + \tfrac12\beta^2 - \tfrac12\beta^3 + \tfrac38\beta^4 - \cdots$ , and it is tempting to point at the $\tfrac12\beta^2$ and call it "the time-dilation term". It is not — it is a residue. The classical source-moving expansion is $1-\beta+\beta^2-\beta^3+\cdots$ , with coefficient $+1$ at second order; time dilation multiplies by $1/\gamma = 1 - \tfrac12\beta^2 - \cdots$ , subtracting $\tfrac12\beta^2$ and leaving $+\tfrac12$ . The time-dilation contribution is the difference between the classical and the relativistic predictions, $-\tfrac12\beta^2$ , and it is the quantity an experiment must isolate. That is exactly what Ives and Stilwell did in 1938 (see aq-sr-transverse-doppler-effect), and it is why the test is a second-order test.

Common trapCosmological redshift is not this formula

Equation (4) inverts a redshift into a velocity, and for a star or a jet that is the right thing to do. For a distant galaxy it is not. Cosmological redshift is not a Doppler shift accumulated at a moment of emission; it is the integrated stretching of a wavelength by the expansion of space along the photon's whole path, $1 + z = a(t_{\text{obs}})/a(t_{\text{emit}})$ , and there is no single relative velocity to plug in — indeed for distant sources there is no frame-independent way to define one. Using $\beta = [(1+z)^2-1]/[(1+z)^2+1]$ on a cosmological redshift gives a number, and that number means nothing. Peculiar velocities on top of the expansion do obey (4).

Worked examples

1 · The quasar 3C 273 — reading a real redshift three ways

In 1963 Maarten Schmidt identified the Balmer lines in the spectrum of the radio source 3C 273, shifted redward by $z = 0.158$ . Read that shift as a pure special-relativistic radial recession and find $\beta$ . Compare with the naive answer.

Step 1 — identify. We are given $\lambda_{\text{obs}}/\lambda_{\text{src}} = 1 + z = 1.158$ , i.e. $1/\mathcal{D} = 1.158$ , so $\mathcal{D} = 0.863558$ . We want $\beta$ , so invert (4).

Step 2 — apply. \beta = \frac{(1+z)^2 - 1}{(1+z)^2 + 1} = \frac{1.158^2 - 1}{1.158^2 + 1} = \frac{1.340964 - 1}{1.340964 + 1} = \frac{0.340964}{2.340964} = 0.145651. So $v = 0.145651\,c = 4.367\times10^{4}\ \mathrm{km\,s^{-1}}$ .

Step 3 — the naive answer, and the size of the error. Setting $\beta = z = 0.158$ — the first term of the expansion $z = \beta + \tfrac12\beta^2 + \cdots$ — overstates the speed by $(0.158 - 0.145651)/0.145651 = 8.48\%$ . At $z = 0.158$ that is already thousands of kilometres per second.

Step 4 — the rapidity. $\varphi = \ln(1+z) = \ln 1.158 = 0.146694$ . Notice how close $\varphi$ and $\beta$ are at this speed ( $0.1467$ against $0.1457$ ) — they agree to third order, $\varphi = \beta + \beta^3/3 + \cdots$ . Rapidity is what "velocity" ought to have meant.

Step 5 — the boundary number, and the honest caveat. $\beta = 0.1457$ is what you get if the shift is a special-relativistic radial Doppler shift. It is not. 3C 273's redshift is cosmological: the wavelength was stretched by the expansion of space, not by a velocity through it. What $z = 0.158$ really buys is a distance — a Hubble-law estimate $d \approx cz/H_0 = (0.158)(299\,792.458)/70 = 677\ \mathrm{Mpc}$ , taking $H_0 = 70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$ — and that was the 1963 shock. Schmidt's lines were not exotic; they were ordinary hydrogen, absurdly far away, and therefore absurdly luminous. The kinematic $\beta$ computed above is an exercise in this article's formula and nothing more. The formula is exactly right for a star, a jet, or an ion beam; it is exactly wrong for the expansion of the universe.

2 · Two ships, two shifts, one multiplication

A relay ship recedes from Earth at $0.6c$ . A probe recedes from the relay ship at $0.6c$ , in the same direction. The probe emits at $600\ \mathrm{nm}$ in its own frame. What wavelength arrives at Earth?

Step 1 — identify. Two collinear recessions. The lazy route is velocity addition then one Doppler factor; the fast route is two Doppler factors multiplied. Both must agree.

Step 2 — the fast route. $\mathcal{D}(0.6) = \tfrac12$ each, so $\mathcal{D}_{\text{tot}} = \tfrac12\times\tfrac12 = \tfrac14$ . Wavelengths go as $1/\mathcal{D}$ , so $\lambda_{\text{Earth}} = 600\ \mathrm{nm}\times 4 = 2400\ \mathrm{nm}$ — deep in the infrared. In rapidity: $\varphi_{\text{tot}} = \ln2 + \ln2 = \ln4$ , $\mathcal{D} = e^{-\ln4} = 1/4$ .

Step 3 — the slow route, as a check. Einstein composition: $\beta_{\text{tot}} = (0.6+0.6)/(1+0.36) = 15/17 = 0.882353$ . Then $\mathcal{D} = \sqrt{(1-15/17)/(1+15/17)} = \sqrt{(2/17)/(32/17)} = \sqrt{1/16} = 1/4$ ✓. Same number, more work, and an exact fraction you had to carry.

Step 4 — the boundary number. $z_{\text{tot}} = 1/\mathcal{D} - 1 = 3$ . Note that $z$ did not add: $z_1 = z_2 = 1$ , but $z_{\text{tot}} = 3$ , because $(1+z)$ multiplies. Nor did $\beta$ : $0.6 + 0.6 \ne 0.882$ . Only $\varphi$ added — $\ln2 + \ln2 = \ln4 = 1.386294$ — and it added exactly.

Step 5 — read the structure. $N$ identical recessions of $\beta = 0.6$ give $\mathcal{D} = 2^{-N}$ , so ten relay ships redshift $600\ \mathrm{nm}$ to $614\ \mu\mathrm{m}$ — while $\beta_{\text{tot}} = \tanh(10\ln2) = 0.999998$ has barely moved. The velocity saturates; the redshift does not.

Physics of the portrait

Origins · 2 minChristian Doppler and how this was discovered

Christian Doppler

1803–1853 · Salzburg · Prague · Vienna

Born in Salzburg to a family of stonemasons and too frail for the trade, Christian Doppler spent his thirties in genteel academic poverty in Prague, and on 25 May 1842 read a paper to the Royal Bohemian Society of Sciences — before an audience of five members — titled Über das farbige Licht der Doppelsterne und einiger anderer Gestirne des Himmels. In it he stated the principle that carries his name, and then applied it, boldly and wrongly, to the colours of double stars: he supposed that a star's colour was its motion made visible. It is not. Stellar colours are set by temperature, and the velocities involved are some four orders of magnitude too small to shift a star's hue. The principle was right and the application was a fiasco — a distinction his contemporaries were not always careful to make. Christophorus Buys Ballot set out in 1845 to disprove him, hiring a brass band, loading trumpeters onto an open railway carriage on the Utrecht–Maarssen line, and stationing musicians with perfect pitch beside the track; the approaching note came in about a semitone sharp, and Buys Ballot confirmed what he had meant to demolish. Three years later Hippolyte Fizeau, who had not read Doppler, made the correction that mattered: the effect must be sought not in a star's overall colour but in the displacement of its spectral lines — which is why the French still call it the Doppler–Fizeau effect, and why William Huggins, in 1868, was able to point a spectroscope at Sirius and read off a radial velocity. Doppler, dead of tuberculosis in Venice at forty-nine, never saw a line shift measured.

History

1842

— On 25 May, Christian Doppler reads Über das farbige Licht der Doppelsterne to the Royal Bohemian Society of Sciences in Prague, stating the principle and misapplying it to stellar colours.

1845

— Buys Ballot tests the acoustic effect with trumpeters on a railway carriage between Utrecht and Maarssen, and confirms it while trying to refute it.

1848

— Hippolyte Fizeau, independently, points out that for light the effect must be sought in the displacement of spectral lines, not in overall colour.

1868

— William Huggins measures the radial velocity of Sirius from the shift of a spectral line, the first such measurement of a star.

1905

— Einstein, in section 7 of Zur Elektrodynamik bewegter Körper, derives the relativistic Doppler formula, with the time-dilation factor that Doppler and Fizeau could not have known about.

1963

— Maarten Schmidt identifies the Balmer lines of 3C 273 at a redshift of 0.158, and quasars begin.

Going further

Texts. Rindler, Relativity: Special, General and Cosmological, 2nd ed., §4.3 · Taylor & Wheeler, Spacetime Physics, 2nd ed., ch. 8 · French, Special Relativity, ch. 5 · Einstein, "Zur Elektrodynamik bewegter Körper", Annalen der Physik 17, 891 (1905), §7 · Schmidt, "3C 273: A Star-Like Object with Large Red-Shift", Nature 197, 1040 (1963).

Provenance. Signature $(+,-,-,-)$ ; angle convention band-locked to $\mathcal{D} = 1/[\gamma(1-\beta\cos\theta)]$ , $\theta$ measured to the photon's propagation direction. All numbers recomputed in verify_aq-sr-longitudinal-doppler-effect.py. Base defect repaired (D6): the base printed the angular formula with a plus sign, $1/[\gamma(1+\beta\cos\theta)]$ — self-consistent only under a line-of-sight convention, and irreconcilable with the sibling articles. Base defect found here: the base's second derivation ("Doppler from the Lorentz transformation of phase") reaches the right answer by two cancelling sign errors — it asserts that a source receding along $+\hat{x}$ emits a wave that "propagates in $+x$ " towards an observer it is receding from (geometrically impossible: the ray travels in $-\hat{x}$ , so $k_x = -\omega/c$ ), and then applies the boost with the wrong sign too. Right number, wrong physics. That card is 1907's payload in any case, and is demoted to a by-name recall here.

Scope. The covariant derivation (the wave four-vector, phase invariance, the master formula) is owned by aq-sr-doppler-four-momentum. The transverse case and the Ives–Stilwell test are owned by aq-sr-transverse-doppler-effect. Rapidity additivity is owned by aq-sr-rapidity-hyperbolic-boosts and aq-sr-rapidity. None of them are re-derived here.

Builds on. Time Dilation · The Lorentz Transformation Equations · Rapidity and the Additivity of Velocities. Leads to. Transverse Doppler Effect · Relativistic Doppler and the Photon Four-Momentum · Doppler-Shift Resolution of the Twin Paradox · Relativistic Aberration of Light.

Why does light have only one Doppler formula when sound has two?

Because sound has a medium and light does not. In air you can meaningfully ask "is the source moving, or am I?" — the wind gauge answers it — and the two cases genuinely give different frequencies, differing at order $\beta_s^2$ . For light there is no such question: the principle of relativity says the two set-ups are the same experiment described from two frames, so they must yield the same number. What is remarkable is where that single number sits: exactly at the geometric mean of the two acoustic answers, $\sqrt{(1-\beta)/(1+\beta)}$ , with each classical formula off by precisely one factor of $\gamma$ in opposite directions.

Is the redshift z the same thing as the velocity in units of c?

Only to first order, and the error grows fast. Exactly, $1+z = e^{\varphi}$ , so $z = \beta + \tfrac12\beta^2 + \cdots$ . At $z = 0.158$ (3C 273) the true kinematic $\beta$ is $0.1457$ — reading $\beta = z$ overstates it by $8.5\%$ . At $z = 1$ the naive reading gives $\beta = 1$ , nonsense; the correct kinematic value is $\beta = 0.6$ . And $z$ is not additive: two chained $z = 1$ shifts give $z_{\text{tot}} = 3$ , not $2$ , because $(1+z)$ multiplies. The additive variable is $\varphi = \ln(1+z)$ .

Can the redshift be arbitrarily large?

Kinematically, yes — $1+z = e^{\varphi}$ and $\varphi$ is unbounded, so $z \to \infty$ as $\beta \to 1^-$ . There is no maximum redshift, only a maximum velocity, and the two are not the same statement. This is the practical reason astronomers quote $z$ and not $v$ : $z$ is unbounded and multiplicative, so it degrades gracefully at high speed where $\beta$ is squeezed uselessly against $1$ . (Observed cosmological redshifts above $z \approx 10$ are not, of course, kinematic — see the warning above.)

Where exactly does the time dilation enter the crest-counting derivation?

In exactly one place: Step 2, where the emission rate in the observer's frame is $f_{\text{src}}/\gamma$ rather than $f_{\text{src}}$ . Everything else in the derivation — the retreating source, the extra light-travel time, the $(1+\beta)$ stretch — is Newtonian geometry and was available to Doppler in 1842. If you delete the $\gamma$ , you get $1/(1+\beta)$ , the classical source-moving formula, which at $\beta = 0.6$ gives $0.625$ instead of $0.500$ . The whole of relativity's contribution to the longitudinal effect is that one factor.

Why do the sibling articles write the angular formula with a minus sign?

Because $\theta$ is tied to the ray, not to the line of sight. The band convention is that $\theta$ is the lab-frame angle between the source's velocity and the photon's direction of propagation, so $\theta = 0$ means the photon is flying along the velocity, straight at you, from a source that is chasing it — an approach, a blueshift, $\mathcal{D} = 1/[\gamma(1-\beta)] = \gamma(1+\beta) > 1$ . If you instead measure $\theta$ from your line of sight towards the source, that vector is anti-parallel to the ray and every cosine flips, giving a plus sign. Both are internally consistent; the AllQFT base used the second and thereby contradicted its own siblings. One corpus, one convention.

Does the composition of Doppler shifts really need no correction term?

None whatever. $\mathcal{D}_{\text{tot}} = \mathcal{D}_1\mathcal{D}_2$ exactly, for any number of collinear shifts, because $\mathcal{D} = e^{-\varphi}$ and rapidities add exactly. The velocity-addition law, by contrast, is a rational function you must apply and re-apply. This is not a coincidence: $e^{\pm\varphi}$ are the eigenvalues of a Lorentz boost on the two null directions, and eigenvalues of composed maps multiply. Every awkward relativistic composition law in this chapter is awkward only because it is written in $\beta$ rather than $\varphi$ .