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The mission result description reports a broad asymmetric iron K-alpha profile with a low-energy wing relative to the approximately $6.4\mathrm{keV}$ rest energy. It discusses gravitational redshift and the relativistic Doppler effect in a disk model. The page presents the original observational result and its interpretation, rather than establishing that every possible absorption or illumination model is excluded.

The Doppler effect is a change between emitted and received frequency caused by motion of the source or receiver. An approaching siren can send more wave crests per second to a microphone than the source emits per second; a receding siren sends fewer. The source need not change its own oscillation. The Doppler shift is $\Delta f=f_o-f_s$, where $f_s>0$ is emitted frequency and $f_o>0$ is received frequency, both measured in hertz.

For acoustic waves, motion must be specified relative to the material carrying the wave. The speed of sound is measured in that material's rest reference frame. Moving the source changes the spacing of successive wavefronts; moving the receiver changes how quickly it crosses those wavefronts. These two mechanisms give different exact factors in the classical Doppler effect for sound, even for the same closing relative velocity.

This makes a microphone a motion probe: a stable tone supplies a repeating clock, and changes in arrival spacing reveal the source's motion toward or away from the receiver. An acoustic Doppler measurement can test the prediction directly. A passing source also illustrates why geometry matters: its speed can stay constant while the received frequency changes continuously, as in Doppler frequency along a trajectory.

The following sound model assumes linear acoustics in uniform stationary material with isotropy, constant speed of sound and negligible wave dispersion. Source and receiver speeds are below the speed of sound, and there is a single direct propagation path. Air motion requires transforming source and receiver velocities into the air's rest reference frame; a nonuniform flow requires a more general propagation model. At the sonic threshold the simple periodic-arrival picture fails; moving-point acoustic wavefronts and the Mach cone describe the relevant geometry, while finite shock waves require further physics. Light in vacuum instead requires the relativistic Doppler effect: the acoustic formula cannot be converted into its exact optical counterpart just by replacing the speed of sound with the speed of light.

A kinematic redshift is the relativistic Doppler effect from relative motion of emitter and receiver. With positive radial recession speed $v$, its small-velocity limit is $z_{\mathrm{kin}}=v/c+O(v^2/c^2)$. A peculiar velocity gives such a factor in addition to cosmological redshift.

A mirror receding at velocity $u=\beta c$ from normally incident light produces two applications of the relativistic Doppler effect, one into the mirror rest frame and one back to the observer. In that rest frame, reflection preserves frequency. Consequently ${\omega }_r={\gamma }^2(1-\beta {)}^2{\omega }_i={\omega }_i(1-\beta )/(1+\beta )$. Equivalently, continuity of wave phase on the moving surface $z=ut$ gives ${\omega }_i-k_{i}u={\omega }_r+k_{r}u$, with $k_i={\omega }_i/c$ and $k_r={\omega }_r/c$. For $|u|\ll c$, the fractional frequency change is $-2u/c$ to first order.

Assume homogeneous space and time, isotropic space, inertial frames related linearly, the relativity principle and a common isotropic speed $c$ for light. Take coincident origins at zero and parallel axes, with $S^{\prime }$ moving at $v$ along $x$. The primed origin requires $x^{\prime }=A(x-vt)$. Write $t^{\prime }=Bx+Dt$; mapping each light ray $x=\pm ct$ into $x^{\prime }=\pm c{t}^{\prime }$ gives $D=A$ and $B=-Av/c^2$. Rotational symmetry about the boost axis gives a common transverse scale $B(v)$; reversing the boost direction by spatial isotropy gives $B(v)=B(-v)$. Reciprocity then gives $B(v{)}^2=1$, and continuity and the common orientation select $B(v)=1$. Thus transverse directions are unchanged. Reciprocity requires the inverse to have the same scale $A$ with $v$ replaced by $-v$. Composing gives $A^2(1-v^2/c^2)=1$; continuity from $A=1$ at $v=0$ selects the positive root. Hence the Lorentz transformation isThe printed wavelength formula uses the angle of the line of sight from observer towards star, opposite to photon propagation. Choose the transverse axis so the ray lies in the $xy$ plane. With that directed-angle convention the photon four-momentum isThe inverse coordinate boost takes a rest-frame four-vector to $S$, giving $p^0=\gamma (p^{\prime 0}+\beta p^{\prime 1})$ and $p^1=\gamma (p^{\prime 1}+\beta p^{\prime 0})$, where $\beta =v/c$. Thus $\nu =\gamma {\nu }^{\prime }(1-\beta \cos {\theta }^{\prime })$, and the aberration of light relation is $\cos \theta =(\cos {\theta }^{\prime }-\beta )/(1-\beta \cos {\theta }^{\prime })$. More directly, the boost from $S$ to $S^{\prime }$ gives $p^{\prime 0}=\gamma (p^0-\beta p^1)$, hence ${\nu }^{\prime }=\gamma \nu (1+\beta \cos \theta )$. Since wavelength is $c/\nu$, the relativistic Doppler effect givesIf the angle were instead measured along the actual photon momentum, both cosine signs would reverse and the same physical relation would have a minus sign. Specifying this orientation avoids confusing the two versions.

For $\cos \theta =1$, $\lambda /{\lambda }^{\prime }=\sqrt{(1+\beta )/(1-\beta )}$: a receding star is redshifted. For $\cos \theta =-1$, the ratio is $\sqrt{(1-\beta )/(1+\beta )}$: an approaching star is blueshifted. For $\cos \theta =0$, $\boxed{\lambda /{\lambda }^{\prime }=\gamma }$: the transverse redshift is the time-dilation effect. This transverse condition refers to the observer's angle, not ${\theta }^{\prime }=\pi /2$ in the source frame.

Choose the transverse axis so that the photon travels in the $xy$ plane. Its energy-momentum four-vector is $p^{\mu }=(E/c,E\cos \theta /c,E\sin \theta /c,0)$. A Lorentz boost of speed $v$ along $x$, with $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$, givesBecause a photon has zero invariant mass, $|{\mathbf{p}}^{\prime }|=E^{\prime }/c$. ThereforeThe first is the relativistic Doppler effect in angular form, and the second is relativistic aberration.

Take $v>0$ for recession. By the definition of kinematic redshift, $1+z_{\mathrm{kin}}={\nu }_e/{\nu }_0$. Inverting the supplied relativistic Doppler effect and expanding with $\beta =v/c$ givesThusA negative radial velocity gives a blueshift at this order.

A kinematic redshift is the relativistic Doppler effect arising from relative motion of emitter and receiver. In a local inertial frame, recession decreases the received frequency and approach increases it. A cosmological redshift instead compares photons emitted and received by comoving observers at different cosmic times: expansion of the scale factor stretches their wavelengths. It exists even when both observers have zero peculiar velocity. Both are frequency ratios measured by observers, but a large cosmological redshift should not be inserted into a flat-spacetime radial-speed formula to define the galaxy's unique recession speed.

Choose the photon direction in the $x^{\prime }$–$y^{\prime }$ plane. Its four-momentum in the stellar rest frame isLet $\beta =v/c$, $\gamma =(1-{\beta }^2{)}^{-1/2}$. The Lorentz transformation to $S$ givesUsing $p^0=h\nu /c$ and $p^x=(h\nu /c)\cos \theta$ yields the relativistic Doppler effect and relativistic aberration:

For the oppositely directed transverse photons, their rest-frame spatial momenta cancel and their total energy is $E=2h{\nu }^{\prime }$. Boosting this combined four-vector gives laboratory energy $\gamma E$ and momentumThe star loses rest mass $E/c^2$. At unchanged speed its momentum therefore decreases by $\gamma (E/c^2)v$, exactly the photon momentum. Its laboratory energy loss is likewise $\gamma E$. There is no rest-frame recoil from the symmetric emission, and constant speed is consistent with four-momentum conservation because the stellar mass decreases.

Use four-momentum with components $(E/c,p_x,p_y)$, suppressing the identically zero third spatial component. The photon in the primed frame hasIts energy is $h{\nu }^{\prime }$ and its spatial momentum has magnitude $h{\nu }^{\prime }/c$. Apply the same Lorentz transformation as for the spacetime coordinates. With $\beta =v/c$,The time component gives the relativistic Doppler effectThe aberration of light follows by dividing spatial components by $E/c$:In particularwhere the tangent expression is understood only when its denominators exist. The sine and cosine equations determine the correct quadrant and also cover directions where the tangent/secant expression is undefined.

For the photon-number half-cone under relativistic aberration, the forward half of the primed emission angles is $-\pi /2\le {\theta }^{\prime }\le \pi /2$. The aberration of light is monotone on this interval, and its endpoints have $\cos \theta =\beta$. Thus exactly half of a uniform angular photon distribution lies within the lab-frame coneAs $v\to c^{-}$, $\alpha =\arcsin (1/\gamma )\sim 1/\gamma$, so the cone becomes narrow. Here “half” counts emitted photons or ray directions; it is not a claim that half of the observed energy or power lies there, since the relativistic Doppler effect changes the photon energies with angle.

Take $u>0$ to mean that the mirror recedes along the incident plane electromagnetic wave. Put $\beta =u/c$, with $|\beta |<1$. The moving-mirror Doppler shift can be obtained directly by matching wave phase on the moving surface. Write the incident wave phase as $k_{i}z-{\omega }_{i}t$ and the reflected wave phase as $-k_{r}z-{\omega }_{r}t$, where $k_i={\omega }_i/c$ and $k_r={\omega }_r/c$ are positive wavenumbers. At the mirror, $z=z_0+ut$, the two phases must have the same time dependence; their constant offset incorporates the reflection phase shift. ThusConsequently the observed reflected angular frequency and ordinary frequency areThis is a redshift for a receding mirror and a blueshift for an approaching mirror. At small velocity, ${\omega }_r\simeq {\omega }_i(1-2u/c)$.

Equivalently, the relativistic Doppler effect gives ${\omega }_i^{\prime }=\gamma (1-\beta ){\omega }_i$ in the mirror rest inertial frame. Reflection there reverses propagation and preserves frequency. Transforming the backward-going reflected wave to the laboratory gives another factor $\gamma (1-\beta )$, so ${\omega }_r={\gamma }^2(1-\beta {)}^2{\omega }_i={\omega }_i(1-\beta )/(1+\beta )$. A single Doppler factor would miss the effect of the return path.

Use the inverse coordinate transformation,Substitution into the wave phase $\mathbf{k}\cdot \mathbf{r}-\omega t$ and comparison of the coefficients of ${\mathbf{r}}^{\prime }$ and $t^{\prime }$ givesThe first is the relativistic Doppler effect; the changed direction of the wavevector describes relativistic aberration. Equivalently,The electric-field wave amplitude follows from part (b), after rotating the boost axis to $\mathbf{v}$:For a vacuum plane wave, Maxwell equations give ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$ and $\mathbf{k}\cdot {\mathbf{E}}_0=0$. Thus an expression using only the given electric wave amplitude and wavevector isEquivalently, ${\mathbf{E}}_0^{\prime }=\gamma ({\mathbf{E}}_0+\mathbf{v}\times {\mathbf{B}}_0)-[{\gamma }^2/(\gamma +1)]\mathbf{v}(\mathbf{v}\cdot {\mathbf{E}}_0)/c^2$. The wave amplitude is not a scalar carried unchanged by the wave phase transformation; its direction must remain transverse to the transformed wavevector.

The numerical coordinates assigned to an event can be changed without changing the event. It does not follow that every relation between coordinate systems is devoid of physical content. A standard inertial frame includes a family of resting measuring standards and synchronized clocks. Its coordinates are normalized so that the Minkowski metric has the same standard form. The relation between two such descriptions therefore records properties of the standards and of spacetime, not just an arbitrary choice of names.

Write $x^{\mu }=(ct,x,y,z)$ and use metric signature $(-,+,+,+)$. A Lorentz transformation satisfiesAn origin translation can also be included without changing the Minkowski interval. The displayed condition preserves the interval between events, including the null cone and proper time along a timelike curve. Its physical significance lies in the claim that comparisons made with actual idealized rods and clocks obey this same metric structure.

A more general passive relabeling is allowed, but then the metric components must change with the labels. For example, let $t^{\prime }=2t$ and leave $x,y,z$ fixed. In the new coordinates,Nothing physical has changed, but $t^{\prime }$ no longer has the original clock-normalized meaning of $t$. Retaining the old metric components as well as the new labels would change the mathematical model. This distinction is part of the physical content of a passive Lorentz transformation: the transformation between standard inertial coordinate systems preserves a specified metric form, whereas arbitrary relabelings need not preserve that form.

For an operational example, a clock moving at velocity $v$ between two events on its world line accumulatesin the first frame. In its own rest frame the same result is simply the interval between its local readings. The equality is a physical agreement about one clock, derived using the preserved Minkowski interval. It is not a claim that the two coordinate-time differences are identical. Similarly, an observer measures a light angular frequency through $\omega =-k_{\mu }{u}^{\mu }$, where $k^{\mu }$ is the wave four-vector and $u^{\mu }$ the observer's four-velocity. Transforming both sets of components leaves this scalar unchanged. Comparing two different observers gives the physical relativistic Doppler effect rather than a disagreement over one observer's measurement.

There is a further distinction between passive transformations of the same situation and active changes to a situation. Re-expressing a particular experiment in different coordinates is always possible if all geometric fields are transformed consistently. Preparing a second, moving copy of the apparatus is a different physical arrangement. Its relation to the first is governed by a dynamical symmetry only if the relevant laws and background structure have the appropriate invariance. An active Lorentz transformation of a vector field makes the changed field configuration explicit.

This is why mere Lorentz covariance of notation must not be confused with the Lorentz invariance of the dynamics. A physically preferred timelike vector field can be transformed as a vector under passive relabeling while distinguishing particular states of motion. Holding that preferred field fixed can destroy the active boost symmetry of the matter dynamics. The covariant equation with a preferred timelike vector is a concrete example.

Coordinates are therefore labels, but the Lorentz transformation relates labels tied to shared physical standards and a definite metric structure. Its measurable consequences follow from that structure and the dynamical assumptions, not from the bare act of changing labels.^[ref]

Choose the positive axis from the approaching source towards the observer. The photon that reaches the observer is emitted forward, with source-frame four-momentum $(E/c,E/c,0,0)$. Boosting to the observer frame gives $E_{\mathrm{obs}}=\gamma E(1+\beta )$, where $\beta =v/c$. Since photon energy is $h_{\mathrm{P}}\nu$,This is the approaching-source relativistic Doppler effect. A photon emitted in a general source-frame direction would instead have the factor $\gamma (1+\beta \cos {\theta }_{\ast })$; forward emission is the geometry of the photon directed to the observer here.

Let the source's proper emission period be $T_0$, and let it approach a stationary observer directly with speed $v=\beta c>0$. In the observer's inertial frame, consecutive emissions are separated by $\gamma T_0$ through time dilation. The second crest starts $v\gamma T_0$ nearer the observer, reducing the arrival interval toSince light travels at $c$ in each inertial frame, the longitudinal relativistic Doppler effect isRecession reverses the sign of $\beta$. More generally, if $\Theta$ is the angle between the source velocity and the photon propagation direction in the observer's frame, a Lorentz transformation of the photon four-momentum gives ${\nu }_0=\gamma {\nu }_{\mathrm{obs}}(1-\beta \cos \Theta )$.

The common Doppler scaling of spectral-line separations follows because every line receives the same wavelength factor $D$. The supplied wavelengths have rest separation $50\mathrm{nm}$; the measured separation gives $D=51/50=1.02$. Defining ${\beta }_{\mathrm{rec}}>0$ for recession, inversion givesSource contradiction: a strictly approaching source has $D<1$ and must reduce the wavelength separation below $50\mathrm{nm}$, so no positive approach speed satisfies the stated $51\mathrm{nm}$. For example, approach at $0.02c$ gives a separation of $49.0098\mathrm{nm}$. The speed inferred from the numerical data is therefore about $0.0198c$ receding; retaining the word “approaching” would require a negative signed approach velocity. The two observed wavelengths consistent with the given separation are $357$ and $408\mathrm{nm}$.

The photon propagates along $-x$ in the tunnel frame and has energy $E_{\gamma }=hc/{\lambda }_0$ and momentum $p_x=-E_{\gamma }/c$. A Lorentz transformation into the approaching runner's frame givesSince $E_{\gamma }^{\prime }=hc/{\lambda }^{\prime }$, the approaching-source relativistic Doppler effect isIt is a blueshift. With $q={\lambda }^{\prime }/{\lambda }_0=500.0/589.1$, solving $q^2=(1-\beta )/(1+\beta )$ givesThis shift depends on the relative velocity along the light ray, not on the tunnel length or the location of detection.

Let $S^{\prime }$ move at $+v$ along $x$ relative to $S$. The Lorentz transformation of energy and momentum, from primed to unprimed components, isThe inverse equations have $v$ replaced by $-v$ and the primes interchanged. A photon making angle ${\theta }^{\prime }$ with $+x$ in $S^{\prime }$ has $p_x^{\prime }=(E^{\prime }/c)\cos {\theta }^{\prime }$. Thus its general Doppler factor isThe requested blueshifted expression assumes propagation parallel to the boost, ${\theta }^{\prime }=0$:The propagation-direction assumption matters: for an oppositely travelling photon, $D=\gamma (1-\beta )$, the reciprocal of the displayed factor. Relative frame speed alone does not select the longitudinal relativistic Doppler effect.

First work in the receiver's rest frame. If a source recedes at speed $\beta c$, two successive emissions separated by source proper time $T_0$ occur $\gamma T_0$ apart in this frame. During that interval the source retreats by $\beta c\gamma T_0$, so the arrival interval is $T_R=\gamma T_0(1+\beta )$. The longitudinal relativistic Doppler effect is consequentlyA negative $\beta$ describes approach and gives a blueshift.

Interpret $633\mathrm{nm}$ as the laser's rest-frame wavelength and the stated detector velocity as measured in the initial laboratory frame. Choose $+x$ from source to sensor, with detector velocity $-0.2c$. The required relative recession speed follows fromBy relativistic velocity addition, the source velocity ${\beta }_{S}c$ in the laboratory obeys $({\beta }_S+0.2)/(1+0.2{\beta }_S)=-{\beta }_{\mathrm{rec}}$. HenceThe laser must move away from the sensor at about $0.5784c$ in that laboratory frame. Although the sensor moves toward the source's initial location, the retreating source produces the required net redshift. The frame qualification matters: a velocity permanently fixed relative to the moving source would specify the Doppler factor already, rather than leave it adjustable.

For a particle of rest mass $m_0$ and velocity $\mathbf{u}$, the relativistic energy and momentum areFor a photon of wavelength $\lambda$ travelling along $\hat{\mathbf{n}}$,Here $h$ is Planck constant and $c$ is the speed of light.

Let $\epsilon$ and ${\epsilon }^{\prime }$ be the incident and scattered photon energies. The stationary electron initially has energy $m_e{c}^2$. By four-momentum conservation, its final energy and momentum areUse the electron's relativistic energy-momentum relation, rather than assuming its recoil is nonrelativistic:Cancelling common terms givesSince $\lambda =hc/\epsilon$, the Compton wavelength shift isThe word elastic here means that the final electron has the same rest mass as initially and the complete photon-electron system conserves total energy; the photon alone loses energy to recoil.

For backscattering, $1-\cos \theta =2$, and henceChoose the incident direction as $+x$, so that the scattered photon travels in the $-x$ direction. An observer moving with velocity $+\beta c\hat{\mathbf{x}}$ meets that photon head-on. The Lorentz transformation of four-momentum gives the observed energyThis is the approaching-observer relativistic Doppler effect. Set $D=\epsilon /{\epsilon }^{\prime }=2.511/0.511$ and solve:The observer must move opposite to the scattered photon, along the original incident direction, at approximately $0.9205c$. Moving with the scattered photon would redshift it instead.

Take positive velocity towards $B$, and interpret $100\mathrm{MHz}$ as the emitted frequency in the probe's rest frame. The relativistic velocity-addition formula gives the probe's station-frame velocity:For reception at the station, photons propagate backwards while the probe recedes. The longitudinal relativistic Doppler effect givesBefore the probe overtakes $B$, their separation decreases. The probe's velocity in $B$'s inertial frame isThe approaching-source relativistic Doppler effect therefore givesAn independent frame calculation gives the same result: forward photons have station-frame frequency $300\mathrm{MHz}$, and a receiver moving at $0.4c$ measures $300\sqrt{0.6/1.4}\mathrm{MHz}$. If reception occurs after the probe passes $B$, backwards photons instead give $f_B=100\sqrt{7/27}\mathrm{MHz}\simeq 50.92\mathrm{MHz}$. The direction of propagation and the reception geometry are essential to choosing the Doppler factor.