Gravitational Redshift in Schwarzschild Spacetime

Gravitational redshift is the simplest physical prediction one can read straight off the Schwarzschild metric, and it is the cleanest expression of the idea at the heart of general relativity: gravity is the geometry of time. Set two observers at rest at fixed Schwarzschild radii $r_{\mathrm{emit}}$ and $r_{\mathrm{rec}}$ , let the inner one emit light of proper frequency $\nu_{\mathrm{emit}}$ , and let the outer one measure it. The received frequency is lower — the light is reddened — for two reasons that turn out to be one. Clocks deep in the well run slow relative to clocks higher up, so a wave ticking a fixed number of crests per emitter-second arrives spread over more receiver-seconds. Equivalently, a photon costs energy to climb, and since $E=h\nu$ it pays that toll in frequency. Both statements live in one metric component, $g_{tt}=-(1-2GM/c^2 r)$ . The derivation rests on two facts about static spacetime. First, Schwarzschild geometry is time-independent, so it carries a timelike Killing vector $\xi=\partial_t$ , and the contraction $E=-p_\mu\xi^\mu$ is conserved along any geodesic — including the photon's null worldline. Second, the frequency a given observer measures is not $E$ but the projection $h\nu=-p_\mu u^\mu$ of the four-momentum onto that observer's four-velocity, which for a static observer is normalized by the local lapse $\sqrt{-g_{tt}}$ . Combining the two gives a ratio that depends only on $g_{tt}$ at the endpoints; the photon's path drops out entirely. That is why the result is so robust — radial or bent, the trajectory never enters — and why the divergence at the horizon, where $g_{tt}\to 0$ , is a statement about static observers rather than curvature. The weak-field limit recovers $\Delta\nu/\nu\approx-\Delta\Phi/c^2$ , tested by Pound and Rebka in a tower, by Gravity Probe A with a maser, and corrected for daily inside GPS.

Key takeaways

A photon climbing out of the Schwarzschild well is redshifted: its frequency measured by a static observer decreases with increasing radius, with no motion, scattering, or matter required.
For two static observers the frequency ratio is $\nu_{\mathrm{rec}}/\nu_{\mathrm{emit}}=\sqrt{(1-2GM/c^2 r_{\mathrm{emit}})/(1-2GM/c^2 r_{\mathrm{rec}})}$ , a purely geometric factor set by $g_{tt}$ at the two radii.
It follows from the conserved photon energy $E=-p_\mu\xi^\mu$ along its geodesic ( $\xi=\partial_t$ the timelike Killing vector) projected onto each observer's clock rate; the path never appears.
As the emitter approaches the horizon $r\to 2GM/c^2$ the redshift diverges — signals freeze and fade to infinite redshift — because no static observer exists there.
In the weak field the shift reduces to $\Delta\nu/\nu\approx-\Delta\Phi/c^2$ (the 22.5 m Pound–Rebka tower gives $2.46\times10^{-15}$ , GPS a net $+38\,\mu\mathrm{s/day}$ ); it is the same statement as gravitational time dilation, $\mathrm{d}\tau/\mathrm{d}t=\sqrt{-g_{tt}}$ .
Key idea: the frequency ratio between a static emitter and receiver equals the ratio of the square roots of $-g_{tt}=(1-2GM/c^2r)$ at the two radii, because each static observer's proper time relates to coordinate time by that lapse factor while the photon carries a conserved Killing energy along its path.

Where this lives today

Gravitational redshift has moved from a hard-to-see prediction to a routine tool and a precision probe. It is corrected for continuously inside GPS: an orbiting clock runs fast by about $+45\,\mu\mathrm{s/day}$ from the gravitational term, partly offset by the special-relativistic $-7\,\mu\mathrm{s/day}$ of its speed, a net $+38\,\mu\mathrm{s/day}$ that uncorrected would corrupt positions by $\sim10\,\mathrm{km}$ per day. The 2018 analysis of two Galileo satellites accidentally launched into eccentric orbits turned that liability into a test, confirming the radial dependence to a few parts in $10^5$ . At the other extreme, the GRAVITY interferometer tracked the star S2 through its 2018 perihelion around the four-million-solar-mass black hole Sagittarius A and detected a redshift of order $z\sim2\times10^{-4}$ where Newtonian gravity predicts none — general relativity confirmed at the Galactic Center. Optical-lattice atomic clocks have pushed the effect to the tabletop, resolving the shift across height differences of a single millimeter and even within one millimeter-tall cloud of strontium atoms. The same $\sqrt{-g_{tt}}$ shapes black-hole images directly: the Event Horizon Telescope pictures of M87 and Sgr A* show a dark central "shadow" sculpted in part by the diverging redshift near the horizon that this article computes.

The physical picture

Picture a photon as a tiny clock whose oscillation period is its inverse frequency. As it climbs out of the well the clocks below it run slow relative to those above, so the fixed number of oscillations stamped in at emission spreads over more time when read out higher up — the light looks redder. Energetically the same thing is a toll: the photon cannot slow down, its local speed being always $c$ , so it pays for the climb by shedding energy, and by $E=h\nu$ that means shedding frequency. Neither description involves the trajectory. Because the Killing energy is conserved along the whole geodesic and only the local clock rates differ, a bent path and a radial path give the identical shift — only $g_{tt}$ at emission and reception enter. The first figure freezes this into one picture: a static emitter low in the well, a static receiver high up, and a wave that stretches as it rises.

Figure

A photon climbing out of the Schwarzschild well from a static emitter at r_emit to a static receiver at r_rec loses frequency: the received-to-emitted ratio is √((1−2GM/c²r_emit)/(1−2GM/c²r_rec)) < 1, fixed entirely by g_tt at the two radii

Near the horizon the effect runs away. A distant observer watching an object sink toward $r=2GM/c^2$ sees its light shifted ever redder and dimmer, the source appearing to freeze and fade — the "frozen star" picture — because the static reference clocks there approach a standstill relative to infinity. The second figure plots the single factor that governs everything: the static-clock rate $\sqrt{1-r_s/r}$ , which is also the redshift factor to infinity, versus radius in units of $r_s=2GM/c^2$ .

Figure

The static-clock and redshift factor √(1−r_s/r) versus r/r_s with r_s = 2GM/c²: it falls from 1 far away to 0 at the horizon, so the surface redshift z = 1/√(1−r_s/R) − 1 diverges — a 1.4 M_⊙, 10 km neutron star sits at r/r_s ≈ 2.4 with factor 0.77 and z ≈ 0.31

The core equations

Everything begins with the metric, because $g_{tt}$ is the only ingredient that survives into the final answer. In Schwarzschild coordinates the line element for the vacuum outside a spherical mass $M$ is static and spherically symmetric, and the redshift is written entirely in terms of its time–time component.

\mathrm{d}s^2=-\left(1-\frac{2GM}{c^2 r}\right)c^2\,\mathrm{d}t^2+\left(1-\frac{2GM}{c^2 r}\right)^{-1}\mathrm{d}r^2+r^2\,\mathrm{d}\Omega^2 \qquad (1)

where: $M$ — mass of the central body; $r,t$ — Schwarzschild radial and time coordinates; $\mathrm{d}\Omega^2=\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\varphi^2$ — metric on the unit 2-sphere; $G,c$ — Newton's constant and the speed of light. Reading it: the factor $1-2GM/c^2 r$ vanishes at the horizon $r=2GM/c^2$ and tends to 1 far away. A static observer sits at fixed $r,\theta,\varphi$ , so only the time component of its four-velocity is nonzero. Normalizing that four-velocity to $g_{\mu\nu}u^\mu u^\nu=-c^2$ forces the lapse factor into $u^t$ .

u^\mu_{\mathrm{static}}=\left(\frac{1}{\sqrt{1-2GM/c^2 r}},\,0,\,0,\,0\right) \qquad (2)

where: $u^\mu$ — observer four-velocity (in units where the timelike norm fixes the lapse). Reading it: the closer the observer sits to the horizon, the larger $u^t$ must be — its proper time crawls relative to coordinate time by exactly $\sqrt{-g_{tt}}$ , and standing still there costs unbounded proper acceleration. Because the metric does not depend on $t$ , the vector $\xi=\partial_t$ is a timelike Killing vector, and its contraction with the photon four-momentum is a constant of the motion along the null geodesic.

E\equiv-p_\mu\,\xi^\mu=-p_t=\text{const along the photon geodesic},\qquad \xi=\partial_t \qquad (3)

where: $p_\mu$ — photon four-momentum; $\xi^\mu=(1,0,0,0)$ — timelike Killing vector; $E$ — conserved Killing energy. Reading it: $E$ is the energy "at infinity"; it is the same number at emission and reception, which is precisely why the trajectory cancels out. The frequency a particular observer assigns to the photon is the energy it reads in its own frame — the projection of $p_\mu$ onto its four-velocity, divided by Planck's constant.

\nu=-\frac{1}{h}\,p_\mu u^\mu \qquad (4)

where: $\nu$ — frequency measured by an observer with four-velocity $u^\mu$ ; $h$ — Planck constant, which cancels in every ratio. Reading it: two observers passing the same photon read different $\nu$ because they carry different $u^\mu$ , even though $E$ is shared. Feeding the static four-velocity (2) and the conserved $E$ (3) into the projection (4) and taking the ratio of receiver to emitter gives the central result of the article.

\boxed{\dfrac{\nu_{\mathrm{rec}}}{\nu_{\mathrm{emit}}}=\sqrt{\dfrac{1-2GM/c^2 r_{\mathrm{emit}}}{1-2GM/c^2 r_{\mathrm{rec}}}}} \qquad (5)

where: $r_{\mathrm{emit}},r_{\mathrm{rec}}$ — radii of emitter and receiver; $\nu_{\mathrm{emit}},\nu_{\mathrm{rec}}$ — emitted and received proper frequencies. Reading it: for $r_{\mathrm{rec}}>r_{\mathrm{emit}}$ the numerator is smaller, so the ratio is below 1 — a redshift. Spectroscopists quote the same effect as a fractional redshift $z$ , the change in wavelength over the emitted wavelength.

z\equiv\dfrac{\lambda_{\mathrm{rec}}-\lambda_{\mathrm{emit}}}{\lambda_{\mathrm{emit}}}=\dfrac{\nu_{\mathrm{emit}}}{\nu_{\mathrm{rec}}}-1=\sqrt{\dfrac{1-2GM/c^2 r_{\mathrm{rec}}}{1-2GM/c^2 r_{\mathrm{emit}}}}-1 \qquad (6)

where: $z$ — fractional redshift; $z>0$ climbing out, $z\to\infty$ as $r_{\mathrm{emit}}\to 2GM/c^2$ . Reading it: for a receiver at infinity this gives the surface redshift $z=1/\sqrt{1-2GM/c^2 R}-1$ . Finally, the whole phenomenon is a comparison of clocks: a static clock's proper time ties to coordinate time through the same lapse, so the redshift factor and the time-dilation factor are literally the same function.

\frac{\mathrm{d}\tau}{\mathrm{d}t}=\sqrt{1-\frac{2GM}{c^2 r}} \qquad (7)

where: $\tau$ — proper time of a static observer; $t$ — Schwarzschild coordinate time, shared by all static observers. Reading it: a periodic signal is a clock, so (7) at two radii reproduces (5) with no reference to photons at all.

Derivation ARedshift from the conserved Killing energy

The goal is to obtain the exact ratio (5) directly from geometry, showing that the photon's conserved energy cancels and the trajectory never appears, so that only the values of $g_{tt}$ at the two endpoints matter. We work in geometric units $G=c=1$ , abbreviate the lapse-squared as $f(r)=1-2M/r$ , and restore units at the end; the argument uses only that the metric is static and that both observers are static, so it holds for radial and non-radial photons alike.

Step 1. Because $\xi=\partial_t$ is a Killing vector of the static metric, the quantity $E=-p_\mu\xi^\mu=-p_t$ is constant along the photon's geodesic, taking the same value at emission and reception. Why: Killing's equation guarantees $p_\mu\xi^\mu$ is conserved along any geodesic, so the "energy at infinity" is one number for the whole flight.

Step 2. Any observer measures the photon energy as the projection onto its own four-velocity, $h\nu=-p_\mu u^\mu$ ; this is what a local frequency counter or an atom's absorption actually registers. Why: frequency is a frame-dependent readout, and the invariant that becomes it is the contraction of $p_\mu$ with the observer's $u^\mu$ , not $E$ itself.

Step 3. For a static observer only $u^t=f(r)^{-1/2}$ is nonzero, so the projection keeps a single term, $h\nu=-p_t\,u^t=E/\sqrt{f(r)}$ , using $p_t=-E$ . Why: standing still kills the spatial components of $u^\mu$ , leaving the lapse $\sqrt{f}$ as the sole geometric factor that separates $E$ from the locally seen $\nu$ .

Step 4. Form the receiver-to-emitter ratio and watch $E$ cancel: $\nu_{\mathrm{rec}}/\nu_{\mathrm{emit}}=\big(E/\sqrt{f(r_{\mathrm{rec}})}\big)\big/\big(E/\sqrt{f(r_{\mathrm{emit}})}\big)=\sqrt{f(r_{\mathrm{emit}})/f(r_{\mathrm{rec}})}$ . Restoring units sends $2M/r\to 2GM/c^2 r$ , giving the boxed result (5). Why: because the same conserved $E$ appears in numerator and denominator, the photon's path integrates out completely, leaving a pure ratio of endpoint lapses.

Go deeper · Derivation B · 1 minRedshift as a comparison of static clocks

Derivation BRedshift as a comparison of static clocks

The goal is to reach the same ratio (5) from a wave-crest bookkeeping argument that never mentions momentum, making explicit that gravitational redshift and gravitational time dilation are one phenomenon seen twice. We use only that the metric is independent of $t$ — so a wavefront's coordinate-time offset is preserved as it propagates — and that proper time along a static worldline is set by the lapse; this route generalizes cleanly to any static spacetime.

Step 1. For a static worldline only $g_{tt}$ contributes to $\mathrm{d}s^2$ , so proper time and coordinate time are related by $\mathrm{d}\tau=\sqrt{-g_{tt}}\,\mathrm{d}t=\sqrt{1-2GM/c^2 r}\;\mathrm{d}t$ . Why: with $\mathrm{d}r=\mathrm{d}\theta=\mathrm{d}\varphi=0$ , the line element reduces to a single term, and the square root of its coefficient is the local clock rate.

Step 2. Two successive wave crests leave the emitter separated by coordinate time $\Delta t$ and arrive at the receiver separated by the same $\Delta t$ , because the metric does not depend on $t$ and both observers hold fixed spatial coordinates. Why: staticity makes coordinate time a shared bookkeeping label, so a coordinate-time interval is preserved as the crest pattern propagates outward.

Step 3. Convert the shared $\Delta t$ to each observer's proper period, $\Delta\tau=\sqrt{-g_{tt}}\,\Delta t$ , and use $\nu=1/\Delta\tau$ to get $\nu_{\mathrm{rec}}/\nu_{\mathrm{emit}}=\Delta\tau_{\mathrm{emit}}/\Delta\tau_{\mathrm{rec}}=\sqrt{-g_{tt}(r_{\mathrm{emit}})}/\sqrt{-g_{tt}(r_{\mathrm{rec}})}$ , identical to (5). Why: equal coordinate-time intervals map to unequal proper periods, and the frequency is the inverse proper period, so the redshift is nothing but the ratio of the two clock rates.

Go deeper · Derivation C · 1 minThe weak-field limit and the Pound–Rebka prediction

Derivation CThe weak-field limit and the Pound–Rebka prediction

The goal is to expand the exact ratio (5) to first order in the small quantity $2GM/c^2 r$ , recover the Newtonian result $\Delta\nu/\nu\approx-\Delta\Phi/c^2$ , then specialize to a uniform-field tower to predict the Pound–Rebka shift. We assume $2GM/c^2 r\ll 1$ at both radii, which holds at the Earth's surface where $2GM_\oplus/c^2 R_\oplus\approx 1.4\times10^{-9}$ , so only the linear term is kept.

Step 1. Taylor-expand each square root: $\sqrt{1-2GM/c^2 r}\approx 1-GM/c^2 r$ , valid to first order. Why: the redshift factor is a smooth function of the small parameter, and its leading correction is exactly minus the Newtonian potential per $c^2$ .

Step 2. Form the ratio to linear order, $\nu_{\mathrm{rec}}/\nu_{\mathrm{emit}}\approx 1-GM/c^2 r_{\mathrm{emit}}+GM/c^2 r_{\mathrm{rec}}$ , so that $\Delta\nu/\nu=(\nu_{\mathrm{rec}}-\nu_{\mathrm{emit}})/\nu_{\mathrm{emit}}\approx-\big(GM/c^2\big)\big(1/r_{\mathrm{emit}}-1/r_{\mathrm{rec}}\big)$ . Why: keeping only linear terms turns the ratio of square roots into a difference of potentials, the form Einstein wrote in 1911 before the exact metric existed.

Step 3. Identify the Newtonian potential $\Phi(r)=-GM/r$ , giving $\Delta\nu/\nu=-\Delta\Phi/c^2$ with $\Delta\Phi=\Phi(r_{\mathrm{rec}})-\Phi(r_{\mathrm{emit}})$ ; light rising to higher (less negative) potential loses frequency. Why: this exposes the effect as a pure potential difference, independent of the field's detailed profile, which is why it can be tested in any gravity, not just a black hole's.

Step 4. For a tower of height $H$ in effectively uniform surface gravity $g=GM/R_\oplus^2$ , the potential difference is $\Delta\Phi=gH$ , so $\Delta\nu/\nu=-gH/c^2$ . With $g=9.81\,\mathrm{m/s^2}$ and $H=22.5\,\mathrm{m}$ this is $9.81\times22.5/(2.998\times10^8)^2=2.46\times10^{-15}$ . Why: this is the number Pound and Rebka set out to measure, and reproducing it from the metric is the bridge between exact geometry and a bench-top experiment. *This gives $\Delta\nu/\nu=gH/c^2=2.46\times10^{-15}$ for $H=22.5\,\mathrm{m}$ , the value Pound & Rebka (1960) set out to measure.*

Limiting cases

The exact ratio $\nu_{\mathrm{rec}}/\nu_{\mathrm{emit}}=\sqrt{(1-2GM/c^2 r_{\mathrm{emit}})/(1-2GM/c^2 r_{\mathrm{rec}})}$ specializes cleanly in every regime of interest.

Case	Result	Meaning
Weak field	$\dfrac{\Delta\nu}{\nu}\approx-\dfrac{\Delta\Phi}{c^2}=-\dfrac{GM}{c^2}\left(\dfrac{1}{r_{\mathrm{emit}}}-\dfrac{1}{r_{\mathrm{rec}}}\right)$	For $2GM/c^2 r\ll 1$ the redshift reduces to a difference of Newtonian potential over $c^2$ — the Pound–Rebka, Gravity Probe A, and GPS regime.
Receiver at infinity	$\dfrac{\nu_\infty}{\nu_{\mathrm{emit}}}=\sqrt{1-\dfrac{2GM}{c^2 r_{\mathrm{emit}}}}$	With $r_{\mathrm{rec}}\to\infty$ , $g_{tt}\to-1$ , so the distant frequency is suppressed by the lapse at the emitter alone — the surface-redshift formula.
Horizon limit	$\displaystyle\lim_{r_{\mathrm{emit}}\to 2GM/c^2}\dfrac{\nu_{\mathrm{rec}}}{\nu_{\mathrm{emit}}}=0$	As the emitter nears the horizon the received frequency tends to zero: infinite redshift, $z\to\infty$ . No static emitter exists at or below $r=2GM/c^2$ .
Blueshift (falling in)	$\dfrac{\nu_{\mathrm{rec}}}{\nu_{\mathrm{emit}}}>1\quad(r_{\mathrm{rec}}<r_{\mathrm{emit}})$	A signal sent inward toward the mass is blueshifted by the inverse factor — the symmetric partner of climbing out.

Common trapThe horizon divergence is a Doppler blow-up or a curvature singularity

It is neither. The divergence as $r_{\mathrm{emit}}\to 2GM/c^2$ is not a relative-motion Doppler effect — both observers are static — and it is not a curvature singularity, since every curvature invariant such as the Kretschmann scalar $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}=48G^2M^2/c^4 r^6$ stays finite at the horizon. What breaks is the family of static observers: no observer can hover at or inside $r=2GM/c^2$ without infinite proper acceleration, so the emitter in (5) simply ceases to exist. An observer falling freely across the horizon measures perfectly finite frequencies the whole way. The infinite redshift is a property of the static frame — a coordinate-like statement — not of the light or the geometry.

Worked examples

1 · The Pound–Rebka tower (fully worked)

In 1960 Pound and Rebka shone $14.4\,\mathrm{keV}$ gamma rays from $^{57}\mathrm{Fe}$ up and down the $22.5\,\mathrm{m}$ Jefferson tower at Harvard and used the Mössbauer effect to detect the tiny shift. Predict the fractional frequency shift for a photon rising the full height.

Step 1. The tower is minute compared with the Earth's radius, so gravity is uniform over it and the weak-field tower result applies: $\Delta\nu/\nu=-gH/c^2$ , with the minus sign meaning the upward photon is redshifted.

Step 2. Insert $g=9.81\,\mathrm{m/s^2}$ , $H=22.5\,\mathrm{m}$ , $c=2.998\times10^8\,\mathrm{m/s}$ : $|\Delta\nu/\nu|=gH/c^2=9.81\times22.5/(2.998\times10^8)^2=220.7/8.99\times10^{16}$ .

Step 3. Evaluate: $|\Delta\nu/\nu|=\mathbf{2.46\times10^{-15}}$ . This is a shift of about $2.5$ parts in a thousand trillion — the Mössbauer linewidth of $^{57}\mathrm{Fe}$ is narrow enough (about $10^{-12}$ in fractional terms) to resolve it only by scanning the source velocity through a few $\mathrm{mm/s}$ .

Step 4. The 1960 experiment confirmed this to about $10\%$ ; the refined 1964 Pound–Snider version reached $1\%$ . The predicted rise-shift is $\Delta\nu/\nu=-2.46\times10^{-15}$ , a redshift confirmed to the percent level.

2 · The gravitational redshift of Sirius B (fully worked)

Sirius B is a white dwarf of mass $M\approx1.02\,M_\odot$ and radius $R\approx5850\,\mathrm{km}$ . Predict the gravitational redshift of a spectral line emitted at its surface and observed far away, and express it as an equivalent velocity.

Step 1. The receiver is effectively at infinity, so the surface-redshift formula applies: $z=1/\sqrt{1-2GM/c^2 R}-1$ . First form the compactness $2GM/c^2 R$ .

Step 2. The Schwarzschild radius of $1\,M_\odot$ is $2GM_\odot/c^2=2.95\,\mathrm{km}$ , so for $M=1.02\,M_\odot$ it is $3.01\,\mathrm{km}$ . Then $2GM/c^2 R=3.01/5850=\mathbf{5.15\times10^{-4}}$ — still a weak field, since it is far below 1.

Step 3. Because the compactness is small, $z\approx\tfrac12(2GM/c^2 R)=2.58\times10^{-4}$ (the exact formula gives the same to three figures). As an equivalent velocity, $cz=2.58\times10^{-4}\times2.998\times10^5\,\mathrm{km/s}=\mathbf{77\,km/s}$ .

Step 4. The Hubble Space Telescope measurement (Barstow et al., 2005) found $80.4\pm4.8\,\mathrm{km/s}$ , agreeing within the mass and radius uncertainties. Predicted $z\approx2.6\times10^{-4}$ , or $\approx77\,\mathrm{km/s}$ , matching the measured $\approx80\,\mathrm{km/s}$ . After Adams' controversial 1925 attempt, this is the effect that turned white-dwarf spectra into a mass–radius test.

3 · A neutron-star surface, where the weak field fails

A canonical neutron star has $M=1.4\,M_\odot$ and $R=10\,\mathrm{km}$ . Find the surface redshift and the observed energy of a $6.4\,\mathrm{keV}$ iron line, and check whether the weak-field approximation is still adequate.

Step 1. Compactness: $2GM/c^2 R=(2.95\times1.4)/10=4.13/10=\mathbf{0.414}$ — no longer small, so the exact formula is mandatory. This puts the surface at $r/r_s\approx2.4$ , the green point in the second figure.

Step 2. Exact surface redshift: $z=1/\sqrt{1-0.414}-1=1/\sqrt{0.586}-1=1/0.766-1=\mathbf{0.31}$ . The weak-field guess $\tfrac12\times0.414=0.207$ underestimates it by a third — the linear approximation has broken down.

Step 3. A $6.4\,\mathrm{keV}$ rest-frame iron line is received at $E_{\mathrm{obs}}=6.4/(1+z)=6.4/1.31=\mathbf{4.9\,keV}$ . The surface redshift is $z\approx0.31$ , shifting the iron line to $\approx4.9\,\mathrm{keV}$ ; the weak-field formula is off by $\sim30\%$ — strong fields demand the full $\sqrt{-g_{tt}}$ .

Physics of the portrait

Gravitational redshift was on the table before general relativity was finished. Albert Einstein deduced it from the equivalence principle in 1907 — years before he had field equations or Schwarzschild had a metric — reasoning that an accelerating elevator and a gravitational field must treat clocks and light identically.

Origins · 1 minAlbert Einstein and how this was discovered

Albert Einstein

1879–1955 · the equivalence principle and the first prediction of gravitational redshift

Long before the Schwarzschild solution existed, Albert Einstein extracted gravitational redshift from a single thought experiment. In his 1907 review article he imagined a uniformly accelerated reference frame and argued, by the equivalence of acceleration and gravity, that a clock lower in a gravitational potential must run slow relative to one higher up — and therefore that light climbing out of a well must shift toward the red. In 1911, in "On the Influence of Gravitation on the Propagation of Light," he sharpened this into a number, deriving $\Delta\nu/\nu\approx-\Delta\Phi/c^2$ and predicting a shift of about two parts per million for sunlight, alongside a starlight-bending value he would later correct by a factor of two once the full 1915 theory supplied the spatial curvature. Crucially, the redshift is a pure consequence of the equivalence principle and survives untouched into general relativity: when Schwarzschild wrote down the exact vacuum metric in 1916, the factor $g_{tt}=-(1-2GM/c^2 r)$ made Einstein's heuristic exact. Solar-line measurements stayed ambiguous for decades, so decisive confirmation waited until Pound and Rebka's 1960 tower — but the physical content, that gravity is the geometry of time, was Einstein's from the start.

History

1907

— Einstein predicts gravitational time dilation from the equivalence principle

1911

— Einstein derives the solar redshift Δν/ν ≈ −GM/c²R and predicts light bending

1916

— Schwarzschild's exact vacuum solution fixes g_tt = −(1 − 2GM/c²r)

1960

— Pound and Rebka confirm the tower shift to about 10 percent

1976

— Gravity Probe A verifies the maser shift to one part in ten thousand

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §25.4 · Wald, General Relativity (1984), §6.3 · Carroll, Spacetime and Geometry (2004), §5.4 · Hartle, Gravity (2003), §9.2 · Schutz, A First Course in General Relativity, 2nd ed. (2009), §10.5 · Pound & Rebka, Phys. Rev. Lett. 4, 337 (1960). The static four-velocity component $u^t=(1-2GM/c^2 r)^{-1/2}$ follows from the normalization $g_{\mu\nu}u^\mu u^\nu=-1$ , and the boxed ratio (5) follows from $h\nu=-p_\mu u^\mu$ with conserved $p_t$ . Key numerical values used throughout: Pound–Rebka $\Delta\nu/\nu=gH/c^2=2.46\times10^{-15}$ for $H=22.5\,\mathrm{m}$ (Pound & Rebka 1960); Sirius B $z\approx2.6\times10^{-4}$ , $\approx77\,\mathrm{km/s}$ predicted versus $80.4\pm4.8\,\mathrm{km/s}$ measured (Barstow et al. 2005); neutron-star surface $z\approx0.31$ for $M=1.4\,M_\odot$ , $R=10\,\mathrm{km}$ ; GPS net offset $\approx38\,\mu\mathrm{s/day}$ (Ashby, Living Rev. Relativity 2003); Gravity Probe A precision $\sim10^{-4}$ (Vessot et al. 1980); $r_s=2GM_\odot/c^2=2.95\,\mathrm{km}$ .

Prerequisites. The Schwarzschild Metric · Killing Vectors and Conserved Quantities in Schwarzschild · Schwarzschild Radius and the Geometric Mass. Leads to. Photon Orbits and the Photon Sphere · Coordinate versus Curvature Singularities · Geodesic Equations in Schwarzschild Spacetime.

Ask the tutor: Ask me to walk through why the photon's Killing energy makes the redshift path-independent, or to work the surface redshift of any star, white dwarf, or neutron star you name.

Is gravitational redshift a Doppler effect?

No — and this is the point most worth getting straight. A Doppler shift comes from relative motion between source and observer, but in gravitational redshift both observers are static, holding fixed Schwarzschild radii, with zero relative velocity in any local sense. The shift arises purely because their clocks tick at different rates, set by the local value of $\sqrt{-g_{tt}}$ . There is a genuine and beautiful connection through the equivalence principle: in a uniformly accelerating rocket, a light signal sent from floor to ceiling is Doppler-shifted because the ceiling has picked up speed during the light's transit, and the equivalence principle says a static gravitational field must do the same thing. But that is a heuristic bridge, not an identity — the exact effect is geometric, encoded in the metric, and needs no motion at all.

Why does the photon's path not matter, only the endpoints?

Because the quantity a static spacetime conserves is the photon's Killing energy $E=-p_\mu\xi^\mu$ , the same number at every point along the geodesic. Each measured frequency is $E$ divided by the local lapse $\sqrt{-g_{tt}}$ , so the shared $E$ cancels and only the two endpoint lapses survive. Physically, whatever energy the photon spends bending or looping it gains back, so the net toll depends only on the potential difference between where it started and stopped. It is the same reason a ball's change in kinetic energy on a frictionless track depends only on the height difference, not the shape of the track — conservation of a potential-like quantity makes the process path-independent.

What actually goes wrong at the horizon?

The redshift formula (5) sends $\nu_{\mathrm{rec}}/\nu_{\mathrm{emit}}\to0$ as the emitter approaches $r=2GM/c^2$ , but nothing physical becomes infinite there. Curvature invariants like the Kretschmann scalar stay finite at the horizon; the real singularity is at $r=0$ . What breaks down is the assumption that the emitter is static. Holding a fixed radius just outside the horizon demands ever-larger proper acceleration, diverging exactly at $r=2GM/c^2$ , so a hovering emitter cannot exist there. If instead the emitter falls freely across the horizon, it measures finite frequencies the entire way, and a distant observer simply sees its last signals stretched toward infinite wavelength and dimmed to invisibility — the "frozen star" appearance. The infinity is a property of an impossible observer, not of the spacetime.

How is this the same thing as gravitational time dilation?

They are two names for one fact. A static clock deep in the well obeys $\mathrm{d}\tau=\sqrt{-g_{tt}}\,\mathrm{d}t$ , so it ticks slowly compared with one higher up — that is time dilation. But any periodic signal, including a light wave, is itself a clock: its frequency is ticks per second. So comparing the frequency emitted below with the frequency counted above just compares the two clocks' rates, giving the ratio $\sqrt{-g_{tt}(r_{\mathrm{emit}})}/\sqrt{-g_{tt}(r_{\mathrm{rec}})}$ . Redshift is time dilation read off a light wave rather than a wristwatch; the GPS clock correction and the reddening of a photon are the identical calculation.

Does gravitational redshift cause the cosmological redshift of galaxies?

No, they are distinct effects, though both stretch wavelengths. Gravitational redshift, as derived here, comes from a static metric and a difference in gravitational potential between two fixed radii — the effect for light escaping a star, a white dwarf, or a black hole's neighborhood. The cosmological redshift of distant galaxies instead comes from the expansion of space in the time-dependent Friedmann–Robertson–Walker metric, where wavelengths grow in proportion to the cosmic scale factor during the light's flight. One is depth in a fixed potential well; the other is the stretching of space over cosmic time. They can coexist — a distant galaxy's photon carries a small gravitational shift from leaving that galaxy plus a large cosmological shift from the intervening expansion — but the underlying physics differs.

Could you power a perpetual-motion machine by cycling photons up and down?

No, and checking why is a good consistency test. Suppose you drop a photon into the well, letting it blueshift and gain energy, then convert it back to mass at the bottom, haul the mass up, and release another photon. The bookkeeping closes exactly: lifting the mass against gravity costs precisely the energy the photon gained falling, because the blueshift factor and the potential-energy change are both governed by $\sqrt{-g_{tt}}$ . Einstein used this cycle in reverse to argue for gravitational redshift — if light did not lose energy climbing out, you could build a perpetual-motion machine, which is forbidden. Far from violating energy conservation, gravitational redshift is demanded by it once you accept $E=h\nu$ and $E=mc^2$ .