research38 min studyLesson 828 in Physics

Hawking radiation from horizon redshift

Connect the exponential late-time ray relation to Hawking occupation while specifying the quantum state, greybody filtering, and semiclassical regime.

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What you’ll understand

  • Identify the redshift relation responsible for the thermal frequency factor.
  • Apply a greybody transmission probability to one outgoing bosonic channel.
  • Distinguish collapse, equilibrium, and static-vacuum state choices.

The Hawking calculation compares a quantum field's early incoming modes with its late outgoing modes in a spacetime containing collapse. The crucial geometric input is an exponential relation between an early affine null coordinate and late retarded time. Positive frequency at early times then mixes with negative frequency in the late-mode comparison. The resulting thermal factor is a prediction for specified asymptotic modes and a specified state, within quantum field theory on a classical or slowly evolving background. It does not require an observer-independent local cloud of particles at the horizon.

Exponential redshift produces the frequency ratio

Retain signature (−,+,+,+) and geometric G=c=1, with the late stationary horizon generator normalized by the clock at infinity. Let u be retarded time at future null infinity and U an affine incoming coordinate at past null infinity. Rays escaping at increasingly late u passed increasingly close to the last ray that could escape. To leading late-time order their relation is exponential, with a positive constant A that depends on the collapse history but not the universal temperature factor.

Tracing an outgoing positive-frequency wave backward gives an oscillatory power of UH−U. Its Fourier decomposition in the incoming affine coordinate has positive- and negative-frequency pieces with a universal magnitude ratio. Wavepackets make the mode normalization and late-time localization well behaved. For a bosonic channel, canonical normalization converts the ratio into a Planck occupation. The asymptotic relation concerns late outgoing rays; it is not a claim that a single geometric-optics ray already contains a quantum particle.

U≃UH−Ae−κu,∣βω∣2∣αω∣2=e−2πω/κ,kBT∞=ℏκ2π(c=1)U\simeq U_H-Ae^{-\kappa u},\qquad \frac{|\beta_\omega|^2}{|\alpha_\omega|^2}=e^{-2\pi\omega/\kappa},\qquad k_BT_\infty=\frac{\hbar\kappa}{2\pi}\quad(c=1)
U,u,A have length units, while the geometric surface gravity κ and angular frequency ω have inverse-length units. The coefficient ratio schematically denotes the late-time packet comparison before exterior transmission. Restoring SI units gives kBT∞=ħ c κgeo/(2π); for Schwarzschild κgeo=1/(4Mgeo).↔ Scroll wide equations horizontally.

Specify the state and propagate through the exterior

For a nonrotating Schwarzschild hole, the curvature potential partly reflects outgoing modes. The occupation reaching infinity in one angular channel is multiplied by a greybody transmission probability Γ between zero and one. Γ is an intensity or flux probability, not a transmission amplitude, and it depends on frequency and angular momentum. Summing channels and including their energy weights gives a luminosity; one mean occupation alone does not determine the total power. Fermionic fields instead have a plus sign in the thermal denominator.

The collapse state is modeled by an Unruh-type state: no incoming thermal bath from past infinity, regular behavior across the future horizon, and late outgoing flux. The Hartle–Hawking state on an eternal stationary background describes thermal equilibrium with both incoming and outgoing radiation. The Boulware state is a static vacuum at infinity but is singular on the Schwarzschild horizon. These choices are physically different. A statement about a static equilibrium thermometer must not silently be transferred to every freely falling detector in a collapse state.

⟨Nωℓmout⟩=Γωℓe2πω/κ−1,0≤Γωℓ≤1,Tlocal=T∞1−2M/r\langle N_{\omega\ell m}^{\mathrm{out}}\rangle=\frac{\Gamma_{\omega\ell}}{e^{2\pi\omega/\kappa}-1},\qquad 0\leq\Gamma_{\omega\ell}\leq1,\qquad T_{\mathrm{local}}=\frac{T_\infty}{\sqrt{1-2M/r}}
The occupation is the late-time bosonic result for Schwarzschild with the incoming vacuum condition; rotational superradiance is outside this formula. The final Tolman relation separately applies to static observers in stationary thermal equilibrium, such as the Hartle–Hawking state, rather than to every detector in the outgoing collapse flux.↔ Scroll wide equations horizontally.

Keep the semiclassical approximation visible

The standard derivation uses a free or sufficiently weakly interacting quantum field with the appropriate short-distance, Hadamard behavior on a regular background. In an adiabatic evaporation treatment, changes of the surface gravity must be slow, for example ∣∂uκ∣/κ2≪1|\partial_u\kappa|/\kappa^2\ll1 with u normalized at infinity. The outgoing spectrum is approximately stationary over the measurement interval, and backreaction is treated in a controlled expansion. Small curvature for a large hole helps justify that regime; it does not establish the detailed microscopic fate of a Planck-scale endpoint.

Backward propagation of a late outgoing mode reaches extremely high incoming frequencies. The calculation therefore has a short-distance assumption, often discussed as the trans-Planckian issue. This identifies where robustness or ultraviolet completion must be investigated, rather than by itself disproving the low-energy thermal result. Greybody filtering, finite formation transients, and slowly changing parameters also limit an exactly blackbody interpretation. A detector response, a renormalized stress tensor, and an asymptotic occupation are related calculations with different specifications; none should be substituted for another without checking those specifications.

Filter one outgoing bosonic channel

In a late stationary Schwarzschild calculation, take one bosonic mode with 2πω/κ=ln 2 and greybody probability Γ=1/4. Compute its occupation at infinity. Then double its frequency, assuming solely for this toy comparison that Γ remains 1/4, and compare occupations.

  1. Before exterior filtering, the mean occupation is 1/(eln⁡2−1)=11/(e^{\ln2}-1)=1. Multiplication by Γ=1/4 gives an outgoing occupation of 1/4.

  2. At twice the frequency, the exponent becomes 2 ln 2, so the unfiltered occupation is 1/(4−1)=1/31/(4-1)=1/3. With the explicitly fixed toy Γ, the transmitted occupation is 1/12.

  3. The high-to-low occupation ratio is (1/12)/(1/4)=1/3(1/12)/(1/4)=1/3. Real Schwarzschild greybody probabilities vary with frequency, so this ratio is not a general black-hole spectral law.

The result

The occupations are 1/4 and 1/12, with ratio 1/3 only under the declared equal-transmission assumption.

Explore the model

Interactive lab

Interactive lab · probability

Separate transmission from the source distribution

For each fresh comparison, Reset and set BOTH Trials n=4 and Success probability p. Compare p=0.25 and p=0.50, predicting means 1 and 2 and variances 0.75 and 1. Interpret success as transmission of each of four independently prepared incident quanta through an illustrative channel. The exact binomial model does not generate the thermal incident occupation, quantum bunching, greybody coefficients, a Hawking state, or a black-hole geometry.

Separate transmission from the source distributionThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA transmission probability filters a specified input; it does not by itself establish the statistics or temperature of the source.

Check your understanding

Which state describes stationary thermal equilibrium with both incoming and outgoing radiation on an eternal Schwarzschild background?

Try it yourself

Use the Hartle–Hawking equilibrium state, explicitly rather than the outgoing collapse flux. A static observer sits where f=1−2M/r=1/4. Find r/M, the local temperature relative to T∞, and the local energy of a fixed-frequency photon relative to its asymptotic energy. What happens to E/(kBT)?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Apply the same inverse-square-root redshift factor to the photon energy and to the equilibrium Tolman temperature.

Compare with the solution

The radius satisfies 2M/r=3/4, so r/M=8/3. The lapse is sqrt(f)=1/2; therefore Tlocal=2T∞T_{\mathrm{local}}=2T_\infty and Elocal=2E∞E_{\mathrm{local}}=2E_\infty. Their ratio remains Elocal/(kBTlocal)=E∞/(kBT∞)E_{\mathrm{local}}/(k_BT_{\mathrm{local}})=E_\infty/(k_BT_\infty). This is a static equilibrium relation for the stated state and worldline. It does not establish that a freely falling observer has the same response, or that a nonequilibrium collapse flux is an exactly isotropic local thermal bath.

Take it with you

  • Late-time exponential redshift gives a thermal factor for a specified incoming state and outgoing mode choice.
  • Greybody probabilities modify the asymptotic spectrum and are distinct from amplitudes.
  • State choice, observer motion, and the semiclassical regime remain part of every Hawking-radiation claim.

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