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Sandbox Physics
From distant light to a galaxy census

L2 · Light, images and selection

How redshift turns UV luminosity into received flux

Sandbox Physics ·

An observed flux density per unit frequency carries a redshift factor in its numerator. Luminosity distance already describes the dilution of total radiation. The remaining factor converts the width of the frequency interval. Keeping those operations distinct makes the sign of the correction easier to understand.

Define each “per” before doing the conversion

Total luminosity is emitted energy per second, in watts. Emitted spectral density Lν,eL_{\nu,e} is also per unit emitted frequency, in watts per hertz. Observed spectral density fν,of_{\nu,o} is received energy per second, per square metre and per observed hertz. The subscripts distinguish emitter and observer.

Redshift lowers a photon’s frequency and compresses a narrow emitted frequency interval into a narrower observed interval. A density quoted per hertz has to account for that change in interval width.

νe=(1+z)νo,dνe=(1+z) dνo\nu_e=(1+z)\nu_o,\qquad d\nu_e=(1+z)\,d\nu_o
At redshift ten, an eleven-hertz emitted interval occupies one observed hertz.

Reading: Hogg

Use all radiation to define luminosity distance

Bolometric means integrated over the whole spectrum. Luminosity distance is defined through bolometric luminosity and flux. Energy loss per photon and the change in arrival rate are already included in that distance convention.

Fbol=Lbol4πDL2,Fbol=∫fν,o dνo,Lbol=∫Lν,e dνeF_{\rm bol}=\frac{L_{\rm bol}}{4\pi D_L^2},\qquad F_{\rm bol}=\int f_{\nu,o}\,d\nu_o,\qquad L_{\rm bol}=\int L_{\nu,e}\,d\nu_e
Bolometric flux is in watts per square metre. Use a length unit consistent with its area factor.

Reading: Hogg (1999)

Return to one narrow spectral interval

An emitted slice and its redshifted counterpart must give the same bolometric relation when all slices are added. The observed slice is narrower, so its density needs the compensating factor.

fν,o(νo)=(1+z)Lν,e((1+z)νo)4πDL2f_{\nu,o}(\nu_o)=\frac{(1+z)L_{\nu,e}((1+z)\nu_o)}{4\pi D_L^2}
Evaluate the emitted spectrum at the corresponding emitted frequency. The dilution remains in luminosity distance.
∫fν,o dνo=14πDL2∫Lν,e (1+z)dνe1+z=Lbol4πDL2\int f_{\nu,o}\,d\nu_o=\frac{1}{4\pi D_L^2}\int L_{\nu,e}\,(1+z)\frac{d\nu_e}{1+z}=\frac{L_{\rm bol}}{4\pi D_L^2}
Integrating catches a reversed redshift factor by testing the bolometric definition, including dimensionless factors.

Reading: Hogg (1999) · Hogg

An AB magnitude specifies a spectral-density scale

AB magnitude is logarithmic in frequency-density flux. This example explicitly adopts a zero point of 3631 janskys. One jansky is ten to the minus twenty-six watts per square metre per hertz. A smaller magnitude means more flux.

An absolute UV magnitude places the source at ten parsecs and specifies the normalization at rest-frame 150 nanometres. Inferring stellar mass would also require information about stellar ages, composition and other possible light sources. The magnitude here sets one part of the UV spectrum.

mAB=−2.5log⁡10 ⁣(fν3631 Jy),Lν,1500=4π(10 pc)2(3631 Jy)10−0.4M1500m_{\rm AB}=-2.5\log_{10}\!\left(\frac{f_\nu}{3631\,\mathrm{Jy}}\right),\qquad L_{\nu,1500}=4\pi(10\,\mathrm{pc})^2(3631\,\mathrm{Jy})10^{-0.4M_{1500}}
Convert janskys to SI spectral-density units when evaluating the right-hand luminosity in SI. A broad passband still requires its response-weighted integral.
Lλ∝λβ,Lν=Lλλ2c∝λβ+2L_\lambda\propto\lambda^\beta,\qquad L_\nu=L_\lambda\frac{\lambda^2}{c}\propto\lambda^{\beta+2}
The two densities use different spectral intervals. A wavelength slope of minus two gives a flat unattenuated frequency spectrum.

Reading: Hogg

Work an unattenuated flat-spectrum example

Choose redshift ten, absolute UV magnitude minus twenty and a flat frequency spectrum. Adopt a flat matter-plus-Lambda cosmology with a Hubble constant of seventy kilometres per second per megaparsec and matter fraction three tenths, omitting radiation. These specified parameters turn the conversion into a numerical example.

Its luminosity distance is about 103.843 billion parsecs. The observed flux density is about 37.04 nanojanskys and its AB magnitude 27.48. Dust and intergalactic absorption are switched off for this check. Each passband seeing that same flat frequency spectrum must return the same mean.

mAB=M1500+5log⁡10 ⁣(DL10 pc)−2.5log⁡10(1+z)m_{\rm AB}=M_{1500}+5\log_{10}\!\left(\frac{D_L}{10\,\mathrm{pc}}\right)-2.5\log_{10}(1+z)
This short relation applies to the specified flat frequency spectrum and AB reference. A general spectrum through different filters needs the spectral arguments and passband integral.
Distance in the same adopted cosmologyAt redshift ten / Mpc
Radial comoving distance9440.25
Angular-diameter distance858.205
Luminosity distance103842.746

Reading: Hogg (1999) · Hogg · Sandbox Physics

Flux and angular size use different distances

Luminosity distance is defined by brightness; it is not light-travel time multiplied by the speed of light. It converts total luminosity into flux. Angular-diameter distance converts physical radius into an angle on the sky. Their units match, but their physical jobs differ.

DL=(1+z)2DA,θ1/2=r1/2DAD_L=(1+z)^2D_A,\qquad\theta_{1/2}=\frac{r_{1/2}}{D_A}
The angle is in radians after radius and distance are put in the same length unit. At redshift ten, 0.3 kiloparsecs subtends about 0.0721 arcseconds.

Reading: Hogg (1999) · Sandbox Physics

Try the observation

Change intrinsic light and pixel distribution separately

  1. Open luminosity and spectrum, hold redshift fixed and brighten the absolute magnitude by one. Predict the total-flux ratio.
  2. Restore the magnitude and increase physical half-light radius. Compare input total flux with measurement uncertainty.
  3. Change redshift and inspect both the spectral gap and angular image size. Identify the role of each distance.

The example uses a prescribed AB continuum, screen, sharp IGM cutoff and one adopted radiation-free cosmology; UV magnitude fixes its spectral normalization.

Exercises and explained answers

If the redshift factor in the frequency-density relation were divided instead of multiplied, how far too small would the redshift-ten flux be?

The ratio is (1+z)2=121(1+z)^2=121. The wrong magnitude would be about 5.21 magnitudes fainter. Integrating the wrong spectrum also fails the bolometric definition.

At fixed redshift and spectral shape, change absolute magnitude from minus twenty to minus twenty-one. How much does flux increase?

By 100.4≃2.51210^{0.4}\simeq2.512. The flat-spectrum example rises from about 37 to 93 nanojanskys. One magnitude is a ratio of fluxes, not a fixed added quantity.

Double physical radius at fixed distance and total UV luminosity. What happens to angular radius and total received flux?

Angular radius doubles. With spectrum and attenuation fixed, the total flux is unchanged. The distribution over pixels changes, which affects measurement precision.

Primary sources and calculation notes

  1. Hogg (1999), Distance measures in cosmology

    Luminosity-distance section: bolometric flux, angular/luminosity distance and spectral-density transformations. The numerical examples use the explicitly adopted radiation-free model.

  2. Hogg, Baldry, Blanton & Eisenstein (2002), The K correction

    Equations 1–7 and the AB-reference definition. Keep emitted and observed frequency arguments distinct; the tutorial fixes its reference to 3631 Jy.

  3. Sandbox Physics, specified synthetic imaging examples

    Original synthetic examples with exact parameters, distances, native-band totals and saved noisy measurements. No observed galaxy catalog, foreign source ZIP, executable Engine or measured cosmic LF is included.

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