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For an isotropic point source in free propagation, measured flux is luminosity spread over a sphere at the observer’s distance.
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Formula
L in W; F in W/m2; d in m
| Symbol | Meaning | Unit |
|---|---|---|
| Total radiated power of the source | W/m2 | |
| Received energy per unit area and time | W | |
| Distance from source to observer | m |
F = \frac{L}{4\pi d^2}At radius d, isotropic radiation crosses a sphere with area 4πd2, giving F=L/(4πd2). This assumes no absorption, scattering, beaming or cosmological correction and that luminosity and flux use compatible spectral bands (or both are bolometric). Dust extinction must be corrected before using the relation. Flux alone does not determine both luminosity and distance; one of them must be known or independently estimated.
Applies to isotropic point-source emission in free propagation; absorption, beaming and cosmological effects are excluded.
Total power spreads over a sphere of radius d.
Distance
Use compatible bolometric luminosity and flux.
Luminosity
Correct for extinction before substituting.
If distance doubles, how does F change?
F scales as 1/d2, so it falls to one quarter.
For L=3.8·1026 W at 1 AU, estimate F.
F=L/(4πd2)≈1360 W/m2, assuming isotropic emission and no attenuation.
Confusing total luminosity with flux per square metre.
L is W, F is W/m2; the spherical area connects them.
Using the simple relation unchanged despite absorption or beamed emission.
Correct for source geometry and transmission or use a fuller model.
Formula cluster
Flux alone determines the third quantity only if luminosity or distance is known.
Using L=3.8×1026 W and d=1 AU gives F≈1.35×103 W/m2 under the isotropic, unattenuated approximation.
Relate a source’s intrinsic power to observed brightness and estimate distance when luminosity is independently known.
Here is how to work through a typical Luminosity, flux and distance (F = L/(4πd²)) task step by step:
Task
If distance doubles, how does F change?
Solution path
F scales as 1/d2, so it falls to one quarter.
Task
For L=3.8·1026 W at 1 AU, estimate F.
Solution path
F=L/(4πd2)≈1360 W/m2, assuming isotropic emission and no attenuation.
Luminosity L is the total energy emitted per time by a source, measured in watts. Flux F is energy per time and area received at the observer, measured in W/m2. A luminous source can still produce a small flux when far away because its power spreads over a large spherical area. L is the source’s total emitted power; F is power per area at the observer. The simple spherical relation assumes no absorption, scattering or beaming and compatible passbands. Apparent magnitude is logarithmic and is not automatically bolometric flux. Correct for dust and instrument response before comparing the quantities.
For isotropic emission, power spreads uniformly over a sphere of radius d. Its surface area is 4πd2, so flux falls inversely with the square of distance. Doubling d leaves one quarter of the original flux. The spherical geometry produces the factor 4π. L is the source’s total emitted power; F is power per area at the observer. The simple spherical relation assumes no absorption, scattering or beaming and compatible passbands. Apparent magnitude is logarithmic and is not automatically bolometric flux. Correct for dust and instrument response before comparing the quantities.
If L is known and F measured, rearrange to d=√(L/(4πF)). Use SI units so L is in W and F in W/m2. Without an independent luminosity estimate, distance cannot be determined from F alone; many combinations of L and d produce the same flux. L is the source’s total emitted power; F is power per area at the observer. The simple spherical relation assumes no absorption, scattering or beaming and compatible passbands. Apparent magnitude is logarithmic and is not automatically bolometric flux. Correct for dust and instrument response before comparing the quantities.
Correct the relation when dust absorbs or scatters light, emission is beamed, or cosmological effects matter. The observed flux is then not simply unattenuated power per spherical area. Also account for spectral band and detector calibration. The formula remains useful as a first approximation when such effects are small. L is the source’s total emitted power; F is power per area at the observer. The simple spherical relation assumes no absorption, scattering or beaming and compatible passbands. Apparent magnitude is logarithmic and is not automatically bolometric flux. Correct for dust and instrument response before comparing the quantities.
Yes, but only with a defined magnitude convention and a reference flux in the relevant band. Magnitudes are logarithmic: a flux ratio of 100 corresponds to five magnitudes. Bolometric flux and a band magnitude are not interchangeable. State the photometric filter definition alongside the flux measurement. L is the source’s total emitted power; F is power per area at the observer. The simple spherical relation assumes no absorption, scattering or beaming and compatible passbands. Apparent magnitude is logarithmic and is not automatically bolometric flux. Correct for dust and instrument response before comparing the quantities.
Quanta exam set
Curated exam set for "Luminosity, flux and distance":
Question (front)
Which formula describes Luminosity, flux and distance?
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How do you rearrange Luminosity, flux and distance for Distance?
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Question (front)
Which common mistake happens with Luminosity, flux and distance?
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