Understand the relationship
The reasoning behind the result
Power spreads over a spherical area
F0=L/(4πd²)
L is source energy emitted per second, while F0 is energy received per second per square metre before extra attenuation. Isotropic emission spreads the power over a sphere whose area is 4πd². Doubling distance gives one-quarter of the flux; it does not change source luminosity.
For a beamed or anisotropic source this arithmetic describes an isotropic-equivalent luminosity rather than automatically recovering its actual emitted power. The entered geometry and interpretation must support the model.
Attenuation changes the received fraction
T=10^(−0.4A); F=T F0
A is a nonnegative entered attenuation in magnitudes for the actual integrated quantity being compared. At A=2.5 the received fraction is one-tenth. Recovering luminosity from the received flux therefore divides by T.
No dust law, passband correction or absorption coefficient is supplied. A single-band attenuation must not be used as a bolometric correction without an external justification. Scattered or reprocessed light is outside this simple loss model.
Integrated flux is not spectral flux density
Watts per square metre and erg per second per square centimetre measure integrated energy flux. Jansky measures energy flux per unit frequency and cannot be inserted into this formula as if it were total energy flux.
One erg is 10⁻⁷ joule and one square centimetre is 10⁻⁴ square metre, so one erg/s/cm² equals 10⁻³ W/m². The nominal solar luminosity unit used here is exactly 3.828×10²⁶ W; it is a conversion reference, not a live measurement of the Sun.
Luminosity distance preserves the flux identity
A cosmological luminosity distance is defined so the bolometric relation F=L/(4πdL²) holds. That distance already includes cosmological dimming in its definition. Multiplying by an extra redshift factor here would double-count a correction unless a different specifically justified quantity is being used.
This workbench accepts a supplied luminosity distance but does not infer one from redshift. Band-limited rest and observed quantities additionally require compatible passband and spectral corrections outside this model. It does not convert luminosity distance into elapsed travel time.
Follow the numbers
A 2.5-mag loss transmits one-tenth
- Let L=100 W and d=1 m. Before attenuation, F0=100/(4π)=7.957747155 W/m².
- At A=2.5 mag, T=10^(−0.4×2.5)=0.1, giving F=0.7957747155 W/m².
- The inverse recovers L=4π×1²×0.7957747155/0.1=100 W.
The attenuation is applied once, after geometric spreading, and reversed once in the inverse solve.
Quick guide
How to use this calculator
- Choose the missing quantity and the distance interpretation.
- State whether power and flux are bolometric or correspond to the same integrated band.
- Select units independently and enter an applicable attenuation, or zero for none.
- Inspect source power, unattenuated geometrical flux and received flux separately.
Calculation method
Calculation and interpretation
Keep source power separate from the energy flux received per unit area.
F=L·10^(−0.4A)/(4πd²); L=4πd²F·10^(0.4A); d=√[L·10^(−0.4A)/(4πF)].
Worked example
A 2.5-mag loss transmits one-tenth
The attenuation is applied once, after geometric spreading, and reversed once in the inverse solve.
F=L·10^(−0.4A)/(4πd²); L=4πd²F·10^(0.4A); d=√[L·10^(−0.4A)/(4πF)].
Supported inputs
Precision and limits
Positive-source model
Power, distance and received flux must be positive. Zero or negative background-subtracted measurements do not define a finite positive-source inverse in this model.
No inferred spectrum or cosmology
No spectral density integration, redshift-distance model, beaming factor, lensing magnification or extinction law is inferred.
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