How to convert mW/m^2 to W/m^2/nm?

[AdrianD]

I have a vizier catalog with H-alpha fluxes as logF(Haplha) in units as mW/m^2. See sample data here: http://vizier.cds.unistra.fr/viz-bin/VizieR-3?-source=J/MNRAS/431/2/fluxes How do I convert mW/m^2 to W/m^2/nm or erg/s/cm^2/A? The first entry is logF(Halpha) = -12.03 mW/m^2

 

[scottdave]

These are not the same dimension. You have mW (milliwatts) per square meter (Power divided by Length²). then you have Watts per square meter per nanometer (Power divided by Length³)

 

[AdrianD]

These are not the same dimension. You have mW (milliwatts) per square meter (Power divided by Length²). then you have Watts per square meter per nanometer (Power divided by Length³)

Ok, but there must be a way, otherwise there is no point to measure/report the value in the catalog if it's useless for computing flux or flux density

 

[WernerQH]

You need to know something about the spectrum. Or at least the width of the line.

 

[AdrianD]

You need to know something about the spectrum. Or at least the width of the line.

I think I made some progress.
The article for the Vizier data base is here: https://academic.oup.com/mnras/article/431/1/2/1036908
On pg. 12 in the pdf "
with the exception of McCullough et al. (2001) who
determined a flux for Abell 36 of F(Hα) = 2.8 × 10−11 erg cm−2 s−1,
which is 28 per cent lower than our own determination of 3.9 ×
10−11 erg cm−2 s−1.
"
If I look up Abell 36 in their database I find a logF = -10.41 mW/m^2
Then:
[CODE lang="python" title="conversion"]import numpy as np
from scipy.integrate import simps

# Given logFlux in mW/m²
logFlux_mW_per_m2 = -10.41

# Convert logFlux to regular flux in mW/m²
flux_mW_per_m2 = 10 ** logFlux_mW_per_m2

# Load filter transmission data from file (assuming two columns: wavelength, transmission)
filter_data = np.loadtxt('H-alpha-3nm_2.txt')
wavelengths = filter_data[:, 0] # Wavelengths in Ångströms
transmission_values = filter_data[:, 1] # Filter transmission values

# Interpolate the filter transmission data to match the flux wavelengths (if needed)
# Assuming the filter and flux data have matching wavelengths

# Calculate the product of flux and filter transmission
flux_times_transmission = flux_mW_per_m2 * transmission_values

# Integrate the product over the wavelength range to get erg/s/cm²
erg_per_s_per_cm2 = simps(flux_times_transmission, wavelengths)

# Convert the result to erg/s/cm²/Å by dividing by the bandwidth (in Ångströms)

erg_per_s_per_cm2_per_Angstrom = erg_per_s_per_cm2 / 30 #bandwidth_Angstrom

print("Flux in erg/s/cm²/Å:", erg_per_s_per_cm2_per_Angstrom)
f = erg_per_s_per_cm2_per_Angstrom * units.FLAM
f.to(u.W / (u.m**2 * u.nm), u.spectral_density(Hydrogen_filter))[/CODE]
Print result:
[CODE lang="python" title="Result"]Flux in erg/s/cm²/Å: 3.90462998411593e-11[/CODE]

Get the same value as them but in my case I had to divide by the filter bandwidth so the result is in erg/s/cm²/Å, but they report the result in erg/s/cm². Don't know what to make of it

 

[WernerQH]

Why do you need a wavelength-dependent quantity? What you start with is an integral over wavelengths. What's the width of the line?

 

[AdrianD]

Why do you need a wavelength-dependent quantity? What you start with is an integral over wavelengths. What's the width of the line?

The bandwidth of the H-alpha filter is 30 A or 3 nm. I need the wavelength dependent quantity because I will compute the counts (electrons/sec) from the object. The general formula being: Flux/PhotonEnergy(λ) * Filter_bandwidth (nm) * Aperture (m^2) * QE * PlateScale^2(arcsec^2). For broadband BVRI photometric filters and Vega mag things are easy but narrowband drives me crazy

 

[Vanadium 50]

Why don't you turn it around? Instead of trying to make one table look like the other, figure out which one you want?

 

[Drakkith]

Is it not as simple as dividing by the filter bandwidth, in this case 3 nm, and then converting mW to W?

 

[Drakkith]

The bandwidth of the H-alpha filter is 30 A or 3 nm. I need the wavelength dependent quantity because I will compute the counts (electrons/sec) from the object. The general formula being: Flux/PhotonEnergy(λ) * Filter_bandwidth (nm) * Aperture (m^2) * QE * PlateScale^2(arcsec^2). For broadband BVRI photometric filters and Vega mag things are easy but narrowband drives me crazy

Is it possible that filter bandwidth is irrelevant here since the spectral line width is much smaller than the filter bandwidth? I believe the HA line width is less than 0.5 Angstroms, or 0.05 nm, or so this lecture tells me.

 

[WernerQH]

The bandwidth of the H-alpha filter is 30 A or 3 nm.

Filter bandwidth is not the same as linewidth. Assuming a rectangular line profile of 3 nm width, I get from a total flux $F \approx 3.9 × 10^{-11}~{\rm mW\,m^{-2}}$ a spectral flux of $$
F_\lambda \sim {F \over \Delta \lambda}
\approx 1.3 × 10^{-14}~{\rm W\,m^{-2}\,nm^{-1}}
= 1.3 × 10^{-4}~{\rm erg\,s^{-1}\,cm^{-2}\,A^{-1}} \ .
$$

The general formula being: Flux/PhotonEnergy(λ) * Filter_bandwidth (nm) * Aperture (m^2) * QE * PlateScale^2(arcsec^2).

Sorry, I can't make sense of your formula and have no idea what you are trying to calculate.

 

[Ken G]

You say you want the count rate in the H alpha line, and you have the H alpha flux per area. So you only need the effective area of the detector (so that's an area times a QE), and the energy per photon (which is h*nu), and you have the counts. In other words, the flux you already know is the first two terms in that formula you gave, so they are already multiplied together for you, there's no need to separate them and remultiply them. Also, I don't think you need that last term, the plate scale in angular area, because that should be if you are given a resolved brightness, but instead you have a total flux per area. In short, the data you now have has already been integrated over both the linewidth, and the angular size, so you don't need to know either of those things in your formula to get the count rate. (Note also that the formula you gave would give you the rate of energy detection, not the rate of photon detection, so you still have to divide by the photon energy, which is simple.)