<p>You can't with any precision.</p>
<p>mW/m<span class="math-container">$^2$</span> is a measure of flux, presumably integrated over a H<span class="math-container">$\alpha$</span> emission line.</p>
<p>W/m<span class="math-container">$^2$</span>/nm would be the units of the spectrum flux density - i.e. the quantity that would be integrated over wavelength to give you a total flux. This is not a single value, but a spectrum.</p>
<p>If you want to know the <em>average</em> spectrum flux density at the wavelength of H<span class="math-container">$\alpha$</span> is, about the best you could do is divide the flux in the H<span class="math-container">$\alpha$</span> line by what you think its full width half maximum is in nm. This might be set by the properties of the source, but will at a minimum be set by the spectral resolution of the instrument that measured the spectrum.</p>
<p>Thus in your example, suppose the line had a FWHM of 10 nm (very broad!), then we would subtract 3 from the log (divide by 1000) to put
mW into W and then subtract 1 more (divide by 10) to spread that flux across 10 nm, and the <em>rough, average</em> spectrum flux density over the H<span class="math-container">$\alpha$</span> line would be <span class="math-container">$10^{-16.03}$</span> W/m<span class="math-container">$^2$</span>/nm.</p>

CC BY-SA 4.0
2023-08-14T07:24:55.283Z
2023-08-14T07:24:55.283Z
ProfRob
<p>I have a vizier catalog with H-alpha fluxes as logF(Haplha) in units as mW/m^2. See sample data here: <a href="http://vizier.cds.unistra.fr/viz-bin/VizieR-3?-source=J/MNRAS/431/2/fluxes" rel="nofollow noreferrer">http://vizier.cds.unistra.fr/viz-bin/VizieR-3?-source=J/MNRAS/431/2/fluxes</a>
How do I convert <code>mW/m^2</code> to <code>W/m^2/nm</code> or <code>erg/s/cm^2/A</code>?
The first entry is logF(Halpha) = -12.03 <code>mW/m^2</code></p>

CC BY-SA 4.0
2023-08-13T15:34:20.303Z
You'll need some spectrographic information about the flux, since these units are not consistent with each other.
James K
CC BY-SA 4.0
2023-08-13T15:37:14.130Z
Do you know what kind of spectra info? The H-alpha is around 656 nm
Adrian
CC BY-SA 4.0
2023-08-13T15:43:50.763Z
No, but on the face of it this question is like asking to convert force to pressure (it can't be done, unless you know the shape or area)  The "per nm" part of the unit is is because a wideband flux with the same flux per nm would carry greater power than a narrowband flux.  The total flux is integrated over the frequency range... but this is now rapidly going above my paygrade.
James K
<units>
How to convert mW/m^2 to W/m^2/nm?
Adrian
# How to convert mW/m^2 to W/m^2/nm?

(Asked by: username_0 on 2023-08-13)

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`

**Tags:** units
> **Comments:**
> **username_1**: You'll need some spectrographic information about the flux, since these units are not consistent with each other.  
> **username_0**: Do you know what kind of spectra info? The H-alpha is around 656 nm  
> **username_1**: No, but on the face of it this question is like asking to convert force to pressure (it can't be done, unless you know the shape or area)  The "per nm" part of the unit is is because a wideband flux with the same flux per nm would carry greater power than a narrowband flux.  The total flux is integrated over the frequency range... but this is now rapidly going above my paygrade.  

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## 1 answers

### Answer by username_2 on 2023-08-14. Score: 2

You can't with any precision.

mW/m$^2$ is a measure of flux, presumably integrated over a H$\alpha$ emission line.

W/m$^2$/nm would be the units of the spectrum flux density - i.e. the quantity that would be integrated over wavelength to give you a total flux. This is not a single value, but a spectrum.

If you want to know the *average* spectrum flux density at the wavelength of H$\alpha$ is, about the best you could do is divide the flux in the H$\alpha$ line by what you think its full width half maximum is in nm. This might be set by the properties of the source, but will at a minimum be set by the spectral resolution of the instrument that measured the spectrum.

Thus in your example, suppose the line had a FWHM of 10 nm (very broad!), then we would subtract 3 from the log (divide by 1000) to put
mW into W and then subtract 1 more (divide by 10) to spread that flux across 10 nm, and the *rough, average* spectrum flux density over the H$\alpha$ line would be $10^{-16.03}$ W/m$^2$/nm.


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