Unit Conversion of Flux: Jansky to Erg/s/cm²/Å - Simplified Guide

AI Thread Summary
The discussion focuses on converting flux measurements from Janskys to erg/s/cm²/Å. The conversion requires understanding the relationship between wavelength and frequency, specifically that λν = c, leading to the need for factors of ν² or λ² depending on the conversion direction. A participant clarifies that the negative sign in the equations indicates the inverse relationship between wavelength and frequency, which is typically ignored in practical calculations. The correct conversion formula is provided, showing how to cancel units effectively. Ultimately, the original poster confirms their understanding through a practical example involving photon flux.
Astro Student
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Hello,

I am struggling a little bit with what I believe to be a simple unit conversion. For this problem, I have many fluxes given in units of Janskys. I would like to convert them from their original units of

Jy = 10-23 erg s-1 cm-2 Hz-1

to units of

erg s-1 cm-2 Angstrom-1

When I try to do a conversion using the simple λv = c equations, the units do not work out properly. I assume there is some sort of integral I must take?

Thanks,

Astro Student
 
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You need to realize that since λ ν = c, this means that λ = c /ν, so dλ = -c/ν^2 dν, and likewise dν = -c/λ^2 dλ. So, depending on which direction you are converting, you need a factor of ν^2 or λ^2 in the denominator. This should solve your problem. Let me know if it doesn't.
 
phyzguy said:
You need to realize that since λ ν = c, this means that λ = c /ν, so dλ = -c/ν^2 dν, and likewise dν = -c/λ^2 dλ. So, depending on which direction you are converting, you need a factor of ν^2 or λ^2 in the denominator. This should solve your problem. Let me know if it doesn't.

Thanks for the reply. I am still very confused; this is giving me units of per meter per second. Should I be integrating this then over all frequencies/wavelengths? I also now have a factor of -c with which I do not know what to do, since the flux should not be negative.
 
The minus sign is just telling you that increasing wavelengths represent decreasing frequencies and vice-versa. Normally this would be ignored. As for the unit conversions, I think the following is correct:
\rm 1 \frac{erg}{s \, cm^2\, \unicode{x212B}} = 1 \frac{erg}{s \,cm^2 Hz} \times \frac{3.0E8 (m/sec)}{\lambda^2 (\unicode{x212B}^2)}\times 1E10 \frac{\unicode{x212B}}{m} = \frac{3.0E18}{\lambda^2 (\unicode{x212B}^2)} \times 1 \frac{erg}{s \,cm^2 Hz}= \frac{3.0E-5}{\lambda^2 (\unicode{x212B}^2)} Jy

These units work out, because the Hz in the denominator cancels with the sec-1 in the numerator. Does this make sense?

Edit: adding a web site with more detail.
 
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phyzguy said:
The minus sign is just telling you that increasing wavelengths represent decreasing frequencies and vice-versa. Normally this would be ignored. As for the unit conversions, I think the following is correct:
\rm 1 \frac{erg}{s \, cm^2\, \unicode{x212B}} = 1 \frac{erg}{s \,cm^2 Hz} \times \frac{3.0E8 (m/sec)}{\lambda^2 (\unicode{x212B}^2)}\times 1E10 \frac{\unicode{x212B}}{m} = \frac{3.0E18}{\lambda^2 (\unicode{x212B}^2)} \times 1 \frac{erg}{s \,cm^2 Hz}= \frac{3.0E-5}{\lambda^2 (\unicode{x212B}^2)} Jy

These units work out, because the Hz in the denominator cancels with the sec-1 in the numerator. Does this make sense?

Edit: adding a web site with more detail.
This makes a lot of sense. I was able to check using some knowledge from the textbook that a zero-magnitude star receives 1000 photons at 550nm per second per cm^2 per Angstrom. The energy of these photons was equivalent to the flux of a zero-magnitude star in the V band (3640 Jy). Thank you!
 
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