Converting flux from Janskys to erg/s/cm²/Å

[Astro Student]

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

 

[phyzguy]

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.

 

[Astro Student]

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.

 

[phyzguy]

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.

 

[Astro Student]

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!