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

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Discussion Overview

The discussion revolves around the conversion of flux units from Janskys to erg/s/cm²/Å. Participants explore the mathematical relationships and necessary adjustments for unit conversion, including the implications of frequency and wavelength in the context of these units.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion regarding the conversion from Janskys to erg/s/cm²/Å, suggesting that an integral may be necessary due to unit discrepancies.
  • Another participant explains the relationship between wavelength and frequency, indicating that a factor of ν² or λ² is needed depending on the conversion direction.
  • A subsequent reply reiterates the need for careful handling of the negative sign in the conversion, clarifying that it reflects the inverse relationship between wavelength and frequency.
  • One participant proposes a specific conversion formula, asserting that the units work out correctly when properly applied, and provides a detailed breakdown of the conversion process.
  • A later reply confirms understanding of the conversion process by relating it to a known example involving a zero-magnitude star's flux, indicating a practical application of the discussed concepts.

Areas of Agreement / Disagreement

Participants demonstrate varying levels of understanding and agreement on the conversion process, with some expressing confusion and others providing clarification. The discussion does not reach a consensus on the necessity of integration or the handling of the negative sign.

Contextual Notes

Participants mention potential complications in the conversion process, such as the need for integration over frequencies or wavelengths and the implications of the negative sign in the context of unit conversion.

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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