This is a hypothetical scenario in an attempt to test my understanding of the color term when performing photometric calibrations:

Imagine we are observing stars through a B filter with an offset towards the blue end of the spectrum (let's say about 50 Angstroms bluer than the standard B filter). Naturally, the stars we observe will appear to be bluer than they really are (there will be a systematic offset). In this case, a correction term would need to be added in order to effectively "dim" the star. We would want to add a positive number to make the magnitude larger (dimmer).

This is where I am a bit confused. Say this offset is 0.25 magnitudes between the "our" hypothetical bluer B filter and a real blue filter for a star. Would the color term be negative or positive?

I would think it would be positive since the color term appears on the right as follows: m_instrumental - m_B = C_B (B-V) + Constants (ZP + extinction)

ie. We want to add a positive number to the instrumental magnitude so it would be negative. Is this correct or am I missing something?

Here is a good resource link and or a refresher for those that want to take a look at what I am referring to:

Update: for arguments sake, let's pretend all the stars appear bluer.


  • 1
    $\begingroup$ Whether the star appears bluer or redder will depend on the stellar spectrum. Most stars observed through a bluer filter will appear fainter. $\endgroup$
    – ProfRob
    Jul 24, 2021 at 23:45
  • $\begingroup$ Updated to add, for arguments sake, let's pretend all the stars do indeed appear bluer. Thanks for bringing that point up! $\endgroup$
    – Astroturf
    Jul 24, 2021 at 23:56
  • $\begingroup$ I don't get the question. A blue filter shows things blue, a green one green. There is no general correction to obtainr the brightness of objects as seen through one filter to brightness as seen through another. Otherwise using different filters would be pointless $\endgroup$ Jul 25, 2021 at 0:30
  • $\begingroup$ If we observe young blue stars through a B filter that is offset to the bluer part of the spectrum (a central wavelength shifted to shorter wavelengths) it will appear brighter in the blue than a standard B filter without this offset as B is a standard... so we would need to correct for this. $\endgroup$
    – Astroturf
    Jul 25, 2021 at 0:44
  • $\begingroup$ The point of asking this question is because filters are never perfect so this correction needs to be done to do precision photometry. I'm just posing a hypothetical scenario to provide context for the question $\endgroup$
    – Astroturf
    Jul 25, 2021 at 1:04

1 Answer 1


I think I have an answer to my question. This website seems to explain it best: How to measure the color terms for your camera

If we measure a variety of standard stars (bluer and redder stars) using b and v magnitudes, we can compare them to the what their magnitudes should be using the following equation:

m_instrumental - m_B = C_B (B-V) + ZP

We can plot the data we gather from our observation and find a relationship that may look similar to this:

photometric calibration

In this case, the redder stars are more strongly affected than the bluer stars in that their magnitude differences are greater (between instrumental and true). The slope is the first order color term and will be positive in this case. This happens because standard filters are not "perfect" and they will not exactly represent same profile as the original filter the standards were measured.

Going back to the initial problem I asked, our B filter is shifted further towards the shorter wavelengths which would cause the bluer stars to appear brighter in the blue filter while red stars appear dimmer (capturing less red light). This can be understood by looking at the diagram below:

Example stellar spectra

I would argue that the color term for my example would be positive as well for the following reasons:

  1. The redder stars will appear dimmer so their difference will warrant a positive number.
  2. The bluer stars will appear even bluer in the offset filter (look at blue star's spectra) so their differences will be negative.

These differences will be added to the constant offset ZP. Regardless, we are left with lower numbers on the left hand said of the x axis and higher numbers on the right hand side which accounts for a positive slope.

  • $\begingroup$ The plot x-axis and the equation are inconsistent. I think it should be $B-V$ on the plot x-axis. $\endgroup$
    – ProfRob
    Aug 24, 2021 at 8:04

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