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Following Mo et al. 2010 (page 479), the quantity $E(B-V)$, called the color excess between B and V bands, is equal to:

$$E(B-V) = A_B - A_V = (B - V) - (B - V)_0 = (m_B - m_V) - (M_B- M_V)$$

where $m_B, m_V$ are the apparent magnitude in B and V bands, and $M_B, M_V$ are the absolute magnitude in B and V bands.

Keep in mind the relation between apparent and absolute magnitudes, with $d=10$ pc:

$$m = M - 5 \log(d) + 5$$

If we are at redshift $z>0$, the color excess should be $E(B-V)>0$, because the extinction due to dust is greater in the B band than in the V band, and apparent magnitudes are greater than absolute magnitudes.

I'm wondering if it is correct to have $E(B-V)=0$ if the redshift is $z=0$ and $M=m$. The apparent magnitude is equal to the absolute magnitude and the relation above mentioned should become:

$$E(B-V) = A_B - A_V = (B - V) - (B - V)_0 = (m_B - m_V) - (M_B- M_V) = (M_B- M_V)_0 - (M_B- M_V)_0 = 0$$

Is that correct? Is it possible that we don't have color excess for galaxies at $z=0$, what am I missing?

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    $\begingroup$ Welcome to Astronomy SE! I've converted your equations to MathJax which is the standard way to render math in Stack Exchange. Can you take a look to make sure you're okay with it? You (or I) can roll it back if not. $\endgroup$
    – uhoh
    Aug 26 at 23:29
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$(B-V)_0$ is the intrinsic (unreddened) colour.

Reddening is not directly related to redshift because some of it occurs due to the dust in our Galaxy. Thus there is usually some reddening for nearby galaxies that depends on the sightline to them through our Galaxy.

The apparent magnitude does not equal the absolute magnitude at $z=0$. Absolute magnitude is the apparent magnitude at 10 pc.

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    $\begingroup$ I think I was confusing rest-frame apparent&absolute magnitude (where z=0, and d=D for apparent and d=10 pc for absolute magitude), and observed-frame apparent&absolute magnitude (where z>0, but d=D for apparent and d=10 pc for absolute mag). Can you confirm it? $\endgroup$
    – fslack
    Aug 27 at 10:24

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