<p>The definition of a magnitude is something like
$$ B = -2.5\log_{10} f_B + Z_B,$$
where $f_B$ is a physical flux in whatever unit system you are using and $Z_B$ is a zeropoint for the magnitude system and the minus sign is there to ensure small magnitudes are brighter.</p>

<p>A colour index is therefore something like
$$ B-V = -2.5\log_{10} f_B + 2.5\log_{10}f_V + Z_B - Z_V = -2.5\log_{10} \frac{f_B}{f_V} + Z_{BV}$$</p>

<p>Now, let's think about what you are asking.</p>

<p>$$BV = (-2.5\log_{10} f_B + Z_B)(-2.5\log_{10} f_V + Z_V)$$
$$BV= -2.5Z_v\log_{10} f_B -2.5Z_B\log_{10} f_V +Z_{B}Z_V -2.5 \log_{10} f_B^{-2.5\log_{10}f_V}$$</p>

<p>This numerology has no physical significance at all and, because the fluxes themselves are distance-dependent, then like $B$ and $V$ individually, $BV$ and $B/V$ would be distance-dependent and therefore have no relationship to anything physical that is intrinsic to the star.</p>

CC BY-SA 4.0
2018-08-17T11:14:39.897Z
2018-08-17T11:14:39.897Z
ProfRob
<p>Consider for example these bright stars (V and B-V values from <a href="http://cdsarc.u-strasbg.fr/viz-bin/Cat?V/50" rel="nofollow noreferrer">Yale Bright Star Catalog</a>).
Near V=0, B/V is ill-behaved, and B*V is near 0.</p>

<pre><code>Name       B     V     B-V    B/V   B*V
Arcturus  1.19 -0.04  +1.23 -29.7  -0.05
Vega      0.03  0.03   0.00   1.00  0.00
Capella   0.88  0.08  +0.80  11.0   0.07
Rigel     0.09  0.12  -0.03   0.75  0.01
</code></pre>

<p>Now imagine that each of those stars were 10 times as far away.
Neglecting <a href="https://en.wikipedia.org/wiki/Extinction_(astronomy)" rel="nofollow noreferrer">interstellar extinction</a>, this would decrease the apparent brightness by a factor of 100 and increase the apparent magnitude by 5.0.
As B and V increase together, B/V approaches 1 and B*V approaches V^2, telling us nothing meaningful about a given star.</p>

<pre><code>Name       B     V     B-V   B/V    B*V
Arcturus  6.19  4.96  +1.23  1.25  30.7
Vega      5.03  5.03   0.00  1.00  25.3
Capella   5.88  5.08  +0.80  1.16  29.9
Rigel     5.09  5.12  -0.03  0.99  26.1
</code></pre>

<p>B-V, on the other hand, has stood the test of time as a distance-independent metric of a star's color and surface temperature.</p>

CC BY-SA 4.0
2018-08-17T05:44:22.070Z
2018-08-17T05:44:22.070Z
Mike G
<p>The magnitude is logarithmic, so taking the difference between two values is effectively the same as taking the ratio between the two unlogged values.</p>

CC BY-SA 4.0
2018-08-16T23:34:01.957Z
Which is pretty much what I said. My question remains: is there any physical meaning in the ratio or product of the magnitudes?
Jim421616
CC BY-SA 4.0
2018-08-17T07:03:18.587Z
There is no such physical significance. @Jim421616
Kartik Chhajed
CC BY-SA 4.0
2018-08-16T23:25:59.983Z
2018-08-16T23:25:59.983Z
Ingolifs
<p>The <strong>color index</strong> in the photometric system is defined by the differences in magnitudes between two wavelength filters, which gives the ratio of intensities. For example, to determine the B-V index, you'd need the magnitudes in the B band (which is centered on $\lambda = 445$nm) and in the V band, centered on $\lambda = 551$nm.</p>

<p>But what about other operations? For instance, would there be any significance in taking, say, the ratio $B/V$ or $B\times V$?</p>

<star><observational-astronomy><magnitude>
Ratios of magnitudes instead of differences
Jim421616
# Ratios of magnitudes instead of differences

(Asked by: username_0 on 2018-08-16)

The **color index** in the photometric system is defined by the differences in magnitudes between two wavelength filters, which gives the ratio of intensities. For example, to determine the B-V index, you'd need the magnitudes in the B band (which is centered on $\lambda = 445$nm) and in the V band, centered on $\lambda = 551$nm.

But what about other operations? For instance, would there be any significance in taking, say, the ratio $B/V$ or $B\times V$?

**Tags:** star, observational-astronomy, magnitude


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## 3 answers

### Answer by username_1 on 2018-08-17. Score: 2 | Accepted answer

The definition of a magnitude is something like
$$ B = -2.5\log\_{10} f\_B + Z\_B,$$
where $f\_B$ is a physical flux in whatever unit system you are using and $Z\_B$ is a zeropoint for the magnitude system and the minus sign is there to ensure small magnitudes are brighter.

A colour index is therefore something like
$$ B-V = -2.5\log\_{10} f\_B + 2.5\log\_{10}f\_V + Z\_B - Z\_V = -2.5\log\_{10} \frac{f\_B}{f\_V} + Z\_{BV}$$

Now, let's think about what you are asking.

$$BV = (-2.5\log\_{10} f\_B + Z\_B)(-2.5\log\_{10} f\_V + Z\_V)$$
$$BV= -2.5Z\_v\log\_{10} f\_B -2.5Z\_B\log\_{10} f\_V +Z\_{B}Z\_V -2.5 \log\_{10} f\_B^{-2.5\log\_{10}f\_V}$$

This numerology has no physical significance at all and, because the fluxes themselves are distance-dependent, then like $B$ and $V$ individually, $BV$ and $B/V$ would be distance-dependent and therefore have no relationship to anything physical that is intrinsic to the star.


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### Answer by username_2 on 2018-08-17. Score: 3

Consider for example these bright stars (V and B-V values from Yale Bright Star Catalog).
Near V=0, B/V is ill-behaved, and B*V is near 0.
Name       B     V     B-V    B/V   B*V
Arcturus  1.19 -0.04  +1.23 -29.7  -0.05
Vega      0.03  0.03   0.00   1.00  0.00
Capella   0.88  0.08  +0.80  11.0   0.07
Rigel     0.09  0.12  -0.03   0.75  0.01

Now imagine that each of those stars were 10 times as far away.
Neglecting interstellar extinction, this would decrease the apparent brightness by a factor of 100 and increase the apparent magnitude by 5.0.
As B and V increase together, B/V approaches 1 and B*V approaches V^2, telling us nothing meaningful about a given star.
Name       B     V     B-V   B/V    B*V
Arcturus  6.19  4.96  +1.23  1.25  30.7
Vega      5.03  5.03   0.00  1.00  25.3
Capella   5.88  5.08  +0.80  1.16  29.9
Rigel     5.09  5.12  -0.03  0.99  26.1

B-V, on the other hand, has stood the test of time as a distance-independent metric of a star's color and surface temperature.


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### Answer by username_3 on 2018-08-16. Score: -1

The magnitude is logarithmic, so taking the difference between two values is effectively the same as taking the ratio between the two unlogged values.

> **Comments:**
> **username_0**: Which is pretty much what I said. My question remains: is there any physical meaning in the ratio or product of the magnitudes?  
> **username_4**: There is no such physical significance. @username_0  

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