<p>Both $B-V$ and $U-B$ are a reflection of how hot the surface of a star is. To first order (inhomogeneous surface temperatures do exist) this has a single value for a given stellar photosphere and so there ought to be a direct relationship between $B-V$ and $U-B$. Here it is, shown below as a graph appropriate for main sequence stars with a composition similar to the Sun. You can reproduce this yourself using data found (for example) in the <a href="http://ads.harvard.edu/cgi-bin/bbrowse?book=hsaa&amp;chap=2&amp;page=0068" rel="nofollow noreferrer">Handbook of space Astronomy and Astrophysics (pp68-69)</a> compiled by M.V. Zombeck.
<a href="https://i.sstatic.net/32k8z.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/32k8z.jpg" alt="colour-colour plot"></a></p>

<p>You can see that the relationship is complicated; this is because stellar surfaces are not blackbodies and the absorption by various chemical species cuts out chunks of the spectrum, particularly in the blue and ultraviolet. However, given a $B-V$ one could look up the appropriate $U-B$. Note that this is not straightforward the other way around, where there may be two values of $B-V$ possible for a given $U-B$.</p>

<p>In practice things are much more complicated. The plot above assumes the stars are on the main sequence. Stellar atmospheres change a bit as stars evolve and get bigger, with lower surface gravity. Thus changes the relationship a little. The relationship also changes if the composition is not similar to the Sun, especially around the bump at $B-V \simeq 0.4$. Finally, this relationship is <em>intrinsic</em>, it assumes you are not viewing the star through extinction caused by dust. If that is not the case then the whole relationship slides redward in both colours (the vector direction of which is shown  in the plot above for about 2 magnitudes of visible extinction) but also changes shape slightly, depending on how much extinction there is.</p>

CC BY-SA 4.0
2018-08-17T07:24:22.903Z
2018-08-17T07:24:22.903Z
ProfRob
<p>You can convert from B-V to U-B by assuming spectral energy distribution (SED).</p>

<p>For example, if you assume blackbody SED (BBSED). The BBSED is determined by only the temperature (+ scaling constant which is irrelevant here). Therefore, there is a mapping from B-V to the certain temperature. Once you get the temperature, you can find U-B by applying the temperature to the BBSED (again, there is a mapping).</p>

<p>Becareful in the situation where the object's SED is not BB (e.g., galaxy). Or, the object's SED in the range from U to V has issues such as strong line emission or absorption (especially 4000-A break). In these cases, you cannot assume BBSED.</p>

<p>People might provide you more informative idea how to deal with the problem if you specify what kind of object you are dealing with.</p>

CC BY-SA 4.0
2018-08-17T18:22:37.600Z
No real astrophysical objects are even approximately close to blackbodies for the purpose of this conversion - see the dashed line in the plot in my answer. The closest might be a very hot white dwarf - but even here the blue continuum can be heavily absorbed by the Balmer series.
ProfRob
CC BY-SA 4.0
2018-08-17T14:00:28.850Z
2018-08-17T14:00:28.850Z
Kornpob Bhirombhakdi
<p>In photometry, we determine the <strong>color index</strong> (or <em>colour</em> index if you're from my part of the world) as the difference in magnitudes between two wavelength filters. The reason that we use the difference, is because the magnitude system is logarithmic, so using the difference tells us the  ratio of the intensities of the two wavelength bands.</p>

<p>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>To find the U-B index, you'd also need the U-magnitude, at $\lambda = 365$nm.</p>

<p>So, as @zephyr said, there's no way to convert between the two color indices.</p>

CC BY-SA 4.0
2018-08-16T22:41:10.450Z
2018-08-16T22:41:10.450Z
Jim421616
<p>Is there a formula to convert the B-V index to U-B index or are they totally different and separate observations and can't be convert like you can with Fahrenheit->Celsius?  (Updated)  </p>

<p>I know F->C is not astronomy related directly but there is a formula to convert between the two measurements.</p>

CC BY-SA 3.0
2017-05-26T19:41:40.670Z
Not sure what F->C is, but a B-V index means you measured with both the B and V filters and subtracted the magnitudes. A U-B index implies you have U filter magnitudes. There is no way of knowing U-B if you only know B-V.
zephyr
CC BY-SA 3.0
2017-10-30T20:26:14.597Z
Ok, thanks Zephyr, F->C meant Fahrenheit to Celsius
MiscellaneousUser
<star><spectral-type>
B-V to U-B colour index
MiscellaneousUser
# B-V to U-B colour index

(Asked by: username_0 on 2017-05-26)

Is there a formula to convert the B-V index to U-B index or are they totally different and separate observations and can't be convert like you can with Fahrenheit->Celsius? (Updated)

I know F->C is not astronomy related directly but there is a formula to convert between the two measurements.

**Tags:** star, spectral-type

> **Comments:**
> **username_1**: Not sure what F->C is, but a B-V index means you measured with both the B and V filters and subtracted the magnitudes. A U-B index implies you have U filter magnitudes. There is no way of knowing U-B if you only know B-V.  
> **username_0**: Ok, thanks Zephyr, F->C meant Fahrenheit to Celsius  

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## 3 answers

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

Both $B-V$ and $U-B$ are a reflection of how hot the surface of a star is. To first order (inhomogeneous surface temperatures do exist) this has a single value for a given stellar photosphere and so there ought to be a direct relationship between $B-V$ and $U-B$. Here it is, shown below as a graph appropriate for main sequence stars with a composition similar to the Sun. You can reproduce this yourself using data found (for example) in the [Handbook of space Astronomy and Astrophysics (pp68-69)](http://ads.harvard.edu/cgi-bin/bbrowse?book=hsaa&chap=2&page=0068) compiled by M.V. Zombeck.
[![colour-colour plot](https://i.sstatic.net/32k8z.jpg)](https://i.sstatic.net/32k8z.jpg)

You can see that the relationship is complicated; this is because stellar surfaces are not blackbodies and the absorption by various chemical species cuts out chunks of the spectrum, particularly in the blue and ultraviolet. However, given a $B-V$ one could look up the appropriate $U-B$. Note that this is not straightforward the other way around, where there may be two values of $B-V$ possible for a given $U-B$.

In practice things are much more complicated. The plot above assumes the stars are on the main sequence. Stellar atmospheres change a bit as stars evolve and get bigger, with lower surface gravity. Thus changes the relationship a little. The relationship also changes if the composition is not similar to the Sun, especially around the bump at $B-V \simeq 0.4$. Finally, this relationship is *intrinsic*, it assumes you are not viewing the star through extinction caused by dust. If that is not the case then the whole relationship slides redward in both colours (the vector direction of which is shown in the plot above for about 2 magnitudes of visible extinction) but also changes shape slightly, depending on how much extinction there is.


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### Answer by username_3 on 2018-08-17. Score: 1

You can convert from B-V to U-B by assuming spectral energy distribution (SED).

For example, if you assume blackbody SED (BBSED). The BBSED is determined by only the temperature (+ scaling constant which is irrelevant here). Therefore, there is a mapping from B-V to the certain temperature. Once you get the temperature, you can find U-B by applying the temperature to the BBSED (again, there is a mapping).

Becareful in the situation where the object's SED is not BB (e.g., galaxy). Or, the object's SED in the range from U to V has issues such as strong line emission or absorption (especially 4000-A break). In these cases, you cannot assume BBSED.

People might provide you more informative idea how to deal with the problem if you specify what kind of object you are dealing with.

> **Comments:**
> **username_2**: No real astrophysical objects are even approximately close to blackbodies for the purpose of this conversion - see the dashed line in the plot in my answer. The closest might be a very hot white dwarf - but even here the blue continuum can be heavily absorbed by the Balmer series.  

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### Answer by username_4 on 2018-08-16. Score: 0

In photometry, we determine the **color index** (or *colour* index if you're from my part of the world) as the difference in magnitudes between two wavelength filters. The reason that we use the difference, is because the magnitude system is logarithmic, so using the difference tells us the ratio of the intensities of the two wavelength bands.

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.

To find the U-B index, you'd also need the U-magnitude, at $\lambda = 365$nm.

So, as @username_1 said, there's no way to convert between the two color indices.


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