Talk:Absolute magnitude

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[edit] Equations

Uh, should not be absolute magnitude be

  • M_{star}=m_{star}+5(1-\log_{10}(parallax_{star})) \,

rather then

  • M_{star}=m_{star}+5(1+\log_{10}(parallax_{star})) \,

?

No, the latter is correct.

Imagine 2 stars, the first, star A, is 10 parsecs away; the second, star B, is 100 parsecs away. The Both have the same apparent magnitude as seen from Earth, m.
Star A's parallax is 0.1" (seconds of arc). Star B's parallax is 0.01".
Star A's absolute magnitude is m (by definition of absolute magnitude)
Star B is 100 times brighter intrinsically than star A, and hence will have a lower (brighter) absolute magnitude;
M_{Star B} = M_{Star A} - 5 = m - 5 \,

Try putting the numbers in to the above. The latter equation gives;
M_{Star A} = m + 5 (1 + \log_{10} (0.1)) = m \,
M_{Star B} = m + 5 (1 + \log_{10} (0.01)) = m - 5 \,

The first of your equations gives;
M_{Star A} = m + 5 (1 - \log_{10} (0.1)) = m + 10 \,
M_{Star B} = m + 5 (1 - \log_{10} (0.01)) = m + 15 \,

Hope that's clear enough that the latter equation (the one that appears in the article) is the correct one. Richard B 16:07, 7 October 2005 (UTC)

[edit] Phase Integrals?

Some questions about the planetary apparent magnitude calculation:

1) What happens when the phase angle is 180 degrees (pi radians)? The phase integral in this case is zero, which means you're dividing by zero in the apparent magnitude equation. In reality, this situation would be when the body is between the sun and the observer, so you wouldn't see anything at all (like seeing a 'new moon'). But this apparently breaks the equation, since an invisible sphere wouldn't have an apparent magnitude. :)

I don't think there's a problem really.
As less and less of the objects surface is illuminated as we see it, it will get increasingly faint - so its apparent magnitude will increase.
The formula predicts that if the phase integral tends to 0, then the apparent magnitude will tend to +infinity.
If an object did have an apparent magnitude of +infinity - then this equates to an object which reflects no light - so that's fine and agrees with the formula.Richard B 23:36, 27 October 2005 (UTC)
OK, that's what I thought. I think I was confused because when I see a formula divided by zero I don't think "will tend to infinity", I think "divide by zero so you get a broken equation" :).
Though this seems odd: If the Body-Observer distance is very small and the phase angle is 180 degrees, the apparent magnitude is very bright. I'm using this equation to calculate planetary magnitudes in Celestia and I'm seeing an apparent magnitude of -3.41 when I'm about 25,000km over the dark side of Venus with a phase angle of 179.72 degrees. But when I look at the same view of Venus from Earth (ie Body-Observer distance is large, 0.289AU) I get an apparent magnitude of 12.36, which is more reasonable. Shouldn't the apparent magnitude be large (ie dim) when you're right up close to the planet too?
The apparent magnitude will go down (get brighter) when you're closer to an object. This is intuitive if you think about it - you wouldn't expect the Moon to be very faint because it's so close, nor would you expect extra-solar planets (planets orbiting other stars) to be really bright because they're so far away. If you assume that something has a constant surface brightness (measured in magnitudes per square arcsecond or similar), then you will get an object getting brighter as you get closer, because the apparent angular surface area is getting larger as you get closer.
The apparent magnitude of an extended object, such as a planet from 25000km away, is actually the total integrated brightness from its entire visible surface. I wouldn't have thought that -3.41 is that unrealistic when you're considering that the planet has an angular diameter of about 25º Richard B 10:09, 28 October 2005 (UTC)
Also, from what source/reference does that p(\chi)\!\, Phase Integral equation actually come from? I can't find any others phrased that way. In fact, all I can find are constants (see below), not equations whose results vary depending on phase angle.

----------------------------

2) The Ideal Diffusing Sphere equation shown here doesn't really work well in practise for the planets in our solar system, since they're not actually lambertian reflectors. Does anyone know the actual phase integral equations used to calculate the real apparent magnitudes for planets? (I'm assuming the numbers we see in astronomy magazines for the apparent magnitude of Mars or Venus or Jupiter must actually be calculated somehow...) I've found some values for the phase integral for other types of reflector (4/3 for a Raleigh atmosphere, 16/3(1 - ln2) for a lommel-seeliger reflector) but these are constants and not equations like the one shown here - where can I find the equations for these types of reflector that depend on phase angle? Or do these constants just replace the (2/3) at the start of the Lambert phase integral equation?

[edit] Magnitude (astronomy)

When amateurs refer to magnitude in astronomy, they probably don't know the difference between apparent magnitude and absolute magnitude. This also creates a problem for the disambigation link repair: I, for example, have no idea whether to pipe magnitude link from (exampple) All Sky Automated Survey article to absolute or apparent. Could sb create the article in the heading explaining the basic magnitude idea in astronomy and comparing the specific types, so the current two articles would become subarticles? Note that currently it is a redirect to magnitude disambig - far from a pefect solution.--Piotr Konieczny aka Prokonsul Piotrus Talk 17:47, 9 January 2006 (UTC)

[edit] AB magnitude

Note that we have a 'redlink' from Jansky to AB magnitude, if anyone wants to oblige with a stub. Bob aka Linuxlad 23:07, 8 February 2006 (UTC)

Since nobody has responded to Bob in 8 months, I'll assume that the AB is just an abbreviation for "absolute" and act accordingly. Gene Nygaard 18:31, 6 October 2006 (UTC)

[edit] Help me!

I still don't get what Absolute magnitude is! —The preceding unsigned comment was added by Janie71991 (talk • contribs).

It's just a way to compare the brightness of two different objects. If you had two candles and you wanted to know which was brighter, you'd have to put them next to each other to see. It's the same with stars. — RJH (talk) 21:04, 24 August 2006 (UTC)
I think the point she's trying to make is that the article isn't clear as to what Absolute Magnitude is, at least to a layperson. I'm something of a space buff, I know about magnitude, I know about apparent magnitude, yet I read this article and I'm none the wiser as to what absolute magnitude is. Is the luminosity distance fixed or does it change? I'd guess (because the maths is well beyond me) that it's fixed, in which case why not state what said distance is? And if it's not fixed you get an idea as to how little the article has helped me. In fact Common sense gives me a better idea of what absolute magnitude is than this article as I gathered it was something like RJHall stated. If there was a WP:Clarity tag (I can't find one) I'd use it on this article. (User_talk:Moriarty 10:33, 29th Oct 2006 UTC)

Wow. I didn;t get that at first, but now I do. Thanks I forgot even after my science teacher taught me!! user:lexispice

[edit] Help me!

On my computers I get a box where a minus sign should be?

Also, Absolute Magnitude and similar related topics wrt stars are hard to understand. Some give symbols without defining, bad, bad, bad....... LR

LR:

In response to discussion of m vs. M. The page is incorrect: absolute magnitude is NOT the apparent magnitude. Simply, the apparent magnitude (m) is how bright the star appears. That is, how our eyes react to a stars luminescence. Absolute magnitude (M) is an actual measure of luminosity – its true brightness in space at a standard distance. This is how I explained it in a recent paper (2008©).

Phithetakappa. —Preceding unsigned comment added by Phithetakappa (talk • contribs) 23:18, 24 January 2008 (UTC)

[edit] References

Since the equation listed for the apparent magnitude of a body under different conditions

m = H + 2.5 \log_{10}{(\frac{d_{BS}^2 d_{BO}^2}{p(\chi) d_0^4})}\!\,

is not a standard equation found in most astronomy books, is there a reference for the equation?

129.123.104.5 00:16, 12 October 2006 (UTC)Nathan

[edit] Units

In the equations for calculating apparent magnitude from absolute magnitude (either for planets or for stars) what units are to be used for distance? Kaiguy 16:23, 16 May 2007 (UTC)

[edit] Bolometric versus absolute

Someone needs to define more precisely the difference between bolometric and absolute magnitudes. What do the two things measure? Dicklyon 00:32, 27 February 2007 (UTC)

  • Bolometric magnitude measures the integrated flux. Magnitudes are defined for different filters (e.g., the visual filter, V-band, is the most common and usually the basis used when determining extinction), which have a set range of wavelengths for which they transmit light. The bolometric magnitude would be the magnitude for all light, none filtered out. The sun's bolometric magnitude is +4.75, according to [1] and [2]. A 1938 Kuiper paper gives 4.62. [3] gives 4.72. I'm looking for other sources, but I didn't find any in a brief ADS search. Someone's textbook ought to have this with a reference... --Keflavich 22:13, 1 May 2007 (UTC)

[edit] The absolute magnitude of a galaxy...

...is equal to the absolute magnitude of an imaginary point source of the total luminosity of the galaxy's whole luminous mass, am I right? Because you can't be "ten parsecs away" from a galaxy. The article does not make this clear. - ∅ (∅), 17:35, 19 May 2007 (UTC)