1
$\begingroup$

EDIT: I am still following up, but here's the latest response from Wolfram:

Paclet servers are updated and optimized for each version of Mathematica, so I suspect you may be working on an older version which we no longer update. Many times we get reports that data from the paclet data for version 9 was incorrect, only to find that it has already been fixed for version 10.

According to Mathematica and several other sources, the known information about HIP31978 (also known as "S Monocerotis" and "15 Monocerotis") includes:

  • The star system is about 101.06 light years from our own.

  • The star's absolute magnitude is -2.79 (some sources say even lower).

  • The star's visual magnitude from our solar system is 4.66.

The star is slightly variable, but not enough to explain the following discrepancy:

If the star's magnitude at 32.6 light years is -2.79 (definition of absolute magnitude), it would be about 9.6 times fainter at 101.06 light years (it's actual distance from us). However, that's only about 2.5 magnitudes fainter, nowhere near enough to bring it down to magnitude 4.66.

What's happening here, and is this a one-off special case, or does this happen a lot?

For reference, my calculation of the apparent magnitude:

-2.79 - Log[100, (32.6/101.06)^2]*5 == -0.333192

Improve this question
$\endgroup$
4
  • $\begingroup$ Further, a star that is 3.1 times further away is NOT 3.1 times fainter, it is 9.61 times fainter. $\endgroup$
    – ProfRob
    Jan 12, 2016 at 15:02
  • $\begingroup$ Are you sure that's true for magnitudes, not just intensity? It seemed counter-intuitive to me too, but I think magnitude is a measure of light intensity at a given point, not the total light intensity emitted. Example: Sirius has apparent magnitude -1.44, absolute magnitude +1.45 (so apparent is 10^(2.9/5) = 3.8 times brighter). Its distance is 8.60 light years, which is about 1/3.8 of the 10 parsecs used for absolute magnitude. $\endgroup$
    – user21
    Jan 12, 2016 at 17:49
  • $\begingroup$ Not sure what your point is. 3.1 times further away is 9.61 times fainter, corresponding to an increase of 2.46 magnitudes. No debate. $\endgroup$
    – ProfRob
    Jan 12, 2016 at 21:25
  • $\begingroup$ My mistake. I incorrectly computed 5 magnitudes = 10 times brighter, not 5 magnitudes = 100 times brighter. Will fix my post shortly $\endgroup$
    – user21
    Jan 12, 2016 at 21:54

1 Answer 1

Reset to default
0
$\begingroup$

The absolute magnitude you quote appears to correspond to the Hipparcos parallax distance estimate, which is $\sim 1000 \mbox{ ly}$, not the $\sim 100 \mbox{ ly}$ distance you have used.

Improve this answer
$\endgroup$
4
  • $\begingroup$ This makes the numbers work, so I'll accept the answer, but it still bugs me. The whole point of having an absolute magnitude is to compare relative brightness. Using two different definitions seems inconsistent with this goal. $\endgroup$
    – user21
    Jan 12, 2016 at 17:50
  • $\begingroup$ @barrycarter There is only one definition of absolute magnitude - the magnitude the object would have at 10 pc. $\endgroup$
    – ProfRob
    Jan 12, 2016 at 22:49
  • $\begingroup$ Oh, OK, I get it. @ConradTurner is saying that the distance estimate I have is way off from the distance estimate used to compute the absolute magnitude... by a factor of 10. So, Mathematica is using one source for distance estimate and another for absolute magnitude, and the two sources don't agree closely on distance. $\endgroup$
    – user21
    Jan 12, 2016 at 23:28
  • $\begingroup$ @barrycarter More likely is that Mathematica has a typo in the distance, or some other error, rather than another source for the estimate. $\endgroup$
    – Conrad Turner
    Jan 13, 2016 at 7:54

You must log in to answer this question.