I was hoping someone could explain (gently, to a non-numbers type) "msin i" as a planet mass boundary. The link in question is http://adsabs.harvard.edu/abs/2003A&A...403.1077K and the sentence: "Throughout the habitable zone around Barnard's star, i.e. 0.034-0.082 AU, we exclude planets with msin i〉 7.5 MEarth and m〉 3.1 MNeptune."

>> msin i >

think "minimum mass" and "inclination of orbit"

Hum,
have a look at this
http://www.astronomycafe.net/qadir/q2441.html

__________________
`Irony` actually does mean `metal like`...

I had seen that but, um, not enough info for a fellow who's forgotten what sine means...

Hum,
don`t worry about it...

Read more (PDF)

__________________
`Irony` actually does mean `metal like`...

I'll take a crack at it without numbers, I'm pretty sure if I slip, someone will point and chuckle, then correct me.

M = mass
i = orbit inclination

Msin(i) expresses the minimum possible mass of a planet orbiting a star with an unknown inclination.

Thanks, but surely the maximum mass if anything?

By seeing alternating red- and blue-shift in a star's spectral lines, we detect that it is wobbling alternately towards us and away from us. From that, we infer that a planet exists, and that the star is wobbling around its common centre of gravity with the planet.
But we can't see the planet, and we can't see any other movement of the star: all we know is the movement in the line of sight. So we don't know if the wobble we see is all the wobble there is.
The measured towards-and-away wobble will only reflect the total movement of the star if the planet happens to orbit in a plane that's edge-on to our line of sight (conventionally this orientation is taken to be an inclination of 90 degrees, since it's measured relative to the plane of the sky). If this were the situation, all the star's wobble is being revealed to us, we can see the total gravitational effect of the planet.
But if the planet's orbit is tilted relative to our line of sight, then the red- and blue-shift will show only part of the planet's influence on the star. So we underestimate the planet's true mass. At the limit, if the planet happened to orbit precisely in the plane of the sky (inclination zero), it could be hugely massive and we would detect no red- or blue-shift, because none of the induced wobble would be along our line of sight.
So a given amount of wobble may be due to a low-mass planet with an inclination of 90 degrees, a moderate-mass planet with an inclination of 45 degrees, or a hugely massive planet with an inclination of 1 degree. The sin(i) term makes the conversion from the true mass of the planet (M) to the amount of wobble we see.

To put some figures on it. Suppose the observed wobble suggests that M.sin(i) is two Jupiter masses (2J). This degree of wobble could be due to a 2J planet inclined at 90 degrees [since sin(90) = 1, and 2J x 1 = 2J]; or it could be due to a 4J planet inclined at 30 degrees [since sin(30) = 0.5, and 4J x 0.5 = 2J]; or it could be due to an 8J planet inclined at 14.5 degrees [since sin(14.5) = 0.25, and 8J x 0.25 = 2J].

Grant Hutchison

Mass is determined by a ratio of spectral shift and distance. In order to cause a given amount of tug, its got to be X mass and Y distance. The minimum is MUCH easier to lock down since it can't be but so far from a star, because the maximum is theoretically infinite, given infinite distance from a star. A third constraining factor is orbital velocity, the further out you go, the faster you've got to orbit. Dealing with the closest distance, the least mass, and the lowest orbital speed, and you can document that hard edge to the error bar, the other side of the range limits gets a little fuzzy.

You can make some reasoned judgement calls on what the maximum mass, given that beyond a certain point, you've got a brown dwarf or a star on your hands (which will stand out themselves against the star), but the distance and velocity are harder to judge, because we just don't know enough about what the real world limits are to planetary behavior.

To Doodler. Your initial statement should be maximum, I think: insofar as a tug does not NOT exist at such-and-such, no planet greater than such-and-such may exist (i.e., a tug we are not seeing is possible with a planet at -1X or below, but X is the MAXIMUM boundary). Or put more practically, studying a star with Hubble isn't going to tell us that a planet half-Mercury is impossible (a minimumum), but rather that a planet twice Jupiter is (a maximum).

To Grant: "The sin(i) term makes the conversion from the true mass of the planet (M) to the amount of wobble we see." Could you place that in the terms of the abstract I cited. I think I'm almost there (just a few more slugs of beer).

The value given is the value for the closest possible distance, with the minimum possible value of mass. If a candidate planet is further out, it would require greater mass to have the same effect (to say nothing of moving MUCH faster).

At least that's how I understand it. Its what surprised the teams that first discovered the epistellar jovians, because the minimum mass required for planets orbiting so quickly around their stars was so high (and so close).

These fellas have detected a level of apparent wobble in their chosen star, but they ascribe some of the line-of-sight movement to convective movement of the star's visible surface, rather than mass movement of the star as a whole. This leaves them with an unexplained bit which might be wobble caused by a planetary companion. Since they know the wobble does not exceed a certain amount, they know the planetary companion cannot exceed a certain value of M.sin(i).
If a planet were as close as 0.017 AU, M.sin(i) would have to be less than 0.12J, or they would have detected a larger wobble in velocity; out at 0.98 AU, M.sin(i) would have to less than 0.86J: a more massive object is required to make the star exceed the critical "wobble threshold".

The M.sin(i) factor means they've excluded an object at 0.98 AU with a mass larger than 0.86J in an orbit inclined at 90 degrees; an object with a mass larger than 1.72J in an orbit inclined at 30 degrees; an object with a mass of 3.44J in an orbit inclined at 14.5 degrees; and so on, using sin(i) to work out the mass limit at each inclination.

So you can see that M.sin(i) does give a minimum value. What's perhaps confusing is that the paper deduces the maximum possible value of that minimum!

Grant Hutchison

OK, thx both. I think I get the basics. Here is the page http://en.wikipedia.org/wiki/Barnard's_star that prompted my to ask BTW.