Is my analysis of star observations in Orion accurate?

[omoplata]

I'm trying to get an expression for how many stars below a certain magnitude $$m$$ I will see in the entire field of view of a telescope.

First, consider only one spectral type with absolute magnitude $$M$$. The distance $$r$$ to a star of apparent magnitude $$m$$ is given by,

$$r=10^{3(m-M+5)/5}$$

$$N=n \cdot V$$

$$N$$ is the number of stars observed that are brighter than magnitude m
$$n$$ is the space density of stars (number of stars per cubic parsec)
$$V$$ is the volume of the spherical cone out to distance $$r$$, which is the view cone of the telescope.

But,
$$V= \frac{4 \pi}{3} r^{3} \times \frac{\Omega}{4 \pi}= \frac{\Omega r^{3}}{3}$$

Where $$\Omega$$ is the solid angle subtended by the telescope.

Therefore,
$$N= \frac{\Omega}{3} \cdot n \cdot 10^{3(m-M+5)/5}$$

Now, in order to account for more than one spectral type, let $$N_{i}$$ be the number of stars of spectral type $$i$$ visible below magnitude $$m$$, let $$n_{i}$$ be the space density of stars of spectral type $$i$$ and $$M_{i}$$ be the absolute magnitude of stars of spectral type $$i$$. Then,

$$N_{i}= \frac{\Omega}{3} \cdot n_{i} \cdot 10^{3(m-M_{i}+5)/5}$$

For all the spectral types,

$$\displaystyle \sum_{i} N_{i}= \displaystyle \sum_{i} \frac{\Omega}{3} \cdot n_{i} \cdot 10^{3(m-M_{i}+5)/5}$$

$$\displaystyle \sum_{i} N_{i}= 10^{3m/5} \displaystyle \sum_{i} \frac{\Omega}{3} \cdot n_{i} \cdot 10^{3(-M_{i}+5)/5}$$

Let $$c = \displaystyle \sum_{i} \frac{\Omega}{3} \cdot n_{i} \cdot 10^{3(-M_{i}+5)/5}$$, which is a constant w.r.t. $$m$$.

So,

$$\displaystyle \sum_{i} N_{i}= c \cdot 10^{3m/5}$$

Did I make a mistake here?

The problem is that when I count the number of stars below a magnitude $$m$$ in some astronomical images that I have of orion, and fit a curve to it, what I get is $$\displaystyle \sum_{i} N_{i} \propto 10^{0.3m}$$. According to this analysis it should be 0.6 instead of 0.3. Why is this?
The instrument I'm using has a sensitivity limit of relative magnitude of about 16. So I was told to ignore the effects of extinction. If there is nothing wrong with my math, is there something wrong with my physics?

 

[Ken G]

Why did you start out with the assumption that r scaled like 10 to 3m/5? I would have thought you'd want 10 to the m/2, since the apparent brightness scales like r^-2 in flux units.

 

[omoplata]

Oh, sorry. It should be $$r = 10^{(m-M+5)/5}$$. Thanks for catching that. But I had it right in the original calculation and made the mistake when typing. So there must be something else wrong.

The magnitude scale is defined to be,

$$100^{(m_{1}-m_{2})/5}=\frac{F_{2}}{F_{1}}$$

Where the $$m$$'s are relative magnitudes and $$F$$'s are fluxes.

But,

$$L=4 \pi r^{2} F$$ , where $$L$$ is the luminosity.

$$100^{(m_{1}-m_{2})/5}=\frac{L_{2} \cdot r_{1}^{2}}{L_{1} \cdot r_{2}^{2}}$$

If it's for the same star, $$L_{1}=L_{2}$$

So,

$$10^{(m_{1}-m_{2})/5}=\frac{r_{1}}{r_{2}}$$

Let $$r_{2}$$ be 10 parsecs. Then $$m_{2}$$ is the absolute magnitude $$M$$.

$$10^{(m_{1}-M)/5}=\frac{r_{1}}{10pc}$$

$$10^{(m_{1}-M+5)/5}=r_{1}$$ , in parsecs.

Let $$r_{1}=r$$ and $$m_{1}=m$$.

So,

$$r=10^{(m-M+5)/5}$$

I found this derivation in "Introduction to Modern Astrophysics" by Carroll and Ostlie.

It is unlikely that the observational data is wrong. Someone else had made a fit to another set of data from another instrument, and they got the same fit constant. 0.3.

 

[Ken G]

Yes, you're right about the magnitude variable, and you say that wasn't the problem. Then I would look at your V assumption. You are not looking at all the stars within a spherical volume of radius r away from you, you are looking in a given solid angle. That wouldn't make any difference if the stars were uniformly distributed, for then that solid angle would be like any other. But you say you are looking at Orion, and that's a star forming region, so you should have a whole bunch of stars at roughly the same distance. The number of stars you can see, above some m, in a star-forming region should depend on the initial mass function of that star forming region, not any issues that have to do with varying distance.

In other words, the idea that c is a constant requires a spatially homogeneous distribution of stars. In that case, you would be right-- the distribution of stellar Ms wouldn't matter, it would only be how the r changes as you change m. But when you have a cluster, the opposite things matter-- now the r is all the same, so that doesn't affect what you see when you change m, but the different number of stars at each M is what will matter. That means it is changes in the M cutoff that goes into c that control the situation, not changes in r. Hopefully that will resolve your problem-- you are observing the M distribution in a cluster at the same r, not the volume variations in stars with the same M distribution.

 

[omoplata]

Thanks for the suggestion. I'll try to find images pointing in other directions and see if they give the same fit constant.

But what I got, 0.3, is less than the anticipated value of 0.6. Since Orion is a star forming region, that means the density of stars is higher in that direction? Shouldn't the fit constant be more than 0.6 in that case? Or at least, increase before it decreases, giving a sort of a peak in the graph of (number of stars observed brighter than magnitude m) vs. magnitude?

 

[Ken G]

Thanks for the suggestion. I'll try to find images pointing in other directions and see if they give the same fit constant.

You might try a field in the "Milky way" swath, because that should have lots of stars and a pretty homogeneous dstribution not too far from the Sun, IIRC.

But what I got, 0.3, is less than the anticipated value of 0.6. Since Orion is a star forming region, that means the density of stars is higher in that direction? Shouldn't the fit constant be more than 0.6 in that case? Or at least, increase before it decreases, giving a sort of a peak in the graph of (number of stars observed brighter than magnitude m) vs. magnitude?

If all you see is the star-forming region at some fixed r, then all you'll get is the M distribution of the stars, which in principle could be larger or smaller than 0.3m. For example, if you are seeing the Saltpeter "initial mass function", then the cumulative distribution goes sort of like 10^0.2m. I got that by saying the cumulative distribution with mass goes like mass to the -1.3, and the luminosity goes like mass to the 3 or so, so the cumulative distribution with luminosity goes like luminosity to the -0.5, and that converts into going like 10^0.2m. Maybe if you throw in some field stars that are going like 10^0.6m, it pushes it up to 10^0.3m, I don't know. But a clear prediction here is, if you find a more uniformly populated field of view, that exponent should push higher.

ETA: the conceptual point here is that according to your analysis and mine, going to a telescope with a better magnitude limit is more helpful for seeing more stars in a spatial distribution that extends from the Sun to a long way away, than it is for seeing more stars in a single distant cluster.

 

[omoplata]

I took the point sources from 1 degree radius view cones 2MASS star catalog in 6 different directions. If my original direction was +x, I also took -x, +y, -y, +z and -z directions. I still get a curve that goes like [tex]10^{0.3}[\tex] instead of one that goes like [tex]10^{0.6}[\tex].

It's probably not relativistic effects, right?

Is it because I'm using the J band magnitudes instead of the bolometric magnitudes? I'll try some other catalogs and see what happens, like DSS for optical.

I'll also try looking in the "Milky way" swath.

 

[Ken G]

Hmm, that does surprise me. I would have expected a young star cluster to have a very different magnitude dependence than a line-of-sight that is volume-limited for any given absolute magnitude.