# Homework Help: Magnitude of Stars - Flux, Photon Counts

1. Jan 29, 2006

### dekoi

For an assignment, we were told to use a program titled "Photoelectric Photometry of the Pleiades", located at this website.

It is basically a simulation of a telescope, in which we can "measure" the apparent magnitude of the stars in the Pleiades cluster.
My question is as follows:

Show that the program is at least faking the results continuously. Measure the V magnitudes of any two stars which are close together. Also record the counts for each star. Verify that the program is correctly convreting flux ratios to magnitude differences. To do this, you will have to determine how the flux is related to the photon count. (Don't forget the sky.)

photon count is proportional to flux

Therefore,
$$\frac{P_2}{P_1}=\frac{F_2}{F_1}=\frac{1067491}{1976639}$$
(I used the average photon counts from both stars).

Theoretically, the difference in magnitudes should be,
$$m_2 - m_1 = 6.43 - 5.76 = 0.67$$
(I used the apparent magnitudes that I "recorded" with the telescope).

Using the flux ratio, the difference in magnitudes should be,
$$m_2 - m_1 = -2.5log(\frac{1067491}{1976639}) = 0.669 = 0.67$$

I obviously obtained the correct results. However, I did not consider anything about the sky, which was a "hint" given in the question.
Also, what is the purpose of finding the magnitudes of 2 stars which are close to each other?

I do not understand how this shows that the program is faking the results, if they are converting the flux ratios to magnitude differences correctly.

Thank you for any help -- Sorry for the long post.

2. Jan 30, 2006

### SpaceTiger

Staff Emeritus
Your calculation looks fine. Does the program give a value for the sky? If not, you should be able to get a rough value by looking at the photon counts in a "blank" part of the sky (no stars around). If your detector has a linear response (was this discussed at all?), then you have to subtract the sky from photon counts for the stars. This subtraction will depend on the sky area over which the program counted photons in order to derive the star's photon counts...and that, presumably, will depend upon the point spread function of your instrument.

This all can get rather complicated and I'm not sure what level of sophistication your professor is expecting. Perhaps you could elaborate more on what you discussed in class and how you performed the measurements.

Different parts of the detector generally have different sensitivities, so if this has not been automatically corrected for, then the photon counts of a given star will depend on the part of the detector its light hits. Two stars close together are hitting roughly the same part of the detector, so their relative photon counts should be a good approximation to their relative brightnesses.

3. Jan 30, 2006

### dekoi

In order to perform the measurements, I:
1- Took a photon count for the background sky.
2- Took a photon count on the star.
3- We were told that the program automatically subtracts the "sky count" from the "star count".

We didn't discuss much in class... a few equations were mentioned, and that is all I know. That is basically why I'm having trouble.

Thank you for responding.

4. Jan 30, 2006

### SpaceTiger

Staff Emeritus
Is it the count per pixel? To get a real star photon count, one would have to integrate over the point spread function. Maybe this complication is being disregarded.

Then I don't see why you would have to worry about the sky.

5. Jan 30, 2006

### dekoi

6. Jan 30, 2006

### SpaceTiger

Staff Emeritus
From what you've told me, it sounds like you should just go with what you have.

7. Jan 30, 2006

### dekoi

I think so to.

Thank you very much for responding so quickly-- you are one of the few very helpful people I have found on this forum. :)

8. Jan 30, 2006

### SpaceTiger

Staff Emeritus
Nah, there just aren't that many astronomers hanging around.

9. Jan 27, 2010

### seto6

isn't the equation is" m2-m1=log(f2/f1)"

now i am confused
can any one tell me where i went wrong

10. Nov 8, 2011

### Jenab6

Sky, star, sky, comparison star, repeat.

You have to subtract the sky.
You also need to subtract the system dark current.

m = −2.5 log F − 18.88

m = apparent magnitude
F = flux in watts per square meter