# Calculating Quasar Variability Using CCD Camera and Magnitude Comparison

• StillLearningToronto
In summary: Yes, you are off to a good start. :smile:By the way, earlier in Post #2 I said,"power 'flux' here is power per unit area."But I meant to say power "flux density" is power per unit area. To get the "flux," you multiply the flux density by the given area. Other than that, what I said was OK. :smile:
StillLearningToronto

## Homework Statement

You have decided to use a CCD camera to check if a 16th magnitude quasar is variable.
With your telescope/camera combination, you know that a star with a magnitude of 0 would deliver 1 × 109 photons/second to one pixel, so this allows you to work out the photons/second from the quasar delivered to one pixel.

SO- taking out the numbers i need: My quasar is the 16th magnitude.
The star I am comparing it to has a magnitude of 0.
The 0 mag star gives out 1*10^9 photons/second (1 pixel)

## Homework Equations

m2-m1=-2.5log(F2/F1)

## The Attempt at a Solution

Im not sure if "flux" and "photons/second" are the same thing, BUT if they are, then:
m2-m1=-2.5log(F2/F1)
16-0=-2.5log(F2/1*10^9photons/second)
F2=3.98*10^2 photons/second.**Im not looking for the answer because I really want to figure this out, I just need to know if I am going in the right direction, thank you in advance ! (:

StillLearningToronto said:

## Homework Statement

You have decided to use a CCD camera to check if a 16th magnitude quasar is variable.
With your telescope/camera combination, you know that a star with a magnitude of 0 would deliver 1 × 109 photons/second to one pixel, so this allows you to work out the photons/second from the quasar delivered to one pixel.

SO- taking out the numbers i need: My quasar is the 16th magnitude.
The star I am comparing it to has a magnitude of 0.
The 0 mag star gives out 1*10^9 photons/second (1 pixel)

## Homework Equations

m2-m1=-2.5log(F2/F1)

## The Attempt at a Solution

Im not sure if "flux" and "photons/second" are the same thing, BUT if they are, then:
m2-m1=-2.5log(F2/F1)
16-0=-2.5log(F2/1*10^9photons/second)
F2=3.98*10^2 photons/second.**Im not looking for the answer because I really want to figure this out, I just need to know if I am going in the right direction, thank you in advance ! (:
You're approach looks good to me.

By the way, power "flux" here is power per unit area. And the area in question is the area of a single pixel. So in the end, you're really just comparing powers. Power is energy per unit time.

And another "by the way," this problem seems to have an inherent assumption that the photons from the reference star and the photons from the quasar have approximately equal energies, on average. Or to put it another way, it assumes that the color of the quasar is the same color as your magnitude 0 reference star, as seen from the telescope (including the effects of red-shifts -- most quasars are significantly red-shifted due to expanding space). This might be a perfectly good assumption depending on the star and quasar in question, but I thought I'd mention it.

Last edited:
collinsmark said:
You're approach looks good to me.

By the way, power "flux" here is power per unit area. And the area in question is the area of a single pixel. So in the end, you're really just comparing powers. Power is energy per unit time.

And another "by the way," this problem seems to have an inherent assumption that the photons from the reference star and the photons from the quazar have approximately equal energies, on average. Or to put it another way, it assumes that the color of the quasar is the same color as your magnitude 0 reference star, as seen from the telescope (including the effects of red-shifts -- most quasars are significantly red-shifted due to expanding space). This might be a perfectly good assumption depending on the star and quasar in question, but I thought I'd mention it.

Thank you so much ! Its a part of a bigger question I am working on, but i just wanted to make sure i was off to a good start (:

collinsmark
StillLearningToronto said:
Thank you so much ! Its a part of a bigger question I am working on, but i just wanted to make sure i was off to a good start (:
Yes, you are off to a good start.

By the way, earlier in Post #2 I said,

"power 'flux' here is power per unit area."​

But I meant to say power "flux density" is power per unit area. To get the "flux," you multiply the flux density by the given area. Other than that, what I said was OK.

## 1. What are magnitudes?

Magnitudes refer to the size or quantity of something. In science, magnitudes are often used to measure physical properties such as length, mass, or temperature.

## 2. How do scientists compare magnitudes?

Scientists use various tools and techniques to compare magnitudes. This can include using instruments such as rulers, scales, or thermometers, as well as mathematical equations and calculations.

## 3. Why is it important to compare magnitudes in science?

Comparing magnitudes allows scientists to gather data and make observations about the physical world. It also helps in understanding relationships between different variables and in making predictions about future outcomes.

## 4. What is the significance of the SI system in measuring and comparing magnitudes?

The SI (International System of Units) is a standardized system used by scientists to measure and compare magnitudes. It ensures consistency and accuracy in scientific measurements and allows for easy communication and understanding among scientists worldwide.

## 5. Can magnitudes be measured and compared across different scientific fields?

Yes, magnitudes can be measured and compared across different scientific fields. This is possible because the SI system provides a universal standard for measuring and comparing physical properties, regardless of the specific field of study.

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