Solve Supernova Problem: Apparent Magnitude at Max Luminosity

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In summary, Shishkabob solved the problem of finding the supernova magnitude in a flash by taking the two magnitudes you were given and plugging them into an equation. He got a relation from which it should be easy to extract the value of Fsn as a factor of Fg. Froddo solved the problem by taking the two magnitudes you were given and plugging them into an equation. He got a relation from which it should be easy to extract the value of Fsn as a factor of Fg.
  • #1
Froddo
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Hello community, I really need help with few assignments!

First one would be about Supernova:
In a distant galaxy a supernova flash occured. When supernova reached luminosity maximum, galaxy apparent magnitude
was 17,6. Before the supernova flash the galaxy apparent magnitude was 18,0. What was the supernova's apparent magnitude at it's luminosity maximum?

Could you help me solve this and explain it to me ?
 
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  • #2
So basically you've got the following:

mg, msn, and mg + sn

where mg = 17.6 and mg + sn = 18.0

So we know that mg + msn is NOT mg + sn

but we do know that

Lg + Lsn = Lg + sn

Do you have any equations that relate luminosity to magnitude? Or even a relationship between flux and magnitude, since F is proportional to L.
 
  • #3
Don't really have any equation or even a far-near one that would help in this problem. There is no actually luminosity mentioned as the numbers only state for apparent magnitude . I searched the whole net for relation between galaxy and supernova apparent magnitude, though did not find anything useful.

By the way , this is not really a homework or something, trying to learn some astroP on my own :)
 
  • #4
Of course there's the basic m1 - m2 = -2.5 log10 ( F1 / F2)

Where we can inputed m1 and m2 , though not sure which would I put in first. Like m1=mg and m2=mg+sn.

The final question is what was the Supernova's magnitude in the flash. Not sure this formula is useful in that way.
 
  • #5
you want to do something like

[itex]\frac{F_{sn+g}}{F_{g}} = 10^{\frac{m_{sn + g} - m_{g}}{-2.5}}[/itex]

I just solved your equation for the ratio of the fluxes.

Now we also know that we can just add up the fluxes, i.e.

Fsn + g = Fg + Fsn

does this help? Remember that if you get a number for the magnitude, it should be something comparable to the two magnitudes you've already got.
 
  • #6
Thread closed temporarily for Moderation...

Thread re-opened. Froddo -- check your PMs.
 
Last edited:
  • #7
Done berkman :)

On-topic : I've tried to solve the problem whole night with no results, I think I'm missing something important. Anyone with experience could give me some heads up ? Not sure where to begin..
 
  • #8
Let's try doing it in a more step-by-step fashion.

Look at the equation you provided in post #4. You needn't throw your condescension at it by calling it "basic", as this is your workhorse here. You won't be doing much more than playing with it.

Shishkabob solved the first step for you in post #5.
He took the two magnitudes you were given, as defined by yourself in post #4, and after plugging them into the equation and using the definition of the logarithm(http://en.wikipedia.org/wiki/Logarithm#Definition), he got a relation from which it should be easy to extract the value of Fsn as a factor of Fg.
In other words, it'll tell you how many times the supernova was brighter/dimmer than the galaxy.


Notice that it doesn't matter which goes first and which goes second when you plug the values here:

m1 - m2 = -2.5 log10 ( F1 / F2)

the end result is the same as if it were:

m2 - m1 = -2.5 log10 ( F2 / F1)

Now, use the same equation again, only this time with the values for the galaxy(mg, Fg) and the supernova(msn, Fsn). Plug in the value for Fsn from the previous calculations.
The flux will disappear from the equation, and you'll be left with a common logarithm, whose value you can check in the tables, e.g. here:
http://www.sosmath.com/tables/logtable/logtable.html

In the end, you should end up with the ratio of fluxes Fsn=~0,45Fg,
and supernova magnitude around 18,9.
Which is good, as we know that one magnitude difference corresponds to two-and-a-half times difference in flux, and here we've got something in that range.
 

What is a supernova?

A supernova is a powerful explosion that occurs when a massive star reaches the end of its life and collapses in on itself. This explosion can briefly outshine an entire galaxy and release vast amounts of energy and matter into space.

What is the apparent magnitude of a supernova at maximum luminosity?

The apparent magnitude of a supernova at maximum luminosity can vary, but on average it is around -19.3. This means that it can be billions of times brighter than our Sun.

How is the apparent magnitude of a supernova at maximum luminosity calculated?

The apparent magnitude of a supernova at maximum luminosity is calculated by comparing its brightness to that of a standard star with a known magnitude. The difference in brightness between the two is then converted into a logarithmic scale to determine the apparent magnitude.

Why is the apparent magnitude at maximum luminosity important in studying supernovae?

The apparent magnitude at maximum luminosity is important because it provides information about the energy and size of the supernova explosion. This can help scientists understand the physics behind the explosion and the properties of the star that caused it.

How do scientists use the apparent magnitude at maximum luminosity to study supernovae?

Scientists use the apparent magnitude at maximum luminosity to study supernovae by observing and measuring the changes in brightness over time. They can also use this information to classify and compare different types of supernovae and gain insights into the evolution of stars and galaxies.

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