Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pair-instability supernova calculation

  1. Jun 15, 2012 #1
    Usual supernova 6 billion times brighter then ancestor star. Pair-instability supernova 100 times brighter then supernova 600 billion times brighter then ancestor star. If ancestor star apparent magnitude is 0 then apparent magnitude of pair-instability supernova from that star is = - 2.5* lg(6*10^11) = -29.5 Is my calculation correct?
     
  2. jcsd
  3. Jun 15, 2012 #2
    Yes, it is correct. I assure you, but the concept is not
    the apparent magnitude depends upon the distance also. You also need to compare the distance of the ancestor star from you with the distance of supernova from you. Something like this....
    Let the distance from you to ancestor star be d, and distance from you to supernova be x.
    Then the apparent magnitude will be -2.5*log((6*10^11)*d^2/x^2), since it is inversely proportional to the square of distance.
    That is why you are getting such an unusually high magnitude. In fact, higher than the Sun. It is -26.7.
     
  4. Jun 17, 2012 #3
    It's hard to imagine that star located several thousand light earth from solar system with so small apparent diameter just 0.00001'' can potentially create on Earth Mercury weather
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Pair-instability supernova calculation
  1. Pair-Instability SN (Replies: 4)

  2. Jeans instability (Replies: 1)

Loading...