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Blackholes and Supernovae

  1. Apr 10, 2012 #1
    I have a question about the formation of black holes.
    Correct me if I am wrong but as I understand it, if you have a massive enough star that when its used up all its fuel and collapses, it will overcome the exclusion principle and crush itself into a black hole.

    My question is, during this collapse, doesn't the core of the star get hot enough for it to become a supernova?
    Basically, while the core is collapsing, on its way to becoming a black hole, shouldn't the star also go supernova? and if it does, how do we end up with a black hole after the star went bang?

    Or is there another size bracket there so that if a star is within a certain range for mass, it will go supernova but if its bigger then it just collapses into a black hole faster than it has a chance to become a supernova?

    Thanks
     
  2. jcsd
  3. Apr 10, 2012 #2
    Great question.
    Short answer, take a look at this schematic:
    http://startswithabang.com/wp-content/uploads/2008/04/star2evo.jpg
    Keep in mind, it is very schematic.

    The two main categories of supernovae are core-collapse, and thermonuclear. In general core collapse supernovae occur from stars more massive than about 8 solar masses, and less than about 30. These generally look like 'Type II' supernovae, and produce neutron-star remnants.

    Thermonuclear supernovae are the explosions of white-dwarfs (which form from stars between about 2 and 10 solar masses), which (we think) look like 'Type I' supernovae, and leave no remnant (i.e. just a nebula).

    Black-holes are produced by more massive stars, i.e. those above about 30 solar masses via either "direct collapse" or "fall-back". For stars right around 30 solar masses, exactly what you guess happens---as they collapse, they get so hot that they explode.... but the explosion doesn't have enough energy for the material to escape, and it ends up falling back to the center and forming a black-hole. Direct-collapse is believed to happen in very massive stars which huge iron cores. The gravity of these stars is so strong that their material is completely sucked in, with little to no emission.

    The details of these processes aren't entirely clear---especially when it comes to the electromagnetic signatures expected from their collapse. During the formation of a black-hole, so much material is consumed that there isn't really enough stuff released to make a bright supernovae. In the end, no matter how hot the material gets---there's just too much gravity.

    For something of a summary of these stellar-fates:
    http://hera.ph1.uni-koeln.de/~heintzma/SNR/bilder/SN_NS.gif [Broken]
     
    Last edited by a moderator: May 5, 2017
  4. Apr 11, 2012 #3

    Chronos

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    Nice one, zhermes.
     
  5. Apr 11, 2012 #4
    One of the unique properties of iron is that it takes more energy to fuse iron into another element than is released by the fusion process. So when the core of a massive star fuses to iron the core stops radiating energy. The outer layers of the star collapse at something like a quarter of the speed of light and crush the iron core. The iron does not explode, for the above reason, but becomes a neutron star or a black hole.


    Also, a white dwarf star does not simply become a Type Ia SN unless it accretes mass from a companion star or perhaps collides with another white dwarf.

    http://www.optcorp.com/edu/articleDetailEDU.aspx?aid=1665 [Broken]
     
    Last edited by a moderator: May 5, 2017
  6. Apr 11, 2012 #5
    thanks for the reply zhermes, that makes much more sense to me now.
    thanks arch for the link, i had a very very basic understanding of type Ia supernova, just that white dwarfs steal matter from a companion star and go pop, it was a very good read and i enjoyed learning how it works.

    another follow up question with regards to the gravitational field around the black hole, specifically how it changes.
    lets say there's a planet in a stable orbit some distance away from a given star that collapses and forms a black hole. lets assume for a minute that the planet survives and is otherwise not blown to bits.
    now whenever dealing with black holes, people generally picture incredible amounts of gravity. but the total mass of the star/blackhole hasnt really changed. its just been compressed into a very tiny volume.
    would a planet that used to have a stable orbit around the star, still have a stable orbit around the black hole?

    from very close to the star (or perhaps inside the star) i can imagine feeling significant changes in the gravitational field. what used to be a fairly spread out mass pulling you in different directions now is pulling you all in one direction (cuz its small) but from a distance, would you notice something as happened strictly by feeling the gravitational pull of the star?
     
  7. Apr 11, 2012 #6
    If you replaced the star with the same mass black-hole, then yes - the planet's orbit would not only still be stable---it would be (virtually) exactly the same. The complicating factor is the formation process of the black-hole, which, in particular, would involve some amount of mass loss. The amount of mass loss determines how much it would perturb the orbit. For the planet to become unbound, the star would have to lose at least half of its mass (I think that's right...). One final complicating factor is that black-holes might get 'kicks' when they form --- the explosion during the collapse of the progenitor star might not be perfectly symmetric, and could act like a rocket - giving the black-hole some velocity. We don't know if/how-much this happens, but most likely it wouldn't have much of an effect (I don't think).

    Excellent question. The story goes that Newton invented* calculus to solve (essentially) this problem. He wanted to know how you can add together the effects of every individual piece of an astronomical body to get the resulting gravitational force (which he new was the 'inverse square law' --- what we call Newton's law). What he found was that it doesn't really matter what the matter distribution is like, as long as its roughly symmetrical --- you can pretend its a 'point-mass' and the gravity will be the same.

    Thus, as I said above, if you replaced the star with a BH (of the same mass) the effects of gravity would be the same.

    * - I'm not sure how accurate this 'story' is; but one important thing to note is that Liebniz seems to have also independently invented calculus. Generally mathematicians will say Liebniz, and physicists Newton.
     
  8. Apr 12, 2012 #7
    Thanks for the replies. I think I understand it a little bit better now.

    Cheers.
     
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